gfale Bicentennial Duplications THE MECHANICS OF ENGINEERING THE MECHANICS OF ENGINEERING VOLUME I KINEMATICS, STATICS, KINETICS, STATICS OF RIGID BODIES AND OF ELASTIC SOLIDS BY A. JAY DuBOIS, C.E., PH.D. PROFESSOR OF CIVIL ENGINEERING IN THE SHEFFIELD SCIENTIFIC SCHOOL OF YALE UNIVERSITY FIJtST EDITION FIRST THOUSAND NEW YORK JOHN WILEY & SONS LONDON: CHAPMAN & HALL, LIMITED I 95 Copyright, 1902, BY A. JAY DuBOIS. Librar V. Hetrtcatton With the approval of the President and Fellows of Yale University, a series of volumes has been prepared bv a number of the Professors and Instructors, to be issued in connection with the Bicentennial Anniversary, as a partial indication of the character of the studies in which the University teachers are engaged. This series of volumes is respectfully dedicated to erirrf By ALBERT S. COOK, Ph.D., L.H.D., Professor of English. 8vo, $4.00, net, Shakespearean ^ars: I. Shakespeare as a Dramatic Artist. By THOMAS R. 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JAY DuBois, of Sheffield Scientific School, with Additions showing Ameri- can Practice by R. H. BUEL, C.E. 8vo. viii-fssg pages, 818 figures. Cloth, $5.00. PREFACE. THIS work is presented as the first volume of a series dealing with the applications of Mechanics to engineering problems. % It is issued uniform with " Stresses in Framed Struc- tures " by the same author, which will be revised as soon as possible, so as to form the second volume of the series. The present volume opens with a carefully considered presentation of the fundamental principles of Kinematics, Statics, and Kinetics. Then follow in proper order the practical applications. Throughout the work numerous problems and illustrative examples are given in direct connection with each important mechanical principle. The method of presentation is the result of much thought and teaching experience. The author believes that this method will commend itself to teacher and student. Thus the first chapters are devoted to the preliminary discussions of mass and space only, without refer- ence to time, motion, or force. Under this head are treated not only mass, density, and centre of mass, but also moment of inertia, considered simply as a mass and space quantity. The experience of the author has convinced him of the advantages of this method of presen- tation. The student in his preparatory study of Geometry has already become conversant with those space relations of which he must make future use. He should now become equally familiar with those mass and space relations of which he must also make future use, and it is both proper and advantageous that this subject should be treated, like Geometry, as a prepara- tory study. Then the student is in a position to take up not only the subject of Kinematics, which deals with relations of space and motion, but also the subject of Kinetics, which deals with mass, space, and motion. In the chapters on Kinematics and Kinetics the method of presentation has been much abridged from the author's " Elementary Principles of Mechanics, " and will be found simpler, more logical and direct. The numerous examples introduced here will be found of service both to teacher and student. In the applications the author has included the results of his own work in this direction, and he believes that the professional reader will find here new and valuable discussions of engineering problems, especially in the chapters on Masonry Walls and Dams, the Strength of Long Columns, the Swing Bridge, the Metal Arch, the Suspension System, and the Stone Arch. In all these chapters the entire treatment and many results are different from those already given by the author in "Stresses in Framed Structures" and "Elementary Principles of Mechanics." By the application of the principle of least work many new and simple results and methods have been obtained. Especially is this the case as regards the stone arch. Treatises on this subject are prolix, overburdened with mathematical discussions and involved formulae, limited at best in their application, diverse in their assumptions, and discouraging ix x PREFACE. alike to the student and engineer. The method of solution here given is simple, direct and practical, and adapts itself with equal ease to any form of arch and any surcharge. The size of this volume is due to its scope and plan. It treats with thoroughness and with direct reference to practice, of several topics for which the student must otherwise have recourse to separate treatises. It thus offers in one volume and in proper sequence Courses on Elementary Mechanics, Kinematics, Statics, Kinetics, Framed Structures, Graphical Statics. Walls and Dams, Retaining Walls, Mechanics of Materials, Swing Bridge, Metal and Stone Arches, and Suspension System. GENERAL CONTENTS. INTRODUCTION. CHAPTER I. PAGE Mechanics. Kinematics. Dynamics. Statics. Kinetics. Measurement i CHAPTER II. Position. Terms and Definitions. . u MEASURABLE RELATIONS OF MASS AND SPACE. CHAPTER I. Mass and Density 14 CHAPTER II. Centre of Mass 20 CHAPTER III. Moment of Inertia. . . . ; 31 KINEMATICS OF A POINT. GENERAL PRINCIPLES. CHAPTER I. Linear and Angular Displacement 51 CHAPTER II. Resolution and Composition of Linear Displacements 54 CHAPTER III. Resolution and Composition of Angular Displacements 58 xii GENERAL CONTENTS. CHAPTER IV. PACK Linear and Angular Speed and Velocity 6 1 CHAPTER V. Linear and Angular Velocity 65 CHAPTER VI. Linear and Angular Rate of Change of Speed. Linear Acceleration 73 CHAPTER VII. Angular Acceleration v 82 CHAPTER VIII. Moments. Moment of Displacement, Velocity and Acceleration 88 KINMEATICS OF A POINT. APPLICATION OF PRINCIPLES. CHAPTER I. Motion of a Point. Constant and Variable Rate of Change of Speed 91 CHAPTER II. Uniform Acceleration 100 CHAPTER III. Motion under Variable Acceleration in General. Central Acceleration. Central Acceleration inversely as the Square of the Distance 109 CHAPTER IV. Central Acceleration directly as the Distance. Harmonic Motion t 125 CHAPTER V. Constrained Motion of a Point 136 KINEMATICS OF A RIGID BODY. CHAPTER I. Angular Velocity and Acceleration Couples. Angular and Linear Velocity Combined 143 CHAPTER II. Rotation and Translation. Analytic Relations 153 GENERAL CONTENTS. xii DYNAMICS. GENERAL PRINCIPLES. CHAPTER I. PAGE Force. Newton's Laws of Motion 1 69 CHAPTER II. Resolution and Composition of Forces 176 CHAPTER III. Moment of a Forcer Resolution and Composition of Moments 181 CHAPTER IV. Force Couples. Effect of Force Couple on a Rigid Body 185 CHAPTER V. Centre of Parallel Forces. Centre of Mass 189 CHAPTER VI. Non-Concurring Forces in General. Analytic Equations 192 CHAPTER VII. Force of Gravitation. Centre of Gravity 203 STATICS. GENERAL PRINCIPLES. CHAPTER I. Equilibrium of Forces. Determination of Mass 209 CHAPTER II. Work. Virtual Work. . 215 CHAPTER III. Static Friction. KINETICS OF A PARTICLE. CHAPTER I. : _ ; . , Deflecting Force 242 CHAPTER II. Tangential Force. Momentum. Impulse 255 xiv GENERAL CONTENTS. CHAPTER III. rxc Work. Power 260 CHAPTER IV. Kinetic Friction 263 CHAPTER V. Kinetic and Potential Energy. Law of Energy. Conservation of Energy. Equilibrium of a Particle 271 CHAPTER VI. The Potential. . . 286 KINETICS OF A MATERIAL SYSTEM. CHAPTER I. General Principles 297 CHAPTER II. Equilibrium of a Material System 310 CHAPTER III. Rotation about a Fixed Axis 315 CHAPTER IV. Compound Pendulum. Centre of Oscillation and Percussion. Experimental Determination of Moment of Inertia 336 CHAPTER V. Impact 342 CHAPTER VI. Rotation about a Translating Axis 367 CHAPTER VII. Rotation about a Fixed Point 375 CHAPTER VIII. Rotation and Translation 389 GENERAL CONTENTS. STATICS OF RIGID BODIES. CHAPTER I. PACK Framed Structures 397 CHAPTER II. Graphical Statics. Concurring Forces 404 CHAPTER III. Graphical Statics. Non -Concurring Forces 412 CHAPTER IV. Walls. Masonry Dams 424 CHAPTER V. Retaining Walls. Earth Pressure. Equilibrium of Earth 454 STATICS OF ELASTIC SOLIDS. CHAPTER I. Elasticity and Strength for Tension, Compression and Shear 473 CHAPTER II. Strength of Pipes and Cylinders. Riveting 486 CHAPTER III. Strength of Beams. Torsion 493 CHAPTER IV. Work of Straining. Deflection of Framed Structures. Principle of Least Work. Redundant Members. Beams fixed Horizontally at the Ends 515 CHAPTER V. Deflection of Beams 530 CHAPTER VI. Shearing Stress 553 CHAPTER VII. Strength of Long Columns 559 xvi GENERAL CONTENTS. CHAPTER VIII. FACE The Pivot or Swing Bridge 571 CHAPTER IX. The Metal Arch 578 CHAPTER X. The Stone Arch 597 CHAPTER XL The Suspension System , ... 609 TABLE OF CONTENTS. INTRODUCTION. CHAPTER I. MECHANICS. KINEMATICS. DYNAMICS. STATICS. KINETICS. MEASUREMENT. PACK Physical Science i Mechanics. Kinematics. Dynamics. Statics. Kinetics i Measurement. Unit. Statement of a quantity. Derived unit 2 Homogeneous equations - 3 Unit of time, length, mass 4 Unit of angle. Radian 6 Unit of conical angle. Square radian. Curvature 7 Table of measures , 9 CHAPTER II. POSITION. TERMS AND DEFINITIONS. Point. Point of reference. Position of a point n Plane polar co-ordinates. Space polar co-ordinates 1 1 Cartesian co-ordinates 12 Direction cosines 13 MEASURABLE RELATIONS OF MASS AND SPACE. CHAPTER I. MASS AND DENSITY. Mass and space relations. Mass. Unit of mass. Measurement of mass 14 Mass independent of gravity 14 Notation for mass. Density 15 Unity of density. Specific mass , 16 Determination of specific mass. Table of specific mass 17 xvni TABLE OF CONTENTS. CHAPTER II. CENTRE OF MASS. PACK Elementary mass or particle 20 Material surface, line. Centre of mass 20 Position of centre of mass in general 21 Moment of mass, volume, area, line 22 Centre of gravity 22 Plane and axis of symmetry. ...... 23 Material line, area, volume 23 Determination of centre of mass 24 CHAPTER IIL MOMENT OF INERTIA. Moment of inertia 31 Radius of gyration 32 Reduction of moment of inertia 32 Moment of inertia relative to an axis 33 Polar moment of inertia for a plane area 33 Moment of inertia relative to a point 34 Moment of inertia for any axis in general 34 Ellipsoid of inertia 34 Principal axes 35 Determination of moment of inertia 38 KINEMATICS OF A POINT. GENERAL PRINCIPLES. CHAPTER I. LINEAR AND ANGULAR DISPLACEMENT. Kinematics 51 Path of a point 51 Angle described by a point , 51 Linear and angular displacement 52 Line representative of linear and angular displacement 52 Vector quantities 52 Displacement in general 52 CHAPTER II. RESOLUTION AND COMPOSITION OF LINEAR DISPLACEMENTS. Resolution and composition of linear displacements 54 Trian^ ic and polygon of linear displacements 54 Rectangular components 55 Components of resultant 55 Relative displacement 55 TABLE OF CONTENTS. xix PACK Notation 56 Triangle and polygon of relative displacements 56 CHAPTER III. RESOLUTION AND COMPOSITION OF ANGULAR DISPLACEMENTS. Resolution and composition of finite successive angular displacements 58 Resolution and composition of angular displacements in general 59 CHAPTER IV. LINEAR AND ANGULAR SPEED AND VELOCITY. Mean linear and angular speed 61 Mean linear and angular velocity 61 Line representatives of mean linear and angular velocity 61 Distinction between mean speed and velocity 62 Instantaneous linear and angular velocity 63 CHAPTER V. LINEAR AND ANGULAR VELOCITY. Speed and velocity in general 65 Uniform and variable linear velocity 65 Uniform and variable angular velocity 65 Resolution and composition of linear velocity 66 Rectangular components of linear velocity 67 Analytic determination of resultant linear velocity 67 Resolution and composition of angular velocity 68 Rectangular components of angular velocity 69 Analytic determination of resultant angular velocity 69 Linear in terms of angular velocity 70 CHAPTER VI. LINEAR AND ANGULAR RATE OF CHANGE OF SPEED. LINEAR ACCELERATION. Change of speed 73 Instantaneous rate of change of speed 73 Mean linear acceleration 75 Instantaneous linear acceleration 75 Resolution and composition of linear acceleration 76 Rectangular components of acceleration 77 Analytic determination of resultant acceleration 77 Tangential and central acceleration 77 Uniform and variable acceleration 78 The Hodograph 79 Axial and radial or deflecting and deviating acceleration 80 CHAPTER VII. ANGULAR ACCELERATION. Mean angular acceleration 82 Instantaneous angular acceleration 83 xv TABLE Of CONTENTS. FAGR Resolution and composition of angular acceleration 83 Rectangular components of angular acceleration 83 Analytic determination of resultant angular acceleration 83 Axial and normal angular acceleration 83 Uniform and variable angular acceleration 84 Linear in terms of angular acceleration 84 Homogeneous equations 86 CHAPTER VIII. MOMENTS. MOMENT OF DISPLACEMENT, VELOCITY, ACCELERATION. Moment of a vector quantity 88 Resolution and composition of moments 88 Moment about an axis 88 Moment of resultant 89 Moment of displacement 89 Moment of velocity 89 Moment of acceleration 89 Moment of angular velocity 90 Moment of angular acceleration 90 KINEMATICS OF A POINT. APPLICATION OF PRINCIPLES. CHAFFER I. MOTION OF A POINT. CONSTANT AND VARIABLE RATE OF CHANGE OF SPEED. Rate of change of speed zero 91 Rate of change of speed constant 91 Rate of change of speed variable 92 Graphic representation of rate of change of speed 93 Rate of change of angular speed zero 96 Rate of change of angular speed uniform 96 Rate of change of angular speed variable 97 Graphic representation of rate of change of angular speed 98 CHAPTER II. UNIFORM ACCELERATION. Uniform acceleration, motion in a straight line 100 Value of g ioo Body projected vertically up or down 101 Uniform acceleration motion in a curve 103 " " " " " equation of the path 103 " " " "" " velocity in the path ; 104 " " " " " " time of flight and range 105 " " " " " " displacement 105 " " "." " angle of elevation 105 " " " " " " envelope of all trajectories 106 TABLE OF CONTENTS. xxi CHAPTER III. MOTION UNDER VARIABLE ACCELERATION IN GENERAL. CENTRAL ACCELERATION. CENTRAL ACCELERATION INVERSELY AS THE SQUARE OF THE DISTANCE. PAGE Motion under variable acceleration in general 1 09 Central acceleration m Central acceleration inversely as the square of the distance 112 " " " " " " " " " path rectilinear n^ " " " " " " " " " curved path 114 Planetary motion. Kepler's laws 120 Velocity of a planet 121 Periodic time 122 Value ofy^ for planetary motion 123 CHAPTER IV. CENTRAL ACCELERATION DIRECTLY AS THE DISTANCE. HARMONIC MOTION. Harmonic motion 125 Simple harmonic motion 125 Periodic time 127 Epoch, phase 128 Compound harmonic motion 129 Resolution and composition of harmonic motions 129 Graphic representation 133 Blackburn's pendulum 135 CHAPTER V. CONSTRAINED MOTION OF A POINT. Motion on an inclined plane, uniform acceleration 136 Motion in a curved path, uniform acceleration 137 Motion in a circle, uniform acceleration 138 Motion in a cycloid, uniform acceleration 139 KINEMATICS OF A RIGID BODY. CHAPTER I. ANGULAR VELOCITY AND ACCELERATION COUPLE. ANGULAR AND LINEAR VELOCITY AND ACCELERATION COMBINED. Angular velocity couple 143 Moment of anguW velocity couple 143 Angular acceleration couple . 1 44 Angular and linear velocity combined 144 Instantaneous axis of rotation 145 Spin. Screw-spin 145 Spontaneous axis of rotation 145 Angular and linear acceleration combined z ^ xxii TABLE OF CONTENTS. PAGE Instantaneous axis of acceleration 151 Twist. Screw-twist 152 Spontaneous axis of acceleration 152 CHAPTER II. ROTATION AND TRANSLATION. ANALYTIC RELATIONS. Components of motion 153 Components of change of motion 153 Motion of a point of a rigid body. General analytic equations 153 Rotation, centre of mass fixed 154 Rotation, any point fixed 154 Rotation, translating axis 154 Resultant velocity 154 Resultant angular velocity 155 Moment of velocity 155 Velocity along axis 155 Velocity normal to axis 155 Instantaneous axis of rotation 156 Invariant for components of motion 157 Change of motion of a point of a rigid body. Analytic equations 157 Tangent acceleration 157 Central acceleration 158 Deflecting and deviating acceleration 159 Resultant acceleration 159 Resultant angular acceleration 159 Moment of acceleration 1 60 Acceleration along axis 160 Acceleration normal to axis 1 60 Instantaneous axis of acceleration 1 6 1 Invariant for components of change of motion 1 6 1 Euler's geometric equations 167 DYNAMICS. GENERAL PRINCIPLES. CHAPTER I. FORCE. NEWTON'S LAWS OF MOTION. Dynamics 169 Material particle ! 169 Impressed force 1 69 Newton's first law of motion 169 Inertia 169 Force proportional to acceleration 170 Force proportional to mass 170 Unit of force 170 Weight of a body 170 TABLE OF CONTENTS. xxiii PACK Gravitation unit of force 171 Momentum and impulse 172 i, Newton's second law of motion 172 \, Measurement of mass ' 173 Mass independent of gravity *. . 173 Notation for mass . . '. 1 74 Newton's third law of motion 1 74 Remark's on Newton's Laws 174 Stress 174 Motion of centre of mass 174 CHAPTER II. RESOLUTION AND COMPOSITION OF FORCES. Line representative of force 176 Resolution and composition of forces 176 Examples 176 CHAPTER III. CONCURRING FORCES. MOMENT OF A FORCE. TWO NON-CONCURRING FORCES. Concurring forces 179 Resultant for concurring forces , 179 Analytic determination of resultant 179 Moment of a force 181 Line representative of moment 181 Resolution and composition of moments 181 Significance of force moment 182 Unit of force moment 182 Resultant of two non-concurring forces 182 Resultant of two parallel forces 183 CHAPTER IV. FORCE COUPLES. EFFECT OF FORCE COUPLE ON RIGID BODY. Force couple ' 185 Moment of force couple 185 Line representative of force couple 185 Resolution and composition of couples 186 Resultant of force couple 1 86 Effect of force couple on a rigid body 187 Resultant for non-concurring forces acting on a rigid body. . 187 CHAPTER V. CENTRE OF PARALLEL FORCES. CENTRE OF MASS. Centre of parallel forces 189 Properties of centre of mass 190 xxiv TABLE OF CONTENTS. CHAPTER VI. NON-CONCURRING FORCES IN GENERAL. ANALYTIC EQUATIONS. AG Resultant for non-concurring forces 1 92 Effect of any system of forces acting on a rigid body 192 Wrench*. Screw-wrench 193 Dynamic components ot motion 193 General analytic equations 194 Resultant force at centre of mass 194 Resultant couple or wrench at centre of mass 194 Resultant couple or wrench at any point 195 Moment at centre of mass along resultant force 195 Moment at centre of mass normal to resultant force 195 Force at centre of mass along axis 196 Force at centre of mass normal to axis 196 Position of resultant force of the screw-wrench 196 Components of motion 197 Screw-wrench 197 Position of force normal to axis 197 The invariant 197 CHAPTER VII. FORCE OF GRAVITATION. CENTRE OF GRAVITY. Force of gravitation 203 Attraction of a homogeneous shell or sphere 203 Value of constant of gravitation 205 Astronomical unit of mass 206 Centre of gravity 207 STATICS. GENERAL PRINCIPLES. CHAPTER I. EQUILIBRIUM OF FORCES. DETERMINATION OF MASS. Statics 209 Equilibrium of concurring forces 209 Static, molar, dynamic and molecular equilibrium 210 Equilibrium of non-concurring forces 211 Determination of mass by the balance 212 CHAPTER II. WORK. VIRTUAL WORK. Work 215 Work of resultant 215 . TABLE OF CONTENTS. xxv PAGE Work, non-concurring forces 215 Principle of virtual work 216 CHAPTER III. STATIC FRICTION. Friction 220 Adhesion 220 Kinds of friction 220 Coefficient of friction. % 221 Limiting equilibrium 221 Coefficient of static sliding friction, experimental determination 221 Cone of friction 222 Laws of static friction 222 Limitations of the laws 223 Values of coefficient of static sliding friction 223 Static friction for pivots 224 Solid flat pivot 224 Hollow flat pivot ' 225 Conical pivot 225 Pivot a truncated cone 226 Pivot with spherical end ., 227 Static friction of axles. 227 Axle partially worn bearings 228 Axle triangular bearing 228 Axle new bearing 229 Friction wheels . 230 Static friction of cords and chains 2-51 Rigidity of ropes 234 Hemp ropes ^5 Wire ropes ; ;5 Static rolling friction 235 KINETICS OF A PARTICLE. CHAPTER I. DEFLECTING FORCE. Kinetics 242 Kinetics of a particle . . . ' 242 Impressed and effective force 242 D'Alembert's principle 242 Deflecting force 243 Simple conical pendulum 244 Centrifugal force 245 Deflecting force at the earth surface. . 246 xxvi TABLE OF CONTENTS. PAGE Particle moving on earth surface 251 Deviation of a falling body owing to earth rotation 252 CHAPTER II. TANGENTIAL FORCE. MOMENTUM. IMPULSE. Tangential force 255 Significance of momentum 256 Impulse 257 Relation between impulse and momentum 257 CHAPTER III. WORK. POWER. Work 260 Unit of work 261 Rate of work. Power 262 CHAPTER IV. KINETIC FRICTION. Kinetic friction 263 Coefficient of kinetic friction 263 Angle of kinetic friction 263 Kinetic friction of pivots, axles, ropes 263 Experimental determination of coefficient of kinetic sliding friction 263 By sled and weight 264 By sled on inclined plane 264 By friction brake 265 Friction -brake test 266 Work of axle friction 266 Coefficients of kinetic sliding friction 267 Efficiency. Mechanical advantage 267 CHAPTER V. KINETIC AMD POTENTIAL ENERGY. LAW OF ENERGY. CONSERVATION OF ENERGY. EQUILIBRIUM OF A PARTICLE. Kinetic energy 271 Kinetic energy of a rotating body 272 Potential energy 274 Total energy 275 Law of energy 275 Conservation of energy 276 Equilibrium of a particle 277 Stable and unstable equilibrium of a particle 277 Change of potential energy 278 For uniform force 278 For central force 278 For central force constant 2 79 For central force proportional to distance 279 For central force inversely proportional to distance .... 280 TABLE OF CONTENTS. xxvil CHAPTER VI. THE POTENTIAL. PAGB The potential 286 Principle of the potential 286 Equipotential surface 287 Lines offeree 287 Tubes of force 287 Gravitation potential 287 Determination of potential. Examples 290 KINETICS OF A MATERIAL SYSTEM. CHAPTER I. GENERAL PRINCIPLES. Material system 297 Internal and external forces 297 Impressed and effective forces 297 D'Alembert's principle 297 Velocity of centre of mass. Momentum of a system 298 Acceleration of centre of mass 299 Motion of centre of mass 299 Conservation of centre of mass 299 Conservation of momentum 300 | Moment of momentum 300 Moment of momentum for a system 300 \. Acceleration of moment of momentum 301 > Conservation of moment of momentum 302 Invariable axis and plane 302 Conservation of areas , 303 Kinetic energy of a system 303 Potential energy of a system 303 Law of energy 303 Conservation of energy 303 Perpetual motion 303 Law of conservation of energy general 303 Examples 305 CHAPTER II. EQUILIBRIUM OF A MATERIAL SYSTEM. Equilibrium of a material system 310 Stable equilibrium 311 Principle of least work - 311 Stability in rolling contact 312 TABLE OF CONTENTS. CHAPTER III. ROTATION ABOUT A FIXED AXIS. MM Rotation about a fixed axis. Effective forces > 315 Moments of effective forces 316 Origin of the term moment of inertia 317 Momentum of a rotating body 318 Moment of momentum 318 Pressures on fixed axis 319 Conservation of moment of momentum. . . 320 Kinetic energy 320 Analogy between equations for rotation and translation 321 Reduction of mass 322 Examples 322 CHAPTER IV. COMPOUND PENDULUM. CENTRE OF OSCILLATION AND PERCUSSION. EXPERIMENTAL DETERMI- NATION OF MOMENT OF INERTIA. EXPERIMENTAL DETERMINATION OF g. Simple pendulum 336 Compound pendulum 337 Centre of oscillation 337 Significance of the term radius of gyration 338 Centre of percussion 339 Experimental determination of moment of inertia 340 Experimental determination of g 341 CHAPTER V. IMPACT. Impact 342 Direct central impact. Non-elastic 342 " " " Perfectly elastic 344 Coefficient of elasticity 346 Modulus of elasticity 347 Direct central impact. Imperfect elasticity 347 Earth consolidation 350 Pile-driving ^ 351 Oblique central impact 352 Friction of oblique central impact 35 ^ Strength and impact 356 Impact of beams 358 Impact of rotating bodies 360 Impact of an oscillating body 361 Ballistic pendulum 362 Eccentric impact 363 CHAPTER VI. ROTATION ABOUT A TRANSLATING AXIS. Effective forces 367 Moments of effective forces 367 TABLE OF CONTENTS. xxix PAGE Momentum 368 Moment of momentum 368 Pressures on axis 369 Conservation of moment of momentum 369 Kinetic energy 370 Instantaneous axis , 371 Examples. 371 CHAPTER VII. ROTATION ABOUT A FIXED POINT. Effective forces 375 Moment of effective forces 375 Momentum 376 Moment of momentum 377 Pressure on fixed point . . . 377 Conservation of moment of momentum 377 Invariable axis 378 Kinetic energy 378 Examples 379 CHAPTER VIII. ROTATION AND TRANSLATION. Effective forces 389 Moments of effective forces 389 Momentum ". . . , 390 Moment of momentum 390 Conservation of moment of momentum 390 Invariable axis 391 Kinetic energy _ . . . . 391 Spontaneous axis 392 Instantaneous axis 392 Examples 392 STATICS OF RIGID BODIES. CHAPTER I. FRAMED STRUCTURES. Stress 397 Tensile stress and force _ 397 Compressive stress and force 397 Shearing stress and force 397 Framed structures 400 Determination of stresses '. 400 By resolution of forces 400 By moments 401 xxx TABLE OF CONTENTS. PACK Superfluous members t 402 Criterion for superfluous members 402 CHAPTER II. GRAPHICAL STATICS. CONCURRING FORCES. Graphical statics 404 Concurring co-planar forces 404 Notation for framed structures 405 Character of the stresses 405 Application to a frame 406 Apparent indetermination of stresses 407 Remarks upon method 408 Choice of scales 408 CHAPTER III. GRAPHICAL STATICS. NON-CONCURRING FORCES. Non-concurring forces 412 Equilibrium polygon 413 Graphic construction for centre of parallel forces 414 Properties of equilibrium polygon 414 Application to parallel forces 415 CHAPTER IV. WALLS. MASONRY DAMS. Definitions 424 Weight and friction of masonry 424 Stability of masonry joint 425 Greatest unit pressure 425 Middle-third rule 426 Stability of a wall in general ) 426 Stability for sliding 1 427 Stability for rotation 427 Stability for pressure 428 High and low wall 428 Design of low wall 429 Design of high wall 429 Design of wall in general 430 Water-pressure 430 Ice- and wave-pressure 432 Stability of a dam 432 Design of dam. Trapezoid section 434 Low dam, bottom base 434 Low dam, top base 434 Low dam, back batter angle 434 Low dam, economic section 435 High dam, bottom base 435 High dam, top base 435 High dam, back batter angle 435 High dam, economic section 436 TABLE OF CONTENTS. xxxi PAGE Design of dam. Economic section 440 Arch dam 450 CHAPTER V. RETAINING WALLS. EARTH-PRESSURE. EQUILIBRIUM OF EARTH. Retaining wall , 454 Point of application of earth-pressure 454 Magnitude and direction of earth -pressure. Graphic determination 454 Magnitude and direction of earth-pressure. Analytic determination 458 Case i. Earth-surface horizontal 461 Case 2. Earth-surface horizontal. Back vertical 462 Case 3. Earth-surface horizontal. Back ang'le 90 ^ 462 Case 4. Earth-surface inclined at angle of repose 463 Case 5. Earth-surface inclined at angle of repose. Back vertical 463 Values of angle of friction, coefficient of friction, density of earth 464 Cohesion of earth 466 Equilibrium of a mass of earth 467 Angle of rupture 468 Coefficient of cohesion. 468 Height of slope. 468 Angle of slope 469 Curve of slope 469 STATICS OF ELASTIC SOLIDS. CHAPTER I. ELASTICITY AND STRENGTH FOR TENSION, COMPRESSION AND SHEAR. Elasticity 473 Prismatic body 473 Stress and force 473 Strain 473 Law of elasticity , 474 Set and shock 475 Determination of elastic limit and ultimate strength 475 Coefficient of elasticity 476 Strain due to weight 479 Stress due to change of temperature 480 Coefficients of expansion 480 Working stress. Factor of safety 481 Variable working stress 48 i Combined tension or compression and shear. . 484 CHAPTER II. STRENGTH OF PIPES AND CYLINDERS. RIVETING. Strength of pipes and cylinders 486 Theory and practice of riveting 487 xxxii TABLE OF CONTENTS. PACk Kinds of riveted joints 487 Size and number of rivets 488 Pitch of -rivets 489 Rivet table 490 CHAPTER III. STRENGTH OF BEAMS, PINS AND EYEBARS. TORSION. Flexure or binding stress 493 Assumptions upon which theory of flexure is based ' 493 Neutral axis of cross-section 493 Bending moment 494 Resisting moment 496 Designing and strength of beams 499 Crippling load. Coefficient of rupture 499 Comparative strength 502 Beams of'uniform strength '. 503 Theory of pins and eyebars 506 Bearing 506 Diameter of pin 506 Maximum bending moment 5*7 Sizes for pins 508 Torsion 509 Neutral axis 509 Twisting and resisting moment . 510 Coefficient of rupture for torsion 510 Coefficient of elasticity for shear determined by torsion 511 Work of torsion 512 Transmission of power by shafts 512 Combined flexure and torsion 512 CHAPTER IV. WORK OF STRAINING. DEFLECTION OF FRAMED STRUCTURES. PRINCIPLE OF LEAST WORK. REDUNDANT MEMBERS. BEAMS FIXED HORIZONTALLY. Work of straining 515 Work and coefficient of resilience 515 Deflection of a framed structure 516 Principle of least work 517 Redundant members 520 No economy due to redundant members 521 Work of bending 521 Beams fixed horizontally at ends 522 Case i. Beams fixed horizontally at one end. Concentrated load 522 Case 2. " " " Uniform load 524 Case 3. " " " " both ends. Concentrated load 525 Case 4. " " " " " " Uniform load 527 Bending and tension or compression combined 528 CHAPTER V. DEFLECTION OF BEAMS. Deflection of a beam , 530 Beam fixed at one end ; loaded at the other 533 " " " " " uniformly loaded 535 TABLE OF CONTENTS. Xxxiii PAGH Application to metal springs 537 Beam supported at both ends. Concentrated load 542 " " " " " Uniform load 544 " " " one end, fixed at other. Uniform load 544 " " " " '" " " " Concentrated load 545 " fixed at both ends. Uniform load 547 " " " " " Concentrated load 548 CHAPTER VI. SHEARING STRESS. Shearing stress in beams 553 Work of shear in beams 554 Influence of shear on deflection -. 554 Determination of coefficient of elasticity for shear 555 Influence of shear on reaction 556 Maximum internal stresses in a beam . 557 CHAPTER VII. . STRENGTH OF LONG COLUMNS. The ideal column 550 Theory of the ideal column 559 Deportment of the ideal column 562 Experimental verification 563 Euler's formula 563 Diagram for ideal column 564 Actual column 564 Practical formulas for long columns 565 Straight-line formula 565 Parabola formula 565 Rankine" s formula 567 Gordon's formula 568 Merriman's formula 568 Allowable unit stress 569 CHAPTER VIII. THE PIVOT OR SWING BRIDGE. Pivot or swing bridge 571 Centre span without bracing 571 General formulas ' 574 Solid beam, uniform cross-section 576 CHAPTER IX. THE METAL ARCH. Kinds of metal arch 578 Framed arch hinged at crown and ends 578 Framed arch hinged at ends only 579 Temperature thrust 581 Solid arch hinged at ends only 583 xxxiv T/IBLE OF CONTENTS. rAC Solid semi-circle hinged at ends only 5^5 Temperature thrust, solid arch hinged at ends 5^5 Framed arch hinged at ends 5^6 Temperature stress 59 1 Solid arch fixed at ends 59 2 Temperature thrust 595 CHAPTER X. THE STONE ARCH. Definitions 597 Reduced surcharge 598 Pressure curve 598 Condition of stability 599 Determination of H, M and V 600 The straight arch 607 CHAPTER XI THE SUSPENSION SYSTEM. Suspension system 609 Horizontal pull of cable .' 610 Shape of cable 6 1 o Length of suspenders 610 Length of cable segment ' 6 10 Stress in cable segment 61 1 Deflection of cable due to temperature 611 Deflection of truss for uniform load 612 Temperature load for truss 612 Old theory of suspension system 612 New theory of suspension system 620 Work on suspenders 620 Work on cable 620 Work on truss . 621 GREEK ALPHABET. In all mathematical works Greek letters are used. The reader should therefore familiar with these letters, so that he can write them and recognize them at sight and < them by their names. Letter. Transliteration. Name. a a Alpha ft b Beta r, Y g Gamma J, <" length. CHAP. I.] DIMENSIONS OF A DERIVED UNIT. 3 The statement [^4] = [] 2 is a statement of the dimensions of the unit of area. Again, let [Z] denote the unit of length, [7^] denote the unit of time, and [V"\ denote the unit of velocity. Then if D ~ d\L\ is any displacement and T= t[T~\ is the time and V= z>[F] is the mean velocity, we have We shall then have the numeric equation v = , or the number of units of velocity is equal to the number of units of displacement divided by the number of units of time, provided we-. have that is, provided the unit of velocity is the unit of length per unit of time. This, then, is a statement of the dimensions of the unit of velocity. Meaning of " Per." It will be observed that the statement m is read, " the unit of velocity is equal to the unit of length per unit of time." The word per is indicated by the line of division between [Z] and [jT]. Now we can divide the numeric d by the numeric t and write v = , because these are abstract numbers. But it would be nonsense to speak of dividing length by time, or a unit of length by a unit of time. We therefore avoid such a statement by the use of the word per. If, then, we give to the symbol of division this new meaning, we can treat it by the rules which apply to the old meaning, and thus avoid the invention of a new symbol by using an old one in a new sense. The sign of division stands then for actual division so far as numerics are concerned, but so far as units are concerned it stands for the word per. Whenever, then, the word PER is used it can be replaced by the sign of division. Homogeneous Equations The letters in all equations or statements of the relations of quantities always stand for the numerics, and the units are always understood but not written in. Thus such an equation as v = or d = vt is a numeric equation, where the units are understood and must be supplied when interpreting the equation. When the units are thus supplied, all terms on both sides of any equation, which are combined by addition or subtraction, must always denote quantities of the same kind. Such an equation is called HOMOGENEOUS. If any numeric equation is not thus homogeneous, it is incorrectly stated. It is also evident that all algebraic combinations of such homogeneous equations must always produce homogeneous equations. If not, some error must have been made in the algebraic work. Thus in the equation d = vt, if we supply the omitted units, we have 4 . MECHJNICS.-1NTRODUCT1ON. [CHAP. I. The equation is therefore homogeneous, since the unit of length is to be understood in both terms, and we have a distance equal to a distance. Again, suppose the result of some investigation is expressed by ioz/. Then, without any reference whatever to the various steps by which this result may have been obtained, we can say at once that it is incorrect. For if we insert the units we have Zdfeet + 2 / seconds = \< * The English standard yard is I part in 17230 shorter than the U. S. copy. 6 MECHANICS -INTRODUCTION. [CHAP. 1. Unit of Angle. There are two units of angle in use, the DEGREE and the RADIAN. The degree is that angle subtended at the centre of any circle by an arc equal in length to - part of the circumference of that circle. It is subdivided sexagesimally into degrees (), minutes ('), and seconds ("). The seconds are subdivided decimally. Minutes and seconds of time are distinguished by being written min., sec. The radian is that angle subtended at the centre of any circle by an arc equal in length to the radius. It is subdivided decimally. If then the length of any arc is s[L'], or s units of length, and the length of the radius is r[], or r units of length, and if the angle subtended at the centre is a radians, we have s[L] s = ~> r '"=* The number of radians in any angle is then found by dividing the number of units of length in the subtending arc by the number of units of length in the radius, and this number is independent of the partictdar unit of length adopted, whether feet or centimeters. If the subtending arc is the entire circumference, the number of radians is = 2n. Hence 2ft radians correspond to 360 degrees, or I radian corresponds to - - = 27* n = 57-29578 degrees = 57 17' 44". 8. Any angle expressed in radians may then be converted into degrees by multiplying the 180 number of radians by - = 57.29578 degrees = I radian. Any angle expressed in degrees may be converted into radians by multiplying the number of degrees by =0.0174533 radians = I degree. I oO . Examples. (I) Express I2 n 34' 56" in terms of radians ; and 3 radians in terms of degrees. ANS. 0.2196 radians; 171 53' I4".424. (2) The radius of a circle is 10 feet ; what is the angle subtended at the centre by an arc of 3 feet? , ANS. radian, or 17 u 19 .44. 10 (3) How much must a rail 30 feet lone; be bent in order to fit into a curve of half a mile radius ? ANS. Tfrr radian, or o 39' 3"-92. 88 (4) Express 45 degrees in terms of radians, and 4 j radians in terms of degrees. ANS. * radians = 0.7854 radians; 257 49' 51 ".636. (5) The angle subtended at the centre of a circle by an arc whose length is 1.57 feet is 7j/ what is //// radius f i8o I5?r (6) What is the sin ^ raff tans ; cos g radians ; co* radians ; tan radians? ANS. 0.5; C:IAP. I.] MEASUREMENT UNIT OF CONICAL ANGLE. ^ (7) Express in degrees and in radians the angle made by the hands of a clock at 35 minues past 3 o'clock. ANS. 102.5 degrees ; 1 -79 radians. Unit of Conical Angle. Let the area of any portion of the surface of a sphere be A[A], or A units of area, and let the square of the radius be r\A\, or r* units of area. If lines are drawn from the centre C of the sphere to every point of the area, they form a cone, and the angle subtended at the centre C by the area we 'call a CONICAL ANGLE. . The conical angle subtended at the centre of a sphere by a portion of its surface whose area is equal to the square of its radius we call a SQUARE RADIAN. If we denote the conical angle subtended by the area A by a square radians, we have The number of square radians in any conical angle is tJius found by dividing the number of units of area in the subtending area by the number of units of area in the square of the radius, and this number is independent of the unit of area adopted. If the subtending area is the entire surface of the sphere, the number of square radians is 2 4/r. Hence the surface of a sphere subtends a conical angle of 4?r square radians. [The terms solid angle and solid radian are usually employed in place of conical angle and square radian as defined, but as they seem in no way descriptive, we have employed the latter terms as more expressive.] Curvature. The direction of a plane curve at any point is that of the tangent to the curve at this point. Thus the direction of the curve AB at the point A is that of the tangent A T. The change of direction between any two points of a plane curve is the angle between the tangents at these two points, and is called the INTEGRAL CURVATURE. Thus the angle a, or change of direction be- tween the tangents at A and B, is the integral curvature for the curve between A and B. The integral curvature for any portion of a plane curve divided by the length of that portion is the mean curvature. Thus if the length from A to B of the curve is s[L], or s units of length, the mean curvature is r r -\> Since or is given in radians, the unit of curvature is one radian per unit of length of arc. When we say, therefore, that the mean curvature is - , we mean radians per unit of length of arc. The limiting value of the mean curvature when the two points are indefinitely near is called the curvature. 8 MECHANICS-INTRODUCTION. [CHAP. I. The curvature, therefore, is the limiting rate of change of direction per unit of length of arc. Its unit is one radian per unit of length of arc. Curvature of a Circle. If the curve is a circle, the angle at the centre between the radii at A and B will be equal to the angle a between the tangents at A and B. We have then = - radians. The mean curvature is then = radians per unit of r s r length of arc. Since this is independent of s, the curvature at every point of a circle is constant and equal to the mean curvature for any two points, viz., - radians per unit of length of arc. Curvature of any Curve. For any curve whatever a circle can always be described whose curvature is the same as that of the given curve at the given point. This is the circle of curvature of the curve at that point. Its radius is the radius of curvature of the curve at that point. If then p is the radius of curvature of a curve at any given point, the curvature of the curve at that point is - radians per unit of length of arc. Since curvature then depends only upon the radius of curvature, the circle is the only curve whose curvature is constant. 1 i ) A circle has a radius of 10 feet. What is its curvature ? Ans. ^ radian per foot of arc, or 5. 73 per foot of arc. (2) If the radius is 10 yards, what is the curvature ? Ans. -jij. radian per yard of arc, or 5. 73 per yard of arc, or -fa radian per foot of arc, or i 91 per foot of arc. Dimensions of Unit of Curvature. If C is the curvature and c the number of a[a] units of curvature, we have by definition c[C] = fry. where [C] is the unit of curvature, and [or] is the unit of angle, [L] the unit of length, and , s the number of units of angle and length. a ["1 We shall always have c = -, provided we take [C] = ryr, that is, provided the unit of curvature is equal to the unit of angle divided by the unit of length. This is a statement of the dimensions of the unit of curvature. The unit of curvature is then one unit of angle per unit of length oS arc, as, for instance, one radian per foot of arc, or one degree per foot of arc. A railway curve has a length of one mile, the curvature is uniform, and the integral curvature is jo degrees. What is the curve, the curvature, and the radius of curvature / Ans. A circle; 0.5236 radian per mile arc; 1.9 miles radius. Tables of Measures. We shall deal in the course of this work with many other derived units, which will be explained as they occur. It will be useful to collect here for conven- ience of reference a number of such units. CHAP. I.] MEASUREMENT TABLES OF MEASURES. I. MEASURES OF SPACE. A. LENGTH. ) CENTI vIETERS 10 ) INC 1 HES \ TABLE 1. i centimeter = 0.3937079 inch ( 39-37079 inches ( 3.2809 feet (0.62137 mile I meter i kilometer = (0.535987 nautical mile TABLE 2. I inch = 2.539954 centimeters i foot = 30.479449 " I yard = 0.91438347 meter I mile = 1.60935 kilometers I nautical mile, or 6080. 26 ft. = 1.85327 kilometers The following are approximate : The centimeter is about f inch. The meter is about 3 ft. 3f inches. The decimeter is about 4 inches. One kilometer is about f of a mile. Distance from pole to equator is about 10000000 meters. Earth's polar radius is about 500000000 inches. TABLE 3. i sq. centimeter = 0.155006 sq. inch I sq. meter = 10.7643 sq. feet i sq. hectometer, ) [ = 2.47114 acres or i hectare ) i sq. kilometer = 0.38611 sq. mile B. AREA. TABLE 4. I sq. inch = 6.45137 centimeters I sq. foot = 928.997 " I sq. yard 0.836097 meter I acre = 0.404672 hectare I sq. mile = 2.58989 sq. kilometers TABLE 5. i cubic centimeter = 0.0610271 cu. inch I liter, or i cubic | = 61.0271 cu. inches decimeter ) = 1.76172 pints i cubic meter = 35.3166 cubic feet C. VOLUME. TABLE 6. i cubic inch = 16.3866 cubic centimeters I cubic foot = 28.3153 liters I cubic yard = 0.764513 cubic meter I pint = 0.567627 liter i gallon = 4.54102 liters II. TABLE 7. I centigram = 0.154323 grain i gram = { '5-4323 grains (0.035373902. I kilogram = 2.20462 Ibs. MEASURES OF MASS. TABLE 8. I grain = 0.064799 gram i oz. = 28.3496 grams i lb., or 16 oz. = 0.453593 kilogram i ton, or 2240 Ibs. = 1016.05 kilos io MECHANICS-INTRODUCTION. [CHAP. I. I gram = mass of I cubic centimeter of pure water at 4 C. I kilogram = I liter of pure water at 4 C. I gallon = 277.274 cubic inches. The gallon contains 10 Ibs. of pure water at 62 F. I cubic foot of water contains about 1000 oz., or 62 Ibs. The pint contains 20 fluid oz. Acceleration of gravity at London = 32.182 feet-per-second per second = 980.889 cen- timeters-per-second per second. Average value 32^ feet-per-second per second or 980.3 centimeters-per-second per second. i dyne = force which will give a mass of I gram an acceleration of I centimeter-per- second per second = about -g T weight of gram = weight of about I milligram. i poundal = force which will give a mass of I pound an acceleration of I foot-per- second per second = about weight of oz. CHAPTER II. POSITION. TERMS AND DEFINITIONS. It is there- position. Point. A mathematical POINT has neither length, breadth, nor thickness, fore without dimensions and indicates position only. Point of Reference. When we speak of a point as having position, some other point or points must always be assumed, by reference to which the position is given. Such a point is a POINT OF REFERENCE. It is also called a POLE, or ORIGIN. Position, then, is always relative. We know nothing of "absolute Thus the position of the point C is known with respect to A when we know the length of the line AC and the angle BAC or the direction of the line AC. The points A and B are points of reference, by means of which C is located. Position of a Point. The position of a point with reference to t other assumed points is then known when we have sufficient data to locate it. These data give rise to two methods of location : ist, by polar co-ordinates; 2d, by Cartesian co-ordinates, so called because first employed by Descartes. Plane Polar Co-ordinates. The data necessary for locating a point by polar coordi- nates, when the point is situated in a given plane, consist of a distance and an angle; if the point is not in a known plane, of a distance and two angles. Thus, if the point P, in the plane of this page, is to be located, we first assume a line OA in the plane, as a line of refer- ence. Then the position of P with reference to O is given by the angle A OP and by the distance OP. The assumed point O is called the POLE ; OA is the LINE OF A REFERENCE; the distance OP is called the RADIUS VECTOR, and its magnitude is usually denoted by r; the angle AOP is the DIRECTION ANGLE; its magnitude is denoted by 0, and it is measured around from OA to the left. The polar co-ordinates for a point in a given plane are therefore r and 0, or a distance and an angle. These are PLANE polar co-ordinates. Space Polar Co-ordinates. If the point P is not in a given plane, we assume as before a pole O, and a reference line OA in space. Through this line we assume any plane, as the plane of this page, OABC, and let OB be the intersection of this plane with a plane OPB, perpendicular to it and passing through OP. The location of P is then given by the length OP or the radius vector r, the angle A OB or 0, and the angle BOP or 0. The polar co-ordinates for a point not in a given plane are therefore r, 0, and ft, or a distance and two angles. These are SPACE polar co-ordinates. If O is a point on the earth's surface, and the reference line OA is a north and south line in the plane of u*c ii astronomical altitude of the point P. the angle would be the MECHANICS-INTRODUCTION. [CHAP. II. ! ' 2 D Y 1 p B +rt y 6 3 4 -y Cartesian Co-ordinates --Plane. The data necessary for locating a point by Cartesian co-ordinates, if the point is in a known plane, consists of two distances, parallel to two assumed lines of reference in that plane, passing through the point of reference, which is called the ORIGIN. The assumed lines of reference are usually taken at right angles. Thus if the point P is known to be in the plane of this page, we assume any origin O and draw two reference lines OX and OY through O in this plane and at right angles. These two lines are called the AXES OF CO-ORDINATES, the horizontal one the axis of x, or the x axis, the other the axis of y, or the y axis. The distance BP or OA is denoted by x and called the ABSCISSA of the point P. The distance AP is denoted oy y and called the ORDINATE of the point P. The abscissa x is positive to the right, negative to the left of the origin, while the ordi- nate y is positive when laid off above, and negative when below, the origin. Any point in the plane is thus located with respect to O. If a point is in the first quad- rant, its co-ordinates are -{- ,r, -j- y\ if in the second quadrant, x and -\- y\ if in the third quadrant, x and y\ if in the fourth quadrant, -j- x and y. The axis OX is always taken so that OA = -f- x is towards the rig/it when we look along O Y from O towards Y. When the point is in a known plane, the co-ordinates are PLANE co-ordinates. Cartesian Co-ordinates Space. If the point P is not in a known plane, we take three axes through the origin, all at right angles. Two of these we denote by .A' and Fas before; the third, at right angles to the plane of XY, we call the axis of z or the z axis. Thus the position of the point Pis given by the distance OA = x, the distance AB or OC y, and the distance BP 2. These are the SPACE co-ordinates of the point P. The signs prefixed to the co-ordinates indicate the quadrant in which the point is located as before. Thus -+- ** 4- jand 2 denote a point in the first quadrant either above or below the plane of XY\ - x % +/, z, a point in the second quadrant above or below the plane of XY\ x, y, . z, and + * }'> -. points in the third and fourth quad- rants above or below the plane of XY. The plane of XYls usually taken as horizontal, so that OZ points to the zenith, and the axis OX is always taken so that OA - - + x is towards the right when we look along O Y, f row t towards Y. Thus if XOY is the plane of the horizon, Z is the zenith. If, then, O\ - towards the north point of the horizon, OX is towards the east. Angles about O measured in the plane XY are always measured round from OX tow- ards OY; in the plane YZ from OY towards OZ\ in the plane ZX from OZ towards OX, as indicated by the arrows in the figure. CHAP. II.] DIRECTION COSINES. 13 Direction Cosines. If we join the origin O and the point P by a line and denote the angle of OP with the X axis by a, with the F axis by ft, and with the Z axis by y, we have the relations x =. OP cos a, y = OP cos ft, z = OP cos y. These cosines are called the DIRECTION COSINES of OP, and they always have the same sign as the co-ordinates x, y and z respectively. Since OP is the diagonal of a parallelopipedon, we have OP 2 = x 2 4- j 2 -f- ,s 2 OP\cos* Since 6 = -,, where m is the mass and V the volume of the body, we have Since y is the mass of a unit of volume of water, yV is the mass of a volume of water equal in volume to the body. Hence the specific mass of any body is equal to the ratio of its mass to the mass of an equal volume of water. In the English system the mass of one cubic foot of pure water at 4C., or the point of maximum density, is nearly 1000 ounces, or 62.5 Ibs. (more exactly 998.6 ounces). The density of water is then about 62.5 Ibs. per cubic foot, or _ 62. 5 Ibs.. " i cub. ft. If, then, Fis one cubic foot, we have, from (3), m Ibs. where m is the mass in Ibs. of one cubic foot of any body. CHAP. I.] DETERMINATION OF SPECIFIC MASS. ^^ In the C.G.S. system, the mass of one cubic centimeter of pure water at 4 C. is very nearly one gram, and was intended to be so exactly. The density of water by this system is then I gram = i cub. c." If, then, Fis one cubic centimeter, we have, from (3), m grams 6 ^ m I gram where m is the mass in grams of one cubic centimeter. That is, the mass in grams of one cubic c'entimeter gives at once the specific mass, while in the English system the mass in Ibs. of one cubic foot mnst be divided by 62.5. Or inversely the specific mass of anybody gives at once the mass in grams of one cubic centimeter, while it must be multiplied by 62.5 to obtain the mass in Ibs. of one cubic foot. Determination of Specific Mass. A body totally immersed in water displaces its own volume of water. It is a well-known physical fact that a body so immersed is buoyed up by a force equal to the weight of the volume of water displaced. If, then, a body is weighed in air and then weighed again while wholly immersed in water, the weight in air is proportional to the mass of the body, and the loss of weight is proportional to the mass of the displaced water. When very great accuracy is required the body should be weighed in a vacuum, or else allowance must be made for the buoyant force of the air. But in all cases of practical mechanics this is an unnecessary refinement. To determine the specific mass, then, we have only to divide the weight of the body by its loss of weight in water, or, since weight is proportional to mass, we divide the number of units of mass of the body by the number of units of mass of an equal volume of water. Table of Specific Mass. In the following table the density-ratios, or specific mass, or so-called "specific gravity" relative to water, of a few substances are given. The exact value in many cases will depend on the temperature and the mechanical process, such as hammering, etc., to which the bodies have been subjected in their manu- facture. Air at o C 0.0012759 Alcohol at o C 0.791 Turpentine at o C o. 870 Ice 0.92 Sea-water at o C 1.026 Crown glass . . . . , 2.5 Flint glass 3.0 Aluminum 2.6 Zinc 7.0 Tin 7-4 Iron 7.7 Copper 8.8 Silver 10.5 Lead 11.4 Mercury at o C 13. 596 Gold. 19.3 Platinum 21.5 Examples. (i) The mass of a piece of limestone is 310 grams. When immersed in water it balances a mass of 188.5 grams. Find the specific mass. ANS. The mass of an equal volume of water is 310 188.5 = 121.5 grams. Hence the specific mass is 310 1 8 MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. I (2) In ordtr to find the specific mass of a piece of oak, a piece of wire which lost 10.5 grams when ivci^ hcd in water was wrapped around the wood, the mass of which was 426.5 grams. The compound mass was 484.5 grams lighter in the water than in the air. Find the specific mass. ANS. The loss of the wood alone was 484.5 10.5 = 474 grams. Hence the specific mass is^ 2 --'-?- = 0.9. (3) An iron vessel filled with mercury has a mass 0/300 Ids. and when weighed in water shows a loss of 40 Ibs. If the specific mass of the iron is 7.7 and of the mercury 13.6, find the mass of the -vessel and of the mercury. S m ANS. Since specific mass e = -, where 5 is density and y is density of water, and since 5 = , where m is mass and v is volume, we have e = , or v = . Let nil be the mass of the iron and m of the mercury, and m the combined mass. Then for the volume of the iron we can write v\ = -, for the volume of the mercury z/ = - -*, and for m the combined volume v = Since the combined volume is equal to the sum of the volumes, we have *L +*!=, or "! + ?-' = 5. . I Since the combined mass is equal to the sum of the masses, we have m, + m, = m. (2) Combining (i) and (2) we obtain - _ !_ -L m, = m , m = m . ii ii fl ft t \ In the present case we have e = 5 , e, = 7.7, ?, = 13.6 and m = 500 Ibs. Hence m, = 57{| Ibs., m, = 442^ Ibs. NOTE. This is called the problem of Archimedes, because first solved by him for an alloy of gold and silver. Its application to alloys or chemical compositions is, however, limited, as in general in such cases there is a change of volume, so that the combined volume is not equal to the sum of the volumes of the components, and equation (i) does not hold. (4) To find the specific mass of a mixture, given the volume or mass and specific mass of each constituent. ANS. We must assume the volume of the mixture equal to the sum of the volumes of the constituents. This is not invariably the case, especially where there is chemical union. mi, m,., m 3 , etc., be the masses of the constituents; *i. f , ^9, " " " specific masses of the constituents ; v\, v t , v t , volumes of the constituents. Let m, v and = be the mass, volume and specific mass of the mixture. Let y be the density or mass of a unit of volume of water. Then nii + m, + m s + etc. = m. But m, = f t rv t . 5, = f . v ,,. e tc. Hence. y + etc. = evy, or e = g|Z/ + f *>* + etc t We have also v = z/, + v , + Vt + e tc. Therefore ~ v^ + v, + etc. W CHAP. I.] EXAMPLES. Again we have z/ = i-, v, = , etc. Hence _ mi + m a + etc. . - + -- + etc. (5) A fiat bar of iron 4f inches wide and inch thick has a linear density of 9.91 Ibs. per foot. Find the mass of a bar of iron i inch square and i yard long. ANS. 10 Ibs. (6) From the preceding example state a rule for finding the mass per foot of a bar of iron of constant area of cross-section ; also for finding the section if the mass per foot is given. ANS. To find the mass per foot in Ibs., multiply the area of cross-section in square inches by 10 and divide by 3. To find the area of cross-section in square inches, multiply the mass per foot in Ibs. by 3 and divide by 10. (7) If a railroad rail weighs jo Ibs. per foot of length, -what is its area of cross-section^f ANS. 15 sq. inches. " CHAPTER II. CENTRE OF MASS. Elementary Mass or Particle. We can consider the entire volume V of any body as subdivided into an indefinitely large number of indefinitely small ELEMENTS, each of equal volume v, so that V 2v. We can carry this subdivision as far as we please, until each equal elementary volume can be treated mathematically as a point. If, then, 6 is the density of such an elementary volume v, its mass is m = dv, and the entire mass m of the body is then m = 26v = 2m. We thus consider the body as composed of elementary masses or PARTICLES of masses ;//, , ;// 2 , M S , etc., if the body itself is not homogeneous, or of elementary masses or particles each of mass m if the body is homogeneous. This conception lies at the basis of all our mathematical treatment, and the student should especially note that it involves no theory whatever as to the actual constitution of matter or as to what matter really is, i.e., as to whether a body is really composed of " atoms " or "molecules" or " particles," or as to whether matter is "continuous" or " discontinuous." We commit ourselves, then, to no theory when we say that, whatever matter is, any body can be divided into equal portions so small that each portion can be considered as homogeneous and treated as a point. Whenever we speak of a " particle," it is to such a portion that we have reference. Whatever, then, the constitution of -matter may really be, we can consider and treat a body as composed of an indefinitely large number of indefinitely small homogeneous elements or material particles of equal volume, so small that each may be treated as a point. We denote the mass of such an element by m. Material Surface. A MATERIAL SURFACE, then, is a body whose uniform thickness is small compared to its other two dimensions. We can consider the entire area A of such a surface as made up of indefinitely small elements of area a, so that A = 2a. Let 6 be the surface density (page 1 5) of such an element. Then its mass is m = da, and the entire mass m of the material surface is m = 2Sa. Material Line. A MATERIAL LINE is a body whose length is large compared to its other dimensions, and whose sectional area at right angles to the length is constant. We can consider the length L of such a line as made up of indefinitely small elements of length /, so that L = 2f. Let 6 be the linear density (page 15) of such an element. Then its mass is m = dl, and the entire mass m of the material line is m = 2dl. Centre of Mass We consider, then, a material body, surface or line as composed of an indefinitely large number of indefinitely small particles. The centre of mass of such a body, surface or line is that point whose distance from any plane is equal to the average distance of all the particles from that plane. CHAP. II. ] CENTRE OF MASS. Thus suppose a body composed of particles whose masses are m v m. 2 , m y etc. the first to contain a number n l of smaller particles of equal mass ;, the second a number n 2 of the same mass m, and so on. Then 21 Suppose m, njn, etc. . * m Let x lt x z , x^, etc., be the distances of m^ , m 2 , ;;/ 3 , etc., m," from the plane FZ, each x being taken with its proper sign. Let the entire number of equal particles of mass m be N, so that the total mass of the body is m = Nm. Then we have for the average distance of all the particles from the plane FZ, that is for the distance of the centre of mass x from the plane N If we multiply numerator and denominator by m t we have m z x z -f etc. " _ * ~~ - m m In the same way we obtain for the distance y of the centre of mass from the plane and for the distance z of the centre of mass from the plane XY Position of Centre of Mass in General If, then, z; is the volume of an element of a body, $ the density, and m the mass, we have m = 6v. The entire^mass m is m = 26v, and we have, from the preceding equations, m in (I) if the body is homogeneous, ft is constant for every element. It therefore cancels out, and we have for a homogeneous body where V is the entire volume. 22 MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. II. In the same way if a is the area of an element of a material surface, 8 the surface density, and m the mass, we have m = da. The entire mass m is m = 28a, and we have for a material surface 28 ax 2 day 28az * = ~' = ' Z= -- If the surface is homogeneous, 8 cancels out and we have - 2ax - 2ay - 2az where A is the entire area. So also, if ds is the length of an element of a material line, and 8 the linear density, we have m = S.ds. The entire mass is m = 26.ds, and we have for a material line 28.ds.x - 28.ds.y - 2S.ds.z / TTT ^ * = :.*- ' = -20^' * = -^5T (III) If the line is homogeneous, 8 cancels out and we have - _ 2ds.x - _ 2ds.y - _ 2ds.z , * s s s where s is the entire length. Moment of Mass, Volume, Area, Line. We call the product of any mass m, volume V, area A or line S by the distance of its centre of mass from any plane, the MOMENT of that mass, volume, area or line, relative to that plane. We can then express all the preceding equations by the simple statement that the moment of the total mass, volume, area, or length relative to any plane is equal to the algebraic sum of the moments of all the elementary masses, volumes, areas or lines relative to the same plane. In taking the algebraic sum, distances are to be taken with their proper signs, positive in the directions OX, OY, OZ, negative in the opposite directions. Centre of Gravity. We shall see hereafter (page 190) that the centre of mass of a body coincides with the point of application of the resultant of a system of parallel forces. The attraction of the earth for a body is the resultant of a system of forces acting upon the particles of a body, each directed towards the centre of the earth. We have thus a system of forces not strictly parallel. But practically the deviation from parallelism is insignificant, since the greatest dimension of any body on the earth with which we have to do is insignificant in comparison with the diameter o*f the earth. Hence the resultant force of gravity passes practically through the centre of mass. This resultant is the weight of the body. The weight of the body acts practically, therefore, at the centre of mass. To determine the centre of mass by experiment, we have then only to suspend it suc- cessively from two different points of its surface. The two lines of suspension, if produced, must intersect practically at the centre of mass. The centre of mass is therefore often called the "centre of gravity." This term is, however, strictly speaking, incorrect. Centre of mass is not dependent at all upon gravity any more than mass itself. We can make use of gravity in finding it, just as we do in measuring mass (page 14). CHAP. II.] ORIGIN AT THE CENTRE OF MASS. 23 The term " centre of gravity" can only be properly applied t6 that point at which if t^lie entire mass of the body were concentrated, it would attract and be attracted, for all positions of the body, just the same as the body itself. In this sense, as we shall see hereafter (page 207), only a few bodies possess a centre of gravity. But all bodies possess a centre of mass. The two conceptions are therefore entirely distinct. Origin at the Centre of Mass. In taking the algebraic sums of the elementary moments 2wx, *2my, 2mz, we must, as already noted, take x, y, z for each element with their proper signs. If, then, we take the origin of co-ordinates at the centre of mass, we must have x = o, y = o, z = o. Hence in this case o, "2my = o, 2mz = o ....... (i) If we take polar co-ordinates and take the pole at the centre of mass, we have 2mr = o, ............ (2) where r is the distance of any particle from the pole. That is, the algebraic sum of the moments of all the elements relative to the centre of mass is zero. Plane and Axis of Symmetry. A body is symmetrical with respect to a plane when for each element on one side of the plane there is an equal, similarly situated element on the other side. In such case the centre of mass is in this plane. A body is symmetrical with respect to an axis when it is symmetrical with respect to two planes passing through that axis. In such case the centre of mass is in that axis. If a body is symmetrical with respect to two axes, the centre of mass is at their inter- section. Material Line, Area, Volume. There is of course a certain inconsistency in speaking of the centre of mass of geometrical lines, areas and volumes. The expression is, however, allowable, since we are understood to mean a material line, area or volume, 8 being linear, surface or body density, as defined on page 15. In the case of a homogeneous line, area or volume, # 'cancels out, as we have seen in the equations of page 22, and we can then allowably speak of the centre of mass of the line, area or volume itself. Determination of Centre of Mass. We have already seen how to determine the posi- tion of the centre of mass by experiment (page 22). The position of the centre of mass for many homogeneous bodies may often be told at once, by applying the principle of symmetry just given. Thus the centre of mass of a homogeneous straight line is at its middle point. For a homogeneous circle, or circular area, or sphere, or spherical shell it is at the centre of figure. For a homogeneous parallelogram it is at the intersection of the two lines through the middle points of opposite sides, or at the intersection of the two diagonals, since these are lines of symmetry. So also for a homogeneous parallelopipedon the centre of mass is at the intersection of its diagonals, or any two lines through the centre of mass of opposite sides. For a homogeneous circular cylinder with parallel ends, at the middle point of the axis. For a homogeneous triangle the centre of mass is at the intersection of any two lines from an apex to the middle point of the opposite side, since these are lines of symmetry. For bodies in general we may find the centre of mass by applying the equations of page 22, or more generally by the application of the Integral Calculus to effect the summations indicated. It is not necessary or desirable to occupy space here with such applications. It is of MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. II. far more importance that the student should understand clearly what the centre of mass is, as defined on page 20, and what use is made of it, and how it enters into mechanical prob- lems, as will be explained later (pages 175 and 191), than that he should spend much time in finding its position. In the following examples we show how to apply the equations of page 22 to a few cases. Examples. (i) Find the centre of mass of a homogeneous circular arc. ANS. Let ADD be a homogeneous circular arc with centre at O', and let the axis of X pass through (7 and the centre D of the arc. A Then O'D is an axis of symmetry, and the centre of mass O is on this axis. Let the chord AB = c, and the length of arc ADB = s, and r be the radius. Take an indefinitely small element ad oi length ds whose cen- tre of mass is at e, and let ab be the vertical projection of ad = ds. Drop eN perpendicular to O'X. Then we have, by similar triangles, ds:ab::r\ O'N ; -X or, since O'N = x, the abscissa of the poi nt e, we have x . ds = r x ab. By equation (3), page 22, we have then for the distance O'O = x oi the centre of mass - _ 2.t . ds r'Sab But 2 P C A PPM C To find the distance x = Ap, let M be the middle point of the base, and draw BM and drop the perpen- diculars Op and BP. Then we have AP h cot A, and for the distance MP MP = h cot A , or MP h cot A, according as A is the smaller or larger base angle. We have then Alp = \MP, or h cot A b or and hence in both cases, if A is acute, - b h cot A b + h cot A BY CALCULUS. Take an elementary strip parallel to the base and at a distance y above it. Let the length of this strip be x. Its thickness will be dy, and its area xdy. We have then h h But we have A = , and, by proportion, Substituting these values of A and x, we have k h ~ 2 /' ^ ^ r ~ f (/t - b h cot A ~3 + 3 (3) Find the centre of mass of a homogeneous trapezoid. ANS. Let A BCD be the trapezoid. Denote the top or smaller base BC by ,, arid the bottom or larger base AD by bi, and the altitude by h. We have then for the area of the trapezoid ' /f r>/~/-i (^' "I" b-i\h area ABLD We can divide the trapezoid into two triangles, ABD and BDC, by the diagonal BD. The area of the triangle ABD is and, as just proved (example 2), the distance of its centre of mass above AD is \h, and on the right of A, bi + h cot A The area of the triangle BDC is 26 MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. II. and the distance of its centre of mass above AD is f/*, and on the right of A, from equation (2), example (2), fa + fa h cot A fa + fa + 2 h cot A h cot A + - - We have then for the distance/ of the centre of mass O above AB area ABCD x / = area ABD x + area BDC x h t and for the distance ~x of the centre of mass O on the right of A area ABCD x x = area ABD x * + A + area BDC x *' + * * C0t A . Inserting the values for the areas, we obtain, if A is acute, -_ (2fa + fa)h -_.* + Ma + V + A (20. + fa) COt /? For a parallelogram b\ = fa and y h, x I- h cot A. For a rectangle A = 90 and/ = ^, JT = b. Graphical Construction. ist. Draw lines Dm\, Dm* from apex D to the middle points m\ and nit of the opposite sides. Then the centre of mass O t of the triangle ABD is on the line Dtn\ at a distance DO\ equal to f of Dm\. The centre of mass O* of the triangle BDE is on the line Din* at a distance DO* equal to f of Dm*. If we join O\ and O* thus found, the centre of mass must be on the line O\O*. " Draw m*m through the middle points of the parallel sides. This line is an axis of symmetry, since it passes through the centre of all elements parallel to BC and AD. The centre of mass must therefore lie in the line tu t ti. It is therefore at the intersection C'of m*m% and 2d. We have, from the values of x and/ just obtained, fa + x _ 2bi + fa + A cot A y h Now if we lay off CG parallel to AD and make it equal to b* , and lay off ///"parallel to #6" and make it equal to b\ , and join F and G, the line FG must pass through O. For we have, by proportion, 0i + * _ fa + h cot A + fa + fa y h Hence the intersection of FG with the axis of symmetry m,m, gives the centre of mass O. 3d. Another convenient construction is as follows : Draw the diagonals AC and DB intersecting at /. Lay off along AC the distance Ae = 1C, and along b f c DB the distance Db = IB. Bisect the diagonals at ;//, and w 3 ,and join n>,<* a d fft-f. The intersection O is the centre of mass. /'roo/.Ry construction the sum of the horizoi t; ! m * ^ projections of Ae and Db is equal to l>, , and the points c ;:. >J b are both at the same distance above AD. Hence the line eb is horizontal and We have then, by construction, wiw, also horizontal and CHAP. II.] CENTRE OF MASS EXAMPLES. We have also, by proportion, if ji is the vertical from /on BC, = - -, or y\ = T~' y\ h y\ bi + o\ By similar triangles, \iy is the distance of O above AD, eb m \rn-i y-y* ~-v -> + Inserting values, &*-6i _ W* - h) bJi ~ h (261 y- just as already found. (4) Find the centre of mass of a homogeneous trapezium. ANS. Let ABCD be the trapezium. Denote AD by 6,, Cby di , AB by d and DC by a. Also let ft be the angle of b\ with horizontal, and take AD horizontal. We have then for the area of the trapezium area ABCD = M *'" A + *** S ''" C . ^ bl We can divide the trapezium into two triangles, ADS and BDC, by the diagonal BD. The area of the triangle ABD is area ABD = V/sin A, the distance of its centre of mass above AD is , and on the right of A (example (2)) d co The area of the triangle BDC is area BDC b\a sin C . the distance of its centre of mass above AD is d sin A + sin , and on the right of A, cos fi -\- bi + id cos We have then for the distance y of the centre of mass O above AB area ABCD x y = area & ^sin .7 . Then the centre of mass O must be on the line OiOt. Again, find the centre of mass O 3 of the triangle DEC. and O, of Z>/?.4. The centre of mass must be on the line O 3 O* . It is therefore at the intersection of OiO* D" and 6> 3 <9 4 . MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. II. (5) Find the centre of mass of a homogeneous circular sector, A ANS. Let O'ACBO' be the sector with centre at O' and radius a A =r. The sector can be divided into an indefinitely large number of indefinitely small triangles. The centre of mass of each triangle is at a distance of Jr from O'. These centres form then a homogeneous circular arc AiCiBi , and the centre of mass of the sector is the centre of mass of this arc. F*rom example (i), then, the centre of mass O lies upon the radius of symmetry O'C and at a distance O'O = x from the centre given by _ chord AiJS t 2_ arc A,L\B, ' J r ' Let c be the length of chord AB of the sector, and s the length of arc ACB of the sector, then chord and we have arc A t C\B t 2 re If is the central angle AO'B in radians, c = 2r sin , s = r& and For the semicircle, c = 2r, s = itr or 6 = it, sin = i ; hence For a quadrant For a sextant Student should solve by Calculus. (6) Find the centre of mass for a homogeneous segment of a circle. ANS. Let ACB be the segment. The centre of mass O is in the radius of symmetry O'C. Let the radius O 'A = r, and the length of arc ACB = s. Then the area of the sector O'ACBO' is A, = -. The distance O'Oi of the centre of mass 0, of the sector has just been found in example (5) to be (70, = , y where c is the length of chord AB. The height of the triangle 0AB is. I r' . Its area is then The distance O'O* of the centre of mass 0, of the triangle is CHAP. II.] CENTRE OF MASS EXAMPLES. 29 The area A of the segment is A = (Ai At). Let O'O = x be the distance of its centre of mass. Then we have the moment of the sector equal to the sum of the moments of the triangle and segment, or Ai . O'Oi = At . O'Ot + (Ai A*)x~. Substituting the values of A\ t A 3 , O'Oi, and O'Ot, we obtain 12 A If 6 is the central angle AO'B in radians, we have s = rQ, c = 2r sin -, and hence c* For a semicircular segment, Q = if, cos -= o, c = 2r, and hence x = , just as we have already found in example (5). Student should solve by Calculus. (7) Find the centre of mass for a homogeneous circular ring. ANS. Let the outer radius O'A t be r\ , and the inner radius O'At be r ,. Let the length of the outer arc A,Ci t be si , and its chord AiBi be ci\ the length of the inner arc A t CiB* be s*, and its chord AtBt be c t . Then, from example (5), we have for the distance of the centre of mass of the sector O'AiBiO' and its area 2 The distance of the centre of mass of the sector O 'A t B*O ' is Xt = : and its area is The area of the ring is Let O'O = x be the distance of its centre of mass. Then we have the moment of the outer sector equal to the sum of the moments of the inner sector and ring, or Aixi = AtXt + (A i A*)x. Substituting the values of Ai, At, x\, x*, we obtain _ - or, since Si = riQ, where 6 is the central angle A\OBi in radians, and ci = 2ri sin -, . 4 *'" r,,. , _ x = (8) Find the centre of mass for the homogeneous surface of a cylinder, or of any prism. ANS. The centre of mass for the homogeneous surface of a cylinder is at the middle point of the axis. For all the elements of the surface, obtained by taking slices parallel to the base, have their centres of mass upon the axis. The centre of mass of the surface is then the centre of mass of the axis. MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. II. For the same reason the centre of mass for the homogeneous surface of a prism is at the middle of the line which unites the centres of mass of its bases. (9) Find the centre of mass for the homogeneous surface of a right cone or pyramid. ANS. The centre of mass for the homogeneous surface of a right cone is in the axis at two-thirds its length from the apex. For the curved surface can be divided into an indefinite number of indefinitely small triangles. The centres of mass of all these triangles form a circle at a distance of two-thirds of the axis from the apex, whose centre of mass is in the axis. The same holds for any right pyramid. (10) Find the centre of mass of a solid homogeneous prism or cylinder with parallel bases. ANS. The centre of mass of a solid homogeneous prism is at the middle of the line joining the centres of mass of its bases. For by passing planes parallel to the bases we divide it into equal slices whose centres of mass lie in the axis. The same holds for a cylinder. (M) Find the centre of mass for a homogeneous pyramid or cone. ANS. Let ABCD be a homogeneous triangular pyramid. Take b at the middle point of BC, and draw Ad and Db. Take a point a on Ad, so that ab = \Ab, and a point e on Db, so that eb = %Db. Join ae, and draw Da and Ae. Then Da and Ae are axes of symmetry and the centre of mass O is at their intersection. But ae is parallel to AD and equal to \AD, and the triangle aOe is similar to AOD. Hence aO = OD, or \OD = 3 . aO, and >C therefore aD = 4 . aO, or aO = . aD. The centre of mass is then on the line joining a vertex with the centre of mass of the opposite base, and at a distance from the vertex of the length of this line. Since any pyramid or cone is composed of triangular pyramids with a common vertex, the same holds true for the centre of mass. We can therefore determine the centre of mass of a pyramid or cone by passing a plane parallel to the base at a distance from the base of \ the altitude, and finding the centre of mass of this section, or the point where it is pierced by the line from the apex to the centre of mass of the base. CHAPTER III. MOMENT OF INERTIA.* Moment of Inertia. The product of any indefinitely small elementary mass or area or volume by the square of its distance from any given point, line or plane is called the MOMENT OF INERTIA of the elementary mass or area or volume relative to that point, line or plane. If, then, a is an elementary area and r its distance from any point, line or plane, ar z is the moment of inertia. If m is the mass of a particle, mr z is the moment of inertia. If v is the volume of an element, vr* is the moment of inertia. The summation for any volume, area or mass of all these products is the moment of inertia of the entire volume, area or mass, relative to the assumed point, line or plane. Thus for any area the moment of inertia is 2ar z , for any volume 2vr z , for any mass 2mr*. The point chosen is always the centre of mass of the entire body, mass, volume, or area, unless otherwise specified. The plane or line chosen is always a plane or line through this centre of mass, unless otherwise specified. We use the term " centre of mass of an area" or of a volume in the sense already defined (page 23). The line relative to which the moment of inertia is determined is the AXIS. We always denote the moment of inertia relative to the centre of mass, or an axis or plane through the centre of mass, by /. We have then for the moment of inertia of an area relative to the centre of mass, or any axis or plane through the centre of mass, f For the moment of inertia of a volume we have I For the moment of inertia of a mass, Significance of the Term. The term " moment of inertia" is not well chosen according to the terminology of modern science. As we see from the definition, it has nothing to do with "inertia" (see page 169). It has been proposed to call it ''second moment " of area or mass. The term " moment of inertia " is, however, of such general use that it is hardly worth while to try to introduce such % change. It is sufficient for the student to note that it is the arbitrary name for a certain quantity. This quantity occurs so often in mechanical problems that it is desirable to have a special name for it, and to discuss it thoroughly in advance of its use. * This chapter may be omitted here if thought desirable, but should be taken before Kinetics of a Material System, page 297. 31 3 2 MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. III. Radius of Gyration. That distance at which, if the entire volume, area or mass were concentrated in a point, the moment of inertia would be the same as for the volume, area or body itself, is called the RADIUS OF GYRATION for the volume, area or body. We denote the radius of gyration relative to the centre of mass, or a line or plane through the centre of mass, by k. We have then, by our definition and notation, for a body of mass m, for an area A, or = . = m m T- At* &- 1 - 2aV * -A--/T for a volume V, Moment of Inertia and Radius of Gyration in General. Unless otherwise specified the term moment of inertia always signifies the moment of inertia for a body of mass m. The letter / always means this moment of inertia for the centre of mass, or an axis or plane through the centre of mass. The letter k denotes the corresponding radius of gyration. Any other point, axis or plane is called an ECCENTRIC point, axis or plane. The moment of inertia is denoted in such case by /', and the corresponding radius of gyration by'. Everything in what follows which holds for / and k, I' and k' ', as thus defined, fora body of mass m, holds good also for a surface of area A or a volume V, unless otherwise specified. Reduction of Moment of Inertia. If, then, / is the moment of inertia for any axis through the centre of mass, and /' the moment of inertia for any parallel axis at the distance d, we can easily prove the relation That is, the moment of inertia of a body relative to any eccentric axis is equal to the moment of inertia relative to a parallel axis through the centre of mass plus the product of the mass by the square of tlie distance between the two axes. The same holds for the moment of inertia of a surface if we replace m by the area A of the surface, or for a volume if we replace m by the volume V. We have then 7' = 7 + ^, T'=I+V(P. This is called the theorem of moment of inertia for parallel axes. By means of it we can find /' for any axis, if 7 for a parallel axis through the centre of mass and the distance d between these axes are given. Conversely, we can find 7 if 7' and d are given. The proof is simple. Let m be the mass of a particle of a body whose distance from an axis through the centre of mass O of the body is r. Let r' be its distance from a parallel axis at O', the distance O' O between the axes being d. Let V be the angle of r with O'O. Then we have for any body r - 2mr f * and 7 = 2tnr*. CHAP. III.] But we have Hence REDUCTION OF MOMENT OF INERTIA. d* 2rdcos0. '* = 2mr* + d*2m 2d2mr cos 0. 33 But mr cos 6 is the moment of m relative to the axis through the centre of mass O. Therefore, as we have seen (page 23), 2mr cos 6 = O We have then 2mr' z = 2mr* + d* 2m, or, since 2m = m, the entire mass of the body The same holds for an area if we replace m by A and 2m by 2a = A, or for a volume if we replace m by V and 2m by 2v = V. We have then also &'* = P -f d z . It is evident, then, that the moment of inertia relative to any axis through the centre of mass is less than for any parallel axis, and the radius of gyration relative to any axis through the centre of mass is less than for any parallel axis. Moment of Inertia relative to an Axis. Let O'X be any axis and XO' Y, XO' Z any two rectangular planes passing through that axis. Then for any particle of a body of mass m whose co-ordinates are x,y, #, we have the moment of inertia relative to O'X mr* = my* -f~ m ^ m(y* -f~ &*}. Summing the moments of inertia for all the par- ticles of the entire body, we have for the moment of inertia of the body relative to the axis O'X 2mr* = 2my z +2mz* = 2[m(y*+* f )]. But 2mr z is the moment of inertia of the body relative to the axis O'X. We denote it, therefore, by I' x . Also, 2my* is the moment of inertia of the body relative to the plane ZX, and 2mz* relative to the '. plane XY. We denote them, therefore, by /and I' xy . X We can therefore write In the same way we have or, the moment of inertia of any body with reference to a line is equal to the sum of the moments of inertia for any two rectangular planes passing through that line. Polar Moment of Inertia for a Plane Area. For any plane area, as XO' Y, we have l xy = o, and hence /.,' = / and I, = 1^ ; hence MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. III. or, the moment of inertia of any plane area relative to an axis perpendicular to the plane is equal to the sum of the moments of inertia for any two rectangular lines in the plane through the foot of the perpendicular. The moment of inertia for a plane area relative to an axes at right angles to the plane is called the POLAR moment of inertia relative to that line. Moment of Inertia Relative to a Point. Let O' be any point, O'X, u O'Z three co-ordinate planes. Then for any particle of mass m whose co-or- dinates are x, y, z, we have for the moment of inertia with reference to O' mr* = mx* -f- my* -\- mz*. Summing the moments of inertia for all the particles, we have for the moment of inertia of the body with reference to O' = /o is the moment of inertia of the body reference to the point O', and 2mx* = I' r , = I' zx , 2ms 2 = I' xy , are the moments of inertia of the body with reference to the co-ordinate planes FZ, ZX, XV. Hence /o = /;+/;,+/;,. But we have just seen that /,', + / = I' x > Hence That is, the moment of inertia with reference to any point is equal to the sum of the moments of inertia for any three rectangular planes through that point ; or, is equal to the sum of the moment of inertia for any line through the point and a plane through the point at right angles to this line. Moment of Inertia for an Inclined Axis, in General. Let M be the mass of any particle of a body whose co-ordinates are x, j, z for any assumed origin O' and co-ordi- nate axes O'X, O' Y, O'Z, and let O'o be any line through the origin O 1 ', making the angles it, ft, y with the co-ordinate axes. Let r be the perpendicular from m upon O'o. Then we have the distance O'o = x cos at -\- y cos /? -f- z cos y ; also, the square of the distance O'm, Hence r 2 = ~O7n Wo , or r 2 = (x 3 -f y + z*) - (x cos a -f y cos ft -f- z cos y) 2 . The moment of inertia of a particle of mass m relative to any axis O'o is, then, mr* = mix* -f f -f j 3 ) - tnix cos a -f- y cos ft -\- z cos yf. CHAP. III.] MOMENT OF INERTIA PRINCIPAL AXES. . Summing the moments of inertia for all the particles of the body we have, since cos 2 a -\- cos 2 /? -J- cos 2 ^ = I, 2(mr*) = 2\in(** + f + z*} (cos 2 a + cos 3 /?-fcos 2 y)] - 2\_m(x cos a + y cos + z cos Multiplying out and reducing, we can write this 2(mr*) = 2[m(y* + ^ 2 )] cos 2 a -f 2[w(^ -f ^ 2 )] cos 2 /? + 2[m(x* + /)] cos 2 y - 2 cos or cos ft 2(mxy) 2 cos /3 cos y 2(myz) 2 cos 7 cos a But, as we have seen, page 33, are the moments of inertia of the body relative to the axis of X, Y, and Z. Also, 2(mr 2 } is the moment of inertia /' relative to the axis O'o. Hence I /' = I' x cos 2 a -{- I y cos 2 ft -\-T M cos 2 y 2 cos a cos ft ^(mxy) 2 cos /? cos 2 cos 7 cos 2(mzx) Equation (i) gives the moment of inertia of a body for any axis O'o making any given angles a, ft, y with the co-ordinate axes. Principal Axes. If for any origin O' we take the rectangular axes X, V, Z in such directions that for any body = o, myz = O, *2mzx = O, these axes are called PRINCIPAL AXES for the point O' , the moments of inertia I' x , I' y , I' z relative to these principal axes are called PRINCIPAL MOMENTS OF INERTIA, and the corre- sponding radii of gyration k' x , k' y , k', are called PRINCIPAL RADII OF GYRATION. If we take the origin at the centre of mass, the principal axes are called CENTRAL PRIN- CIPAL AXES, the corresponding moments of inertia /,, f y , I t are central principal moments of inertia, and the corresponding radii of gyration k x , k y , k s are central principal radii of gyration. Properties of Principal Axes. The three conditions for principal axes are then = o, myz = o, mzx = o. Introducing these conditions in (i) we have . /' = I' x cos 2 a -f I' y cos 2 ft 4- /; cos 2 y ........ (2) . That is, the moment of inertia of a body relative to any line is equal to the sum of the prod- ucts obtained by multiplying the moments of inertia for the principal axes for any point of the line, by the square of the cosines of the angles which the line makes with these principal axes. For any plane material area, taking the plane as that of ZX, we have for any point of the area given by x and z, y = o. Hence, no matter where the origin nor what direction the axes of x and y may have, we have = o, 2myz = o. 36 MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. Ill Also, if a body has a plane of symmetry, then, taking this as the plane of ZX, we have for any origin in the plane, for any particle given by x, z and -f- y above the plane, an equal particle given by x, z and y below. Hence, no matter where the origin nor what direc- tion the axes of x and y may have, we have 2ntxy = o, 2myz = o. Introducing these conditions in (i) we obtain /' I x ' cos 8 a -f- Iy cos 2 ft + I* cos 2 Y 2 cos Y cos ^ mzx. This reduces to (2) when ft = o, and hence a = 90, y == 90, and we have in such case P = Iy. That is, the perpendicular to the plane is a principal axis for the point of intersection. It is not, however, in general a principal axis for any other point of the perpendicular. For if we take some other point on y at a distance d from the origin, then if the axis of y is a principal axis for this point also, we must have 2mx (y d) = o, 2mz (y d) = o. These conditions can only be satisfied when 2mx = o and 2my = o, that is when the perpendicular passes through the centre of mass. Hence, if any two of the three conditions for principal axes are fulfilled, we have either a plane area or a plane of symmetry. Any perpendicular to a plane area or a plane of symmetry is a principal axis for the point of its intersection with the plane. If this point of intersection is the centre of mass, the per- pendicular is a principal axis for any of its points. A line cannot be a principal axis at more than one of its points, unless it passes through the centre of mass, in which case it is a principal axis at every one of its points. Let a body have two planes of symmetry at right angles. Taking one plane as the plane of ZX, we have, as already proved, 2mxy = o, ~2myz = o, and any perpendicular to this plane is a principal axis for its point of intersection. Taking the other plane as the plane of YZ, we have = o, "2,mzx o, 'and any perpendicular to this plane is a principal axis for its point of intersection. We have then all three conditions = o, ^myz = o, 2mzx = o fulfilled, and hence for two rectangular planes of symmetry the principal axes for any point on the line of intersection of these two planes are this line of intersection and the two perpendiculars to it at the point in each plane. If a body has three planes of symmetry at right angles, their point of intersection is the centre of mass (page 23). Hence for three rectangular planes of symmetry any line of intersection is a principal avis for any of its points. The principal axes for any point on a line of intersection are this line and the tivo perpendiculars to it at the point in each plane. The three lines of intersection are principal axes for the centre of mass, or central principal axes. Ellipsoid of Inertia. Since the mass m of a body multiplied by the square of its radius of gyration k for any axis gives the moment of inertia / for that axis, we have, by dividing equation (2) by m, CHAP. III.] ELLIPSOID OF INERTIA, 37 In the preceding Fig., page 34, take a point P on the line O'o at any convenient distance O 'P '= I from O' '. Denote the product Ik' by j 2 , so that lk'=s\ hence / 2 cos 2 a / 2 cos 2 /3 Substituting in (3) we have Let the co-ordinates of the point P be x* ', y' ', z' . Then and we have S* = I (4) ' 2 ' 2 ' 2 * K y K t The equation of an ellipsoid referred to its major and minor axes is 2*+^ +Z" I where ^4, B and (7 are the semi-axes. We see, then, that (4) is the equation of an ellipsoid whose semi-axes are principal axes given by where s may have any convenient value. Hence if we lay off on every line O'o through the origin O' a distance O' P I -r, , where the distance s may be taken any convenient distance, all the points P thus de- termined will lie on the surface of an ellip- soid whose equation is (4), whose axes are the principal axes for the point O', and whose semi-axes are given by the values of A, B and C just found. The square of the reciprocal of any semi diameter / is, then, l-^l 2 !?~ s* ' This ellipsoid is called the ELLIPSOID OF INERTIA for the point O', because the square of the reciprocal of any semi-diameter O'P= I multiplied by the mass m of the body, or -^ . m/' 2 , is proportional to the moment of inertia /' = m/' 2 of the body relative to an axis coinciding with this semi-diameter, and if we take s= i, will be equal to this moment of inertia. MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. III. The axes of this ellipsoid arc principal axes for the point O'. We see at once that the principal moments of inertia must include the greatest and least of all the moments of inertia for the point O', the least corresponding to the longest semi-diameter, O'A, and the greatest to the least semi-diameter, either O'B or O'C. For any point O', then, there must be at least one set of rectangular axes, OA, OB, OC, which arc principal axes. Determination of Moment of Inertia. The moment of inertia of a body can be determined experimentally, as will be explained hereafter (page 340). For many bodies it is readily determined by a simple application of the Integral Calculus. It is not necessary or desirable to occupy space here with such applications. It is of much greater importance that the student should understand clearly what the moment of inertia of a body is, as defined on page 31, and what use is made of it and how it enters into mechanical problems, as will be explained later (page 321). In the following examples we show, how the moment of inertia may be found for a number of simple cases, with and without the use of the Calculus. Examples. (l) Find the moment of inertia for a homogeneous straight line. ANS. Let the \\neab coincide with the axis of Y, and let O be its centre of mass. Then by the principle of symmetry OX, OY, OZ are central principal axes (page 35). jY Let the length of the line be L, and S the linear density. Then the mass m is m = dL. Divide the line above and below O into an indefinitely large number n of short elementary lines. The mass of each is <5 , and the 2/1 distances of these elementary masses above and below OX or OZ are , 2 -, 3 , etc. Multiply the mass of each element by the "2n 2fi "Zn square of its distance, and summing the products, we have for the moment of inertia /* or /, relative to the central principal axes OX and O Y m/.' --(i + 4 + 9 + . . .'). .. The sum of the series i + 4 + 9 + ... = 2* + 3* + As n increases, this sum approaches the limit -. Hence for n indefinitely large /.-/.-* The moment of inertia / for the axis O Y is ft = o. These are the moments of inertia for tin* central principal axes. Let Of be any inclined axis in the plant* AT making the angles a. ft, y with the central principal axes. Then cos y = o and cos the same as for a system of three material particles consisting of one sixth the mass of the line at each end and two thirds the mass of the line at the centre. BY CALCULUS. The mass of any element of the line is Sdy. The moment of inertia of the element relative to OI is then Sdy x y sin 2 ft. Hence, since m = <5Z, /+T =/. / = J 8y*dy . sin' ft = ^- sin' ft. (2) Find the moment of inertia for a system consisting of three parallels equidistant homogeneous straight lines of equal length I, the ends being in parallel straight lines, the mass of the centre line being ;;/, and of each of the outer lines m\. ANS. Let aa' , bb' , cc' be the lines of length /, of mass mi , m\ and ; a respectively, and let d be the length of the line ab through the ends, and A be the angle abb' . _ ; The centre of mass O of the system is evidently at the middle of the centre line. The moment of inertia of aa' relative to any axis through its centre of mass in the plane of the lines making the angle with aa' is, from ex- ample (i), 111 1 7 1 9 -/'srn' . d For a parallel axis through the centre of mass O, since the distance between the axes is sin (A 6), we have (page 32) We have the same result for the line bb' . For the line cc' we have The moment of inertia of the system relative to any axis in the plane of the lines through the centre o' mass O, making the angle 9 with the linos, is then 1 sin' (A -0} + ^/'sin'0. FOI any parallel axis in the plane at. a distance 6 we have, then (page 32), (2 MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. III. We can write this in the equivalent form L. We see from this that the moment of inertia for the system is the same as for a particle of mass ;, at the middle of each outer line, a particle of mass -Uii + r ~\ at each end of the centre line, and a Particle of mass -(; mi) at the centre of mass of the system. SPECIAL CASES. If m t = 4#/,, we have m\ at the middle of each outer line, m\ at each end of the centre line, and 2;i at the centre of mass of the system. If ///, = 2;i, we have m\ at the middle of each outer line, |r, at each end of the centre line and f;, at the centre of mass of the system. If Wi =* m\, we have m\ at the middle of each outer line and ]fin\ at each end of the middle line. (3) Find the moment of inertia for a homogeneous parallelogram and rectangle. ANS. We have seen that the moment of inertia for a homogeneous straight line for any axis is the same as for a particle of one sixth its mass at each end and of four sixths its mass at the middle point. Since the parallelogram ABCD is generated by a straight line AD moving parallel to itself, the particles at A and D will generate lines AB and DC, the mass of each being one sixth the mass of the parallelogram, and the particle ^ -C at d will generate the line dc, whose mass is four sixths the / / mass of the parallelogram. / / We have then a system of three parallel lines, the mass d/. . y c m\ of each outer line being m\ = ^m and the mass of the centre / / line being m, = |m, where m is the mass of the parallelogram. / / We have then m*=.\)n\, and, as just proved in the A g preceding example, the moment of inertia of the parallelogram is the same as for a particle of mass m\ = ^m at the middle of each outer line and at the ends of the centre line, and a par I tele of |m at the centre of mass. For the axis J b through the centre of mass parallel to the base- AB = b we have then I b = rf sin" A. For the axis /<* through the centre of mass parallel to the side AD = [CHAP. III. We thus know the position of the principal axis of A' and the principal moments of inertia I x , I y The moment of inertia for any axis through the centre of mass is then given by / = /, cos* a + I y cos' ft + I, cos' y. For a rectangle we have A = 90, and hence for any axis equation (2) becomes 12 For the axis Oft, p = o, 6 = o and For the axis OI d , p For the axis OZ o, ff = 90 and fd= - In this case m = 8M, and we have from our general formula k} = , or I y = . Hence cos 7 = o, or tt= 90. Hence the axes of I b , Id and /, are central principal axes, as they should be, since the planes of I 6 f d and IdOZ are planes of symmetry (page 36). BY CALCULUS. The mass of an element parallel to AB at a- distance y from OX is m = &bdy. Its moment of inertia for an axis Of making the angle 6 with AB is, from (i), vib* sin 1 6 , and transferring to a parallel axis through the centre of mass O at the distance ^ S - , we have sin A ml? sin" ~~'\2 Hence sin' (A - 0) Js -/:: - 0) or, since &bd sin in, I = 77 [^ a sin' + rf sin 8 (A - 6)], as already found. (4) T-YW /// /// w , , fnr a particle cf one third the mass of the triangle at the middle point of each side. CHAP. III.] MOMENT OF INERTIA EXAMPLES. Let ABC be a triangle whose angles are A, B, C and sides a, b, c. Take any axis AI' a through the vertex A in the plane of the triangle, making the angle 6 with the base b. Drop the perpen- diculars pi and pt iron) B and C, the perpendicular p 3 from the middle point of the side a, and the perpendicular^* from the centre of mass O. Then we have pi = c sin (6 A), = b sin 0, -I- p, b sin + c sin (0 A) If pi is less than - sin 0, we have or = If /i is greater than - sin 0, we have In all cases, then, -sin0 + 1^, -^ sin 0V or / = !(/, + sin 0). = - [ b sin + c sin (9 A) ]. If, then, we take one third of the mass m at the middle point of each side, we have for the moment of inertia for the axis I' a l' a = - [F sin" + c* sin" (8 A) + be sin sin (8 - A)]. Reducing to a parallel axis through the centre of mass O by subtracting m^' (page 33), we have sin"0 + f sin a (9 A) fc sin sin (0 A)] (3) For any axis in the plane of the triangle, then, if p is the perpendicular distance from the centre of mass to the axis, /' = 78 J sin' + c* sin 8 (0 - A) - be sin sin (0 - A)\ + mp\ For the axis /* through the centre of mass O parallel to the base b we have p = o, = o, and equation (3) becomes A. For the axis IH through the centre of mass perpendicular to the base b we have p = o, 6 = 90, and equation (3) becomes I k - [P + c 1 cos' A be cos A\ 18 For the axis /<- through the centre of mass parallel to the side c we have^ = o 9 = A, and equation (3) becomes 44 MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. III. For the axis OZ through the centre of mass at right angles to the plane of the triangle we have (page 34) (P + c * - be cos A), 18 or. since 2 be cos A = b* + c* a\ we can write *-* The axis OZ is a principal axis, but the axes of h and A are not principal axes. Principal Axes. For the axis Db through the vertex B and middle point of AC we have^ = o and A equal to the angle ABb, and hence c sin (8 A) - sin 6, and equation (3) becomes I B = P sin 1 6. We also have Bb sin 8 = c sin A, or sin s O = Hence, substituting this value of sin* 6, In the same way we have for the axis Cc and for the axis Aa c* sin' A c* sin* A 9 sin 3 A 0V 216 sin 1 A I m* 216 216 0~a* We see. then, that the moments of inertia relative to Bb, Cc and Aa are proportional to . -= , -=5 . Ob Oc Oa Hence if we inscribe an ellipse in the triangle tangent at b, c and a, it will be the central ellipse of inertia, and the reciprocal of the square of any semi-diameter will be proportional to the moment of inertia for the coincident axis. If, then, we draw the semi-diameter Oe parallel to AC, we have _, _ mJV sin* A _ Oe = 2i6/ 6 12' Now Oe and Ob are conjugate semi-diameters, and since the parallelogram upon two conjugate semi-diameters is equal to the rectangle of the principal semi-axes, /T /i x Oe sin 6 = /T7i Ob . . sin 8 6 Inserting the value of sin 1 and of Oe* already found, we have JV,in", * y j;2 CHAP. III.] We have also MOMENT OF INERTIA EXAMPLES. 45 Solving these two equations we obtain i2^V sin 2 A, *A\. We thus know the central principal moments of inertia I x = and / If the principal axis of X makes the angle a. with the base b = AC, we have I b = J x cos 2 a + I y sin 2 a; or ^ c* sin 2 A = in/% cos" a + mfc 2 y sin 2 Hence cos 2 a= c* sin 2 A 18 We thus know the position of the principal axis of X and the principal moments of inertia I x , I , I z . The moment of inertia for any axis through the centre of mass is then given by / = /. cos 2 a + I y cos 2 ft + I z cos 2 y. (5) Find the moment of inertia for a homogeneous circular disc. ANS. Let b be the base AC of an isosceles triangle ABC, and r,r, the equal sides and h the altitude Then h? = r 9 and, from example (4), we have for the axis OZ through the centre of mass at right angles to the plane of the triangle For a parallel axis BZ' through the vertex B we have Now the circular disc may be considered as composed of an indefi- A / ^ nitely large number of isosceles triangles all having a vertex at the centre ^ of mass O, and an altitude h equal to r. We have therefore, for a circular disc, for the axis OZ through Y the centre of mass O at right angles to the plane of the disc Any three rectangular planes through the centre of mass are planes of symmetry. Hence (page 36) any three rectangular axes with origin at centre of mass O are central principal axes. We have also J z I x +f y ; or, since /* and I y are evidently equal, (5) For any axis in the plane of the disc we have, then, if p is the distance from the centre of mass to the axis, = cos 2 4 We can write this in the form /' = ~(r cos a + pY + ^(r cos a p)* + ~(r sin a p? + -g(r sin a + pY + MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. III. Hence the moment of inertia for a homogeneous circular disc is the same as for a particle of one eighth the mass of the disc at the extremities of any two rectangular diameters, and a particle of one half the mass of the disc at the centre. BY CALCULUS. Take an elementary circular strip of radius p and thickness dp. The area of this strip is ittpdp, and its mass is m = -zditpdp. Its moment of inertia for the axis OZ is, then, mp* = 2ditp*dp. For the disc, then, we have But Snr* is the mass m of the disc. Hence as already found. (6) Find the moment of inertia for a homogeneous hollow circular disc. ANS. Let r\ be the outer and r t the inner radius. Then, if 8 is the surface density, we have for the mass of a disc of radius /-, m, = *rr% and, from example (5), i _ r, , r, /, = m, = SitrS . . For the mass of a disc of radius r we have ma = 8itr*, and Hence the polar moment of inertia for the hollow disc is ~ But Sn(ri* ri 1 ) is the mass m of the hollow disc. Hence S t + r,'). Hence for any diameter (r, i BY CALCULUS. We have, as in the preceding example, 7 - f' 1 - as already found. CHAP. III.] MOMENT OF INERTIA EXAMPLES. 47 (7) Find the moment of inertia for a homogeneous elliptical disc. ANS. Let the semi-transverse axis Oa be a and the semi-conjugate axis Ob be b. described about the ellipse so that its radius Oa is equal to the semi-transverse axis a. Then we have for the ratio of the mass of any ele- ment ee of the ellipse to that of the corresponding ele- ment EEoi the circle ee _ bb b ~EE ~~BB = ~a' Hence the moment of inertia of the ellipse relative to the principal axis O Y is times that of the circle. In the same way the moment of inertia of the ellipse relative to the principal axis OX is -7- times that of the circle. We have then, from example (5), since Sna* is the mass of the circle and Snab = m is the mass of the ellipse, Let a circle be and for the axis OZ r-. _ a 1 + C Hence the moment of inertia for a homogeneous elliptical disc is the same as for a particle of one eighth the mass of the ellipse at the extremities of the two principal axes, and a particle of one half the mass of the ellipse at its centre of mass. (8) Find the moment of inertia for a homogeneous right parallelopipedon. ANS. We have seen, from example (3), that for a parallelogram we have one sixth of the mass at the middle of each side and two sixths the mass at the centre. If this parallelogram moves parallel to itself it describes the parallelopipedon, and the centre and middle points describe straight lines, aa', bb' , dd' , ed ', each of mass m\ = m and the straight line cc 1 of mass nti = ^m, where m is the mass of the parallelopipedon. We have, then, two systems of three lines each, viz., the system aa', dd' , each of mass m\ = ^m, and cc 1 of mass m t = ^m, and the system bb' , ee' , each of mass m\ = ^m, and cc 1 of mass m* = ^m. From example (2), page 39, we have, for the first system tn\ = ^m at the middle points of aa' and dd' , and \m\ = y^S at c and c 1 . For the second system we have #/, = m at the middle points of bb' and ?vbe the radius of gyration of the base for any line in the plane of the base. Then the radius of gyration of the slice for any parallel line in its plane is y^. The moment of inertia of the slice ft relative to a parallel line through the vertex V is then mfkl , _ SAktydy SAy'dy ~tf~ ~~h 4 + ~~W~' Integrating between the limits y = h and y = o, we have, since m = - is the mass of the cone Or pyramid, for an axis through the apex A at right angles to the axis of the cone or pyramid For a parallel axis through the centre of mass O h = I' b - m(f^) 2 = |mJ + ^gtf' ............ (I) Let k v be the radius of gyration of the base for the vertical axis of the cone or pyramid. Then the radius of gyration of the slice for this axis \&jk v . The moment of inertia of any slice relative to this axis is then Integrating between the limits y = k and/ = o, we have for the moment of inertia relative to the geometric axis . . ...... .... (2) Equations (i) and (2) are general and hold good for any base. We have only to substitute for fo and k v their values in any case. MEASURABLE RELATIONS OF MASS AND SPACE. [CHAP. III. CASE i. For a right pyramid with parallelogram base we have (example 3) for a line through the centre of mass o of the base parallel to AB = b if. _ 4- . Example. Let a point move in a circle of radius r ft. through a half circumference, from a t to a. Take the pole O in the perpendicular OC through the centre C of the circle, so that Oa\ = Oa = I ft. The successive positions of the radius vector are then elements of a cone. If we develop this cone, we have a circular sector a t Oa, whose radius is / ft., length of arc s itr ft., and whose developed angle a } Oa= is the angle described. Hence = itr, or Tfr ,. radians. Linear and Angular Displacement. Let a l and a, as before, be the initial and final positions of a point moving in any path. Then, as already defined, the distance s from a v to a, measured along the path, is the distance described relative to flj, and the angle swept over by the radius vector is the angle described relative to O. Now draw the straight line a^a from the initial position tfj to the final position a. This line a^a is -the LINEAR DIS- PLACEMENT of the moving point. We denote its length by D. If it is indefinitely small, the two points a l and a be- come consecutive, and the displacement D = ds is then an element of the path. Denote the angle a^Oa subtended at O by the linear displacement D by 6. This angle is the ANGULAR DISPLACEMENT of the moving point relative to O. If it is indefinitely small, we have dQ = d. Linear displacement is then measured in units of length, as feet, and angular displacement in radians. Example. In the preceding example we have seen that the distance described is 5 = itr feet, and the angle described is = radians. The linear displacement is, however, D = 2r ft. in a direction from , to a\ and the angular displacement is given by r _ . e /sin 2 r, or sin = -7" Line Representative of Linear and Angular Displacement. We see, then, that the linear displacement a^a D is a straight line. It has therefore both magnitude and direc- tion. Thus a straight line a^a represents by its length the magnitude D ft. of any linear displacement, and the arrow indicates the direc- tion. The linear displacement is thus completely represented in both magnitude and direction by a straight line and arrow. CHAP. I.] VECTOR QUANTITIES. 53 In the same way we can represent angular displacement. Thus if a l and a are the initial and final positions of a point and O the pole, so that a^Oa 6 is the angular displacement, then a straight line AO through the pole at right angles to the plane of a^Oa represents by its length the magnitude of the angular displacement, and the arrow indicates that if we look along this line in the direction of the arrow, the rotation of the radius vector is always seen clock- wise. In the figure, for instance, from Oa l to Oa. The line representative AO coincides, therefore, with the axis about which the radius vector turns. By " direction " of angular displacement we always mean the direction of its line repre- sentative as indicated by its arrow. Thus if the radius vectors Oa l and Oa in the figure lie in a vertical east and west plane, the direction of the angular displacement is north, because the axis is a north and south line, and if we look north along the axis we see rotation of the radius vector clockwise. Example. Let a point move in a circle of radius r ft. in an east and west plane. The point starts from the top and moves east-ward through a quadrant. Find the distance described, the angle described, the linear displacement, the angular displacement, for pole at centre. ANS. The distance described is s = ft. The angle described is = radians. The linear displacement is a^a = r Vz ft., making an angle of 45 with the initial radius Oa\. The angular displacement is 6 = radians north, because the line representative AO must have its arrow pointing north in order that rotation of radius vector may be seen clockwise when we look in the direction of the arrow. Note that "angle described" 0= radians has magnitude only, while "angular displacement" has in this case the same magnitude and also direction, and is 9 = ~ radians north. Vector Quantities. All quantities which have magnitude and direction so that they can be represented by straight lines and arrows are called VECTOR quantities. Linear and angular displacement are thus vector quantities. In order that they may be known, therefore, magnitude and direction must be given. Magnitude alone is not sufficient. Thus, in the preceding example, it is not sufficient to say that the linear displacement is r 1/2 ft. This is the magnitude only. The direction must also be stated. So, also, it is not sufficient to say that the angular displacement is 6 = - radians. In this case this is the angle described, but it is only the magnitude of the angular displacement. The direction must also be stated. Thus the angular displacement is = radians north. Displacement in General. The term "displacement" always signifies linear displacement unless otherwise specified. CHAPTER II. RESOLUTION AND COMPOSITION OF LINEAR DISPLACEMENTS. Resolution and Composition of Linear Displacements. Suppose a point to move by any path from A to B, so that AB is the line representative of the linear displacement. Then let it move by any path from B to C, so that BC is the line representative of the second displacement. It is evident that AC is the line representative of the RESULTANT displacement, when the two displacements AB and BC are thus successive. These two displacements AB and BC are called COMPONENT DISPLACEMENTS. Suppose, however, that the two component displacements, instead of being successive, are simultaneous. That is, while the point goes from A to B, let the line AB move parallel to itself to DC, so that the end B arrives at C at the same instant that the point arrives at B. Then it is evident that AC is still the resultant displacement. We see, then, that if a point has two component displacements AB, BC, either simul- taneous or successive, the resultant displacement AC due to combining them is easily found. This combination is called COMPOSITION of displacements. Also, if we have any given displacement AC, we can find the equivalent component displacements, simultaneous or successive, in any two directions we please. This is called RESOLUTION of displacement. We can RESOLVE a displacement into equivalent components, or we can COMBINE the components into an equivalent resultant. Triangle and Polygon of Linear Displacements We can express this composition or resolution as follows: If two sides of a triangle, AB, BC, taken the same way round, as shown by the arrows, give the linear displacements, simultaneous or successive, of a point, then the third side, AC, taken the opposite way round, as shown by the arrow, gives the resultant displacement. Inversely, any displacement AC can be resolved into two components AB and BC, simultaneous or successive, in any two desired directions, by completing the triangle ABC, and taking AB and BC the opposite way round. This is called the principle of the " triangle of displacements." Again, if we had a third displacement , CD, the resultant of AC already found, and CD is AD. But AC is the resultant of AB and BC. Hence AD is the resultant of AB, BC and CD, either simul- taneous or successive. Hence if any number of displacements, simultaneous or succes sive, are given by the sides of a polygon A B CD, etc., taken tht same way round, the line AD which closes the polygon, taken the other way round, gives the resultant displacement. 54 CHAP. II.] LINEAR DISPLACEMENT. EXAMPLES. 55 This is called "the principle of the " polygon of displacements" The triangle of dis- placements is evidently only a special case. Rectangular Components. When a displacement is resolved into two components at right angles, the components are RECTANGULAR COMPONENTS. Unless otherwise specified, when we speak of the components of any displacement, rectangular components are to be understood. Component of the Resultant equal to the Algebraic Sum of the Components of the Displacements. It is evident that the resultant of any two given displacements is equal to the algebraic sum of their components along the resultant. For if AB and BC are the given displacements, the re- sultant AC is equal to the sum of the components Ad and dC. So also for any number of displacements, the resultant AE is equal to the algebraic sum of the components of the displace- ments along AE. The component in any direction of the resultant itself is then equal to the algebraic sum of the components of the dis- placements in the same direction. Thus the projection ae of the resultant AE upon any line OP, that is, the component of AE along OP, is the same as the algebraic sum of the components of AB, BC, CD, DE along OP. That is, the component of the resultant in any direction is equal to the algebraic sum of the components of tlie displacements in that direction. Examples. (i) A point has three displacements, N. 60" E., 40 ft.; S. soft.; W. jo" N., 60 ft. Find the resultant displacement. ANS 10 4/3 ft. W. (2) Three component displacements have magnitudes represented by I, 2 and j, and directions given by th^ sides of an equilateral triangle. Find, the magnitude of the resultant. ANS. 4/3. (3) Show that the resultant of two equal displacements of magnitude a, inclined 60, is equal to the resultant of a and 2a inclined 120 . (4) To a man in a balloon the starting-point bears N. 20 E., and is depressed jo below the horizontal. A point at the same level as the starting-point and 10 miles from it is vertically below him. Find the compo- nent displacements of the balloon relative to the starting-point. ANS 9.39 miles south ; 3.42 miles west ; 5.77 miles high. Relative Displacement. Since the displacement of a point is change of position, it can only be determined by reference to some chosen point of reference (page 1 1). Thus if a v and a are the initial and final positions of a moving point A, and B is some chosen point of reference, the positions a 1 and a relative to B are known, if we know the radius vectors Ba^ and Ba and the angle a^Ba = 0. To an observer at B the point A is seen to move from a l to a through the angle #, and a^a is then the displacement of A relative to B. A line ab, then, parallel and equal to a^a ivitJi arrow at b gives this displacement a^a in magnitude and direction, and it also indicate: by the end letters that the line ab with arrow at b is the displacement of A relative to B. KINEMATICS OF A POINT. GENERAL PRINCIPLES. [CHAP. II. Now to an observer on the moving point A we see, by completing "the parallelogram, that the point B would appear to the observer on A to move from B to B H , through the same angle 0, and BB n parallel, equal and opposite to a l a gives the displacement of B rela- tive to A. Aline ba, then, parallel and equal to BB Ht with arrow at a, gives the displacement />/> in magnitude and direction, and // also indicates by the end letters that the line ba with arrow at a is the displacement of B relative to A. Hence any change in the relative position of any two points A and B may be regarded as a displacement of A relative to B, as indicated by the line ab with arrow at b, or as an equal and opposite displacement of B relative to A, as indicated by the line ba with arrow at a. Notation. The student should notice carefully the preceding notation. A relative displacement is given in magnitude and direction by a 'straight line and arrow. The letter at the arrow end denotes the point of reference : that at the other end the moving point. Thus the line ab with arrow at b gives the magnitude and direction of the displacement of A relative to B. The equal parallel line with arrow at a gives the displacement of B relative to A. Triangle and Polygon of Relative Displacements. Except for notation we have changed nothing, and the principles already established still hold. Thus let a moving point A have a displacement ab relative to some point B, so that it moves from a to b, and at the same time, or afterwards, let the point B have the displace- ment BB H relative to some point C. Then the line be through b, parallel and equal to BB n , gives the displacement of B relative to C, and it is evident that A moves from a to c, and hence ac is the resultant displacement of A relative to C. Hence if two sides ab, be of a triangle abc, taken the same way round, give the displacement of A rela- tive to B, and B relative to C, either simultaneous or successive, then the third side, ac, will give the resultant displacement of A relative to C if taken the opposite way round, from a to c. same way round, from c to a, it gives the displacement of C relative to A. This is the principle of the triangle of relative displacements. Again, let ab, be, cd, etc., be the line representa- tives of the displacements of A relative to B, B rela- tive to C, and C relative to D, etc. Then if we lay off these displacements we obtain the polygon abed. As we have just seen, ac gives the displacement of A relative to C. The line dv/ gives then the displacement of A relative to D. Also any line in the polygon, as bd t gives the displacement of B relative to D. Hence if any number of relative displacements, simultaneous or successive, are given by the sides of a polygon abed, etc., the line which closes the polygon, taken the opposite way round, gives the displacement of the first point relative to the last ; taken the same way round, the displacement of the last relative to the first. If taken the CHAP. II.] RELATIVE DISPLACEMENT EXAMPLES. 57 The triangle of relative This is the principle of the polygon o f relative displacements. displacements is evidently a special case. Examples. (i) A circle of radius r rolls on a horizontal plane until it turns through a quarter revolution. Find tJie displacement of the point of the circle initially in contact with the plane relative to the point diametrically opposite. ANS. Let O be the initial point of contact on the plane, and A the corresponding point of the circle, initially at O, and B the point of the circle diametrically opposite. When the circle rolls through a quadrant, A has moved to A n , and B to B n . The displacement of A relative to the fixed point O of the plane is given by the line from A to A n . A parallel and equal line aO with arrow at O gives, then the displacement of A relative to O. The displacement of B relative to O is the line BB n . A parallel and equal line bO with arrow at O gives then the displacement of B relative to O. The same line Ofrvfith arrow at b gives the displacement of O relative to B. If, then, we lay off these displacements in order, aO, Ob, the closing line ab gives the displacement of A relative to B. This displacement we then easily find to be 2r ^2, making an angle of 45" with the vertical. The same line ba taken in the opposite direction gives the displacement of B relative to A. (2) Two railway trains A and B run, one north- east a distance d, the other southeast the same distance. Find the displacement of A relative to B. ANS. The displacement of A relative to some fixed point E on the earth is represented by the line ae with arrow at e. The displacement of B relative to the same point E is represented by the line be with arrow at e. The same line with arrow at b gives, then, the displacement of E relative to B. Laying off these displacements ae, eb in order, we find ab' the displacement of A relative to B, to be d ^2 in a direction north. (3) A point A moves jo ft. in a given direction relative to a fixed point P. Another point, B, moves relative to P 40 ft. in a direction at right angles to the direction of A. Find the displace- ment of A relative to B. ANS. 50 ft. in a direction inclined to the direction of A by an angle whose tangent is |-. (4) The displacement of a point A relative to a point B is a distance 6 ft. south, and relative to a point C 5 ft. west. If C is initially a distance 4ft. south of B, find the final position of C relative to B. , ANS. Distance of Cs final position from B is 5 4/5 ft., and the direction from B to the final position of C is east of south by an angle whose tangent is . CHAPTER III. RESOLUTION AND COMPOSITION OF ANGULAR DISPLACEMENTS. Resolution and Composition of Finite Successive Angular Displacements about Different Axes. Let O be a fixed point or pole, and let OA^ and OB 2 be two given axes, intersecting at O and making the angle A 1 O 2 = ex. Suppose successive finite angular displacements about these axes in the following order: ist. An angular displacement 9 l about OA r 2d. An angular displacement 2 about OB V It is required to find the resultant angular displacement. CONSTRUCTION. Draw through O an axis OB l making the angle A l OB l a with OA l , and the angle B l -OB Z ^ with OB V Let the axis OA t and O l be rigidly fixed relative to each other, so that when the angular displacement 6 l takes place about OA^ , the axis OB l will turn into the position OB r Take the distances OA { = OB l = OB r Then if we have the angular displacement 0, about OA l , and next 0. 2 about OB 2 , the points B l and A^ move on the surface of a sphere of radius OA r Join A l B l , B 1 B 2 and B 2 A l by great circles of this sphere. Then the arc B l A l = a, arc B 2 A l = a, arc B^BI = 6 l , and in the spherical triangle B 1 A 1 B 2 the spherical angle at A x is 6 r Bisect this angle by a great circle A^D meeting B^B 2 at D. Draw a great circle B 2 R through B 2 inclined to B Z A^ by the spherical angle # 2 and meeting A X D at R. Then the resultant angular displacement is about the axis OR. PROOF. Draw the arc B 2 R' making the same angle, # 2 , with B 2 A l on the other side, and make the arc B 2 R' = B 2 R. Then A^R' will equal Aft, and the angle B Z A^R' = $0, = B^A.R. Now when the angular displacement 0, takes place about OA^ , the axis OB l moves to OBi through the angle B^A^B^ = B v , and the axis OR, if rigidly fixed relative to OA 1 , will move to OR' through the angle RA^R' = 0,. Next, when the angular displacement 2 takes place about OB 2 through the angle R'B^R 3 , the point R' evidently moves back to R. Hence the axis OR has the same position before and after the angular displacements. The resultant angular displacement is then about the axis OR. t MAGNITUDE AND POSITION. For the position of this resultant axis and the magnitude of the resultant angular displacement about it we have in the spherical triangle A X RB 2 the side A^B 2 = a, the angle A V B^R = $0 2 and the angle RA& = %6 r Also the angle B 2 RA, = 1 80 0. Hence we have for the magnitude of the resultant angular displacement 6 the equation cos = cos t cos i0 2 sin 0, sin $0, cos a (I) 58 CHAP. III.] ANGULAR DISPLACEMENTS IN GENERAL. 59 Also for the direction of OR sin ROA l _ sin ROB Z _ sin a ,,,,. sin %6 2 ' sin \B^ ~ sin %V From (I) we can find 6, and from (II) we can find the angles ROA^ and ROB 2 which the axis OR makes with the axes OA l and OB^. Examples. (i) The telescope of a transit instrument initially horizontal and pointing north is first turned to an altitude of 60 and then turned to the west. Find the resultant angular displacement. Axs. We have 6. = 60, a = 90, a= 90. Hence, from (I), cos |0 = -1. ^, or 6 = 104 28' 39", or 6 = 1.823 radians. Hence sin |Q = |4/|, and we have, from (II), for the position of the resultant axis sin ROAi = ^^ sin 45 = 2 y^, or ROA, = 63 26' 5.8"; sin ROB* = S ^^ sin 3 o r " = Vl or KOB* - 39 13' 53-4". (2) In the preceding example let the successive displacements be taken in reT.>erse order. ANS. We have then 9i = 90, 6 3 = 60", a. 90. Hence cos ^9 = \tf\ and sin = \^\, just as before. But for the position of the resultant axis i = 4/J and sin ROB* = 2|/f Resolution and Composition of Angular Displacements in General. We have just seen how to find the resultant for finite successive angular displacements about intersecting axes. For all other cases of angular displacement about intersecting axes we can combine and resolve angular displacements just like linear displacements, as explained in tlie preceding chapter, This can be shown as follows : All other cases fall under the head of (a) Finite or indefinitely small successive or simultaneous angular displacements about the same axis; (b] Indefinitely small successive or simultaneous angular displacements about different intersecting axes. (c} Finite simultaneous angular displacements about different intersecting axes. Let us consider these cases in order. (a) FINITE OR INDEFINITELY SMALL SUCCESSIVE OR SIMULTANEOUS ANGULAR DIS- PLACEMENTS ABOUT THE SAME AXIS. If the axis of all the angular displacements is the same, the plane of rotation of the radius vector does not change, and all the, line repre- sentatives lie in the same straight line. The resultant is then given by the algebraic sum of the line representatives, and this evidently holds whether the angular displacements are successive or simultaneous, finite or indefinitely small. (b) INDEFINITELY SMALL SUCCESSIVE OR SIMULTANEOUS ANGULAR DISPLACEMENTS ABOUT DIFFERENT INTERSECTING AXES. Let the angular displacements be indefinitely small and successive in the order ft l , # 2 . Then we have sin X = ^6 V sin # 2 = # 2 , sin fl = %0. Hence, from equations (II), : 6 l : O z : : sin a : sin ROB 2 : sin ROA l ....... (i) 6o KINEMATICS OF A POINT. GENERAL PRINCIPLES. [CHAP. III. Let OA l be the line representative of lf OBi the line representative of & 2 , and the angle A V OB Z = a. Complete the parallelogram and draw OR. Then in the triangle ROB 2 we have OR : 0, : 2 : : sin : sin ROB 2 : sin ROA r . (2) Comparing (2) with (i) we see that OR = 6, or the result- ant is the third side of the triangle. Hence we have the " triangle of angular displace- **, menfs," just as for linear displacements, page 54. The same evidently holds for simultaneous angular displacements indefinitely small. (c) FINITE SIMULTANEOUS ANGULAR DISPLACEMENTS ABOUT DIFFERENT INTERSECTING AXES. The same evidently holds for finite simultaneous angular displacements about the same or different axes. For we can divide each finite displacement into a number of indefinitely small displacements, and treat each pair as in the previous case. Hence for all cases except finite successive angular displacements about different intersecting axes, which case we have discussed on page 58, we can combine and resolve angular displacements by means of their line representatives, just like linear displacements in the preceding chapter. We have then the "triangle and polygon of angular displacements," just as for linear displacements, page 55. Also, just as on page 55, the component of the resultant in any direction is equal to the algebraic sum of the components of the displacements in that direction. Also, \ve can find relative angular displacement in precisely the same way (page 56). Examples. (i) A body has two simultaneons rotations of 2 radians and 4 radians about axes inclined 60 to each other. Find the resultant. ANS. 2 1/7 radians about an axis inclined to the greater component at an angle whose sine is _ . (2) A sphere with one of its superficial points fixed has tiuo simultaneous rotations, one of 8 radians about a tangent line and one of 75 radians about a diameter. Find the axis of the resultant angular displacement, the resultant angular displacement, and the number of revolutions about the resultant axis. ANS. Axis inclined to greater component at an angle whose tangent is . Resultant angular displacement 17 radians. Number of revolutions = about 2.75 revolutions. (3> A pendulum suspended from a point in the prolonged polar axis of the earth swings in a plane through the axis. Find the angular displacement of this plane relative to the earth in t2 hours. ANS. The angular displacement of the earth relative to the plane is EP = it radians north. That is, \i we look north along the polar axis the rotation is seen clockwise. The angular displacement of the plane relative to the earth is then PE = n radians south. Chat is, an observer on the earth looking south along the polar axis sees the plane rotate clockwise through 180 in 12 hours. (4) Let the pendulum be suspended from a point in the prolonged radius of the earth at a place of latitude A. and siving in a meridian plane. Find the angular displace ment of the plane relative to the earth in 12 hours. ANS. The line representative of the angular displacement of the earth relative iy the plane is still EP = it radians north. The component of this along the radius of the place is Ep = it sin A radians. The angular displacement of the plane relative to the earth is then pE. That is, an observer at A looking towards the centre of the earth would see the plane shift clockwise through Jt sin A radians in 12 hours. CHAPTER IV. LINEAR AND ANGULAR SPEED AND VELOCITY. Mean Linear and Angular Speed. Let a point move in any path from the .initial position a l to the final position a in the time /. Let s be the distance described (page 51). Then -, or the distance described per unit of time, is the MEAN LINEAR SPEED. Let be the angle described (page 51). Then , or the angle described per unit of time, is the MEAN ANGULAR SPEED. Linear speed is then measured in feet per second, and angular speed in radians per second. Example. Let a point move in a circle of radius r ft. through a half circumference, from a\ to a, in / = 2 seconds. Then the mean linear speed is = ft. per sec. Take the pole O in the perpendicular OC through the centre C of the circle, so that Oai Oa /ft. a, Then (page 52) the angle described is ^ jadians, and the mean an- gular speed is f~rr radians per sec. Mean Linear and Angular Velocity. In the preceding figure a^a = D is the linear displacement (page 52), and the angle afla = is the angular displacement (page 52). Let t be the time of moving from a^ to a. Then , or the a linear displacement per unit of time, is the MEAN LINEAR VELOCITY, and -, or the angular displacement per unit of time, is the mean angular velocity relative to (9. Linear velocity is measured, then, like linear speed, in feet per second, and angular velocity, like angular speed, in radians per second. Example, In the preceding example we have seen that the mean linear speed is S ft. per sec., and the mean angular speed is ^ = ^ radians per sec. The mean linear velocity is, however, = ft. per sec. in a direction from , to a, and the mean angular velocity for pole at O is radians per sec., where is given by /sin |0 = r, or sin ^0 = -j , and has also direction as explained in the next article. Line Representatives of Mean Linear and Angular Velocity. Since mean linear velocity is linear displacement per unit of time, \ve can represent it by a straight line, just like linear displacement itself. 61 KINEMATICS OF A POINT. GENERAL PRINCIPLES. JC'IIAP. IV. Thus the straight line ao represents by its length the magnitude - of any mean linear velocity, and the arrow represents the direction. In the same way, since mean angular velocity is angular displacement per unit of time, we can represent it by a straight line like angular displacement itself. Thus the straight line AO at right angles to the plane of afia represents by its length a the magnitude of any mean angular velocity, and looking in the direction of the arrow from A to O the rotation of the radius vector is always seen clockwise. fey direction of angular velocity we always mean the direction of its line representative. Thus if the initial and final positions Oa l and Oa of the radius vector lie in a vertical east and west plane a^Oa, the direction of the mean angular velocity is north, because in such case the line representative AO would point north, so that looking in the direction of the arrow from A to O we should see the rotation of the radius vector clockwise. Distinction between Mean Speed and Velocity. Let the initial position of a point at the time / t be at a distance s l measured along the path from any fixed point in the path, taken as an origin or point of reference, and at any other time / greater than t l let the dis- tance measured along the path from this same fixed point be s. Then (s s v ) is the dis- tance described in the interval of time (t / t ), and the mean linear speed is In similar notation we have (0 0, ), the angle described in the interval of time (/ and the mean angular speed is - If s is greater than j,, the distance described is away from the origin and the mean linear speed is positive. If s is less than s lt the distance described is towards the origin and the linear speed is negative. So, also, if is greater than lf the angle described is away from the initial line of reference and the mean angular speed is positive. If is less than 0,, the angle described is towards the initial line of reference and the mean angular speed is negative. Thus if the distance from the origin is increasing the linear speed is positive, if decreas- ing it is negative. If the angle described from the initial radius is increasing the angular speed is positive, if decreasing it is negative. Mean linear speed, then, is mean time-rate of distance described, and mean angular speed is mean time-rate o r angle described. Both possess magnitude and sign, according to whether the distance or angle described increases or decreases with the time, but both are independent of direction of the path. Such quantities which have sign and magnitude but are independent of direction are called SCALAR quantities. They cannot be represented by straight lines. But, as we have seen, mean linear velocity is mean time-rate of linear displacement, CHAP. IV.] INSTANTANEOUS LINEAR. AND ANGULAR SPEED AND VELOCITY. e and mean angular velocity is mean time-rate of angular displacement. Both possess not only magnitude but direction. Such quantities are VECTOR quantities. They can be repre- sented by straight lines. We see, then, that mean speed and velocity, although measured in the same units, are quantities of a different kind, and the number of units are in general different for each. Example. Let a point move in a circle of radius r 10 feet. Let the plane of the circle be an east and west plane. Let the point move from the top a\ eastward through a half circumference, from a\ to a, in the time / = 3 seconds. Take the pole O in the perpendicular OC through the centre C of the circle, so that Oa t = On = / = 20 ft. a Then the distance described is s = itr 31.416 ft., and the mean linear speed is = -f 10.472 ft. per sec. The angle described is (page 52) = ~ = 1.5708 radians, and the mean angular speed is = + 0.5236 radians per sec. On the other hand, the linear displacement is D ir = 20 ft. in the direction from ai to a, and the mean linear velocity is = 6.66 ft. per sec. in the same direction. The angular displacement for pole at O is given by / sin |6 = r, or sin J0 = = , or 6 = radians A if * north, and the mean angular velocity isy = radians per sec. north. We see, then, that the mean linear speed and velocity, although measured in the same units, are in this case not only different in magnitude, but one is independent of direction, while the other has direction. So, also, for mean angular speed and velocity. Instantaneous Linear and Angular Speed and Velocity. The limiting magnitude of the mean linear or angular speed when the interval of time is indefinitely small is the INSTAN- TANEOUS linear or angular speed. These are SCALAR quantities having sign and magnitude but independent of direction, just like mean linear and angular speed. The limiting magnitude and direction of the mean linear or angular velocity when the interval of time is indefinitely small is the INSTANTANEOUS linear or angular velocity. Instantaneous linear and instantaneous angular velocity, then, are vector quantities, having both magnitude and direction, and can therefore be represented by straight lines, just like mean linear and mean angular velocity. Thus, let the interval of time be indefinitely small and denoted by dt. Then the initial and final positions a l and a of a moving point will become consecutive points of the path, so that the distance a^a can be denoted by ds. We see, then, that in this case the distance described is ds. That the linear displacement is ds in magnitude, and its direction is tangent to the path, so that a^a is its line representative. That is, in this case, distance described is the magnitude of the linear displace- ment. If, then, we denote the instantaneous linear speed by v, we can write ds and this will also be the magnitude of the instantaneous linear velocity, the direction of which is always tangent to the path. 64 KINEMATICS OF A POINT. GFNER4L PRINCIPLES. [CHAP. IV. Similarly, in the indefinitely small time dt, the indefinitely small angle described is a^Oa, which can be denoted by d. The indefinitely small angular displacement dO will be the same in magnitude, and its direction will be at right angles to the plane of rotation a^Oa, so that AO is its line representative. That is, in this case, angle described (*/0) is the magnitude of the angular displacement (dff). If, then, we denote the instantaneous angular speed by co, we can write dd = -dt> ana" this will also be the magnitude of the instantaneous angular velocity, the direction of which, as indicated by its line representative AO, is always at right angles to the plane of rotation, so that looking in the direction of its arrow we see the rotation of the radius vector clock-wise. The line representative AO coincides, therefore, with the axis about which the radius vector rotates. Examples. (i) Let the distance described by a moving point be given by the equation s = y/ + 8f>, (i) where s is the number of feet described in any number of seconds t. Then for any other number of seconds /, we can write J. = 7A + 8/, (2) Suppose /i to be less than /. Then subtracting (2) from (i) we have for the distance s Si described in the interval of time / t\ j-j, = 7(t - /,) + 8(/ 2 - /,) = ;(/ - /,) + 8(/ + /!)(/ - /,). The mean linear speed is, then, for this interval of time 7 + 8(/ + /,) , (3) We see from (3) that as the interval of time / t\ decreases, t\ approaches equality with / and the mean linear speed given by (3) approaches the limiting value 7 + i6/. We have then for the instantaneous speed, or the magnitude of the instantaneous velocity, v = ~ = 7 + i6/. . . (4) In order that the linear velocity maybe completely known we must specify, in addition to its magnitude as given by (4), its direction, which is always tangent to the path at the instant. Students familiar with the Calculus will note that (4) is obtained directly from (i) by differentiating s with reference to /. If in (4) we make / = 2, we have v = 39 ft. per sec. This does not mean that the point has described 39 feet in the preceding second, nor that it will describe 39 feet in the next second. But it means that at the instant (2 seconds from the start) it is moving at such a rate that, if that rate did not change, the point would describe 39 feet in the next second. It is the instantaneous speed. (2) Let a point move in a circle in a vertical east and west plane, and let the angle described by the radius vector from a fixed point to the moving point be given by the equation = 7 t + Bf, where is the number of radians described in any number of seconds t, the origin being at the centre. Find, as in the preceding example, the mean and instantaneous angular speed and velocity, ANS. The mean angular speed is \^j i = 7 + 8(/ + /,). The instantaneous angular speed is This is the magnitude of the instantaneous angular velocity, whose direction is north if the point i moving eastward in the plane from the top point. CHAPTER V. LINEAR AND ANGULAR VELOCITY. Speed and Velocity in General. The terms "speed" and "velocity" always signify instantaneous linear speed and velocity unless otherwise specified. The terms linear and angular speed and velocity always signify instantaneous linear and angular speed and velocity unless otherwise specified. Uniform and Variable Linear Velocity. When the linear velocity has the same magni- tude and direction whatever the interval of time, it 'is UNIFORM. If either magnitude or direction change, it is VARIABLE. If, then, the velocity is uniform, the line representative has always the same magnitude and direction ; and since velocity is tangent to the path, we have uniform speed in a straight line. The velocity in this case is the same as the mean velocity for any interval of time. If only the direction changes, we have uniform speed in a curve. In this case the mag- nitude of the velocity is the same as the mean speed for any interval of time. If only the magnitude changes, we have variable speed in a straight line. If both magnitude and direction change, we have variable speed in a curve. Examples. (i) A point moves with uniform speed in a circle of radius r = 6ft. and makes a half revolu- tion in the time t = j sec. Find the mean speed, the mean velocity, the instantaneous speed and the instant an- eous velocity. ANS. The mean speed is = lit ft. per sec., and the mean velocity is = 4 ft. per sec. in the direction of the diameter through the starting-point. Since the speed is uniform, the instantaneous speed must be the same as the mean speed, or zit ft. per sec. The magnitude of the instantaneous velocity is the same as the instantaneous speed, or v = lit ft. per sec., but the direction at any instant is tangent to the path at the position of the point at that instant. (2) Criticise the statement, "a point moves in a circle with ttniform velocity." ANS. A point can move in a circle or in any path with uniform speed. But if the velocity is uniform, it can only move in a straight line with uniform speed. A point cannot move in a curve with uniform velocity. Uniform and Variable Angular Velocity. By direction of angular velocity we always mean the direction of its line representative (page 62). When the angular velocity has the same magnitude and direction whatever the interval of time, it is UNIFORM. If either magnitude or direction change, it is VARIABLE. If, then, the angular velocity is uniform, the line representative has always the same magnitude and direction, and we have uniform angular speed in an unchanging plane. The angular velocity in such case is the same as the mean angular velocity for any interval of time. If only the direction changes, we have uniform angular speed, but a changing plane of rotation. In this case the magnitude of the angular velocity is the same as the mean angular speed for any interval of time. 65 66 KINEMATICS OP A POINT. GENERAL PRINCIPLES. [CHAP. V. If only the magnitude changes, we have variable angular speed in an unchanging plane. If both magnitude and direction change, we have variable angular speed and a changing plane of rotation. Examples. (i) A point moves with uniform angular speed in a circle of radius r = 6ft. in a vertical east and west plane, and makes a half revolution in the time t = 2 sec. Find the mean angular speed, the mean angular velocity, the instantaneous angular speed and velocity, origin at the centre. ANS. The mean angular speed is radians per sec. The mean angular velocity, if the point moves eastward from the top, is also - radians per sec. north. Since the angular speed is uniform, the instantaneous angular speed is the same as the mean speed. The magnitude of the instantaneous angular velocity is the same, but its direction, if the point moves east- ward from the top, is north. (2) Criticise the statement, " a point moves in a circle the plane of which is constantly changing, with uniform angular velocity." ANS. A point can move in a circle the plane of which is changing, with uniform angular speed. But if the angular velocity is uniform, its line representative does not change either in magnitude or direction, and the plane therefore cannot change. A point cannot move in a changing plane with uniform angular velocity. Resolution and Composition of Linear Velocity, Since linear velocity is linear dis- placement per unit of time when the interval of time is indefinitely small (page 63), it has magnitude and direction, and can be represented by a straight line, just like linear displace- ment itself. Therefore all the principles of Chapter II, page 54, which hold good for linear displacements hold good also for linear velocities. We have, then, the " triangle and polygon of velocities" and can combine and resolve velocities just like displacements. We also have relative velocity with similar notation as for displacements (page 55). Examples. (i) A ship sails N. jo* E. with a speed of 10 miles an hour. Find its velocity east and north. ANS. 5 miles per hour east, 5l/3~miles per hour north. (2) Find the vertical velocity of a train moving up a i-per-cent O- * gradient at a speed of 30 miles per 'hour. ANS. 0.3 miles per hour. C > P (3) A circle rolls on a horizontal plane. Its centre moves with a velocity v towards the east. Find the velocity of the top and C- b bottom points relative to the plane. ANS. The velocity of the top point a relative to the centre C *< *P >s given by the line ac v towards the east ; of g G c , the centre C relative to the plane by cP = v tow- c > p ards the east; of the bottom point b relative to the centre C by be v towards the west. If, then ,we lay off ac and cP in order, we have the velocity of a relative to the plane P given by aP 2v towards the east. The top point has, relative to the earth, twice the velocity of the centre. If we lay off b relative to C, and C relative to P, we have b relative to P, zero. The G bottom point is at rest relative to the earth. B (4) A ball let fall in an elevator has a velocity relative to the ground of 32 feet per sec., while the elevator has a velocity relative to the ground of T2 feet per sec. Find the velocity of # the ball relative to the elevator when it is rising and falling. ANS. The velocity of the ball relative to the ground is given by BG = 32 downwards. If elevator is rising, we have elevator relative to ground given by EG = 12 upwards. If elevator is falling, EG = 12 downwards. In the first case, laying off ft relative to G. and G relative to E, we have B relative to 44 feet per sec. down. In the second case, laying off B relative to G, and G relative to E, we have B relative to E 20 ft. per sec. downwards. CHAP. V.] RECTANGULAR COMPONENTS OF VELOCITY. Rectangular Components of Velocity. Let a point P be given by its co-ordinates x, y, z, and let it have the velocity v, whose line representative makes the angles <*, ft, y, with axes through P parallel to the co-ordi- nate axes OX, OY, OZ, respectively. Let the components of v parallel to these axes be v x , v y , v t , respectively. Then we have dx v x = " v cos a > v y = - = v cos ft, dz = =v cos y. Analytic Determination of Resultant Velocity. If the point P has several simulta- neous velocities, v 2 , etc., making the angles (ar p ft^, yj, (<* 3 , p y y 2 ), (a a , fl s , y 3 }, etc., then we have for the resultant components dx . . . = ^v cos a, v y = -j- = v l cos P l -j- z/ 2 cos /? 2 -(- z/ 3 cos ^ 3 -J- . . . = 2v cos /?, dfe = -^^cos y. In these summations we must take components in the directions OX, OY, OZ positive, in the opposite directions negative. We have, then, for the magnitude of the resultant velocity v V=~- + Vv x *+ V? + V? (2) This resultant passes, of course, through P, and its direction cosines are cos or = , cos ft = , z; ' z> cos Y = Since cos 2 a -|- cos 2 ft -\- cos 2 y I, we have also, from (3), v = v x cos a-\-v y cos ft -j- v, cos y. (3) (4) In determining these cosines we take v always positive and measure all angles a, ft, y in such directions that their projections on the co-ordinate planes XY, YZ, ZX shall be positive from OX around to OY, from OY around to OZ, from OZ around to OX, as indicated by the arrows in the figure. If all the velocities are in one plane, as, for instance, the plane of XY, we make y = 90, and hence v z = o in (i), (2) and (3). Examples. (i) A fiotnt has the component velocities in the same plane XY, Vi =J2, Vt = 24, v 9 = j6, V* = 48 ft. per sec. , making angles with OX of i = + 60, a* = + fjo , (* = + 240, c- t = Jjo. Find the resultant velocity. 68 KINEMATICS OF A POINT. GENERAL PRINCIPLES. [CHAP. V. ANS. We have a ft = 90. Hence ft\ 30, /* = + 60, fi = + 150, /0 = + 240. Hence z/, = 6 - i2f/3~~ l8 + 24VT= + 8.784 ft. per sec., v y = + 6 ^3 + 12 181/3 24 = 32.784 ft. per sec., The resultant is v = + 33.94 ft. per sec., making the angle a with OX given by cos a = i = 0.2588, or a = about 75' T 33-94 (2) ^4 /0. //*/ the resultant. ANS. We find the angles y by. the equation, page 13, cos" Y = - cos (a + ft) cos (a- ft). Hence y t = + 135, y* = + 45, y t = 90. . We have, then, v x 20 + 25 + 30 = + 75 ft. per sec., z/y = 20 + 25 + 30 4/3"= -f 96.96 ft. per sec., v, = -^L + -5?_-r- o = -f- 7.071 ft. per sec. The resultant is v = 1 19.6 ft. per sec., and its direction cosines are cos a = .' cos ft = , cos y = , or a = 51 ,o', /J = 35 50', y = 86" 37'. Resolution and Composition of Angular Velocity. Since angular velocity is angular displacement per unit of time when the interval of time is indefinitely small (page 63), it has magnitude and direction and can be represented by a straight line, just like angular displace- ment itself. But since the time is indefinitely small, and therefore the angular displacement indefinitely small, we have the case (), page 59. Therefore all the principles of Chapter II, page 54, hold good also for angular velocities whether simultaneous or successive, and we have the "triangle and polygon" of angular velocities, and can combine and resolve them just like linear displacements (page 55). We have also relative angular velocity with similar notation as for linear displacements (page 56). Examples. (i) A point on a sphere is rotating uniformly about a diameter at the rate of 10 radians per minute. Find the component angular "velocity about another diameter inclined 30 to the first. ANS. 5 4/3 radians per min. (2) A point has two simultaneous or successive angular velocities of 2 radians per sec. and 4 radians per sec. about axes inclined 60 to each other. Find the resultant. ANS. 2 y^ radians per sec. about an axis inclined to the greater component at an angle whose sine is (3) A sphere with one of its superficial points fixed has two angular velocities, either simultaneous or successive, one of 8 radians per sec* about a tangent line, and one of 15 radians per sec. about a diameter. Find the resultant angular velocity. ANS. 17 radians per sec. about an axis inclined to the greater component at an angle whose tangent is A. (4) A pendulum suspended from a point in the prolonged polar axis of the earth swings in a plane through the axis. Find the angular velocity of this plane relative to the earth. See example (j), page 60. ANS. 27t radians per day south. That is, an observer looking south along the polar axis sees the plane rotate clockwise at this rate. (5) Let the pendulum be suspended from a point in the prolonged radius of the earth at a place of latitude A. Find the angular velocity of the plane of s^i'iit^ relative to the earth. See example (4}, page 60. CHAP. V.] RECTANGULAR COMPONENTS OF ANGULAR VELOCITY. 69 ANS. -2.it sin A radians per day in a direction towards the centre of the earth. That is, an observer looking towards the centre of the earth would see the plane rotate clockwise at this rate. - sm The time of a complete revolution of the plane would then be = - days. If A is 60 the time oi sin A ' revolution would be 7- = i.iSS days. At the equator sin A = o, and the plane does not rotate relatively to the earth. At the poles sin A = i, and the plane makes a revolution in i day. Rectangular Components of Angular Velocity. Let a point P be given by its co- ordinates x, y, z, and let it have the angular velocity GJ, whose line representative passes through O and makes the angles a, ft, y with the axes OX, O Y, OZ, respectively. Then the components of GO along the axes are oo x oo cos a, oo y = GO cos ft, &, = GO cos y. These components are positive in the direc- tions OX, OY, OZ. Since rotation is always seen clockwise when we look along a line representative in the direction of its arrow, we have, then, positive rotation in the plane XY from OX around to OY; in the plane. FZ from O Y around to OZ; in the plane ZX from OZ around to OX, as shown by the arrows in the figure. Analytic Determination of Resultant Angular Velocity. If the point P has several simultaneous angular velocities, ca lf co 2 , 6? 3 , etc., all passing through O and making the angles (a v /? 1( , y^, (a z , ftj, y 2 ), (ar 3 , /? 3 , j/ 3 ), etc., then we have for the resultant components GO X = co l cos ct l -\- oo 2 cos oc 2 -\- 6? 3 cos G0 y w l COS fi l -f- OL> 2 COS ft 2 -f- < s COS co z = oj l cos &> l -f- GJ Z cos y 2 -f- ce? 3 cos -{-= 2oj cos a ) + . . . = ^GO COS ft, -}- . . . = 2 on cos y. In these summations we must take components in the directions OX, OY, OZ positive, in the opposite direction negative. We have, then, for the magnitude of the resultant GO = -f V<*>* 2 -\- GO* -f w? This resultant passes, of course, through O, and its direction cosines are COS Of = - COS ft , G? OJ GO- cos y = - (3) Since cos 2 a -(- cos 2 ft -\- cos 2 y = i, we have also, from (3), GO GJ^ cos a -f- Go y cos ft -\- GJ Z cos y. (4) In determining the cosines, we take GO always positive and measure all angles a, ft, y in such directions that their projections on the co-ordinate planes XY, YZ, ZX shall be posi- ?o KINEMATICS OF A POINT. GENERAL PRINCIPLES. [CHAP. V. tive from OX around to OY, from OY around to OZ, from OZ around to OX, as indicated by the arrows in the figure. If all the velocities are in one plane, as, for instance, the plane of XY, we make y = 90 and hence , = /.?, a, a = 24, o>> = j6. o4 = 48 radians per sec., making angles with OX, a, = -f- 60, a = + /jo, a = + 240", <* 4 = + Jjo. Find the resultant velocity. (See example (/), page 68). ANS. oo x = + 8.784 radians per sec.; = v, or GO = - = ^^_ ra dians per sec., and its direction is north. CHAPTER VI. LINEAR AND ANGULAR RATE OF CHANGE OF SPEED. LINEAR ACCELERATION. Change of Speed. When the speed of a point varies in magnitude, the difference for any interval of time between the final and initial instantaneous speeds is the change of speed for that interval of time. Mean Rate of Change of Speed. The change of speed pr unit of time is the MEAN RATE OF CHANGE OF SPEED. Thus, if the linear speed of a point at any time / is v, and at an earlier time t l it is t\ , then the change of linear speed is (v z/J for the interval of time (t t^, and the mean rate of change of linear speed is This is positive if v is greater than v lt or if the linear speed increases with the time, and negative if v is less than v lt or if the linear speed decreases with the time. Similarly, if the angular speed of a point at any time t is &?, and at an earlier time, t x , it is co lt then the change of angular speed is (GO GO^ for the interval of time (t / x ), and the mean rate of change of angular speed is GO Gflj * *l ' This is positive if &) is greater than < lt or if the angular speed increases with the time, and negative if oj is less than o^, or if the angular speed decreases with the time. Rate of change of speed, whether linear or angular, is, then, like speed itself, a SCALAR quantity having magnitude and sign but independent of direction (page 62). The unit of rate of change of linear speed is evidently one unit of linear speed per unit of time, or I ft.-per-sec. per sec. The unit of rate of change of angular speed is one unit of angular speed per unit of time, or i radian-per-sec. per sec. Instantaneous Rate of Change of Speed. The limiting value of the mean rate of change of speed when the interval of time is indefinitely small is the INSTANTANEOUS RATE OF CHANGE OF SPEED. We have, then, the instantaneous rate of change of linear speed given by dv and the instantaneous rate of change of angular speed given by doa ~dt' 74 KINEMATICS OF A POINT. GENERAL PRINCIPLES. [CHAP. VI. If the instantaneous rate of change of speed is zero, the speed itself is uniform and therefore the same as the mean speed for any interval of time, which is also uniform. If the instantaneous rate of change of speed does not vary in magnitude, it is uniform and therefore the same as the mean rate of change for any interval of time, which is also uniform. If the instantaneous rate of change ot speed varies in magnitude with the time, it is variable. In such case the mean rate of change is also variable. ExampleSi (i) Let the distance described by a mowing point be given by the equation s = jt + St*, where s is the number of feet described in any number of seconds t. This is the same equation as in example (i), page 64, and we find, as before in that example, the instantaneous linear speed v = ~ = 7 + i6/ ............... (i) For any other time, /, , we can write v, = 7 + i6/, ................. (2) Suppose /, less than /. Then subtracting (2) from (i), we have for the change of speed for any interval of time (/ /,) v -vi = i6(/ - /.) ................. (3) The mean rate of change of speed is then v Vi - = 1 6 ft.-per-sec. per sec. * ii Since this does not change with the time, it is uniform and is the same as the instantaneous rate of change . dv of speed, -jj. (2) Let the instantaneous linear speed of a point be given by v = 7 -f ibt + 6t* ............... (i) For any other less time, /i , we have v, = 7 -f i6A -|- 6/1. Subtracting (2) from (i). we have for the change of speed for any interval of time (/ A) v vi- i6(/ - /,) -|- 6(/ - /') = i6(/ - /,) -f 6(/ -f /,) (/ - /,). The mean rate of change of speed is then This, we see, is variable and changes with the time. As the interval of time (/ /,) decreases, /i approaches equality with t and the mean rate of change of speed given by (3) approaches the limiting value 16 + i2/. We have, then, for the instantaneous rate of change of speed ' -,6+,*. This, we see, changes with the time and is therefore variable. (3) Let the angular speed of a point be g'iven by o> = 7 + i6/ + 6A where oo is the angular speed in radians per sec. for any instant t. CHAP. VI.] MEAN LINEAR ACCELERATION OF A POINT. Then we have, just as before, the change of angular speed for any interval of time (/ - /,) < - i = i6(/ - A) + 6(/ 2 - t\). The mean rate of change of angular speed is, then, 00 ft}, ~-/7 = l6 + 6(/ + /,), and the instantaneous rate of change of angular speed is da) Mean Linear Acceleration of a Point. Let a v and a be two positions of a point moving in any path and passing from a l to a in the time interval t /\. Let v l and v be the corre^ spending instantaneous linear velocities. FlG ,j FlG , Each velocity, Fig. (a), is tangent to the path at the corresponding point, and is equal in magnitude to the speed at that point, so that (v v v ) is the change of speed, v v, and - is the mean rate of change of speed (page 73). In Fig. (b) take any point B as a pole and lay off Bb^ and Bb, the line representatives of 7,'j and v, and join b l and b by a straight line. Then the straight line b v b is the line representative of the change of velocity, and J_ is the 0*Az# rate of change of velocity for the interval of time / / r / t l This is called the MEAN LINEAR ACCELERATION of the point for the interval of time. It is therefore a vector quantity given in magnitude and direction by a straight line parallel to b^b, having the direction from b l to b as shown by the arrow in Fig. (), and b.b equal in magnitude to . / t l Mean linear acceleration can be defined, then, as mean time-rate of change of velocity, whether that change takes place in the direction of the velocity or not. The unit of mean linear acceleration is evidently one unit of linear speed per unit of time, or I ft.-per-sec. per sec., just as for rate of change of linear speed (page 73). Instantaneous Linear Acceleration. The limiting magnitude and direction of the mean linear acceleration when the interval of time is indefinitely small is the INSTANTANEOUS LINEAR ACCELERATION. Thus if the interval of time dt is indefinitely small, the two positions a l and a of the mov- ing point, Fig. (a), are consecutive. FlG - () FIG. (6). In Fig. , then, Vr is the fit instantaneous linear acceleration. It is therefore a vector quantity given in magnitude and direction by a straight line parallel to /, having the direction from b^ to b as shown by the arrow in Fig. (b), and b,b equal in magnitude to ~-r- . 76 KINEMATICS OF A POINT. GENERAL PRINCIPLES. [CHAP. VI. Instantaneous linear acceleration can then be defined as limiting time-rate of change of velocity, whether that change (~y ) take place in the direction of the velocity or not. Its unit is evidently one unit of speed per unit of time, or i ft.-per-sec. per sec., just as for mean linear acceleration. We shall always denote instantaneous linear acceleration by the letter f. The term "acceleration" always signifies instantaneous linear acceleration unless otherwise specified. We shall see later how to determine it in magnitude and direction (page 78) ; it is sufficient to note here that if the path is a straight line, bfi coincides with v, and we have / = -r . That is, the magnitude of the acceleration in this case is the instantaneous rate of change of speed (page 73). Resolution and Composition of Linear Acceleration. Since acceleration is thus time- rate of change of velocity when the interval of time is indefinitely small, and can be represented by a straight line, it follows that all the principles of Chap. II, page 55, for displacements hold good also for accelerations. We can then combine and resolve linear accelerations just like displacements (page 56), and we have the triangle and polygon of accelerations just as for displacements. We have also relative acceleration with similar notation as for displacement (page 56). Let Bb^ and Bb be the initial and final velocities z>, and v of a point in an indefinitely small time dt, so that - is a the acceleration /. Draw BA and ba at right angles to any line Aa through b l in any given direction. Then -^- is the component/, in the direction Aa of the acceleration /. But b,a Aa Ab. ~dt and Aa and Ab^ are the components in the direction Aa of the velocities v and , Hence the component acceleration in any direction is equal to the time-rate of change of velocity in that direction. Examples. (i) A ball let fall in an elevator has an acceleration downwards relative to the ground of 32 ft.-per-sec. per sec. , while the elevator has an acceleration relative to the ground of 12 ft.-per-sec. per sec. Find the acceleration of the ball relative to the elevator when the acceleration of the elevator is upwards and down- wards. ANS. Solution precisely the same as for example (4), page 66. When acceleration of elevator is upwards, speed of the elevator increases if it is going up, decreases if it is going down, and acceleration of ball relative to elevator is 44 ft.-per-sec. per sec. downwards. When acceleration of elevator is downwards, speed of the elevator increases if it is going down, decreases if it is going up, and acceleration of ball relative to elevator is 20 ft.-per-sec. per sec. downwards. (2) Two trains move in straight lines making an angle of bo". The one. A, is increasing its speed at the rate of 4 ft.-per-min. per min. The other, B, has the brakes on and is losing speed at the rate of 8 ft.-per-min. per min. Find the relative accelerations. ANS. A relative to B, 4^7 ft.-per-min. per min. inclined to the direction of A at an angle whose sine is V CHAP. VI.] RECTANGULAR COMPONENTS OF ACCELERATION. 77 (3) The initial and final "velocities of a moving point during an interval of two hottrs are 8 miles per hour E. 30 A 7 ., and 4 miles per hour N. Find the change of velocity and the mean acceleration. ANS. 41/3 miles per hour W. ; 2^/3 miles-per-hour per hour W. (4) A point has the simultaneous accelerations, bo ft.-per-sec. per sec. N.; 88 ft.-per-sec. per sec. W. 30 S.; 60 ft.-per-sec. per sec. E. 30 S. Find the magnitude and direction of the resultant, ANS. 28 ft.-per-sec. per sec. W. 30 S. Rectangular Components of Acceleration. The same equations and conventions hold for the components of an acceleration in three rectangular directions as for velocity (page 66). We have therefore only to replace v in the equations and figure of page 67 by /"with the proper subscripts. Analytic Determination of Resultant Acceleration. So, also, the same equations and conventions hold for finding the resultant acceleration as for finding the resultant velocity, page 67. We have therefore only to replace v in equations (i), (2), (3), (4), page 67, by/ with the proper subscripts. Examples. Students should solve examples (/) and (2), page 68, for accelerations instead of velocities. Tangential and Central Acceleration. We have seen (page 75) that if v l and v are the velocities at a l and a, Fig. (a), and if the interval of time dt is indefinitely small, so that a l and a are consecutive points of the path, then, in Fig. (), -~- is the instantaneous linear a, acceleration /at a. This has magnitude and direction, can be represented by a straight line, and combined p and resolved just like displacement. We can therefore resolve the acceleration bfi cb f = -j- at a into a component -=- along the FIG. (a). dt dt velocity v, and -j- at right angles to the velocity v. Since v is tangent to the path at a, we call the first component, -y-, the TANGENTIAL ACCELERATION and denote it by f t , the subscript / denoting that the acceleration is along v and therefore tangent to the path. The second component, -^-, is perpendicular to v and must evidently always act towards the centre of curvature C of the path along the radius of curvature aC p. We therefore call it the CENTRAL ACCELERATION and denote it by/p, the subscrpit p denoting that the acceleration is along the radius of curvature p and therefore towards the centre of curvature. Now, since the points a l and a are consecutive, we have cb = v v l = dv, and hence cb dv ~dt = ~di' dv But -j- is the instantaneous rate of change of linear speed (page 73). Hence the magnitude of the tangential linear acceleration f t at any instant is the dv instantaneous rate of change of linear speed -j- at that instant. We see also from Fig. (b) that, since v l and v are perpendicular to the radius of curvature 7 8 KINEMATICS OF A POINT. GENERAL PRINCIPLES. [CHA1-. VI. Ca v and Ca in Fig. (a), the indefinitely small angle a^Ca = dO is equal to b^Bb. We have, then, when , and b are consecutive, b v c zy/# vdti, and hence V _ <*0 / '~ dt ~ V ~dt' jn But -i- is the angular velocity <*> for the centre of curvature C, and if p is the radius of curvature, we have (page 71) pw v. Therefore 2, f p = VGO = pa? = '. . . . . . (2) Hence the magnitude of the central acceleration f f at any instant is given by the product voo of the linear velocity v and the angular velocity a? for the centre of curvature at that instant ; or by the product pco 3 of the radius of curvature p and the square of the angular velocity a? for the centre of curvature at that instant ; or by the quotient of the square of the linear velocity v 3 divided by the radius of curvature p at that instant. The direction of the central acceleration is always towards the centre of curvature. Since the acceleration /is the resultant of/ p and/,, we have for the magnitude of/ /= Vft+S! ............ (3) and for its direction cosine with f f (4) Example. A point moves in a vertical circle of radius r = 21 ft. in an east and -west plane with a velocity given at any instant by v 7 + i6t + 6P, where t is the interval of time from the start. At the end of t = 2 sec. the point is at the top of the circle moving eastward. Find o, ft and fp at this instant. ANS. We find, just as in example (2), page 74, or for / = 2 sec. ft = 40 ft.-per-sec. per sec. towards the east. Also, for / = 2 sec. v = 63 ft. per sec. east, v* f and hence at this instant / p = = 189 ft.-per-sec. per sec. towards the centre, and o> = = 3 radians per sec. north. The acceleration /at the instant is, then, / = ^ft + fl = 6 3 i / o"ft.-per-sec. per sec., making with the radius through the top point an angle whose cosine is = ^ / Vio Uniform and Variable Acceleration. If the acceleration /has the same magnitude and direction, whatever the time, it is UNIFORM. If either magnitude or direction change, it is VARIABLE. If / is zero, there is no change of velocity and we have uniform speed in a straight line, or uniform velocity. If/ p is zero but /, is not, we have variable speed in a straight line. The velocity is variable (page 65). CHAP. VI.] THE HODOGR.APH. 79 lift is zero but/ p is not, we have uniform speed in some curved path. The velocity is variable (page 65). If at the same time/ p is constant in magnitude, the curved path is a circle, For/~ p = , and if f p is constant and v is constant, p must also be constant. If f t and f f are both variable, we have variable speed in some curved path. The velocity is variable (page 65). Example. Criticise the statement, " a point moves in a circle with uniform acceleration." ANS. If the acceleration is uniform, it does not change either in magnitude or direction. But if a point moves in a circle, we must have a radial acceleration f p at every instant. The acceleration is therefore not uniform. If the magnitude of f p is constant, then we have uniform speed in the circle. If not, we have varying speed in the circle. A point can move in a circle with constant central acceleration, but not with uniform acceleration. The Hodograph. Let a point moving in any path have the consecutive positions a l , # 2 , e * c - The extremi- ties of these lines will form a polygon b^b^b^, and if the points a l , a 2 , a z are consecutive, the points b lt # 2 , b z will also be consecutive, and the polygon ^/ 2 3 becomes a curve also. Thus as the point a describes the path a^fa, Fig. (a), we can conceive a point b to describe a curve b^b 2 b z , Fig. (b). This curve is called the HODOGRAPH. The point B is the POLE of the hodograph ; the points b lt b 2 , b 3 of the hodograph are the points corresponding to a lt a 2 , a 3 of the path. When the point a moves from a^ to a 2 in the indefinitely small time dt, the corre- sponding point b moves in the same time from to b z . Now in such case dt velocity in the path, and -~~ is, then, the corresponding velocity in the hodograph. the But we have seen (page 75) that -^ is the acceleration f at a r Hence any radius vector Bb in the hodograph is the line representative of the velocity v at the corresponding point of the path, and the velocity at any point of the hodograph is the acceleration f at the corresponding point of the path. Let Ca = p be the radius of curvature, and GO the angular velocity of the point relative to C, so that the radius of curvature p turns about C with the angular velocity oa. Then, since v is perpendicular to p, the angular velocity of Bb in Fig. (b) is also GO. and hence VGO = / p , where f p is the acceleration perpendicular to v and therefore along the radius of curvature p towards the centre of curvature C. We have then = vao, just as already found (page 78). 8o KINEMATICS OF A POINT. GENERAL PRINCIPLES. [CHAP. VI. Examples. (i) A point moves with uniform velocity. What is the hodograph? ANS. A point. (2) A point moves with uniform acceleration. What is the hodograph ? ANS. A straight line. (3) A point moves with unfiorm speed v in any plane path. What is the hodograph f ANS. A circle of radius v. Axial and Radial or Deflecting and Deviating Acceleration. Let a point P have the angular velocity oj about an axis CA through the centre of curvature C, so that PC = p is the radius of curvature. Let v = pco be the velocity of P. Then f f = VGO. We can resolve co at any point O of the axis CA into a component ca r about an axis OP through the point P, and a component a> a about an axis Oa at right angles to OP. We can also resolve the central acceleration f p into a com- ponent f r along OP, and a component f a parallel to Oa. The component oo r does not affect the velocity of P, since it passes through P. Hence, if r is the distance OP t we have v = roo a , and the plane of rotation is the plane POX. The component co r causes this plane to rotate about the axis OP. We see, then, that if a point rotates with angular velocity &> about an axis through the centre of curvature, we can resolve the motion at any point of that axis into rotation /;/ a rotating plane. If the radius of curvature in this plane is r, we have v = rco a . The central acceleration in this plane is then / r =T>. = ro (i) We call this the radial or deflecting acceleration because it is in the direction of the radius and causes change of direction of i> in the plane of rotation POX. But the rotation oa r about OP causes the plane of rotation POX to change direction. We have, then, the acceleration at right angles to the plane of rotation POX f a = voo r = (2) We call this the axial or deviating acceleration because it is in the direction of the axis of rotation Oa and causes deviation of the plane of rotation POX. The radial or deflecting acceleration f r then causes change of direction of v in the plane of rotation, that is P moves in a curve in this plane. The axial or deviating acceleration /, causes change of this plane of rotation. If f r is zero, the path is a straight line in a rotating plane. If/, is zero, the path is a curve in a constant plane. If /, and /. are zero, the path is a straight line. In all cases, since /, and /. are com- ponents of the central acceleration/,, we have (3) CHAP. VI.] DEFLECTING AND DEVIATING ACCELERATION EXAMPLES. 81 Example. A point moves in a vertical circle of radius r = 21 ft. in an east and west plane with a velocity given at any instant by where t is the interval of time from the start. At the end of t - 2 sec. the point is at the top of the circle moving east. At the same instant the plane of the circle rotates about the vertical diameter with an angular velocity of 2 radians per sec. upwards. Find oo,f t , p, andfp andf. ANS. We have, just as in example, page 78, for rotation in the plane the tangential acceleration or for / = 2 sec. ft = 40 ft.-per-sec. per sec. towards the east. Also for / = 2 sec. v = 63 ft. per sec. east, and hence at this instant the deflecting or radial acceleration is f r = = 189 ft.-per-sec. per sec. towards the centre and &> a = -- = 3 radians per sec. north. We have also due to rotation of the plane, since oa r = 2 radians per sec., the axial or deviating acceleration f a = voo r = 126 ft. per sec. north. Hence oo = \/ GO,? -\- ar 2 = 4/ J 3 radians per sec., making an angle with r in the plane of r and K> a whose cosine is cos/S = 4=+ - cosr=-^ = - ^ f 230 CHAPTER VII. ANGULAR ACCELERATION. Mean Angular Acceleration of a Point. Let ^ and & be the initial and final angular 5 velocities of a point for any interval of time t t r v , >v Take any point O as a pole and lay off Ob^ and Ob, the line representatives of GJ, and GO, and join b\b by a straight line. Then the straight line bj> is the line representative of the change of angular velocity, and - is the mean rate of change of angular velocity for the interval * ~~ *i of time / /,. This is called the MEAN ANGULAR ACCELERATION of the point for the interval of time. It is therefore a vector quantity given in magnitude and direction by a straight line parallel to b^b, having the direction from b^ to b as shown by the arrow in the figure, and equal in bJ> magnitude to - . Mean angul-ir acceleration can be defined, then, as mean time-rate of change of angular velocity, whether that change takes place in the direction of the angular velocity or not. The unit of mean angular acceleration is evidently one unit of angular speed per unit of time, or I radian-per-sec. per sec., just as for rate of change of angular speed (page 73). Instantaneous Angular Acceleration. The limiting magnitude and direction of the mean angular acceleration when the interval of time is indefinitely small is the INSTAN- TANEOUS angular acceleration. Thus if the interval of time is dt or indefinitely small, then V is the instantaneous angular acceleration. o- It is therefore a vector quantity given in magnitude and direction by a straight line parallel to bj), having the direction from b l to b as shown by the bb figure and equal in magnitude to -4-- Instantaneous angular acceleration can then be defined as limiting time-rate of change of angular velocity, whether that change take place in the direction of the velocity or not. Its unit is evidently one unit of angular speed per unit of time, or i radian-per-sec. per sec., just as for mean angular acceleration. We shall always denote instantaneous angular acceleration by the letter a. The term "angular acceleration" always signifies instantaneous angular acceleration unless otherwise specified. We shall see later how to determine it in magnitude and direction (page 84); it is sufficient to note here that if the point moves in an unchanging plane, bj> coincides with 82 CHAP. VII.] RESOLUTION AND COMPOSITION OF ANGULAR ACCELERATION. 83 a?, and we have OL -y-. That is, the magnitude of the angular acceleration in this case is the instantaneous rate of change of angular speed (page 73). Resolution and Composition of Angular Acceleration. Since angular acceleration is time-rate of change of angular velocity when the interval of time is indefinitely small, it has magnitude and direction and can be represented by a straight line, just like angular velocity (page 64). Therefore all the principles of Chapter II, page 54, hold good also for angular accelerations whether simultaneous or successive, and we have the "triangle and polygon" of angular accelerations, and can combine and resolve them just like linear displacement. We have also relative angular acceleration with similar notation as for linear displace- ments (page 55). Example, (r) A sphere has angular acceleration about a diameter at the rate of 10 radians-per-sec. per sec. Find tJie component angular acceleration about another diameter inclined 30 to the first. ANS. 51/3 radians-per-sec. per sec. (2) A body has two sitmdtaneous or successive angular accelerations of 2 radians-per-sec. per sec. and 4 radians-per-sec. per sec. about axes inclined 60 to each other. Find the resultant. ANS. 2 Vj radians-per-sec. per sec. about an axis inclined to the greater component at an angle whose (3) A sphere with one of its superficial points fixed has two angular accelerations, either simultaneous or succe-ssive, one of 8 radians-per-sec. per sec. about a tangent line, and one of 15 radians-per-sec. per sec. about a diameter. Find the resultant. ANS. 17 radians-per-sec. per sec. about an axis inclined to the greater component at an angle whose tangent is T 8 r . Rectangular Components of Angular Acceleration. The same equations and conven- tions hold for the components of an angular acceleration in three rectangular directions as for angular velocity, page 69. We have therefore only to replace &> in the equations and figure of page 69 by a with the proper subscripts. Analytic Determination of Resultant Angular Acceleration. So, also, the same equa- tions and conventions hold for finding the resultant angular acceleration as for finding the resultant angular velocity, page 69. We have therefore only to replace oa in equations (i), (2), (3) and (4), page 69, by a with the proper subscripts. Examples. Students should solve examples (/) and (2), page 70, for angular accelerations instead of velocities. Axial and Normal Angular Acceleration. We have seen that if oo l and oa a are the initial and final angular velocities of a point during an indefinitely small interval of time dt, then ~- is the instantaneous angular acceleration a. bb We can resolve this angular acceleration a = into a component -j- along <*> a , and -~ normal to co a . O Since &? a coincides with the axis of rotat ! on of the radius vector, we call the first component the AXIAL angular acceleration and denote it by a a , the subscript a denoting that the acceleration is along the axis. The second component, ~- normal to &7 a , we denote by <*, and call it the NORMAL angular acceleration. KINEMATICS OF A POINT. GENERAL PRINCIPLES. [CHAP. Vll. Now let < r be the angular velocity of the line representative Ob = o? fl about an axis perpendicular to the plane of bflb. Then, just as on page 80, we have CO (2) But - is the instantaneous rate of change of angular speed. Hence the magnitude of the axial angular acceleration a a is the instantaneous rate of c/tange of angular speed. The magnitude of the normal angular acceleration a n is the product co a w r of the angular velocity v a about the axis and the angular velocity oo r of the axis itself. Since the acceleration a is the resultant of a a and a n , we have for the magnitude of a a = and for its direction cosine with the axis (3) cos a = ^ a (4) Uniform and Variable Angular Acceleration. If the acceleration a has the same magnitude and direction whatever the time, it is UNIFORM. If either magnitude or direction change, it is VARIABLE. If a is zero, there is no change of angular velocity, and we have uniform angular speed in an unchanging plane. If is zero but ot a is not, we have variable angular speed in an unchanging plane. If a a is zero but a n is not, we have uniform angular speed in a changing plane. If a a and a n are both variable, we have variable angular speed in a changing plane. Linear in Terms of Angular Acceleration. -Let O be any pole, and Oa r the radius vector to any point a moving in any path. We have proved, on page 71, that rco = sin e, where a? is the angular velocity and v is the linear velocity making the angle e with Oa. We can therefore write rdco = dv sin e, and dividing by dt, dao dv . r- = sin e. dt dt But is the magnitude of the axial angular acceleration a a , and^ is the magnitude of the tangential acceleration f t . Hence we have ra a =/ / sinc=/ l , .......... (i) where / is the component of f normal to r. CHAP. VII.] LINEAR IN TERMS OF ANGULAR ACCELERATION. Hence the product of the radius vector r by the axial angular acceleration a a gives the linear acceleration f n normal to the radius vector in the plane of f t (or v) and r. If the pole O is taken at the centre of curvature C so that Ba is the radius of curvature p, then e = 90, and we have P a .=ft (2) If the pole O is taken anywhere in the plane through p perpendicular to/, we still have e 90. and in this case '*.=/. . V (3) Hence, in this case, the product of the radius vector by the axial angular acceleration gives the tangential acceleration. If, then, a point P has the angular acceleration a a about an axis AO, so that the radius vector OP = r is in a plane perpendicular to / through the radius of curvature, we have If at the same time the plane of rotation has the angular acceleration oc n , we have (4) But, as we have seen, page 84, if ao a is the angular velocity about the axis and <*> r is the angular velocity of the axis about OP, we have a n ^a^r' Hence we have /. = rv r , just as already found on page 80. We have already found (page 80) the radial acceleration /= real. The acceleration / is, then, the resultant of f r ,f a and/,, or of/ p and f t , since / p is the resultant of f r and/,. Example.^ point moves in a vertical circle of radius r 2 ft. in an east and west plane with an angular velocity given at any instant by &) a = 2t + 4?. At the end of t = % sec. the point is at the top of the circle moving east, and at the same instant the plane of the circle is rotating about the radius through the point with the angtilar velocity oo r = j radians per sec. Find a a , a, a, v, fr, fa., fp, ft and f. ANS. We find, just as in example (2), page 74, the axial angular acceleration For / = J sec. we have, then, o. t = 6 radians-per-sec. per sec. north. The angular velocity for / = \ sec. is oo a = 2 radians per sec. north' The normal angular acceleration is a,, = oo a aa r = 6 radians-per-sec. per sec. west if oo r has the direction given in the figure. 86 KINEMATICS OF A POINT. GENERAL PRINCIPLES. L C " A1 '- V11 ' The angular acceleration a is then given by a = v'aj + a* = 6 vTradians-per-sec. per sec., and its direction cosine with the axis Is cos ft = = =. a A/2 The linear velocity is v = ra> a = 4 ft. per sec. east. The tangential acceleration is/<= r**= 12 ft.-per-sec. per sec. east. The radial acceleration is/ r = - = roal - voa a = 8 ft.-per-sec. per sec. towards the centre C. The axial acceleration is/, = roo A oa r = 12 ft.-per-sec. per sec. north. The central acceleration is/, = Vfr +/I = 4 V~i3 ft.-per-sec. per sec., and its direction cosine with the radius is / 4/13 The resultant acceleration /is given by /= t//7 +/r +/- = 1/22 ft.-per-sec. per sec., and its direction cosines with/,,/, and/- are 4= ;L, cos0= = -/L, cosr=4 = -^. / 4/22 / i/22 / i/22 Homogeneous Equations. We have already called attention (page 4) to the fact that the units in all numeric equations are always understood, and when these units are inserted the equation must be HOMOGENEOUS, th=it is, every term must express a quantity of the same kind. When this is not the case the equation is impossible, and some error must have been made in its derivation. Thus suppose that the result of some investigation is expressed by s -j- at = bv, where a and b are numbers only, s is a number of feet, t a number of seconds, and v a number of feet per second. Without reference to the various steps of the investigation by which this result has been reached, we can say at once, from inspection, that it is incorrect and some error has been committed. For we cannot have a number of feet plus a number of seconds equal a number of feet per second. If, however, a stands for a number of feet per second, and b stands for a number of seconds, the equation is homogeneous and possible. For if we insert the units, we now have s (fU + a (J^- ) X / (sec.) = b (sec.) v (^). or s (ft.) +, then, as we have seen (page 71), the moment roo relative to any point a gives the linear velocity V H = v sin e of that point at right angles to the plane of the axis OA and the radius vector Oa, in such a direction that, looking along OA in its direction, the rotation of the radius vector is seen clockwise. The moment of an angular velocity is, then, a linear velocity, and its unit is one foot per second. O Moment of Angular Acceleration. If OA represents an angular acceleration, a then, as we have seen, page 84, the moment roc relative to any point a gives the linear acceleration f n of that point at right angles to the plane of the axis OA and the radius vector Oa, in such a direction that, look- ing along OA in its direction, the rotation of the radius vector is seen clockwise. The moment of an angular acceleration is, then, a linear accelera- tion, and its unit is one foot-per-sec. per sec. Example. The distance in feet s described by a point in any time t sec. is given by s = a + b?. The path is a circle of radius r feet in a vertical east and west plane. The point when t = o is at (he top nd it morses (wards the east. (.1) State the units of a and b. (b) Find the linear velocity, the tangential, radial and resultant accelera- tion at any instant, (c) The angular velocity and acceleration at any instant. (d) The area/ velocity and acceleration of the radius at any instant. At the end of j sees, find (e) the distance described, the mean speed, the angular and linear displacement. ANS. (a) The unit of a must be i ft., of b \ ft.-per-sec. per sec. (b) v = -r = 2bt ft. per sec. tangent to the path at the instant ; / r c=~ = 2b ft.-per-sec. per sec. tangent to the path at the instant ; / p = = -^- ft.-per-sec. per sec. towards the centre; V* ft.-per-sec. per sec., making an angle with the radius whose tangent 2bt (c) = p = ^f radians per sec. north; a a = = radians-per-sec. per sec. north. (d) = brt sq. ft. per sec. north; ^f t -=br sq. -ft.-per-sec. per sec. north. (c) When / = o, j, = a. When / = 3, s = a + 9*. Hence the distance described is s si = gb ft. The mean speed = = 3^ ft. per sec. Angular displacement 8 0, - = radians. Linear displacement = 2r sin in a direction from the initial to the final position. KINEMATICS OF A POINT. APPLICATION OF PRINCIPLES. CHAPTER I. MOTION OF A POINT. CONSTANT AND VARIABLE RATE OF CHANGE OF SPEED. Rate of Change of Speed Zero. As we have seen, page 79. if the tangential acceleration / ( is zero, the resultant acceleration / is the same as the central acceleration f p and therefore always at right angles to the velocity v. In such case we have in general uniform speed 7 ,2 in some curve, and f = f p = (page 78). If the central acceleration f f = f is constant in magnitude, then, since v is also constant in magnitude, we must have p constant in magnitude and the path is a circle. If / p is zero, then p is infinity or the path is a straight line. Whatever the path may be, let s l be the initial distance of the moving point measured along the path from any given fixed point of the path taken as origin, and s the final distance of the moving point from that origin, measured in the same way, at the end of any interval of time t, so that the distance described in this interval of time / is s s r Now if the tangential acceleration f, is zero, the speed in the path does not change and hence the instantaneous speed at any instant is equal to the mean speed for any interval of time. We have, then, This equation is general whatever the path, provided f t is zero. If s is greater than s lt v is positive. If s is less than s l , v is negative. Hence a positive (-{-) value for v denotes that the distance from the origin increases as the interval of time increases, a negative ( ) value for v denotes that the distance from the origin decreases as the interval of time increases, without regard to direction of motion. Rate of Change of Speed Constant. If the tangential acceleration f t is constant in magnitude, we have in general motion in a curved path with uniform rate of change of speed. If in this case the central acceleration f p = o, the path is a straight line. Whatever the path may be, let v^ and v be the initial and final speeds for any interval of time /. Then, since for constant/, the instantaneous rate of change of speed at any instant is equal to the mean rate of change of speed for any interval of time, we have 91 92 KINEMATICS OF A POINT. APPLICATIONS. [CHAP. 1. The value of/) is positive (-}-) when v is greater than ?',, that is when the speed increases as the interval of time increases, and negative ( ) when the speed decreases as the interval of time increases, without regard to direction of motion. From (2) we have (3) The average speed during the interval / is (4) The distance described (s j,) between the initial and final positions, measured along the path, is the mean speed multiplied by the time, or Inserting the value of / from (2) we have also (5) Hence V* = ^ 2 + 2/ t (S - S,} .......... (7) These equations are general whatever the path, provided f t is constant. In applying them/, is positive (+) when the speed increases, and negative ( ) when the speed decreases, during the interval of time /, without regard to direction of motion. So, also, i> is positive (-[-) when the distance from the origin increases, and negative ( ) when the distance from the origin decreases, during the interval of time /, without regard to direction of motion. Rate of Change of Speed Variable. If the rate of change of speed is variable, we have from (i), in Calculus notation, for the velocity at any instant ds and from (2) for the tangential acceleration at any instant "= ................ and from (8) for the distance described s s t = J vdt. ............... (10) We have also, from (9), fids = vdv t and hence / These equations are general, and the preceding equations can be deduced from them. Thus if// is zero, we have uniform speed and CHAP. L] GRAPHIC REPRESENTATION OF RATE OF CHANGE OF SPEED. lift is constant, we have, by integrating (9). v=f t t+C, where C is the constant of integration. When / = o, v equals fi and hence C = v\. Hence v = v l +f t t. Inserting this value of v in (8) and integrating, we have s**vj+ Iff + C. But when / = o we have s = s\, hence C ;= s\. Therefore 93 1 V \ b c d N V , B Graphic Representation of Rate of Change of Speed. If we represent intervals of time by distances laid off horizontally along the axis of x, and the corresponding speeds by ordinates parallel to the axis of y, we shall have in general a curve for which the change of y with x will show the law of change of speed with the time. (a) RATE OF CHANGE OF SPEED ZERO. Lay off from A along AB equal distances, so that the distances from A to I, i to 2, 2 to 3, etc., are all equal and represent each one second of time, and let AB represent the entire time /. Then at A, i, 2, 3, and B erect the perpen- diculars AM, \b, 2c, -$d, BN, and let the length of each represent the speed at the corresponding instant. Since there is no change of speed, these per- pendiculars will all be of equal length, we shall have AM = i b = 2c = ^d = BN = v, and the speed at any interval of time will be given by the ordinate at that instant to the line MN parallel to AB. The space described in any time is given by s s l = vt. This is evidently given by N the area AMNB in the diagram. Therefore the area corresponding to any time gives the space described in that time. (b) RATE OF CHANGE OF SPEED CONSTANT. Lay off as before the time along AB, and at A, i, 2, 3, B, the corresponding speeds, so that AMis the initial speed v lt and BN the final speed v. Draw MC, be' , cd', parallel to AB. Then bb' will be the change of speed in the first sec., cc' the change of speed in the next sec., and so on. Since these are to be constant, NM is a straight line, the ordinate to which at any instant will give the speed at that instant. The rate of change of speed is then =. = = f t . But = - 1 I sec. i sec. I sec. i sec. t = f t . Hence the rate of change of speed is the tangent 'of the angle NMC which the line MN Hence f t , or NC f t t. Y *^ \^ * < ^^ ^_. d, I c M ^^ v b< ) i ^ * Ic & A - t A t makes with the horizontal. The distance described in the time t is, from equation (5), given by But this 9-1 KINEMATICS OF A POINT. APPLICATIONS. [CHAP. I. is the area of AMNB. Therefore the area corresponding to any time gives the space described in that time. We have then directly from the figure, since NC = f t t, X / = If v, is greater than 7% a will be negative, and the line MN is inclined below the horizontal MC. [(r) Rate of Change of Speed Variable]. If the rate of change of speed is not constant, we shall have in general a curve MNn. The tangent to this curve at any point N makes an angle with the axis of X, whose tan- gent is - =//, equation (9), or the rate of change of speed. The elementary area BNnb = vdt = ds, equation (8), and the total area AMNB = / vdt = s Si, equa- //=0 tion (to). dV ftt Bl, V {S When -j7> or ~j-t = . or ft = o, the tangent to the curve is horizontal at the corresponding point, and we have the speed at that point a maximum or minimum according as the curve is concave or convex to the axis of X. Examples. (i) The speed of a point changes from 50 to 30 ft. per sec. in passing over 80 ft. Find the con- stant rate of change of speed and the time of motion. = 10 ft.-per-sec. per sec. The minus sign indicates decreasing speed. 2(j Si) 2 X oO ANS. // (2) Draw a figure representing the motion in the preceding example, and deduce the results directly from it Ans. Average speed = = 40 ft. per sec. Hence 40* = 80, or / = 2 sec. Also/ = ^ = 10 ft.-per-sec. per sec. (3) A point starts from rest and moves with a constant rate of change of speed. Show that this rate is numerically equal to twice the number of units of distance described in the first, second. ANS. We have / = i and z/i = o ; hence, from eq. (5), \ =-/'. or// = - , , i sec. hich is numer- i sec. 2 ically equal to z(s s t ). (4) In an air-brake trial, a train running at 40 miles an hour was stopped in 62 5.6 ft. If the rate of change of spted was constant during stoppage, what was it ? ANS. From eq. (6) we have for v = o, s - s, = 625^6, and v t ~ ~ T o 6 ~~ (6ox6o) f X2X 625.6 = ~ 2<75 ft< -P er - sec - P cr sec - The (-) s '8 n shows retardation. (5) 4 point starts with a speed v\ and has a constant rate of change of speed //. When will it come to rest, and what distance does it describe? ANS. From eq. (3), when v = o, we have r/, -/,/ = o. or / = ! From eq. (5), s - s, = -/ = ft 2 2ft. (6) A point describes /jo ft. in the first three seconds of its motion and 50 ft. in the next two seconds. If the rate of change of speed is constant, when will it come to rest f I Vhen will it have a speed of 30 ft. per sec. f ANS. From cq. (5) we have for s - s, =-. \ 50 and / = 3, 1 50 =3^, + | /,; and for s - s, = 200 and / = 5, too = $ Vl + "l/i. Combine thes (c) From eq.(6) the distance between the initial and final positions, when the speed is 90 ft. per sec., is AC- s = gLn^L = 8|0 ~ 36o = - 56.25 ft. The minus sign shows that C is on the left of A. The a/i 80 total distance described from the start is then 56.25 + 90 = 146.25 ft. The time, from eq. (5), is oo + 60 _ 56.25 = ^~- /, or / = 3.75 sec. ( and from (2), for the axial angular acceleration at any instant, d<*> Ota = and from (8), for the angle described, a " = -di = Q -0! = r*-30 40/ = 80, or / = 2 sec. Also a = to radians-per-sec. per sec. (3) A point moving in a plane has an initial speed of 60 radians per sec. about an assumed point and a rate of change of speed of + 40 radians-per-sec. per sec. Find the speed after 8 sec.; the time required to describe 300 radians ; the change of speed while describing that angle ; the final speed. ANS. Sec example (9), page 95. (4) If the motion in the last example is retarded, find (a) the angular revolution from the start to the turn- ing-point ; (/*) the angle described from the start after 10 sec.; the speed acquired and the angle between the final and initial positions ; (c) the angle described during the time in which the speed changes to go radians per sec., and this time ; (d) the time required by the moving point to return to the initial position. ANS. See example (ro), page 95. (5) A point moving in a plane describes about a fixed point angles of 120 radians, 228 radians and 336 radiant in successive tenths of a second. Show that this is consistent with uniform rate of change of angular speed, and fin. t this rate. ANS. a 10800 radians-per-sec. per sec. (6) Two fioints A and B mor>e in the circumference of a circle with uniform angular speeds GO and a/. The angle between them at the start is a. Find the time of the nth meeting, the angles described by A and R, and the interval of time between two successive meetings. ANS. Time of the th meeting, /* = a + <"^ 2 *. oo ao' Angle described by A is = the linear velocity of a point on its circumference ? Also find its linear central acceleration. ANS. it radians per sec. ; 2.5* ft. per sec., tangent to circ.; 2.5^" ft.-per-sec. per sec. (8) Find the linear and angular speed of a point on the earth's equator, taking radius 4.000 miles ; also the linear centr.il acceleration. ANS. 1 535.9 ft. per sec. ; ^ radians, or 15 per hour; o.i 12 ft.-per-sec. per sec. (9) The angular speed of a wheel is -it radians per sec. Find the linear speed of points at a distance of 3 ft., 4ft. and ro ft. from the centre, also tht linear central acceleration. ANS. ^w. 3*. 7-5* ft- per sec- |je\- **. ^ir ft.-per-sec. per sec. (10) Tf thf linear speed of a point at the equator is v, find the speed linear and angular at any latitude \ ANS. v cos A; radians per hour, or 15 per hour. CHAP. I.] RATE GF CHANGE OF ANGULAR SPEED EXAMPLES. gg (11) A point moves with uniform velocity v. Find at any instant its angular speed about a fixed point whose distance from the path is a. ANS. -^ radians per sec., where r is the radius vector. Uniform velocity means uniform speed in a straight line (page 65). Hence the angular speed of a point moving with uniform speed in a straight line is inversely proportional to the square of the distance of the point from a fixed point not in the line. (12) The speed of the periphery of a mill-wheel 12 feet in diameter is 6 feet per stc. How many revolu- tions does the wheel make per sec. ? ANS. revolutions. 27T (13) Deduce the equivalent of longitude for one minute of time and for one second of time. ANS. 15' to i min., 15" to i sec. (14) 77/i? diameter of the earth is nearly 8000 miles. Required the circumference at the equator and the linear speed at latitude 60. ANS. 25000 miles; 521 miles per hour. (15) The wheel of a bicycle is 52 inches in diameter and performs 3040 revolutions in a journey of 63 minutes'. Find the speed in miles per hour ; the angular speed of any point about the axle ; the areal velocity of a spoke ; the relative velocity of the highest point with respect to the centre. ANS. 12 miles per hour; 8.12 radians per sec. ; 19.06 sq. ft. per sec.; 12 miles per hour. (16) In going 120 yards the front wheel of a carriage makes six revolutions more than the hind wheel. If each circumference were a yard longer, it would make only four revolutions more. Find the cireumference of each wheel. ANS. 4 yards and 5 yards. (17) If the velocity of a point is resolved into several components in one plane, show that its angular speed about any fixed point in the plane is the sum of the angular speeds due to the several components. (18) A point moves with uniform speed v in a circle of radius r. Show that its angular speed about any point in the circumference is . (19) A point starting from rest moves in a circle with a constant rate of change of angular speed of 2 radians-per-sec. per sec. Find the angular speed at the end of 20 sec. and the angular displacement ; also the linear speed and distance described and the mimber of revolutions ; also the linear tangential acceleration and the central linear acceleration at the end of 20 sec. ANS. 40 rad. per sec.; 400 radians; 40^ ft. per sec. ; 4oor ft.; - revolutions ; 2r ft.-per-sec. per sec. 27t tangential acceleration; i6oor ft.-per-sec. per sec. central acceleration. (20) A point moving with uniform rate of change of angular speed in a circle is found to revolve at the rate of 8| revolutions in the eighth second after starting and "j\ revolutions in the thirteenth second after starting. Find its initial angular speed and its uniform rate of change of angular speed ; also the initial linear speed and rate of change of speed ; also the initial central acceleration. ANS. 20.27T radians per sec.; o.^it radians-per-sec. per sec.; 2o.2?rr ft. per sec. ; o.^itr ft.-per-sec. per sec. ; 4o8.o4^V ft.-per-sec. per sec. (21) A point starts from rest and moves in a circle with a uniform rate of change of angular speed of 18 radians-per-sec. per sec. Find the time in which it makes the first, second and third revolutions. ANS. ^, 1^*-*, fa- *rt S ecs. CHAPTER II. UNIFORM ACCELERATION. Uniform Acceleration Motion in a Straight Line. If the central acceleration/, of a moving point is zero, the path must be a straight line and the resultant acceleration / is equal to the tangential acceleration /,. If, then, the acceleration / of a moving point is uniform, so that it does not change in magnitude or direction, and if its direction coincides with that of the velocity, we have motion in a straight line in a given direction with uniform rate of change of speed, and equations (2) to (6), page 92, apply. In applying these equations we should note that sign now indicates direction. If v or/ are positive (-J-), they act away from the origin; if negative (92), they act towards the origin. The most common instance of such motion is that of a body falling near the earth's surface. All points of such a body move in parallel straight lines with the same speed at any instant, so that the body has motion of translation only, and may be considered as a point. The acceleration due to gravity is known to be practically uniform and is always denoted by the letter^. Value of g. The value of g is usually given in feet-per-sec. per sec. or in centimeters- per-scc. per sec. It has been determined by much careful experiment and found to vary with the latitude \ and the height // above sea-level. The general value is given by g= 32. -173 0.0821 cos 2* o. 0000037*, where // is the height above sea-level in feet, and g is given in feet-per-sec. per sec. , or g = 980.6056 2. 5028 COS 2\ 0.000003/*, where // is the height above sea-level in centimeters, and g is given in centimeters-per-sec. per sec. It will be seen that the value of g increases with the latitude, and is greatest at the poles and least at the equator. It also decreases as the height above sea- level increases. The following table gives the value of g at sea-level in a few localities : g g Latitude. F. S. Units. C. S. Units. Equator o o' 32.091 978.10 New Haven 4118 32.162 980.284 Latitude 45 45 o 32.173 980.61 Paris 4850 32.183 980.94 London 5140 32.182 980.889 Greenwich 5129 32.191 981.17 Berlin 5230 32.194 981.25 Edinburgh 5557 32.203 981.54 Pole 90 o 32.255 983.11 CHAP. II.] BODY PROJECTED VERTICALLY UP OR DOWN. 101 For calculations where great accuracy is not required it is customary to take^-= 32 ft.- per-sec. per sec. or g 981 cm.-per-sec. per sec. For the United States g = 32^ is a good average value and is therefore very often used. In exact calculations the value of g for the place must be used. Body Projected Vertically Up or Down. For a body profected vertically upwards in vacua, taking the origin at the starting-point, we have then v l positive, since it is away from the origin, and g negative, since it is towards the origin. If the body is projected vertically downwards, taking still the origin at the starting-point, we have v l positive and g also positive, since both act away from the origin. We have then simply to replace f t in equations (2) to (7), page 92, by .g, taking the plus sign for the falling body and the minus sign for the rising body. When the final velocity for a body projected upwards is zero, we have from (3), page 92, for the time of rising to the highest or turning point, by making v = o and f t = g, g For the time of rising to the highest or turning point and then returning to the starting- point we make s s l = o in (5), page 92, and/, = g, and obtain g Hence the times of rising and returning are equal for motion in vacua. The distance from the starting-point to the turning-point is found from (6), page 92, v% by making v = o and f t g, to be . v 2 v* The distance or is called the height due to the velocity v l or v; that is, the distance a body must fall from rest in order to acquire the velocity i\ or v. v i rf When the distances in rising and falling are equal we have ss l = o or = or v^ = v. That is, for motion in vacua the velocity of return is equal to the velocity of projection. If the time of rising is less than T = , the displacement s s l is equal to the distance described. But if the time is greater than T , the body reaches the turning-point and c> then falls from rest, and the entire distance described is V 2 V Z Z' 2 \ Z? distance described = -^- + \g(t 7)' l --- s= ~ . The student will note that the motion is supposed to take place in a vacuum. That is, the effect of the resistance of the air is neglected. As a matter of fact this resistance has a great influence, and hence the following examples have little practical value except as illus- trating the application of the equations, 102 KINEMATICS OF A POINT-APPLICATIONS. [^HAP. II. Examples. (Unless otherwise specified g = 32.2-ft.-per-sec. per sec. or 981 cm.-per-sec. per sec. All bodies supposed to move in vacuum.) (1) A point moves with a uniform velocity of a ft. per sec. Find the distance from the starting-point at tht end of one hour. ANS. 7200 ft. Motion in a straight line. (2) Two trains have equal and opposite uniform velocities and each consists of 12 cars of 50 ft. They are observed to take 18 sec, to pass. Find their velocities. ANS. 22.73 miles per hour. (3) Two points move with uniform velocities of 8 and 15 ft. per sec. in directions inclined 90. At a given instant their distance is toft, ami their relative velocity is inclined 30" to the line joining them. Find (a} their distance when nearest ; (A) the time after the given instant at which their distance is least. ANS. (a) 5 ft. ; (b) ^- ^sec. (4) A body is projected vertically upwards with a velocity of 300 ft. per sec. Find (a) its velocity after 2 stc. ; (b) its velocity ) when it is moving with a uniform downward acceleration of 9.81 >n.-per-sec. per sec. ; (c) wlnn moving with a uniform downward acceleration of 4.905 m.-per-sec. per sec. ; (a) when moving with a uniform upward acceleration of 4.905 m.-per-sec. per sec. ANS. (a) I sec.; (b) co ; (c) f'Jsec. ; (d) -J/isec. CHAP. II.] UNIFORM ACCELERATION -MOTION IN A CURVE. 103 FIG. (a). FIG. (b). Uniform Acceleration Motion in a Curve. If the acceleration / is uniform, so that it d )es not change either in magnitude or direction, and if~its direction doe-s not coincide with that of the velocity, we have motion in a curve with uniform acceleration. A common case of such motion is that of a body projected with any given velocity in any given direction at the surface of the earth, neglecting the resistance of the air. In such case the accelera- tion due to gravity is practically uni- form, acts downward and is equal to g ft.-per-sec. per sec. The curve or path in such case is called the TRA- JECTORY. EQUATION OF THE PATH. Let the uniform acceleration y be vertical and act downwards. (In the case of gravity f = g.} Take the origin O at the initial position of the point as shown in Fig. (a), and let the initial velocity v.^ make the angle a l with the horizontal. Let the co-ordinates of any point P of the path or trajectory be OB x and BP = y. Let the angle POB - 9. If in Fig. (b} we lay off from Q the line representative Qb l = v^ of v l , the line repre- sentative Qb = v of v, etc., we see that the hodograph b^, etc., is a straight line. (See example (2), page 80.) It is at once evident that the horizontal component v x of the velocity at every point is constant and equal to , v x = v^ cos a^ (i) In any time /, then, the horizontal distance described is x = i\t cos a l (2) The vertical component of the initial velocity 7\ is 7' x sin ae l upwards. But the uniform acceleration f is downwards. Hence the vertical velocity at the end of the time / is, from equation (3), page 92, v y =v l sina l /f (3) The mean vertical velocity during the time / is, then, z\ sin a l -J- v y 2v l sin cr l ft = v, sin a, The vertical distance passed over in the time t is, then, y = v^t sin at l \ff j If we combine (2) and (4) by eliminating /, we have for the equation of the path (4) y x tan of l 2V? cos" a, This is the equation of a. parabola. The time of reaching the highest point C is the time of describing the vertical distance DC. Denote this time by T v . Since at this point the vertical velocity is zero, we have, by making v y = o in (3), _ r I0 4 KINEMATICS OF A POINT APPLICATIONS. [CHAP. II. If we substitute this for / in (2) and (4), we have for the co-ordinates of the vertex C of the parabola ec^fjgA... - (8) The parameter of the parabola is then The directrix is parallel to OD at a distance above the vertex C equal to one half the parameter, or v* cos 2 a l 7'* or at a distance of , above O. That is, the distance of the directrix above O is the height due to tht velocity v r If we transfer the origin to the vertex C, then the horizontal velocity at C is v l cos <* } , the horizontal distance described in any time / is x = i\t cos ^ is the distance of the directrix above any point P, and is the height due to the velocity v (page ioiV ftcnce for a body acted upon by gravity the speed at any point is the same as that acquired by a body falling from the directrix to that point. To find the direction of the velocity v at any point P, the magnitude of which is given by (10) or (11), let a be the angle which it makes with the horizont.il. Then we have directly from the hodograph (Fig. (), page 103). and from equations (3) and (i), v sin a Vy - ^ s in . Let a be the angle between / and /. Then we have ?ds f cos a = ff. ^^ Let ds be the element of the path at P, and dp the / projection of this element upon /. Then we have ^ . dp = ds . cos a. Multiplying these two equations we have fdp=f t ds. That is, the acceleration / multiplied by the elementary displacement in the direction of /is equal to the tangential acceleration / multiplied by the elementary displacement in the direction of/. Since this holds at every point of the path, we have for the entire path CO Let v l and v 2 be the velocities at two consecutive points P l and P t of the path. Then for the indefinitely small time r The average velocity for this time is -, and the distance ds described is We have then IOQ 1 10 KINEMATICS OF A POINT APPLICATIONS. [CHAP. 111. For the next two consective points, P a and P 3 , we have then /*-*f* ' .y' For the next two, P 3 and P 4 , *-*?*. :' I Adding all these, we have, if v l is the initial and v the final velocity, Equation (i) then becomes ? ......... (I) Equation (I) is general, as is seen by its derivation, whatever the path and whatever the acceleration f in magnitude, and however it may change in direction. It will therefore in all cases give us the relation between v and f or f t . In performing the summations in (I) we should take f or f t positive when acting away from the origin, and negative when acting towards the origin. ILLUSTRATIONS. Thus for a point moving in any path with ___ uniform rate of change of speed we have./) constant in magnitude, and positive if away from the origin O, as shown in the figure. Hence 2f t ds = f t 2ds = f t (s - sj, where s is the final and s { the initial distance of the point measured along the path from any point O of the path assumed as origin. We have then, from (I), (2) which holds no matter whether the distance from O is increasing or decreasing. If increas- ing, s is greater than s v If decreasing, s is less than s r This is equation (7), page 92. lif t acts towards the origin, we should take/ negative and obtain * = V* - 2f t (s - 5l ), and this again holds whether the distance from O is increasing or decreasing. Again, for a point moving with uniform acceleration in a straight line, we have for acceleration towards the origin/ = f t negative. Hence = -f ( 2ds = - f t (s - and, from (I), If // g^ this is the case of a body under the action of the earth's attraction, either projected upwards or falling, origin at surface of earth (page 100). If projected upwards, s is greater than s v If falling, s is less than s r CHAP. III.] CENTRAL ACCELERATION. til Again, if the acceleration/" is uniform and does not coincide with the velocity, we have for origin at O, as shown in the figure, /"negative if p downwards. Hence 2f . dp fSdp = fy, and, from (I), which is equation (i i) (page 104). We see, then, that equation (I) is general and applies to all cases. Central Acceleration. If the acceleration of a moving point is always directed towards or away from a fixed point or CENTRE OF ACCELERATION, the acceleration is said to be CENTRAL. The velocity v = Bb of the moving point at any instant is the resultant of the velocity ^ Bb l at the preceding instant, and of the change of velocity (page 75). But if b^b is always directed towards or away from a fixed point, its moment relative to that point is zero. Since the mo- ment of the resultant is equal to the algebraic sum of the moments of the components (page 89), and since in this case the moment of one of the two compo- nents b^b is always zero, it follows that the moment of any velocity v relative to the centre of acceleration is for central acceleration always constant. Conversely, if the moment of the velocity of a moving point relative to any fixed point is constant, the acceleration must be central. But the moment of the velocity is twice the areal velocity of the radius vector (page 89). Hence in all cases of central acceleration the radius vector describes equal areas in equal times. If v l is the known initial velocity at any given instant and / x is its lever-arm, and if at any other instant the velocity is v and lever-arm /, we have for constant moment If r l is the initial radius vector and e 1 is the angle of i\ with r l , we have / x = r l sin e r In the same way / = r sin e. Hence vl vr sin e = v l r l sin e t . . . (3) If a?! is the initial angular velocity, r^oo^ = Vi sin e r In the same way r&) v sin e. Hence GO r 2 f N f*ca = r^ , or = - (4) That is, in all cases of central acceleration the angular velocity is inversely as the square of the radius vector. Cases of Central Acceleration. Two of the most important cases of central accelera- tion are : ist. Central acceleration varying inversely as the square of the distance from the centre of acceleration. 2d. Central acceleration varying directly as the distance from the centre of acceleration. We shall discuss in this chapter the first case. lia KINEMATICS OF A POINT APPLICATIONS. [CHAP. III. Central Acceleration Inversely as the Square of the Distance. Let r t be the initial distance from the centre of acceleration O, the velocity at the point P l being v l , and r the distance to any point P at which the velocity is v. Let / be the known acceleration at a given distance r , and / the acceleration at any distance r. Then This gives the magnitude of the central acceleration. If / P o is towards the centre O, it is negative and we have f - liL j- + If /is away from the centre O, it is positive. (i) ACCELERATION TOWARDS THE CENTRE. From (I), page no, we have v 2 v* fr z 2f.dp= V ^-, where / = - &-. Let />, and P t be two consecutive points at distances r v and r a from O. Then, if the points are consecutive, dp = r 2 r l , and r 2 = r 1 r 2 . Hence >*--('. -')-/* (7 "7). r,r 2 \ r 2 r t / For the next two consecutive points, P 2 and P 3 , we have, in the same way, For the next two consecutive points, P 3 and P 4 , >*-/* - and so on. Summing up, if r l is the initial and r the final distance, we have and hence, from (I), If the distance is increasing, r is greater than r r If decreasing, r is less than r r Equation (i) holds in both cases if the acceleration is towards the centre. (2) ACCELERATION AWAY FROM CENTRE. In this case we have / = +&, and proceeding just as before, we obtain / 1 t\ (2) CHAP. III.] CENTRAL ACCELERATION INVERSELY AS SQUARE OF DISTANCE VELOCITY. 1 13 This also holds whether r is increasing or decreasing. We see that in order to obtain (2) we have simply to change the sign of f Q in (i). COR. It is evident that the same equations hold for motion in any path, if we take the pole O in the path and measure all distances along the patk, if the tangential acceleration f t is inversely as the square of the distance. We have, in such case, for f t towards = v* 2frf. Then the tangent to the hodograph at Q is the direc- tion of the acceleration /at Pand is parallel to OP = r (page 79). Since the angular velocity of r, Fig. (a), is GO, the angular velocity of the tangent at Q, Fig. (), is also co. Let C, Fig. (), be the centfe of curvature of the hodograph, so that CQ is perpen- CHAP. III.] CENTRAL ACCELERATION INVERSELY AS SQUARE OF DISTANCE PATH. 115 dicular to the tangent at Q, and CQ is the radius of the hodograph. Then, since the accel- vj\_ii._y \JL 5^, x- ig. \t/J) we Hdvcy ix ^ . u/ Ul L' J^ But G?^ 11 -^, and /=* " eration/of P, Fig. (a), is the velocity of Q, Fig. (&), we have/= CQ . GO or CQ = ~. /r 2 <7<2 = ^f- 2 - = a constant quantity. rfco, The radius of curvature of the hodograph is therefore constant, and the hodograph is a circle. (6) PATH. Draw O'R, Fig. (), at right angles to CQ and therefore parallel to r. O'R is the component of the velocity v along the radius vector. Draw QN perpendicular to O'C. Then QN is the component of v at right angles to the line O'C, or the diameter A'B' of the hodograph. But by similar triangles OR _QN O'R _ O'C 0'C~ CQ' QN ~ CQ' But O'C is a constant quantity by construction, and CQ, as we have just proved, is also constant. Hence O'R - = a constant quantity = c. That is, the ratio of the velocity along the radius vector to the velocity at right angles to the diameter A'B' of the hodograph is a constant ratio c. If, then, r^ and r 2 are the initial and final values of r for an indefinitely short time t, and d^ d 2 are the corresponding distances of P from any line AB, Fig. (a), parallel to A'B' , we have for the velocity along the radius vector and for the velocity at right angles to AB Hence O'R Since this holds wherever we take the line AB, let us take the initial distance = l or c c = -r. Then we have That is, the ratio ^ = c of the distance r of the point P from a fixed point O to its d distance d from a fixed line AB is constant. n6 KlNEM/tTlCS OF A POINT APPLICATIONS. [CHAP. Hi. This is the property of a conic section for a focus at O and directrix AB. When, therefore, a point has a central acceleration inversely as the square of the radius vector, it must move in a conic section with the centre of acceleration at a focus, and, as we have seen (page 1 1 1) in all cases of central acceleration, the radius vector describes equal areas in equal times. Conversely, if the radius vector describes equal areas in equal times, the acceleration must be central ; and if the path is a conic section and the centre of acceleration is at a. focus, the acceleration is inversely as the square of the radius vector. If c = i or r = d, then the path is a parabola, O'R QN, and the pole O' is on the circumference of the hodograph. The parabola becomes a straight line when the directrix passes through the focus. If c is less than unity, or r is less than d, then the path is an ellipse, O'R is less than QN, and the pole O' is inside the hodograph. If d is infinity, the path is a circle and the pole O' coincides with the centre of the hodograph. If c is greater than unity, or r is greater than d, then the path is an hyperbola, O'R is, greater than QN, and the pole O' is outside the hodograph. General Equation of a Conic Section. The general polar equation of a conic section is by Analytical Geometry, i -\- e cos ' where r is the radius vector of any point making the angle 6 with an axis, and A and e are constants. If e = i, we have a parabola, r is the radius vector from the focus F, as shown in the figure, is the angle AFP of the radius vector with the axis measured round from the apex A. A straight-line path is a special case of the parabola when the directrix passes through the focus. If e < i, we have an ellipse* r is the radius vector from a focus F, and 6 is the angle AFP of the radius vector with the major axis CA measured round from the nearer vertex A, as shown in the figure. In this case A in equation (i) is the semi-major axis CA, CF and e is the eccentricity, or e = ^ . The circle is a special case when e = o. If e > i, we have an hyperbola, r is the radius vector from a focus F, and 6 is the angle AFP of the radius vector with the major axis AC measured round from the nearer vertex A, as shown in the figure. In this case A in equation (i) is the semi-major axis CA, and e is the eccentricity, or e = -^ . Central Acceleration Inversely as the Square of the Dis- tanceEquation of the Path. We have for the path in gen- eral a conic section, as proved on page 115, and for the general polar equation of a conic section, as just explained, + e cos 9 ' (0 CHAP. III. ] CENTRAL ACCELERATION INYERSEL Y AS SQUARE OF DISTANCE EQUATION OF PATH. Let the notation be as in the figure. Thus r v v l and 0j are the given radius vector, velocity and angle for the initial position P v v l making the given angle e x with r r F)r any position P we have r, v and 6. For the apex A we have = o, r x and v y . For = 180 we have r' x and v y . Let / be the known central accelera- y tion at a known distance r . For acceleration towards the centre we have, as already proved, page 112, or, solving for r, r i If in (i) we make 6 V ^ becomes v v r becomes r lt and we have A(i-e*) = ri (i+ecosOJ ......... (3) Substituting in (i) we have ^(i+'cosgj i + ^ cos e If we make = o, r becomes r x , v becomes v y , and we have, from (2) and (4), If we make = 180, r becomes r' x , v becomes v' y , and we have, from (2) and (4), T _ - We have also, by the principle of equal areas (page III), V* = v y r * = v fi sin e i From (7) and (5) we obtain sn e t i - From (7) and (6) we obtain v l sin e t (i - e) , , i + e coslT" ' ' ' ' ' . ' ' ' Substituting (8) in (5), and (9) in (6), we have, after reduction, 2 2 i ei cos fy = r^ 2 sin 2 e/i + r) 2 + (i + e cos e cos ) = r 2 sin 2 ei - e? + (l + e cos n8 KINEMATICS OF A POINT-APPLICATIONS. [CHAP. III. From these two equations we obtain r.?'. 2 sin 2 e. -' ' L 7o r o (10) Substituting (10) and (i I in (4) we have for the general equation of the path, in terms of known quantities, r*v? sin 2 e 1 7>o 2 -f- cos (12) where / is the known central acceleration at a known distance r , v l is the given initial velocity at a given distance r,, making the angle e l with r p and r is the radius vector for any angle 6. The quantity under the radical is the value of i, or when * t > 3&i, or V| > y/^'. We see, then, that the form of the path depends solely upon the magnitude of tJie initial velocity and not upon its direction, tliat is simply upon the initial speed. Prom (2), page 117, making v = o and r infinity, we have for the velocity of projection which would carry the point from r, to rest at an infinite dis- tance, or which the point would acquire in moving from rest at an infinite distance to ;,. The path is an ellipse, parabola or hyperbola according as ?', is less than, equal to or greater than this. For acceleration away from the centre we should change the sign of / in equation (2), as pointed out page 112. The value of e, equation (n), in such case is always greater than unity. Hence for acceleration away from the centre the path is always an hyperbola. CHAP. III.] PATH A PARABOLA. 119 Path a Parabola. If the path is a parabola, e= i, v? = ^J>_ and equation (12^ r i becomes 2r, sin 2 e. i+costf* (13) We have also, from (10), for the angle O l of the initial radius vector ^ with the axis cos ^ = 2 sin 2 e 1 i. . . . ( I4 ) Fig. (a). Fig. (S). The path is then completely determined. The equation referred to the axis and vertex as origin is = 4 sn Path an Ellipse or Hyperbola. If the path is an ellipse or hyperbola, the eccentricity in either case is given by (11), viz., . /'. *.*.' sin' e t (r,'V~2/ f '). ....... (16) V ~7W~ From (16), (3) and (10) we have for the semi-major axis of the ellipse or the semi-trans- verse axis of the hyperbola A = -g& l ........... (17) 2 So r o r \ v \ where the (-{-) sign is taken for the ellipse and the ( ) sign for the hyperbola. We see that this is independent of the angle e l or the direction of the initial velocity v^ , and is the same for *\ and r l constant in magnitude no matter what the direction. The semi-conjugate axis is given by B A^\ e~ for the ellipse or B = A^e 2 I for the hyperbola. From (16) and (17) we have, then, n = __^Li in e i _ /, 8 N " *W*f~ri*ff- where the (-J-) sign is taken for the ellipse and the ( ) sign for the hyperbola. We have, from (10), 'for the angle l of the initial radius vector r l with the axis cos 0, = _ r jrf si " 2 6 * --^ (19) ^/oV + ^i* sin ' i(^'* - 2 /o'o 2 ) The elliptic or hyperbolic path is thus determined. The equation of the path referred to the centre and axes is A*B\ ......... (20) where the'(+) sign is taken for the ellipse and the ( ) sign for the hyperbola. 120 KINEMATICS OF A POINT APPLICATIONS. [CHAP. III. If the path is a circle, we have e = o, e, = 90, / =/, r^ = r, v^ = vr, and hence, from (16), as should be (p. 79). Planetary Motion Kepler's Laws. By long and laborious comparison of the observa- tions which TYCHO BRAHE had made, through many years, of the planets, especially of Mars, KEPLER discovered the three laws of planetary motion which are known as KEPLLR'S LAWS. He gave these laws simply as the expression of facts which seemed established by the observations. The three laws are as follows: I. Tke planets describe ellipses, the sun occupying one of the foci. II. The radius vector of each planet describes equal areas in equal times. This is known as the law of equal areas. III. The squares of the periodic times of the planets are proportional to the cubes of their mean distances from the sun. The second law, as we have seen (page in), is a necessary consequence of tf// central acceleration. The first law, as we have just seen, follows if the central acceleration is inversely as the square of the radius vector. The third law is also a direct consequence of such central acceleration, as we shall see in the next article. Verification by Application to the Moon. If co l and ' \ e 2 . Hence the periodic time is sin e ( 24) or, substituting the values for A and (e}, equations (17) and (16), ^ _ (25) ' From (25) we see that the periodic time is independent of the angle e v or the direction of the initial velocity ?',. We can write (24) in the form *r* sin 3 T* " A(\ -f 3 ) ' or, from (21), ?', 2 r, 5 sin 2 e. CHAP. III.] PL/tNETARY MOTION. 123 But, by Kepler's third law, we have for two different planets of periodic times T and T l T% /J3 /13 A 3 -L Si fi. jfi. A*> Hence is a constant quantity for all the planets. Now/o is the acceleration at a distance r , and, since the acceleration /at any distance r is ^"~ r ;' we see that / Q r 2 is the magnitude of the acceleration at the distance unity, or r I. Hence it follows, from Kepler's third law, that for all the planets the acceleration would be tJu' same at the same distance from tJie sun.* Value of f for Planetary Motion. In all our equations for central acceleration we see that it is necessary to know the acceleration / at some known distance r . It will be proved hereafter (page 206) that if M is the mass of the sun and m the mass of a planet, the value of / at a distance r equal to the mean radius of the eartJi is given by M + m where m is the mass of the earth and g the mean acceleration of a body at the earth's surface. If the two bodies are the earth and a small body of mass ;', then / = - g, or, m o since m is insignificant with respect to ; , / = g. If, in the preceding article, we had used the value of f given by (i), we should have obtained M+m 47t*A s M+m, gr* = ~2- and -- -grf = m Q l 11 o Hence r 2 _ Jf+m l A^ ' A ' We see, then, that Kepler's third law is not strictly exact. The value of / O r a , or the acceleration at units distance, is not strictly constant. The more accurate expression is that the squares of the periodic times are directly as the cubes of the semi-major axes and inversely as the sum of the masses of sun and planet. The error from this source is insignificant, the mass of Jupiter, the largest of the planets, being less than a thousandth part of that of the sun. "Of all the laws," says Sir John Herschel, "to which induction from pure observation has ever con- ducted man, this third law of Kepler may justly be regarded as the most remarkable, and the most pregnant with important consequences. When we contemplate the constituents of the planetary system from the point of view which this relation affords us, it is no longer mere analogy which strikes us. no longer a general resem- blance among them as individuals independent of each other, and circulating about the sun. each according to its own peculiar nature, and connected with it by its own peculiar tie. The resemblance is now perceived to be a true family likeness; they are bound up in one chain; interwoven in one web of mutual relation and harmonious agreement; subjected to one pervading influence, which extends from the centre to the farthest limit of that great system, of which all of them, the earth included, must henceforth be regarded as members." Outlines oj Astronomy, 124 KINEMATICS OF A POINT APPLICATIONS. [CHAI-. 111. The motion of translation of the planets is not affected by their rotation on their axes, and we may treat them as material points at their centres of mass, so far as translation is concerned. The centre of mass of the sun is not strictly a fixed point, but both sun and planet move in orbits about a common centre of mass. The sun is also affected by the other planets, and the planets by each other. The attraction of the planets for each other sensibly modifies their orbits. Kepler's laws are thus approximate. If we had but two bodies, one fixed and the other free to move, then Kepler's first two laws would be accurate, and the third would approach accuracy as the mass of the moving body becomes insignificant with respect to the mass of the other. Examples. (i) Find the speed and periodic time of a body moving in a circle at a distance from the earth's centre of n times the earth's radius, the central acceleration being inversely as the square of the distance. ANS. v = . n. r = . (2) A body at a distance of r t from the centre of the earth is projected in a direction which makes an angle of e, = bo" with r t with a speed v\ , which is to the speed acquired by falling from an infinite distance as i to. ^3. Find the path, the major axis, eccentricity and periodic time. I I^Ti ANS. We have e, = 60, sin e, = , v t = -=> l :5 , A = if. \4 4/3 \ '* The path is an ellipse. From (17), iA = r,. From (16), e = |_i. l~ 3 From (25) T 3-^J. P'_' ( where r u is the radius of the earth. 4'*o ^V JT From (12) we have, by making = o, the perihelion distance FA equal to - r\ ( i I ). From (19) we have for the angle AFPi 61 The centre of the earth is at the focus F. (3) In the preceding example, let e, =90, so that tfo body is projected in a direc- tion at right audits to r . ANS. From (ij),2A = r,, as before. From (i 6), e = . From (25), T= 3- , y '. as before. From (12) we have, by making = o, the perihelion distance FA equal to -r,. 2 From (19) we have for the angle AFPi cos 0i = - i, or 0, = 180*. CHAPTER IV. CENTRAL ACCELERATION DIRECTLY AS THE DISTANCE. HARMONIC MOTION. Harmonic Motion. The motion of a point moving in any path, the acceleration being always directed towards a fixed point and varying directly as the distance from that point, is called HARMONIC motion. If the path is a straight line, the motion is SIMPLE HARMONIC ; if the path is a curve, the motion is COMPOUND HARMONIC. The vibrations of such bodies as a tuning-fork or a piano-wire are approximate ex- amples of such motion, hence the term "harmonic" The vibrations of an elastic body, such as a spring or the air, are also examples of such motion. It is also (page 207) the motion of a body under the action of gravitation, within a homogeneous sphere. The motion of the piston of a steam-engine when moved by a crank and connecting- rod approximates the same motion if the rotation of the crank is uniform, the approxi- mation being closer the longer the connecting-rod. Harmonic Motion Velocity. Let/ be the known acceleration at a given distance r , and / the acceleration at any distance r. Then f:/ ::r:r , or / = -/o- r o This gives the magnitude of the central acceleration. If /is towards the centre of accelefation, it is negative and we have From (I), page no, we have 2f. dp L , where /= / , 2 r o Let P l and P 2 be two consecutive points at distances r l and r z from the centre of accel- eration. Then if the points are consecutive, dp = r z r r and r = -. Hence For the next two consecutive points, P z and P 3 , we have in the same way 125 126 KINEMATICS OF A POINT-APPLICATIONS For the next two consecutive points, P 3 and P 4 , [CUAP. IV. and so on. Summing up, if r, is the initial and r the final distance, we have Hence If the distance is increasing, r is greater than r l ; if decreasing, r is less than r r Equation (i) holds in both cases. COR. It is evident that the same equation holds for motion in any path if we take the centre of acceleration at some fixed point of the path and measure all distances along the path, if the tangential acceleration f t is directly as the distance. We have, in such case, (2) where / is the tangential acceleration at a distance J , s l is the initial and s the final distance from the fixed point of the path, all distances measured along the path. Simple Harmonic Motion. If the central acceleration varies directly as the distance from the centre of acceleration and coincides with the direction of the velocity, the motion is rectilinear and we have simple harmonic motion. In this case equation (2) still holds. Let a point Q move with uniform speed r<& in a circle of radius CQ = r with uniform angular speed w- Then the radial acceleration^ is towards the centre C and equal to /P = roo\ The projection of/ p upon a diameter CA is rat . cos QCA. A But r cos QCA is the distance CP = s, if P is the projection of Q upon the diameter. The acceleration of P along the diameter towards C is then jci 2 , or directly proportional to the distance s. The motion of P is then simple harmonic. Hence, if a point Q move in a circle with uniform speed, its projection P upon a diameter moves -with simple harmonic motion along the diameter. If v\ is the initial velocity at the initial position P^ so that CP l = s lt equation (2) becomes (3) CHAP. IV.] SIMPLE HARMONIC MOTIONS. 127 The point P starts from rest at A at the distance CA = r. If then we make v = o in (3) and s = r, we have /o s o and substituting this in (3), v*= (r 2 .y 2 ) (4\ V We see from (4) that the velocity increases as the distance s decreases till P arrives at C, where the velocity is a maximum and equal to hence = ^ ....... (5) Then the velocity decreases and finally becomes zero when P arrives at A' at the distance r on the other side of C. From A to A' is called a VIBRATION, from A to A' and back to A is called an OSCILLA- TION. The radius r is called the RANGE or AMPLITUDE of an oscillation. PERIODIC TIME. The time from A back to A, or the time of an oscillation, is called the PERIODIC TIME. Since the uniform speed is ra> r-, the periodic time is S Q If /is the acceleration at any distance s, we have for harmonic motion == Hence The periodic time depends, then, only upon the constant ratio -? = 3, and is independent of the range r or amplitude of oscillation. For this reason the oscillations are said to be ISOCHRONOUS, or made in equal times, no matter what the range or amplitude. COR. Since the motion of a body under the action of gravity in a homogeneous sphere is harmonic (page 207), if we put g for/ , and r , the radius of the earth, for j , we have from (3), for a body falling under the action of gravity in a well or shaft assuming the earth to be a homogeneous sphere and neglecting the resistance of the air, 128 KINEMATICS Of-' A POINT-APPLICATIONS. [CHAP. IV. This can be written If the fall takes place near the surface, for a short distance compared to r , we have s l -f- s practically equal to 2r , and hence which is the same as for uniform acceleration g (page 101). We obtained the same result (page 101) for a body external to the earth. The equa- tions of page 101 hold good, then, in all practical cases, whether the fall takes place above the earth or within the earth, for small fall near the surface, neglecting air resistance. Epoch Phase. If P l is the initial position or the position of P at zero of time, the time of passing from A to P l is called the EPOCH. The 9 epoch may also be defined with reference to the auxil- iary circle as the angle ACQ V in radians. This is the I epoch in angular measure. The epoch in angular meas- ure is, then, the angle described on the auxiliary circle in the interval of time defined as the epoch. The epoch locates the initial position of P. The fraction of the periodic time in passing from A to any- position "P is called the PHASE. Measured on the circle is the ratio of the angle ACQ radians to 2n radians. The phase locates the position of P at any instant. It therefore varies with the time or with the position of P. The phase at the initial position P l multiplied by 2n gives, then, the epoch in angular measure, and multiplied by the periodic time gives the epoch in time. Examples. (i) A point whose motion is simple harmonic has velocities 20 and 25 m. per sec. at distances 10 and 8 m. from the centre of acceleration. Find the period and acceleration at units distance. ANS. We have 400 = f-(r* 100) and 625 = ^Vr 1 64). Therefore s t Jo = / We have also *-ss -^-,or/= j-. Making s = i,/= J = 6.25 ft.-per-sec. per sec. (2) The period of a simple harmonic motion is 20 sec. and the maximum velocity is to ft. per sec. bo Find the velocity at a distance of //. from the centre. ANS. We have = = 20 sec. Therefore - = . When s = o, v = 10 and 100 = r>, orr = *ft. //. s t 100 100 * J. Hence Vi = ^( 1 ? -Sr)- or " = 8ft.persec (3) Find the mean speed of a point in simple harmonic motion during the time of moving from one to the other extremity of its range, tts maximum speed being s ft. Per sec. CHAP. IV.] COMPOUND HARMONIC MOTION. I2 9 & ANS. The distance is 2r. The time . The mean speed ^L. When s = o, we have t/ 4L *#* Jo 25 = r*, or Therefore the mean speed is ft. per sec. (4) If T is the period and r the amplitude of a simple harmonic motion, v the 'velocity and s the distance from the centre at any instant, show that (5) ^4 /0// has simple harmonic motion whose period is 4 min. 12 sec. Find the time during which its phase changes from ^ to \ of a period* ANS. 21 sec. Compound Harmonic Motion. If the central acceleration varies directly as the distance from the centre of acceleration and does not coincide with the direction of the velocity, we have motion in a curve and the motion is COMPOUND HARMONIC. Any Compound Harmonic Motion may be Resolved into Two Simple Harmonic Motions at Right Angles. Let C be the centre of acceleration, and P the position of the moving point at any instant. Let the velocity v of P make an angle a. with the axis of X, and let the motion of P be harmonic so that the acceleration of P is r, where / is the acceleration at a r o known distance r , and r is the distance CP. The velocity v may be resolved into v cos a and V sin a in the directions CX and CY, and the acceleration may be resolved c ^rcosPCA or^CA, r n r n and r cos PCB or -CB, in the same directions. ' '9 '0 The component accelerations are therefore directly as the distances CA and CB, and the component velocities are in the directions of CA and CB. The compound harmonic motion of P, whatever the direction of the velocity v, is therefore the resultant of two simple harmonic motions in the lines CA and CB at right angles. If, then, any compound harmonic motion is resolved into two components at right angles, the component motions are rectilinear harmonic. Conversely, the resultant of two rectilinear harmonic motions at right angles is a com- pound harmonic motion. Composition of Simple Rectilinear Harmonic Motions in Different Lines. Let the point Q move in a circle AQA' of radius r = CA = CQ with a constant angular velocity GO. Then the motion of the projection P in the line AA' is simple harmonic (page 126). Let the point Q l move in the circle CBQ l of radius r l = CB = CQ^, with constant angular velocity co l . Then the motion of the projection P l in the line CB is simple harmonic. Let the angle BCA between the planes of the circles be a. FIG. i. 130 KINEMATICS OF A POINTAPPLICATIONS. [CHAP. IV. Let the time count from the instant when Q l is at B t so that the epoch of 1\ is zero (page 128). At this instant let the epoch of P be e. Then e is the difference of epoch, or, in angular measure, the angle of Q above or below A at the beginning of the time. In any time /, Q l will have moved from B through the angle aoj measured from CB, and Q through the angle cat e measured from CA. By the preceding article we can resolve the harmonic motion of P t into a simple rectilinear harmonic motion at right angles to CA, and another along CA. The displacement of /*, from C for any time / is r t cos ( 2 is then (page 136) Let t be the time in which the point would with the same acceleration and with initial speed zero move from D to C. Then CD = \P. Hence 72 -CD = CD(CN, - CNj). Now by a property of the cycloid 4CD . CN, = CQ? and $CD . CN 2 = CQ*. Hence f&jtfW'+ZQfi. 140 KINEMATICS OF A POINT-APPLICATIONS'. [CHAP. V. Now t* = is a constant. Hence the motion of Q in the cycloid is harmonic (page 126), where - =^ , / being the tangential acceleration of Q at the distance J measured along the curve. If T is the time of a complete oscillation, we have If /' is the time occupied in moving from Q l to C, ' = or the time of a pendulnm whose length is 2 CD, or 4 times the radius of the generating circle. As this involves only constant quantities, the time is the same whatever be the position of the starting-point Q l , or the oscillations are isoch- ronous. Hence the cycloid is called a tnutochrone. This result is rendered of practical importance by one of the properties of the cycloid, viz., that if a flexible and inextensible string AB is fixed at the end A and wrapped tightly round the semi-cycloid AC, the end B of the string as it unwinds will describe another semi-cycloid. If, then, AC and AD are fixed semi- cycloids, symmetrical with reference to the vertical AB, and AB is a simple pendulum, B will describe a cycloid, and its oscillations will be isochronous whatever their extent. [Application of the Calculus. To Determine the Motion of a Point Constrained to Move in a Cycloid, the Acceleration being Constant, in the Direction of the Axis and towards the Vertex.] By the application of the general formulas of page 92 we can deduce the results already obtained. Let the axis CD = ir, where r is the radius of the generating circle DP'C. Let the acceleration / act downward. Let CN = y, NP = x and the length of arc CP = s. Let the initial position he />, at the height CN, h above C, and the speed at />, be v, = o. We have for the speed at any point given by CN y. Whenj/ = o, we have, at the lowest point C, v = which is the same as that due to the vertical height //. By the property of the cycloid we have s = arc CP Hence We have then ds dy j/HT. = 2 chord Cf. CHAP, v.] CONSTRAINED MOTION. Integrating, since for / = Q,y h, we have 141 (0 For the time of descent to the lowest point where y = o, or for the time of one quarter of a complete oscillation, The periodic time is then or the same as a simple pendulum (page 138) whose length is 4 times the radius of the generating circle DP'C. The time is independent of h and is the same no matter what the position from which the point begins to descend. The oscillations are therefore isochronous and hence the cycloid is called the tautochrone. The reason of this remarkable property is easily seen by considering the tangential acceleration. In the cycloid the chord CP' is always parallel to the tangent TP. The tangential acceleration or tangential component of y is then The tangential acceleration is therefore directly proportional to the distance from the vertex measured along the path, and the motion of P is simple harmonic (page 125). The periodic time is then (page 127) If in (i) we makej' = , we have - |/ -j. , or half the time from Pi to C. The time, therefore, in descending through half the vertical space to C is half the time of passing from P\ to C. [To Find a Curve such that a Point Moving on it under the Action of Gravity will Pass from any one Given Posi- tion to any Other in Less Time than by any Other Curve through the Same Two Points.] This is the celebrated problem of the "curve of swiftest descent" first propounded by Bernoulli. The following is founded upon his original solution. If the time of descent through the entire curve is a minimum, that through any portion of the curve is a minimum. It is also obvious that between any two contiguous equal values of a continuously "varying quantity, a maximum or minimum must lie. This principle, though simple, is of very great power, and often enables us to solve problems of maxima and minima, such as require not merely the processes of the Differential Calculus but those of the Calculus of Variations. The present case is a good example. Let, then, PQ, QR and PQ', Q' R be two pairs of indefinitely small sides of a polygon such that the time of descending through either pair, starting from P, may be equal. Let QQ be horizontal and indefinitely small compared with PQ and QR. The curve of swiftest descent must lie between these paths, and must possess any property which they have in common. Hence if we draw Qm, Qn perpendicular to RQ , PQ, and let v be the speed down PQ or PQ (supposed uniform) and i/ that down QR or Q 1 R, we have for the time from P to R by either path QR PQ QR z> or Qn Q'm 14* KINEMATICS OF A POINT- APPLICATIONS. [CHAP. V. Now if 6 be the inclination of FQ to the horizontal, and tf that of QR, we have Qn = QQ cos 6, Q'n - QQ cos 6'. Hence cos 8 _ cos 6' This is true for any two consecutive elements of the required curve, and therefore throughout the curve we have, at any point, v proportional to the cosine of the angle which the tangent to the curve at that point make/ with the horizontal. But 7/ 1 is proportional to the vertical distance // fallen through. Hence the curve required is such that the cosine of the angle it makes with the horizontal line through the point of departure varies as the square root of the distance from that line. Now in the figure of page 140 we have, from the property of a cycloid, cos CfN = cos TPN = cos CDP = DC The curve required is therefore the cycloid. The cycloid has received on account of this property the name of Brachistochrone. KINEMATICS OF A RIGID BODY. CHAPTER I. ANGULAR VELOCITY AND ACCELERATION COUPLES. ANGULAR AND LINEAR VELOCITY AND ACCELERATION COMBINED. Angular Velocity Couple. Two simultaneous, equal, parallel and opposite angular velocities, not in the same straight line, we call an ANGULAR VE- & LOCITY COUPLE. / Thus, if the point P has an angular velocity + GO about an p / v axis Oa, so that Oa = -f~ GO is its line representative, and at the same time has an angular velocity oj about an axis O'b, so that O'b = GO is its line representative, then, since the line represen- / tatives are equal, parallel and opposite, and do not coincide, they constitute a couple. Moment of Angular Velocity Couple. We have seen (page 90) that the moment OP X GO of the angular velocity Oa = -j- co relative to P, gives the linear velocity Pa = -j- v at right angles to the plane of Oa and OP, due to angular velocity about the axis Oa. The direction of + v is such that when we look along Oa in its & * ~* direction, -f- v is seen as clockwise rotation, or towards the reader in the preceding figure. In the same way O'P X GO gives the linear velocity Pb r of P in a direction away from the reader, at right angles to the plane of the couple. The resultant linear velocity of P is then v = (O'P - OP] GO= O'O X GO, where O'O is the perpendicular distance between the line representatives Oa, O'b of the couple. This resultant velocity is at right angles to the plane of the couple and in such a direc- tion that when we look along its line representative, rotation as indicated by the arrows of the couple is seen clockwise. With this convention, if the distance O'O = /, we have V = I GO This result holds no matter where the point P may be taken in the plane of the couple. If, then, P is a point of a rigid body, every point of this body in the plane of the couple must have the same velocity v in the same direction. That is, the body has a velocity of translation. 143 144 KINEMATICS OF A RIGID BODY. [CHAP. Hence, if a rigid body is acted upon by an angular velocity couple, the result is a velocity of translation for every point of the body. We denote velocity of translation always by v. Angular Acceleration Couple. Two simultaneous, equal parallel and opposite angular accelerations not in the same line we call an ANGULAR ACCELERATION COUPLE. We have then, in precisely the same way and with the same convention as to direction, the acceleration for every point of the body /=/, where a is the angular acceleration and / is the linear acceleration of translation of the body. Hence, if a rigid body is acted upon by an angular acceleration couple, the result is a linear acceleration of translation for every point of the body. We denote acceleration of translation always by/. Angular and Linear Velocity Combined. Let a rigid body have an angular velocity GO about an axis O'a, so that O'a = GO is the line representative. /* v * Let O be any point of the body. If at this point we apply two ,/^ equal and opposite angular velocities Oa = GO and Ob = GO, both parallel to O a = GO, the previous motion of the body is evidently not affected. We see, then, that the single angular velocity O'a = GO about an axis O'a can be reduced to the same angular velocity Oa = GO about a parallel axis Oa through any point O, and an angular velocity couple O'a and Ob. Let / be the distance between the parallel axes. Then, as we have just seen, the couple causes a velocity of translation ~v n = loo at right angles to the plane of the couple, so that looking along the line representative of v n in its direction, the arrows of the couple indicate clockwise rotation. In the figure v n is at right angles to the plane of the couple and away from the reader. Hence (a) A single angular velocity GO of a rigid body about a given axis, can be resolved into an equal angular velocity about a parallel axis through any point O of the body at a distance /, and a normal velocity of translation V H = IGO of this axis in a direction at right angles to the plane of the two axes. (&) Conversely, the resultant of an angular velocity GO of a rigid body about a given axis and a simultaneous velocity of translation V H normal to that axis, is a single equal v m angular velocity about a parallel axis at a distance / = , the plane of the two axes being perpendicular to v n . (c) If, then, a rigid body has any number of angular velocities, each one about a different axis through a different point, then by (a) we can reduce each one to an equal angular velocity about an axis through any point O we please, and a normal velocity of translation of this axis. All the angular velocities at this point O can then be reduced to a single resultant angular velocity GO about a resultant axis by the polygon of angular velocities (page 68), and all the normal velocities of translation can be reduced to a single resultant velocity of translation v, not necessarily normal to the resultant axis, by the polygon of linear velocities (page 66). The motion of a rigid body in general can then be reduced at any instant to an angular CHAP. I.] INSTANTANEOUS AXIS OF ROTATION. 145 velocity GO about an axis through any point O we . please, and a velocity of translation v of this axis. This velocity v of translation is not necessarily normal to the axis. The angular velocity GO has the same magnitude and direction no matter what point is taken, but the velocity of translation v varies in magnitude and direction with the position of this point. (d) This velocity of translation v is not necessarily normal to the axis and can in general be resolved into a component ~v a along the axis through O, and a component v n normal to this axis. But by (b] we can reduce GO and ~v n to the same angular velocity GO about a parallel axis at a distance / = . CO Instantaneous Axis of Rotation. This axis is called the INSTANTANEOUS AXIS OF ROTATION because it is the axis without translation about which at a given instant angular velocity takes place. Hence, in general, the motion of a rigid body, at any instant, can be reduced to an angular velocity GO about an axis through any point of the body, a velocity of translation v a along this axis, and a normal velocity v n of translation of this axis. Or to an angular velocity GO about a parallel instantaneous axis at a distance / = and a velocity of transla- tion v a along this axis. Spin. Screw-Spin. Angular velocity of a rigid body about any axis we call a SPIN about that axis. Angular velocity of a rigid body about any axis together with velocity of translation along that axis we call a SCREW-SPIN. The velocity of translation v a along the axis we call the VELOCITY OF ADVANCE. Hence the motion of a rigid body at any instant can be reduced in general to a spin or a screw-spin about an instantaneous axis, or to a spin or screw-spin about a parallel axis through any point together with normal velocity of translation v n of that axis. Spontaneous Axis of Rotation. The axis of rotation through the centre of -mass of a rigid body at any instant is called the SPONTANEOUS axis of rotation. If it has normal velocity of translation v n , the instantaneous axis of rotation is parallel to it at the distance / = V J*-, the plane of these two axes being perpendicular to v n (page 144). The velocity of any point is that due to angular velocity about the instantaneous axis of rotation, or angular velocity about the parallel translating spon- taneous axis (page 144). If the spontaneous axis of rotation has no normal velocity of translation v n , the spontaneous and in- stantaneous axes coincide. Examples. (i) A vertical circle of radius r = 2 ft. rotates about a fixed horizontal axis through the centre of mass at right angles to the plane of the circle "with angular velocity of j radians per sec. Find the velocity and central acceleration of the top, bottom, forward and rear points, and of any point in general. ANS. Take co-ordinate axes as shown in the figure. Then r =. 2 ft., K> y = + 3 radians per sec. Since the spontaneous axis O Y is without translation, the instantaneous axis coincides with it. We have then at the top point a the velocity 2/^=4- roo y = + 6 ft. per sec. At the bottom point KINEMATICS OF A RIGID BODY. [CHAP. 1 b the velocity v x = rat y = 6 ft. per sec. At the forward point c the velocity t/, = roo y = 6 ft. per sec. At the rear point the central acceleration fp, = + rao, + 18 ft.-per-sec. per sec. At the forward point f the central acceleration f ftx = roo' = 18 ft.-per-sec. per sec. At the rear point d the central acceleration f px = + 18 ft.-per-sec. per sec. Let + JT, + y, +s be the co-ordinates of any point P. Then we have the component velocities of that point given by v x = zw y , v y = o, v t = - xa>y (I) The resultant velocity is = OOy (2) and its direction cosines are cos ft = o, cos Y = (3) The central acceleration is /J> = reoj towards the centre. The radius r has the direction cosines cos ot= cos = o, cos ^ = . The component central accelerations are then /puc = - /p- = - The resultant central acceleration is and its direction cosines are f P y=0, (4) (5) /p - -, cos /S (6) (2) /? vertical circle of radius r = ?//. rolls on a horizontal straight line. The centre moves parallel to that line with a velocity of 6 ft. per sec. Find the angular velocity ; the velocity and central acceleration of the toft, bottom, forward and rear points, and of any point in general. ANS. Take co-ordinate axes as shown in the figure. Then r = 2 ft., v x = + 6 ft. per sec. The spon- taneous axis is O Y. Since the circle rolls, the instantaneous axis b Y' passes through the bottom point b parallel to the spon- taneous axis. We have then for velocity of translation v x = r} , fpy = o, f(# = - za>l (4) The resultant is and its direction cosines are j> * __ (6) These equations are the same as (4), (5), (6) of the preceding example. (3) Let a vertical circle of radius r = 2 ft. roll on a horizontal plane. The centre moves with a "velocity of 6 ft. per sec. At the same time let the plane of the circle rotate about the vertical diameter with an angular velocity of 2 radians per sec. downwards. Find the angular velocity about the horizontal axis ; the velocity and central acceleration of the top, bottom, forward and rear points, and of any points in general. ANS. Take the co-ordinate axes as shown in the figure. We have then r = 2 ft., o> : = 2 radians per sec.,^* = + 6 ft. per sec. Since the circle rolls, we have roo y = ~v x or oo y = = +3 radians per sec., and the instantaneous axis passes through the bottom point b. The velocity of any point is that due to translation and angular velocity ooy about OF and < about OZ, or to translation and angular velocity GO about the spontaneous axis OA, or to angular velocity oo only about the instantaneous axis bA' parallel to OA. We have then at the top point a the velocity v x = ~v x + rooy = 2roo y = +12 ft. per sec. At the bottom point b the velocity is v x = v x - rca y = o. At the forward point c we have the component velocities v x = v'x = rooy = + 6 ft. per sec., v y = roo z = 4 ft. per sec., v z = rw y = 6 ft per sec. The resultant f Vy + V a = 2 |/22 velocity is then v = \v\ sec., and its direction cosines are given by v x ? cos a = - = + -4=. per KINEMATICS OF A RIGID BODY. [CHAP. 1. At the rear point d we have the component velocities v x = v x = r y - jcool , The resultant is f f = V The direction cosines of/ p are cos a =S' X . JP These equations reduce to (4). (5). (6) of = Z00yt0t f ft = - zoo, =, JP Jo nrerrding example if ao t = o. (4) (6) CHAP. I.] ANGULAR AND LINEAR VELOCITY COMBINED EXAMPLES. 149 (4) Let a vertical circle of radius r = 2 ft. roll on a horizontal plane with an angular velocity of j radians per sec., while its centre describes a horizontal circle of radius j ft. with angular velocity about a fixed vertical axis. Find the angular velocity about the fixed axis ; the velocity and central acceleration of the top, bottom, forward and rear points, and of any point in general. ANS. Take the co-ordinate axes as shown in the figure. Let O ' Z' be the fixed axis. Then r = 2 ft., x. = 3 ft., < = + 3 radians per sec. Since the circle rolls, the bottom point b is on the instantaneous axis. The instantaneous axis must also pass through the point O 1 . Hence Ob is the instantaneous axis, and OA parallel to it is the spontaneous axis. We have then rtxty - - yco z , or rooy oo z = =~ 2 radians per sec. oo z is therefore negative, as shown in the figure. The velocity of translation of -^_ the centre is given by v x = raoy = -\- 6 ft. per sec. The velocity at any point is that due to translation and angular velocity co about the spontaneous axis OA, or to angular velocity K> only about the instantaneous axis O'b. We have then at the top point a the velocity v x = z/* + roo y = + 2ry = + 12 ft. per sec. At the bottoii) point b the velocity is v x = v x ra) y = o. At the forward point c we have the component velocities v x = v x = rooy = + 6 ft. per sec. v y = roo z = 4 ft. per sec., v, = roo y = 6 ft. per sec. The result- ant velocity is then v = \'v\ + v y + vl = 2 4/22 ft. per sec., and its direction cosines are given by V x 3 cos a = + v 4/22 |/22 cos Y = = T=. V 4/22 At the rear point d we have the component velocities v x = v x = roa y = + 6 ft. per sec., v y = rco 2 = + 4 ft. per sec., v z = + roo y = + 6 ft per sec. The resultant velocity is, as before, v = 2 ^22 ft. per sec., and its direction cosines are cos a = = H ^= 3_ |/22 4/22 v 4/22 The central acceleration of any point is that due to angular velocity about OA considered as without translation (page 145). or to anglar velocity coy about O Y and GO Z about OZ, without translation. In either case O and hence every point has acceleration XGO\ towards O'Z', owing to rotation about OZ'. We have then at the top point a, the deflecting acceleration (page 80) f rz = mo* = 1 8 ft.-per-sec. per sec. due to rotation about OY, the deviating acceler- ation (page So) fay = raoyao, = 12 ft.- per-sec. per sec. due to rotation of the plane about OZ, and the deflecting accel- eration f^ = - xv>l = - 12 ft.-per-sec. per sec. due to rotation about O'Z'. The comconent central accelerations are then f pz = - ra) ) = - 1 8 ft.-per-sec. per sec. and/p, =f ry + fay = my* - J** = - 24 ft.-per-sec. per sec. The resultant is then / p = 4// p 2 + fpy = 30 ft.-per-sec. per sec., and its direction cosines are cos a =. o, cosy = = - 0.6. yp 150 KINEMATICS OF A RIGID BODY. [CHAP. I. At the bottom point b the deflecting acceleration due to rotation about OY is fn = + roa y . The deviating acceleration due to rotation of the plane about OZ is fay = r t . The deflecting acceleration due to rotation about O'Z' is/ry = - xoo\. The component central accelerations are then ff, = rtOy = + 1 8 ft.-per-sec. per sec.,/^ = f ay +/ry = rooyco, -yv>\ = o. The resultant central accelera- tion is then/p,. At the forward point c the deflecting acceleration due to rotation about O Y isf rx = rooiy, and due to rotation about O'Z, fry = yv>l. The deflecting acceleration due to rotation of the plane about OZ is frs = rooj. The component central accelerations are then f px = ra>y real = 26 ft.-per-sec. per sec., f n = - To; = 12 ft.-per-sec. per sec. The resultant acceleration is then f a = ^fp$ + f p % = 2 4/205 ft.-per-sec. per sec., and its direction cosines are fa x 13 fay 6 cos a = ->- = =, cos ft = -- = , cos y = o. /p 4/205 yp 4/205 ' At the rear point d we have, in the same way, f px = + rao} + roo\,f w yu>\, or/p* = + 26 ft.-per- sec. per sec., f n = 12 ft.-per-sec. per sec.,/p = 24/205 ft.-per-sec. per sec., and the direction cosines are cos a = H L==, cos ft = =, cos y = o. 4/205 4/205 Let + x, + y, + z be the co-ordinates of any point P. Then we have the component velocities The resultant velocity is where r' is the radius vector of the point from b. The direction cosines of v are cos a = , cos /$=, cos y = -- (3) Those equations are the same as in the preceding example. For the component deviating accelerations due to rotation of the plane about OZ we have f ax = O, f a y = i-G^H , faz = O. For the component deflecting accelerations due to rotation about OY, OZ, and O'Z' we have fr* = XODy* ~ -fO^ 8 , fry = - *<*)?, f rz = - TO?/. Hence y (4) The resultant is fa = 4/ypTl-/P>+yV. (5) and its direction cosines are yp yp yp These equations reduce to (4). (5), (6) of the preceding example if x = o. Angular and Linear Acceleration Combined. Let a rigid body have an angular acceleration at about an axis O'a so that O'a = a is the line representative. Let O be any point of the body. If at this point we apply two equal and opposite angular accelerations Oa = a and Ob or, both parallel and equal to O'a = a, the motion of the body is evidently not affected. | We see, then, that the single angular acceleration O'a = a about an axis O'a can be reduced to the same angular accoJera- tion Oa a about a parallel axis Oa through any point O, and an angular acceleration couple O'a and Ob. I g * g Let / be the distance between the parallel axes. Then CHAP. I.] ANGULAR AND LINEAR ACCELERATION COMBINED. 151 (page 144) the couple causes acceleration of translation f n = la at right angles to the plane of the couple, so that looking along the line representative of f n in its direction, the arrows of the couple indicate clockwise rotation. In the figure/, is at right angles to the plane of the couple and away from the reader. Hence (a) A single angular acceleration a of a rigid body about a given axis can be resolved into an equal angular acceleration about a parallel axis through any point of the body at a distance /, and a normal acceleration of translation f n = la of this axis in a direction at right angles to the plane of the two axes. () Conversely, the resultant of an angular acceleration a of a rigid body about a given axis and a simultaneous acceleration of translation/, normal to that axis, is a single equal angular acceleration about a parallel axis at a distance /= , the plane of the two axes being perpendicular to/",,. (c) If, then, a rigid body has any number of angular accelerations, each one about a different axis through a different point, then by (a) we can reduce each 'one to an equal angular acceleration about an axis through any point we please, and a normal acceleration of translation of this axis. All the angular accelerations at this point can then be reduced to a single resultant angular acceleration a about a resultant axis by the polygon of angular accelerations (page 83), and all the normal accelerations of translation can be reduced to a single resultant acceleration of translation / not necessarily normal to t;he resultant axis, by the polygon of linear accelerations (page 76). The change of motion of a rigid body in general can then be reduced, at any instant, to an angular acceleration n about an axis through any point we please, and an acceleration of translation / of this axis. This acceleration /of translation is not necessarily normal to the axis. The angular acceleration a has the same magnitude and direction no matter what point is taken, but the acceleration of translation f varies in magnitude and direction with the position of this point. (d) This acceleration of translation f is not necessarily normal to the axis and can in general be resolved into a component/ along the axis of or and a component f n normal to this axis. But by (fr) we can reduce a and/, to the same angular acceleration a about a parallel fn- axis at a distance / =z a INSTANTANEOUS Axis OF ACCELERATION. This axis is the INSTANTANEOUS AXIS OF ACCELERATION because it is the axis -without acceleration about which at a given instant angular acceleration takes place. Hence, in general, the change of motion of a rigid body at any instant can be reduced to an angular acceleration a about an axis through any point of the body, an acceleration of translation/, along this axis, and a normal acceleration/, of translation of this axis. Or fn to an angular acceleration a about a parallel instantaneous axis at a distance / = ~ and an acceleration of translation /, along this axis. IS* KINEMATICS OF A RIGID BODY. [CHAP. I. Twist Screw-Twist. Angular acceleration of a rigid body about any axis we call a TWIST. Angular acceleration about any axis together with acceleration of translation along that axis we call a SCREW-TWIST. Hence the change of motion of a rigid body at any instant can be reduced in general to a twist or screw-twist about an instantaneous axis ; or to a twist or screw-twist about a Parallel axis through any point together with a normal acceleration of translation f H of that axis. SPONTANEOUS Axis OF ACCELERATION. The axis of acceleration through the centre of mass of a rigid body at any instant is called the spontaneous axis of acceleration. If it has normal acceleration of translation / , the instantaneous axis of translation is parallel to it at the distance / = , the plane of these two axes being perpendicular to^, (page 151). The acceleration of any point is that due to angular acceleration about the instan- taneous axis of acceleration, or angular acceleration about the parallel translating sponta- neous axis (page 151). If the spontaneous axis of acceleration has no normal acceleration of translation f H , the spontaneous and instantaneous axes coincide. CHAPTER II. ROTATION AND TRANSLATION ANALYTIC RELATIONS. Components of Motion. We have seen (page 144) that the motion of a rigid body at any instant can be reduced to a resultant angular velocity GO about an axis through any point of the body and a resultant velocity of translation v of this axis or of the body. The motion of the body at any instant is then known, if we know at that instant the velocity of translation of the axis through some given point of the body and the angular velocity of the body about this axis. We always take the given point at the centre of mass of the body, and denote the com- ponents of the velocity of translation v of the axis through it, along the co-ordinate axes, by ~v x , T> y , D z , and the components of the angular velocity w by GO X , oo y , co z . The motion of the body at any instant is then known when these six quantities are known : These six quantities are therefore called the COMPONENTS OF MOTION of the body. Components of Change of Motion. We have seen (page 151) that the change of motion of a rigid body at any instant can be reduced to a resultant angular acceleration a about an axis through any point of the body and a resultant acceleration of translation / of this axis. The change of motion of the body at any instant is then known, if we know at that instant the acceleration of translation of the axis through some given point of the body and the angular acceleration of the body about this axis. We always take the given point at the centre of mass of the body, and denote the components of the acceleration of translation f of the axis through it, along the co-ordinate axes, by f x , f y , f z , and the components of the angular acceleration a by a x , M about OZ. In the same way we have, parallel to OY, xoo, positive and zw x negative, and parallel to OZ, yoo, positive and xoo y negative. If, then, v x , Vy, v, are the component velocities of the velocity v of the point Pdue to rotation about the axis Oa through the fixed point O, we have for rotation only about an axis Oa through the fixed centre of mass V M = zooy y&>. , Vy = xoo, ZGO X , v g = yoo x xw y (i) ROTATION ANY POINT FIXED. Let any point O' of the body be fixed, and take the co-ordinate axes OX 1 , O' Y , O' Z 1 , parallel to OX, OY, OZ. Let the point P have the angular velocity GO about an axis O'a' through the fixed point O' , and let the components of along the co-ordinate axes X' , Y' , Z' be ai x , co y , ov Let the co-ordinates of the centre of mass O relative to O' be x, y , ~s. Then the co-ordinates of any point P will be x -\- x, ~y _|_ y t -3 _j_ z, where x, y, z are the co-ordinates of P relative to the centre of mass O. We have then, from (i), for the component velocities of the velocity v of the point P due to rotation only about an axis O'a' through any fixed point O' of the body = (s -h *), - ( J+ y)w f . v, = (x + x)w M - (J + 2)00, , v, - (y+y}oo x -(* + *),. (2) If, in these equations, we make x = o, y = O, z = O, we have for the component velocities v x , v 1t v, of the centre of mass O due to rotation about the axis O'a' through the fixed point O' v x = ~ZGo y y y (3) ROTATION ABOUT TRANSLATING Axis THROUGH CENTRE OF MASS Let the axis Oa through the centre of mass O have the velocity of translation v in any direction, and let ^* > Vy, v t be the components of ~v. The point P will have, then, component velocities of translation in addition to the component velocities given by equations (i ) for rotation only about an axis through the fixed centre of mass. Hence, for rotation and translation combined, we have the. component velocities of P : v, = v x +ZGo y yoo f , v y Vy-\- xw t z* /o\ cos a = , cos 8 = -^, cos y = (8) GO ' GO' GO MOMENT OF VELOCITY. Let the moment of the velocity of P about the axis O' a' through any point O' be M' v . Then for the component moments M' vx , M' vy , M' vz about the axes X , Y' , Z' of the component velocities v x , v y , v z of the point P due to translation and rotation we have about O'X' M^ x = v z (y+y) v y ( z~ -\- z), ~] where v x , v y , v z are given by equations (4). Substituting these values, we have for rotation and translation - x(y * (* + s)<,- y(s- ^(^ -f ^)w,- 4^ For rotation only ~v x = o, ^ = o, ^ = o. For axis through centre of mass ~x = o, J 7 o, F o. VELOCITY ALONG THE Axis OF ROTATION. For the velocity of translation v a along the axis of rotation we have a = v x cos a -- v y cos y or, from (8), - = y,, + r,,+ W GO This is called the axial velocity or velocity of advance. It is the velocity with which the body moves along the axis of rotation. VELOCITY NORMAL TO THE Axis OF ROTATION. The components, along the axes, of the axial velocity are T> a cos a, ~v a cos /3, v a cos y. If we subtract these from the components v x , v y , v t of the velocity v of translation, we have the components of the velocity of translation v n normal to the axis of rotation: VGfl x I v nx = v x v cos a v v v ny v y v a cos ft = v y ~- y _ v nz = v z v a cos y = v z '5* KINEMATICS OF A RIGID BODY. [CHAP. II. Equations (7), (8) and (i i) give the screw-spin (page 145) about the axis of rotation through the centre of mass. Equations (7) and (8) give the angular velocity (about the axis of rotation and the direction of this axis, and equations (u) give the velocity v a along this axis. Equations (12) give the components of the velocity of translation normal to this axis. INSTANTANEOUS Axis OF ROTATION. Let x t y, z be the co-ordinates of any point of the instantaneous axis of rotation. Since the normal velocity of any point of this axis is zero, we have, from equations (4) (page 1 54), These are the equations of the projections of the instantaneous axis of rotation on the three co-ordinate planes. Let JT,, y t . 2, be the intercepts of these projections on the co-ordinate axes. Then in equations (13), making y = o and z = o in the last two, we have making x = o and z = o in the first and last, Vnx y* = w. making y = o and x o in the first two, (H) GO X Let the perpendicular from the centre of mass O upon the axis of rotation be />, and let /* Pyt P t De its projections on the co-ordinate axes. Let the intersection of p with the axis be O', so that 00' =p. ^ Let us consider the projection of the axis on the plane YZ. In the figure p, Ob and p y Oc. We have, then, where s l and j/, are the intercepts Oa and Od given by equations (14). We have also the distance Hence Substituting the value of p y , we obtain A Substituting the values of , and _y, from (14), we have = P-M, _ _Vn * * HI* I /V) Z ' '1 ' ft" - CHAP. II.] THE INVARIANT FOR COMPONENTS OF MOTION. 157 In the same way we can find p x and fl x , p x and p y on the other two co-ordinate planes. We thus have /,= - Py=~ 05) Equations (15) give the position of the instantaneous axis of rotation. We have also, from (13) and (8), v x =v a cos a a)(p x cos ft p y cos y) , GO X = GO cos a, v y = v a cos ft Go(p x cos Y p z cos a), oo y = GO cos /?, v f = v a cos Y Go(p v cos a p x cos ft), G? = GO cos y. (16) When, therefore, the components of motion, v x , v y , v, , .,., c^ , GO Z , are given for the centre of mass O, we have GO from (7), the direction of the instantaneous axis of rota- tion from (8), and the position of this axis from (15)- We have also the velocity v a along this axis from (11), and the normal velocity v n from (12). On the other hand, if the position and direction of the instantaneous axis of rotation are given, together with the velocity v a along it and the angular velocity GO about it, the compo- nents of motion are given by equations (16). THE INVARIANT FOR COMPONENTS OF MOTION. From (11) we have V a GO = V X GO X -\-VyG0y -}-^V z GO z But whatever point we take as origin, GO does not change, and the velocity v a along the instantaneous axis of rotation does not change. This quantity (17) is therefore called the invariant of the components of motion. If the invariant is zero in any case, we must evidently have either v a or GO zero. If GO is not zero but the invariant is zero, then the velocity v a along the axis must be zero. In this <--- x * case the condition V X GO X ~h ^y^v + V,<0, = O is the condition for a spin, or angular velocity only, about the instantaneous axis of rotation. If v " a is not zero but the invariant is zero, then GO must be zero, and we have translation only. Change of Motion of a Point of a Rigid Body General Analytic Equations. Let O be the centre of mass of a rigid body. Take any co- ordinate axes X, Y, Z through the centre of mass O, and let the co-ordinates of any point P be x, y, z. Let the point P have angular acceleration a about any axis Oa, and let the components of a along the co-ordinate axes be a Let the axes X, Y, Z be fixed in the body and f Ax >/> inith it X -a' -Y' move with it. I 5 8 KINEMATICS OF A RIGID BODY. [CHAP. II. TANGENT ACCELERATION. Let the axis Oa be fixed. Then zciy is the tangent acceleration of P parallel to OX in a positive direction due to angular acceleration a y about OY, and ya n is the tangent acceleration parallel to OX in a negative direction due to angular acceleration a t about OZ. In the same way we have, parallel to OY, xa t positive and sa M negative, and parallel to OZ, yn x positive and xa y negative. If, then, ftxiftyt f tt are the component accelerations of the tangent acceleration f t of the point P due to angular acceleration a about the fixed axis Oa, we have = sa, ya. f t , = xa t za x , f t . = yet, - xa y . For any other axis through any point O' we have x -f- x t "y-\- y, z -\-z\n place of x,y, z, and hence / te = (* 4- *K - CP /*== V These equations are general. For axis through the centre of mass we have x = o, y = O, 2 = O. Wz CENTRAL ACCELERATION. The central acceleration is due to rotation only, and is not affected by translation. ^"*<" 2 Let, then, v x , v y , v t be the component velocities of the point P due to rotation only about an axis O'a' through any point O'. Then v t co y is the central acceleration of P parallel to O'X' in a positive direction, and v y Go x is the central acceleration parallel to O'X' in a negative direction. In the same way Y' we have, parallel to O' Y' , v x , /p, are the component central accelera- tions of the point P due to rotation only about an axis O'a' through any point O', we have where v xt v y , v t are given by equations (i) page 154. Substituting these values of v x , t' y , v nt we have for axis through centre of mass ffy = ZOOyGO, 4 XOO y CO x yw* yco*, /P, = xw,v x +y( l ov> y zoo* sco*. For axis through any point O' we have only to put ~x 4- x, y + y, z \- z in place of x, y, z, and we have then, in general, - (x + x)<*> y a> x (y+y}co* (y + y)<*l, fp, = (* 4 *)<.<, 4- ( y +?)<*>,<*>, - ( * 4- -)! - ( z + z)coj. These equations are general for axis through any point. For axis through centre of mass we have x o. y = o, 2 = o. CHAP. II.] DEFLECTING AND DEVIATING ACCELERATIONS. 159 DKFLECTING AND DEVIATING ACCELERATIONS. We see, from page 80, that in equations (2) all terms containing &?J, o^, col give the component deflecting accelerations, the other terms give the component deviating accelerations. If, then,/ r is the deflecting or radial, and f a the deviating or axial, acceleration, we have for the components of the deflecting acceleration (4) (3) frt ~ ( ^4~ 2 ) <*% ( ' and for the components of the deviating acceleration fa* = (y -i-f)*, + ( *. . fay = ( Z + Z)B>y<0, + ( X f az =(x+*}<*>,GO x + All these equations are general. For axis through centre of mass we have ~x= o, y = o, RESULTANT ACCELERATION. Let the centre of mass O have the acceleration of translation f and the components f x , f y , f z . Then the components of the resultant acceleration /are /, =Z +/P, +/, = f* +f* -h / + A, f, =7, +/<> +A =7, +/ +/ +A; or substituting the values oif ax , f ay , fa*, fr** fry, / , /<* , A' /< f rom equations (4), (3) and (i), we have f* = 7* + ( fy = f y + (*~+ *)/. +'(*+*)^^ ^ These equations are general. For axis through centre of mass we have x = o, = o, 5 = o. In any case the resultant acceleration of P is given by /= Vf** +fy Z +/? ' . .- -. (6) Its line representative passes through P, and its direction cosines are ....... (7) t , RESULTANT ANGULAR ACCELERATION. The resultant angular acceleration is given by 'a? + +^? ..... ..... (8) 160 KINEMATICS OF A RIGID BODY. [CHAP. II. Its line representative passes through O, and its direction cosines are cos a = ^, cos = - r , cos y = - ........ (9) MOMENT OF ACCELERATION. Let the moment of the acceleration of P about the axis O'a' through any point O' be M' r Then for the component moments M' fx , M' fy , M ft about the axes X ', V, Z' Ave have about O'X' ...... M',*= " 0'Y> ...... ^= " O'Z' ...... M*= where f x , / /, are given by equations (5). Substituting these values, we have for rotation and translation M' fx = f.(y +y)-ffc + **)_+(?+ y)C* + *)(.<* - ,)-(+*)( f + *)(,<,+ 1 4- (* For rotation only f x = o, ^ = o, yj = o. For axis through centre of mass J o, J = o, 5"= o. ACCELERATION ALONG THE Axis. For the acceleration of translation /, along the axis of angular acceleration we have f a = f x cos a + f y cos or, from (9), This is called the axial acceleration or the acceleration of advance. It is the acceleration with which the body moves along the axis of angular acceleration. ACCELERATION NORMAL TO THE Axis OF ANGULAR ACCELERATION. The com- ponents along the co-ordinate axes of the axial acceleration are f a cos a , ft cos fi, f a cos y. If we subtract these from the components f x , f y , f, , we have the components of the acceleration of translationy,, normal to the axis of angular acceleration, = A - fa cos a = f jc f a ^, /=/ -/.cos /*=/,-/ .- =/.-/- cos y = /,-/, ' (12) CHAP. II.] INSTANTANEOUS AXIS OF ACCELERATION. 161 Equations (8), (9) and (11) give the screw-twist about the axis of angular acceleration through the centre of mass. Equations (8) and (9) give the angular acceleration and the direction of the axis of angular acceleration, and equation (i i) gives the acceleration f a along this axis. Equations (12) give the components of the acceleration of translation normal to this axis. INSTANTANEOUS Axis OF ACCELERATION. Let*, y, z be the co-ordinates of any point of the instantaneous axis. Since the normal acceleration for any point of this axis is zero, we have 7n* + My-?"* = fny + XOi * ~ *<** = O, f nz + y Uje - XOi y = O. (13) These are the equations of the projections of the instantaneous axis of acceleration on the three co-ordinate planes. Let x^ y v , z l be the intercepts of these projections on the co-ordinate axes. Then, in equations (13), making^ = o and z = o in the last two, we have _<_^. a' making x = o and z = o in the first and last, ^ = ^. = / ctr making y = o and x = o in the first two, fny 04) Let the perpendicular from the centre of mass O upon the axis of angular acceleration be/, and let/^, p y , p z be its projections on the co-ordinate axes. Let the intersection of/ with the axis be O', so that OO' = p. Then, just as on page 157, we have P* = ~ Py ~ ~ fny<* (15) ot + ct- y " a* + a* We have also, from (13) and (9), /* = fa cos a a(p z cos /? p y cos y), f y =f a cos /3 <* y , <* z are given for the centre of mass O, we have a from (8), the direction of the instantaneous axis of acceleration from (9), and the position of this axis from (15). We have also the acceleration f a along the axis from (n), and the normal acceleration/, from (12). On the other hand, if the position and direction of the instantaneous axis of acceleration are given, together with the acceleration f a along it and the angular acceleration a about it, the components of change of motion are given by (16). THE INVARIANT FOR COMPONENTS OF CHANGE OF MOTION. From (n) we have f a a = f x a x + f y a y (17) 162 klNM/1TICS Of A RIGID BODY. fCHAP. 11. But whatever point we take as origin, a does not change and the acceleration /, along the instantaneous axis of acceleration does not change. This quantity (17) is therefore called the invariant of the components of change of motion. If in any case the invariant is zero, we must evidently have either/, or a zero. If is not zero but the invariant is zero, then the acceleration /, along the axis must be zero. In this case the condition *(** -\-fytXy = o is the condition for a twist, or angular acceleration only about the instantaneous axis of acceleration. If f a is not zero but the invariant is zero, then a must be zero and we have no angular acceleration. Examples. ( i) A baseball rotates about an axis through its centre of mass, with angular velocity GO, and its centre of mass has a -velocity v making an angle with oo. Find the resultant motion. ANS. Take v coinciding with the axjp of Y, and the plane of v and GO as the plane of XY. The components of motion are v x = o, v y = v, v, = o, v a v cos ft, 00 X = 00 COS a, GOy = 00 COS ft', CO, = O, where a and ft are the angles of GO with the axes of X and Y. For the position of the instantaneous axis of rotation we have, from equations (12), page 155, and equations (15), page Y 157. Vnx = ~ V COS ft, V,,y =:~V V COS 4 ft, V nt = O, Show how this can be. where v a is the component of the velocity along G^ or v a = z7 cos a. The motion of the ball is then a screw-spin consisting at any instant of rotation GO about an axis O'a' parallel to the axis of rotation at the centre of mass O at the distance p z from the centre of mass, and a velocity of translation W a along this axis. Or rotation to about the axis Oa, througli the centre of mass, velocity v a along this axis and z7 of translation of this axis, where ~-v n "v cos ft. Disregarding the action of gravity, the centre of mass O moves in the resultant of z/" a and V H , or along OY with uniform velocity v. Since, owing to gravity, the ball falls vertically, the centre of mass O moves in a vertical curve, the plane of which intersects the plane XY in a straight line OY. (2) Kail-players assert that this intersection is not a straight line, but a curve. ANS. No account has been taken of the resistance of the air. ist. CURVATURE NORMAL TO THE PLANE OF GO andv. Owing to rotation ao about the axis Oa and the velocity v n of translation of this axis, the velocity of all points of the advancing quadrant Obc will be greater than for correspond- ing points of the receding quadrant Ocd. The normal pressure on the advancing quadrant Obc will then be greater than the normal pressure n' on the re- ceding quadrant Ocd. The resultant pressure N at the centre O makes an angle with T- n and its component N H normal to V H and to the plane of oo. and v is always away from the advancing quadrant Obc and causes curvature in this direction. It is generally supposed that the curvature is due to the ball rolling upon a cushion of compressed air CHAP. II. J EXAMPLES. 163 in front (5f it, but, as we see, the direction of N n is always opposite to the direction in which the ball would thus tend to roll. 3d. CURVATURE IN THE PLANE OF a> AND v. The air causes a retardation of v~ a and 17 n , that is of translation along the axis and translation of the axis. If these retardations were proportional to the velocities there would be no curvature in the plane of GO and v'. But these retardations are more nearly proportional to the squares of the velocities. Hence the greater component is retarded proportionally more than the lesser, and we have curvature in the direction of the lesser. If, then, ~v makes an angle with GO less than 45, we have curvature Oa in the direction of z7 M ; if greater than 45, we have curvature Ob in the direction of z> a - Thus by giving the ball a spin the pitcher is able to make it curve right or left in the plane of GO and v, and at the same time a curve at right angles to this plane has also been proved. We havejthen, an acceleration/ right or left in the plane, and an acceleration / parallel to Nn in the first figure at right angles to this plane. The curvature will always, then, be opposite in direction to the direction in which the ball would tend to roll on the forward cushion of compressed air. So far as any such action exists it decreases the curvature. If, then, TJ is at right angles to oo, or makes an angles of 45 with it, there is no curvature in the plane of v and on at that instant, but in all cases and at every instant there is curvature at right angles to this plane opposite to the direction in which the ball would tend to roll on the forward cushion of compressed air, and diminished more or less by this tendency. (3) A vertical circle of radius r = 2 ft. rotates about a fixed horizontal axis through the centre of mass at right angles to the plane of the circle with angular velocity of j radians per sec. Find the velocity and central acceleration for any point in general. ANS. (Compare with example (i), page 146.) Take co-ordinate axes as shown in the figure. Then r = 2 ft., a> = oo y = + 3 radians per sec., oo x = o, o> 2 = o, v x = o, ~v y = o v z = o. The components of motion, then, are l^nown. Let x and z be the co-ordinates of any point P. For all points y = o. Then, from equations (i), page 154, we have V x = ZOOy , Vy = Q, V z = XOOy (i) The resultant velocity is v = 4/Vx + Yand its direction cosines are j + v\ = V x Z a Vy cos a = + , cos p = = o, v r v V, X COS Y = - = v r (2) (3) / These are the same equations as on page 146. From equations (15), page 157, we see that the axis OY is the instantaneous axis, or the spontaneous and instantaneous axes of rotation coincide. From equations (17), page 157, we see that the invariant is zero, and we have a spin only about the instantaneous axis. , From equations (2), page 158, we have for the component central accelerations, since J = o, y = o, z = o, and/ o, f ftx XK>1 , fpy = O, fpz = ZGOy. (4) The resultant central acceleration is (5) and its direction cosines are cos a = , = , cos ft - fy- = o, IP (6) OF A RIGID BODY. [CHAP. It. These are the same equations as in example (i), page 146. From equations (4), page 159, we see that the component deviating accelerations are /., = o, f ay = o, /, = o, or the plane of rotation does not change in direction. (4) A vertical c irclt of radius r = 2 ft. rolls on a horizontal straight line. The centre moves parallel to that line with a velocity of 6 ft. per sec. Find the angular velocity and the ^elocity and central acceleration at any point. ANS. (Compare example (2), page 146.) Take co-ordinate axes as shown in the figure. Then r = 2 ft. and the components of motion are = + 6 ft. per sec., v y = o, = o, oo y = These are the same equations as in example (4), page 149. From equations (3), page 159, we have the component deflecting accelerations (4) (5) (6) and from equations (4), page 159, we have the component deviating accelerations fa, = O, f ay =. ZGO y W t , f at = O. The plane of rotation therefore changes in direction. CHAP. II.] EULER'S GEOMETRIC EQUATIONS. 167 Euler's Geometric Equations. Let OX, OY, OZ be rectangular co-ordinate axes fixed in the body and therefore rotating with it, and let the body rotate about some axisjfcm/ in the body and therefore making invariable angles with these axes, so that the component angular velocities are o? r , co v , GJ Z . Let OX V , OY lt OZ l be rectangular co-ordinate axes whose directions in space are invariable. For instance, the axis OZ l may be always directed towards the north pole or parallel to the earth's axis, then X l Y 1 is the plane of the celestial equator. Let the point O be taken as the centre of a sphere of radius r. Let X l , Y l , Z iy X, Y, Z be the points in which this sphere is pierced by the fixed and moving axes. Let the axes OX, OY, 6>Z have the initial positions OX lt OY lt OZ r Turn the body, 1st, about OZ^ through the angle X^Z^P = $, so that OX l moves to OP, and OY l to ON. 2d, about ON through the angle PNE = 6, so that OP moves to OE, and OZ^ to OZ. 3d, about OZ through the angle EOX = 0, so that OE moves to OX, and ON to OY. It is required to find the geometric relations between 0, 0, i/> and <& x , oo y , oo z . These geometric relations are called EULER'S GEOMETRIC EQUATIONS. The line O N is called the line of nodes y 0js the obliquity and $ the precession. The angular velocity of Z perpendicular to the plane of ZOZ^ or about OZ^ at any d>b instant is -=-. This is called the angular velocity of precession. The angular velocity of Z 7 f\ along ZZ^ or about ON is at the same instant -j-. This is called the angular velocity of nutation. The angular velocity of X relative to E, or Y relative to N, at the same instant d Draw ZD perpendicular to OZ r Then ZD = r sin 0, and the linear velocity at any d& instant of Z perpendicular to the plane of ZOZ^ is r sin . j-, and along ZZ^ at the same instant it is r -r The linear velocity at the same instant of Z along YZ is roa x , and along ZX it is roo . We have, then, directly from the figure, dB r r- roo v cos 4- rco, sin 0, dt r sin ti . j- = roOy sin d> roo^ cos 0, or, since r cancels out, de . , = co y cos -f- K) X sin y sin oo x cos 0. (0 1 68 KINEMATICS OF A RIGID BODY. Combining these two equations, we have oo, = -77 . sin -7- sin cos 0, a/ at [CHAP. II. (2) - , . cos + - sin sin 0. We have also the linear velocity of E perpendicular to the plane of Z^OE, equal to dtp dd> r cos '-ft, and of X relative to E along EX, r-, f . We have, then, for the velocity of X along XY dtp QS e ^d r( * ~ T dt ~* dt ' or (3) Equations (2) and (3) are Euler's Geometric Equations. They give the relations between w x , co y , co t and the angular velocity -,- of OZ about OZ' ', -, of OZ about ON, d(f> the line of nodes, ,-.- of O Y relative to ON. at Auxiliary Angles. From the spherical angles of the figure, page 167, considering A 7 as a vertex in each, we have for the direction cosines of the moving axes, with reference to the fixed, , cos XOX l sin if} sin -{- cos ?/' cos cos ft, cos YOX l = sin ip cos cos /? sin cos 0, cos ZOX^ = sin 6 cos ?/;, cos XOY l = cos if; sin -|- sin cos cos ft, cos F<9K, cos ?/ cos sin ^ sin cos 0, cos Z0K, = sin H sin ^-, cos XOZ^ = sin cos 0, cos YOZ X = sin sin 0, cos Z<9Z, = cos 0. For the angles which the axes Z, , Z and ON make with the axes X, Y and Z we have cos ZflX = cos sin 0, cos Z,(9K= sin sin 0, cos Z,(9Z = cos 0, cos ZOX = o, cos Z6>F ~ o, j. /-\ cos ZOZ= i, cos A^A' = sin 0, cos AY? Y cos 0, cos NOZ = o. DYNAMICS. GENERAL PRINCIPLES. CHAPTER I. FORCE. NEWTON'S LAWS OF MOTION. Dynamics. We have given in the preceding pages the principles of KINEMATICS, or the measurable relations of space and time only, that is of pure motion. But we have to deal in nature with material bodies and force. That science which treats of the measurable relations of force and of those measurable relations of matter, space and time involved in the study of the change of motion of bodies due to force, is called DYNAMICS (Svva^is, force]. Material Particle. We have already seerf(page 20) that, whatever the constitution of matter may be, we can consider a' body as composed of an indefinitely large number of in- definitely small PARTICLES, so small that each may be treated as a point. We denote the mass of such a particle by in, and the sum of the masses of all the par- ticles of a body or the entire mass of a body by in, so that m = 2m. Impressed Force. It is a fact of universal experience that no particle of matter is able of itself to change its own motion. If, then, a particle is at rest it must remain at rest unless acted upon from without. If it is moving at any instant in a given direction with a given speed, it cannot change either its speed or direction, that is, its velocity is uniform, unless acted upon from without. This action from without to which the change of velocity in any case can always be attributed we call IMPRESSED FORCE. Newton's First Law Of Motion. A body is a collection of particles. If there are no impressed forces, each particle must then be at rest or move with uniform speed in a straight line, and hence the body itself has motion of translation in a straight line. This fact was expressed by Newton as follows : Every body continues in its state of rest or of uniform motion in a straight line, except in sj far as it may be compelled to change that state by impressed forces. This is known as " Newton's first law of motion." It implicitly defines force as that which causes change of motion of matter. Inertia. We may also express this law by saying that all matter is inert, that is, has no power of itself to change its state of rest or motion. This property of matter we call INERTIA, and Newton's first law we may call the " law of inertia." We recognize, then, not only extension and impenetrability, but also inertia as essential properties of matter. That is, matter occupies space, two bodies cannot occupy the same space at the same time, and all matter is inert. 169 170 DYNAMICS. GENERAL PRINCIPLES. [CHAP. I. Force proportional to Acceleration. Acceleration / of a point we have already illus- trated and defined (page 76) as time-rate of change of velocity when the interval of time is indefinitely small. We can only measure force by its effects. These effects are apparently different in different cases, but in all cases when analyzed they are found to consist either in change of velocity of particles or changes of form or volume of a body. As change of form or volume of a body is due to change of relative position and therefore change of velocity of the particles, we see that in all cases the effect of force is to cause acceleration. The impressed force on a particle must then be proportional to the acceleration / of tjie particle, and have the same direction. Force proportional to Mass. Consider a body composed of a number N of particles, and let the acceleration of each particle be /and in the same direction, so that the body has motion of translation. Then the force on each particle is proportional to /and the entire force on the body is proportional to N f and in the direction of f. But the number N of particles is proportional to the entire mass m of the body (page 20). The force on the body is then proportional to the mass as well as the acceleration. We have, then, for a particle of mass m or 'a translating body of mass m the impressed force F in the direction of /and given by F=cmf. (i) where c is a constant. Unit of Force. Equation (i) expresses the fact that force is proportional both to mass and to acceleration. We see from (i) that we shall always have >=m/, (2) if we take c = equal to unity and [/J] = [,] X [/]. That is, equation (2) holds, provided we take as our unit of force that uniform force which will gii-e one unit of mass one unit of acceleration in the direction of the force. This is called "Gauss's absolute unit" or the "absolute unit of force," because it furnishes a standard force in any system, independent of the force of gravity at different localities. In the foot-pound-second or " F. P. S." system, then, the absolute unit of force is that uniform force which will give a mass of one Ib. a change of velocity in the direction of the force of one ft.-per-sec. in a second. This has been called by Prof. James Thompson the POUNDAL. It is, then, the English absolute unit of force. The French absolute unit of force is that uniform force which will give one kilogram a change of velocity in the direction of the force of one metre-per-second per second. In the centimeter-gram-second or " C. G. S." system the absolute unit of force is the uniform force which will give one gram a change of velocity in the direction of the force of one centimeter-per-sec. per sec. This is called the DYNE. Weight of a Body. The weight of any body is the force with which the eartli attracts it. This force must vary, then, with the acceleration g due to gravity, and this, as we have seen (page 100), varies with the locality. The weight of a body, then, varies with the locality, while its mass of course remains invariable. If, then, m is the mass of a body and IV its weight, we have from (2), for the weight in absolute units, W '= tt absolute units. CHAP. I.] GRAVITATION UNIT OF FORCE. 171 But if m is taken as one unit of mass, then W\s numerically equal to g, or ONE LB. WEIGHS g POUNDALS, ONE GRAM WEIGHS g DYNES, according to the system we use. Since g (page 100) is about 32 ft.-per-sec. per sec., the average weight of one Ib. is about 32 poundals, or ONE POUNDAL IS THE WEIGHT OF ABOUT HALF AN OUNCE. Strictly speaking, it is the weight of th part of a Ib., where g must be taken for the C> ^ locality. Example. An athlete throwing a hammer of 100 Ibs. at New Haven and at Edinburgh throws a heavier weight at the latter place, by about 4 poundals, or the weight of 2 ounces more. The mass is the same and would " weigh " the same on an equal armed balance in both places. But a spring-balance would show a greater strtech at Edinburgh. Gravitation Unit of Force. We have seen (page 170) that the absolute unit of force is that force which will give one unit of mass one unit of acceleration. -In the F. P. S. system this is the poundal; in the C. G. S. system the dyne. It is the unit used in physical measure- ments and generally when small quantities are of importance and great accuracy is desired. In ordinary mechanical problems this unit is inconveniently small. Also very great accuracy is not a requisite. It is therefore usual in mechanics to express a force at any locality by comparing it with the weight of the unit of mass at that locality. The weight of the unit of mass at tlie locality is then taken as the unit of force. Such a unit is evidently not constant, but varies with the locality. The variation is so small, however, that it can be disregarded in ordinary mechanical problems. Thus if the mass of a translating body is m Ibs. and the acceleration of the centre of mass is/", we have for the force, if /"is taken in ft.-per-sec. per sec., F = m/" poundals. But since at any locality where the acceleration of gravity is g the weight of one Ib. is g poundals, if we take this as the unit of force we have m/ F = pounds, , or / = ^ = ^ = g ft,per-sec. per sec. i 7 8 DYNAMICS. GENERAL PRINCIPLES. [CHAP. II. Or we may obtain the same result as follows : The moving force is the weight of P or thejittraction of gravity for P, or Pg poundals, or the weight of 28 Ibs., as in Ex. 1 1. The mass moved is P -f m. Hence (P + m)/ = Pg, or / = -p^jg We have for uniform acceleration (page 192) , = !/(. or 5 = 2 - The tension on the rope is m/ or P(g f) or t or / = 5.051 sec. Pm ^J^= - 224 x - 3 - poundals, or the weight of . = == = 27!? lbs. P + m 81 /*+ (13) Two masses P = 2 '4 lbs. and Q = 2212 lbs. are hung by means of a perfectly flexible inextensible robe < or / = -^p-^r = -JTfjj ft.-per-sec. per sec. Or we may obtain the same result as follows : The weight of P is Pg poundals. The weight of Q is Qg poundals. The moving force is Pg Qg or (P Q)g poundals, or the weight of 28 lbs., as in Ex. u and 12. The mass moved is P + Q. Hence Since s = -/f, we have 5 = r x or/ = 7 ' 04 sec - The tension on the rope is Q(g + /) or P(g f) or -jr~Q poundals, or the weight of -p~rn = 222 5-9 lbs - NOTE. The moving force in Ex. 11, 12, 13 is the weight of 28 lbs. In Ex. 11 the mass moved is m 2240 lbs., hence 28a the body is evidently not affected. Hence the single force O'a' = F can be replaced by the same force, Oa F, acting at any given point O, and a force couple O'a' and Ob. Let p be the arm of the couple. Then the moment of the couple is M Fp, and its line repre- sentative is OM at right angles to the plane of the couple. When we speak of the direction of a moment or force couple we always mean the direction of its line representative. Hence (compare page 151) we have the following principles: (a) A single force /"acting at any point of a rigid body can be replaced by an equal and parallel force .F acting at any given point O and a couple whose moment is M = Fp and whose line representative through O is at right angles to the plane of the couple. CHAP. IV.] DYNAMICS. GENERAL PRINCIPLES. 187 (#) Conversely, the resultant of a couple M and a force F in the plane of the couple is M a single equal and parallel force in that plane at a distance /=-.. (c] If any number of forces act upon a rigid body, each acting at a different point and in a different direction, then by (a) we can replace each one by an equal and parallel force acting at any point O, and a couple whose line representative through O is at right angles to the force. We can reduce all the forces at O by the polygon of forces (page 176) to a single resultant force jpat O, and all the couples at O by the polygon of moments (page- 88) to a single resultant couple M whose line representative is not necessarily at right angles to F. Hence, generally, Any system of forces acting on a rigid body can be reduced at any point O to a single resultant force F and a single resultant couple M ivhose line representative is not necessarily at riglit angles to F. CHAPTER V. CENTRE OF PARALLEL FORCES. CENTRE OF MASS. Centre of Parallel Forces. Let F lt F t , F 3 , etc., be any number of parallel forces acting at the points A lt A 2 , A 3 , etc., of a rigid body and given by the co-ordinates (x l , y v , ^), (^ 2 , j/ 2 , ~ 2 ), etc., and making the angles a, /?, y with the co- ordinate axes. Then the resultant F is parallel to the forces and equal to their algebraic sum, or F=F l + F t + F 9 +... = 2F. . . (i) In taking the algebraic sum in (i), forces acting in one direction are taken as positive, in the other di- rection as negative. If 2F is not zero, we have a single resultant. It remains to find its position. For the moments M* , M y , M z of the forces about the axes of X, Y, Z we have M x =. 2F cos y . y ^F cos /? . z = cos y2Fy cos fi2Fz, M y = 2F cos a . z 2F cos y . x = cos a2Fz cos y2Fx, ... (2) M, = 2F cos ft , x 2F cos a . y = cos fi^Fx cos a2Fy. j Let the point of application O' of the resultant be given by the co-ordinates ~x, y, ~z. If we transfer the origin O to this point, and take co-ordinate axes X', V, Z' at this point parallel to X, Y, Z, we can replace x, y, z in equations (2) by x Ic, y ~y, z 2, and we have for the moments of the forces about the axes X' t Y', Z' through the point of applica- tion O' of the resultant M x ' = cos y2F(y - y) - cos ftZF(z - 2), j M,' = cos a2F(s -z)- cos y2F(x - *), j. ...... (3) M.' = cos ft2F(x - ) - cos a2F(y - y). j But the moment of the resultant is equal to the algebraic sum of the moments of the components, and for the point of application of the resultant its moment is zero. Hence x, y, z must have such values as to make equations (3) zero. We have then 2F(x -*)= = o, or x = =o, or = o, or z - . (4) 1 88 CHAP. V.] PROPERTIES OF CENTRE OF MASS. 189 Equations (4), then, give the position of the point of application for the resultant of a system of parallel forces for any assumed origin and co-ordinate axes. This point is called the CENTRE OF PARALLEL FORCES. We see that its position is independent of the direction of the forces and depends only upon the forces and their points of application. Properties of Centre of Mass. When a body has motion of translation all particles of the body move in parallel paths with the same velocity, in the same direction, at the same instant. The acceleration of every particle is therefore the same and in the same direction at any instant. Let m l , m 2 , m 3 , etc. , be the masses of par- ticles of a translating body, each one therefore having the same acceleration / in the same direction. The forces on the particles are m l f, m zf> m 3/> etc - an d these forces constitute a system of parallel forces. Let m = 2m be the mass of the entire body, and F the resultant force. Then, by the preceding article, we have F=m/ F = m,f + m t f + mj + . . . = f2m = m/, . and for the point of application O' of this resultant we have the co-ordinates f2mz (0 _ ~~ or, since /is constant, _ m/ 2 my z = ~ (2) But we have seen, page 21, that these values of x , y, ~z give the position of the centre of mass of the body. Hence the center of mass of a body coincides with the point of application of the resultant of that system of parallel forces which acts upon all the particles of the translating body. If m is the mass of the body and f the common acceleration, the resultant F is given by F = in/. Conversely, if a force F act at the centre of mass of a body of mass m, it will cause in every particle the same acceleration f, in the same direction, given by f = g That is, it will cause acceleration of translation only. We have seen, page 188, that any number of forces acting in any directions on a body can be reduced to a force .F acting at any point and a couple. Take this point as the centre of mass. Then we have a force F acting at the centre of mass and a couple. But we have seen, page 187, that the effect of a couple is to cause angular acceleration about an axis through the centre of mass. It does not, then, affect the motion of the centre of mass itself. We have also just seen that the effect of the force F acting at the centre of mass is to cause F in every particle an acceleration /in the same direction given by/= -^.- Since, then, the acceleration of the centre of mass is not affected by the couple, we have for the acceleration 190 DYNAMICS. GENERAL PRINCIPLES. [CHAP. V. _ p f of the centre of mass for any number of forces in any directions, =/ = . Hence F m/, where (page 188) F is the resultant of all the forces acting upon the body, each one con- sidered as transferred to the centre of mass, without change in magnitude or direction. Hence the acceleration f of the centre of mass of a body is the same as if all the forces acting on the body were applied to the entire mass m concentrated there. It is this property which makes the centre of mass of such importance in Mechanics. So far as the motion of the centre of mass of a body is concerned, we can always consider it as a particle of mass equal to the mass of the body and acted upon by all the forces which act upon the body, unchanged in magnitude and direction. We have also the following properties of the centre of mass : 1. The attraction of the earth for a body whose longest dimension is insignificant com- pared to the earth's radius is practically a parallel force mg on every particle of the body of mass m. The entire weight of the body, mg. acts, in such case, practically at the centre of mass for all positions of the body, and a body acted upon by its weight only has motion of translation. Hence the centre of mass is often called erroneously the " centre of gravity " (page 207). 2. If a rigid body at rest is supported at its centre of mass and is acted upon by gravity only, it will remain at rest in all positions. CHAPTER VI. NON-CONCURRING FORCES IN GENERAL. ANALYTIC EQUATIONS. Resultant for Non-Concurring Forces. We have seen (page 188) that, generally, any system of forces acting on a rigid body can be reduced at any point O to a single resultant force F and a single resultant couple M, whose line representative is not necessarily at right angles to F. The force F does not change in magnitude or direction no matter where the point O is taken. ' The couple M, however, changes in magnitude and direction with the position of the point O, The line representative of the couple M at the centre of mass is the spontaneous axis of angular acceleration (page 152). Since M is not necessarily at right angles to F, we can resolve it at any point O into a component M a along F and a component M n normal to F. But by (fr), page 188, we can reduce F and M n to an equal and parallel force at a distance p ^. Since the couple M a r at O can be replaced by the same couple along F (page 186), we have a force .Fand a couple M a whose line representative coincides with F. Hence, generally, Any system vf forces acting on a rigid body can be reduced at any point O to a resultant force F, a couple M a along F and a couple M n normal to F. Or to a force F at a distance M p = -=? and a couple M a along F at this distance. Again, since Mis not necessarily at right angles to F, we can resolve F at any point O into a component F a along M at that point and a component F n normal to M at that point. But by (&), page 188, we can reduce F n and M to an equal and parallel force F n at a distance M p = -p. Hence, generally, Any system of forces acting on a rigid body can be reduced at any point O to a couple M, a force F a along M and a force F n normal to M. Or to a force F a at O and a normal force F n at a distance p = . ft As M changes in magnitude and direction with the point O, F a and F M will change also. But the resultant F is unchanged in magnitude and direction no matter where O is taken. 'The line representative of M at the centre of mass is the spontaneous axis of angular acceleration, and the instantaneous axis is parallel to it (page 150). Effect of any System of Forces acting on a Rigid Body. Since we can take the point O where we please, let us always take it at the centre of mass, We have seen (page 190) that a force acting at the centre of mass of a rigid body causes acceleration of translation /of the body in the direction of the force. Also (page 187) that 191 I9 2 DYN/1MICS. GENERAL PRINCIPLES. [CHAP VI. a force-couple acting on a rigid body causes angular acceleration a of the body about the line representative through the centre of mass as axis. Therefore the force F m along M at the centre of mass causes acceleration of translation f a of the body along the spontaneous axis of angular acceleration. The normal force F n at the centre of mass causes acceleration of translation / of this axis. The couple M at the centre of mass causes angular acceleration at about the spontaneous axis of accelera- tion. But we have seen (page 151) that f a ,f and a reduce to f a and a about the instan- taneous axis of acceleration, or to a screw-twist. Hence, generally, The effect of any system of forces acting on a rigid body at any instant is to cause a screw- twist ', or angular acceleration about tlie instantaneous axis and acceleration of translation along this axis. Wrench. Screw-Wrench. A force-couple acting on a rigid body we call a wrench. It causes angular acceleration about its line representative through the centre of mass (page 187). The line representative is the axis of the wrench and may be taken anywhere parallel to itself (page 186). A force-couple or wrench, and a force parallel to the axis of the wrench, we call a scr civ -wrench. The preceding principles may then be expressed as follows: Any system of forces acting on a rigid body reduces in general to a resultant force Fat the centre of mass, and a resultant couple or wrench M with axis at the centre of mass. Any system of forces acting on a rigid body reduces in general to a screw-wrench M a about an axis coinciding with F at a distance from the centre of mass given by /> = -^, where M n is the component of M normal to F. Any system of forces acting on a rigid body reduces in general to a screw-wrench M with axis at the centre of mass, a force F a coinciding with this axis and a force F n normal to it. Or to a force F a at the centre of mass, and a normal force F n at a distance from the M. centre of mass given by/> = ~ ^ ' H The effect of any system of forces acting on a rigid body is, generally, to cause a screw- twist about the instantaneous axis of acceleration. This axis is parallel to the spontaneous axis of acceleration, that is to M at the centre of mass. Dynamic Components of Motion. We have just seen (page 191) that any number of forces acting upon a rigid body reduces to a single resultant force F at the centre of mass and a resultant couple M at the centre of mass. Let us take co-ordinate axes through the centre of mass O, and let the components of Fbe F Mt F r , F x , and the components of M be M x , M yt M . The motion of the body under the action of forces is then known if these six quantities are known. These six quantities F mt F y , F,, M x , M,, M,, with reference to 'co-ordinate axes through the centre of mass, are therefore called the dynamic COMPONENTS OF MOTION of the body. CHAP. VI.] GENERAL ANALYTIC EQUATIONS. 193 General Analytic Equations. Let any number of forces F lt F z , F^ act at points A 19 A z , A 3 of a rigid body given by the co-ordinates^, y^ ^), 0*2 yv "2)' e ^ c * -^ e ^ ^i ma ke with the co-ordinate axes the angles (or p /? x , y^, F 2 the angles (a v >5 2 , y^, etc. Take the origin O at the centre of mass. We can replace each force by an equal parallel force in the same direction at O and a couple with axis through (page 187). All the forces can then be reduced to a single resultant F, and all the couples to a single couple whose moment is M. RESULTANT FORCE AT CENTRE OF MASS. We have for the algebraic sum of all the components along the co- ordinate axes through the centre of mass O F x = F l cos a l -J- F 2 cos ar 2 -|- . . . = 2Fcos a, F y = F l cos A + ^2 cos 2 + - =2Fcos/3, F z = F l cos y l 4- F 2 cos y 2 -|- . . . = 2Fcos y. ^ The resultant force is then (2) The line representative passes through the centre of mass O, and its direction cosines are F. F x F y COS Of = t COS p -ff- , (3) These equations give in any case the components of motion F x , F y , F tt and the magnitude and direction of the resultant force .pof the screw-wrench. They hold good, evidently, no matter where the origin is taken, whether at the centre of mass or not. RESULTANT COUPLE OR WRENCH AT CENTRE OF MASS. For the moments about the co-ordinate axes through the centre of mass we have M x = 2F cos y .y 2F cos /? . z, My = 2F cos a . z 2F cos y . x, M, = 2F cos /3 . x 2F cos a . y. The moment of the resultant couple M at the centre of mass is then (4) (5) Its line representative passes through the centre of mass O, and its direction cosines are cos = (6) These equations give in any case the components of motion M x , M , M t , and the magnitude and direction of the resultant moment M at the centre of mass O. If we take the i 9 4 DYNAMICS. GENERAL PRINCIPLES. [HAI'. VI - origin at any other point, the moment M changes in direction and magnitude. These equations hold good, therefore, only for origin at the centre of mass. RESULTANT COUPLE OR WRENCH AT ANY POINT. Instead of taking the origin at the centre of mass O, let us take it at any point O' not at the centre of mass. Take co-ordinate axes O 'X' t O'Y' t O'Z' at the origin O' parallel to OX, OY, OZ at the centre of mass O (see figure, page 193). Let the co-ordinates of the centre of mass O be ~x, y, ~z. Then equations (4) still hold if we put x -\- x, y -(- y, ~z -\- z in place of x, y, s. We have, then, for the moments about the co-ordinate axes at any point O' ' = M, + FJ>- M: = M.+ F~ X - F x y. The moment of the resultant couple M' is then M' = (7) Its line representative passes through the origin O' , and its direction cosines are These equations give the components of motion M x ', M y ' ', M t ' and the magnitude and direction of the resultant moment M' for any point O'. If we take the origin at the centre of mass, they reduce to equations (4), (5) and (6). MOMENT M a AT CENTRE OF MASS ALONG THE RESULTANT FORCE. Let the compo- nent of the moment M at the centre of mass along the line representative of the resultant force F be M a . Then we have M a = M x cos a-\- My cos ft -f M t cos y, where cos a, cos ft, cos y are given by equations (3). We have then The resultant force F is independent of where we take the origin, and the moment J/, at the centre of mass remains the same no matter where we take the origin. Equation (8) therefore holds good no matter where we take the origin, whether at the centre of mass or not. MOMENT M n AT CENTRE OF MASS NORMAL TO THE RESULTANT FORCE. Let the component of the moment M at the centre of mass normal to the line representative of the resultant force F be AT n . The components of M a along the co-ordinate axes through the centre of mass O are M a cos a, M a cos /3, M a cos y, CHAP. VI.] FORCE F a AT CENTRE OF MASS. 195 where cos or, cos/3, cos y are given by equations (3). .If we subtract these components from the components M x , M y , M z of the resultant moment M at the centre of mass, we have the components of the normal moment M at the centre of mass O M nx = M X - M ny =M y - cos a cos M M * F * =. M x --- M nz = M z - M a cos y = M z - (9) FORCE F a AT CENTRE OF MASS ALONG M. Let the component of the force F along M at the centre of mass be F a . Then we have F a = F x cos a. -f- F y cos ft -}- F, cos y, where cos a, cos ft, cos y are given by equations (6). We have then The resultant force F is independent of where we take the origin, and the moment M at the centre of mass remains the same no matter where we take the origin. Equation (10) therefore holds good no matter where we take the origin, whether at the centre of mass or not. FORCE F n AT CENTRE OF MASS NORMAL TO M. Let the component of the force F normal to the axis of M at the centre of mass be F H . The components of F a along the co-ordinate axes through the centre of mass O are F a cos or, F a cos F a cos y, where cos a, cos ft, cos y are given by equations (6). If we subtract these components from the components F x , F y , F z of the force F, we have the components of F n i , = F , - F a cos y F z - M (n) POSITION OF RESULTANT FORCE F OF THE SCREW-WRENCH M a . We have from (8) the moment of the screw-wrench M a , and from (2) the force /''along its axis, and from (3) the direction of the axis. It remains only to find the position of F relative to the centre of mass. The moment M n at the centre of mass is equal to Fp, where/ is the distance of F from the centre of mass O. Let x, y, z be the co-ordinates of any point on the line representative of F. Then we have F y z = M nx , Fx = M n F x y =M nz . (12) 196 DYNAMICS. GENERAL PRINCIPLES. [CHAP. VI. These are the equations of the projection of F on the three co-ordinate planes. Let .r,, y v s l be the intercepts on the co-ordinate axes of these projections. Then we have from equations (12), making^ = o and s = o in the last two, _M_ Hy _M^ ~ ~ "" making x o and z = o in the first and last, x making^ = o and x = o in the first two, 03) ^_^ F. F. Let the distance of the force F from the centre of mass O be /, and let p x , p y , p t be its projections on the co-ordinate axes. Let the intersection of/ with F be O', so that OO' = /. Let us consider the projection of F on the plane KZ. In the figure A = Ob and p y = Oc . We have then where , and j/, are the intercepts Oa and Od given by equations (13). We have also the distance ab =^. Hence Substituting the value of p y , we obtain Substituting the values of ^r, and _y, from equations (13), we have In the same way we can find p m and /, , /, and p y on the other two co-ordinate planes. We thus have * "- /7_i_ /ra ' A = ~ *f i * A = - . (14) CHAP. VI.] COMPONENTS OF MOTION. 197 The screw-wrench M a is thus completely determined if the components of motion F x , F y , F z , M x , My , M z are given. From (2) and (3) we have the magnitude and direction of the resultant force F t and from (14) its position. From (8) we have the moment M a . COMPONENTS OF MOTION. Suppose, on the other hand, the screw-wrench M a is given, that is, we have M a and F and the position and direction of /''given. From equations (12) and (3) we have M x M a cos a F(p s cos /3 / cos y}, F x = F cos a; My = M a cos /3 F( p x cos y p z cos a), F y F cos fi ; M z = M a cos Y F(p y cos a y x cos /?), F 2 = F cos y. If, then, the screw-wrench M a is given, we have from (15) the components of motion. SCREW-WRENCH M. The screw-wrench M is already completely determined. Its axis passes through the centre of mass, and its direction is given by equations (6), its moment by (5), and the force F a along its axis by (10). POSITION OF THE FORCE F n . We have seen (page 193) that the force system reduces to a force F a along M at the centre of mass, or the spontaneous axis of angular acceleration, M and a force F* normal to this axis, at a distance p . For any point on the line representative of Fn given by the co-ordinates x, y, z we have then F nz y - F ny z = M x , F nx z - F nz x = M y , F ny x - F nx y = M z . These are the equations of the projections of F M on the these co-ordinate planes. Let x v , y^ z v be the intercepts on the co-ordinate axes of these projections. Then we have, just as on page 196, Proceeding, then, just as on page 196, we have for the co-ordinates/,, p y , p* giving the position of F n F nz M y F ny M z P* 7T2 i c*2 rrz i 772 WT <* ^*-r^ ".-r^ , ^i ^>, ^ 6) Equations (16) give the position of F n . The values of M x , M y , M z are given by equations (4) ; the values of M and F a by equations (5) and (10). THE INVARIANT. From (8) and(io) we have FM a = F / 3 = + 3 " 23 = 4 4 " /% = 1 5 " -r 4 = + 3 " y* = + 4 " z* = + 5 " /^ 6 = + 200 " xs, = + 4 " y& + 5 " 2-6 =+ 6 " ANS. F = + 90 Ibs., p x = + 2f ft., A = + 3 ft., />* = 4 3i ft. If the forces are parallel to the axis of Y, we have M x = + 31 5 lb.-ft., M z = + 240 lb.-ft., M r = 396 Ib.-ft. The line representative making the angles with the axes of X, Y, Z given by +H, or a = 322 41' 41", /S = 90, r = 5 2 4i' 4i'. If we look along the line representative towards the origin, the rotation is seen counter-clockwise. (3) Let a rigid body be acted upon by the coplanar forces F! = 50 Ibs., F* = 30 Ibs., F 3 = 7o Ibs., /\, = 90 Ibs., F s = 120 Ibs. icting at the points given by Xl = 4- 5 ft., j, = + 10 ft. ; x* = + 9 ft., ^ 2 =-fi2 ft.; jr. = 4- 17 ft., j 3 = 4- 14 ft. ; :r = 4- 20 ft., j 4 = 4 13 ft. ; jr 6 = + 15 ft.,j 6 = 4 8 ft. Let the forces make angles with the axes of X and Y given by i = 70, /?! = 20 ; a, = 60, ft* = 1 50 ; a 3 = 120, /&, =30 ; 4 = 1 50, /? 4 = 120 ; 6 = 90, /3 5 = o. Find the resultant, etc. ANS. We have for the components parallel to the axes X and Y: F x = 50 cos 70 4 30 cos 60 70 cos 60 90 cos 30 = 80. 842 Ibs. ; F y = 50 cos 20 30 cos 30 4 1 20 4 70 cos 30 90 cos 60 = 4 1 56.626 Ibs. ; F z = o. The resultant is given in magnitude by F = \/F* x + ^ = 176.259 Ibs., and its direction cosines by cos a = , F 176.259 cos ft = $ = + I f- 626 . or ft = 27 18' i. F 176.259 We have from equation (4), page 193, SFx cos ft = 4 50 cos 20 x 5 30 cos 30 x 9 4 70 cos 30 x 17 90 cos 60 x 20 4 120 x 15 = 4 1931.670 lb.-ft. ; 2Fycos a = 4 50 cos 70 x 10 4 30 cos 60 x 12 70 cos 60 x 14 90 cos 30 x 13 = 1152.245 lb.-ft. M* = o, My = o, M, = 2F.v cos ft 2Fy cos a= 4 3083.915 lb.-ft. Since, then, equation (18), page 198, F X M X 4 FyMy 4 F,M, = o, is satisfied, the forces reduce to a single resultant force. The moment of this resultant force relative to the origin is M = ^ Ml + My + Ml = M z = + 3083.915 lb.-ft. 200 DYNAMICS, GENERAL PRINCIPLES. [ C " AP - V1 - Its lever-arm is M == ft< f /.' - 176.259 The equation of the line of direction of the resultant is y - FjL x _ ^ = - I.95J- + 38.14. The co-ordinates of the point of application of the resultant are given from equation (10), page 195 : (4) Find the resultant, etc., for the force system acting on a rigid body given by F t = $o IDS., a, = 60, ft t = 40, ^i acute ; F, = 70 " a t = 65, / = 45. ^ 3 obtuse ; F = 90 " a = 70, ft 3 = 50, y 3 acute ; Ft = 120 " = 75. /* = 55. y< obtuse. Xi = 0, J'l = O, ^1 = 0; JCt = + I ft., j, = + 4 ft., zt = + 7 ft. ; x. = + 2 / = + 5 " *3 = + 8 " -r. = + 3 " /4 = + 6 " z* = + 9 " ANS. We find the angles y b V the formula, page 13. cos 3 Y cos (a +#) cos (a ft). Then, from page 193, we have F, = + 116.423105., ^= + 214.480^5., F, = -51.057 Ibs. Therefore the resultant is F = ^Fl + Fl + F1= + 249.325 Ibs., and its direction cosineS are given by C osa = ^, cos ft = -p-, co*Y = ~, or = 62 9' 48", ft = 30 39' 20", Y = 101 49'. We also have for the moments from equation (4), page 193, M x = - 1838.604, My = +928.947, M, = - 86.903 Ib.-ft. The resultant moment about the origin is M= 4/yJ/| + M\ + Ml = + 2061.789 Ib.-ft., and the direction cosines of its line representative are given by M~ M 9 M* or a = 153 5' 40", ft = 63 14' 15", Y 92 24' 56". Looking along this line representative towards the origin, the direction of rotation is seen y construction. These triangles are therefore similar and we have AB _ CA AP CP. 20 4 DYNAMICS. GENERAL PRINCIPLES. L CHAP - VII Now let As represent any small elementary area of the spherical surface, and An its projection normal to AB. Let oo square radians (page 7) denote the conical angle subtended at B by An. Then Al? . co the area denoted by An is equal to Alf . co, and the area denoted by As is equal to CQS ^ , since the angle nAs = BAC=B t and the angle snA is a right angle. The mass of the elementary area denoted by As is then ' , and the attraction of this mass for the particle of mass m at P is, by Newton's law, m. dA?. GO and acts in the line AP. If we draw.4/4' perpendicular to CP, we have evidently the same attraction between the equal elementary mass at A' and the particle of mass m at P acting in the line A' P. We can resolve each of these equal forces into a component along the line CP and at right angles to CPat P. Since the angles A PC and A' PC are each equal to B, the two com- ponents at right angles to CP at Pare equal and opposite and therefore produce no effect upon P. The resultant attraction of the two elements at A and A' upon the particle of mass m at P acts then in the line CP and is equal to m . 8AJ?. GO , or, since -r- n -^ tne resultant attraction is At C.J in . fiCA . But CA . co is the area of the elementary area at A or A', and 2k ~^z is constant for all pairs of elements A and A' . The total attraction of the shell for the particle of mass m at Pacts, then, in the line CP and is equal to where the summation is to be taken for an entire hemisphere. But 2CA . co for a hemi- sphere is 2nCA*, and hence the attraction is equal to 4rr6CA z . m w/m where m = 4*6 C A is the total mass of the spherical shell. We see, then, that the spherical shell attracts a particle of mass m at any outside point P, just as if its entire mass were condensed at the centre of the shell. If instead of a homogeneous spherical shell we have a solid homogeneous sphere, we may consider it as composed of an indefinite number of concentric homogeneous spherical shells, each of which attracts the mass at Pas if its entire mass were condensed at its centre. CHAP. VII. ] ATTRACTION OF A HOMOGENEOUS SHELL OR SPHERE. 205 Hence the attraction of a homogeneous spherical shell or of a homogeneous sphere upon a particle at any outside point is the same as if the entire mass of the shell or sphere were con - densed in a point at the centre. We can therefore consider a homogeneous shell or sphere as a particle of equal mass at the centre, so far as its attraction upon an outside particle is concerned. COR. If the sphere is not homogeneous, but the density of every point at the same distance from the centre is the same, we may still consider the sphere as composed of homo- geneous spherical concentric shells, each one of which attracts an outside mass as if its entire mass were condensed at the centre. Hence the same holds true for the sphere. Value of Constant of Gravitation. We have seen that Newton's law is expressed by where the constant k must be determined by experiment. This determination we are now able to make. Thus for a body of mass m^ at any point on the earth's surface where the acceleration is g, we know that the force of gravity is F = m l g poundals. Let m Q be the mass of the earth, and let r Q be the radius of the earth at the place of experiment. If we consider the earth as a sphere whose density is either constant or the same at all points at equal distances from the centre, then, as we have just seen, we may consider it as a particle of equal mass at the centre of mass so far as its attraction upon any outside particle is concerned, and the centre of mass is the centre of figure. The earth is not strictly spherical, but its deviation from sphericity is very slight. Also, the density is not constant nor strictly the same at points equidistant from the centre. But the small distance between the centre of mass and that point at which in any case of attraction we may consider its mass condensed is insignificant compared to its radius. So far as its attraction for any outside particle is concerned, we may then still consider it as a particle of equal mass at the centre of mass, and the centre of mass as the centre of figure. Also, since the dimensions of any body with which we experiment at the earth's surface are insignificant compared to the radius, we may consider any such body as a particle. Hence in equation (i) we can take r as the radius of the earth r Q , and m 2 as the mass ; of the earth, and we then have Hence we have *=* ........... (*) "*0 Substituting this value of /fin (i), we have Equation (3) gives the force of attraction F between two masses m l and m z at any dis- tance r, in absolute units. Thus if m , m^, m 2 are taken in Ibs., r and r in feet, and g in ft.-per-sec. per sec., equation (3) gives Fin poundals. If we take mass in grams and distance in centimeters, we have F in dynes. zo6 DYNAMICS. GENERAL PRINCIPLES. [CHAP. VII. If we divide by^-, we have in gravitation units f= j y b - r ~ . ; . ( 4 ) If M is the mass of the sun and m the mass of a planet, we have, from (3), for the mutual force of attraction ., Mm r* F= ~^~ -m/' Since force equals mass X acceleration, if we divide this by M, we have for the acelera- tion SO of the sun relative to a fixed point O s :*u If we divide by ;, we have for the acceleration PO of the planet relative to a fixed point O M_ rj_ *2 M S I /'*fl opposite in direction to'SO. We have then for the acceleration PS of the planet relative to the sun PO + OS, or s *~ ~~O P M+ m r_l r> ' m/' At a distance r we have, making r r , the acceleration / of the planet relative to the sun jyi ( m i \ / = . g. (5) m This is the value of f to be used in our equations for planetary motion, page 123. Astronomical Unit of Mass. That mass which attracts an equal mass at unit distance with unit force is called the ASTRONOMICAL UNIT OF MASS. Let [F] denote the unit of force, and [] the unit of length. Then, from equation (3), we have for the astronomical unit of mass [/] r/n M' r " v MZj-.i* Hence we obtain for the astronomical unit of mass [#/] Equation (6) gives by definition the astronomical unit of mass. The mass w of the earth is about I 1920 X io 21 Ibs. The mean radius of the earth r is about 21 X 10* ft. Taking these values and = 32 ft.-per-sec. per sec., we have, from (6), for the astronomical unit of mass [m] = 29063 Ibs. Taking w equal to 6.14 X io 27 grams, r equal to 6.37 X io 8 cm. and^-= 981 cm.-per- sec. per sec., we have \nt] = 3928 grams. CHAP. VII. J ASTRONOMICAL UNIT OF MASS. 207 If we adopt the astronomical unit of mass, we can then write simply the numeric equation m^ ~^' where m l and m. 2 are the number of astronomical units of mass in the two particles, r the number of units of length in the distance between them, and F the number of absolute units of force in the attraction. Centre of Gravity. When a body attracts and is attracted by all external bodies, whatever their distance and position, as though its mass were condensed in a single point fixed relatively to the body, that point is properly called the CENTRE OF GRAVITY (page 23), A body which has a centre of gravity is said to be centrobaric or barycentric. In gen- eral bodies are not centrobaric if the law of attraction follows Newton's law, that is, if the force is inversely as the square of the distance. As we have seen, page 205, a homogeneous spherical shell or a homogeneous sphere is centrobaric and the centre of gravity is at the centre of mass. So also for a non-homo- geneous sphere whose density is the same at all equidistant points from the centre. In general if a body has a centre of gravity, it must always coincide with the centre of mass, because the attraction of an infinitely distant body upon it constitutes a system of parallel particle forces, and the centre for such a system is (see page 190) at the centre of mass. All bodies, then, have a centre of mass, but, as we have seen, only a few bodies have a centre of gravity. Examples. (i) Show that the attraction of a thin spherical shell of uniform thickness and density upon a particle inside is zero. -" ANS. Let P be the particle of mass m\. Take any point A on the spherical surface. Join AP and produce to A'. If from all points of a small element of the surface at A lines be drawn through P, they will mark off a corresponding element at A'. Both these ele- ments subtend the same conical angle (page 7), oo square radians. The area of the element at A is then AP*. much per cent of his former weight would the weight of a man be increased by being removed to the surface of Jupiter? ANS. By 200 per cent. He would weigh by a spring-balance three times as much as before. The same number of standard pounds would, however, balance him in a lever-balance. The standard pound at Jupiter would be attracted by a force three times as great as the earth's attraction here. The lever-balance weight which gives his mass is unchanged. (10) Find the acceleration due to the attraction of the earth at the distance of the moon. Assume g = 3* at the earth's surface, the diameter of the moon's orbit 480,000 miles, the diameter of the earth 8000 miles. ANS. 0.0089 ft.-per-sec. per sec. STATICS. GENERAL PRINCIPLES. CHAPTER I. EQUILIBRIUM OF FORCES. DETERMINATION OF MASS. Statics. That portion of Dynamics which treats of balanced forces, and of bodies at rest under the action of balanced forces, is called STATICS. Statics is thus a special case of Dynamics which it is convenient to consider separately. Equilibrium of Concurring Forces. Any number of forces acting upon a point or par- ticle are said to be concurring forces. If the resultant of any system of concurring forces is zero, the forces are said to be in equilibrium. The effect of the resultant force is to cause acceleration of the particle. When the forces are in equilibrium, then, the particle has no acceleration, and if the particle is at rest it is said to be in static equilibrium. But if the resultant force is zero, all the concurring forces must evidently reduce to two equal and opposite forces, or, what is the same thing, any one of the forces must be equal and opposite to the resultant of all the others. Hence the algebraic sum of the components of all the forces in any three rectangular directions must be zero. We have then, from equations (i), page 194, for the conditions of equilibrium of con- curring forces F x = 2F cos a = o, F y 2F cos ft = o, F, = 2F cos 7 = 0. We obtain, then, the following obvious results from the condition for equilibrium of concurring forces, which will be found useful in special cases : (1) If two concurring forces are in equilibrium, they must be equal in magnitude and opposite in direction. (2) If three concurring forces are in equilibrium, they must all act in the same plane. For the resultant of any two must act in their plane and be equal and opposite to the third. (3) If three concurring forces are represented in magnitude and direction by the sides of a triangle taken the same way round, the resultant is zero and the forces are in equilibrium. (4) Hence, if three concurring forces are in equilibrium, each one is proportional to the sine of the angle between the other two. (5) If three concurring forces are in equilibrium and their directions are represented by the sides of a triangle taken the same way round, their magnitudes will also be represented by the sides of that triangle, and vice versa. 209 210 STATICS. GENERAL PRINCIPLES. \pUf. 1. (6) If any number of concurring coplanar forces are represented in magnitude am! direction by the sides of a plane closed polygon taken the same way round, they are in equilibrium. If their magnitudes are given by the sides of the polygon, their directions are also given by the directions of the sides. Hut if the directions only of the forces are given by the sides of the plane polygon, it does not follow that the sides of this polygon represent the magnitudes, because any number of plane polygons with parallel sides may be drawn, the magnitudes of the sides varying. (7) If three concurring forces in different planes are represented by the three edges of a parallelopipcdon, the diagonal taken the opposite way round will represent the resultant in direction and magnitude. This is called the para llelopipedon of forces. Examples. (i) Find the resultant of forces of 7, /. /, j units, represented by lines drawn from one angle of a regular pentagon towards the other angles taken in order. ANS. f/74 units. (2) P andQ are two component forces at right angles, whose resultant is R. S is the resultant of R and P. If Q. = aP, what is Sf ANS. S = 2/V7. (3) Component forces P, Q. K are represented in direction by the sides of an equilateral triangle taken the same way round. Find the magnitude of the resultant. ANS. v7* + < + A' 8 - QR -PR- PQ. (4) A weight of 10 tons is hanging by a chain 20 feet long. Find how much the tension in the chain is increased by the weight being pulled out by a horizontal force to a distance of 12 feet from the vertical. ANS. By 2.5 tons. (5) A weight of 4 pounds is suspended by a string, and is acted upon by a horizontal force. If in the position of equilibrium the tension of the string is j pounds, what is the horizontal force ? ANS. 3 Ibs. (6) A mass of 10 Ibs. is supported by strings of lengths 3 and 4 feet attached to two points in the ceiling 5 feet apart. W 'hat is the tension of each string ? ANS. 8 Ibs. and 6 Ibs. (7) A particle is actfd on by a force whose magnitude is unknown, but whose direction makes an angle of 60" with the horizon. The horizontal component of the force is 1.33 dynes. Determine the total force and its vertical component. ANS. 2.7 dynes and 2.34 dynes. (8) Three forces proportional to /, 2, j. act on a point. The angle between the first and second is 60", between the second and third jo". Find Hie angle which the resultant makes with the first. ANS. About 67. (9) Three cords are tied together at a point. One is pulled in a northerly direction with a force of 6 pounds, and another in an easterly direction with a force of 8 pounds. With what force must the third be pulled in order to keep the whole at rest f ANS. 10 pounds, at an angle with the horizon whose tang = -. Static Molar Dynamic and Molecular Equilibrium. Forces acting at different points of a body are said to be non-concurring forces. We have seen (page 188) that any number of non-concurring forces acting upon a rigid body can be reduced to a resultant force and a resultant couple at the centre of mass. The effect of the resultant force (page 190) is to cause linear acceleration of translation of the body. The effect of the resultant couple (page 187) is to cause angular acceleration of the body about an axis through the centre of mass. If both resultant force and couple are zero and the body is at rest, it will remain at rest and then it is said to be in static equilibrium. If every pcint has the same velocity of trans- lation, this velocity of translation will b<_ uniform, and the body is then in molecular equi- CHAP. I.] EQUILIBRIUM OF NON-CONCURRING FORCES. 211 librium. If the body has angular velocity about an axis through the centre of mass, this angular velocity is unchanged and the body is said to be in molar equilibrium. If only the resultant force is zero, but not the couple, the centre of mass is either at rest or moves with uniform velocity, but the angular velocity changes. The body is then said to be in dynamic equilibrium. Equilibrium of Non-Concurring Forces When both the resultant force and couple are zero the forces are in equilibrium. In such .case all the forces must evidently reduce to two equal and opposite forces in the same straight line, or what is the same thing, any one of the forces must be equal and opposite to the resultant of all the others, and lie in the same straight line with it. We have, then, for the equilibrium of non-concurring forces two necessary conditions. ist. TJie algebraic sum of the components of all the forces in any three rectangular direc- tions must be zero. From equations (i), page 194, we have, then, F x 2F cos a = o, F y = 2Fcos ft = o, F z 2F cos y = O. . . (l) When these conditions are complied with, there is no resultant force on the centre of mass of the body, and any one force is equal and opposite to the resultant of all the others but does not necessarily lie in the same straight line with it. 2d. The algebraic sum of the component moments in any three rectangular planes must be zero. From equations (4), page 194, we have, then, M x 2Fcos y. y 2F cos ft . z = O, My 2Fcos of . 2 2F cos Y x Mz = 2Fcos ft . x 2F cos a . y = o. ( 2 ) When these conditions are complied with there is no resultant moment at the origin. But it does not necessarily follow that there is no moment at any other point unless the resultant force is also zero. When, then, both equations (i) and (2) are satisfied the forces are in equilibrium, and the body, if at rest, remains at rest and is in static equilibrium. If every point has the same velocity of translation, it is in molecular equilibrium. If the body has angular velocity about an axis through the centre of mass, it is in molar equilibrium. If only equations (i) are satisfied but not (2), the forces reduce to a couple. The body is then in dynamic equilibrium. If only equations (2) are satisfied but not (i), we have a single resultant force passing through the origin. COR. i. If the forces are all co-planar, let XFbe their plane. Then z = o, y = 90, and we have F x = 2F cos <* = o, F y =2Fcos/3 ....... (0 M z = SFcosfi . x SFcosa .y =o ....... (2) That is ist. The algebraic sum of the components of all the forces in any two rectangular direc- tions in the plane of Ihe forces must be zero. 2d. The algebraic sum of the moments of the forces about any point in this plane must be zero. 212 STATICS. GENERAL PRINCIPLES. [CtiAP. 1. If the first condition only is satisfied, we have a couple. If the second only is satisfied, there is a single resultant passing through the origin. COR. 2. If three forces acting at different points of a rigid body are in equilibrium, their lines of direction must intersect in a common point and the forces must be coplcinar. For the resultant of any two must pass through their point of intersection and lie in their plane. The third must be equal and opposite to this resultant and lie in the same straight line. COR. 3. If the forces are parallel, take their common direction parallel to the axis of y. Then cos a = o, cos y = o, cos ft = I , F x = o, F t = o, F y = 2F, and we have (i) (2) 2Fy = o. ) That is: ist. The algebraic sum of the forces must be zero. 2d. The algebraic sum of the component moments of the forces in any two rectangular planes must be zero. If the first condition only is satisfied, we have a couple. If the second condition only is satisfied, the resultant passes through the origin and coincides with the axis of Y. Determination of Mass by the Balance. Let two masses m, and in~ 2 be suspended from the arms of an equal-armed balance, the fulcrum F being in a vertical through the centre of mass O of the balance. Let the balance before the masses nij and m. 2 are applied be horizontal and at rest. Then the balance is in static equilibrium, and its weight W acting at the centre of mass O must be equal and opposite to the upward pressure of the fulcrum and in the same line. Now let the masses nlj and m 2 be suspended, the length of arm AC = BC being /, and suppose the balance still remains horizontal and at rest. It is then still in static equilibruim and the acting forces are in equilibrium. These forces are the weight W of the balance, the weights m^ and m^g of the masses, and the upward reaction R of the fulcrum. We have, then : ist. Algebraic sum of the forces equal to zero, or R W m l g m 2 -= o, or R = 2d. Algebraic sum of the moments about any point zero. Take F as the point. Then m^ . / rx^gl o, or m, = m 2 . Hence for equilibrium the masses must be equal. (See page 173.) Examples. (i) A circular table -weighing m Ibs. has three equal legs at equidistant points on its circum- ference. The table is placed on a level floor. Neglecting the mass of the legs and considering table and floor and legs rigid, find the mass which, placed anywhere on the table, will just bring it to the point of over- turning. ANS. m Ibs. CHAP. I.] EXAMPLES. 213 (2) If the table has four legs at equidistant points on the circumference, find the mass that will just bring it to the point of overturning. ANS. 2.4 m. (3) The centre of mass of a ladder weighing 50 Ibs. is 12 ft. from one end, which is fixed. What force must a man apply at a distance of b ft. from this end to just raise the ladder? ANS. iop Ibs. (4) A horizontal beam of length I is supported at its ends. It is acted upon by vertical downward force: Fi , Fi , F 3 acting at points of application A\ , A* , A-j , dividing the beam into segments b, c, d, e. Find the resultant pressures R\ , R* at the right and left supports, neglecting the weight of the beam. ANS. A*i and Ri act upwards, and are given by A 5 ) = (5) A mass of 6 Ibs. hangs on the arm of a safety-valve at a distance of 18 inches from the fulcrum. The valve-spindle is attached at i inch from the fulcrum. Disregarding friction and the weight of the arm, fend the steam-pressure for static equilibrium. ANS. 108 pounds. (6) In a wheel and axle the radius of the axle is r, and of the wheel R. A mass Q hangs by a rope wouna round the axle. Find the force P acting tangent to the wheel in order to hold Q suspended, disregarding friction. ANS. P = . A (7) A shopkeeper has correct weights but an untrue balance, one arm of which is a and the other b. He serves out to each of two customers, according to his balance, m Ibs. of a commodity, using first one scale-pan ana then the other. Does he gain or lose, and how much f (a - ANS. Loses m -, ab ., Ibs. (8) The arms of a balance are unequal, and one of the scale-pans is loaded. A body the true mass of which is m Ibs. appears when placed in the loaded pan to weigh ;, Ibs., and when placed in the other m, Ibs. Find the ratio of the arms and the mass with which the pan is loaded. mi m ANS. Ratio of arms = mass = H-P m\ m (9) A balance is in equilibrium and unloaded. A body in one scale-pan is found to balance m\ Ibs. In the other scale-pan it balances m* Ids. Find the true mass. ANS. ymitnt. ( i o) Find the condition for static equilibrium for a screw, neglecting friction. ANS. Let + P, P be the couple applied at the top with lever-arm ac = /; let ab =. r be the radius of the screw, and p the pitch or distance between the threads, so that if the screw be developed we have an in- clined plane, which has a rise p for a base iitr. Let a be the inclination of the screw to the horizontal, so that P tan a -- 2nr ' Let Q be the mass supported. We can resolve Q into a normal to the thread and a horizontal force Q tan a. The normal is balanced by the upward normal pressure A 7 on the threads. The horizontal com- ponents at every two diametrically opposite points of the thread are equal and opposite. We have then a couple Q tan a x r balanced by PI, or gtan ax r= PI; .-. ^ = ^7-1 Inserting the value for tan a, e = 8t. 214 STATICS. GENERAL PRINCIPLES. [CHAP. I. (ii) The differential screw consists of a screw n d which works in a fixed nut CC. The screw is hollow and has a thread cut inside in which a scre^u it ' e works. This screw de is prevented from turning by the rod _, /"/', whose ends slide in vertical grooves. Find the condition for static equilibrium, neglecting friction. ANS. Let pi and /> be the pitch of the screws, and a the common inclination of the threads, and r\ and r t the radii of the screws. When the arm ac = I turns through lit radians screw ad rises a distance p\ and screw de falls a distance p t . We have then just as before, in the preceding example, Q tan a x r, Q tan a x r t = PI, Hence Q(ri - r,) tan a I or, since tan a = P* If p t =o, we have the simple screw. By making p\ and pi nearly equal, we can have P very small compared to Q. (12) Let the force P act normally upon the middle of the back of an isosceles wedge. Find the condition of static equilibrium, neglecting friction. p ANS. The pressure N on each side must be normal. Let a be the angle of the wedge. Then Hence = o, N or P = 2./Vsin . If a is very small, N may be made very great. ( 1 3) In a wheel and axle the radius of the wheel is CA = a, and of the axle CB = b Find the conditions for static equilibrium when a mass P hung from the wheel balances a mass Q hung from the a.rle, neglfcting friction and rigidity of ropes. ANS. The weight of /"is Pg\ of Q, Qg poundals. Let 1\ be the upward pressure of the journal bearings. Then LZ1P R Pg Qg = o, or, Taking moments about the centre C, we have Qgb Pga =o, or Q = (Pg +<2.f. r| poundals, or, in gravitation units, '/? = (> + 0. CHAPTER II. WORK. VIRTUAL WORK. FIG. (i). FIG. (2). Work. The product of a uniform force by the component displacement along the line of the force of its point of application, is called the WORK of the force. Thus let F be a force acting at the point P l , and let the displace- ment be d = P-JP y Let F be uniform during displacement, and let the projection P v n along the line of the force of the displacement be p = d cos 6, where is the angle of the displacement with F. Then the work of F is Fp Fd cos 0. If the projection p is in the direction of the force, the work ir. positive and the force is said to do work. If the projection p is opposite to the direction of the force, the work is negative and work is done against the force. Now F cos d is the projection of the force along the line of the displacement. We can therefore define work generally as the product of a uniform force by the component displace- ment along the line of the force ( F . d cos 0), or the product of the displacement by the com- ponent force along the line of the displacement ( d . F cos #). Work of Resultant. Let F r F v F y etc., Fig I, be any number of forces acting on a point P which has an indefinitely small displacement ds in any given direction. Then during this dis- placement the forces remain unchanged in magnitude and direction. Let F 19 F 2 , /3 , etc., Fig. 2, be the line repre- sentatives of the forces. The resultant is then F r given in magnitude and direction by the closing line of the force polygon. The projection of F r along the line of the dis- placement ds is then evidently equal to the algebraic sum of the projections of all the com- ponents. Hence the , work of the resultant for any indefinitely small displacement, is equal to the algebraic sum of the works of the components. If the forces do not change in magnitude or direction with the displacement, then the same principle will hold for any displacement large or small. Work Non- Concurring Forces. Let a rigid body acted upon by any system of non- concurring forces F l , F 2 , F, , etc. have an indefinitely small displacement. Let A > A > A etc., be the displacement of each point of application along the line of the force at that point. 2*5 216 S7//7/CS. GENERAL PRINCIPLES. [CHAP. II. Then the work, no matter whether the displacement is one of translation or of rotation or of both combined, is given by work = F,p, -4- F t p t + F 3 /> 3 + . . . = 2Fp. . . . (i) In equation (i) each term is to be taken positive or negative according as the projections A A' A etc< are m t ^ lc direction of the corresponding forces or in the opposite direction. Let us suppose, first, that the displacement is one of translation and given by ds. Take any origin O and co-ordinate axes X, Y, Z. Let + ^i ^2' ^s etc - act at Pi nts PI* PV PV etc - given by the \ co-ordinates (*,, j,, 2^, (x 2 , y.^ z 2 ), etc. Let the direction id, cosines of F lt F 2 , F 3 , etc., be (ar lt P r y^> (a 2 , /? 2 , y 2 ), etc. Then the components of F l are F v cos a lt F l cos /?,, -* F l cos y l \ of F 2 , F 2 cos cr 2 , F. z cos fi 2 , F 2 cos y 2 , and so on. Also the components in the direction of the axes are given (page 194) by r i ft* F x FI cos a 1 -\- F 2 cos 2 + = 2F cos a, F y = F v cos /?, -j- F 2 cos /5 2 + . . . = 2F cos /?, F f = ^, cos ^ -j- F 2 cos ^ 2 -j- . . . = ~2F cos y, and the moments about the axes are given (page 194) by M x = 2F cos y . y 2F cos /? . z, ~\ My = 2F cos a . z 2F cos y . x, M t = 2F cos ft . x 2F cos a . y. (2) (3) Let the components of the displacement ds be dx, dy, ds. Since the work of any force is equal to the algebraic sum of the works of its components, we have for translation - (4) F l p l = F l cos j . dx -\- FI cos /?! . dy -\- F l cos ^ . ^, ^2/2 = F 2 cos a 2 . ^ -(- /^ cos ft z . dy -\- F 2 cos ^ 2 . ds, and so on. By addition, we have, from (i) and (2), for the ivork of translation work F x .dx + F y . dy -\- F t .dz (5) Now let us suppose the displacement is one of rotation due to an indefinitely small angular displacement dtt about an axis through O. Let the components of this displacement along the axes be dO x , dft y , dV t . Then we have dx z . dH y y . * N ' where F is the total friction and N the total normal pressure when motion either sliding or rolling is just about to begin, or else when motion either sliding or rolling has taken place. In the first case /* is the coefficient of static friction of sliding or rolling. In the second case ju is the coefficient of kinetic friction of sliding or rolling. We have to do in this portion of the work with static friction only. Limiting Equilibrium. The student should carefully note that does not give the actual resistance of friction in all cases of equilibrium, but only the resist- ance which exists when the surfaces are on the point of motion. Friction acts always in a direction opposite to the force which tends to cause motion, and so long as there is equilibrium it is always equal in magnitude to this force. But when this force has the magnitude l^N motion \sj2ist about to begin, and the body is said to be in LIMITING EQUILIBRIUM. If this force is less than ^V, there will still be equilibrium, what- ever its magnitude, and the body is in NON-LIMITING EQUILIBRIUM. Coefficient of Static Sliding Friction Experimental Determination. Let a body of weight W, acting at the centre of mass C, rest in equilibrium upon a rough plane AB, the surfaces of contact being plane. Then for equilibrium the reaction R of the plane is equal and opposite to Wand in the same vertical line, and the sum TV of all the normal pressures acting at every point of contact must be equal and opposite to the normal component of W, and the sum F of all the frictions at every point of contact must be equal and opposite to the component T of F parallel to the plane. We have, then, when sliding is about to begin, for the coeffi- cient of sliding friction and we see from the figure that -^ is the tangent of the angle which the reaction R makes with the normal when sliding is about to begin. Now the reaction at every point of contact is paral- 222 STATICS. GENERAL PRINCIPLES. [CHAP. III. lei to A' or IV, and sliding begins at all points of contact simultaneously. Hence the angle which R makes with the normal when sliding is about to begin is evidently the same as the angle which the plane makes with the horizontal. Therefore F P = = tan 0. We call the angle which the plane makes with the horizontal when motion is about to begin the ANGLE OF REPOSE. That is, the coefficient of static sliding friction is equal to the tangent of the angle of repose. If, then, we place a body upon a rough plane and then gradually incline the plane until sliding just begins, the inclination of the plane at this instant gives the angle of repose 0. The tangent of this angle gives the coefficient ju of static sliding friction for plane surfaces. We obtain the same result by resolution of forces. Thus let be the inclination of the plane when sliding begins. Then for equilibrium W cos = N, and W sin = F. Hence F M = TF = tan 0. We can thus make use of the inclined plane as an apparatus for determining /< by experiment. Again, if we place a body of weight If on a horizontal plane and measure the horizontal force /''just necessary to cause it to begin to slide, we have M = W = tan where is the angle of the reaction R with the normal when sliding begins, or the angle of repose. Such an apparatus should be so constructed that the friction of the pulley and other resistances due to the string, etc., can be disregarded or else allowed for. Cone of Friction. If we revolve the line representative of R in the figure page 221 about the vertical, it will describe a cone. This cone is called the CONE OF FRICTION. The reaction of R is equal and opposite to the resultant of F and W acting at C. We see, then, that no force acting at C, horuever great in magnitude, can cause sliding to begin if its line representative lies within the cone of friction. Laws of Static Sliding Friction. The following laws of static sliding friction have been established by experiment as holding true within the limits indicated : i . Other things being the same, within certain limits of the normal pressure, static sliding friction is proportional to the total normal pressure and independent of the area of the surfaces in contact. In other words, within the limits of normal pressure referred to, the coefficient of static sliding friction fit is constant for the same two surfaces in the same condition, whatever the area of the surfaces of contact and whatever the total normal nressure. i CHAP. III.] LIMITATIONS OF THE LAW. 223 Thus, if the normal pressure N over a given area is increased or decreased, the friction ^increases or decreases in the same proportion and p = is unchanged. It follows directly that if the area increases or decreases, N remaining the same, the number of points of contact is correspondingly increased or decreased, but the normal pressure at each point, and therefore the friction at each point, is correspondingly decreased or p increased. The sum of all the frictions F remains, then, the same and /* = -- is unchanged. Limitations of the Law. The limitations of normal pressure referred to are as follows: If the normal pressure per unit of area approaches the crushing strength or becomes so great as to break up the film of interposing unguent, the friction F increases more rapidly than the normal pressure, and the law fails. In properly designed structures the normal pressure per unit of area is much less than this limit, and the law applies. Again, if the normal pressure per unit of area is very small, adhesion may constitute the larger portion of the resistance. This adhesion increases with the area of contact (page 220). In all practical cases, however, the influence of adhesion may be neglected. Hence in practical appplications the friction is the only resistance which is considered, and it is assumed that gives the resistance, where /* is in practice a constant for the same two surfaces in the same condition, whatever the area of the surfaces in contact and whatever the total normal pressure N. 2. Other things being the same, within certain limits of the normal pressure, the static sliding friction of greased surfaces is less than that of ungr eased and depends less upon the surfaces than upon the unguent. Here, again, if the normal pressure per unit of area becomes so great as to break up the film of interposing unguent, surface comes in contact with surface and the friction may depend more on the surfaces than upon the unguent. In properly designed structures the normal pressure per unit of area is much less than this, and the law applies. Again, if the normal pressure per unit of area is very small, adhesion may constitute the larger portion of the resistance, and this adhesion is increased by the unguent. In all practical cases, however, the influence of adhesion may be neglected. Hence in practical applications the friction is the only resistance which is considered, and it is assumed that F= nN gives the resistance, where /* is in practice a constant for the same two surfaces in the same condition, whatever the area of the surfaces in contact and whatever the total normal pressure N. Upon these two laws depend the value and use of the values for the coefficient of static sliding friction given in the next article. Values of Coefficient of Static Sliding Friction. The following table gives a few values of /* as determined by experiment for static sliding friction. STATICS. GENERAL PRINCIPLES. COEFFICIENTS OF STATIC SLIDING FRICTION /* = tan 0. [CHAP. III. Substances in Contact. Condition of Surfaces and Kind of Unguent. Dry. Wet. Olive Oil. Lard. Tallow. Dry Soap. Polished and Greasy. {minimum 0.30 0.50 0.70 0.15 O.l8 0.24 O.6o 0.50 0.63 0.80 0.47 0.6 5 0.68 0.71 0.14 0.19 0.25 O.II 0.12 O.22 0.36 0.44 0.30 0-35 0.40 0.15 O.IO 0.28 0.21 maximum i minimum O.II 0.12 O.I') O.IO O.IO O.I2 1 maximum 0.65 0.87 Hemp ropes or plaits on wood - Leather belts over drums minimum mean maximum wood Stone or brick on stone or minimum maximum 0.67 0.75 0.65 0.74 0.40 0.7 too. 3 0.51 0-33 0.25 to i I.O Masonry and brickwork, damp mortar. . . . D 1 d 1 More extensive tables will be found in treatises on Friction. It will be noted that the coefficient of static sliding friction is practically always less than unity. In only one case given in the table, viz., for damp clay on damp clay, is p = i, corresponding to an angle of repose of = 45. Rankine gives for " shingle on gravel " a maximum /* = 1. 1 1, corresponding to an angle of repose = 48. Static Friction for Pivots. In all cases of the sliding of two surfaces we denote the coefficient of static sliding friction by yu and take the value of p as given by the table. We have then, in all cases of sliding friction, for the friction when sliding is about to begin F = ^N=Ntan 0, where N is the total normal pressure and is the angle of repose, and // is given by the table. The direction of the friction is always opposite to the direction of motion if motion were to take place. The application to pivots is then simple. i. SOLID FLAT PIVOT. Let ACB be the base of a solid flat pivot, and A^the total normal pressure upon the base. We have then for the static friction F=pN, ........ (i) where H is given by the table. If we divide the base into a very large number of very small equal triangles, such as ACD, the friction on each can be considered as the resultant of equal parallel forces distributed over the surface. The point of application for each triangle is then at the centre of mass for that triangle. The point of application of the entire friction is then at a The moment of the entire friction with reference to the distance Cs = - r from the centre. axis is then CHAP. III.] STATIC FRICTION FOR PIYOTS. 225 Since for any point s of the base there is a corresponding point s' for which the friction is equal and opposite, the moment of the friction is the moment of a couple and is therefore the same for every point in the plane of the base (page 185). 2. HOLLOW FLAT PIVOT. If the rubbing surface is a flat nngADEB, we have as before F=pN t . .-._;. . ... (i) where N is the total normal pressure on the base, and ju is the coefficient of static sliding friction as given by the table page 224. Let the outer radius be r l and the inner radius r z . Then any small portion of the base is a circular ring for which the length of chord and arc AD may be taken equal. The centre of mass (page 29) for each small portion is then at a distance Cs from the axis given by 3 Hence the moment of the friction with reference to the axis is (2) Since for any point s there is a corresponding point s' for which the friction is equal and opposite, the moment of the friction is the moment of a couple and is therefore the same for any point in the plane of the base (page 185). 3. CONICAL PIVOT. In the case of a conical pivot let R be the pressure along the axis, and let the half angle of convergence ADC be a. If we divide the conical surface into a large number of very small triangles with their vertices at the point D, each will sustain the vertical r> load , and the normal pressure on each will be R If we denote the radius C l A l = C l B l of the pivot at the point of entrance by r lt the resultant normal pressure upon each small elementary triangle acts at a distance r l from the axis. We have then for the total friction R sin a' where p is the coefficient of static sliding friction as given by the table page 224, and the moment of the friction with reference to the axis is ,_ 2 Rr, M = fjt - -, 3 sin a or, since . 1 = the side DA, of the cone of contact = a, we have sin a. M= 2 - (2) ,,0 STATICS. GENERAL PRINCIPLES. [CHAP. 111. This is also the moment of a couple and hence the same for any point in the plane per- pendicular to the axis at a distance above the point D equal to two thirds the height of the cone of contact. 4. PIVOT A TRUNCATED CONE. Let R be the pressure along the axis, and let the half angle of convergence ADC be a. Let R a be the pressure sustained by the flat base, and R v the pressure sustained by the conical surface. Then s^& Also, if r l is the radius C l A l at the point of entrance, and r 2 the radius of the base, r 2 R 3 :R:: nr* : nr?, or R 2 = -\R, 7*1 and hence - r* r 2 We have then as in Case I, page 224, for the flat pivot, the friction F 2 on the base and its moment about the axis For the friction on the conical surface we have, as in Case 3, page 225, for the conical pivot _ R^ __ r? - r 2 2 R sin a r? sin a and for its lever-arm, as in Case 2, page 225, for hollow pivot, 2 r? - r a 8 3 r i 2 r t Its moment, then, about the axis is 3 r, 2 sin a The total friction for the truncated pivot is then and its total moment about the axis is M = M, + M. = - fi- t 3 r t - sin a (0 (2) where n is the coefficient of static sliding friction as given by the table page 224. CHAP. III.] STATIC FRICTION FOR PIVOTS. 227 [Pivot with Spherical End.] Let R bz the pressure along the axis, denote the radius AO of the spherical surface by r, and the radius AC by r t , and let the angle AOC be a, p Then the load per unit of area of horizontal projection is ^75- Take any element of the surface at a, distant ab = x from the axis, and let Ob = y. The horizontal projection of this element is 2itxdx, and the load sustained by it is R 2Rxdx then iitxdx x y A/ r "* _ The cosine of the angle aOb is cos aOb = ^ = 3- The normal pressure on the element at a is then and the static friction is zRxdx zuRr xdx Integrating between the limits of x = o and x = r\ , we have for the total friction F = or, since j/r a r* = r cos a and r\ = sin a, sin' a i + cos a. where ju is the coefficient of static sliding friction as given by the table, page 224. For hemispherical end a= 90 and F = 2juA\ For flat end ct= o and F = uR. The moment about the axis of the friction on an element is 2juRr x*dx Integrating between the limits x = o and x = r*, we have for the total moment of the friction about the M WRrr r* . T r, r, , - -H - r- -sin" 1 -- 4/r"-ri* , r? \_2 r 2 v or, inserting the values of J/V" r\* = r cos a and r\ = r sin a and reducing, fa. \ M nRr\ - cot a 1 \sm 2 a I (2) For hemispherical end a = , sin a = i, cot a = o, and this becomes M = -. Static Friction of Axles. In all cases of the sliding of two surfaces, we denote the coefficient of static sliding friction by yu and take the value of jj. as given by the table, page 224. We have then, in all cases of sliding friction, for the friction when sliding is about to begin F= uN = Af tan 0, where N is the total normal pressure and is the angle of repose, and n is given by the table, page 224. 228 STATICS. GENERAL PRINCIPLES. [CHAP. 111. The direction of the friction is always opposite to the direction of motion if motion were about to take place. The application to axles is then simple. i. AXLE IN PARTIALLY WORN BEARING. Let the bearing be partially worn, then the axle at the moment when sliding begins touches the bearing at a point A, and the resultant pressure R at this point makes the angle of repose with the normal. We have then for the normal pressure N = R cos 0, and for the friction F = N tan = R sin 0, where is the angle of repose as given by the table, page 224. Let r be the radius AC of the axle. Then the moment of the friction with reference to the axis is = Rr sin 0. (2) If the axle is well greased, the angle of repose is very small, and we may take = tan = sin 0. In the practical case of a well-greased axle, then, we have where n is given by the table, page 224. If the wheel AB revolves, as shown, about a fixed axle AC, the friction is the same as before, but the lever- arm of the friction is not the radius of the axle, but the inner radius of the wheel. 2. AXLE TRIANGULAR BEARING. If the bearing is triangular, the axle is supported at two points A and B. The resultant pressure R can be resolved into two components R^ and R 2 , and when sliding begins each of these makes the angle of repose with the normals at A and B. The normal pressure at A is then N l = R t cos 0, and the fric- tion at A is ^"i = NI tan = R v sin 0. The friction at B is in like manner F 2 = R 2 sin 0. The total friction is then Let the angle ACS = 20. Then the angle AOR = ft - 0, and the angle BOR = + 0. We have then /?, : R : : sin (ft + 0) : sin 20, or R t = /? 2 :/?:: sin (0-0): sin 20, or R, =--. sin 2ft Hence the total friction is F= [sin (ft + 0) + sin (ft - 0)]^^- CHAP. 111.] STATIC FRICTION FOR AXLES. 229 But sin (/? + 0) + sin (ft 0) = 2 sin ft cos is very small, and we may take sin 20 = 2 sin , = - .......... (6) In the first case, when the weight is about to sink, we have for the friction / (, - tan ")"\ *,-fl-W- -4il (;) If the weight is about to rise, r-r.-Q-- .-> ......() \('-f> tan -j / CHAP. III.] STATIC FRICTION FOR CORDS AND CHAINS. 2 33 Formulas (5), (6), (7) and (8) are also applicable to the case of a chain composed of links which is passed round a cylindrical surface, where n is the number of links in contact. If the length of each link is AB = /, and the distance CA of the axis A of a link from the centre C is r, we have for the angle of deviation DBL = ACB a A . sin = , 2 2r [If a flexible cord lies in contact with a rough surface, let ACB = a be the arc r s_, of contact. If 7' is the tension at any point of contact D for the indefinitely small portion of the cord Dd, the fric- tion at this point is dT. Let the indefinitely small angle DCa be da. Then, from equation (i), page 231, dT- da 2u T tan i +ju tan But since da is indefinitely small, we may take the arc equal to the tangent and disregard u tan with reference to i. We have then dT = nda. . Integrating between the limits a = o and a, we have, since for a = o, T = Q, and for a = a, T = P, p logn P = //a + logn Q, or logn = jua. We have then, when motion in the direction of P just begins, P = ^ a , where e = 2.3026 = base of Naperian system of logarithms. When motion in the direction of <2'just begins, we have, by interchanging P and Q, Also, inversely, a= 2 -3Q26(log /'-log 2) (9) (10) (n) where common logarithms are taken. If the arc a of the cord is given in degrees instead of radians, we must substitute = ^r- it. If the surface is cylindrical and the number of coils n of the rope is given, we have a = 2icn. We see from (9) and (10) that the friction of a cord, F = P Q or F = Q P, upon a surface does not depend at all upon the radius of curvature, but only upon the arc of contact a, or upon the number of coils, 2Ttn, if the surface is cylindrical. If we take u = -, we have for a cylindrical surface : for - coils,/ 7 1.692; 4 i 2 "2 " P = 65.94^; "4 " P = 4343-56(2. The friction can thus be increased to any amount by increasing the number of coils.] '34 ST4T1CS. GENERAL PRINCIPLES. [CHAP. III. Rigidity of Ropes. When a rope is perfectly flexible it offers no resistance to bending. iVhen a rope is not perfectly flexible it offers a resistance by reason of its rigidity when wound on to a drum, pulley or axle, though none is offered when il is wound off. Thus let a rope whose tension is T be on the point of being wound on to a pulley. Let a AC = BC be the radius of the pulley, and t the thick- ness of the rope. Then the lever-arm of the axis of the rope on the is Cb = a -j -- . The distance Ac from the pulley to the rope on the on side will T+T' depend on the kind of rope and will be less as is greater. Thus for hemp ropes we can put where c l is a constant to be determined by experiment for the kind of rope; and iorwire ropes AC that is, Ac increases with the lever-arm a -\ -- and decreases as T increases. It is also evident that those fibres farthest out on the on side are stretched more than those nearer the pulley. The resultant tension T will therefore act further from the pulley than the central axis of the rope. We denote the distance of T from the central axis by C Y Let the tension along the central axis on the ^side be T -\- T . Then we have for equilibrium, for hemp ropes, or and for wire ropes, or c n T We have then Ct=(T+T')Cd, or Cc = i -f ^}Cb. The rope can be considered, then, as without rigidity if we increase the lever T . . (3) arm of the tension on the on side by the amount T' CHAP. IK J RIGIDITY OF ROPES. Hemp Ropes. For tarred hemp ropes experiment gives 235 IOO +O.222T 1 - -- pounds, where T is to be taken in pounds, and a and / in inches. For new hemp ropes, untarred, 4 +0.0645 7 T T = = pounds, where T is to be taken in pounds, and a and t in inches. Wire Ropes, For wire ropes we have = ,. 08 + pounds, where T is to be taken in pounds, and a and t in inches. Static Rolling Friction. Let ACB be a roller resting on a plane surface. By reason of the pressure N of the roller on the plane, the roller is compressed. Let a force F be applied at the centre C parallel to the plane. When the resultant R of /''and N just passes through the edge D of the base, rolling begins and the force F is equal and opposite to the friction. Let the distance AD d. Then, when rolling is about to begin, the angle A CD is the angle of repose 0. Let r be the radius. Since the compression is small compared to the radius, we have tan = = jt = coefficient of static rolling friction. Hence for equilibrium Fr = Nd, or The distance d depends on the materials in contact. The theory of rolling friction is not yet well established and but few experiments upon it have been made. In all practical cases of rolling we usually have to do with axle friction, which has already been discussed (page 228). Examples (i) A body of mass m rests upon a rough inclined plane which makes an angle a. with the horizontal and is acted ubon by a force P which makes the angle ft with the plane. Find the conditions for equilibrium. (For .smooth plane see ex. (8), page 2 1 8.) ANS. Consider the body as a particle at any point O. Let the angle POD ft be positive when above the plane, and negative when below. i. BODY ON THE POINT OF MOTION UP THE PLANE. In this case we have the component of P parallel to the plane acting up the plane, and the friction uN acting down. We have, then, P, N, m and uNin equilibrium. If we put the algebraic sum of forces along the plane and perpendicular to the plane equal to zero, we have in gravitation measure 236 57//77C5. GENERAL PRINCIPLES. [CHAP. 111. P cos/5 -, sin a -uN = o, or j, * + cos p Psin ft + N m cos a = o, or N = m cos a P sin # where u is the coefficient of friction. From these equations we have _ sin a + u cos a _ sin (a + ) ~ cos ft + // sin fl' m ~ cos (/* - ) m ' where 5s the angle of repose whose tangent is equal to//. If// = o, there is no friction and we have equation (2), example (8), page 219. If ft = 90 - a. P = m and N = o. For any greater value of positive ft, N is negative and no equilibrium is possible. For negative ft we must have ft less than 90 0. If negative ft is greater than this, we have N negative and no equilibrium is possible. Equation (i) holds good, then, for all values of ft between -|- (90 a) and (90 0). The force P is a minimum when cos (ft 0) is a maximum or when ft = . This minimum value of P is then P = m sin (a -t- , which is the same result as in ex. (i), page 217. 27T/ (3) Find the conditions for equilibrium for the differential screw given in example (2), page 217, taking ftiction into account. ANS. /> = g (4) Find the conditions of equilibrium for the wedge given in example (j), page 217, taking friction into account. . P = in f * co.5) - ^ .in (f ,). where the (+) sign is taken for wedge on point of entering, and the ( ) sign on point of sliding out. ct If P is between these values, the wedge is neither on the point of entering nor sliding out. If ~ = , there is no force required to keep the wedge from sliding out. The angle of the wedge should not, then, exceed 20, (5) A rod rests with its ends against a rough vertical and horizontal plane. The mass of the rod is m, and its weight acts at its middle point. Find the conditions of equilibrium. ANS. Let 9 be the angle with the horizontal, and JVi , N, the normal pressures on the horizontal and vertical planes respectively. Then tan 9 = cot. 20, i = m cos 2 0, N* = m sin cos 0. (6) In a wheel and axle the radius of the wheel is a, and of the axle b. Find the conditions for equilibrium, taking into account friction and the rigidity of the rope, when a mass P hung from the wheel just balances a mass Q hung from the axle. (Without friction and rigidity see example (13), page 214.) ANS. We have seen (page 228) that for well-greased axle and small surface of contact we can take in all cases of axle friction the friction F = nR = u(P -j- Q), where n is the coefficient of static sliding friction. Let the radius of the journal be r, and let t be the thickness of the rope. Then when P is just about to fall, we have (page 234) for the lever-arm of Q, \ib H J, and hence for equilibrium p = where (page 235) for hemp ropes T' = c -l^- for wire ropes T' = c\ H *j 2 the values of c\ and c? being given on page 235. 238 STATICS. GENERAL PRINCIPLES. [CHAP. Ill When Q is just about to fall, we have (page 234) for the lever-arm of P, ^i + -^J \a + - J, and hence +1 where (page 235) for hemp ropes T' = ; for wire ropes T' = c\ -\ ; the values of c t and c t being given on page 235. For values of P less than the first and greater than the second we have non-limiting equilibrium, and the wheel and axle is not upon the point of rotating in either direction. >+'- If we neglect friction and rigidity, we have P = -- -Q, or, neglecting the thickness of the rope + i P = -Q, as in ex. (13), page 214. If b = a, we have the case of the single pulley. For partially worn bearing (page 228) we can put more accurately sin

-j- u)(u 2nr) For values of P less than the first and greater than the second we have non-limiting equilibrium and P is not on the point of falling or rising. For wire ropes we have only to substitute Ci\ a + J in place of c\. For partially worn bearing or new bearing we can replace /* by the values given in the preceding example. If we neglect friction and rigidity, we have P = as in ex. (5), page 218. (8) In the system of pulleys shown find the relation between the force P and the mass Q for equilibrium, taking into account friction and rigidity of the rope. (Without friction and rigidity see ex. (6), page 218.) ANS. Let m be the mass of each movable pulley, and the number of movable pulleys. Let r be the radius of the axle of each pulley, a the radius of each pulley, u the coefficient of static sliding friction, / the thickness of the rope, and Ci and c* as given on page 235. For convenience of notation let = a H \- vr + Ci\ w = a -\ jur; v = u + w = 2a 4- / + c-i. Then, from the preceding example, we have, when P is just about to fall, for hemp ropes T\ = : 1- ~ rj = "(7* + m\ 1 S = u( T, + m) c 2 V V and so on. Inserting the values of T\ and T<>, we have in general T = 4- v n "* But from the preceding example we have Hence, since u v = w, P = uTn . d 240 STATICS. GENERAL PRINCIPLES. [CHAP. III. For wire ropes we have only to substitute c\ \a -j- -1 in place of c\. For partially worn bearing or new bearing we replace u by the values given in ex. (6). reduces to P Q + (2" \)m /A \a -f -j, u . whiclj is the same result as given in ex. (6), page 218. (9) In the system of pulleys shown find the relation between the force P and the mass Q for equilibrium, taking into account friction and the rigidity of the ropes. (Without friction and rigidity see ex. (7), page 218.) ANS. Let m be the mass of the lower block. ;ind the number of ropes coming from the lower block. Let r be the radius of the axle of each pulley, n the coefficient of static sliding friction, / the thickness of the rope, and d and ft as given on page 235. Let a be the mean radius of the pulleys. For convenience of notation let u = a + - + ur + ft, w = a H ur. Then we have for hemp ropes, when P is about to descend, For wire ropes we have only to substitute c\(a-\ ] in place of c\. For partially worn bearing or new bearing we replace u by the values given in ex. (6). If we neglect friction and rigidity, we have u=w and <:, = o. The value of P reduces then to P = - ; but if we divide numerator and denominator by u w and then make u = w, we have P= which is the same result as given in ex. (7), page 218. (10) In the system of pulleys shown find the relation between the force P and the mass Q for equilibrium, taking into account friction and the rigidity of the ropes. ANS. Let m be the mass of each pulley and n the number of pulleys. Letr be the radius of the axle of each pulley, n the coefficient of static sliding friction, / the thickness of the rope and c\ , c* as given on page 235. Let a be the radius of each pulley, and for convenience of notation let u = a H \-ur + ft, w = a ^ ur. Then we have, when P is about to descend, for hemp ropes For wire ropes we have only to substitute fJa + - J in place of c\. For partially worn bearing or new bearing we replace n by the values given in ex. (6). If we neglect friction and rigidity, we have u = w and c\ = o, and Q + nm (2* \)m CHAP. III.] STATIC FRICTION-EXAMPLES. 241 (i i) In the differential piilley shown in the figure an endless chain passes over a fixed pulley A, then under a movable pitlley to which the mass Q is attached, and then over another fixed pulley B, a little smaller but co-axial with A. The two pulleys A and B are in one piece and obliged to turn together through the same angle. The two ends of the chain are joined so as to form a loop. The force P is applied to the' right-hand portion of the loop. To prevent the chain from slipping, there are cavities in the circumferences of the upper pulleys into which the links of the chain fit. Find the relation of P to Qfor equilibrium, taking into account friction. ANS. Let a be the radius of the pulley A, and b the radius of the pulley B, m the mass of each pulley above and below, r the radius of each axle, and fi the coefficient of friction. Since the pulley is worked by a chain, we can disregard rigidity and have only friction to take into account. Let T be the tension of the chain. Then for equilibrium Q + 2T=Q + m, or T = Let -Fbe the friction. Then taking moments about C, we have for equilibrium - Pa + Ta Tb + Fr = o. The pressure on each axle is Q + 2m + P. Therefore the friction is F = 2/*(Q + 2m + P). Substituting this value of /'"and the value of T in the preceding equation, we have p _ 2 + 2ur(Q + 2m) 2/j.r For partially worn bearing, or for new bearing, we can replace n by the values given in ex. (6). If we neglect friction and the mass of the pulleys, we have P = ^ ~ -. KINETICS OF A PARTICLE. CHAPTER I. DEFLECTING FORCE. Kinetics. That portion of Dynamics which treats of forces with reference to the change of motion of bodies caused by them is called KINETICS. Kinetics of a Particle. We have seen (page 190) that when a rigid body is acted upon by any number of forces, the motion of the centre of mass is the same as if it were a par- ticle of mass equal to that of the body, all the forces acting upon the body being transferred without change in direction or intensity to this particle. When, therefore, we consider only the motion of translation of a body without reference to its rotation, we can always consider the body as a particle of equal mass concentrated at the centre of mass. It will therefore be convenient to first consider the KINETICS OF A PARTICLE. Impressed and Effective Force on a Particle. Let F be the resultant of any number of forces F lt F 2 , F 3 , etc., acting upon a particle. These forces we have called impressed forces (page 169), and the resultant F is the resultant impressed force. Let m be the mass of the particle, and /"its acceleration. Then /"will be in the direc- tion of F, and we have from equation (2), page 170, F = mf. (i) We call the quantity w/the effective force of the particle. We see from equation (i), then, that the resultant impressed force is equal to the effective force. D'Alembert's Principle Applied to a Particle. We can write equation (i) F-mf=o (2) That is, if we reverse the direction of the effective force it will hold the resultaTit impressed force in equilibrium. Hence the impressed forces acting on a particle and the reversed effective force of the particle constitute a system of concurring forces in equilibrium. This is D'Alembert's principle as applied 'to a particle, it reduces any kinetic prob- lem to one of equilibrium between actual (" impressed"} forces and a fictitious (" reversed effective ") force. Examples. (I) Let a mass m be moved on a smooth horizontal plane oy i rope which passes over the edge of the plane on a pulley and has a mass P hung at its end. Disregarding all friction ami mass of pulley and rope and rigidity of rope, find the acceleration. ANS. This example has already been solved (see example (12). pape 177). We solve it here by D'Alem- bert's principle. 242 CHAP. I.] DEFLECTING FORCE. 243 The impressed force on m is the tension of the rope P(g f). The effective force on m is m/. Revers- ing this, we have, by D'Alembert's principle, for equilibrium P(g~ /) -m/=o, or / = P+ m* (2) Two masses P and Q, P being the greater, are hung by means of a rope over a pulley. Disregarding friction and the mass of the pulley and rope, find the acceleration. ANS. This example has already been solved (see example (13), page 178). P(9 r -/j We solve it here by D'Alembert's principle. The impressed forces on Q are Qg downwards and the upward tension of the string P{g /). The effective force of Q is Qf upward. Reversing we have, by D'Alembert's principle, for* equilibrium P + Q ' Deflecting Force. We have seen (page 77) that when a particle moves in a curve whose radius of curvature is p with a velocity v at any instant, there must be a central accel- eration f p always directed toivards the centre of curvature and given by This central acceleration causes no change of speed, but only change of direction of the velocity. If m is the mass of the particle, we must then have a central force always directed towards the centre of curvature given by mv* This, then, is the force which causes the particle to move in a curve. If this force did not act, the particle would move in a straight line. We therefore call this force the DEFLECTING FORCE, and denote it by Ty Since the centre of mass of a. body moves as if the whole mass m were concentrated at its centre of mass, we have, when the centre of mass of a body moves in a curve with velocity v, the deflecting force for the entire body (i) where p is the distance from the centre of curvature to the centre of mass, and GO is the angular velocity, so that pco = v. If the path is a circle, p is constant and equal to the radius r and we have F == -- = = mvco. If there is no tangential acceleration / there will be no change of speed and v will be constant in magnitude, changing only in direction. In such case, if T is the time of revolution in the circle, we have KINETICS Of A P ARTICLE [CHAP. 1. 27T 271T = -~> or v = -~-t and (3) All equations give the force in poundals. For gravitation units divide by g (page 171). Simple Conical Pendulum. The simple conical pendulum consists of a particle of mass m attached to a fixed point P by a massless inextensible string of length /, and moving with uniform speed v in a circular path about a vertical axis through the fixed point. In this case the particle is acted upon by two forces, its weight mg vertically downwards and the stress 5 of the string directed towards the fixed point P. If the particle moves with uniform speed v in the circle whose radius is r I sin 6, where is the inclination of the string to the vertical, the vertical com- ponent 5 cos ff of the stress S must balance the weight mg, and imfl the horizontal component S sin 6 of the stress 5 must be equal to the deflecting force necessary to make the particle move in a circle with uniform speed. We have then mg -j- 5 cos = o, or 5 cos 6 = mg, and from equation (2), page 243, mv z mv* = mr ~ (i) (2) where co is the angular speed. We can find 5 from either (i) or (2). Squaring (i) and (2) and adding, we have also 5 = Also dividing (2) by (i), we have = m (3) For 5 in gravitation units we must divide (3) by g as usual. Let // be the distance of the fixed point P above the plane of motion, or the height of the pendulum. Then, since h tan = r, we have, from (4), (5) If, then, GO is the angular speed of the particle about the centre 0, roo = v, and, from (5), V . If / is the time of a revolution, This is the same as the time of oscillation of a simple pendulum of length h (page 138). 245 CHAP. I.] DEFLECTING FORCE. COR. i. If is indefinitely small, // and /are equal and <' and we have the case of the simple pendulum. COR. 2. Since GO = |/ , we see that the greater the angular ve- locity the less h, and, as / is constant, the greater r. This fact is taken advantage of in the steam-engine governor. As the piston speed increases, the spindle PO revolves more rapidly, the balls separate and the slide at B rises and by means of levers acts upon the valves of the engine to diminish the supply of steam. Centrifugal Force. Let us now solve the preceding problem of the simple conical pendulum by applying D'Alembert's principle (page 242). The impressed forces acting upon the particle are the weight mg and the stress 5 of the string. The effective force is mv z the deflecting force acting towards O. This is an actual force. If now we reverse the direction of this effective force, we have This is not an actual but a fictitious mv* acting away from u. force. There is in reality no force acting in this direction upon m. The actual force upon m is towards O. But by D'Alembert's principle the impressed and reversed effective forces form a system in equilibrium, have then, by reversing the effective force, We mi? S sin = o. We thus evidently obtain the same results as in the preceding article, where we took correctly as a force acting towards the centre O. This reversed effective force acting away from is often called CENTRIFUGAL FORCE, and as thus used the term is allowable. But it should be borne in mind that it is a fictitious and not an actual force. There is really no force acting on m away from the centre, and there really is a force acting on m towards the centre. If we really had a centrifugal force in equilibrium with S and mg, the particle m would move in a straight line and not in a circle. When, then, a particle moves in a curve, since its direction of motion changes, there must be an unbalanced force acting towards the centre of curvature, that is, a deflecting force. If we consider this force reversed, we can apply D'Alembert's principle, and the term " centrifugal force " only means, then, this reversed force, which is a purely fictitious force not really existing. A particle moving in a curve is often incorrectly represented, however, as possessing an inherent "centrifugal force" by virtue of which it "tends to fly away fiom the centre." Indeed it is sometimes represented that it is acted upon simultaneously by such a force, and 2 4 6 KINETICS OF A PARTICLE. [CHAP. I. also by an equal and opposite deflecting force towards the centre.* Both views are erroneous and misleading. The student would do well, therefore, to discard the term "centrifugal force," since it answers no real purpose. It is sufficient in all cases to take all forces, both impressed and effective, as they really are, and then to apply D'Alembert's principle. Deflecting Force at the Earth's Surface. Suppose the earth to be a homogeneous sphere of radius r and centre C. Let WME represent the equator, NS the axis, and let P be any point of the surface on the meridian PMS, so that the latitude is A = PCM. Let a particle of mass m rest on the surface at P. Let G be the actual acceleration of gravity acting towards the centre C along the radius PC. Then mG is the actual force of attraction towards C acting upon the particle. Let g be the observed ac- celeration of gravity at P. Then the observed weight of the particle at P is mg. The pressure of the earth on the particle is then mg acting away from C. The difference between w/6"and mg, or mG mg, is that part of the earth's attraction necessary to keep the particle on the earth without pressure. This force mG nig acting towards C can be resolved into a component F t tangent to the meridian at P, and a component F in the direction PO. This latter com- ponent is the deflecting force which makes the particle move in the latitude circle whose radius is PO = r cos A This, then, is the effective force, while m(G g) towards C and F t are impressed forces. If we reverse the effective force, we have, by D'Alembert's principle, m(G g) -f- F Q cos A = o, F sin A F t = o. But (page 244) since the angular velocity GO = -=-,, where Tis the time of rotation, Hence F = tnrar cos A = cos A. mg = mG mra? cos 2 A = mG 3 cos 2 A (i) f,, 2 n*ntr sin 2A = y- 2 sin 2 A. (2) " When I was about nine years old, I was taken to hear a course of lectures by an itinerant lecturer in a country town, to get as much as I could of the second half of a good, sound philosophical omniscience. . . . "'You have heard what I have said of the wonderful centripetal force, by which Divine Wisdom has retained the planets in their orbits round the sun. But, ladies and gentlemen, it must also be clear to you that if there were no other force in action, this centripetal force would draw our earth and the other planets into the sun, and universal ruin would ensue. To prevent such a catastrophe, the same Wisdom has implanted a centrifOgal force of the same amount, and directly opposite.' . . . " I had never heard of Alfonso X. of Castile, but I ventured to think that if Divine Wisdom had just let the planets alone it would come to the same thing with equal and opposite troubles saved." DE MORGAN, Budget of Paradoxes. CHAP. 1.] CENTRIFUGAL FORCE. 247 If the earth were a homogeneous sphere, the effect of this tangential component F t upon liquid particles on the surface would be to force them towards the equator and thus increase the equatorial and diminish the polar diameter. The fact that the earth is not a sphere thus indicates that the now solid portions may once have existed in a plastic condition. The equatorial diameter is found to exceed the polar by about 26 miles. The ratio of this dif- ference to the equatorial diameter, called the ELLIPTICITY of the earth, is about ^-^. The earth is considered, then, as an ellipsoid of revolution with this ellipticity, so that the direction of the observed force of gravity, or of the plumb-line ZP, is always normal to the surface and hence does not pass through C except at the equator and poles. The force of attraction mG acting towards C is then resolved into two components, one normal to the surface along PC and one F along PO. This latter is the deflect- ing force necessary to keep the particle on the latitude circle whose radius is PO r cos A. The former is balanced by the pressure of the earth upon the particle. There is then no tangential force F t , and no tendency of the particle at P to move towards the equator. The effective force is then F^ as before acting towards O, and the impressed forces are mG acting towards C and nig acting towards Z. Let be the angle CPC' '. If we reverse the effective force, we have, by D'Alembert's principle, mg -|- F cos (A -(- 0) mG cos = o, or, substituting for F its value, mg = mG cos mra? cos A cos (A -{- 0) = mG cos ^ cos A cos (A -f- 0). Since in the case of the earth the deviation from a sphere is small, the angle is very small and this equation reduces practically to (i). We can then treat the earth as a sphere of mean radius r and neglect the tangential component F t . COR. i. If we take the mean radius r = 3960 miles and T = 86164 seconds for a sidereal day, we have 4?r 2 r rGi3 2 ___- 0.111255 ft.-per-sec. per sec (3) We have then, from (i), for the total acceleration of gravity G at any point P in latitude A G ^-+0.11 1255 cos 2 A (41 At the poles A = 90 and G g, or the observed acceleration of gravity g at the pole , is equal to the total acceleration of gravity G of the earth. At the equator A = o and here the observed value of ^at sea-level is found to be about g = 32.09022 ft.-per-sec. per sec. Hence from (4) we obtain, assuming the earth as a sphere, G 32.20148 ft.-per-sec. per sec. The resultant central force mG mg acting towards C is then, from (4), , mG ( i } = o. 1 1 1 2 5 5 m cos 2 A = mr<*? cos 2 A = ~ 2 cos 2 A. \ (j ' J- 248 KINETICS OF A PARTICLk. [CHAP. I. I At the equator \ o, and I - ~ = g-. Hence at the equator (5) That is, the deflecting force at the equator is about ^ of the total force of gravity. COR. 2. To find the time of rotation T of the earth in order that a particle at any point P may have no observed weight, i.e., exert no pressure on the surface, we have from (i), by putting mg = o and T= T , cos 2 A, or T 9 = A - 9 = A V But, from (5), we have for the actual time of rotation T I 4n 2 r 289 X 4rr z r _ = , or G= -. Substituting this value of G, wex obtain At the equator we have A = o and T = T. In order, then, that a body at the equator may have no weight, the earth should rotate in about one seventeenth of a day. If the earth were to rotate faster than this, bodies at the equator would not stay on the surface. Examples. f I) A string just breaks with a weight of 20 pounds. It is fastened to a fixed point at one end and at the other to a mass of 5 Ibs., which revolves round t/te point in a horizontal plane. Find the greatest number of complete revolutions in a minute before the string breaks, if the radius to the centre to mass is j /"/. ) ANS. To break the string requires a force of 2og poundals. Let n be the number of revolutions per ~. 2itrn nn . mv* minute. Then v = = ft. per sec. The stress in the string is - = 20^-, or v Hence n = - = 4 complete revolutions and a fraction over. (2) Suppose the mass revolves in a vertical plane f ANS. Then at the lowest point the stress is ^- + mg poundals and - + mg = 2qf , or v = ^ \ tf= -g- Hence n = ^ *& = 41 complete revolutions and a fraction over. (3) What portion of their weight do bodies lose at the equator, taking the radius of the earth 4000 miles and g = 32 ft. -per- sec. per sec. ANS. About ^ (4) Find the length of day in order that a body in latitude 60" may possess no weight. ANS. One thirty-fourth of the present day. (5) A skater moves in a curve of 100 feet radius with a speed of 40 feel per tec. Find his inclination to the ice. (g = 3*-) ANS. 60 degrees. CHAP. I.] DEFLECTING FORCE EXAMPLES. 249 mg (6) Find the necessary elevation of the outer rail on a railroad-track on a curve of radius r, so that an engine weighing m Ibs. moving with a speed v can pass witliout lateral pressure on the rails by the wheel-flanges. Also find the pressures on the rails. ANS. Let C be the centre of mass of the engine and. h = CO the distance of the centre of mass above the rails. We have acting at Cthe weight mg, downwards. Since there is no lateial pressure on the rails, pw the rail pressures are A\ and R, at right angks to AB, as shown in the '\ figure. The impressed forces are then mg,Ri and R* , and the effective \ f Q force is F = '^- horizontal and acting towards the centre of curvature. By D'Alembert's principle, if we reverse F we have a system of forces .R, in equilibrium. Let a be the angle of elevation BAD, and w the width of track AB. We have then, putting the algebraic sum of vertical components equal to zero, Ri cos a + R* cos a mg = o, or (A'i + AY) cos a = mg. . . (i) Putting the algebraic sum of the horizontal components equal to zero, A'i sin a R, sin a + " = o, or (A'i + AY) sin a = '^~. . (2) r r Taking moments about A and putting the algebraic sum of the moments equal to zero, m/v R^isj _ m g\ cos a h sin a From (i) and (2) we have )fnv* / , TV . \ ( h cos a + sin a\ and substituting this in (3), we obtain and hence, from (i), v* tan a = ; m 2 cos a mg 2 cos a. Equations (5) and (6) give Ri and /?, in poundals. For gravitation measure we divide by^ and obtaii Ri = R* = in From (4) we obtain /' V* cos a (4) (5) (6) in (7) 7/* We have then for the elevation DB = w sin a. DB= (8) and for Ri and /?- , in pounds, A", (9) Equations (8) and (9) are accurate. They admit of practical simplification. Thus if we take the maxi- mum speed a mile per minute or 88 feet per second, and a ten-degree curve for which r= 573- 6 9 ft., we have for A r = 32, = L, or ~ a = ~-, and this value is far greater than it will ever actually be. We see, then, 6 , v that ^ can be disregarded in (8) and (9), and we have practically 2 5 KINETICS Oh' A PARTICLE. and for the elevation of the outer rail the practical formula [CHAP. 1. or, taking^ = 32 and iv 4 fj. 8$ in., DB = 7 -?- inches, nearly, (10) where r is to be taken in ft. and v in ft. per sec. If r is taken in ft. and v in miles per hour, we have DB = ~ inches, nearly (n) 4 r Equations (10) and (n) are then practical formulas giving the elevation in inches for r in feet, and v in ft. per sec. or in miles per hour. (7) Find the elevation of outer rail for radius of 300 yards and speed of 45 miles per hour. ANS. 8.43 inches. (8) Find the speed v of an engine on a curved /eve/ track of radius r and gauge w when it is just on the Point of overturning, the centre of mass being h above the rails. ANS. v=\/^. (9) A vessel containing water revolves with uniform angular velocity (a about a vertical axis through the centre of mass. Find the curve of the surface of the water. ANS. Let a particle of mass /// be at any point P of the surface whose co-ordinates are ON =y and NP = x. The impressed forces are the weight mg of the particle and the pres- sure A* of the surface upon the particle. The surface at P must be at right angles to A'. Hence the tangent PT is at right angles to R. The effective force is mxv? acting towards N. If we reverse this force, we have, by D'Alembert's principle, mxa? R sin NPT = o, R cos NPT - mg = o. Hence we have g Produce the direction of R to S, so that SP is the normal and 5A r is the subnormal at P. Then SN X tan NPT = x, or SN = -^. The property of a parabola is that the subnormal is constant and equal to the parameter. Therefore \\\t curve is a parabola and its equation is (10) A vessel containing water is moved horizontally with an acceleration f. Find the inclination of the water-surface. ANS. Let a particle of mass m be at any point P of the surface. The impressed forces are the weight mg of the particle and the pres- sure R of the surface upon the particle. The surface must be at right angles to R. Let a be the angle of the surface with the horizontal. The effective force is mf in the direction of/. If we reverse this force, we have, by D'Alembert's principle, Hence R cos a tng = o, R sin mf = o. tan ft = . CHAP. I.] PARTICLE MO'/ING ON THE EARTH'S SURFACE. Particle Moving on the Earth's Surface. Let the particle P, instead of being at rest on the surface of the earth, have a velocity v relative to the earth at any instant in any direction tangent to the earth's surface. Take the point P as origin, the axis of X towards the east, the axis of F towards the north, the axis of Z along the radius through P. Let v x and v y be the components of v along the axes of X and Y, so that v x is positive towards the east and negative towards the west, and v y is positive towards the north and negative towards the south. Let P^P 2 v y t be the distance south along the meridian described by P in north latitude, in an indefinitely small time t. If there were no rotation, P-^P. 2 would coincide with the meridian through P r But owing to the rotation of the earth this meridian moves to P{M, while P l moves to P 2 ', so that if P^'O' is parallel to the axis NO, the angle MO'P^ = otf, where 03 is the angular velocity of rotation. The angle O'P{P^ = A = the latitude of P r We have then O'P 2 ' Vyt sin A and MP 2 ' = v y t sin A . cot. But if f x is the acceleration due to rotation of P with reference to the meredian M, we have (page 92) MP Z ' = l -f x t* = v y cot* sin A. Hence we have for the acceleration of P with reference to the meridian M, due to rotation and the velocity v yj f x = sin A, (I) Equation (i) is general if we take v y positive towards the north and negative towards the south, and A positive or negative according as the point is north or south of the equator. Thus in the figure v y is south or negative, and hence for north latitude f x is negative or towards the west. Again, let P^P Z = v x t be the distance east described by P in north latitude in an indefinitely small time t. The meridian moves to M while P l N moves to P z , so that the angle MP P out. We have then and its projection on the meridian is MP Z sin A = cov x t 2 sin A towards the south. If f y is the acceleration of /'with reference to the meridian, due to rotation and the velocity v x , we have sin A, or f y = 2< A. (2) 252 KINETICS OF A PARTICLE. [CHAP. I. Equation (2) is general if we take v x positive towards the east, negative towards the west, latitude north positive, south negative, and f y positive towards the north, negative towards the south. Again, we have in the preceding figure for the projection of MI\ along the radius r, J//*, cos A = oov^ cos A upwards. If f t is the radial acceleration due to rotation and the velocity v x of P with reference to the meridian, we have -/,/* = y positive towards the north, negative towards the south, A positive for north, negative for south latitude, and f, positive upwards, negative downwards. Deviation of a Falling Body by Reason of the Rotation of the Earth. Let a particle be projected upwards along the radius of the earth with a relative velocity v z , and let P^P 2 = v s t be the distance described in an indefinitely small time t. If there were no rotation, / > 1 /" > 2 would coincide with the radius through P r But owing to the rotation of the earth the meridian moves to P^M while />, moves to />,', so that if PjO" is parallel to the axis NO, the angle MO"P 2 ' = cot, where GO is the angular velocity of rotation. The angle O"P^P.^ = 90 A, where A is the latitude of P r t t . cos A and v t t cos A . oat: \Ve have then O"P 2 ' = But if/, is. the acceleration due to rotation of P with reference to the meridian, we have MP 2 ' = -//=- z' z / cos A. Hence we have for the acceleration in longitude of P with reference to the meridian, due to rotation and the velocity r t , f x 2es on a perfectly smooth horizontal plane. Disregarding resistance of the air, find the acceleration and the distance described in S4 KINETICS OF A PARTICLE [CHAP. I. latitude and longitude in 4 seconds (a) when the velocity is north ; (b) south ; (r) east ; (if) west, (r = 3960 mites.) ANS. (<*)/* = 0.00436 ft.-per-sec. per sec. east, distance 240 ft. north and 0.035 h- ca* 1 - (6) / = 0.00436 ft.-per-sec. per sec. west, distance 240 ft. south and 0.015 ft. west. (c) f y = 0.05236 ft.-per-sec. per sec. south, distance 240 ft. east and 0.42 ft. south. (d)f, = 0.05236 ft.-per-sec. per sec. north, distance 240 ft. west and 0.42 ft. north. (6) A cannon-ball is fired in latitude jo 9 north with a -velocity of 1440 feet per sec. Neglecting resistance of the air. find the acceleration and distance described in latitude and longitude in 4 seconds (a) when the velocity is north ; (b) south ; (c) eas^t ; (a) west. ANS. (a)f x = 0.10472 ft.-per-sec. per sec. east, distance 5760 ft. north and 0.84 ft. east. (b) f x = 0.10472 ft.-per-sec. per sec. west, distance 5760 ft. south and 0.84 ft. west. (f) f y = 0.15272 ft.-per-sec. per sec. south, distance 5760 ft. east and 1.22 ft. south. (d) f y 0.15272 ft.-per-sec. per sec. north, distance 5760 ft. west and 1.22 ft. north. (7) A particle in latitude 60 north fulls from rest a distance of 1296 ft. to the ground. Find the deviation in latitude, disregarding resistance of the air. ANS. 0.2827 ft. towards the east. (8) In latitude jo north, find the angular velocity of rotation of the plane of a pendulum. ANS. 0.0000363 radians per sec. in a direction clockwise to one facing the north. The plane rotates through 180* in 24 hours. (9) A locomotive weighing 32 tons runs at the rate of 45 miles per hour in latitude jo north in a direction S. 30" E. on a curve of one mile radius in a counter-clockwise direction to one looking north. Find the pressure on the outer rail. ANS. 1868 pounds. If we disregard rotation of the earth, the pressure would be 1848 pounds. (10) In the preceding example suppose the direction is clockwise to one looking north. ANS. 1825.6 pounds. CHAPTER II. TANGENTIAL FORCE. MOMENTUM. IMPULSE. Tangential Force We have seen (page 77) that when a particle P moves in a curve whose radius of curvature is p, with a velocity v and an acceleration / at any instant, this acceleration /can be resolved into a central acceleration p v^ dv dv f p = and a tangential acceleration f t = r , where -=- is the p at at rate of change of speed. If m is the mass of the particle, then the force F acting on it is F mf in the direction of/", and this force can be re- solved into a central or deflecting force F p mf p = , which causes change of direction of dv motion and a tangential force F t = mf t = m ^- , which causes change of speed. The deflecting force F p we have discussed in the preceding chapter. Let us now consider the tangential force F t . Momentum. Let the mass of a particle be in, and its velocity v. Then we call the quantity mv the MOMENTUM of the particle. Hence the momentum of a particle is the product of its mass and velocity. Momentum, then, has magnitude and direction, and we can represent it by a straight line, just like velocity, and all the principles which hold for velocity hold also for momentum. We can therefore combine and resolve momentums, and we have the triangle and poly- gon of momentum just the same as for velocity or force. Hence the momentum of a particle in any direction is the product of its mass and velocity in that direction, We can also have the moment of a momentum, just the same as for a velocity, and the same principles apply. The unit of momentum is, then, evidently the momentum of one unit of mass moving with one unit of velocity. We may call a unit of velocity, or one unit of length per unit of time, a " velo." The English unit of momentum is then \he pound-veto (Ib.-velo), or the momentum of a mass of one Ib. moving with a velocity of one foot per second. If, then, a particle of mass 40 Ibs. has a velocity in any given direction at any instant of 8 ft. per sec., the momentum in the direction of the velocity is 320 Ib.-velos. In the C. G. S. system, we would have, then, the gram-velo, or kilogram-velo, that is, the momentum of a mass of one gram or one kilogram moving with a velocity of one centimeter per second.* * A committee of the British Association have proposed for this the name "dole." This has, however, never come into use. 255 KINETICS OF A PARTICLE. ' [CHAP. II. Significance of Momentum. Let a particle of mass m move in any path from the posi- v, tion /> to P in the time /. Let the velocity at /\ be v v and at ^ be We have (page 255) the tangential force F t given by dv Now if F t is constant in magnitude, the tangential acceleration f f =~ J s also con- dt stant in magnitude. In such case the instantaneous rate of change of speed is equal to the mean rate of change of speed -y-^ 1 for any interval of time / (page 74). We have then for F t constant in magnitude mv mv. 0) Now mv mv l is the change of momentum in the path in the time /, (v v^) is the chajige of speed in the time /, and * is the time-rate of change of momentum in the path. Hence, whatever the path and however the actual tangential force may vary in magnitude during the time, the time-rate of change of momentum in the path gives a tangential force of constant magnitude which, acting for that time, would cause the change of speed in the path. Again, let us resolve v^ and v into components in any given direction P^a and at right angles to this direction. Let z\ make the angle a^ and v the angle a with the direction P^a. Then the component ~ velocities in this direction are i\ cos a l , v cos a, and the mean time-rate of change of speed in this direction is v cos a v. cos a. - . If the force F in this direction were constant, the instantaneous rate of change of speed would be the same as the mean for any interval of time, and we should have _ H mv cos mv cos or (2) Hence the time-rate of change of momentum in any direction gives the uniform force F u in that direction which, acting for that time, would cause the change of velocity in that direction. No matter, then, how the actual force in the given direction may vary during the time /, we can find from (2) in any case that equivalent uniform force F H which, acting for that time in that direction, would produce the change of velocity in that direction. From equation (2) we see that as the time / decreases the force F H increases for the same change of momentum. If / is zero, F H becomes infinitely large. That is, change of momentum requires time, and the less the time the greater the force. CHAP. II.] RELATION BETWEEN IMPULSE AND MOMENTUM. 257 If the time is one second and the initial velocity v l cos a l in any direction s zero, we have, from (2), mv cos a Jr.. = -. I sec. If the time is one second and the final velocity v cos a in any direction is zero, we have, from (2), mv l cos a * = I sec. ' That is, the momentum in any direction is NUMERICALLY equal to that uniform force which, acting in the direction of the momentum, would give the particle starting f rout rest its velocity in that direction in one second ; or which, acting opposite to the direction of momentum, would bring the particle to rest in one second. If we take mass in Ibs. and velocity in feet per sec., the force in equations (i) and (2) is given in poundals. For gravitation measure divide by g (page 172). If, then, a particle of mass 40 Ibs. has a velocity in any given direction at any instant of 8 ft. per sec., 320 the momentum in that direction is 320 Ib.-velos (page 255), and a uniform force of 320 poundals or pounds acting in the direction of the velocity would give the particle starting from rest its given velocity in one second. Acting opposite to the velocity, it would bring the particle to rest in one second. Impulse. The product Ft of a uniform force F by its time of action / is called the IMPULSE of the force. Hence impulse is the product of a uniform force by its time of action, and it acts in the direction of the force. Impulse, then, has magnitude and direction, and we can represent it by a straight line just like force, and all the principles which hold good for force hold also for impulse. We can therefore combine and resolve impulses and have the triangle and polygon of impulse just the same as for force. We can also have the moment of an impulse just the same as for force, and the same principles apply. The unit of impulse is then, by definition, the impulse of one unit of force acting for one unit of time. The English unit is, then, the poundal-sec. or the pound-sec. The C.G.S. unit is the dyne-sec. If, then, a uniform force of 320 poundals acts for 3 sec., the impulse is 960 poundal-sec. or pound-sec. Relation between Impulse and Momentum. Let a particle of mass m move in any path through the positions P l , P 2 , jP 3 , etc. Let v lt v 2 , v s , etc., be the corresponding velocities, t lt t z , ( 3 , etc., the corresponding times in passing from point to point, and F lt F z , F^, etc., the correspond- ing forces in any given direction, as P v a. Let the points P lt P 2 , P 3 be consecutive, so that the times t lt / 2 , / 3 are indefinitely small. Then, how- ever the force in the direction P^a may vary, we can consider F l as constant for the time t r Its impulse is then Fj r The impulse of F 2 is F z t z , and so on. S 5 8 KINETICS OF A t>ART\CL. But from equation (2), page 256, we have /r^ = mv t cos a a im\ cos a v F a t t = *t- 8 cos tf s w 2 cos a 2 , ^5/3 = mi'i COS 4 W2/ s COS (X y [CHAP. 11. If we sum these equations and let the final velocity be i>, making the angle a with the direction P l a t we obtain 2 ( F i'\ 4- ^'2 4- ^s's 4" etc -) = mv cos a ~ mv i cos a i ..... ( ! ) Let / be the ei.tire time, so that Let F m be the equivalent uniform force in the direction /\0 , whose impulse F H t for the entire time / is the same as the sum of all the actual impulses during that time, so that We have then, from (i), FJ = iX/Vi + F A 4- ^3 + etc.). etc.). = tf**' cos a- (2) This is equation (2; of page 256. Hence the change of momentum in any direction for any time t gives the sum of the impulses in that direction during that time, no matter liow the force in that direction may vary. Examples, -(i ) A ball-player catches a ball moving with a velocity of 50 ft. per sec. The mass of the ball is jj oz. If the time of coming to rest is fa sec., find the equivalent uniform pressure. ANS. We have m = Ib. The equivalent uniform pressure is then F n = r- = = 859 - poundals, or, taking^- = 32 ft.-per-sec. per sec., 26.85 pounds. This is a uniform force which would stop the ball in the given time. The actual force acting at any instant we cannot tell without a full knowledge of the law of variation of the pressure with the time. If we assume the pressure at first contact to be zero and to increase directly with the time, then the final pressure would be twice as great as the equivalent uniform pressure. (2) An So-ton gun fires a shot of 5b Ibs. with a horizontal muzzle velocity v of 1800 fl. per sec. Find the velocity of recoil V. ANS. Mass of gun M = 80 x 2240 = 179200 Ibs. If the velocity of the shot is imparted in the time /, the sum of the impulses on the shot during that time is mv = 56 x 1800 = ioo8oo-lb. velos. Since action and reaction are equal at every instant, the sum of the impulses on the gun during the time is the same. Hence MV=mv, or V = = - ft .per sec. 179200 1 6 (3) A horizontal stream of water "those cross-section is a and velocity v\ meets a surface moving in the same direction with a velocity v. Find the pressure exerted on the surface. ANS. Let the water pass off the surface in a direction making the angle a with the direction of motion. The volume of water in any time / is av\t. If y is the density or mass of a unit of volume of water, the mass in the time / is yavj = m. The velocity relative to the surface just before impact is vi v. After impact the velocity relative to the surface, in the direction of motion is (z>, v) cos a. We have then the impulse Ft = mv\ m (v, v) cos a, or ,._ f(r', - i>) md't v) cos a . CHAP II.] MOMENTUM EXAMPLES. Inserting the value of m = yav\t, F= yav*(vi v)(i cos a). If the surface is plane and at right angles to the stream, a = 90, cos a = o and F = yavi(vi v). If the surface is curved so that the water is reversed in direction, a = 180, cos a = i and the pressure is twice as great, or F = 2yav^(vi - v). These values of F are given in poundals. For gravitation measure divide by^. (4) A rifle-bullet of one ounce mass is shot into a block of wood of 53 Ibs. and gives the block a velocity of 2 ft. per sec. Find the velocity of the bullet. ANS. We have mv for the bullet equal to (JM-\-m)v for the combined mass of block and bullet, or -TV = ( 53 + -7)2, or v= 1698 ft. per sec. 16 16 (5) A mass moving with a velocity of 'j ft. per sec. is brvught to rest by a uniform opposing force of one pound in 2 sec. Assuming g = 32 ft.-per-sec. per sec., find the mass. ANS. A force of one pound is i Ib. Kg, or g poundals. We have then F= ^, or g = -, or m = = 2ii Ibs. 3 (6) A uniform force of i o pounds acts for 2 sec. upon a mass of 10 Ibs. and then ceases. With what velocity will the mass continue to move in the direction of the force ? ANS. A force of 10 pounds is log poundals. We have then F= , or \og = , or v = 2g ft. per sec. (7) A particle of 10 Ibs. mass has an initial velocity of 20 ft. per sec. north and is acted upon by two uni- form forces, one of 3 pounds in a direction northeast and the other of the same magnitude in a direction north- west. Find its velocity after one minute. _ANS. The resultant force is 3^2 pounds, or 31/2 .g poundals, north. We have then Ft = m(v v t ), or 34/2.^x60 \o(v 20). Hence v = 18 1/2 . ^ + 20, or for,^ = 32. v = 834.46 ft. per sec. north. (8) A particle of mass m is moving east with a velocity v. Find the uniform force necessary to make it move north with an equal velocity in t seconds. ANS. In the time / the velocity east is zero. The impulse towards the west is then mv. In the same time we must also have_an impulse mv towards the north. The resultant impulse is then Ft = mv . |/2 northwest, or F = ^- northwest. CHAPTER III. WORK. POWER. Work. Let a uniform force Fact upon a particle which moves in any path from P l to Let the projection Pji of the path along the line of the force be denoted by/. Then the product Fp is called WORK. If the projection/ of the path is in the direction of the force, work is done by the force and Fp is positive. If the projection / is opposite to the direction of the force, work is done against the fofce and Fp is negative. Let the displacement P V P 2 = d make the angle ft with F. Then we have / = d cos 0. The work of F is then if (.he projection Pji = p is in the direction of the force F, and W= - Fp= - Fdcose if the projection /*, =/ is opposite to the direction of F. We have then, in general, W= F.dcosB= But d cos 6 is the displacement along the line of the force, and F cos 0" is the force along the displacement. We can then define WORK generally as follows: Work is the product of a uniform force by the component displacement along the line of the force (W F. dcos #); or, the product of the displacement by the component force along the line of the displacement. ( W = d . F cos 0.) COR. i. It is evident that if the displacement is at right angles to the force, the work is zero. COR. 2. We see that work is independent of time. A given uniform force and displace- ment give the same work no matter what the time occupied by the displacement. COR. 3. The work done in raising a body is equal to the weight of the body acting at the centre of mass, multiplied by the vertical displacement of the centre of mass. This work is done against the weight and is therefore negative. Also, the work of lowering a body is equal to its weight multiplied by the vertical dis- placement of the centre of mass. This work is done in the direction of the weight and is positive. 260 CHAP. Ill ] UKIT op u/ORK. EXAMPLES. 261 If 31 is the mass of the body, then mg is its weight. If d is the vertical displacement, then W mgd, and this is the same for the same weight and vertical displacement whatever the time or whatever tlic path. Unit of Work. The unit of work is then evidently the work of one unit of force with one unit of displacement. The English unit is then the foot-poundal, or a uniform force of one poundal acting through one foot; or, in gravitation units, the foot-pound, or the weight of one Ib. acting through one foot. ' This latter is of course variable, since the weight of one Ib. varies at different localities. If, then, we take-in feet and m in Ibs., gives work in foot-poundals. For foot-pounds we divide by^- and have . W = md foot-pounds. The C. G. S. unit of work is the uniform force of one dyne acting through one centi- meter. It is called an ERG. A multiple of this equal to loooooooergs, or io 7 ergs, is used in electrical measure- .ments and called a JOULE, after Dr. James Prescott Joule. Examples. (i) Find the work expended in raising 16000 Ibs. through a distance of 20 feet. ANS. 320000^ ft.-poundals, or, if g = 32, 10000 ft. -pounds, or the work of raising 10000 Ibs. one foot at a locality where _ is 32 ft.-per-sec. per sec., whatever the time or the path. (2) A body of 80 Ids. mass moves along a rough horizontal plane with a speed of 30 ft. per sec. If the retarding force of friction is constant and equal to 20 pounds, find the work done against friction in the first second and in coming to rest. ANS. For motion in a straight line with uniform acceleration, we have (page 92) for the distance described in anv time / arid for the distance described in coming to rest The force Ft in the path is . F t = mf t . In the present case this force is a retarding force of 20 pounds, and therefore Ft = 20^ poundals. Since m = 80 Ibs., we have The work, then, in any time / is - 2cg = Soft, or ft = *> F t s = 20?- (7/1 / + -ft?) ft'.-poundals, or 2o(z/i/ -ft?) ft. -pounds, and the work in coming to rest is v* v a Fts = l/Qg-- ft.-poundals, or 20 ^ ft. -pounds. 2 Jt 2ft Taking g 32 ft.-per-sec. per sec., Vi = 50 ft. per sec. and f t as found, we have for the work in one second F t s = 920 //. -pounds, and for the work in coming to rest; F$ = *- 3125 //. -pounds. a6z KINETICS OF A PARTICLE. [CHAP. III. (3) A mass of * tons (2*40 Ibs.) is pulled up too feet of an incline which rises t foot in 25 ft. of length. Taking the resistance of friction at 'jo pounds for each ton of weight, find the work done. ANS. Let a be the angle of inclination of the plane, and in the mass. The weight is then nig, and the component of the weight down the plane is mg sin a poundals. The friction is 300^- poundals also down the plane. The total force down the plane is then (tng sin a + yyog) poundals, or (; sin a + 300) pounds. The work is then FtS = (tn sin a + 300)100 ft. -founds. In the present case m = 4480 Ibs., sin a , and hence Fa = 47920 ft. -pounds. (4) Find the work done by a crane in lifting the materials for a stone wall 100 feet long, 10 feet high and 3 feet thick, the average density being fjo Ibs. per cubic foot. ANS. i 500 ooo ft. -pounds. Rate of Work Power. Work, as we have seen, is independent of time. If we raise one Ib. one foot, we do the same work whether the time of raising is one second or one minute. But the rate at which the work is done is not the same. If then we take time into consideration, the time-rate of work is called POWER. We exert more power when we raise one Ib. one foot in one second than when we raise one Ib. one foot in one minute, although we perform the same work in both cases. The unit of power is then one unit of work per unit of time. The English absolute unit of power is then one ft.-poundal per sec., and the C. G. S. absolute unit is ona erg per sec. In gravitation units we have, in English measures one foot-pound per sec., and in French measures one meter-kilogram per sec. These units are, however, inconveniently small in most cases. The gravitation unit employed in English engineering calculations is therefore taken at 550 foot-pounds per sec. This is called a HORSE POWER and is denoted by H. P. In French engineering calculations the gravitation unit employed is 75 meter-kilograms per sec. This is Called FORCE DE CHEVAL. In electrical measurements the unit adopted is io 7 ergs per sec. This is called a WATT, after James Watt. The watt is therefore one joule per sec. (page 261). Examples. (i) The area of the piston of a steam-engine is A sq. inches, the length of stroke L feet, the steiini pressure in pounds per sq. inch is P, the number of strokes per minute N. Find the work per minuti and the horse-power. ANS. Work per min. = P. L. A. N. H. P = R L " A> N . (i\ Find the work done against grtivitv, neglecting friction, in pulling a car of a. 5 tons (2240 Ms.) loaded with 30 passengers weighing *54 Ibs. each, up an incline the ends of which differ in level by 120 feet ; also the horse-power if the time is half an hour. ANS. i 226400 ft. -pounds; 1.24 horse-power. ($) In the transmission of pou an incline of r in $00 with a uniform speed of 40 miles an hour. Assuming friction to be equivalent to 6 pounds per ton, find the rate at which the engine is working. ANS. 70.4 H. P. CHAPTER IV. KINETIC FRICTION. Kinetic Friction. The friction which justs prevents motion, that is, the friction which exists when motion is just about to begin, we have called STATIC FRICTION, and we have fully discussed it already (page 220). The friction which exists after motion has taken place is called KINETIC FRICTION. Coefficient of Kinetic Friction. We have the same laws for kinetic as for static friction (page 222). The ratio of the total friction to the total normal pressure when motion is just about to begin we have called the coefficient of static friction. The same ratio after motion has taken place is the coefficient of kinetic friction. We denote the coefficient of friction in general by /*. We have then, in all cases, P = > or F=pN, where F is the total friction and N the total normal pressure. Angle of Kinetic Friction. If N is the normal pressure for a body moving on a rough surface, then ^N is the friction. The resultant reaction of the sur- face is then R, making an angle with the normal given by We call the angle which the reaction R makes with the normal the ANGLE OF FRIC- TION. For static friction it is the ANGLE OF REPOSE (page 222). Hence the coefficient p of kinetic friction is equal to the tangent of the angle of friction 0. Kinetic Friction of Pivots, Axles, Ropes, etc. The application of the equation F /sN = Ntan to pivots, axles, ropes, etc., is then precisely the same as for static friction (page 224). We have only to let JA stand for the coefficient of kinetic instead of static friction. With this change we have in each case the same value for the friction and moment of the friction as already given for static friction, Experimental Determination of Coefficient of Kinetic Sliding Friction. We may determine the coefficient of kinetic sliding friction by means of various contrivances, some of which we shall now describe. 263 264 KINETICS OF A PARTICLE. [CHAP. IV. I. BY SLED AND WEIGHT. Let a sled rest upon a horizontal plane and be dragged along _ ^ j _._, ky means of a string passing over a pulley, to jimg< ^' m ^ > C ^\ t ^ ie enc * * wmcl1 a weight is hung. In order to obtain coefficients for different substances, the runners and plane can be of the materials desired. In such an apparatus the mass of string and pulley, and friction of string and pulley, as well as rigidity of the string, should all be insignificant, or else they must be allowed for. Let us suppose them insignificant and let m be the mass of the sled, and P the suspended mass. The normal pressure is then m^, the friction ywm^, and the weight of the suspended mass is Pg. Let the acceleration of the masses P and m be f. The impressed forces acting on m are then the friction //m^, and the horizontal tension of the string P(g f}. The effective force on m is m/. By D'Alembert's principle (page 242), if we reverse the effec- tive force we have - m/ = o, m But for uniformly accelerated motion we have the distance described in any time falling from rest (page 92) I 2S Substituting this in (i) we have m mgf* If then we observe the fall s of P starting from rest in any time /, we have jw, from equation (2), for kinetic friction. If motion is just about to begin, f = o, and we have, from (i), for static friction We see, then, that the coefficient for kinetic friction is less than the coefficient for static friction. 2. BY SLED ON INCLINED PLANE If we place the sled on an vN inclined plane and the sled slides down with an acceleration f, we have the impressed forces mg sin a clown the plane, the friction ^N /.ttog cos up the plane, and the effective force taf down the plane. If we reverse the effective force, we have, by D'Alembert's T^_ principle (page 242), /*m- cos a mg sin a -f- mf o, or CHAP. IV.] BY SLED ON INCLINED PLANE. The distance s described in the time t is, as before, 265 or /=_. Substituting in (i), we have = tan or * 2S gr cos o If then we observe the distance s described by the sled starting from rest in any time , we have //, from equation (2), for kinetic friction. If motion is just about to begin, f o and we have, from (i), for static friction = tan a = -7 = tan 0, where // is the height and b the base of the plane, and

sin sin 0' where /? is the bearing angle aOb. Let r be the radius of the axle. Then we have for equilibrium Hence we have for the coefficient of kinetic friction (P- 0V sin /? A* = (P+ Q ft 266 KINETICS OF A PARTIBLE. [CHAP. IV. Friction-brake Test. The friction brake can be used for measuring the work done by an engine when working uniformly. Thus suppose the axle is driven by an engine, and that by means of a crank on the axle some machine, as, for instance, a pump, is worked. We first count the number of revolutions n per minute while the pump is in action. If then we disconnect the pump, we shall find that the axle revolves much more rapidly than before, since the only work now done by the engine is against the friction of the axle- bearing. We now apply the brake and load it at the ends with P and Q until it rests hori- zontally, and the axle is slowed up to its former speed of n revolutions per minute. In this condition the work now done against the brake friction is equal to the work before consumed by the pump, provided the engine works uniformly. But the friction F is given by or, in gravitation measure, by We have then for the work done in one revolution 2nrF, and in n revolutions per minute the work per minute is 2nrnF. Taking Fin gravitation measure or in pounds, and r in feet, this is foot-pounds per minute. If we divide by 33000, we obtain (page 262) horse-power. Hence = nrnF = nn(P - Q}1 16500 16500 ' where F, P and Q are in pounds, / and r in feet, and n is the number of revolutions per minute made while the pump was connected. Work of Axle- friction. The friction upon an axle in any case when /i is known is given on page 228. Thus for a neiv bearing we have (page 229) where R is the resultant pressure on the axle, and ft is the bearing angle. If we substitute this in the place of Fin the preceding article, we have the work per minute and for the horse-power ft 2 nrnF = 2 ituRrn - - a , sin p HP-.. rri ' Krit ft 16500 sin ft ' where R is taken in pounds, r in feet, n in revolutions per minute. If the bearing angle is small, we have ft = sin ft nearly. CHAP. IV.] COEFFICIENTS OF KINETIC SLIDING FRICTION. 267 Coefficients of Kinetic Sliding Friction. The following tables give a few values of the value of fj. as determined by experiment for kinetic sliding friction and axle-friction. COEFFICIENTS OF KINETIC SLIDING FRICTION, yU = tan 0. Substances in Contact. Condition of Surfaces and Kind of Unguent. Dry. Wet. Olive Oil. Lard. Tallow. Dry Soap. Polished and Greasy. O.o6 0.07 0.07 0.07 0.09 O.II 0.07 0.07 0.08 O.o6 O.O7 0.08 0.07 O.OQ O.II O.O6 O.O8 0.10 0.14 0.15 0.16 O.O8 O.I2 0.15 O.II 0.13 0.17 O IO 0.14 0.16 Wood on ) rj 0.36 o 48 0.25 wood 1 i] ea " o 18 0.06 O.O7 0.08 0.05 0.06 0.08 Metal on ] ;, metal ) ^ ea ? 0.24 0.31 O.2O O.26 Wood on ( Minimum J Mean 0.42 0.62 0.45 0.24 0-33 metal ) ;i . ' ' (. Maximum Hemp ropes ( On wood 0.19 O.2O Leather belts ( Raw 0.54 0.30 0.36 0.16 on wood or < Pounded metal ( Greasy 0.25 0.34 0.31 0.14 0.14 COEFFICIENTS OF AXLE-FRICTION. Dry or Slightly Greasy. Oil, Tallow, or Lard. Damp and Greasy. Ordinary Lubrication. Thorough Lubrication. O OQ7 0.049 0.054 0.054 0.054 0.054 Wrought iron on bell-metal 0.251 0.075 0.075 0-075 0.075 0.189 Cast iron on cast iron 0-137 o. 161 " " " bell-metal .... 0.194 Comparing the values in the tables just given with those in the table given on page 224, we see that the coefficient of kinetic is always less than tJie coefficient of static sliding friction. We see also that for axle-friction, in general we have for the coefficient of kinetic friction : for ordinary lubrication JJL = 0.070 to 0.080; for tJiorougJi lubrication /* = 0.054. .. Efficiency Mechanical Advantage. In a machine we have in general a "moving force" F acting at some point called the "point of application," and at another point, called the " working point," we have a " useful resistance" F' overcome through a certain distance. If there is no friction, the rate of work of the moving force must always equal the rate of work of the resistance. Owing to friction it must always be greater. The ratio of the rate of work of the useful resistance to the rate of work of the moving force is called the EFFICIENCY of the machine. It must always be a fraction less than unity, and it approaches unity the more perfect 268 KINETICS OF A PARTICLE, [CHAP. IV. the machine and the less the friction. If we denote it by e, and let v be the velocity of the moving force F, and v' the velocity of the resistance F', we have If there is no friction, e= I and Fv' = F'v. The ratio e ^ is called the MECHANICAL ADVANTAGE of the machine. If F' is greater than F, v' must be less than v in nearly the same proportion, or, if /'" v friction is disregarded, in exactly the same proportion, that is, = ^ ,. Hence the familiar maxim that " What is gained in force is lost in speed." Examples. (i) A body of So pounds mass is projected along a rough horizontal plane with a speed of 50 ft. per sec. It slides 155^8 ft. in coming to rest. Find the coefficient of kinetic sliding friction, the t etarding force of friction, and the work done against friction in coming to rest. ANS. u = ^, or, if g = 32.2 ft.-per-sec. per sec., u 0.25. Retarding force of friction is 20 Ibs.; work done, 3105.6 ft.-lbs. (2) A tody of So pounds mass is dragged along a rough horizontal plane by means of a mass of 186 pounds attached to a string passing over a pulley (page 264). // is observed to slide to feet in the first second starting from rest. Disregarding rigidity of string and mass and friction of string and pulley, find the coefficient of kinetic sliding friction, (g = J2.) ANS. u 0.25. (3) A body placed upon a rough inclined plane whose height is i foot and base 16 inches is observed to slide 6.4 inches in I he first second starting from rest. Find the coefficient of friction, (g = 32.} ANS. u = kinetic; u = 0.75 static. 3 (4) A friction brake of 'm = /j Ibs. mass, and length of 4 feet, is balanced on a rotating shaft of radius r = 6 inches, by masses of Q = fo Ibs. and P jo Ibs. to oz. Find the coefficient of kinetic friction and the friction. Also, if the shaft makes 60 revolutions per minute, find the rate of work of the friction. ANS. u = 0.07, F = 2.5 Ibs. Rate of work of friction - 7.854 ft.-lbs. per sec., or 0.01428 horse-power. (5) A screw of radius r = / inch is acted upon by a force of P - Ib. with a constant lever-arm of I i ft- and overcomes a resistance of Q = j Ibs. If the angle of the thread is a =45, find the coefficient of kinetic sliding f fiction if the number of revolutions per minute is 60. Also find the efficiency, and the acceleiation of P. Disregard the mass of the screw, and take g = 32 J ft.-per-sec. per sec. ANS. Let P be the force applied at the end of the arm /, and let the radius of the screw be r, the pitch /, and the resistance Q. If N is the sum of the normal pressures and a the inclination of the thread to the horizontal, we have J _ . N = - ' - , and the friction F= uN ~^, where u is the coefficient of friction. cos a cos a Let/ be the acceleration of P. Then the moving force is P(g /) poiiiidiils. If s is the distance passed through by P in any time /, then the work of the moving force is P(g f)s ft.-poundals. */ The resistance Q is overcome through the distance -.s. The work of over coming the resistance is then _ ^ . J^ r ft.-poundals. The friction is overcome through the distance . - . The work of overcoming the friction is then uQ rs COS a / cos a CHAP. IV.] KINETIC FRICTION-EXAMPLES. 269 The minus sign is used because work is done against friction and the resistance. The work of P(g /) must be equal and opposite to the work done against friction and the resistance. Hence the algebraic sum must be zero, or From this we have, since -^- = tan a and f = , ^ (I) and from (i), for the coefficient of kinetic friction, Pt , / 2SPI \ > = 77- cos a sin acos a i + (2) Or \ gfQr tan a) For the efficiency we have 2TCl I 7. . . . (3) 2jcl /cos* a sin a cos a If/ = o, we have equilibrium, and from (i) we have in this case or the same as already found, ex. (2), page 237. In this case (2) becomes the coefficient of static friction, PI H = -Q- cos a sin a cos a. We see from (3) that the efficiency is a maximum when sin a cos a is a maximum, or when sin a = cos a or a = 45. If is the number of revolutions per minute, the distance s described in one minute is 2itln. We have then 25 _ ytln _ itln gf ~~ 60 x 60 x g ~ Inserting in these equations the values /= i ft., r = ft., P =-lb., Q = 5 Ibs., a. = 45, n = 60, g- = 32^ ft.-per-sec. per sec., we have /i = 0.096, e = 0.84, / = 0.007, g = 0.225 ft.-per-sec. per sec. (6) A train runs on a horizontal track with the speed v\ , and by the application of brakes to the driving- wheels of the locomotive the speed is reduced to the speed v. Find the distance and time of running during the reduction of speed, disregarding all resistances other than those due to the action of the brakes. ANS. Let m be the mass of the train in pounds, Vi the initial and v the final speed in feet per second, s the distance in feet, and / the corresponding time in seconds. Let n be the number of driving-wheels braked, and R the pressure of each braked wheel on the rails in pounds. Let n>, be the coefficient of kinetic sliding friction, and /^ the coefficient of static sliding friction. ist. Let tJie brakes be set so that the wheels do not tttrn. In this case the retarding force due to friction is n/^kK pounds, or njn k Rg poundals. We have then for the retardation/ mf = nUkRg, or / = From page 92, for uniformly retarded motion the distance described is 2/ 2 7 KINETICS OF A PARTICLE. [CHAP. IV. 2d. Ltt the brakes be set so that the wheels are just on the point of slipping. In this Case the retarding force due to friction is nptKg poundals. We have then or Hence the distance described is If the train is brought to rest, we have i< = o in these equations. Now the coefficient u, for static sliding friction is always less than the coefficient * for kinetic sliding friction (page 267). We see. then, that the train unit be stopped in the least distance whtn the drakes are applied so that the wheels are just on the point of slipping but do not slip. CHAPTER V. KINETIC AND POTENTIAL ENERGY. LAW OF ENERGY. CONSERVATION OF ENERGY. EQUILIBRIUM OF A PARTICLE. Kinetic Energy. Let a particle of mass m move in any path and have the initial velocity v l at P l . In the indefinitely small timer let the particle move to P 2 , so that P l and P 2 are consecutive points, and let the velocity at . ^L^P*_ P 2 be Then the tangential acceleration is and the tangential force is v -4- v. The mean speed is - , and hence the distance described is Let F be the force on the particle. This force can be resolved into the tangential com- ponent F t already found, and the force F p in the direction of the radius of curvature. The work of the force F during the passage from P l to P 2 is equal to the sum of the works of the components (page 215). But since the path ds is at right angles to F p , the work of F p is zero, and hence the work of F is the same as the work of F t . We have then for the work done in passing from P l to P 2 when P l and P 2 are consecutive In the same way, in passing from P 2 to P s when P 2 and P 3 are consecutive z For the next two consecutive points *~ 2 and so on. 271 272 KINETICS OF A PARTICLE. [CHAP. V. If we add together all these works and denote the final velocity at P by v we have for the entire work W, if s l and s are the distances of P l and P measured along the path from any point O of the path, (i) If the final velocity v is greater than the initial velocity v lt work is done on the particle in giving it increased velocity and is positive. If v l is greater than v, work is done against the particle and is negative. In the one case we have a tangential force F t acting in the direc- tion of motion. In the other case F t is opposite to the direction of wotion. We see that equation (l) is independent of the time and path. Hence, whatever the time or path, and however the tangential force may vary, the work done in giving a particle of mass m an increase of velocity (v v^) is equal to one half the product of the mass m and the difference of the squares of the final and initial velocities. If the initial velocity is zero, then v l = o and is the work (positive) done in giving a particle of mass m the velocity v starting from rest, no matter what the time or path, or however the accelerating force may vary. Conversely, W = - -mv* is also the work (negative) done on the particle, or the work which a particle of mass M moving with a velocity v can do while being brought to rest by opposing force, no matter what the time or path or however the retarding force varies. The work which a particle or body is capable of doing is called its ENERGY. Since - mi? is, then, the work which a particle of mass m and velocity v is capable of doing by reason of its velocity, we call it the KINETIC ENERGY of the particle. We denote kinetic energy by the letter K. If then the initial velocity is *',, the initial kinetic energy is *, = \rnvf. If the final velocity is v, the final kinetic energy is i K = -mv*. 2 The work given by equation (i) can then be written ^m/Js = 3C 3C, r:= - m (ir* - V fi ( 2 ) That is, the work (positive] done on the particle is equal to the gain of kinetic energy, and inversely the work (negative) done by the particle is equal to the loss of kinetic energy. CHAP. V.] KINETIC ENERGY OF A ROTATING BODY. 273 Kinetic Energy of a Rotating Body.* Let a body have the angular velocity a. about any axis. Then the velocity of any particle at a distance r from the axis is v = ra>. The kinetic energy of the particle is then mr 2 ^. Since all the particles have the same angular velocity about the axis, we have the kinetic energy of the entire body, 3f= GT^wr*. But 2mr* is the moment of inertia /' of the body relative to the axis (page 3 1). We have then for the kinetic energy of a rotating body (3) Since the motion of a body at any instant consists in general (page 145) of translation and rotation about an axis through the centre of mass, we have for the kinetic energy of a rigid body in general (4) In all equations 5C is given in ft.-poundals if we take m in Ibs. and v in ft. per sec. For foot-pounds divide by g. Examples. ( i) A fly-wheel has a mass of 30 tons and radius of 8 ft. What work can it do in coming to rest, considering it as a disc, if it is making 20 revolutions per minute f ANS. For a disc we have for a principal axis through the centre of mass (page 45) /= - . r*. The angular velocity is oo = 2 * 2g . The kinetic energy is then 3C = * . **L = 30 x- 2240^64 _ 4x9.87 = 47l6$44 ft ._ pounda]S) 49 4 9 or, taking^- = 32 ft.-per-sec. per sec., 3C = 147392 ft.-pounds. (2) In the preceding example suppose the fly-wheel starts from rest and acquires its speed in one minute iinder the action of a constant force applied at the extremity of a crank 18 inches long. Find this force. ANS. The mean angular speed is 10 revolutions per minute. The distance s passed through by the constant force is then, since the length / of the crank is \\ ft., S= IOX27T/ = 307T ft. The kinetic energy is as before 147392 ft.-pounds. Hence the force is I473Q2 6 pounds. 10 x 27t/ y>it (3) A ball-player catches a ball moving with a velocity of jo ft. per sec. Tkt mass of the ball is S\ oz. If the space in which the ball is brought to rest is 6 inches, find the average opposing pressure and the time of stoppage. ANS. We have F t s = ^ mv*. or F t = " = IIX ^^ X2 = 859! poundals. Taking g = 32 ft.-per-sec. per sec., we have F t = 26.85 pounds. The mean velocity is = 25 ft. per sec. and * = /, or t = sec. (See example (i), page 258.) (4) If an ounce bullet with a velocity of Soo ft. per sec. strikes a man in " bullet-proof" armor and the bullet is brought to rest in a space of six inches, find the average pressure and time of stoppage, (g = 32.) ANS. Pressure = 1250 pounds; time = ^ sec. * This article properly belongs to the discussion of the kinetics of a rigid body and not of a particle. It Is, however, such a direct result of the preceding that we give it here. a 74 KINETICS OF A PARTICLE. [CHAP. V. (5) Find the tension of a rope which pulls a car of S tons up a smooth incline of i in 5 and causes a uni- form acceleration of 3 ft.-per-sec. per sec. If the rope breaks when the speed is 48 ft. per sec. , how far will the car continue to move up the incline f (f = 32.} ANS. 5264 pounds; 1 80 feet. (6) A bullet of a\ os. leaves a gun with a velocity of iSSoft. per sec. The length of barrel is a\ ft. Find the average force of the powder. ( g = 3*. ) ANS. 2346 pounds. (7) A train of 200 tons, starting from rest, acquires a speed of 40 miles an hour in three minutes. Find the effective moving force, assuming it uniform. ANS. 2.03 tons. (8) An engine exerts on a car weighing 20000 Ibs. a net pull of 2 !bs. per ton of 2000 Ibs. Find the energy of the car after going 2\ miles. If shunted onto a level side track where friction is 10 pounds per ton, how far will it run f If this side track has a one per cent grade, how far will it run ? ANS. 264000 ft.-lbs.; one-half mile; one sixth mile. Potential Energy. Let a particle be constrained to move in any path from />, to P. Let F be the resultant of all those forces acting upon the particle which depend solely upon the position of the particle. For an indefinitely small displacement ds in the path F can be considered uniform, and its work is F . dp, where dp is the component displacement along F. We can resolve F into a tangential component F t and a normal component ./y In the displacement ds from P l to P v the work of F f is zero. Since the work of the resultant is equal to the algebraic sum of the works of the components (page 2 15), we have F . dp - F t . ds. This can also be proved as follows: If is the angle of F with the tangent, we have dp = ils cos t) and F . dp F . ds cos 0. But /''cos 6 = F t . Hence F . dp = F t . ds. The force F and hence F t depends solely upon position. It is in general variable and changes in magnitude and direction as the particle changes its- position. For any two consecutive points we have, however, F uniform. If, then, j, is the distance of 7^, and s of P, measured along the path from any point O,> we have for the entire work on the particle in passing from P l to P, due to change of position only, ^F. dp = 2*F t .ds (5) The work which a particle is thus capable of doing by reason of change of position only is called its POTENTIAL ENERGY. The potential energy of a particle or body, then, is the work it can do by reason of change of position or configuration under the action of forces which depend solely upon such change. Thus a mass suspended at a distance above the earth can do work if released, by reason of the force of gravity, which is a force depending solely upon position. This work is potential energy. A bent spring can do work when released, by reason of the elastic forces between its particles, which depend solely upon configuration. This work is potential energy. CHAP. V.] TOTAL ENERGY. 275 We denote potential energy by the letter 2. The initial potential energy of the particle at /\ in the figure, page 274, is then \ = 2^ F . dp = y ' F t . ds. The final potential energy at P is 2 = 2 s F. dp = 2* F t . ds. The work done on the particle due to change of position only is then \-* = 2 t F.dp-^F t .d S ' . . (6) That is, the work (positive) done on the particle by reason of change uf position only is equal to the loss of potential energy, or, inversely, the work (negative) done by the particle by reason of change of position only is equal to the gain of pi tential energy. Total Energy. The total energy of the particle at any instant is the sum of its poten- tial and kinetic energy at that instant. Let us denote the total energy by &. Then the initial energy is t and the final energy is = % -f -mv 2 = -f- 3C. Law of Energy. When a particle or body moves it generally encounters resistance due to friction, resistance of the air, etc. Such forces do not depend upon position solely, but upon velocity or upon the character of the surfaces in contact, upon pressure between these surfaces, etc. They always resist change of motion, and we call them therefore resistances. Let, then, in be the mass of a particle constrained to move in any path from P l to P, and let F be the resultant of all those forces acting on the par- ticle which depend solely upon its position, N the normal pressure of the path upon the particle and R the resultant of all the resistances. We can resolve Finto F t and F p tan- o gent and normal to the path. So, also, R can be resolved into R t and R p , and R t is always opposite to the direction of motion. These are impressed forces. If the particle has the acceleration /, this can be resolved into f t and f p , and the components of the effective force ;;//~are then mf t and mf p . Now by* D'Alembert's principle (page 242) the impressed and reversed effective forces .an be treated as a system of forces in equilibrium, Hence F t - R t - mf t = o. Multiplying by ds, we have F t ds = mf t . 276 KINETICS Of A PARTICLE. [CHAP. V If, then, 5, is the distanceof /> lf and s olP, measured along the path from any point O, we have for the entire work done by F irom P l to P, whatever the path, and however the forces may vary in magnitude or direction, But we have seen from equation (6) that 2' t F. dp = X,^ . fc = 2, the loss of potential energy. We have also seen from equation (2), page 272, that or the gain of kinetic energy. We have then 5 1 2=3C 3Cj+ 2 tl R t .ds ......... (8) That is, the lass of potential energy is equal to the gain of kinetic energy plus the work of overcoming all resistances. Since, as we have seen, we can write Sj = B, -f 3C, , & = 2 + M. Equation (8) can be written & l -& = 2 f R t .ds ........... (9) That is, the loss of energy is equal to the work done in overcoming resistances. This is known as the law of energy. -. Conservation of Energy. If there are no resistances, equation (9) becomes j = o. or 2, - $ = K - 5C, ........ (10) That is, if there are no resistances, but only forces which depend solely upon position, there is no loss of energy, or the loss of potential is equal to the gain of kinetic energy, and con- versely, the gain of potential is equal to the loss of kinetic energy. This is called the law of conservation of energy, and hence forces which depend solely upon position are called conservative forces, because for them the law of conservation holds. The force of gravity upon a particle depends solely upon the position of the particle and is therefore a conservative force. So is the elastic force of a spring which depends upon con- figuration only. So, also, are the forces of nature generally. For such we have the law of conservation of energy as given by equation (10). Forces which do not depend upon position are called non-conservative forces. The resistance due to friction, and in general all resistances to motion, do not depend upon position and are therefore non-conservative. The resultant of such forces always acts opposite to the direction of motion. For such we have the law of energy as given by equa- tion (8). CHAP. V.] EQUILIBRIUM OF A PARTICLE. 277 Equilibrium of a Particle. If a particle is at rest under the action of forces, it is said to be in static equilibrium. If it moves with uniform speed in a straight line, that is, with uniform velocity, it is said to be in molecular equilibrium. Let F x , F y , F z be the components in any three rectangular directions of all the con- servative forces acting upon a particle ; let R t be the resultant non-conservative force, and jR A , R y , R z its components. Then for any possible indefinitely small displacement ds, real or virtual, we have the work of the conservative forces, or the loss of potential energy 2j - 2 = F x dx + F y dy -f F s dz, and by the law of energy, equation (8), $! - 2 = F x dx + F y dy + F s dz = -m(v* - ^ 2 ) + R t ds. But R t ds R x dx 4- R v dy -f R z dz. Hence 2, - * - R t ds = (F x - R x }dx + (F y - R y }dy + (F, - R^dz = m(v> - vf). For static equilibrium v = o, 7^ = o, and for molecular equilibrium v = v r Hence in all cases of equilibrium, static or molecular, (F x - R x yx + (F y - R y }dy + (F. - R^dz = o. Since dx, dy, dz are not zero, we must have F x - R y = o, F y - R y = o, F 2 - R 2 o. That is, in all cases of equilibrium, static or molecular, the forces, conservative and non-con- servative, acting upon the particle must constitute a system of forces in equilibrium. Since in all cases of equilibrium, static or molecular, m(i? z^ 2 ), or the change of. kinetic energy during any indefinitely small displacement is zero, the kinetic energy is zero for static equilibrium, and either a maximum or a minimum when the particle is in molecular equilibrium. Stable and Unstable Equilibrium of a Particle. The equilibrium of a particle is said to be stable if, when at rest or when supposed to be at rest, it would return after any possi- ble indefinitely small displacement to its original position. Now in thus returning work must be done by the conservative forces, and hence potential energy lost. In stable equilibrium, then, the potential energy is a minimum. If the particle would not return to its original position, the equilibrium is said to be unstable. In such case work must be done against the conservative forces in order to bring it back, and potential energy would be gained. In unstable equilibrium, then, the potential energy is a maximum. If potential energy is neither lost nor gained, the equilibrium is said to be indifferent, and the particle in its new position is still in equilibrium. Now when potential energy is lost, kinetic energy is gained, and conversely. Hence if a particle is in equilibrium, static or molecular, all the acting forces must con- stitute a system of forces in equilibrium. If the potential energy is a minimum, the equilib- 2 7 8 KINETICS OF A PARTICLE. [CHAP. V. rium is stable and the kinetic energy, if any, a maximum. If the potential energy is a maximum, the equilibrium is unstable and t/ie kinetic energy, if any, a minimum. If the potential energy does not change, the equilibrium is indifferent and the kinetic energy, if any, does not change. Illustration. Take the case of a pendulum. Let / be the length and m the mass of the bob. Disregard the mass of the string. When the bob is at the lowest point the potential energy is mg . //. For every possible displacement h increases and the poten- tial energy increases. The potential energy is then a minimum. This, then, is a position of stable equilibrium. If the bob is at rest, it is in stable static equilibrium. If the pendulum swings, when it arrives at its lowest point the kinetic energy is a maximum and the potential a minimum, and at the instant it is in stable kinetic equilibrium. When the bob is at the highest point the potential energy is a maximum. This, then, is a position of unstable equilibrium. If the bob is at rest, it is in unstable static equilibrium. If in motion, when it arrives at the highest point the kinetic energy is a minimum and the potential a maximum, and at this instant it is in unstable kinetic equilibrium. Change of Potential Energy. In order to apply equations (8) or (10) it will be con- venient to find the value of the change of potential energy 5, 2 in special cases. (a) FORCE UNIFORM. If a particle is acted upon by a uniform conservative force F, the work done during a change of position from P l to P is, by equation (6), page 275, ^ - 2 = F.p= F. ,, P 2 , P 3 . . . Pbc any path from I\ to P, and let the posi- tions />,, 7> 2 , />,, etc., be consecutive. Draw OP l , OP 2 , OP 3 , etc., d/\ \ and with O as a centre describe arcs of- circles through P 2 , P 3 , etc., intersecting OP } at b, c, d, etc. , p The force F z at P 2 acting towards O may be considered uniform / * for the indefinitely small displacement P 2 to P 3 . P The work, then, from P 2 to P s is F 2 X P 2 n = F 2 X be. Every element of the path may be treated in the same way. Thus the work from P { to P 2 is ^, X P,b; from PS to P, F, X cd, and so on. The entire work from />, to P is then F, X P v b + F 2 X be + F^ X cd-\ etc. = 2F. dr, where dr is the elementary distance along the radius vector OP { between any two arcs. CHAP. V.] CHANGE OF POTENTIAL ENERGY. 279 The work, then, necessary to move the particle from P l to P by any path, under the influence of a central force always directed towards O, is equal to that necessary to move it from P l to d in the straight line OP r This work is independent of the path and depends only upon the initial and final posi- tions and the magnitude of the force. If this force depends only upon position, this work is the loss of potential energy and we have in general ... (2) If the central force is opposite to dr, we have gain of potential energy, or <% - \=^F .dr. (c] CENTRAL FORCE CONSTANT. If the magnitude of the central force Fis constant, we have, from (2), V-. * = F2dr = F(r, - r), . . . ' ..... (3) where r l is the initial and r the final radius vector from O. If r l is less than r, we have gain of potential energy, or / % - \ = F(r - r,}. (d] CENTRAL FORCE PROPORTIONAL TO DISTANCE FROM CENTRE. If the magnitude of the central force varies directly as the distance from the centre O, let F Q be its known magnitude at a given distance r . Then the force at any other distance r is given by F:F ::r:r , or F = - F c o- We have then at P l the force F l = F , at P 2 the force F 2 = F , at P 3 the force P , and so on. ; 'o The average force between P l and P 2 is then --- 2 = . - - - , and the work from 2 fg 2 P l to P 2 is then, from (2), % 2 ^ * In the same way we have for the Work from P 2 to P 3 and so on. The entire work from P l to P is then the loss of potential energy given by where r l is the initial and r. z the final radius vector from O. 2 So KINETICS OF A PARTICLE. If r t is less than r, we have gain of potential energy, or [CHAP. V. 2r o f>) CENTRAL FORCE INVERSELY PROPORTIONAL TO THE SQUARE OF THE DISTANCE FROM CERTRE. If the magnitude of the central force varies inversely as the square of the distance from the centre, let F be its known magnitude at a given distance r . Then the force F at any point is given by r * r* We have then at P l the force F l = ^F Qt and at P t the force F 2 = ~F Q , and so on. For the indefinitely small displacement from P l to P 2 , r, and r 2 are equal, and we have r, 2 = r,r 2 and r 2 2 = rj v Hence the force at P l can be written F l = -2 ?, and the force at n 77 __ 0. 2 i 'j ~ The work, then, from (3), from F l to F t is From P 2 to P 3 we have in the same way the work and so on. The entire work from P l to P is then the loss of potential energy and is given by If r, is less than r, we have gain of potential energy, or Exam pies. -( i) ervation of energy to a falling body. ANS. Let / he the mass of the body, v\ its initial and v its final velocity. Then the gain of kinetic energy is We have from equation (5), for central force varying inversely as the square of ihe distance, the loss of potential energy. 4, -2 = ^,'(--'-11 \ r r^' where r, is the initial distance and r the final distance from the centre of force, and F, is the known force at a yiven distance r . In the present case CHAP. V.] ENERG Y EXAMPLES. 281 the centre of force is the centre of the earth, r is the radius of the earth, and Fa at the earth s surface :s mg. We have then But ri r is the distance described, (j, s\, if Si is the initial and s the final distance from the centre of the earth. We have, by the principle of conservation of energy, gain of kinetic equals loss of potential energy, or 3C - 3Ci = 2, - This is the same as equation (5), page 113. If we suppose the force to be uniform and equal to mg, as it practically is at the surface of the earth, we have, from equation (i), page 278, Si S = tng(si - s), and hence m . , __ nx , V v? + - s), which is equation (7), page 92. (2) Apply the law of conservation of energy to a projectile. ANS. Let the force be vertical and equal to mg. The initial energy g at O is ^-. The energy g at Jis v 1 kinetic and mgy potential. If there is no loss of energy, &i = , or ^=^ + , or v* = vS-*g y . This is equation (11), page 104. (3) Apply the law of conservation of energy to the simple pendulum. ANS. Let the velocity at P\ be zero!, Then the initial energy at />, is 1 = m glj _ / cos 0,) potential, relative to A. At any point P where the velocity is v we have the energy , ^~ kinetic and mg(l I cos 0) potential. If there is no loss of energy = @, or Hence mg(l - I cos 0.) = ~ + mg(l /cos 9). s 9 cos 61). But /(cos 9 - cos9,) is the fall from Pi to P. Hence the velocity is the same as for a particle falling through the vertical distance from Pi to P. (4) Let a spring whose unstrained length is AB be fixed at the end B and compressed from A to C, where it presses against a body of mass m. Disregarding the mass of the spring, find Jhe motion when released. ANS. Let the force at any distance x from A be F, and at the given distance s = ACirom A be Fi. Then we have F : x : : F\ : s, or F = *" The initial energy is &! = FiS, all potential. The energy g at .F is Fx potential and - kinetic. If there is no loss of energy, F, = Fx + mv* 282 KINETICS OF A PARTICLE. [CHAP. V. Substituting the value of F t we have We see then, from page 126, that the motion is simple harmonic. (5) A vessel containing water has a small orifice whose centre is at a distance h below the surface. The water flows in at top with a vertical velocity v y , and the top area of cross-section is A. The water flows out of the orifice with a velocity v in any direction, and the area of the orifice at right angles to v is a. If water flows in as fast as it flows out, so that the water-level remains constant, find the theoretic velocity of efflux v, disregard- ing friction and all resistances. ANS. The quantity of water flowing in per second is Av y , and the quantity flowing out as av. By the conditions, for constant level we must have Av 3 = av, or v y = -jv. The initial energy &, of a particle of mass m at the top level is mgh potential and - kinetic, or mvl , = mgh + 7 . When this particle reaches the orifice its energy g is all kinetic and = . If no energy is lost, Si = &, or mgh + mv\ mv* Substituting the value of v y , we have (I) If a is very small compared to A, we can practically neglect the fraction ^ relative to i, and we then have The theoretic velocity of efflux in this case is the same as for a particle falling freely through the distance h. This is known as Torricelli's principle. If y is the density or mass of a cubic unit of water, in a very small lime r the mass discharged is m = yavr. The kinetic energy at efflux is then the work done in giving the mass m the velocity v, or mv* yav*r work= =^-. If the mass m acquires the velocity v in a very short time r from rest, the average velocity is - and the distance is . If we divide the work by this distance, we have for the uniform force /''during the short time r in the direction of v or, from (i), F = yav*. F = y ^ \ poundals. For F in gravitation measure (3) CHAP. V.] From (3), for a small relative to A EXAMPLES. 283 F = iyah. or the weight of a column of water whose base is the area a of the orifice and whose height is twice the distance h. If a is the angle of v with the horizontal, we have for the horizontal component of F F x = 2yah . cos a (4) Since action and reaction are equal, this is the horizontal pressure on the side of the vessel in a direction opposite to v cos a. We can also obtain (2) directly by the principle of impulse (page 257). Thus mv fv = mv, or f 1 = . Substituting m = yavr, we have at once F = yav\ (6) In the preceding example let the vessel move horizontally with the uniform velocity v x , while the water flows in at the top with a vertical velocity v y , the top area being A, and is discharged with the velocity z. making an angle a. with the horizontal, the area of orifice at right angles to v being a. ANS. We have for constant level, as before, Av y = av, and, disregarding the motion of the vessel, we have, as in the preceding example, = mgh + mv 2 and therefore, just as in the preceding example, we have for S, = 8 . mvl mv* and for the velocity of efflux relative to the vessel. just as before. Now the absolute velocity at A is given by (1) Hence the total energy at A is &> = + mgh = (v'x + Vy) + mgh. The absolute velocity at a is given by vl = Vx + v* 2w x cos a. Hence the total energy at a is is = = (7/1 + V* 227Z/ X COS a). (2) (3) Equation (2) gives the work the mass m at A can do, and equation (3) the work this mass at a can do after leaving the vessel. The difference i is then the work done upon the vessel. If* then we subtract (3) from (2) and reduce by (i), we obtain for the work done upon the vessel work = mw x cos a. 284 KINETICS OF A PARTICLE. [CHAP. V. In a very short time r we have m = yavT, where y is the density or mass of a cubic unit of water. Hence work = yav*v x T cos a (4) But the distance in the time r is v x r. Hence, dividing (4) by v x r, we have for the horizontal force F K on the vessel Fx = yav* cos a. This is the same result as obtained in the preceding example. Thus, if we put v ^zgh, we have F K = 2yagh . cos a poundals, or, in gravitation measures, Fx = 2yah .cos. a, just as before. (7) A horizontal stream of water whose cross-section is a and velocity 7/1 meets a surface moving in the same direction with a velocity v. Find the pressure, disregarding friction and resistances. ANS. Let the water pass off the surface in a direction fi ' making the angle or with the direction of motion. In any small \ a yy time r the mass of water discharged is m = y.iv^r, where y is the density or mass of a cubic unit of water. The initial kinetic energy of m is then * mv\* yav*r >-* : ~ : ~^~ The velocity of the water as it leaves the surface is v t v y >" v relative to the surface. The velocity of the surface is v. The resultant velocity is then given by v r * = (t/i z>)* + v* + 2(vi v)v cos a. The final kinetic energy of the mass m is then ,,, m yav\r r & = V r ' = ' [zv 2Wi + 2V + 2(V\ V)V COS , and hence = J J i Sr dr cos Q dB d. . . .. (8, Examples. (I) Particles of masses 3.928, 39.28 and 392.8 kilograms are situated at three of the corners of a square -whose side is I metre. Find the potential at the fourth corner. ANS. n = J2 , and the astronomical unit of mass is 3928 grams (page 206). Hence II = 1.087 ergs. (2) Find the potential and attraction of a homogeneous circular ring of radius r upon a point C on the Perpendicular to its plane through its centre O. ANS. Let the distance of the point C from the centre O be x. Then the distance Cm for any particle of the ring is fV*-+ x\ If the linear density of the ring is S, the mass is 2itrS, and therefore the potential H = ==. fV 1 + x* The attraction upon a unit mass at C parallel to the plane of the ring is then - -, taking the astronomical unit of mass (page 206). But r is constant and hence , = o. That is the sum of the component attractions of the elements of the ring in the plane of the ring is zero. The attraction in the direction CO upon a unit mass at C taking the astronomical unit of mass, is F x = j- = - , the minus sign denoting attraction or force towards the centre O. If we multiply the value of II and F x by kM, where M is the mass of any particle at O, and k is % (page 205), we have the result for any mass M at C, using the ordinary unit of mass. When x o. the potential at the centre of the ring is n = 2*8. (3) Find the potential and attraction of a circular arc at its centre. C ANS. Let 6 be the angle subtended by any portion of the arc estimated from its middle point D. The length of any element is rdQ, its mass is rS dQ, where 5 is the linear 0\\ density, and the potential is f + ^ J-a ' where a is the angle A CD. D This is independent of the radius r of the arc. The attraction of any element whose mass is rSdQ for a unit mass at C, using the astronomical unit of mass (page 206). is j-. The component of this at right angles to CD is ; sin 0, and along CD, 5 cos We have then for the resultant attraction at right angles to CD S f** 7 * = - I <& sin 6 = T J - a CHAP. VI.] EXAMPLES. 291 and for the resultant attraction along CD -if*' %y a Fy=-~- I JQcosO = -^sin, the minus sign denoting attraction. For any mass M at C, using the ordinary unit of mass, we multiply by kM, where k = (page 205). (4) Find the potential and attraction of a straight line upon an external point. ANS. Let AB be the line and C the point. Drop the perpendicular CD, take D as origin, and let CD = y. Then for any point P of the line distant DP = x we have CP = r = Vf + ** Let 5 be the linear density. Then the mass of any element is $dx, and the potential is n= Taking this between the limits of x DA = + a and x = DB = + b, we have n = s log + y -? ~L . The component attraction upon unit mass at C in the direction of the line is dTL 8 Introducing the limits + a and + ^. For the component attraction upon the unit mass at C perpendicular to the line we have dy Introducing the limits + a and + b, we have and -.-L\ CB r CB'-~CD~' ~CA = ~CD' *Crf-8b, =CZ?sin# F* = ^^f cos /* ~ c s A /> = -^ (sin a - sii , V d Let the angle Z>C4 = a, DCB = ft, ACB = ft a = r . Then I COS klNETlCS OF A PARTICLE. The resultant force upon unit mass at 6' is then [CHAP. VI. The tangent of the angle which this resultant force makes with the vertical is Ft _ cos ft cos a _ a + ft F 9 sin a sin ft 2 Therefore the resultant attraction bisects the angle ACB. The results are all for unit mass at C and astronomical unit of mass (page 206). For mass M at Cand zrS ordinary unit of mass we have only to multiply F x , F y , R by kM, where k = - (page 205). (5) Find the potential and attraction for a circular disc at a point on the perpendicular to its plane through its centre, ANS. Let^ be the radius and dy the thickness of an elementary ring, and $ the surface density. Then the mass of the elementary ring is 2nSydy. If the distance OC is x, we have for the potential of the disc n = which for the limits^ = R = radius of disk,and^ = o becomes For the centre of the disc this becomes 2itSR. The potential, then, is constant for x constant. The com- ponent force upon unit mass at C parallel to the disc is then -jj^ o. For the component force along OC we have the minus sign denoting attraction. For mass M at <9and ordinary unit of mass we have only to multiply F x by kM, where k = ^- (page 205). (6) Find the potential and attraction at the vertex for a right tvne with circular base. ANS. Let the half angle at the vertex, OCB, of the preceding figure be 0. Then * = cos 6. Hence, from the preceding example, we see that the attraction of all circular elementary slices for a particle at C is the same, and equal to 2*8ax(i cos 6). The total attraction is then F 3t =^=- 2it8x(i - cos 0), which for the limits h and o becomes F M = - 2itSh(\ - cos 6). For mass M at C and ordinary units of mass we have only to multiply by kM, where k = ^~ (page 25). We have then H = - *6x*(\ - cos S), or for limits o and h n *= Tt8h\ i - cos 6). CHAP. VI.] EXAMPLES, 293 (7) Find the potential and attraction of a spherical shell at any point. ANS. Let r be the radius of the shell, / its thickness, p the distance of the point B from the centre C, AB = a = the distance of any point of the shell from the given point B. Take the origin at C, and let BC coincide with the axis of Y. Then the elementary volume is dv = r*t sin and a = 4/r 5 + p' 2rp cos 0. Hence, if $ is the density, H = <5/r* '/V I yr* sin QdQ d 2rp cos + Integrating first with respect to 0, we have and then with respect to 9, H = - (r* - 2rp + When the point B is within the shell p < r, and when it is outside of the shell p > r. In the first case, when B is within the shell, we have P)- (r- p)] = where m is the mass of the shell. The resultant force of attraction is then R = -j =o. This is the same dp result as in example (i), page 207. In the second case, when B is outside the shell, we have where m is the mass of the shell. The resultant force of attraction then is R = = -^ , where the dp p 1 minus sign denotes attraction. If we take the mass M at B and use the ordinary unit of mass, we have R = k . This is the same P result as already obtained, page 205. ( 8) Find the potential and attraction of a thick homogeneous spherical shell at any point. ANS. Let the external radius be r\ , and the internal radius r y . Then in the preceding example we can out / = dr, and we have for the potential of that part of the shell outside of the spherical surface containing the point / 1 4jt$rdr, and for the potential of that part of the shell inside of the spherical surface contain- JP ln -he,point / 2=21-=-. Hence n = rdr 294 KINETICS OF A PARTICLE. [C HAP. VI The mass of the shell is m = ^(r, 1 - r, f ). If the point is wholly within the shell, H and if the thickness is very small, r\ r* = t and r ( + r = 2r, and n = 45/r, as found in the preceding nple. Also, the attraction is R = -^- = dp If the point is wholly without the shell, example. Also, the attraction is fi = = o, as found in the preceding example. and R = j, as found in the preceding example. If the shell becomes a sphere, r a = o and r t = r, and we have for an interior point n = 27T<5r a where m = -nr*S. For an exterior point For p = r we have in both cases _ 3" Hence we see that for a homogeneous sphere we may take the potential and attraction at any external point as though the whole mass were concentrated at the centre, while the attraction at an interior point is directly proportional to the distance from the centre. The first result has been proved, page 205 ; the second in example (2), page 207. (9) Find the potential and attraction for a cylinder of length J^and radius- R for a point on the axis at a distance d from the nearest end, ANS. We have found, example (5). for the component force along the axis of a circular disk 2ird( i -- * -J ; the component at right angles to the axis being zero. If the disk has a thickness dx, we have for a cylinder ^= = F x = - 2x8 ujc ^ or, taking the limits - , etc., are the velocities v l , v 2 , etc. of the particles in this direction. Hence _ WjZ'j -f- in 2 v 2 -\- w 3 ^ 3 -f- etc. ~^~ ^^ or But *,*>,, m 2 v 2 , etc., are the momentums of the particles in any given direction, and 2mi> i$ then the momentum of the system in that direction. Hence, for any material system, the momentum in a,ny direction is the same as for a par- ticle of mass equal to the mass of the system moving with the velocity of the centre of mass in that direction. Also, the velocity of the centre of mass in any direction is equal to the momentum of the system in that direction divided by the mass of the system. From (i) we have 2mv ~v2m = o = But 2m(v v) is the momentum of the system in any direction relative to the centre of mass. Hence the momentum of a material system in any direction relative to the centre of mass is zero. CHAP - L ] ACCELERATION OF THE CENTRE OF M4SS. 299 Acceleration of the Centre of Mass. We find the acceleration of the centre of mass in precisely the same way. Thus if v l is the initial and v the final velocity of the centre of mass of any material system in any given direction during an indefinitely small time r, we have, from (i), Subtracting the first from the second and dividing by r, we obtain If r is indefinitely small, ~ is the acceleration/ of the centre of mass in the given direction, and = is the sum 2mf of all the effective forces in that direction. Hence (2) That is, for any material system the effective force in any direction is the same as for a particle of mass equal to that of the system moving witli the acceleration of the centre of mass in that direction. Also, the acceleration of the centre of mass in any direction is equal to the acceleration of the system in that direction divided by the mass of the system. From (2) we have 2mf f2m o = 2m(f /). But ^m(f /) is the effective force of the system relative to the centre of mass. Hence the effective force of a system in any direction relative to the centre of mass is zero. Motion of Centre of Mass. Let F lt F 2 , F 3 , etc., be the components in any direction of all the impressed or external forces acting upon any material system, and m^f^ m^f 2 ,' > 3 / 3 , etc., the components in the same direction of all the effective forces. Then, by D'Alembert's principle (page 297), we have 2F2mf=o, or 2F=2mf=mf^ ...... (3) Hence the motion of the centre of mass of any material system is not affected by the internal forces or stresses between the particles of the system, but only by the external or impressed forces. Also, the acceleration of the centre of mass is the same as for a particle of mass equal to the mass of the system, all the impressed forces being transferred to this particle without change of magnitude or direction. This principle we have already deduced for a rigid body (page 175). We see that it holds for any material system, rigid or not. Conservation of Centre of Mass. If, then, there are no external or impressed forces, 2F = o, f=o, and the motion of the centre of mass cannot change. If at rest, it remains at rest. If in motion, it moves with uniform speed in a straight line, no matter what the internal forces or stresses may be. 3 KINETICS OF A MATERIAL SYSTEM. J('HAP. I. That is, no material system can of itself and ivithout the action of external forces change the motion of its centre of mass. This is called the principle of conservation of centre of mass. Conservation of Momentum. If z;, is the initial and v the final velocity of a particle in any direction during an indefinitely small timer, the acceleration / in that direction is v if m(v _ v "\ - and the effective force mf in that direction is -- -. We have then, from (3), We see, then, that the change of momentum 2m(v v^) in any direction of a material system is not affected by the internal forces or stresses, but only by the impressed forces. If then the impressed forces are zero, 2m(v v^) is zero, and the momentum of the system in any direction does not change. That is, no material system can of itself and without the action of external forces change its momentum in any direction. This is called the principle of conservation of momentum. Moment of Momentum. The moment of momentum of a particle relative to a given point or axis is the product of its mass by the moment of its velocity 'V (P a S e 89) relative to that point or that axis. Thus let m be the mass of a particle, v its velocity, / the lever- arm or perpendicular from any point O' upon the direction of v. Then the moment of the velocity (page 89) is vl. Or if r .is the radius vector of m, e the angle of v with r, and o> the angular velocity relative to O', rco = v sin e and -v = -r- But /= r sin e. Hence sm e. vl = vr sin e = r?co. The moment of momentum of the particle relative to O' is then mvl = Its line representative is a straight line O' M at right angles to the plane of v and /, whose magnitude is mvl = mr^aa, and whose direction is such that looking along it in its direction, as shown by the arrow in the figure, the rotation is seen clockwise. By " direction " of moment of momentum we always mean the direction of its line representative. We see, then, that the moment of momentum of a particle relative to a point is repre- sented in magnitude and direction by a straight line of definite magnitude and direction through that point, just like moment of velocity (page 89), and the same principles apply. The same holds for moment of momentum relative to an axis. We can therefore combine and resolve moment of momentum just like moment of velocity, or just like velocity, acceleration or force, and can find the resultant for any num- ber of concurring line representatives just as for force. Moment of Momentum for a System. For a material system each particle has its own velocity, and for any point O' its own moment of momentum of definite magnitude and direction, given by its line representative through O', as in the preceding figure. For the entire system we have then a number of line representatives concurring at O' t and the resultant of all these is the moment of momentum of the system relative to '. CHAP. I.] ACCELERATION OF MOMENT OF MOMENTUM. 301 The moment of momentum of a material system relative to any point reduces, then, to a resultant moment of momentum of definite magnitude about a definite axis through that point. Let this axis be the axis of Z', and let O'X', O' Y' be the other two rectangular axes. Let ^, y, ~z be the co-ordinates of the centre of mass O. With O as origin take parallel axes OX, OY, OZ, and let x, y, z be the co-ordinates of any particle P relative to O. Then the co-ordinates of P for origin O' are ^ -(- x, ~y -f- y, ~z -j- z. Let v x , v y , v z be the components of the velocity of the centre of mass O. Then the velocity components of P are v xV* yz v y =i ~Vy~^~ Xa) " where GO,, is the angular velocity relative to the axis OZ or O'Z' . The moment of momentum for a particle of mass m at P is then given by mv y ( * -f- x) mv x ( y+y). Substituting the values of v x ,a.nd v y , we have for the entire system the moment of momentum M = 2mv y ( x + x} 2mv x (y -j- y) , + 2m( xx -j- yy)oo,. But 2m = m = mass of the system, v x and v y are constant, and 2mx = o, 2my = o, since O is the centre of mass. Hence M = mv But 2m(x* -{-y^oo, is the moment of momentum of the system about an axis parallel to the given axis, through the centre of mass O, and mzyir miv x y is the moment of momentum of a particle of mass m at the centre of mass. Hence the moment of momentum of a material system about a given axis is equal to the moment of momentum about a parallel axis through the centre of mass plus the moment of momentum of a particle of mass equal to t/ie mass of the system, situated at the centre of mass. Acceleration of Moment of Momentum. If v l is the initial and v the final velocity of a particle of mass m, in any given direction, during an indefinitely small time r, then - is the acceleration /* that direction, and is the effective force in that direction. The moment of this force relative to any point O' is then _ m(v-vjt where r is the radius vector, GO^ and GO are the initial and final angular velocities and a is the angular acceleration. The line representative is a straight line O' M at right angles to the plane of /"and /, whose magnitude is mfl and whose direction is such that, looking along it in its direction as shown by its arrow in the figure, the rotation is seen clockwise. By " direc- tion " of moment of a force we always mean the direction of its line representative. 302 KINETICS OF A MATERIAL SYSTEM. [CHAP. I. But is the acceleration of the moment of momentum in the direction O'M of the particle relative to O'. For the entire system we have for the acceleration of moment of momentum in this direction But, by D'Alembert's principle (page 297), the external or impressed forces and the reversed effective forces form a system of forces in equilibrium. Hence 2FI = 2mfl, or (4) That is, for a particle or for a material system of particles the acceleration of the moment of momentum in any direction relative to any point is equal to the moment in that direction of all the impressed forces relative to that point. Conservation of Moment of Momentum. If now 2F/= o, we have or the change of moment of momentum in any direction relative to any point is zero. But we have seen (page 301) that the moment of momentum of a particle or of a system relative to any point has a definite magnitude and direction. Its projection, then, in any direction does not change when 2F/ = o, and it is therefore invariable in magnitude and direction. This is called the principle of conservation of moment of momentum, and it may be expressed generally as follows : When the resultant moment 2FI of all the impressed forces acting upon a particle or upon a material system, relative to any point or any axis, is always zero, the moment of momentum relative to tliat point or axis does not change either in magnitude or direction, and its line repre- sentative is then a straight line of constant magnitude and invariable direction, passing through that point or coinciding with that axis. Now 2FI is always zero for all points when there are no impressed forces ; also when all the impressed forces are always in equilibrium. In both cases the principle of conserva- tion of moment of momentum holds for any point, and we have an invariable axis for any point. Also, 2FI is always zero for any one given point when all the impressed forces always pass through that point or when they always reduce to a single resultant through that point. In this case the principle holds for that point, and we have an invariable axis through that point. Also, 2FI is always zero for any axis when all the impressed forces always pass through that axis or are always parallel to that axis. In this case the principle holds for this axis. Invariable Axis and Plane. In all these cases we have an invariable axis either through any point, or through a given point, or coinciding with a given axis. But from the principle of page 301 we see that in all cases we have a parallel axis through the centre of mass, whether the centre of mass is fixed or not. ' This axis through the centre of mass, invariable in direction, is called the INVARIABLE AXIS of the system, and the plane through the centre of mass at right angles to it is the INVARIABLE PLANE of the system. CHAP. I.] CONSERVATION OF AREAS. 33 Thus, in the solar system, if there are no forces external to the system, we have an invariable axis through the centre of mass of the system and an invariable plane at right angles to this axis, through the centre of mass. Also, if for each planet and satellite the force of attraction on every particle always passes through the centre of mass of the system, we have for each planet and satellite an invariable axis through that point. The moment of momentum of each does not change in magnitude, and lies along its oVn invariable axis. The resultant of all is the moment of momentum of the system, and lies along its invariable axis. If we take the axis of Z as the invariable axis of a system for which the moment of momentum is constant in magnitude, the projection on the axes of X and Y will be zero. Hence, for the invariable axis, the moment of momentum of a system is a maximum. Conservation of Areas. The moment vl of a velocity is equal to twice the areal velocity of the radius vector (page 89). The moment of momentum mvl of a particle is then proportional to twice the areal velocity of the radius vector. Hence, from the principle of conservation of moment of momentum, When the t resultant moment 2 Fl of all the impressed forces acting upon a particle, relative to any point or any axis, is always zero, the areal velocity of the radius vector of the particle does not change either in magnitude or direction. For a system the areal velocity of any particle relative to the invariable axis does not change. This is called the principle of conservation of areas. Thus, in the solar system, the areal velocity of the radius vector of any planet does not change. This is Kepler's second law (page 120). Kinetic Energy of a System. The kinetic energy of a material system is the sum of. the kinetic energy of all its particles, or Let v x , ~v y , v f be the component velocities of the centre of mass O relative to the origin O' , and v x , v y , v z the component velocities of any particle relative to the centre of mass O. Then the component velocities of the particle relative to O' are and the kinetic energy of the system is K = ~ Expanding, we can write But (page 298) 2mv x = o, 2mv y = o, 2mv t = o. Also, - 2tn(v^ + v} -|- z/) is the kinetic energy of all the particles moving with velocities equal to their velocities relative to the centre of mass O, and-^w( v* x -\- ~v* y + z ) =-m(z^-f- V*, -\-vfy is the kinetic energy of a particle of mass m equal to the mass of the system moving with the velocity of the centre of mass. 304 KINETICS OF A MATERIAL SYSTEM. [CHAP. I. Hence the kinetic energy of a system is equal to the sum of the kinetic energy of the particles relative to the centre of mass plus the kinetic energy of a particle of mass equal to the mass of the system moving with the velocity of the centre of mass. _m _ !!'<" ^ then / is the lever-arm of the velocity v relative to O, and co is the corresponding angular velocity, we have v = loo, and the kinetic energy ^ml^tf. The kinetic energy of all the particles relative to O is then -2ml*(*P. If v is the velocity of the centre \j 2 of mass, and m the mass of the system, we have then for the kinetic energy of the system 5C = mv 2 -f SmlW .......... (5) Potential Energy of a System. The potential energy of a system is the sum of the potential energies of its particles. Since the potential energy of each particle depends upon its position (page 274), the potential energy of the system depends upon its configuration and position. If no external forces act upon the system, its potential energy will depend upon its configuration only. Law of Energy. Since the law of energy (page 275) holds for every particle of a system, it holds for the entire system. Therefore, if &, is the total initial and & the total final energy, potential and kinetic, of a system, and if W'\s the work done against non-conservative forces, we have or the total loss of energy is equal to the work done against non-conservative forces. Such a system is called a non-conservative system. Conservation of Energy. If there are no non-conservative forces W is zero, and we have or the total energy does not change. Such a system is called a CONSERVATIVE system. Perpetual Motion. When a material system after a series of changes returns to its original position and configuration, its potential energy is not changed ; and if the system is conservative, the kinetic energy is unchanged. Such a system would then go through its cycle of changes for ever if set in motion. If the bodies of the solar system encounter no resistance, we have a system of this kind. If we could abolish friction and all other non-conservative forces, a machine could be constructed which, once started, would run forever, provided no work is done by it. Since friction and other non-conservative forces cannot be thus abolished, such a machine is not possible. Law of Conservation of Energy General. The law of conservation of energy just enunciated , & = o appears as a special case of the law of energy CHAP. L] EXAMPLES. 305 when W the work against non-conservative forces, such as friction, is zero. In general, then, when work is done against such forces we say that energy is " lost." Experiment shows that when energy is thus " lost" heat is always developed, and that heat is a form of energy dependent upon kinetic energy of particles. This heat is the exact equivalent of the energy "lost." If then we take into account heat energy, the law of conservation of energy becomes general, and we have in ajl cases, whether the forces are conservative or non-conservative, That is, the entire energy of an isolated system, including kinetic, potential, and thermal, is constant. Examples. (i) What effect has the bursting of a bomb upon the motion of its centre of mass? ANS. Neglecting air resistance, none whatever. By the principle of conservation of centre of mass* internal forces cannot affect the motion of the centre of mass. (2) Show that the centre of mass of the universe must either be fixed in space or move in a straight line with uniform speed. ANS. By the principle of conservation of centre of mass no material system can of itself and without the action of external force change the motion of its centre of mass. (3) Two particles of mass nti and mi are moving in the same straight line with velocities z/i and v*. Find the velocity of their centre of mass. ANS. From equation (i), page 298, "v = _ . m We have m = m\ + mi and 2mv = m\v\ + miVi if the velocities are in the same direction. * Hence _ _ mivi oz/a ^ ve j oc j t j es m tne same direction and ffti-r mi _ m\v\ m^Vi . t v = it Vi is opposite to v\. nt\ -rnti (4) Two particles of mass mi and mi at a distance s' are at rest on a smooth horizontal plane and are drawn together by a uniform force F. After a time t the mass mi has a velocity Vi. Find the final velocity Vi of the mass mi , the internal force F, the distance s apart at the end of the time t, and the distance of each mass from the centre of mass. ANS. Let v be the velocity of the centre of mass O. Then, since Vi and Vi are opposite in direction, we have, by the principle of velocity of , centre of mass, page 298, (mi + mi)v=mivi-mivi f F J Y * & Hence v = ** 1 ^~^ 7/a . But the centre of mass is initially at rest, or v = o, and by the principle of conservation of centre of mass (page 300) it must remain at rest. Hence (i) We also obtain this result directly from the principle of conservation of momentum (page 300). Since the force F on mi is uniform, the acceleration is uniform and the distance passed over by mi in the time /, starting from rest, is /. The distance passed over by mi is /. The distance s' s is then given by (!_+ ft), = 2*2i 306 KINETICS OF A MATERIAL SYSTEM. [CHAP. 1. By the principle of impulse (page 257) the uniform force Fon mi is tt L^l % and on m t , f "y*. Since these forces are equal and opposite, we have ....... . . .. . . from which we obtain again (i). Again, the initial energy is Si = Fs', potential. The final energy & is Fs potential and m\v* ^ -- kinetic, or & = fr ~s + miVi* H By the principle of conservation of energy, &i = , or Ft = Fs + 1 mivS + ~ m,vS, or F tftlVl * + m * v ? 2(s f - JT~ Inserting the value of s' - s from (2), and of z/, from (i), we obtain again ,.. _ m\v\ _ mtVt ~T~' t ' The distance of m\ from the centre of mass C at the start is mi + m\ ' and at the end of the time / The distance of , from the centre of mass C at the start is mi + and at the end of the time / r t = If mi = 50 Ibs., m t = loo Ibs., v, = 10 ft. per sec., / = ~ sec., / = 3 ft., we have Vi = 20 ft. per sec. ; s = 2.25 ft. ; F = 20000 poundals = pounds; r\ = 2 ft. ; ri = 1.5 ft. ; r,' = i ft. ; r, = 0.75 ft. (5) In the preceding example suppose the particles revolve about the centre of mass with the initial angular velocity o>'. ANS. Take the same notation as before and let a>i be the angular velocity of mi at the distance ri, and o t of m^ at the distance r t . We have the same values for v t , / s, ri,<.t\, rY.and r,as before. It is therefore only required to find F. We have, by the principle of conservation of areas, rjV = r.'oj, and rjV = r, 1 *,, hence CHAP. L] EXAMPLES. 307 or K)I and GO* are equal, as they should be, since, by the principle of conservation of centre of mass, the centre of mass is fixed. The initial energy of the system is & = FS + ~ mirfu* + ~ mir!*P. The final energy of the system is r- * l . ' I =.r.y ^ m\Vi H w a zv -f- 222 By the conservation of energy, a = $, and inserting the values of -z/a, aoi, o>a, r\, r\, r\, r*, we have _ \ _ *ii ^^ , nu . 2m, Hence _ _ miVi mimes'* GO'* (s' + s) / 2(mi + m^s* (6) / a /^ velocity v and the mass the velocity -v. Then the final energy of P is Pg(h s) potential and V* kinetic, and of Q, Qg(h + -J + ~i*- The total final energy is then t~Pf(k -s) + P - + + & (h + ) + g>. The work consumed by friction is F s. By the law of energy the loss of energy is equal to the work consumed by friction, or Substituting the values of i, & and /"already found, we obtain That is, the loss of potential energy, Pgs Q -s, equals the gain of kinetic energy, z/ + =- . v*, plus the work of overcoming friction. Also, the loss of potential energy of P or Pgs equals the gain of potential energy of Q or Qg - s, plus the gain of kinetic energy of the system, plus the work of overcoming friction. If we substitute s = -//*, v =// and solve for/, we obtain the same value for/ as already obtained. The acceleration of Q is then -/ = 0.363^ = 11.66 ft.-per-sec. per sec. The velocity of P at the end of / = 5 sec. is v =// = 8744 ft. per sec., and the velocity of Q v = 58.29 ft. per sec. The distance passed through by P is * = //* = 218.59 ft., and by Q -s = 145-73 ft- CHAP. I.] . EXAMPLES. 309 Tension in string for P is P(g /) = 4-S^ poundals or 4.56 pounds; work of P = 4.56.1 = 996.77 ft.-lbs. = /** -- ; " onQ = 6.82 x i=. 993.88 " = Q-s + -^- The difference of these works, or 2.89 ft.-lbs., is the work consumed by friction. The power of P The efficiency is The power of P is " * 77 = 199.35 ft.-lbs. per sec., or I99 ' 35 = 0.36 horse-power. CHAPTER II. EQUILIBRIUM OF A MATERIAL SYSTEM. Equilibrium of a Material System. We have seen (page 299) that the motion of the centre of mass of a material system is not affected by the internal forces, but only by the external forces. Also, the acceleration of the centre of mass is the same as for a particle of mass equal to the mass of the system, all the external forces being transferred to this particle without change of magnitude or direction. If all these forces constitute a system of forces in equilibrium, and every particle of the material system is at rest, it is said to be in static equilibrium. If not at rest, since the external forces constitute a system of forces in equilibrium, the centre of mass has no acceleration, and the resultant moment 2F! of all these forces relative to any point is zero. But we have seen (page 302) that when 2FI is zero the moment of momentum does not change, and we have then uniform angular velocity of every particle about an invariable axis through the centre of mass, and uniform velocity of translation of this axis. A system in this condition is said to be in molar equilibrium. If there is no rotation about an axis, but simply uniform translation in a straight line, then the forces acting upon every particle are in equilibrium, and the system is said to be in molecular equilibrium. The necessary and sufficient conditions of molar equilibrium are then F x =o, F y = o, F, = o, M x = o, M y = o, M t o. That is, the external forces acting upon the system form a system of forces in equilib- rium, or the algebraic sum of the components of all the external forces in any direction is zero, and the algebraic sum of the moments of these forces about any axis is zero. For molecular equilibrium, we have, in addition to these conditions, the condition that all the forces external and internal on every particle form a system of forces in equilibrium. For static equilibrium we have still the added condition that every particle is at rest. In both molecular and static equilibrium, then, the work for an indefinitely small change of position or configuration must be zero. Let 2zv be the work of all the external and 2iu' of all the internal forces for an indefinitely small change of position or configuration. Then we have 2w -f- ^w' = o, or 2-w = ^v/. If the system is rigid, 2w' = o, and hence 2w = o. ILLUSTRATIONS. Thus a billiard-ball at rest on a table is in static equilibrium. All the external forces acting upon it form a system of forces in equilibrium, and every particle is at rest. Hence the forces acting on every particle must form a system in equilibrium. For an indefinitely small change of position or con- figuration the work is zero. 310 CHAP. II. ] STABLE EQUILIBRIUM. If the ball rotates about a fixed vertical axis with uniform angular velocity, it is in molar equilibrium. All the external forces acting upon it form a system of forces in equilibrium, but the forces acting upon every particle do not. Since the moment of momentum cannot change, the centre of mass is at rest, and the ball must always rotate about the vertical axis with uniform angular velocity. For an indefinitely small change of position or configuration the work is not zero. If the ball rolls so that the centre of mass moves with uniform speed in a straight line, it is also in molar equilibrium. If the ball is projected through the air without rotation, so that every particle has uniform motion of translation only in a straight line, the ball is in molecular equilibrium. All the external forces form a system in equilibrium, and the internal forces on every particle also form a system in equilibrium. For any indefinitely small change of position or configuration the work is zero. If the particles of a mass of water in a horizontal circular dish are rotating about a vertical axis through the centre of mass and are acted upon only by gravity the mass is in molar equilibrium. The centre of mass,, is fixed, and no particle, disregarding friction, can change its moment of momentum about the vertical axis. A beam or spring bent by a load and at rest is in static equilibrium. If every particle has uniform motion of translation only in a straight line, it is in molecular equilibrium. In both cases, for an indefinitely small change of configuration, the work of the load is equal and opposite to the work of the internal forces. Stable Equilibrium. The same conditions must hold for stable equilibrium of a material system as for each particle (page 277). Hence for stable equilibrium the potential energy is a minimum and the kinetic energy, if any, is a maximum. If the potential energy is a maximum, the equilibrium is unstable and the kinetic energy, if any, a minimum. If the potential energy is neither a maximum nor a minimum, the equilibrium is stable for displacements for which the poten- tial energy increases, and unstable for displacements for which the potential energy decreases. If the potential energy is constant, the equilibrium is neutral. If for all possible displacements, large or small, the potential energy is constant, the equilibrium is indifferent. ILLUSTRATIONS. Let a prism stand on a level base. For equilibrium the weight W acting at the centre of mass C, and the resultant upward pressure R on the base at P, must be equal and opposite and in the same straight line. If the prism is so constrained that it can have displacement of translation only along a horizontal plane, the potential energy W x CP does not change for any displacement large or small, and the equilibrium is indifferent. If it is possible to rotate the prism about an edge at Bor A, the potential energy is increased, so long as the weight W falls inside the base, and the equilibrium is stable. If, however, the weight W cuts the base at an edge B, then for rotation about A the potential energy increases and the equilibrium is stable, while for rotation about B the potential energy decreases and the equilibrium is unstable. Again, if a pendulum hangs vertically with the bob below the point of suspension P. the potential energy W x CO is a minimum for all possible displacements and the equilibrium is stable. If the pendulum is reversed, the potential energy is a maximum and the equilibrium is unstable. If the pendu- lum swings, the kinetic energy is a maximum in the first case and a minimum in the second case. Principle of Least Work. The work done by external forces I 1 in changing the configuration of an elastic body can be given back by the body when the external forces are removed. It is therefore potential energy. If such a body is in static or molecular equilibrium, the forces external and internal form a system of forces in equilibrium. If the equilibrium is stable, the work done in causing the change of configuration must be a minimum consistent with the conditions of equilibrium of the forces. This is called the " principle of least work " for elastic bodies. P B W EQUILIBRIUM OF A MATERIAL SYSTEM. [CHAP. II. Stability in Rolling Contact. As an application of the preceding, let us investigate the equilibrium of a body with a curved surface resting and rolling upon a curved surface. Let O be the centre of curvature of the fixed surface APB, and o the centre of curvature of the body aPc resting on it at P. The reaction R at P is for equilibrium, equal and opposite to the resultant R' acting at A, of all the other forces, and in the same straight line. Let aPc be displaced by rolling through an indefinitely small angle so that it comes into the position a' PC', A' and o' being the new positions of A and o. Let the radius of curvature oP = p and OP= p l , and the angle a'o'P' = ft and POP'= a. Let the distance PA = h. We have then PP' = a'P', or pjt = p/3. Take a plane OX at right angles to R' . The equilibrium will be stable when the potential energy is a minimum, or when AO is less than A'O', and unstable when the potential energy is a maximum, or when AO is greater than A'O'. The distance A O = p l -f- //. We have then for stable equilibrium for unstable equilibrium and for neutral equilibrium Now we have p l +h A'O', A'O' = (p + ft) cos a - (p //) cos (a -f /?). v But, by Trigonometry, and cos (a -f- ft) = cos a cos ft sin a sin /?, cos a = i 2 sin 2 -, cos ft = i 2 sin 2 -. Substituting, we have, after reduction, = p l + A 2p l sin 2 a sin h sin a sin ft 2/t sin 2 - 2/1 sin 2 -(\ 2 sin 2 -J. Now when a is very small, 2 sin 2 can be neglected relative to I, and we can take the arc in place of sin a. Hence, since ft = l ar, we can take . * PI . , 2 . ,0 sin a= a, sin p = -a, sm z - = , sur- = r 2 4 ^ CHAP. II.] STABILITY IN ROLLING CONTACT. Making these substitutions, we have We have then for stable equilibrium Pi + hj + j>'. for unstable equilibrium while for neutral equilibrium we have . p l = A, or >fr = If the concavity of either surface is turned the other way, we have the same result except that the sign of the corresponding radius will be changed. If either surface is plane, its radius is infinite. The same results can be obtained more simply as follows : The equilibrium will be stable, unstable or neutral according as A' lies to the left, right or directly over P' , that is, according as the horizontal distance of A' from P is less, greater than or equal to the horizontal distance of P' from P. The horizontal distance of P' from P is p l sin a. The horizontal distance of A' from j is the same as from a' , or h sin (a -f- /?). Hence we have h sin (or -j- /?) = p l sin OL. the same conditions as before. Inserting fi = V, we have for a very small, so that we can take the arc for the sine, Examples. (i) A body made up of a cone and a hemisphere having a common base rests with the axis vertical on a horizontal plane. Find the greatest height of the cone for stable equilibrium. ANS. Let h be the height of the cone, r the radius of the hemisphere, and C the centre of mass. The height required is that height for which PC = r. -> j The volume of the hemisphere is -xr 3 . The volume of the cone is itr^h. The centre of mass of the hemisphere is at a distance above P equal to gV. The o centre of mass of the cone is at a distance above P equal to r + . We have then 2 5 Ttr^h I -Ttr 3 x f-r + x [r + - PC=* L 3__A , =rt or A = , TJ . 3 , ~ ~3~ (2) A prolate spheroid rests with its axis horizontal on a roi'gh horizontal plane. Show that for rolling displacement in its equatorial plane the equilibrium is indifferent, and for rolling displacement in the vertical plane through the axis it is stable. 3*4 EQUILIBRIUM OF A MATERIAL SYSTEM. [CHAP. II. (3) A right circular cylinder of radius r rests with its axis horizontal on a fixed sphere of radius R greattr than r. Show that for rolling displacement the equilibrium is stable or unstable according us the plane of displacement makes an angle with the vertical plane through the axis of the cylinder whose sine is less or greater than y J jr . ANS. Let p be the radius of curvature of the rolling curve at the point of contact. Then the condition for stable equilibrium is -s, 4. _ r R ' Let the plane of displacement make the angle with the vertical plane through the axis of the cylinder. The rolling curve is then an ellipse whose semi-minor axis is r and whose semi-major axis is radius of curvature at the point of contact, that is, at the vertex of the minor axis, is The P = sln~e Hence for stable equilibrium sin'0 or sin 9 < i/ i -. (4) A prolate hemispheroid rests with its vertex on a horizontal plane. Show that for rolling displace- ment the equilibrium is stable or unstable according as the eccentricity of the generating ellipse is less or greater than ANS. Let a be the semi-major and b the semi-minor axis. Then the distance OCto the centre of mass is 50 /~ \i 4 $a a 8 The distance PC then is ^-a. The radius of curvature at P is p = . We have a then for stable equilibrium i But the eccentricity of the generating ellipse is Hence for stable equilibrium e < CHAPTER III. ROTATION ABOUT A FIXED AXIS.* Rotation about a Fixed Axis Effective Forces. We have seen that D'Alembert's' principle (page 297) reduces any kinetic problem to one of equilibrium between actual (impressed] forces and fictitious (reversed effective] forces. In order to apply it, then, we must be able to find in any given case the effective forces and the moments of the effective forces. Let us consider first the case of rotation about a fixed axis. Let O be the centre of mass, and let us take the origin of co-ordinates O' at the point of intersection of the axis of rotation with a plane through the centre of mass O at right angles to this axis. Let ~x, y ', Is be the co-ordinates of the centre of mass O for this origin O' and any co-ordinate axes O'X', O'Y', O'Z'. Equations (i), page 158, give the components of the tangential acceleration for any particle of a rotating body. If we multiply ea,ch term by the mass m of the particle and sum up for all the particles, we shall have the components of the effective tangential forces of the -body. These forces cause change of magnitude of the velocity of each particle in its plane of rotation. In summing up, since x, y, z are taken from the centre of mass O as origin, we have ~2,mx = o, ~2my = o, ~2.mz = o. We have then for the components of the effective tangential forces for all the particles of the body, since 2m = m the mass of the body, where a x , y , a z are the components of the angular acceleration a about the axis of rotation. These forces cause change of magnitude of the velocity of each particle in its plane of rotation. Equations (3), page 159, give the components of the deflecting acceleration for any particle of a rotating body. Here again, multiplying each term by the mass m of the particle and summing up for all the particles, we have the components of the effective deflecting forces for all the particles, 2mf r = niyoo^ inyctf* 2 , } (2) * Before reading this and the following chapters the student should be familiar with the principles of Chap. II, page 153, and Chap. Ill, page 31. 315 3 l6 KINETICS OF A MATERIAL SYSTEM ROTATION FIXED AXIS. [CHAP. III. where a> x , to yt oo, are the components of the angular velocity a> about the axis of rotation. These forces cause change of direction of velocity of each particle in its plane of rotation. For fixed axis the plane of rotation of each particle does not change. There are therefore no deviating accelerations. Adding equations (i) and (2) we have then for the components of the effective forces for a body rotating about a fixed axis, for any co-ordinate axes we please. y <*) t = O, oo y oo x = o, oo t ca x = O, oa t oa y = o. We can also put ^', y ', z in place oi~x-\-x, ~y -\- y, ~z -\- z, and for rotation only we have f x = o,fj, = o, /. = o. If we make these changes in equations (ib), page 160, multiply each term by the mass m of the particle and sum up for all the particles, we shall have the components of the moments of the effective forces for all the particles of the body. In summing up we have where I' x , Ij, /.' are the moments of inertia of the body for the axes O'X', O'Y', O'Z'. We have then, from equations (10), page 160, for the component moments of the effective forces, . . . (2) M' fx = - M' fy = a t 2*nc'y a^mx'y -\- (a?* QC%)2mxJ + I y a y M' ft = CHAP. III.] ORIGIN OF THE TERM "MOMENT OF INERTIA." 317 If the fixed axis coincides with one of the co-ordinate axes, as, for instance, O ' X\ we have Go y = O, GO, = o, a y = o, a, = o, and hence = comx'z' (5) Since we can take any co-ordinate axes we please, let the co-ordinate axes be principal axes at the point O'. Then (page 35) 2mx'y' = o, 2my'z f = o, 2mz'x' = o, and equations (4) become M' fx =I' x a x , M =*;,, M' fz = I' t a s (6) If the axis of rotation is a principal axis, let it coincide with O'X', for instance. Then a y o, n x = o, and from (5) we have M fx = I' x a x , M fy o, M ft = o. That is, when a body rotates about a principal axis with angular acceleration or, the moment of the effective forces relative to the axis of rotation is equal to the moment of inertia /' of the body relative to the axis multiplied by the angular acceleration, or M' f = I'a. If I x , I y , I z are the moments of inertia for principal axes through the centre of mass. O, parallel to O'X', O'Y', O'Z' , we have (page 33) /; = I x + mCy 2 + P), /; = I y + m(P + j?), // = /. + m(^ + 5?), where m is the mass of the body and ^, y, ~z the co-ordinates of the centre of mass Inserting these values in (6), we see that the moment of the effective forces about any fixed axis is equal to the moment about a parallel axis through the centre of mass plus the moment of the effective force of a particle of mass equal to the mass of the body at the centre of mass. If we take distance in feet and mass in Ibs., these equations (4), (5), (6) give moments in poundal-feet. For pound-feet divide by g (page 171). Origin of the Term " Moment of Inertia." Let F be the resultant of all the impressed forces at right angles to the axis of rotation and p its lever-arm, so that Fp is the resultant moment of all the impressed forces relative to the axis of rotation. Then, by D'Alembert's principle, when the axis of rotation is a principal axis FpM f o, or Fp=I'a. The term "moment of inertia" is due to Euler. Euler used the term "inertia" as synonymous with what we call mass. Thus the equation of force, F= mf, would be read in the terminology of Euler Force = inertia X linear acceleration. In the equation Fp = I'a = a^mr 3 Euler called the term ^mr* " moment of inertia" and thus obtained the analogous expression moment of force = moment of inertia X angular acceleration. 318 KINETICS OF A MATERIAL SYSTEM-ROTATION -FIXED AXIS. [CHAP. III. The term " moment of inertia " in modern scientific terminology is an improper expres- sion. Inertia (page 169) is a property of matter like color or hardness, and we cannot properly speak of moment of inertia any more than of moment of color or hardness. The term "second moment " of mass would more correctly describe the product Stnr 2 , and has been used by some recent authors. The expression "moment of inertia," however, has beco'me firmly established by long usage. The student, while using it, should consider it simply as a name for the quantity ZZtnr 2 , which occurs so frequently in dynamic problems that it is convenient to give it a special name. Momentum of Rotating Body. From equations (2), page 154, we have tne component velocities for any particle of a rotating body v M =(s + )a>, (y+y)a>,, v y =(x + x} t (z+z)c*> x , v, = (J+y)co x - (x + x)v y . If we multiply each term by the mass m of the particle and sum up for all the particles, we shall have the components of momentum for all the particles. In summing up, since x, y, z are taken from the centre of mass, we have "Sinx = o, 2my = o, 2mz o. We have then for the components of the momentum of a rotating body, since 2m = m the mass of the body, , TOLZGO X , For axis of rotation through the centre of mass we have ~x = o, ~y = o, ^ = o, and hence 2mv x = o, 2mi> y = o, 2mv t = o. Hence the momentum of a body rotating about an axis is the same as for a particle of mass equal to ttie mass of the body at the centre of mass. Moment of Momentum Rotating Body. Equations (10), page 155, give the component moments of the velocity for any particle of a rotating and translating body. For rotation only we have v x = o, v y = o, v, = o. If we make these changes in equations (10), page 155, multiply by the mass m of the particle and sum up for all the particles, we shall have the components of the moment of momentum for the body, for any co-ordinate axes we please. Let us take these axes principal axes at O'. Then we have 'Smx'y' o, ^my'z' o, 2mz'x' = o. We have also Zm( /* + z"^ = f' x , 2 m (^ + x' z ) - /',, 2t(x'* +/ a ) = /,' , and 2 mx = o, 2i y = o, 2ntz = o. We have then from equations (10), page 155, the component moments of momentum (8) If the axis of rotation coincides with a principal axis, as, for instance, O'X', we have a?, = o, co t = o, and from (8) we have M' vx = J x a> x , M^ = o, M' n = o. That is, when a body rotates about a principal axis with angular velocity a?, the moment of momentum relative to the axis of rotation is equal to the moment of inertia /' of the body relative to that axis, multiplied by the angular velocity, or Ml= Too. If /, , I y , I t are the moments of inertia for principal axes through the centre of mass O, we have /',= /, + ^( ? + **) 7 - = 7 + CHAP. III.] PRESSURES ON FIXED AXIS. 319 Inserting these values in (8), we see that the moment of momentum for a body rotating about an axis is equal to the moment of momentum about a parallel axis through the centre of mass, plus the moment of momentum for a particle of mass equal to the mass of the body at the centre of mass. Pressures on Fixed Axis. Equations (4) give the component effective forces for a body rotating about any axis O'X f , and equations (5) give the component moments of the effective forces for a body rotating about any axis O' X' . Let the axis be supported at the ends a and b so that it cannot change its direction, and let the distances O'a = l lt O'b = / 2 , the plane Y'Z' being a plane through the centre of mass O. Let the component pressures at a and b be R' x , R' y , R' f and R' x , R y> R" t . , Let the components of all the other impressed forces | /v be 2F X , 2F y , and the component moments of these forces about the co-ordinate axes be about O'X' = 2F 2F " O'Z' = We have then from equations (4), by D'Alembert's principle, for any fixed axis O'X' + 3F, = - m^G? 2 , - mj",, ] [. ..... . . (9) where m is the mass of the body ; x, y, J, the co-ordinates of the centre of mass. We have also from equations (5), by D'Alembert's principle, for any axis O'X' (10) = I x 'a x . From the last of equations (10) we can find a x , and then from the first two and the first two of equations (9) we can find R y ' , R y ' , R M ', R n " . This leaves R x and R x " indeterminate, but their sum is given by the last of equations (9). If we take distance in feet and mass in Ibs. , the pressures./? will be in poundals. For pounds divide by g (page 171). In equations (9) the terms mFo?^ 2 and 'myGo x 2 are the sums of the components parallel to Z' and Y f of the effective deflecting forces (page 315) of all the particles, and the terms *&y a x and mza x are the sums of the components parallel to Z' and Y' of the effective tangential forces (page 315) of all the particles. 320 KINETICS OF A MATERIAL SYSTEM-ROTATION-FIXED AXIS. [CHAP. III. In equations (10) the terms cs&Smy'x' and of^mx ' z' are the moments about Z' and V of the effective deflecting forces, and a x 2mx'z f , a^mx'y' are the moments about Z' and Y' of the effective tangential forces, and I' x a x is the moment about the axis of rotation X' of the effective tangential forces. If the axis of rotation passes through the centre of mass we have y = o, ~z = o, and we see from equations (9) that the sums of the components of the effective deflecting and tangential forces are zero, or these forces reduce to a couple. If the axis of rotation is a principal axis, then taking the other two axes as principal axes we have 2my'x' = o, 2mx'z' o, and we see from equations (10) that the moments of the effective deflecting and tangential forces about Y' and Z' are zero. There is then no tendency of the axis of rotation O'X' to turn about the other two co-ordinate axes O' Z' orO'Y'. If, then, the fixed axis of rotation is a principal axis through the centre of mass, there will be no pressure on this axis due to rotation, and in such case we have, from equations (9) and (10), R x r -f- R x ' + 2F X = o, R y ' + R y " + 2F y = o, JR.' + R," + 2F. = o, 2F,v = o, - J?,'/, + *."/, + ^Fjt - SF* = o, That is, for a body rotating about a fixed principal axis through the centre of mass the Pressures on the axis are the same as if the body did not rotate, and are given by the conditions of equilibrium of the impressed forces only. Conservation of Moment of Momentum Fixed Axis. We have, from equations (8), for the component moments of momentum for a body rotating about a fixed axis, taking the co-ordinate axes as principal axes at O', and, from equations (6), for the component moments of the effective forces If, in equations (#), we have oc x = o, a y =. o, a t = o, we shall evidently have oo x , oo y , GO, constant in equations (a). But when a x o, a y = o, a, = o, we have M fx = o, M fy = o, M ft = o. Since, by D'Alembert's principle, the moment of the effective forces is equal to the moment of the impressed forces, we have the moment of the impressed forces zero. Hence, if the moment of the impressed forces about a fixed axis is always zero, the moment of momentum about that axis is constant. Kinetic Energy Fixed Axis. The kinetic energy of a particle of mass m and velocity v is -mv* (page 272). If a particle has the component velocities v x , v y , v tt we have and hence CHAP. III.] EQUATIONS FOR ROTATION AND TRANSLATION. 3 21 From page 154 we have for the component velocities for any particle of a rotating body where x', y 1 ', z' are taken from the origin O', the intersection with the axis of rotation of a plane through the centre of mass O at right angles to the axis of rotation. If we square these component velocities, multiply each term by -m, sum up for all the particles and add, we shall have the kinetic energy for a rotating body for any co-ordinate axes O'X', O' Y' , O ' Z 1 ', we please. Let us take these co-ordinate axes as principal axes of the body at the origin (y. Then we have = O, myz = O, mzx = o. We thus obtain for the kinetic energy of a rotating body K = - But 2m(y* + z"*) = r m , 2m(z* -f *' 2 ) = // Hence *=^+5#aJ+5/>2 ......... (II) But GO X = a? 2 cos 2 a, a$ = a? 2 cos 2 /?, a? 2 = a? 2 cos 2 y, where GO is the angular velocity about the axis of rotation and a, /3, y are the direction angles of this axis. Therefore equation (n) becomes (page 35) (12) where /' is the moment of inertia relative to the axis of rotation. Hence the kinetic energy for a body rotating about an axis is equal to one half the product of the moment of inertia of the body relative to that axis and the square of the angular velociy about that axis. Analogy between the Equations for Rotation and Translation The student should not fail to note the analogy between the equations for rotation and translation. Thus for translation F - m/ = force, while for rotation (principal axis) Fp = I' a = moment of force. For translation mv = momentum, while for rotation (principal axis) /'<= moment of momentum. For translation = uniform force, 322 KINETICS OF A MATERIAL SYSTEM ^ROTATION-FIXED AXIS. [CHAP. ill. while for rotation (principal axis) /'(o> <) - i- = moment of uniform force. For translation - ihz/* = kinetic energy, while for rotation (any axis) 7'oj 2 = kinetic energy. We see that if linear acceleration and velocity are replaced by angular acceleration, and velocity and mass by moment of inertia, force and momentum become moment of force and momentum, and kinetic energy is given in both cases. Reduction of Mass. It is often desirable to be able to reduce a rotating mass to a particle of equivalent mass at any desired distance from the axis. Thus let a body of mass m rotate about a principal axis YY with angular velocity co and angular acceleration a. Then the moment of momentum is I 'GO and the moment of the force is I 'a, and the kinetic energy is -I'd. JNow suppose a particle of mass m at any desired distance d from the axis to have the same angular velocity and acceleration. Then its mo- ment of momentum is md*(*>, the moment of the force is md*a, and the "Y kinetic energy is md*co z . If then we have md z = I' t or w = 8 , the particle m at d will have the same moment of momentum, moment of force and kinetic energy as the body itself. It is therefore equivalent to the body. Hence to reduce a rotating mass to a particle of equivalent mass at any desired dis- tance from the axis we divide the moment of inertia by the square of the distance. Examples. (i) A driving-wheel of a locomotive has a arank-pin and cross-head, etc., of mass m\ =2600 Ibs., the centre of mass being at a distance r\ = 6 inches from centre of wheel. If the wheel has a radius of r = j feet and is running 60 miles an hour, find the pressure on the rail when the crank-pin has its lowest and highest position. (Take^ = 32 ft.-per-sec. per sec.) ANS. Let m be the mass of the wheel alone. A speed of 60 miles an hour is 88 feet per sec. The 88 angular speed is then given by roo = 88, or oo = radians per sec. The impressed forces are the upward pressure fi of the rail and the weight (m. + mi)g of the wheel and crank-pin and cross-head. The effective force is nt t rtn\\g m,riao* = o, or R = (mi + 37555) pounds. The effect is to cause a blow upon the rail every time the crank passes the lowest point. CHAP. III.] EXAMPLES. ROTATION. 323 (2) In the preceding example let the thickness of the wheel be t = 2 inches, and density $ = 480 Ibs. per cubic foot. It is required to balance the crank by filling in between the spokes. Find the inner radius r for any given angle 6 = 60. ANS. Let mi be the mass of the ring, and J the distance of its centre of mass. Then we should have < We have (page 29) w = (r 1 r ) and - Substituting and solving, we have for r 3*in 2<5/sin Inserting numerical values, we have r = 2.79 feet. 3. A sphere of mass m and radius ri has an angular velocity <,, and contracts until its radius r\. Find the final angular velocity GO if there are no external forces. ANS. The moment of momentum cannot change (page 320). Hence /iJ ' * 1' .22=? = , 845jSV .ur n , 4* x ico (- + 25 CHAP. III.] EXAMPLES. RO TA TION. 327 The time necessary to make one turn is x ioo I- + 25 2CP- X 8 = 2.23 sec. (9) A sphere whose mass is m rests upon the rim of a horizontal disc of mass D. A perfectly flexible string passes round the disc and over a pulley and has a mass P at its lower end. Disregarding friction and mass of pulley and string, find the distance described by P in the time t. ANS. Let R be the radius of the disc and r the radius of the sphere. If the sphere moves with the disc as if it were part of it, i.e., rotates about the axis ab in the same time that it rotates about the parallel axis AB, we have the moment of inertia of the sphere relative to the axis AB , a 2 ? mJ In this case we have for the angular acceleration about AB 2 + + j D Hence the acceleration /of P is The distance described by P is then If the sphere does not rotate about the axis ab, as when, for instance, it is hung from the rim, we may consider it as a particle, and its moment of inertia is then m/v*. We have then PgP (2) Hence we have 2/> + 2m + D' If the sphere has an angular acceleration a t , not equal to a, about ad in same direction as the disc, we have for the moment of the force causing this acceleration -mr"a, . Hence by D'Alembert's principle P + m + -- ( 10) A hollow circular disc whose outer radius is a\ and inner radius b\ and thickness t\ revolves about an axis perpendicular to its plane. Find the thickness (* of an equivalent disc whose outer radius is a t and inner radius b^. ANS. If the discs are equivalent, the same force moment should give the same angular acceleration a to each, or I\a = fr,a, or I\ = It . But 7i = mi(ai 2 + of) In the same way 7 = 8itt?(ay* , 4 ). Hence ' = ?V^ J K-'>- 3*8 KINETICS OF A MATERIAL SYSTEM ROTATION-FIXED AXIS. [CHAP. III. (i i) A sphtre of radius r\ rotates about the axis YY at a distance a. Find the height h of an equivalent cylinder of base radius r\ , whose axis is parallel to YY at a distance b. v ANS. The moment of inertia of the sphere of mass OTI relative to YY is The moment of inertia for a cylinder of mass m is We have for equivalence // = /'. Hence if the cylinder and sphere 'jfi rotate about their own axes parallel to YY in the same time they rotate about YY, But if Si is the density of the sphere and S t of the cylinder, Hence n If the cylinder and sphere rotate about the axis YY without turning on their own axes, they can be treated as particles and we have If the cylinder and sphere have the angular velocity or acceleration oa, or a, about their own axes and o> or a about YY in the same or opposite directions, we have fiir*i m,a a < = m 2 <, m,6*ta, Hence 2 _ 5 j, or h (12) Upon a vertical hollow axle whose outer radius isr\ and inner radius r t , and whose length is I, there is fixed a circular disc of radius a at right angles to the axle. Under the action of a force F of constant magnitude acting with the lever-arm a the angular velocity oo, is attained. Find the time t\ of attaining this velocity. If now the force F ceases to act, find the time t of coming to rest and' the number of revolutions in that time. ANS. Let the mass of the axle be A, and of the disc D. Then the moment of inertia of the axle is (page 46) and the moment of inertia of the disc is The total moment of inertia for the centre line of the axle is then and this axis is a principal axis. The pressure on the foot of the axle is (I) + A)g poundals. The moment of the friction for hollow flat pivot (page 225) is where n is the coefficient o( kinetic friction. CHAP. III.] EXAMPLES. RO TA TION. 329 If ai is the angular acceleration, we have the moment of the effective forces Ia\ (page 317). Hence, by UAlembert's principle, Fa M Fa M 7i = o, or ai = . Hence the angular velocity at the end of the time A is a, Fa M ' _ If the force F now ceases, we have the angular retardation M and the angular velocity at the end of any time / from the instant the force F ceases is CD = oo, + at = oo, - . When the apparatus comes to rest oo = o, and the time t of coming to rest is t = ~Jf The number of radians described in this time / is The number of revolutions is then D -] ,-) + -<- + >} Again, the kinetic energy at the instant the force .F ceases is (page 321) -<,*. The work of the friction for one revolution is 2itM. In n revolutions the work is 2itn.\f. We have then, in coming to rest, as before. (13) Suppose a wheel and axle composed of hollow discs for the wheel and axle and a solid cylinder for the journal. The radius of the wheel is a = 3 ft., of the axle b = 2 ft., of the journal r = i inch. Let the mass of the wheel be W 3 Ibs., of the axle A = 3 Ibs., of the journal J = 2 Ibs. Let the moving mass be P = 10 Ibs. and the mass lifted Q= 5 Ibs. Let the string be perfectly flexible, and disregard its mass. Let P start from rest and fall for t = 5 sec. Discuss the motion of the apparatus, taking into account the mass of the wheel, axle and journal, and the friction ; the coefficient of friction being u = 0.07. Take g = 32\ft.-per-sec. per sec. (Compare _p example (6), page 307.) ANS. We have for principal axis through the centre of mass O W Moment of inertia of wheel = (a* + P) = 32.5 lb.-ft. 2 " axle = -(P + r*) = " journal = r 2 i 144 33 KINETICS OF A MATERIAL SYSTEM ROTATION FIXED AXIS. [CHAP. III. Hence moment of inertia of wheel, axle and journal is The impressed forces are the upward reaction R at the centre O, the downward weights Pg , Qg, Wg, Ag t Jg, the friction 4- F and the equal and opposite reaction Foi the bearing. The moment of the fric- tion is then the moment of a couple, and is Fr at any point in its plane (page 185). The effective forces are Pf down and g /up, and the effective forces of the particles of the wheel, journal and axle. The sum of these in any direction is zero (page 316), and their moment about O is la (page 317). If we reverse the effective forces and then apply D'Alembert's principle, we have (i) Pfa +Q-f=o ........... (3) From (2) we have the pressure on the bearing, R = (P + Q+ W + A + f)g \P - Q~jf poundals. We see that the pressure on the bearing is less than the weight of the apparatus. The friction for new bearing is then (page 229) = - sin ft sin ft +J)g -(P- Q-}/~] poundals, \ a/ where u is the coefficient of kinetic friction and ft is the bearing angle. We have also aa =/, or a = /. a Inserting this value for a and Fin (3), we have for the acceleration /at the circumference of the wheel ... (4) If we disregard mass of wheel, axle and journal, we have the same value for /as already found in example (6), page 307. If ft is small, sin ft = ft, and we have for the given numerical values / = 0.4013^ = 12.912 ft.-per-sec. per sec., and a = / 4t , radians 4- 34 a ***** S ec. The acceleration of Q is then = 8.608 ft.-per-sec. per sec. a CHAP. III.] EXAMPLES. ROTATION. 33 1 The velocity of P at the end of the time t = 5 sec. is v = ft = 64.56 ft. per sec., and the angular velocity of the wheel is oo = = 21.52 radians per sec. a, The velocity of Q at the end of the time / = 5 sec. is v 43.04 ft. per sec. a The pressure R on the bearing is R = 22.32426^- poundals = 22. 32426 pounds, whereas the weight of the apparatus is 25 pounds. The friction is F = nR = 1.5627^- poundals = 1.5627 pounds. The moment of the friction is Fr = 0.13022^- poundal-ft. = 0.13022 pound-ft. The moment of the effective forces of the particles of the wheel, axle and journal is fa = 5.1531^- poundal-ft. = 5.1531 pound-ft. The distance s described by P is s = -//" = 161.4 ft. The distance described by Q is s = 107.6 ft. a The tension T r on right-hand string is Tr P(g J) 5.9864^ poundals = 5.9864 pounds. The tension TI on left-hand string is TI = Q(g + -/) = 6.337%- poundals = 6.3378 pounds. Moment of tension on right = 17.9592 pound-ft., " left = 12.6756 Difference = 5.2836 and this difference we see is equal to Fr + fa, the moment of friction and effective forces of wheel, axle and journal, as should be. Work of P = 5.9864 x 161.4 = 966.205 ft.-lbs. = Ps ^~, " on Q = 6.3378 x 107.6 = 681.947 - = o-s + Difference = 284.258 33 2 KINETICS OF A MATERIAL SYSTEM ROTATION-FIXED AXIS. [CHAP. III. This difference must be work on wheel and work of friction. Now work of friction is F^s = 7.005 ft.-lbs. Hence work on wheel = 277.253 ft.-lbs.. and is equal to of (page 321), as should be. The power of P (page 262) = 9 ^' 2 5 = i93- 2 4i ft.-lbs. per sec., or =0.35 horse-power. The efficiency of the apparatus (page 267) is 681.947 e = ,, = 0.70. 966. 205 Note. The total mass reduced to the circumference (page 322) is M a = The weight Qg reduced to the circumference is Q- g, and the force of friction F reduced to the circumference is F-. a The moving force at the circumference is then moving force = I P Q\g F* We have then directly, reduced mass x acceleration = moving force,'or -^k- If we substitute the values of M a and F, we obtain (4). (14) Suppose a wheel and axle composed of hollow discs for the rim or outer circumference C, the hub H, and the axle A ; of a solid cylinder for the journal f, and of four spokes, each spoke S being a bar of uniform cross-section. Let the outer radius of C be a = 20 inches, and the inner radius r\ = 19 inches ; the outer radius of H be r\ = 8 inches and the inner radius b = b inches ; the radius of the* journal J is r = / inch. Let the mass of the rim or outer circumference be C = 40 Ids. , of the hub H = 12 Ibs. , of the axle A = 10 Ibs. , of the journal J = 2 Ibs., and of each spoke S = jf Ibs. Let the moving mass P = bo Ibs. and the mass lifted Q = 1 60 Ibs. Let the string be perfectly flexible, and disregard its mass. Let P start from rest and full for a time t = 3 sec. Discuss the motion of the apparatus, taking into account the mass of the wheel, axle, journal and spokes, and the friction ; the coefficient of kinetic friction being M = 0.07. Take g = j2^ft.-per-sec. per sec. ANS. The moment of inertia of a spoke for axis through its centre of mass at right angles to plane of wheel is (page 38) For parallel axis through centre of wheel it is then (page 33) CHAP. III.] EXAMPLES. ROTATION. For four spokes we have then 333 The moment of inertia of the rim is (page 46) of the hub, of the axle, of the journal, 144 The total moment of inertia for the axis through the centre of the wheel at right angles to its plane is then 7181 / = ; Ib.-ft.*, 64 and this axis is a principal axis. The impressed forces are the upward reaction R, the weights Pg, Qg, Cg, Hg, Ag,Jg, ^Sg, the friction + F and the equal and opposite reaction F of the bearing. The effective forces are Pf down and Q f up and the effective forces of the particles of the wheel. If we reverse these forces and apply D'Alembert's principle, we have . + R - Pg - Qg - Cg - Hg - Ag - Jg - *Sg - - Pga + Qgb + Fr - la + Pfa + Q f = o. From (i) we have the pressure on the bearing, Pf -Q-f=, (i) (2) We see that the pressure on the bearing is less than the weight of the apparatus. Let the mass of the apparatus be M, so that M = C + H+ A+J + 4S+Q + P. Then we can write The friction for new bearing is then (page 229) where J* is the coefficient of kinetic friction and ft is the bearing angle. We have also aa. =f, or a = +-. Inserting these values for a and F \n (2), we have for the acceleration /at the circumference / = P P n*\ ~ Q a) 334 KINETICS OF A MATERIAL SYSTEM -ROTATION-FIXED AXIS. [CHAP. III. If ft is small, sin ft = ft, and we have for the given numerical values / = o.og- = 2.895 ft.-per-sec. per sec., and a = *- = 1.737 The acceleration of O is then f = 0.8685 ft.-per-sec. per sec. The velocity of P at the end of the time / = 3 sec. is v =// = 8.685 ft- P er sec ' and the angular velocity of the wheel is o>= = 5.211 radians per sec, a The velocity of Q at the end of / = 3 sec. is v = 2.6055 ft. per sec. a The pressure R on the bearing is R 297.92^ poundals = 297.92 pounds, Whereas the weight of the apparatus is 299 pounds. The friction is F = fiR = 20.8544^- poundals = 20.8544 pounds. The moment of the friction is Fr = i.73787ir poundal-ft. = 1.737875 pound-ft. The moment of the effective forces for all rotating particles is la = 227.8666 poundal-ft. = 227.866 pound-ft. The distance / described by P is * = -ff = 13-0275 ft. The distance described by Q is -t = 3-90825 ft. The tension T r on the right-hand string is T r = P(g /) = 54-6^ poundals = 54.6 pounds. The tension T t on the left-hand string is Tl = Q(# + ~^f) = l6 4-33f poundals = 164.32 pounds. Moment of tension on right = 91.00 pound-ft.. " left = 82.16 Difference = 8.84 and this difference we see is equal to Fr + la, the moment of friction and effective forces of rotating particles, as should be. Work of P = 54.6 x 13.0275 = 711.3015 ft.-lbs. = Ps - , 3f on Q = 164.32 x 3.90825 = 642.2036 " b_ QPv* Difference = 69.0979 ~ * J f " 2 ^r- This difference must be work on wheel and work of friction. Now, work of friction is F = '3-5841 ft.-lbs. CHAP. III.] EXAMPLES. ROTATION. 335 Hence work on wheel = 55.5138 fi.-lbs., and this is equal to GO* (page 321), as should be. , 711.3015 Ihe power of P = - = 237.1005 ft.-4bs. per sec., or The efficiency of the apparatus is 237.1005 = 0.43 horse-power. 550 642.2036 = 0.93. 711.3015 Note. Just as in example (13), the moving force reduced to the circumference is \P Q \g F. If then the reduced mass (page 322) is M a , we have, just as before, Mt In the present case the reduced mass is ~ + ~i a a If we substitute this and the value for F, we have/ directly. CHAPTER IV. COMPOUND PENDULUM. CENTRE OF OSCILLATION. CENTRE OF PERCUSSION. EXPERIMENTAL DETERMINATION OF MOMENT OF INERTIA. EXPERIMENTAL DETERMINATION OF g. Simple Pendulum. A particle suspended from a fixed point O' by a perfectly flexible inextensible string without mass, and swinging under the action of gravity, is called a simple pendulum. It is then a purely ideal conception. Let the mass of the particle be m, the length of string /, the angle with the vertical at ,any instant 6, the angular acceleration a, and the angular velocity co. The impressed forces are the tension T in the string and the weight . mg. The effective forces are mla at right angles to /, and mla? along /towards O'. By D'Alembert's principle, reversing the effective forces and taking mo- ments about O', \ g X / sin 6 ml*a = o, or a = ^ Sin (i) -mla. Instead of a particle of mass m suppose we have a body of mass in. Let this body have the angular acceleration u' about a principal axis through its centre of mass O, and the centre of mass the acceleration a about a parallel axis at O'. Let the distance OO' be d. Then, by the principle of page 317, the moment of the effective forces of rotation about axis at O' is md*a -f la', where /is the moment of inertia of the body for axis at O, and, by D'Alem- bert's principle, ni- X d sin mW 2 la' = O (2) Now if a' = o, that is, if the body has no angular acceleration about the axis through the centre of mass O, we have, as before, and we still have a simple pendulum of length O'O = d, and can treat the body as if it were a particle of equal mass at O. But if the body has an angular acceleration about the axis through O, we can no longer treat the body as a particle and we have no longer a simple pendulum. 336 CHAP. IV.] COMPOUND PENDULUM. 337 Compound Pendulum. If a' = a, then the body has the same angular acceleration about the axis through O that it has about the parallel axis through O' '. This is the case of a rigid body swinging under the action of gravity about a fixed axis at O' . Such a body is called a compound or physical pendulum. It is an actual pendulum and not a purely ideal conception. We have in such case, from (2), _ mgd sin /0j or, since / = m/c 2 , where K is the radius of gyration for the ax s through the centre of mass O, a = gd sin (3) EQUIVALENT SIMPLE PENDULUM. If now we equate (i) and (3), we shall obtain the length / of the equivalent simple pendulum, that is, the length of a simple pendulum which has at any instant the same angular acceleration as the actual pendulum, and which there- fore swings in the same time. We have then, since m(/c 2 -j- d*) = /' == the moment of inertia relative to the axis at O r t ** + <**_ I (4) That is, the length of the equivalent simple pendulum is equal to the -moment of inertia relative to the axis at O' divided by the moment of the mass relative to this axis. CENTRE OF OSCILLATION. If we lay off this distance / given by (4) along O'O, we obtain a point G. This point is called the centre of oscillation or centre of gyration, because it is the point at which, if the whole mass were concentrated, we should have ^ r equivalent simple pendulum, which would vibrate in the same time. . (5) Let OG = p be the distance of this point from the centre of mass O. we have / / d, or, from (4), =-r, or or __ * ' (5) That is, the radius of gyration K is a mean proportional between the distances p and d ojr the centres of oscillation and suspension from the centre of mass. A /so, the distance p is equal to the moment of inertia I relative to a parallel axis at the centre of mass O divided by the moment of the mass relative to the axis of suspension at O' . Suppose now the body turned end for end and suspended from the point G instead of dX. Then, from (4), the length of the equivalent simple pendulum will be V ;X KINETICS OF A MATERIAL SYSTEM ROTATION FIXED AXIS. [CHAP. IV. or, inserting the value of/ from (5), / == or just the same as before. Hence the centre of suspension and oscillation can be interchanged without changing the time of vibration. TIME OF VIBRATION. The time of vibration of the simple pendulum (page 138) is If we put for / its value from (4), we have for the time of vibration of the compound pendulum I tf + d* I~T = 7t \ I - =7rA/=-- / V gd V m ^ (7) Example. The bob of a heavy pendulum contains a spherical cavity filled with water. Determine the motion. ANS. Let m be the mass of the pendulum, ~i< its radius of gyration for axis through centre of mass O of the pendulum, A: 2 , we see, from (3), that R is positive, or in the same direction as F. \{pd < K Z , we have R negative, or opposite in direction to F. \ipd=* ) or the reaction R of the axis at O is zero. In this latter case we have then The point G given by (5) is called the CENTRE OF PERCUSSION, because if a body is struck at this point so that the force F is at right angles to the plane of O'O and the axis of suspension, there will be no reaction of the axis. We see, from page 337, that the centre of percussion is the same as the centre of oscillation. We can find equation (4) directly from our equations (15), page 161, for the position of the instantaneous axis of acceleration. Thus taking origin at O, and the plane of rotation as the plane of XY, we have /=/ 7ny = O, /~ = O, a x = o, a y = o, a g = a. Therefore, from equations (15), page 161, we have for the position of the instantaneous axis of acceleration P. = o. A = ~' A = o. 340 KINETICS OF A MATERIAL SYSTEM-ROTATION-FIXED AXIS. [(HAP. IV. But In/ = F, if there is no reaction R. Hence /= =. Also, by D'Alembert's prin- Fp ciple, Fp = /or, or a = -j~. Substituting, we have which is the same as equation (4). To prevent a hammer from "jarring " the hand, or reacting upon a fixed axis about which it turns, the direction of the force at the moment of striking should pass through the centre of percussion G at right angles to the line O"O. Suppose we grasp a prismatic rod O 1 A of length /at the end Cf and strike an obstacle with a force F at right angles to the line O"O, where O is the centre of mass. , Then, from (3), the force R at O is A G P O d O In the case of the rod / = (page 38), and hence ** = . and d = -. Hence If then the obstacle is struck by O, p o and R = , that is, the force on the hand is one fourth of that on the obstacle and opposite in direction. / F If the obstacle is struck by A, p = and R = + , or the force on the hand is one half of that on the obstacle and in the same direction. If the obstacle is struck by G, so that p = -7, R o, and there will be no force on the hand. G is the centre of percussion. Experimental Determination of Moment of Inertia. From the principles of page 337 we can determine experimentally the moment of inertia of a body relative to any axis. ist METHOD. Thus we have, from equation (7), page 338, (i) We must first, then, determine the mass m of the body and locate its centre of mass O. Then suspend the body from an axis at O' and measure the distance O' O = d. Then swing the body about this axis and note the time / of vibration. The moment of inertia /' relative to the axis at O is then given by (i). The moment of inertia / relative to a parallel axis through the centre of mass is then given by 1=1' mW 2 . 2d METHOD. From equation (4), page 337, we have for the distance O'G from the axis of suspension to the centre of oscillation (2) We must first, then, determine the mass m of the body and locate its centre of mass O as before. Then suspend the body from an > axis at O' and measure the distance O'O = d. CHAP. IV.] EXPERIMENTAL DETERMINATION OF MOMENT OF INERTIA. 341 Then swing the body about this axis and note the time of vibration. Then turn the body over and suspend it from another point G in the line O'O at such a distance from O, found by trial, that the time of vibration about a parallel axis at G is unchanged, and measure O'G. The point G is the centre of oscillation, and the moment of inertia I' relative to the axis at O is then given by (2). The moment of inertia for a parallel axis through the centre of mass is then, as before, 1 = 1' -md z . This method does not involve knowing the value of g. Experimental Determination of g, Having thus found m, s and I' and the time of vibration t, we have, from equation (7), page 338, 7T 2 /' We can thus determine g by pendulum observations. The quantity -=-j is the pendulum constant which m^ Once known, the observation of / at any locality gives at once the value of g. The quantity -=-j is the pendulum constant which must be accurately determined. CHAPTER V. IMPACT. Impact. When two moving bodies come into collision the straight line normal to the compressed surfaces at the point of contact is the line of impact. If the centre of mass of the two bodies is upon this line, the impact is central ; if not, we have eccentric impact. When we consider the direction of motion, we can distinguish direct impact when the line of impact coincides with the direction of motion, and oblique impact when the line of impact does not coincide with the direction of motion. Thus in Fig. I if the two bodies move in the direc- tions u l and 2 , we have oblique central impact, and in Fig. 2 we have oblique eccentric impact. If in Fig. I the directions of motion u l and ;/ 2 coincided with Ofl^ , we should have direct central impact. If in Fig. 2 the direction of motion ?/, coincided with O^N, and # 2 were parallel, we should have direct eccentric impact. Direct Central Impact Non-Elastic. We can evidently consider the bodies in direct central impact as particles. Let m l and u^ be the mass and initial velocity of one particle before impact, and m 2 , u 2 of the other before impact. Let u l be greater than HI and in the same straight line. Let the direction of u l be positive, the opposite direction negative. When the bodies meet there is a short interval of compression, at the end of which both masses have a common velocity v. If the bodies are non-elastic, they remain in contact with the / _ y ( common velocity v. By the principle of conservation of the centre of mass (page 300), if there are no external forces the motion of the centre of mass is unaffected by the impact. We have then _ = mft ' Or, by the principle of conservation of momentum (page 300), we have m l u l -f- m t u z = (m l + m^v, (i) or the momentum before equals the momentum after impact. In equation (i) a velocity opposite to w t is to be taken as negative. The energy before impact is 342 CHAP. V.] DIRECT CENTRAL IMPACT NON-ELASTIC. 343 and the energy after impact is & = 2 K + "** The loss of energy is then This work causes deformation and rise of temperature. If we take mass in Ibs. and velocity in ft. per sec., equation (2) gives loss of energy in ft.-poundals. If we wish foot-pounds, we must divide by^*, or in foot-pounds We call ~T^~ ^ e h armon * c mean between m^ and m 2 , and *-3 is the height due to the difference of the velocities. Hence the loss of energy in non-elastic impact is equal to the product of the harmonic mean of the masses and the height due to the difference of their velocities. If the mass ; 2 is at rest, we have 2 = o, and _ = If the bodies move toward each other, 2 is negative, and In this case if the momentums of the bodies are equal, or mji^ = mji v v = o, or the bodies come to rest. If, on the contrary, the masses are equal, we have 2 2 ' 2g If the bodies move in the same direction and the mass of the one in advance, m 2 , is infinite, we have or the velocity of the infinite mass is not changed by the impact. If the infinite mass is at rest, or # 2 = o, we have Examples. (i) A non-elastic body of mass ;i = 5 Ibs. moving with a velocity u> = 7 ft. per sec. impinges centrally upon another of mass m, 3 Ibs. moving in the same direction with a velocity of u* = j feet per sec. Find the velocity with which the two move on together after the collision, and the work expended in deformation and heating. ANS. v = 5 ft. per sec. in the direction of u\ \ ft.-pounds. 344 KINETICS OF A MATERIAL SYSTEM-IMPACT. [CHAP. V. (2) In order to cause a non-elastic body weighing no Ibs. to change its velocity from t, to 2 feet per sec., we let a non-elastic body weighing 5 Ibs. strike it. Find the velocity of the latter body, and the work expended in deformation and heating. ANS. 3.2 ft.-per-sec. ; - ft.-pounds. (3) Two non-elastic masses of 3 and 5 tons impinge centrally with velocities of 4 and 5-5 ft- per sec. respectively. Find their final velocity when moving in the same and opposite directions. ANS. 4lf ft. per sec.; i^| ft. per sec. in the direction of the larger velocity. (4) Two non-elastic bodies of 3 Ibs. and i oz. are moving in opposite directions and impinge centrally . The first has a velocity of 3\ and the latter of 9 ft. per sec. In what direction do they move after impact? ANS. In the direction of the first with a velocity 3jj ft. per sec. (5) A non-elastic body whose mass is it> Ibs. moving with a velocity of 23 miles an hour impinges centrally on another moving in the opposite direction. The two come to rest. If the mass of the latter were 28 Ibs. , find its velocity. If the velocity of the latter were 66ft. per sec. , find its mass. ANS. 14^ miles per hour; 8| Ibs. (6) A number of non-elastic balls of masses mi,m t .m m n lie on a straight line at rest. If the first have a velocity of i toward the others, what will be the ultimate velocity of all? ANS. v n (7) If in the preceding example the initial velocities are ,,,,* . . . u H ,find the ultimate velocity. miUi + m*ui + . . . m n u n ANS - " = mi + mt + ... ,, (8) If in a machine ib impacts per minute occur between two non-elastic masses mi = too Ibs. and my = 1200 Ibs. moving in the velocities u\ = 5 and ut = 2 ft. per sec., find the loss of energy. 16 (5 2)* 1000 x 1 200 ANS. ^-. . = 20.3 ft.-pounds per sec. 60 ig 2200 (9) If two trains m\ = 120000 Ibs. and m* = 1 60000 Ibs. come into collision with the opposite velocities u\ = 20 and u t = 15 ft. per sec., find the loss of energy which is expended in the destruction of the cars, considering them as non-elastic. (20 + I?)* 120000 X 160000 ANS. ~~~ 28oooo = i 302000 ft.-pounds. Direct Central Impact Perfect Elasticity. When two bodies collide, and the impact is direct central, there is a short interval of compression at the end of which both masses have a common velocity v. If the bodies are non-elastic, they remain in contact. If, however, they are elastic, there is another short interval of expansion at the end of which m l has the final velocity Z' P and m^ the final velocity v 2 . If the bodies are perfectly elastic, the works of compression and of expansion are equal, and hence no energy is lost. 'By the principle of conservation of the centre of mass (page 300) we have the velocity F of the centre of mass, _ _ ?,! -j- or M l U l + M a U s =ttt l V l + 9Hft ......... (I) That is, by the principle of conservation of momentum the momentum before equals the momentum after impact. Also, since for perfectly elastic impact no energy is lost, the energy before must equal the energy after impact, or *l**+ 1 **f* S *>l+ m * v ........ (2) CHAP. V.] DIRECT CENTRAL IMPACT PERFECT ELASTICITY. 345 From (i) and (2) we have at once = _ , + If we subtract the first of these from the second, we have That is, the velocity of approach (u l 2 ) equals the velocity of separation (v 2 vj. The loss and gain of velocity are _ or /WHY- ! = ! , ^ 2 = 0. That is, the velocity of the impinging body is reversed. Examples. (i) Two perfectly elastic balls weighing to Ibs. and 16 Ibs. collide centrally with the velocities 12 and 6 feet per sec. Find the loss and gain of velocity and the final velocities after collision, if the initial velocities are in the same and in opposite direction. ANS. In the first case the final velocities are Vi + 4^- and v, = + IO T \ ft. per sec. The first body loses, then, 7^ and the other gains 4 T 8 5 ft. per sec. In the second case the final velocities are Vi = 10^ and v 9 + 7H ft. per sec. Each body then rebounds with these velocities. The first body loses 22 T 2 3 and the other gains 13/5 ft. per sec. (2) A number of perfectly elastic balls of masses nti, m-t , m s . . . m n lie in a straight line at rest. If the first have a velocity of u\ toward the others, find their velocities after impact. 346 KINETICS OF A MATERIAL SYSTEM IMPACT. [CHAP. V. ANS. The velocity of the first is (mi m t )ui v\ = . Ml + Mt The velocity of any intermediate ball is ti . m t . VH = The velocity of the last ball is (mi + m,)(m t + m t ) . . . (m n + MH+I) (3) In the preceding example let there be four balls, the mass of the first mi, of the second m t = ami , of the third m t = am t = a*Mi , of the fourth m 4 = am* = a"/i. ' I +a l + a (i + a)"' If, for example, the mass of each ball is one half that of the preceding, i 4 16 64 Vl =-Ui, V* = ~Ui, V* = Ui, V.= ~Ui. Coefficient of Elasticity. No body is perfectly elastic or absolutely non-elastic. We deal with bodies more or less elastic, that is, ^ imperfectly elastic. Let a prismatic body of length / and area of cross-section A be acted upon by a stress 5 in its axis, and let the elongation or compression or, in general, the strain be denoted by A. We know by experiment that within a certain limit, 3< [ ^] - ^5 twice, three times, or four times, etc., the stress S will cause a strain of 2 A, 3A, 4!, etc. The limit up to which this law of proportionality of stress to strain holds true, for any material, is called the elastic limit for that material for the kind of stress under consideration. Thus if we lay off the successive stresses to any convenient scale horizontally, and lay off the corresponding observed strains to any convenient scale vertically, we obtain within the elastic limit a straight line AB. The co-ordinates AD and DP of any point P of this line give the stress and corresponding strain. The point B at which the straight line begins to curve gives the stress AL, and this is the elastic limit stress. We see, then, that within the elastic limit the ratio ^- of A A S 2S D L stress to strain is constant. Now if A is the area of cross-section of the prism, then - is the unit stress, or stress per square inch. Also, if / is the original length, then y is the A * unit strain, or strain per unit of length. If then the experiment were made upon a prism of one unit area and one unit length, we should have within the elastic limit the ratio - + - t or -jj- constant. This constant for any material is called the coefficient of A / An. elasticity for that material, for the kind of stress under consideration. We denote it by E. We have then within the elastic limit and we can define the coefficient of elasticity in any case as the unit stress divided by the unit strain. Values of E for various materials, as determined by experiment, are given on page 478. CHAP. V.] MODULUS OF ELASTICITY. 347 Examples. (i) A wrought-iron rod 30 feet long and 4 square inches in cross-section is subjected to a tensile stress of 4000 pounds. The elongation is found to be o.oi ft. Find E. ANS. The unit stress is = 400 = 1000 pounds per square inch. The unit strain is = - ft A 4 * 3000 per foot of length. We have then E = ry = 30 oooooo pounds per square inch. Ah- (2) Taking E for -wrought iron as thus determined, find the compression of a wrought-iron prism 2 ft. long and 12 square inches cross-section under a stress of 1 50000 pounds. SI 150000 x 24 i . ANS. A = ^ = inch. AE 12 x 30000000 loo Modulus of Elasticity. Let F c be the force of compression when two bodies collide at the end of the period of compression, and F r the force of expansion or force of restitution at the beginning of the period of expansion. If a body is non-elastic, F r is zero. If a body is perfectly elastic, F r = F c . Strictly speaking, no bodies are absolutely non-elastic or perfectly elastic. We deal .with bodies imperfectly elastic where F r is less than F c . The ratio ~ of the force of restitution to the force of compression is found by experiment to be a constant for any body so long as the limit of elasticity is not exceeded. This ratio we denote by e and call it the modulus of elasticity. We have then If we let a sphere of mass m fall from a height h upon a -relatively very large rigidly supported mass of the same material, it will rebound to a height ft , less than the original height. The velocity with which it strikes is u x = M2gh, and the velocity of rebound is v l ^2gli' . If the interval of compression / is very short, we have the force of compression and the force of restitution mv, m t We have then We can thus experimentally determine the modulus of elasticity e for various materials. For perfect elasticity e\ t and for non-elastic bodies e o. The following average values have been thus determined by experiment : Cast iron, e = I nearly. Glass, e = ^. Steel, e = --. Clay, wood, e = o nearly. Direct Central Impact Imperfect Elasticity. Let F e be the force at the end of the period of compression when both bodies are moving with the common velocity 348 KINETICS OF A MATERIAL SYSTEM IMPACT. [CHAP. V. Let A, be the compression of m l , and A 3 of m r Then the work of compression is, since -F t is the average force, This work should equal the energy lost during compression, or -F e (X, + A 2 ) = -mp* + l -m^ - l -(m, + m 2 )v*, or putting for v its value, we obtain for F c in pounds, if we take mass in Ibs. and distance in feet and velocity in ft. per sec., Let F r ' be the force of restitution at the beginning of the period of expansion of m l , and A,' its expansion. Let F r " be the force of restitution at the beginning of the period of expansion of m z , and A 2 ' its expansion. Then the work of expansion is - 2r r . The total energy lost then is, in foot-pounds, L/r(A -f AJ - -F r '\' - -F r "\' = m,u* + *, -- -*n.v* mj>*. (2) 2 A i T t) 2 r i 2 2 2g i i -T 2g 2 2 2g i i 2g * * Now, from page 346, we have J ^' H J ^ Al = 4^ and A2== ^" 2 ' where /j and / 2 are the lengths of the masses ; t and m z , ^4 a and ^4 2 their areas of cross- section, and EI and E 2 their coefficients of elasticity. For the sake of simplicity we put the quantities f = ^ = ", and % = *&=*, ....... (3) Aj ^ A 2 1 2 AE The quantity // = j- in general we call the hardness of a body. The hardness of a body is then the ratio of the stress to strain. We can write, then, F' F e Aj = and A 2 = -^ .......... (4) where H l and ff t are given by (3). We also have from page 347, since stress and strain are proportional, Fr V F r " V y iaB ir *> and :F~ := r = ' 2 ' V *| * f A 2 where ^ and e z are the moduli of elasticity for the masses ;, and ; 2 . We have then F r ' = fl F et F," = f2 F e , V = ^. V = ^r .... (5) CHAP. V.] DIRECT CENTRAL IMPACT IMPERFECT ELASTICITY. 349 If we substitute the values for \, A 2 , F/, F r " , A/ and A 2 ', given by equations (4) and (5), in equations (i) and (2), we obtain 2 e L HI H z J 2g ' 2- 2^ We also have, from the principle of conservation of momentum (page 300), From these three equations we obtain + Equations (7) are general and give the final velocities v v and v 2 of w x and m^. If both masses are of the same material, H^ = H z and e l = e z = e, and we have (8) If the bodies are perfectly elastic, we have e l = e^ = i and equations (7) reduce to equations (3), page 345. If the bodies are non-elastic, e l = o, or, taking^- = 32 ft.-per-sec. per sec., F c = 247 350 pounds. From (4) we have for the compression of the sledge and plate /r 7^ AI= - =0.0016 ft. = 0.0127 in., A a = =0.0147 ft. =0.177 i. Hi fit 350 KINETICS OF A MATERIAL SYSTEM- IMPACT. [CHAP. V. (2) In the preceding example consider the sledge as perfectly elastic and tne plate as inelastic. ANS. We have = o, m\ = o , e\ = i and e t = o. Hence, from (7), = s - That is, the sledge rebounds with this velocity. (3) Find the velocities after impact of two steel masses, if the velocities before impact are *, = 10 and *t = 6ft. per sec., and the masses mi = jo Ids., m* = 40 Ibs., taking e = -. ANS. w = 10 16 . ^f i + -J = - 4.22 ft. per sec., */, = - 6 + 16 ^(i + |j = + 4.665 ftpersec. That is, m\ rebounds and m t rebounds. Earth Consolidation. When a maul strikes a mass of soft earth it compresses the earth with a certain force F. Let s be the depth of penetration, m the mass of the maul, and h the height from which it is let fall. Then the energy of the maul before it is dropped is mh. Since this energy is expended in compressing the soil, we have ff , T? mk Fs = mn, or F = . s If we divide this force F by the cross-section A of the maul, we have for the unit force of compression F _mk ~A~~As The resistance F of soils to the penetration of the maul is generally variable and increases with the depth s of penetration. In many cases we may assume it to increase directly with the penetration. In such case we should have Fs = mh, or F = , and -r = 3 , or twice as much as before. p If A is taken in square inches and s and h in feet or inches and m in Ibs., is the number of pounds per square inch resistance of the soil. Allowing a factor of safety of IO, we could then safely load the compacted soil up to --.-. IO /i Example. A maul whose mass is 120 Ibs. falls from a height of 18 inches, and the earth is compressed one eighth of an inch by the last blow. The cross-section is 16 square inches. Taking a factor of safety of io,find the safe load. ANS. = ^ = = 1080 pounds per sq. inch. Taking a factor of safety of 10, we have 108 i6Xg pounds per sq. inch safe load. CHAP. V.] OBLIQUE CENTRAL IMPACT. 353 making the angle with O t O t given by tan /ffj = sin - - tan (8) For non-elastic bodies e = o, and for perfectly elastic bodies e = I. For non-elastic bodies, then, from (6), (7), (8), we have v l = o, > w l = M I sinar l , tan Pi = oo or ft = 90. That is, the velocity v l along the line of impact is annihilated, that at right angles is unchanged, and the body moves after impact in the direction O^F at right angles to O^O^ with the speed u v sin a r For perfectly elastic bodies v l = x cos a l , w^ = u^ , tan /?j = tan a r That is, the velocity along the line of impact is reversed, the angle of reflection NO^B is equal to the angle of incidence NO^A, and the body moves after impact in the direction O^B with the original speed u r For imperfect elasticity we have, from (8), tan tan /V or the modulus of elasticity is equal to the ratio of the tangent of the angle of incidence to the tangent of the angle of reflection. Example. Two balls, m\ = 30 Ibs., m* = 30 Ibs., impinge with the velocities u\ = 20 and u* = 25ft. per sec., making the angles with the line of impact cti = 21" sf and a, = 65 20'. Find the velocities after impact if the bodies are non-elastic. Hence ANS. ui sin a, = 7.357 ft. per sec., u, sin or = 22.719 ft. per sec. Ui cos a, = 18.598 " " u t cos 0.1 = 10.433 " " " vi = 18.598 - (18.598 - 10.433) 3 = 1 3-495 ^ P er sec.. v t = 10.433 + (18.598 - 10.433) 3= J3-495 ..... ' The resulting velocities are then = V 1 3.495" + 7.357 s = 15.37 ft. per sec., w* = | /I 3-495 2 + 22.719* = 26.42 making the angles ft\ and ft* with the line of impact given by ' or * = = 59 if. Friction of Oblique Central Impact. The pressure between the colliding bodies gives rise to friction. If Pis the pressure due to impact, F the friction and // the coefficient of friction, we have F = JiP. 354 KINETICS OF A MATERIAL SYSTEM-IMPACT. [CHAP. V Let the mass of the impinging body be ; 1 , and the initial and final velocity along the line of impact be u l and i lt and / be the very short time of impact. Then we have for the impulse (page 255) = *,(.,-,,), or F = "-'<"' Hence the friction is That is, the impulse of the friction divided by the mass is equal to // times the change of velocity along the line of impact ; or the change of velocity due to friction, at right angles to the line of impact, is equal to /* times the change of velocity along the line of impact. This change of velocity is always a retardation, since friction is a retarding force. Thus if a mass w, falls vertically with a velocity ;/, upon a horizontal sled of mass m z moving horizontally with the velocity u 2 , and if the velocity u l is entirely lost by the collision, we have for the friction But the retarding force during the time / for both masses in contact, if v is the common velocity after impact, is also Hence we have for the change of velocity of the sled If a body of mass m^ strikes an immovable mass with a velocity x at an angle a lt we have from equations (7), page 349, for the change of velocity along the line of impact, since u 2 = o and m. 2 = 00, / 2 = oo and , cos a l i\ = ;/, cos ari ' 2 Hence the change of velocity due to friction is ^u l cosa,(l +',), and after impact the component w, sin , becomes j sin ! /^M, cosor,(l + ^ 2 ) = [sin ai fit cos ^(i + ^ 2 )] w i For perfectly elastic bodies e 2 = I and (3) becomes (sin a l 2J* cos ,)^ 1 , CHAP. V.] FR.ICT10K OF OBLIQUE CENTRAL IMPACT. 355 and for non-elastic bodies e 2 = o and (3) becomes (sin a 1 /* cos a^)u r The friction often causes bodies to turn about an axis through the centre' of mass, or if before impact rotation exists, that motion is changed. Thus let r be the radius of a sphere, O3 l its initial and GO its final angular velocity during the time t of impact. Then the moment of the friction is / . -- - (page 322). But the moment of the . . m^u cos #,(1 -j- e 2 }r friction is also - ! -- -=- . Hence Equation (4) gives the change of angular velocity. Example. A billiard-ball strikes the cushion with a velocity u\ = 15 ft. per sec., the angle of incidence being ai = 45* . // e t = 0.55 and the coefficient of friction is fj. = 0.2, find the motion after impact. ANS. The velocity after impact along the line of impact is v . = e-iUi cos ori = 0.55 X 15 cos 45 = 5.833 ft. per sec. The velocity parallel to the cushion is i sin tt t jj.ui cos a^i +' The ball also acquires the angular velocity oo z about a vertical axis through the centre of mass. Since / = Wir a , where tn t is the mass and r the radius of the ball : o.2m\ x 15 cos 4; x i.Z^r 8.22 oo z = -- = - ^ -- 32 - 22. = radians per sec. 2 ' r mir* The ball also has angular velocity ao x about a horizontal axis through the centre of mass at right angles to iv\ given by wi 9-36 oo x = = - radians per sec. The resultant angular velocity is then oo = 4/Go x * + aoS = radians per sec. about an axis through the centre of mass in a vertical plane at right angles to w Jt making an angle with the vertical whose tangent is ~? = |^ = 1.138, or 48 42'. '35 6 KINETICS OF A MATERIAL SYSTEM-IMPACT. [CHAP. V. Strength and Impact. Let the mass ** t , moving with the velocity , , impinge on the mass m t which is supported by the rod AB of uniform cross- section A and length /. Let v be the velocity of both masses during impact. Then, by conservation of momentum, / \ u \ ( m \ _ or v = 1 and the work in foot-pounds necessary to bring the combined masses *oi m work = where - = h is the height of fall of m r This work is equal to the work of stretching or compressing the rod. If Fis the force of impact at the end of the compression or stretch, A, is the average force and . A is the work. But, from page 346, within the limit of elasticity we have where E is the coefficient of elasticity. Hence /I 2 2/ *, + **,' From (3) we can find the strain A of the rod caused by the impact. Let the rod be strained up to the elastic limit unit stress S e . Then we have from (2), by putting F= SfA, w and hence, from (3), But ^4/ is the volume of the rod V. The velocity of impact which is necessary to strain the rod up to the limit of elasticity, is then given by The quantity r is called the coefficient of resilience (page 516). We see from (5) that the greater the volume or mass of the rod, the greater the blow it can bear. Hence the mass of bodies subjected to impact should be made as great as possible. Since / t and > 2 fall during impact through the distance A, we have more correctly CHAP V -] STRENGTH AND IMPACT. 357 and hence, instead of (5), we have If, finally, we wish to take into account the mass m^ of the rod, we have, since its centre of mass moves through the distance -A, and hence, instead of (5), we have , . . ^ " s^ ~~<~ ^- If a mass m l moving with a velocity u^ puts in motion another mass, m t , by means of a chain or rope, we have in the same way for the velocity of both bodies during impact and the work in foot-pounds expended in stretching the chain is ME.**, ! work = -*-* ( m where // = is the height due to the velocity r We have then, if the chain is stretched to the elastic limit unit stress S e , S? A ,_ mjn* ' 7-* -ti ^ j * /fr 2h m^ -|- m 2 where A is the cross-section and /the length of the chain. Hence k ^ = ^^ S 4.Al. . (8) 2g m v m z 2E Example. The two opposite suspension rods of a suspension bridge support a constant load of 5000 pounds, S a which is increased by 6000 pounds by a passing load. The coefficient of resilience ^p- for wroitght iron is 7 poitnds per square inch (page jv<5). The length of rod is 200 inches, and cross-section 1.3 square inches. Find the height of fall to stretch the rods to the limit of elasticity. ANS. h = (5QOO + 6000) x 7 x 200 x 1.5 x 2 ^ ^ inches 6000" If then a cart of 6boo pounds should pass over an obstacle of 1.3 inches high, and drop, the rods are in danger of being stretched beyond the elastic limit, '356 KINETICS OF A MATERIAL SYSTEM IMPACT. [CHAP. V. Strength and Impact. Let the mass w t , moving with the velocity u^ , impinge on the mass w 2 which is supported by the rod AB of uniform cross- section A and length /. Let v be the velocity of both masses during impact. Then, by conservation of momentum, m \ u \ = ( w > + t )v, or v m \ l m 1 ~T 2 ' and the work in foot-pounds necessary to bring the combined masses to rest is work = *g w, + w 2 /! -f- w 2 ' (0 # * where L = h is the height of fall of m r This work is equal to the work of stretching or compressing the rod. If Fis the force of impact at the end of the compression or stretch, A, - is the average force and - . A is the work. But, from page 346, within the limit of elasticity we have EA\ (2) where E is the coefficient of elasticity. Hence i EAK m*h 2l or A = m l + mi ' EA ' (3) From (3) we can find the strain A of the rod caused by the impact. Let the rod be strained up to the elastic limit unit stress S e . Then we have from (2), by putting F= S^ , = - (4) and hence, from (3), 2E But Al is the volume of the rod V. The velocity of impact which is necessary to strain the rod up to the limit of elasticity, is then given by ' 2E (5) The quantity ^r is called the coefficient of resilience (page 516). We see from (5) that the greater the volume or mass of the rod, the greater the blow it can bear. Hence the mass of bodies subjected to impact should be made as great as possible. Since m l and m a fall during impact through the distance A, we have more correctly work = i CHAP V -J STRENGTH AND IMPACT. 357 and hence, instead of (5), we have m, + m z Sl_ (m v -f mtf S e l If, finally, we wish to take into account the mass #/ 3 of the rod, we have, since its centre of mass moves through the distance A, + (* + ">>+?,)*, and hence, instead of (5), we have m ' 2E ' ' ~~~ ' ~' If a mass w x moving with a velocity x puts in motion another mass, m 2 , by means of a chain or rope, we have in the same way for the velocity of both bodies during impact and the work in foot-pounds expended in stretching the chain is m.u 2 i m.m 9 u 2 m m work = --- (m -f mXv 2 = - \-*- . -i- . m i m * L 2 * ^ where h = is the height due to the velocity u r We have then, if the chain is stretched to the elastic limit unit stress S e , . 2h m^ -f- m z where A is the cross-section and /the length of the chain. Hence Example. The two opposite suspension rods of a suspension bridge support a constant load of 3000 pounds, which is increased by 6000 pounds by a passing load. The coefficient of resilience ^g for wrought iron is 7 pounds per square inch {page j/d). The length of rod is 200 inches, and cross-section 1.5 square inches. Find the height of fall to stretch the rods to the limit of elasticity. ANS. h = (5QOO + 6000) x 7 _x 200 x 1.5 x 2 = i<2g . nches< 6000" If then a cart of 6000 pounds should pass over an obstacle of 1.3 inches high, and drop, the rods are in danger of being stretched beyond the, elastic limit, 358 KINETICS OF A MATERIAL SYSTEM-IMPACT. [CuAP. V. Impact of Beams. Let a mass m l fall from a height // upon a beam AB of uniform cross-section A and span /, supported at the ends. Let 6 be the density of the beam, and Fits volume. * Then =5? V=Al t and the mass m, of the beam is Let the mass m l strike the beam at the centre C, and let the greatest velocity at the centre be v e , and at any other point of the beam be v. Then the work in foot-pounds necessary to bring the combined masses to rest is work = If y is the deflection at any point, and A the deflection at the centre, we have We have then mtf , *Av* r^dx work = - I - *g 2g J ^ Now if Pis the pressure at the centre, we have, from page 543. Pt s where E is the coefficient of elasticity, / the moment of inertia of the cross-section of the beam relative to the horizontal axis through the centre of mass of the cross-section at right angles to the length of the beam, and x is the distance of any point from the left end. We have then and substituting in the expression for the work and integrating, we have work = . . 2ff 35 2f The distance through which the point C moves in the short time /is -' . /. If we divide the work by this distance, we have the force which brings the bodies to rest. This force should equal -~ ' Hence , 176 Al > mfl ~ mf . +-., or ,,= CHAP. V ] IMPACT OF BEAMS. . 359 If we insert this value of v t , \ve have for the work of bringing the combined masses to rest ^ work = m.-\ l 35 / 35 But this work is also equal to Pd. If the elastic limit is reached, we have, from page 500, and ^= vl \2vE' where S e is the elastic limit unit stress, and v is the distance of the most remote fibre of the cross-section from the neutral axis. Hence \JdAl I - s < n m * h i,-?L !L 35 * * l ~'" Since dAl is the mass m 2 of the beam, // If, for instance, the cross-section of the beam is a rectangle of breadth b and depth d, we have / = bd* and v = -d. Hence for this case 12 2 s. w > * ~a2* 9 ' , 2 Example, Find the height from which a mass of 200 Ibs. must fall in order that, striking the centre of a plate of cast iron j6 inches long, 12 inches wide and 3 inches thick, supported at both ends, it may bend it to the elastic timit. ANS. If we take the coefficient of resilience (page 516), = 1.2 pounds per square inch, iE we have, since S for cast iron is about 0.259 Ibs. per cubic inch, m* = 12x3x36x0.259 = 335.7 Ibs. Hence for height of fall / = 1.2x12x3x36 aoo + ^ 335-7 = 9 40000 3 6 KINETICS OF A MATERIAL SYSTEM-IMPACT. [CHAP. V. Impact of Rotating Bodies. Let two bodies of mass ;, and m 2 rotate about fixed axes at O l and O a and impinge, and let AB be the line of impact. Let the normals O^A = a l and O 2 B = a t . Then (page 322) we can reduce the masses m l and ; 2 to the equivalent masses at A and B w,*-,' 2 w 2 /c/ 1 a ! and -nr"! where */ and /c- 2 ' are the radii of gyration relative to O l and 2 . If then we substitute these masses in the place of ;, and ; 2 in the equations for central impact (page 349), we have for bodies of the same material ' :,'V where , and 2 are the velocities at ^4 and .# before and ^ , ?' 2 after impact, and e the modulus of elasticity. If e, and e 2 are the angular velocities before and 00, , &? 2 after impact, we have, taking counter-clockwise rotation as positive, the origins at (7j and O 2 ; and n l , ti 3 as coinciding with the axes of Y for each origin, = u 2 , a l co l = ?', Hence G7, = 6, ,,' 2 Examples. The moment of inertia of the shaft O\B relative to its axis of rotation at O\ is m\K\'* = 40000 Ib.-ft*, ami that of the trip-hammer BO* relative to its axis of rotation at O, zs w,Ky = 130000 lb.-ft*. The arm O\B of the shaft is a\ = 2 ft., and that of the hammer BO* is tit = 6 ft. The angular velocity of the shaft before impact is f\ = 1.05 radians per sec. Find the velocity after impact and the loss of energy at each impact, supposing both bodies inelastic. ANS. The angular velocity of the shaft after impact is, since ,.o 5 jcj5ooc _ = . 74 , radians per sec . 40000 x 36 + i 50000 x 4 The angular velocity of the hammer after impact is O , 6 x 2 x 1.05 a, = - -= = 0.247 radians per sec. The loss of energy at each impact is, in foot-pounds, = 20 .. 63 foot-pounds. CHAP. V.] IMPACT OF AN OSCILLATING BODY. 3 6t Impact of an Oscillating Body. If a body of mass m l impinges upon a body of mass w/ 2 which is suspended from an axis at O 2 , the equations of the preceding article apply if we put in equations (i) m l in place of ^ , an< ^ a 2 e z m P^ ace f ^ 2 > *i I Og and a. 2 &). 2 in place of v 2 . We have then for the velocity of the mass ;//j after impact, taking clockwise rotation as positive, origin at O 2 , and O 2 'O 2 as axis of Y, if /c/ and /c 2 ' are the radii of gyration for axes at O l m/ and O 2 , and for the angular velocity of the mass m 2 after impact If the mass m 2 were at rest before impact, we have e 2 = o and 07 2 = x ( I -j- ^) (4) If m l is at rest and the oscillating body impinges on it, we have u r = o, and hence (6) yo c *> I v* i *" / -^ I . #a i " The angular velocity &> 2 in the first case, equation (4), or the velocity z/ x of w x in the second case, equation (5), is a maximum when is a maximum, or when is a minimum. Putting the first differential coefficient equal to zero, we have for the value of a 2 when the maximum velocity is imparted (7) 3 62 KINETICS OF A MATERIAL SYSTEM. [CHAI-. V. Hence the maximum velocity imparted to the oscillating body w 2 when at rest and struck by ;//, is given by 2/c, V /. and the maximum velocity imparted to m l when at rest and struck by #/ 2 is (8) (9) REACTION OF THE Axis. Let the force of impact be F, and the reaction of the axis be R. Then, from page 339, (10) If we give to a 2 its value from (7), we have for the reaction of the axis when the maxi- mum velocity is imparted ;n) The centre of percussion (page 339) is at the distance from the axis. If the impact takes place at this distance, there is no reaction of the axis. Ballistic Pendulum. The ballistic pendulum consists of a large mass / 2 hung from a horizontal axis O' ' . It is set in oscillation by a cannon-ball shot against it, and is used to determine the velocity of the ball. In order to render the impact inelastic, the mass ; 2 consists of a box filled with sand or clay, so that the ball enters the mass and oscillates with it. In order to determine the velocity of the ball, the angle of oscillation is measured by a pointer directly below the centre of mass O, which moves on a graduated arc AB. Let m l be the mass of the ball. Then, from equation (4) of the preceding article, making e = O, we have for the angular velocity after impact (O where /c 2 ' is the radius of gyration of the pendulum relative to the axis at O', and a 2 is the distance of the point of impact below this axis. Let / be the length of the equivalent simple pendulum which oscillates in the same time as the ballistic pendulum, and let the angle measured on the arc AB be 0. CHAP. V.] ECCENTRIC IMPACT. 3 6 3 We have then for the simple pendulum (page 336) the angular acceleration g sin If O' O = d is the distance of the centre of mass O of the ballistic pendulum from O' we have (page 317) (m l -f- m^gd sin = I' a = (m^a* + m 2 K^}a, or \n Equating these two values of a, we obtain for the length of the equivalent simple pen- dulum l = m fa + Zjd (2) The height of displacement is a h = I /cos e = 2l sin 2 -. Hence the velocity at the lowest point of swing is 3 v = V2gh = 2 Vgl sin ~> and the corresponding angular velocity is 00 = 7 = 2 V 7 ' sin I- Equating this to (i) and inserting the value of / from (2), we have for the velocity of the ball (m,-\-m<\d ,f 6 , \ u,=2\- 2 -i/^/.sm- . (3) If the pendulum makes vibrations per minute, the duration of a vibration is 77 60 =7f\/- , V g n t=7f\- and hence g Hence Eccentric Impact. Let ^, , m 2 be the masses, 6> t and O 2 their centres m, of mass, BN the line of impact, and BO 2 = p the distance from the line of ( Also, at the instant of impact v, =p* " 4"" The velocity at any point distant x from O after impact is where/ and x are positive towards B and negative towards A. Hence for A striking and for B striking After impact the centre moves in the same vertical with the uniform acceleration^, while the angular velocity &?., remains unchanged. 3 66 KINETICS OF A MATERIAL SYSTEM. [CHAP. V. (2) An inextensible string is wound around a cylinder and has its free end attached to a fixed point. The cylinder falls through a height h, and at the instant of impact the string is vertical and tangent to the cylinder. Find the motion after impact. ANS. v = (3) An iron ball of mass m\ = 65 Ibs. moving with a velocity of jb ft. Per sec. strikes a pine beam of uniform rectangular cross-section in the centre line of a side and at ri^ht angles, at a distance p = t\ft. above the centre of mass. The mass of the beam is 842.4 Ibs., its length j ft. and breadth 2 ft. If the beam is at rest, find the motion after impact, con- sidering the impact as non-elastic. ANS. The moment of inertia (page 38) is the same as for m - concentrated at a corner We have then 2.4i6; a , or 2.416. Hence the velocity of the ball after impact is 4 The velocity of the centre of mass of the beam after impact is 2.364 ft. per sec. The angular velocity of the beam after impact is mipui OJ, = (mi S + rrti 5 = 1.712 radians per sec. (4) A ballistic pendulum weighing 30,000 Ibs. is set in oscillation by a 6-lb. ball, and the angular displacement is 75. If the distance d of the centre of mass from the axis is 5ft., and the distance a t of the point of impact below the axis is 5.5 ft., and the number of oscillations per minute is n = 40, find the velocity of the ball. 3006 1 20 x 32.2 x 5 V ' 40 x 3.14-6 x 55 S1 " 7i = l828 ft> PCr ^ ANS. CHAPTER VI. ROTATION ABOUT A TRANSLATING AXIS. Effective Forces Rotation about a Translating Axis. Equations (5), page 159, give the components of the acceleration for any particle of a rotating and translating body. If the axis of rotation passes through the centre of mass, we have ^ o, y = o, J o. The co-ordinates x, y, s are taken from the centre of mass O. If then we make these changes in equations (5), page 159, multiply each term by the mass m of the particle and sum up for all the particles, we have, since x, y, z are taken from the centre of mass O and hence 2mx = o, 2my = o, 2mz = o, for the component effective forces for a body of mass m = 2m rotating about any translating axis Hence the effective force in any direction for a body rotating about a translating axis is the same as for a particle of mass equal to the mass of the body having the acceleration of the centre of mass in that direction. If we take mass in Ibs. and acceleration in ft.-per-sec. per sec., we have force in poundals. For force in pounds divide by g (page 171). Moments of the Effective Forces Translating Axis. Equations (10), page 160, give the component moments of the acceleration for any particle of a rotating and translating body. If the axis of rotation passes through the centre of mass, we have ~x = o, y = 0,^ = 0, and if it does not change in direction, we have oo x K) y O, GO X OO, = o, oo y oa x = o, <&,<*), = o, ao g co x = O, os. co v o. If we make these changes in equations (10), page 160, multiply each term by the mass in of the particle and sum up for all the particles, we have, since x, y, z are taken from the centre of mass O and hence = o, my = o, mz = o, = I x , SM(** + x^ = I y , 2m(x* -\- /)==/ where I x , I y , I x are the moments of inertia of the body for the axes of X, K, Z through the centre of mass O, M' fx = ni/J' ftt/jj <*y2myx a^mzx + (&?/ oo^'Sntzy -f I x a M' fy - mf x z - mf z x - a z ^mzy - a^mxy -f (^ - , , M fy = I y a y , M fx = I t a t . If we take distance in feet and mass in Ibs., these equations (2), (3), (4) give moments in poundal-feet. For pound-feet divide by g (page 171). Momentum Translating Axis. Equations (4), page 154, give the component velocities for any particle of a rotating and translating body. If we multiply each term by the mass in of the particle and sum up for all the particles, we have, since JT, y, z are taken from the centre of mass O and hence 2mx = o, 2 my = o, 2mz = o 2mv x = mF, , 2mv y = mz^ , 2mv M = mv x ...... (5) Hence the momentum in any direction for a body rotating about a translating axis is the same as for a particle of mass equal to the mass of the body having the velocity of the centre of mass in that direction. Moment of Momentum Translating Axis. Equations (10), page 155, give the com- ponent moments of velocity for any particle of a rotating and translating body. If we multiply each term by the mass m of the particle and sum up for all the particles, we shall obtain the component moments of momentum for any co-ordinate axes we please. Let us take these axes as principal axes at the point O', the intersection with the axis of rotation of a plane through the centre of mass O at right angles to the axis. (Figure, page 153.) Then the parallel axes OX, OY, OZ at the centre of mass O are principal axes, and we have 2mxy = o, 2myz = o, 2mzx = o, 2mx = o, 2nty = o, 2tnz = o, where /,, I y , /, are the moments of inertia of the body for the parallel axes OX, OY, OZ. We have then, from equations (10), page 155, (6) If the axis passes through the centre of mass, we have ^ = o, y = o, z =o and CHAP. VI.] PRESSURES ON TRANSLATING AXIS. 369 Pressures on Translating Axis Permanent Axis. We have just as on page 319 for fixed axis, and using the same notation, x = mf x , R,' + R y " + 2F, = m/7, ..... (7) We have also - mf,x + mf s y - mf y z + I x a x . . (8) For a principal axis through the centre of mass we have ~2mxy = o, 2myz = o, = o and x = o, y = o, J = o, and these equations become x y = o, - ., , j - ,x = o, 2F.y- 2F y z = I x a x . We see from these last equations that if we have always M, = ^F y x - 2F x y = o, M = 2F^ - 2F,x = o, that is, if the moments M z , M y about the axes of Z and Y of the impressed forces F are always zero, there is no rotation of the axis, and hence even if this axis is unconstrained it will not change its direction. The axis is then a permanent axis of rotation. Hence if a body rotates about a principal axis through the centre of mass and M g , M y are always zero, the axis of rotation is a permanent axis and will not change in direction even if the pressures on the axis are removed. M x and M y are always zero when there are no impressed forces ; when the impressed forces always reduce to a resultant force through the centre of mass, or to a resultant force through the centre of mass and a couple whose plane is at right angles to the axis ; when the impressed forces all lie in a plane through the centre of mass at right angles to the axis, or reduce to a resultant force, or force and couple, in this plane. Conservation of Moment of Momentum _ Translating Axis. We have from equations (6) for the component moments of momentum for a body rotating about a translating axis, taking the co-ordinate axes as principal axes at O' ', M vy = mz^J mv^x H- fy. VI. equations (a). But since the motion of the centre of mass is the same as if all the impressed forces were applied to a particle of mass m at the centre of mass (page 299), when f x = o, f, = o,/ t = owc must have the algebraic sum of the components of all the impressed forces in three rectangular directions equal to zero, or 2F X = o, 2F, = o, 2F M = o. Also, since by D'Alembert's principle the moment of the effective forces is equal to the moment of the impressed forces, when M/,, = o, M fy = o, M /t = o we have the moment of the impressed forces for the axis of rotation zero also. If we have at the same time M fm o, Mfy = o, Mf, = o, and 2F, = o, 2F y = o, 2F, = o, the impressed forces then form a system of forces in static equilibrium, and from (b) we see that a x = O, a, O, or, = O. Hence if the impressed forces acting upon a body rotating about a translating axis form a system of forces in static equilibrium, the moment of momentum about that axis is constant. Kinetic Energy Translating Axis. The kinetic energy of a particle of mass m and velocity v is-tntf. If a particle has the component velocities v x , v yt v tt we have v* = v x * + v* + v*, and Lffg^ = - m (v* -)-] Vy * _|_ v v) m Equations (4), page 154, give the component velocities for any particle of a translating and rotating body. If we square these component velocities, multiply each term by m, sum up for all the particles and add, we shall have the kinetic energy for a rotating and translating body for any co-ordinate axes we please. Let us take these axes as principal axes at the point O', the intersection with the axis of rotation of a plane through the centre of mass O at right angles to the axis. (Figure, page 153.) Then the parallel axes OX, OY, OZ at the centre of mass O are principal axes and we have 2mxy = o, 2myz = o, 2mzx = o, 2mx O, 2 my o, 2mz = o, where I x , f y , f, are the moments of inertia of the body for the parallel axes OX, OY, OZ. We have then from equations (4), page 154, for the kinetic energy (9) If the axis of rotation coincides with one of the principal axes, as, for instance, O'Z', we have GO X = o, co y = o, and (10) The kinetic energy is then the sum of the kinetic energy of translation and rotation. If we take mass in Ibs., these equations give kinetic energy in foot-poundals. For foot- pounds divide by g (page I7 1 )- CHAP. VI.] INSTANTANEOUS AXIS ROTATIONTRANSLATING AXIS. 371 Instantaneous Axis. If a body rotates about a translating axis, the instantaneous axis is parallel to the translating axis and passes through a point whose co-ordinates are given by equations (15), page I57 f r the instantaneous axis of rotation, and by equations (i5) page 161, for the instantaneous axis of acceleration. If we take the translating axis as the axis of OX, we have co y = o, co t = o, and for the instantaneous axis of rotation the co-ordinates from the centre of mass are We also have a y = o, or, = o, and the co-ordinates of the instantaneous axis of accel- eration are Examples. (i) A circular disc of mass m and whose plane is -vertical has a force of P pounds applied at the centre, and rolls upon a horizontal plane. Determine its motion. ANS. Let the force P make the angle 9 with the horizontal and act in the plane of the disc. The force is Pg poundals, and the horizontal component Pg cos 0, and the vertical com- ponent Pg sin 0. Let r be the radius, and A the point of contact. The moment of the impressed forces about A is Pg cos 9 . r. Let K be the radius of gyration for axis through the centre of mass O at right angles to the disc, and /' the moment of inertia for parallax axis through A. Then /' = m (tf 2 + r 1 *), and we have the moment of the effective forces fa. By D'Alembert's principle, /'a Pg cos 9 . r = o, _ _ Pgr cos ~ The axis at A is the instantaneous axis. The acceleration at the centre O is then 7--_^_ ~ For a disc, K* = , hence _ iPg cos iPg cos 9 3m Both angular and linear accelerations are then constant if P is constant. (2) A disc of mass m whose plane is vertical rolls (without sliding) down a rigid inclined plane. Determine its motion. ANS. Let the radius be r, the radius of gyration about an axis through the centre of mass O perpendicular to the plane of the disc be K, and the inclination of the plane, and A the point of contact. Then the weight is mg-, the force parallel to the plane is mg sin 0, and its moment about A, rn.gr sin 0. We have then, as in the preceding example, I'a = + r a )a = vagr sin 0, or a = gr sin + r* Also, since A is the instantaneous axis, the linear acceleration of the centre is Since K* = , we have 37* KINETICS OF A MATERIAL SYSTEM. [CHAP. VI. Both linear, tangential and angular accelerations are constant and the velocity after any time may readily be determined. (3) Find the time a rigid homogeneous cylinder will take to roll from rest down a plane 20 ft. long and inclined 30" to the horizon, the axis of the cylinder being horizontal. ANS. f-9j sec. (4) A rigid homogeneous circular disk of mass m and radius r, whose plane is vertical, moves in contact with a smooth inclined plane whose angle is 0. From a point in the same vertical plane as the disc, and at a distance from the inclined plane equal to the diameter of the disc, a string is carried parallel to the inclined plane and is wrapped round the edge of the disc, and its end is fixed in the circumference. Find the tension T in the string, the linear acceleration f of the centre, and the angular acceleration a of the disc. m?K* sin 8 me sin m sin 6 ., ANS. T = ^ = -3E-; poundals or - Ibs. ; radians-per-sec. per sec. (5) A perfectly flexible and inextensible ribbon is coiled on the circumference of a homogeneous circular disc of radius r and mass m, and has its free end attached at a fixed point. A part of the ribbon is unrolled and vertical, and the disc is allowed to fall from rest by its own weight. Find the" acceleration f of the centre and the angular acceleration a. before the ribbon becomes wholly unrolled, and the distance s which the centre "will descend in one second. ~P 3* ~~2 ~3* (6) A homogeneous hemisphere of radius r performs small oscillations on a rough horizontal plane. Find the periodic time. ANS. For the simple pendulum of length /and mass m we have mg x / sin = For the hemisphere let / the cylinder will be stationary for a time t = ^-j- , at the end of which the angular velocity will be zero. (10) A uniform square is supported in a vertical plane with one diagonal horizontal by two supports, one at each extremity of the diagonal. Show that the initial force on one support when the other is removed is equal to one fourth of the weight of the square. (11) A uniform horizontal bar, suspended from any two points in its length by two parallel cords, is at rtst. If one of the cords be cut, find the initial tension in the other. wr ANS. T jr wi where / is the length of the bar, d the distance from its centre of mass to the point of attachment of the uncut cord, and W is the weight of the bar. (ip) A uniform beam of weight IV rests with one end against a smooth vertical wall and the other on a smooth horizontal plane, the inclination to the horizon being 0. // is prevented from falling by a string attached to its lower end and to the wall. Find the force between the upper end and the wall when the string is cut. ANS. W cot 9. (13) A sphere is laid upon a rough inclined plane of inclination 8. Show that it will not slide if the coefficient of friction is equal to or greater than - tan 0. (14) A sphere of radius r whose centre of mass is not at its centre of figure is placed on a rough horizontal plane, coefficient of friction H. Find whether tt will slide or roll. ANS. Let K" be the radius of gyration of the sphere about the line through the point of contact at right angles to the plane of the centres of figure and mass. Then if the initial distance of the centre of mass from a vertical through the centre of figure is greater than - , it will 'begin to slide; if less, to roll. CHAPTER VII. ROTATION ABOUT A FIXED POINT. Effective Forces Rotation about a Fixed Point. Equations (5), page 159, give the components of the acceleration of any particle of a rotating and translating body. For rotation only we should omit all terms containing f x , f y ; f z . If we make these changes in equations (5), page 159, multiply each term by the mass m of the particle and sum up for all the particles, we have, since 2mx = o, 2 my = o, 2mz = o, for the compo- nent effective forces for a rotating body of mass m = 2m for any co-ordinate axes we please through the fixed point O' 1&}>Go x a)y -f- = mJ co y co z -\- mxGo z oa x -\- mj 7 oo z oo y rascal TS^ZGO* -\- mya x If any one of the co-ordinate axes is fixed, as, for instance, O' Z' we have oo z os = o, GO Z GO X o, oa x oj y = o, ooyoo, = o. For O' Y' fixed we have o^ y oo x = o, oo y co, = o, co x &) z = o, ao z Go x = o. For O'X ' fixed we have Go x co y = o, oo x co z = o, Go y K> z = o, (jo z w y = o. For all the co-ordinate axes fixed we have equations (3), page 316, for a fixed axis. If the fixed point is at the centre of mass, we have ~x =o, ~y = o, J = o, and hence ~2mf x = o, ^inf y = o, 2mf z = o. That is, if the centre of mass is fixed, the components of the effective forces are zero. If we take distance in feet and mass in Ibs., equations (i) give force in poundals. For force in pounds divide by g (page I/ 1 )- Moment of Effective Forces Rotation about a Fixed Point. Equations (10), page 160, give the component moments of the acceleration of any particle of a rotating and translating body. For rotation only we should omit all terms containing f x , f y , f x , and put x' ', y' , z' in place of ~x -\- x,~y -\-y,~z -\- z. If we make these changes in equations (10), page 160, multiply each term by the mass m of the particle and sum up for all the particles, we obtain the component moments of the effective forces. In making the summation we have where I' x , /,','/,' are the moments of inertia of the body for the co-ordinate axes O'X', O'Y', O'Z'. We also have 2mx> z = I' yz , 2 my* = !' , 2mz* = I' xy , where 7 xy , f y ' t , I' zx are the moments of inertia of the body for the co-ordinate planes X'Y', Y'Z', Z'X', 375 co t w y 2mx'y' ct^mz'y' ot^mx'y' -f (o^ 376 KINETICS OF A MATERIAL SYSTEM. [CHAP. VII. We have then (2) Since we can take any co-ordinate axes we please, let us take them principal axes at the fixed point X. Then we have "Stnx'y 1 = o, ^nty'z' = o, 2mz'x' = o, and equations (2) become M' x = l' tX OO t OOy I'xyWyCOt + l' X (X X , M' ft = If any co-ordinate axis is fixed, as, for instance, O'Z', we have GO,C*) X o, oo t z o? v = o. If O'X' is fixed, we have a?,oa>, = o, ^G?, = o, cayo?, = o, &) z co y = o. If all are fixed, equations (2 and (3) become equations (4) and (6), page 317, for fixed axis. If the co-ordinate axes are not fixed, we have oo t oa y oo y oo^ oo x co f =oa f co x , oo y co x oo x co y ', and since we have equations (3) become Mf, = (f. ~ fya>.<, + 7 >,' I M f , = (I' x ~ Ow,w. + />,. | ....... (4) M,. = (/; - /;),, + /;.. j If we take distance in feet and mass in Ibs. , all these equations give moments in poundal- feet. For pound-feet divide by^ (page 171). Momentum Rotation about Fixed Point. From equations (2), page 154, we have the component velocities for any particle of a rotating body, v x = (z + z)w y (J If we multiply by the mass m of the particle and sum up, we have, since 2mx = o, my = O, 2mz = o for the components of the momentum , = mxao.-m.3Go,, ..... ... (5) For axis through the centre of mass we have * = o, J = o, ~z = o, and hence x = o, 2fnv y = o, 2mv t = o. Hence the momentum for a body rotating about a fixed point is the same as for a particle of equal mass at the centre of mass. CHAP. VII.] MOMENT OF MOMENTUM -ROTATION ABOUT FIXED POINT. 377 Moment of Momentum Rotation about Fixed Point. Equations (10), page 155, give the component moments of velocity for any particle of a rotating and translating body. If the origin O' is a fixed point, we should omit terms containing v xt v y , v z , and x, J7, J, and put .*', y' r z' in place of ^r, y, 2. If we make these changes in equations (10), page 155, multiply each term by the mass m <>f the particle and sum up for all the particles, we shall obtain the component moments of momentum for any co-ordinate axes we please. Let us take these axes as principal axes at O' ' . Then we have ~2mx'y' = o, ~2my'z' = o, = o. We have then from equations (10), page 155, since 2m(y* -f- z' 2 ) = I x , JC = I' x oo x , M' Ty = f y co y , M' vz = f gC o, ....... (6) Pressure on Fixed Point. Let the component pressures at the fixed point be R x , R y , R z . Let the components of all other impressed forces be 2F X , 2F y , 2F Z . We have then from equations (i), by D'Alembert's principle, R x -j- 2F X myGo x Go y -f- lR-ZGa x Go x m.xc0* IR.XGO? -\- Viz a y mj7a- z R y -j- 2F y T&zK} y cd s -\- mx&) y oj x myoa? mjo?^ 2 -{-mxa x mFa x , }- . (7) R, -j- 2F Z = m.x(*) z GJ X -\- mJ/Ci?, oj y m^ oj x 2 TCLZ oof -f- VLya x ?&xa y Equations (7) give R x1 R y , R,. If the centre of mass is the fixed point, we have ~x = o, y = o, J = o, and hence that is, when the centre of mass is fixed, the pressure on the fixed point is the same as if there were no rotation, and is found by the conditions of static equilibrium. Conservation of Moment of Momentum Rotation about Fixed Point. We have from equations (6) for the component moments of momentum for a body rotating about a fixed point O' t taking the co-ordinate axes as principal axes at O', M^ x =I x co x , M' vy =I y ' ( a y , -0C = /,'<,, ...... (a) and from equations (4) for the component moments of the effective forces If in equations (ff) the co-ordinate axes O' X' , O' Y' , O' Z' are fixed (figure, page 153), we have GO,CO = o, co x oo, = o, GO oo x = O, and the axis of rotation is fixed. If also we have a x = o t a = o, ot z = o, we have M fx = o, M fy = o, M ft = o, and co x , CO Y , oo z in equations (a) are constant. But since, by D'Alembert's principle, the moment of the effective forces is equal to the moment of the impressed forces, we have the moment of the impressed forces zero. 3?8 KINETICS OF A MATERIAL SYSTEM. [CHAI-. \ II. Hence if the moment of the impressed forces about any fixed axis through the fixed point is always zero, the moment of momentum of the body about that axis is constant, and if the moment of inertia about that axis does not change, then the angular velocity about that axis is also constant. Invariable Axis. If the moment of the impressed forces relative to the fixed point O' is always zero, the resultant must either be zero or always pass through O'. In such case the co-ordinate axes do not change in direction, and we have <,<#, = O, co x oa t = O, co y co x = O, and also Jlf f( = o, M fy = o, M fl = O. Hence, from equations (b), a x = O, a, = O, a, = O, and co x , &? y , . in equations (a) are constant. We have then the moment of momentum // trT* ' i T 1 '* ' i /"'* ' CO = V /, CO X + I, G0 y + /, GO, ........ about an axis through the fixed point whose direction cosines are I x 'oo x a I'cOy 7/cw, and this axis is then invariable in direction. Hence when the moment of the impressed forces relative to the fixed point is always zero, or if there are no impressed forces, and a body rotates about any axis through the fixed point at any instant, there will be a certain axis through the fixed point for which the moment of momentum is constant. This axis is invariable in direction, and its direction cosines are given by equations (//). The moment of momentum about this axis is given by equations (c). Kinetic Energy Rotation about a Fixed Point. The kinetic energy of a particle of mass m and velocity v is mv z . If a particle has the component velocities V M , v v , V M , we have v 2 = v x -\- v y * -j- v* and I i . I , . I -mir ~mv* + -mv* -\--mv?. From page 154 we have for the component velocities for any particle of a rotating body v x = z'coy y'oo, , v y X'GO, z'oo x , v t = y'oo, x'oo y , where x ' , y' , z are to be taken from the fixed point O' as origin. If we square these com- ponent velocities, multiply each term by m, sum up for all the particles and add, we shall have the kinetic energy for a rotating body for any co-ordinate axes we please. Let us take these axes as principal axes at O'. Then we have 2mxy = o, 2myz = o, ~2ntzx = o. We have then, since 2 m (y* + z>*} = /,', 2m(z lZ -f x*) = //, ^m(x + y*} = //, for the kinetic energy 5C = ^ / c, ? , 2 +^/;a J / + ^/>, 2 . .......... (7) If we take mass in Ibs., this equation gives kinetic energy in foot-poundals. For foot- pound divide by g. CHAP. VII.] ROTATION ABOUT A FIXED POINT EXAMPLES. 379 9 Examples. (i) A circular disc supported at its centre of mass Oj'otates about the principal axis OZ at right angles to the plane of the disc with an angular "velocity oo z = ^/j radians per second. The plane of the disc makes an angle 9, = jo with the horizontal. If now an angular velocity O0y = i radian per second is given to the disc about a principal axis O Y in the plane of, the disc, find the motion. A\S. From equations (4), page 376, we have for the component momt.its of the effective forces Mf x =. (7 Z Iy) GUfOOy + Ix&X , Mfy (I x 7,)* je , + lyOCy , M fz = (Iy-I x ) x + I z a z . By D'Alembert's principle the moments of the effective forces are equal to the moments of the impressed forces. In the present case the only impressed force is the weight mg acting at O and the equnl ;md opposite reaction R of the support at O. The component moments of the impressed forces are therefore zerc, and hence Jlff x o, Mfy = o, J\ffi = o. We have also a> x = o and a x = o, a y = o, nr z o, and the angular velocities ooy, &? z are therefore constant. By the principle of page 378, we have in this case an invariable axis of rotation OZi fixed in space for which the moment of momentum is constant. Let 7i be the moment of inertia for this invariable axis, and -j- be the angular velocity about it, so that 7i -J- is the moment of momentum. Since 7 2 27, for the disc, we have / t ^= I/AW + //"/ = If is the angle Z<9Z t , we have (0 OOy From page 35, and therefore, from (i) and (2), .. ^V> y 4*a 2 + &>y* T ^ V 1 dt dt 7, = I z cos 2 9 + 7^ sin 1 9 = T y (\ -+ cos" 9). (2) (3) (4) dt i + cos 2 9 Equations (2) give the angle ZOZ\. of OZ with the invariable axis OZ\, and the motion of the disc is the same as if the axis OZ fixed to the disc were to rotate about OZi, always making the angle 9 with it, with the angular velocity -j~ given by (4). Inserting numerical values, we have cos 9 = ? _ . = 0.95958, or 9 = 1 6" 21', 1/13 j- = ^ =1.87 radians per sec. dt 25 2D SOLUTION. From page 168 we have Euler's geometric equations, o = GO X = sin sin 9 cos 0, dt dt dj, = cos -H -77 sin 9 sin , dt KINETICS OF A MATERIAL SYSTEM. [CHAP. VII. where is the angular velocity about OZ\, -j- the angular velocity of OZ about the line of nodes ON (see figure, page 167), -%- the angular velocity of OY relative to ON. From the first two of these equations, eliminating -r-, we obtain , sin sin = sin cos Q sin 6 and, as before, sin 6 = We have then sin = CQ ^ Q , and !!1 _ "> wo " 4 /, , ___ dt sin 9(i + cos' 0) ' dt ~ i + cos'9 V or, putting in the values of sin 9, cos 9, >,* 4- ~dt oo y cos 9 sin 6(1 + cos 3 &) dt - 8oV + ^ Inserting numerical values, we have 9 = 16 21', ^ = 1.87 radians per sec., -j = 3.08 radians per sec., dd> = 0.07 radians per sec. tit (2) Let a vertical disc of mass m = 40 Ibs. and radius r = 3 ft. roll on a horizontal plane with angular velocity to y = -f 2 radians per sec., while its centre O describes a horizontal circle of radius O'O = y = g feet with angular velocity ea about a fixed vertical axis O 1 Z' . At each extremity of a diameter ab let a mass m = 10 Ibs. be attached. Find the pressure on the horizontal plane when these masses are in a vertical and in a horizontal line. Take g = 32 ft.-per-sec. per sec. ANS. Let O be the origin, OZ' the fixed axis, and take x = o and cr x = o, ct y = o. ft* = o. Also, since O'Z 1 is fixed, oflyoa, = o, 03,09, = o, oo f (o x = o, oa,oo y = o. From equations (3), page 376, we have then and since the disc rolls, rooy = J( or o>, = = y Now I'xy , or the moment of inertia relative to the plane X 1 Y', is the same as the moment of inertia for the horizontal axis cd through the centre of mass. Hence if 9 is the angle of Oa with the vertical, mr 1 /', = + 2tnr* cos* 9. Let R be the vertical pressure of the horizontal plane at b. Then, by D'Alembert's principle, Ry~ -(m + im)sy = M fx = -f + 2mr* cos" Q\,, we have /? = (m + 2m)g + ~>-Y^! + vmr* cos 8 y \ 4 If we take distance in feet and mass in Ibs., this equation gives R in poundals. For R in pounds we divide by g and obtain R = m + 2m + -=% ( + 2mr* cos" 91. (0 We see from (i) that the pressure R is greater than the total weight m + 2m. The greatest possible pressure for given r and y is when cos 9 = i or = o, that is, when the masses m are at a and b in the figure. The least possible pressure is when cos 9 = o, or 9 = 90, that is, when the masses m are at c and d in the figure. In the first case, when the masses m are in the vertical line ab, we have 9 = o and R = 60 H 2(90 + 180) = 61.25 pounds. In the second case, when the masses are in the horizontal line cd, we have 9 = 90 and R 60 + ^~- = 60.4166 pounds. If the masses m are removed, we have in all positions R = 60 + ^-^ = 60.4166 pounds. We see, then, that for a disc rolling as in the example the pressure on the horizontal plane is greater than the weight of the disc. (3) In the preceding example let the horizontal plane on which the disc rolls be above instead of below the disc. Find the force R necessary to keep the disc in contact with the plant. ANS. In this case we have roo y = y <*>* , or oo z = + -=*, and, as before, /' = cos* 9, and Mf* = I'xyVy 00 *' We have then, as before, Ry (m + -zm *)gy = M/ x = ( + 2wr 2 cos" 9 )ay, \ 4 / Substituting the value of &>,, we have or for R in pounds R = (m + 2m)g - ^fc- + 2mr* cos* 9J, _l^.pl + 2 ;r'cos'e We see in this case that the force R is less than the total weight (m + 2m). If the masses m are vertical at a and b, we have 9 = o and R = 60 - ~ (90 + 180) = 58.75 pounds. 382 KINETICS OF A MATERIAL SYSTEM. If the masses m are horizontal at c and d, we have = 90 and [CHAP. VII. If in (i) we put /? = o, we have for the value of oa, for *rhich no force K will be requirtd + 2wr* cos 1 6 , (2) If the masses m are removed, we 19.6 radians per sec. If the disc rolls with this angular velocity, R will be zero, that is, the disc is self-supporting and will roll round on the horizontal plane abov<^ i* without any force R being necessary to keep it in contact. If oo y is greater than this, the disc will press against the horizontal plane above it. If o^ is less than this, a force R will be necessary to keep the disc in position. (4) Discuss the action of the Gyroscope and Spinning Top. DESCRIPTION. The gyroscope consists of a disc aa which is set in rotation about an axis in the direc- tion of the diameter of the ring A 1 . This ring is attached to the rod S, which passes through a sleeve at h. This sleeve is pivoted in the fork^ so that S can rotate in a vertical plane, and the fork^ is pivoted at/ so that this rod can rotate horizontally. A sliding counterweight G can be so adjusted that the centre of mass of the apparatus can be made to lie on the same side of the standard as the disc, on the opposite side, or directly over the standard. SOLUTION. Let O'Z' be the axis of the disc, y ........ (i) where (page 167) is the angle of O' Y' with the line of nodes. The initial kinetic energy is /,'&V (page 378). and the initial potential energy relative to a horizontal plane at a distance y below O', if y is the distance O'O of the centre of mass, is mg(~y + y cos 61) when O is on the same side of the standard as the disk, and wg(y y cos 61) when O is on the opposite side of the standard from the disc. The total initial energy is then , = -/,'<, + &g(yy~ cos 61). The final kinetic energy is (page 378) and the final potential energy is mg-(y~ y cos 6). The total final energy is then g = /*' *** + ~fy <>y + -J*' 00 ** + g(y~ 7 COS 9 )' If we disregard friction, we have, by the principle of conservation of energy (page 304), = Si, or / 3t 'ao i * + I,'tDy* = mg7(cos6i cos 6), . . .. . . . . . . . (2) where the (+) sign is to be taken for centre of mass O on same side of standard as the disc, and the (-) sign when it is on the opposite side. We have also, from Euler's geometric equations (page 168), oo x = -3- . sin -j- sin 6 cos V = cos + -r- y dt dt (ill) , d

, (3) 3 8 4 KINETICS OF A MATERIAL SYSTEM. [CHAP. VII. i# ' Sthe an K ular velocity oa, about W, or the angular velocity of precession; -^ is the angular velocity of ffZ' about the line of nodes (see figure, page 167), or the angular velocity of nutation ; -j- is the angular velocity of O 1 Y relative to the line of nodes. Squaring and adding the first two of equations (3), we have Substituting this in (2), we have, since for the disc /,' = //, *' + /,'fLsin*9= imgy (cose.-cose) .......... (4) Substituting the values of o> M , oo y from (3) in (i), we have 7*'^ sin' G = f f 'oo, (cos 6, - cos 6) ........... (5) We have also, from the last of equations (3), + ^cosO=, ............... (6) Equations (4), (5), (6) are the differential equations for the gyroscope. When 6 = 6,, or at the beginning of motion, we have From ,(5) we have the angular velocity about (7 V, or the angular velocity of precession, _ dij) _ I*'oOf cos 61 cos ~ ~dt ~ I x ' ' sin*! ............ (7) and substituting this in (4) we have for the angular velocity of CfZ' about the line of nodes, or the angular velocity of nutation, T' "7"sin'6 v -cos 6)1 (8) where the ( + ) sign is taken for centre of mass O on same side of standard as disc, and the ( ) sign when it is on the opposite side. From (8), taking the( + ) sign, we see that for the centre of mass O on same side of standard as the disc, dt is imaginary when 6 is less than 6|. Taking the ( ) sign, we see that for the centre of mass O on the oppo- site side of standard from disc, is imaginary when 6 is greater than 6,. Hence the centre of mass O always falls from its initial position and can never rise above it. From (7), then, if o>, is positive, that is, if the rotation of the disc when looking from O 1 to Z' is clock- ji wise, -j- is positive, or the rotation about <7 V looking from (7 to V is clockwise if the centre of mass O is on the same side of standard as the disc. If O is on the opposite side and oo t is positive, -^- is negative. If the centre of mass is over the standard, J o and the angle 6, remains unchanged. Hence cos 6, cos 6 = 0, and = , -3 ill at The axis O ' Z' then remains stationary. These conclusions can be verified by the apparatus (figure, page 382) by shifting the counterweight. If we put = o in (8), we obtain the maximum and minimum values of 6. We have '- = o when CHAP. VII.] ROTATION ABOUT A FIXED POINT EXAMPLES. 385 0=0,, and this is the minimum value of 9, for we have just seen that O always falls, and hence cannot be less than 0j. We also have- = o and a maximum when at cos 0i cos = zmgy ^-^ 5 . (9) If we insert this value in (7), we have for the maximum value of on, max. < = ~Y^~. . (io) The minimum value of oa v is when Q = 0i , or min. oa v = o. We see from (io) that tjie maximum value of oo v is independent of 0j and 0, or of the initial and final positions of the disc. From (9) we obtain for the maximum value of Let us put fof brevity and convenience the quantity Then we can write equation (i i) COS0max. = ft* + -f/I T 2/5* COS QI + ft* ........... (12) where the upper signs are for O on same side of standard as disc, and the lower signs for O on the opposite side from disc. We see, from (12), that the maximum value of depends upon ft, and that this maximum value can be o ot 1 80, that is, cos 0max. can be + i or I only when ft = o. But ft can be zero only when oj = o. Hence any velocity of rotation GO, of the disc, however small, is sufficient to prevent the axis CfZ' from reaching the vertical. The self-sustaining power of the gyroscope is thus proved. From (12) we have If ft or oo t is very great, cos 0i cos 0max. is very small. Hence, by increasing the value of oo t , max . 0, can be made less than any assignable quantity. This proves the apparently paradoxical result that the disc when rotating rapidly and set at any angle t with the vertical does not visibly rise or fall. But we see, from (7), that for = 61 , - is zero, and hence the disc must rise or fall in order to generate rotation about at Cf V. If oo f is great, this rise or fall is, however, very small and may not be visible. Let A be the length of the simple pendulum which would oscillate about QfX' or 7 Y' in the same time as the apparatus, when oo x is zero. Then (page 337) we have and equation (8) becomes sin 5 = jf~ sin 5 - 2/y(cos 0, - cos 0)l(cos 0, - cos 6). (14) and equation (7) becomes sin a 0-^ = 2/3A/.(cosQi cos 6) (15) and equation (io) becomes /T (16) 3&6 KINETICS OF A MATERIAL SYSTEM. [CHAP. Vll. The centre of mass then oscillates up and down between the minimum and maximum values of 0i and max. 8 as given by (12), while the angular velocity of the centre of mass about O 1 V varies from -^ o. when the axis is in its initial position, to the maximum value given by (16). The complete solution of the problem requires the integration of the differential equations (4),( 5) and (6). This necessitates the use of elliptic functions. If, however, we assume that the velocity of rotation of the disc <, is very great, and hence cos Oi cos max. or 0, very minute, we may obtain integrals of (4) and (5) which will express the motion with all requisite accuracy. Let us then assume ; ........ (20) 4/2* sin 9, - 4/JV integrating, since when / = o, u = 0, = o, versin-' _ . cos 2ft sin 0, 2ft 'f, _ 4/M \ sin fi,/ Hence CHAP. VII.] ROTATION ABOUT A FIXED POINT EXAMPLES. or, since cos 2A = i 2 sin* A, u = sin 9l si 387 (23) We have from (18), neglecting the square as well as higher powers of u (which may be done without sensible error owing to the minuteness of u, though it could not be done in the foregoing values of dt and /, since /?* is great when u is small), COS 0] COS u sin 0i sin" sin* 0, + 2 sin 0, cos 0/ The greatest possible value of sin 0i cos 0i is for 0, = 45, or sin 0i cos 0j = . 2 Since u is very small, we have then, neglecting 2 sin 0, cos t , cos 0i cos u and substituting in (15), we obtain sin* sin t ' dt r A sin 0i Inserting the value of u from (23), we have Integrating, since dty o when / = o, we obtain Equations (23), (24), (25) give with all requisite accuracy the vertical angular depression of OZ' , w = 0,, the horizontal angular velocity , and the horizontal angle $ at the end of any time /, provided oo x is very great. Referring to (20), we see that // is the differential equation of a cycloid generated by a circle whose angular . sin 0! diameter is -^j- (page 140). Mf When, starting from / = o (and therefore u = o, -j- and ^ = o), u has its greatest value, we have, from (21), (22), (24), (25) After the expiration of the time / = ^y we have and O ' Z' has regained its original elevation and the horizontal velocity is zero. The axis O' Z' then moves as if it were the element of a right circular cone AO'Z 1 ', the angle AO ' Z' being equal to , which rolls on the cone ZO 'A, the angle ZO' A being equal to fl,. (5) A layer of dust of uniform depth d, d being small compared to the radius of the earth, is formed on the earth bv the fall of meteors reaching the earth from all directions. Consider the earth as a homogeneous sphere of radius r and density ^, and let 8 be the density of the layer. Find the change in length of the day. ANS. Let ci be the angular velocity before and 00 after the layer is formed, and /i the moment of inertia 388 KINETICS OF A MATERIAL SYSTEM. [CHAP. VII. of the earth and / that of the layer. Since there are no forces in the system except the mutual action of the particles by the principle of conservation of moment of momentum (page 376), we have /,<, =(/,+; The mass of the earth is A*r*. The moment of inertia of the earth is then 3 5 3 '5 The moment of inertia of the dust-layer is Hence / _ S(r + <*)- 8r* Expanding, and neglecting j- and all higher powers, we have Therefore If the density of the earth is taken at 5.5 as compared to water, and the density of the dust is taken at 2, and d = r, we have 20 10 5.5x20 The length of the day in this case would be ii of 24 hours, or only 22 hours. 12 (6) If the earth gradually contracted by radiation of heat, so as to be always similar to itself as regards its physical constitution and form, show that when every radius vector has contracted an nth part of its length where is a small fraction, the angular velocity has increased a 2nth part of its value. ANS. Let m be the mass of the earth, r\ its initial and r its final radius, and I\ its initial and / its final moment of inertia. Then / 2 * / 2 r* ~~$ mi ~~s mr and /i r\* But r = fi nri = r\(i n). Hence =r7(r^>=(rnr <01 ' Expanding, and neglecting ' and higher powers, (V) = 09, = (I + 2ft)i and oo the angular velocities before and after. Then /= Er*, where r is the radius of the earth, and we have Er Hence, neglecting -^ and higher powers, = _E^_f +$> \ x _ (8) Suppose a mass of ice m to melt from the polar regions for 20 degrees round each pole to the extent of about a foot thick, or enough to give i-faft. over those areas, which spread over the whole globe would raise the sea-level by only some such undiscoverable difference as three fourths of an inch. Show that this would diminish the time of the earth's rotation by one tenth of a second per year. ANS. Let be the angle from the pole, and 8 the density of the ice. Then the mass m is m = 47t$r*( i cos0), where r is the radius of the earth. We have rdB = ds and the mass of a strip is iitSxds. But x = r sin 0, hence the mass of a strip is zxSr* sin fl dQ. The moment of inertia is then 7=2 cos 0(i - cos 0)(i +cos 0), or, substituting the value of m, mr* 1= - COS 0(l +COS 0). If E is the mass of the earth, the moment of inertia of the earth is /= lEr* If o>i is the angular velocity before and GO after melting, we have, by the principle of conservation of moment of momentum, COS (XI +COS >. = The final angular velocity GO is then greater than the initial, and the time of rotation is diminished. The difference of time for one day is 21C 21t Substituting numerical values, the difference of time is easily found. CHAPTER VIII. ROTATION AND TRANSLATION. Effective Forces Rotation and Translation. Equations (5), page 159, give the components of the acceleration of any particle of a rotating and translating body. For origin at the centre of mass we have Ic = o, y = o, T = o. If we make these changes, multiply each term by the mass m of the particle and sum up for all the particles, we have, since 2mx = o, 2 my = o, 2mM = o, for the component effective forces for a rotating and translating body of mass m = 2m, for any co-ordinate axes we please, 2m/, = where,/^, f yt f t are the component accelerations of the centre of mass. That is, the effective force in any direction is the same as for a particle of mass equal to the mass of the body having the acceleration in that' direction of the centre of mass. Moments of Effective Forces Rotation and Translation. Equations (10), page 160, give the component moments of the acceleration of any particle of a rotating and translating body. If we multiply each term by the mass m of the particle and sum up for all the particles, we shall obtain the component moments of the effective forces for any co-ordinate axes we please. Let us take these axes as principal axes at the origin O' . Then we have 2mx = o, 2 my = o, 2ms = o, also 2mxy = o, 2mys o, 2//isx = o. We also have 2m( y* + z>) = f x , 2 m ( z * + **) = I y , where /,, I y , f, are the moments of inertia of the body for the co-ordinate axes OX, O Y, OZ. We also have 2mx* = I ys , 2 my 2 = / , 2,,ix 3 = I xy , where I XJ , I yt , I tx are the moments of inertia of the body for the co-ordinate planes XY, YZ, ZX. We have then from equations (10), page 160, for the component moments of the effective forces for principal co-ordinate axes through the origin O' M' fx mf r y mf^ -f- I^GO.GO, I x ,<** y <*> t + I, a, , X1 v x <# t Iy f <*>, x + lyi*,- *,* ~ ^.w, + f ,<*. > J M' fy = 390 CHAP. VIII.] MOMENTUM- ROTATION AND TRANSLATION. 39 1 where J', J, ~~ are the co-ordinates of the centre of mass and /.,., f y , f z are the component accelerations of the centre of mass. If any co-ordinate axis, as O'X' ', is fixed, we have co x oo y = O, K) X OS Z = O ; if O' Y' is fixed, we have co y oj^ = O, &? y aj, = O ; if O' ' Z' is fixed, we have oo t oo x = O, co z oo y = O. If the co-ordinate axes are not fixed, we have co x oo y 07,07,, co y Go f co z cj y , Go t (a x = co x oo z , and since / - /* = ^*(/ - ^ = I. - J y I*y ~ 4. - ^^ - * 2 ) = I- I., 4 - / = ^(*? -/) = //- /,, equations (2) reduce to Y A - m/J - mF + (/. - />., + /A , ] M' xy = mfjs - mf? + (I x - S z )Go x Go t + /,*, [ ...... (3) J/;, = m/^ - mf x y + (/, - /Ja,^, + 1 A . \ If we take distance in feet and mass in Ibs., these equations give moments in poundal- feet. For pound-feet divide by g. Momentum Rotation and Translation. Equations (4), page 154, give the component velocities for any particle of a rotating and translating body. If we multiply each term by the mass m of the particle, and sum up for all the particles, we have, since ^mx = o, 2my = O, 2mz o, for the components of the momentum for a rotating and translating body of mass m = 2m, for any co-ordinate axes we please, 2mv x = mv x , 2mv y = mv y , 2mv t = m^ r , ..... (4) where v x , v y , v z are the component velocities of the centre of mass. That is, the momentum in any direction is the same as for a particle of mass equal to the mass of the body having the velocity in that direction of the centre of mass. Moment of Momentum Rotation and Translation. Equations (10), page 155, give the component moments of velocity for any particle of a rotating and translating body. If we multiply each term by the mass m of the particle and sum up for all the particles, we shall obtain the component moments of momentum for any co-ordinate axes we please. We have then from equations (10), page 155, since ~2nix = o, 2my = o, 2ms = o, and 2m(y* -j- } = 4, 2m(z* -\- x*) = I y , 2m(x* +./) = / 2 , for the component moments of momentum for a rotating and translating body of mass m 2m M vx = M vy = mz^ ~y -f 7,67, . (5) Conservation of Moment of Momentum Rotation and Translation. We have from equations (4) for the component moments of momentum for a body rotating about a trans- lating axis through the centre of mass, since J = o, jr = o, ^ = o, M vx =S x co x , M v = !,&,, M w = l r , M , - K- 2 <,} K-'OJ, - rvi For any interval of time less than / as given by (2) equations (i) give v and K>, when /< is the coefficient for slipping. For any interval of time greater than this equations (i) also give v and ca, but /i is the coeffi- cient for rolling friction. In equations (3), if rv, K-OJ-. is negative, v is negative. Hence if oji is greater than ~. the hoop at the instant sliding ceases will have translation opposite in direction to the initial translation. Equations (i), (2), (3) hold for a sphere or cylinder also. We have only to substitute the value of x- a in each case. (7) A disc whose plane is vertical rolls without sliding down an inclined plane. Find its motion. ANS. The effective force is m/ parallel to the plane and the moment of the effective force la, where / is the moment of inertia for axis through the centre of mass O at right angles to the plane of the disc. The impressed forces are the weight mg, the normal pressure R = mgcosi, where / is the angle of inclination, and the friction F = M K = urn? cos /acting opposite to the motion. By D'Alembert's principle we have then m/= m- sin / umg cos /, or J g sin / ug cos /, . . (i) la u'nrg cos / urg cos / . . = Fr = umrg cos /, or a = j = ^ , (-) mg where K is the radius of gyration. Now the condition for rolling without sliding \sf-\- ra = o. From (2), then, we have /V " gr 1 cos /' 396 KINETICS OF A MATERIAL SYSTEM. [CHAP. VIII. Substituting this value of // in (i), we have 7 = *!li!L, or a = -2^4. .' .......... (3) J K-* + r* K 1 +r* Equations (3) give the linear and angular accelerations. We see that they are both constant. We have then The equations hold for a sphere or cylinder also. We have only to substitute the value of K* in each case. (8) Find the time a cylinder will take to roll from rest down a plane 20 ft. long, inclined jo" , the axis of the cylinder being horizontal. r* I ANS. We have z/i = oand * = . Hence, since sin * = and taking^ = 32, STATICS OF RIGID BODIES. CHAPTER I. FRAMED STRUCTURES. Stress. We have seen (page 1 74) that the exertion of force upon a body or particle is only one side of the complete phenomenon which consists of the simultaneous action of equal and opposite forces between two bodies or particles. We call these internal forces STRESSES. Stress, then, is a force internal to the body or system considered, while the term force is reserved for external action. We speak, then, of the force on a body or particle, and the stress in a body or between two bodies or particles. Tensile Stress and Force. Let a body AB be acted upon by two equal and opposite forces -j- F l F in the same straight line. These __ r-r ^ ^ forces act to stretch the body, and are therefore called tensile forces. If there is equilibrium, we must have at any two sections A and .5 two equal and oppo- site stresses, 4- S, S, which resist the extension. These are called TENSILE STRESSES. The tensile forces act away from each other, tending to pull A and B apart. The tensile stresses act towards each other, tending to draw A and B together. We have thus a force F and an equal and opposite stress in equilibrium at A, and a force F and an equal and oppo- site stress in equilibrium at B. We see also that if the stresses are tensile, they act away from the ends A and B. Compressive Stress and Force In the same way, if the forces are reversed in direction, we have COMPRESSIVE FORCE and COMPRESSIVE STRESS. The compressive forces act to make A and B move towards each other, and the compressive stresses to move them apart. ^F We see also that if the stresses are compress- ive, they act toivards the ends A and B. Shearing Force and Stress. The algebraic sum of the components parallel to a section of all the external forces on the right of that section tends to make that section slide upon the consecutive section on the left. The algebraic sum of the components parallel to a section of all the external forces on the left of a section tends to make that section slide upon the consecutive section on the right. For reasons to be given later (page 402) we always take the algebraic sum of the components parallel to a section of all the external forces on the left of that section, and call this algebraic sum the shearing force for that .section. We define, then, the shearing force for any section as the algebraic sum of the components parallel to that section of all the external forces on the left. It is resisted by the shearing stress or resistance of the section to sliding on the consecutive section on the right. This shearing stress is equal and opposite to the shearing force. Since for equilibrium the algebraic sum on one side must be equal and opposite to that on the other, we can define the shearing stress for any section as the algebraic sum of 397 -S 398 ST /tTICS OF RIGID BODIES. [CHAP. 1. the components parallel to that section of all the external forces on the right. Thus in the case of a horizontal beam AB acted upon by vertical loads P lt P 2 , P 3 and the vertical upward reactions /?, and 7? 2 , the shearing force for any section at a between the left end and P l is by definition -\- R r For any section at b between P l and PI the shearing force is -j- /?, P r For any section at c between P 2 and P 3 the shearing force is -|- /?, P l P 2 . For any section at d between P 3 and the right end the shearing force is + j?j P l /> 2 P y The ordinates of the shaded area thus give the shearing force to scale. The shearing stress is equal and opposite. If we have a load of w per unit of length, uniformly distributed, we have for the shearing force for any section distant x from the left end shearing force = ivx, which is the equation to the straight line A" B" passing through the centre of the span. The ordinate to this line at any section a gives the shearing force to scale. The shearing stress is equal and opposite. The shearing force at the centre is zero. Equilibrium of a System of Bodies. If a number of rigid bodies are connected by strings, rods, joints, etc., and the system is in equilibrium, each body of the system must be in equilibrium, and the conditions of equilibrium apply to each body as well as to the whole system. Examples. (i) In the system of pulleys sh + $m, T 4 = SP + 7m, T n = 2*- l P + (2*-' i)m, where n is the number of pulleys. Also we have T, + T, + T 3 + . . . Tn = Q. Substituting the values of 7\, 7", etc., P[i + 2 + 4 + . . . 2"] + w[i + 3 + 7 + . . . (2*- 1 i)] CO) Hence , or P = (2-l) [Student should solve by virtual work.] (4) In the differential pulley shown in the figure, an endless chain passes over a fixed pulley A, then under a movable pulley to which the mass Q is hung, and then over another fixed pulley B a little smaller than A. The two pulleys, A and B, are in one piece and obliged to turn together. The two ends of the chain are joined. The force P is applied as shown. To prevent the chain from slipping, there are cavities in the circumferences of the pulleys into which the / links of the chain fit. Find the relation between P and Qfor equilibrium. I Q A ANS. We have 27" Q = o, or T = . Let a be the radius of A, and b the radius of B. Then, taking moments about C, Ta Tb Pa o, or, inserting * LDP By taking a and b nearly equal we can have P very small. (5) The requisites of a good balance are as follows : ist. It should be "true," that is, when loaded unth equal masses the beam should be horizontal. 2d. It should be " sensitive," that is, when the massed differ by a small amount the angle of /,"> beam with the horizontal should be large, jd. It should be " stable," that is, wJien moved from equilibrium it should return quickly. Skow the conditions necessary for these requisites. 400 STATICS OF RIGID BODIES. fCHAP. I. i ANS. Let the masse* be /'and Q, and W the weight in pounds of the balance acting at its centre of mass C. Let the fulcrum be.at F, above C, and draw FD perpendicular to the beam AB at Z>. Let AD - a, BD = b, FC = d, FD = h, and fl the angle of the C _ . beam with the horizontal. Thru for equilibrium, taking moments about F, we have Q(b cos 8 h sin 8) - P(a cos + h sin 6) - Wd sin 6 = 0. Hence tane= (Q + P )h + Wd' I. If the balance is "true," we must have 6 = when P = Q. We see, from (i), that to satisfy this requisite the arms must be equal. We have then for a true balance a = b and tanfl 2. If the balance is to be "sensitive," 9 must be large when Q P is small. We see from (2) that to satisfy this requisite h and //must be small relative to the length of arm a. The sensitiveness is increased, therefore, by either increasing the length of arm or by bringing the centre of mass and point of suspension nearer the beam. 3. If the balance is to be " stable," or to return quickly when disturbed, the moment Wd sin 6 of W about the fulcrum must be large. Hence stability is increased by moving the point of suspension away from the beam. The conditions, then, for stability and sensitiveness are at variance. Stability is gained at the expense of sensitiveness, and vice versa. In scientific measurements, where great accuracy is required, stability is sacrificed to obtain great sensi- tiveness. The balance recovers slowly from a disturbance, and time is required for it to come to rest. For ordinary commercial purposes, where it is desirable to save time, sensitiveness is sacrificed to stability. Framed Structures. A framed structure is a collection of straight MEMBERS joined together at the ends so as to make a rigid frame. The simplest rigid frame is obviously a triangle, because that is the only figure whose shape cannot be altered without changing the lengths of its sides. All rigid frames must consist, therefore, of a combination of triangles. Any point where two or more members meet is called an APEX of the frame. Determination of the Stresses in a Framed Structure. We have external forces acting upon the frame, and tensile or compressive stresses in the members. If >he frame is in equilibrium, the stresses and external forces form a system in equilibrium. Also, since equilibrium must exist at every apex of the frame, all the forces and stresses at any apex form a system of concurring forces in equilibrium. Tensile stress acts away from an apex, compressive stress towards an apex (page 397). We have then two methods of solution : I. METHOD BY RESOLUTION OF FORCES. In the figure we have a frame acted upon by known external forces /*,, P 2 , P 3 , and the upward pres- sures of the supports R l and R 2 at A and B. Take any apex, as a. At this apex we have P l and the stresses in aA, ae, 0/and ab, forming a system of con- curring forces in equilibrium. If we denote these stresses by S lt S 3 , S 3 , S 4 , the angles of the members with the horizontal by ,, o- 2 , a s , *r 4 , and with the vertical by /?,, /3 2 , /3 3 , /3 t , we have then for equilibrium, if 5 is the angle of P l with the horizontal, and /3 5 with the vertical, 5, cos a l + S, cos a 2 -\- S 3 cos s + S 4 cos a 4 -f- PI cos S = o. J ^ COS Since we have thus two equations of condition, this method can be applied at any apex where there are not -more than tivo unknoivn stresses. Components to the right and upward are positive, to the left and downward negative. If then a stress comes out with a minus sign, it denotes that the stress acts towards the apex and is compressive. If it comes out with a plus sign, it denotes that the stress acts away from the apex and is tensile (page 397). 2. METHOD BY MOMENTS. Suppose the frame divided into two parts. Then the stresses belonging to the cut members must evidently hold in equilibrium the external forces on either side of the section. Thus, in the figure, if we suppose ab, af, ef cut, the stresses of these members must hold in equilibrium R and /\. The algebraic sum of the moments relative to any point must then be zero. ^ ^ If then we wish to find the stress in ab, we can take the centre ^g^ I ^ v>v >k. of moments at/". The moments of the stresses in af and ef will 1 then be zero, and if / is the lever-arm for ab, we have R ab . l-\- 2 moments of external forces on left of section = o. Observe that the moment for the stress in ab is to be taken with the sign indicated by the arrow. So if we wish to find ef, we should take moments about a, thus eliminating the moments of ab and af. Or if we wish to find af, we should take moments about A, thus eliminating the moments of ab and ef. The method, then, is in general as follows: Divide the frame and consider only one portion, as, for instance, the left-hand portion. Place arrows on each cut member pointing towards the section. To find the stress in any one of the cut members take the intersection of the other members cut as a point of moments. This will eliminate these members. Place the algebraic sum of the moments of all forces and stresses for this point of moments equal to zero. This method is general and can always be applied to find the stress in any member when all the cut members whose stresses are unknown, except the one whose stress is desired, meet in a point. If two of the cut members are parallel, their intersection is at an infinite distance, but the method still applies. Thus if we wish to find the stress in cb, we take a section cutting ab, be and cd. The intersection of ab and cd'\s at an infinite distance. We therefore have the lever-arm for cb, oo cos /?, where is the angle of cb with the vertical. Hence Jjoo Pjoo P 2 oo -f- cb X oo cos /3 = o, or cb cos ft + (R v /\ PJ = o, cb (R^ P l P a ) sec /S. 402 STATICS OF RIGID BODIES. [ClIAI'. I. The algebraic sum of the external forces (R l P l P a ) is the SHKAKINC PORCH (page 397). For horizontal chords and vertical forces we have then the vertical component of the stress in a brace in equilibrium with the shearing force. It is evident that if the shearing force is upwards or positive, the stress in cb will be compression and negative. If then we take the compression as negative and tension as positive, the sign will denote the character of the stress. It is for this reason that we have taken shearing force as the algebraic sum of all the forces on the left (page 397). Superfluous Members. We see from equations (i), page 401, that we have two equations of condition for equilibrium at any apex. If then at any apex there are more than two members whose stresses are necessarily unknown, the frame has superfluous members. Criterion for Superfluous Members. The simplest rigid frame is a triangle, and all rigid frames must consist of a combination of triangles. Any one member fixes the position of two apices. Every other apex after the first two requires two members to fix its position. If then n is the number of apices, 2(n 2) will be the number of members, lacking one. Let in be the number of members. Then if there are no superfluous members, we must have M I = 2(n 2), or ; = 2n If m is less than 2/1 3, there are not members enough, there are superfluous members. Examples. (i) /// the cases of the three frames represented, show that in the first case there are not enough members and the frame is not rigid; in the second case the frame, is rigid and there are no superfluous members ; in the third case there is one superfluous member. IP IP ,v If nf is greater than 2n 3, ANS. From our criterion we should have m = 2 3. But we have in the first case n = 6. Hence m should be 9. But m is only 8, or one less than the necessary number. In the second case = 6 and m = 9, as should be. In the third case n = 6 and m = 10, or one more than necessary. (2) A roof -truss has a span of 48 feet and a centre height of 18 feet. Each rafter is divided into two equal parts, and the lower tie into two equal parts, and the bracing is as shown in the figure. Find the stresses in the members for a load of 800 pounds at each upper apex. ANS. The reaction at each end is R = 1200 pounds. We find it as follows : Take B as a point of moments. Then we have R x 48 - 800 x 36 800 x 24 - 800 x R= + 1200. 12 = o, or ist METHOD. For the different members we have for the cosines of the angles with the horizontal (cos a) and vertical (COS ft) Aa cos a = 0.8 cos ft = 0.6 0.8 0.6 Ad i o 0.8 0.6 At apex A we have then + 1200 + Aa-x. 0.6 =o, .' a A x 08 -f Ad = o .' Aa 2000 pounds. Ad = -f 1600 CHAP. L] EXAMPLES. 403 The minus sign denotes stress towards the apex, or compression ; the plus sign stress away from the apex, or tension. At apex a we have Aa x 0.6 800 + ab x 0.6 ad x 0.6 = o, Aa x 0.8 + ab x 0.8 + ad x 0.8 = o. Inserting the value of Aa already found and solving for #and ad, we have ab = 13334, ad = 666f. At apex b, since be is evidently the same as ab, we can put 2ab x 0.6 800 bd = o, or bd = + 800. 2d METHOD. The centre of moments for Aa and ab should be taken at d, for ad and bd at A, for We have then for the lever-arms Aa ab Ad ad bd Lever-arm 1.44 14.4 9 14.4 24 Hence we have Aa x 1 4.4 1 200 x 24 = o, Aa = 2000 pounds ; Adx. 9 1200x12 = 0, Ad = + 1600 " #3x14.4+ 800 x 12' I2oox 24 = o, ab = 1333^ " - ad*. 14.4 800x12 = 0, ad 666f " For bd we have bdy. 24 800 x 12 + cdy. 14.4 = o, or, since cd = ad = 666f, bd= + 800 pounds. (3) A bridge truss 100 ft. long is divided into five equal panels in the lower chord and four equal panels in the upper chord. The depth is constant and equal to a b Q 10 ft. The bracing is isosceles. Find the stresses for a /V /\ f\ f\ 7V load of 800 pounds at each lower apex. / \ / \ / \ / \/\ ANS. ^^= + 1600, jC CD = + 4800, ab 3200, = 4800, Aa = 2262.4, a = + 2262.4, 2?<5 = 1131.2, bC= + 1131.2, CHAPTER II. GRAPHICAL STATICS. CONCURRING FORCES. FIG. i. FIG. 2. Graphical Statics. While the solution of statical problems by computation and ana- lytical methods is sometimes tedious and involved, they may often be solved with compara- tive ease and sufficient accuracy by graphic construction. The solution of statical problems by graphic methods gives rise to GRAPHICAL STATICS. We shall consider only co-planar forces. Concurring Co-planar Forces. Let any number of co-planar forces F lf F 2 , F 3 , F 4 , etc., given in magnitude and direction, act at a point A, Fig. i. In Fig. 2, from any point o, lay off to scale the line representative of F l from o to i, then the line representative of F 2 from i to 2, then the line representative of F 3 from 2 to 3, then the line representative of F 4 from 3 to 4, and so on. The polygon 01234 thus obtained we call the FORCE POLYGON. If all these forces are in equilibrium, the algebraic sum of their horizontal and vertical components must be zero. But when this is the case, evidently 4 and o, in Fig. 2, must coincide, or the force polygon must close. We have then the following principle: If any number of concurring forces are in equilibrium, the force polygon is closed. If the force polygon is not closed, the line 04. necessary to make it close gives the magnitude and direction of the resultant R. If we consider this resultant acting at the point of application t 1 in the direction from 4. to o, obtained by following round the polygon in the direction of the forces, it will hold the forces at A in equilibrium. If taken as acting in the opposite direction at A , it will replace the forces. COR. I. The order in which the forces are laid off in the force polygon is immaterial. Thus, in Fig. 2, if we had laid off o i, then the line representative of F 3 from i to 3', and then the line representative of F t , we should arrive at 3 just as before. By a similar change of two and two we can have any order we please. COR. 2. Any line in the force polygon, as o 2, o 3, or I 3, is the resultant of the forces on either side. Thus o 2 is the resultant of F l and F t , and, acting in the direction from 2 to o, holds FI and F 2 in equilibrium and replaces F 3 , F 4 and ./?. 404 CHAP. II. ] STATICS OF RIGID BODIES. 405 FlG COR. 3. If the forces are all parallel, the force polygon becomes a straight line. Thus in Fig. i, if the parallel forces F lt F 2 , F 3 , F 4 , etc., act at the point A, we FlG have the force polygon Fig. 2, 01234, and the closing line 40 is, as before, the resultant R and equal to the algebraic sum of the forces. If taken as acting from 4 to o, it will hold the forces at A in equilibrium. In the opposite direction it will replace the forces. Notation for Framed "Structures. Let the figure represent a roof-truss composed of two rafters, a horizontal tie-rod and intermediate braces consisting of struts and ties. The notation which we adopt in order to designate any member of a framed structure, or any force acting upon the structure, is as follows: We place a letter in each of the triangular spaces into which the frame is divided by the members, and also a letter between any two forces. Any member or force is then denoted by the letters on both sides of it. Thus in the figure AB denotes the force F{, BC denotes the force F 2 , CD denotes the force F 3 , DE denotes the upward pressure of the right-hand support R 2 , EA denotes the upward pressure of the left-hand support R r Also, Aa, Bb, Cd, De denote the portions of the rafters which have these letters on each side. The portions into which the lower tie is divided are in the same way Ea, EC, Ee. The braces are ab, be, cd, de. The student should carefully adhere to this notation for the frame whenever using the grapJiic method. Character of the Stresses. The determination of the kind of stress in a member of a frame, whether tension or compression, is as important as the determination of the magnitude of the stress. In the preceding figure, suppose we know the upward pressure at the left support R l or EA, and we wish to find the stresses in the members Ea and Aa, Fig. I, which meet at the lower left-hand apex. If these FlG . x> FlG . 2 . stresses and R l are in equilibrium, they will make a closed polygon. If then we lay off EA in Fig. 2, upwards, equal to R v , and then from A and E draw lines parallel to Aa and Ea in Fig. I, and produce them till they intersect at a, Fig. 2, evidently the lines Aa and Ea in Fig. 2, taken to the [R* ' R same scale as EA, will give the magnitude of the stresses in Ea and Aa in Fig. I. Thus lines in the force polygon which have letters at each end give the stresses in those members of the frame denoted by the same letters at the sides. Now as to the character of these stresses, the directions Aa and aE in Fig. 2, obtained by following round in the known direction of R lt are the directions for equilibrium (page 404). Since we are considering the concurring forces acting at the left-hand apex, transfer these directions to Fig. I, and we see that Aa acts towards the apex we are considering and thus resists compression, and aE acts away from it and therefore resists tension. The st' ess in Aa is therefore compressive ( ), and in aE tensile (-f-). GRAPHICAL STATICS.-CONCURRING FORCES. [CHAP. II. In general, then, if we take any apex of the frame in Fig. I, and consider the concurring forces acting at that apex as a system of concurring forces in equilibrium, we have the following rule: Follow round the force polygon in Fig. 2 in the direction indicated by any one of these forces already known, and transfer the directions thus obtained for the stresses to the apex in Fig. i under consideration. If the stress in any member is thus found acting away from the apex, it is tension (-(-); if towards the apex, it is compression (). Application of Preceding Principles to a Frame. Let Fig. i be. a frame consisting of two rafters, a horizontal tie-rod and bracing as shown, carefully drawn to a scale of a certain number of feet to an inch. This we call the FRAME DIAGRAM. Let the forces 7^, F 2 , F 3 act at the upper apices, and let the reactions or upward pressures of the supports be ^ and R r Notate the frame and these forces as directed, so that F, = AB, F 2 = BC, F 3 = CD, R 2 = -DE, R^ = EA, while the members are Aa, Bb, Cd, De, Ee, EC, Ea, ab, be, cd, de. stresses in the members. These outer FIG. i. FIG. 2. The outer forces acting upon the frame cause forces must first be all known, or if any are unknown, they must first be found. Lay off these outer forces AB, BC, CD, DE, EA in Fig. 2 to a scale of a certain number of pounds to an inch. Each force in Fig. 2, having letters at its ends, is equal and parallel to those forces in Fig. I which have the same letters at the sides. The polygon formed by AB, BC, CD, DE, EA (in this case a straight line, Cor. 3, page 405) we have called the FORCE POLYGON. If the frame is in equilibrium, this polygon must always close, that is, the outer forces acting upon the frame must be in equilibrium. If it does not close, these outer forces are not in equilibrium and the frame will move. That is, the frame itself, so far as its motion as a whole is considered, may be treated as a point. Having thus drawn and notated the frame Fig. i and constructed the force polygon Fig. 2, we can find the stresses in the members. The forces and stresses at each apex must be in equilibrium, and therefore form a closed polygon. Thus consider first the left-hand apex, Fig. I. At this point we have the reaction EA and the stresses in Aa and Ea, constituting a system of concurring forces in equilibrium. But we already have EA laid off in Fig. 2. If then we draw Aa and Ea in Fig. 2 parallel to Aa and Ea in Fig. I, and produce to intersection' a, the polygon is closed and we have in Fig 2 the stresses in Aa and Ea, to the same scale employed in laying off EA. Since EA acts upwards, if we follow round from E to A, and A to a, and a to E, in Fig. 2, and transfer the directions thus obtained for Aa and aE to the left-hand apex in Fig. i, we have the stress in Aa towards this apex or compression ( ), and the stress in aE away from the apex and therefore tension (-{-) [The student should follow with his own sketch and mark each stress with its proper sign as he finds it.] Let us now pass to the next upper apex, at F lt Fig. i. Here we have F l or AB and the stresses in Aa, ab and Bb in equilibrium. But we already have the stress in Aa and AB laid off in Fig. 2. If then we draw from a and B in Fig. 2 lines parallel to ab and Bb in Fig. i, and pro- CHAP. II.] STATICS OF RIGID BODIES. 407 duce to intersection b, the polygon is closed and we have in Fig. 2 the stresses in ab and Bb. Since AB is known to act downward, we follow round in Fig. 2, from A to B, B to d, d to a, and a to y4, and transfer the directions thus obtained to the apex at F lt Fig. i, under consideration. We thus obtain the stress in Bb towards the apex or compression, the stress in ba towards the apex or compression, and the stress in aA towards the apex or com- pression, just as already found. Note that in the first case, when we were considering the apex at R^ , we found the stress in aA acting towards that apex. Now when we consider the apex at F^ we find the stress in aA acting towards that apex in both cases, then, compression (page 397). Let us now consider the second lower apex, Fig. i. We have here no outer force, but the stresses in Ea, ab, be and cE must be in equilibrium and therefore form a closed polygon. But in Fig. 2 we have already found the stresses in Ea and ab. If then we draw from b a line parallel to be in Fig. i, and produce it to intersection c withEa, the polygon closes, and we have in Fig. 2 the stresses in be and cE. We have already found aE to be tension. It must therefore act away from the apex we are considering. We therefore follow round in Fig. 2, from E to a, a to b, b to^r, and c to E, and transfer the directions thus found to the corresponding members in Fig. i. We thus obtain the stress in Ea tension and the stress in ab compression as already found, and the stress in be tension and in cE tension. Let us now consider the top apex. We have here the force F 2 = BC, and the stresses in Bb, be, cd, and dC, in equilibrium. But in Fig. 2 we have already laid off BC, and we have found the stresses in Bb and be. If then we draw from c and C lines parallel to cd and Cd'm Fig. i, and produce to intersection d, the polygon closes and we have in Fig. 2 the stresses in cd and Cd. Since BC acts downwards, we follow round from to C, C to d, d to c, c to b, and b to B. Transferring these directions to the corresponding members in Fig. i, we obtain the stress in Cd compression and in dc tension, while the stress in cb is tension and in bB compression as already found. We can thus go to each apex and find the stresses in every member. The lines in Fig. 2 which thus give the stresses in the members constitute the STRESS DIAGRAM. Each stress having letters at its ends in Fig. 2 is parallel to that member in Fig. i which has the same letters at its sides. Apparent Indetermination of Stresses. It sometimes happens that a frame has no superfluous members, and yet in applying the graphic method we are unable to find any apex at which all the forces but two are known. In such case the difficulty may be overcome by taking out one or more of the members and replacing them by another member, and then applying the method until we find the stress in some member which is not affected by the change. Or we may find the stress in this member by the method of sections (page 401). Having found this stress, we can replace the members taken out and find the actual stresses. Thus let Fig. i (page 408) be a frame* acted upon by the forces F l , F 2 , F z , F 4 , etc., and the reactions or upward pressures of the supports J? lt R z . Notate the frame and the forces by letters on each side as directed (page 405). Then lay off to scale the outer forces in Fig. 2, thus forming the force polygon ABCD . . . HIA. This polygon is a straight line in this case, because all the forces are parallel, and it must close, that is, the outer forces are in equilibrium. We can now proceed to find the stresses as follows : Consider first the left-hand apex, Fig. i. At this point we have the reaction I A and the stresses in Aa and la constituting a system of concurring forces in equilibrium. But we already * Disregard for the present the dotted member in Fig. I. 4 o8 GRAPHICAL STATICS-CONCURRING FORCES. [CHAP. II. have IA laid off in Fig. 2. If then we draw Aa and la in Fig. 2 parallel to la and Aa in Fig. I, and produce to intersection a, the polygon is closed and we have in Fig. 2 the stresses in Aa and la to the same scale employed in laying off the forces. Since I A acts upwards, we FIG. i. FIG. 2. follow round from 7 to A, A to #, and a to /, in Fig. 2, and transfer the directions thus obtained for Aa and al to the corresponding members in Fig. I. We have then the stress in Aa towards the apex we are considering or compression ( ), and the stress in rz/away from that apex or tension (+). Considering now the next upper apex, we have here the force AB known, the stress in Aa already found, and the stresses in ab and Bb unknown. If then in Fig. 2 we draw ab and Bb, thus closing the polygon, we obtain the stresses in ab and Bb. Since AB acts down, we follow round in Fig. 2 from A to B t B to b, b to a, and a back to A, and transfer the directions thus obtained to the corresponding members in Fig. I. We have then the stress in Bb towards the apex we are considering or compression ( ), the stress in ba towards that apex or compression ( ), and the stress in aA also towards that apex or compression ( ), just as we have already found it. Note that when we were considering the apex at R^ , we found the stress in aA acting towards that apex. Now when we consider the apex at F^ we find the stress in aA acting towards that apex. In both cases, then, compression (page 397). We can now consider the next lower apex, where we have the stresses in fa, ab, fcand cl in equilibrium. We already know la and ab, and if we draw in Fig. 2 be and cl, we obtain the stresses in be tension (+) and in cl tension (+). Thus far there has been no difficulty in the application of the graphic method. But now we cannot consider the next upper or lower apex, because at each we have more than two unknown forces. If we should start at the right end, we should soon come to the same difficulty on the right side. Apparently we can go no farther. The number of members is 27 (we disregard the dotted member in Fig. i). The number of apices is 15. We have then, applying the criterion for superfluous members (page 402), m = 2n 3. There are then no superfluous members. If now we remove the two members de and r/and replace them by the dotted member c'f, where e takes the place in the new notation of the two letters e and d, we have still a rigid frame with no superfluous mcmbcr s . For the number of members is now in = 25, and the number of apices is n = 14. We have then ;;/ = 2n 3. But this change has evidently not affected tlie stress in the member Ig. We can therefore now carry on the diagram until we find the stress in Ig, or we may compute the stress in Ig directly by the method of sections (page 401). CHAP. II.] STATICS OF RIGID BODIES. 49 Thus if we now consider the apex at F 2 , Fig. i, we have at this point the stresses in the members Bb, be, ce' and e C, and the force BC, all in equilibrium. We know BC, Bb and be, and if we draw in Fig. 2 ce' and e'C, we obtain the stresses in e'C compression and in ce compression. We can then pass to the apex at F 3 , Fig. I, where we know all the forces except the stresses in D/and/e'. We draw then Df and / acting at each apex must be zero, we have -{- R -\- Aa cos a o, or Aa = - , where COS IX a is the angle of the rafter with the vertical. We get the same result at once from Fig. 2 by solving the triangle AaE. In the same way we have at once, from Fig.. 2, ab = F l cos ft, where ft is the angle of ab with the vertical. We see also from Fig. 2, page 406, other relations. Thus we see that the stress in ab will be the least possible when ab is perpendicular to the rafter. We also see at a glance how the stress in any member is affected by a change of inclination of the member. Finally, the application of the method is equally simple no matter how irregular the frame may be. If the frame is symmetrical with respect to the centre, and the forces F lt F 3 in Fig. 2 (page 406) are equal, it is evident that the stresses in each half will be the same. We have then Cd = Bb, cd cb, and so on. Choice of Scales, etc. In general the larger the frame is drawn in Fig. I, the better, as it then gives more accurately the direction of the members composing it. The force polygon Fig. 2, on the other hand, should be taken to no larger scale than consistent with scaling off the forces to the degree of accuracy required, so as to avoid the intersection of very long lines, where a slight deviation from true direction multiplies the 4 io GRAPHICAL STATICS. CONCURRING FORCES. [CHAP. II. error. If an error of one twenty-fifth of an inch is considered the allowable limit, the scale should be so chosen that one twenty-fifth of an inch shall represent a small number of pounds, within the degree of accuracy required. The stress polygon Fig. 2 should be completely finished and the signs for tension (-{-) and compression ( ) placed on the frame for each member as its stress is found, to avoid confusion, before the stresses are taken off to scale. A good scale, dividers, straight-edge, triangle, and hard fine-pointed pencil are all the tools required. The work should be done with care, all lines drawn light, points of intersection accurately located and the frame properly notated to correspond with the force polygon. Care should be exercised to secure perfect parallelism in the lines of the frame and stress polygon. Some practice is necessary in order to obtain close results. It should be remembered that careful habits of manipulation, while they tend to give constantly increased skill and more accurate results, affect very slightly the rapidity and ease with which these results are obtained. FIG. i. Examples. (i) A roof-truss has a span of 50 feet and rise of 12.5 feet. Each rafter is divided into four equal panels, and the Imver horizontal tie into six equal panels. The bracing ts as shown in the figure. A weight of 800 Ibs. is sustained at each upper apex. Find the stresses. ANS. Draw the frame in Fig. i to a scale of, say, 12 feet to an inch, and notate it. Then construct the force polygon ABC. ..HI A, Fig. 2. Note that 7v' s or HI and K\ or I A are equal and eacli 2800 Ibs. The force polygon then closes as it should. We can take the scale of Fig. 2 as 3200 Ihs. to an inch. Then an error of 2 ' y of an inch will be about 128 Ibs. We can then find the stresses as shown in Fig. 2. Aa Bb Cd Df la 6280 5816 4700 3580 5624 Ic 4832 le ab + 4024 - 720 be + 720 cd 1060 de ef 928 - 1452 fg + 2400 Ibs. The accurate results as found by computation (page 402) are 6260 - 5813 4696 3577 + 5600 + 4802 + 4003 720 + 720 1081 + 920 1443 + 2401 Ibs. It will be seen that the greatest error is only 30 Ibs. The above results were actually obtained from the diagram, using the scales FIG. 2. given. (2) Sketch the stress diagram for a roof-truss as shown in the following Fig. i, equal forces acting at every upper and lower apex. ANS. The student should note that the reactions DE and GA are each equal to half the sum of the downward forces, or 2$ forces. We lay off then in Fig. 2 AB, BC. CD downwards. Then DE upwards equal to 2J forces. Then EF, FG downwards. Then GA upwards equal to 2| forces, and closing the force polygon. The stresses can now be found as always. FIG. i. FIG. 2. CHAP. II.] STATICS OF RIGID BODIES. 411 (3) We give in the following fi^ttres a number of frames with their stress diagrams* For the sake of generality the outer forces and reactions are often taken inclined as well as vertical. rig. i. Fig. 2. Fig. 3. Fig. 4, Fig. 5. Fig. 6. Fig. 9. *The student should sketch the stress diagrams for himself in each case, putting down, as he goes alcng, the sign ( ) and (-f-) for compression and tension upon each member of the frame as soon as he finds it. CHAPTER III. GRAPHICAL STATICS. NON-CONCURRING FORCES. Non-concurring Forces. Let the co-planar forces F lt F 2 , F^, F 4 , etc., act at the points A l} A t , A^ A 4 of any rigid body, Fig. i. If we lay off the forces to scale in Fig. 2, we have as before the force polygon 01234, and the closing line o 4 gives as before the resultant. If this resultant acts in the direction FIG. i. X B \ 4 o upon the rigid body, it will hold the given forces in equilibrium. If it acts in the direc- tion o 4, it will replace the given forces. We thus know the magnitude and direction of the resultant. But 'imposition in the plane of the forces in Fig. I is as yet unknown. In order to determine this, choose any point O in Fig. 2, and draw the lines Oo and (^4. This point O we call the POLE of the force polygon. Now since every line in the force polygon represents a force, by thus choosing a pole O and drawing lines Oo, 04 to the extremities of the resultant O 4, we have resolved the resultant into the two forces represented by Oo and $4. This is evident from the fact that these two lines make a closed polygon with O 4, and hence taken as acting from 4 to O and O to O, as shown by the arrows, hold the forces /*",, F 2 , F 3 , F t in equilibrium, or replace the resultant 4 o (page 404). As the pole O is taken anywhere we please, we can thus resolve the resultant 4 o for equilibrium into forces in any two directions we wish. Let us then consider the resultant 40 for equilibrium, replaced by the two forces 4er horizontal tie is divided into six equal panels. The bracing is as shown in the figure. Find the stresses in the members, by Hie graphic method of moments, for a weight of 800 Ibs. at each upper apex. ANS. We have computed the stresses (page 402), by the two methods, resolution of forces and moments. We have also found the stresses by the graphic method of resolution of forces (page 410, ex. (i)). We can construct the force polygon Fig. 2, and then the equilibrium polygon Fig. i. This, however, is not advisable for rea- f-jt/ sons already given. It will be FIG 2. FIG. i. more accurate to assume the pole distance as unity, thus discarding the force polygon altogether, and construct points in a parabola from the equation A B C D -._ L E FT G X ^o In the present case the load per foot is, if we suppose half weights of 400 at the ends, = 128 Ibs. = p. Taking x =-1, ^-/, etc., we have ll. ll. /= 17500, 30000, 37500, 40000 Ib.-ft. Laying these off to any convenient scale, we determine very accurately the points a, b, c, d of the equilibrium polygon. The other half of the polygon is precisely similar. 422 GRAPHICAL STATICS NON-CONCURRING FORCES. [CHAP. III. The ordinates to this polygon will give, to the scale adopted, the moment, for any point of the truss, of the outer forces left or right. Thus the moment with reference to k of all forces right or left is km, Fig. i. We find by scale km = 2i666| Ib.-ft. In the same way for the next lower apex we find the moment 35000 Ib.-ft. The moment at the next lower apex or centre of the span is 40000 Ib.-ft. Now by the method of sections (page 401) we have for any member Stress x lever-arm + 2 moments of outer forces = o. The second term is given by the ordinates of the equilibrium polygon to scale. As regards the centre of moments for any member, we must observe the rule (page 401), viz. : Cut the truss entirely through by a section cutting only three members the strains in which are unknown. For any one of these take the point of moments at the intersection of the other two. For the proper sign for the first member of the equation place an arrow on the cut member pointing away from the end belonging to the left-hand portion, and take the moment ( + ) or ( ) according as the rotation indicated by this arrow is counter-clockwise or clockwise. If the stress comes out positive, it indicates tension ; if negative, compression. Take, for instance, the first lower panel, La. The centre of moments must be taken at the first upper apex. The moment for this point is given by the ordinate na of the equilibrium polygon, or 17500 Ib.-ft. We take the minus sign, because the rotation is clockwise. We have then La x 3.125 17500 = o, or La = + 5600 Ibs., where 3.125 ft. is the lever-arm of La. In similar manner we have Lc x 6.25 30000 = o, or I^c = + 4800 Ibs., where 6.25 ft. is the lever-arm of Lc. For Le we have Le x 9.375 37500 = o, or Le = + 4000 Ibs., where 9.375 ft. is the lever-arm of I.e. For the first upper panel, Aa, take the centre of moments at k The moment f< r this point is given by the ordinate from k to the first line of the polygon produced. It is therefore larger tlian km. which gives the combined moment of the reaction and first weight. We find it by scale to be 23333^ Ib.-ft. We have then Aa x 3.727 23333$ = o, or Aa = + 6260 Ibs., where 3.727 ft. is the lever-arm for Aa. In like manner for Bb we have centre of moments at k, and moment km = 21666}. Hence - Bb x 3.727 21666} = o, or Bb = + 5813 Ibs. For Cd we have Cd x 7.454 35000 = o, or Cd + 4691 Ibs., where 7.454 ft. is the arm-lever for Cd. For Df we have Df x 11.151 - 40000 = o, or Df = + 3587 Ibs. For all the braces the point of moments is at the left-hand end. Taking a section through Bb, ab and La, we have acting on the left-hand portion only the weight AB and the reaction. The moment of the weight relative to the left end is the ordinate a'b', or by scale 5000 Ib.-ft. The lever-arm for ab is 6.934 ft. Hence ab x 6.934 5000 = o, or ab = 721 Ibs. For be we have + ab x 6.934 5000 = o, or ab == + 721 Ibs. For cd the moment is a'V + b' the entire base is not brought into action. In such case the joints on the inside tend to open, and in a well-designed wall this should not occur. Also, the greatest unit pressure/ should never exceed the allowable unit stress C as given by the table page 424. In a properly designed wall, then, e must be equal to or greater than b 2 , and / must be less than, or at most equal to, C. If, in any case when e = - 2 , / is less than, or at most equal to, C, the wall is said to be low. If, when / = C, we have e greater than b z , the wall is said to be high. If then in the second of equations (III), we make/ C, we have for e = -b^ _2(W+V) 2 ~ c * For trapezoid section CHAP. IV.] DESIGN OF LOW AXD HIGH WALLS TRAPEZOID SECTION. 429 Hence we have for the limit of low wall Equation (I) gives the limit of /^ for low wall of trapezoid section, for given top base b l and bottom base b z . For /^ greater than this limit the wall is high. For h v less than this the wall is low. For rectangular section we have b z = b^ , and in this case we have b.C - Equation (i) gives the limit of h v for low wall of rectangular section for given base b r For h v greater than this limit the wall is high. For /i l less than this the wall is low. Design of Low Wall Trapezoid Section. For both water and earth pressure h (figure, page 427) is the distance from base to water or earth surface, and d^ as we shall see hereafter (page 43 1), isj/*. BOTTOM BASE. If then we make e~ ~b z inequation (II), page 428, put d^ = -h, and substitute the values of Wand ^ as given by equations (i) and (5), page 427, and solve for b 2 , we have '* 2 =-^+ VBf +E,* (II) where the quantities S l and E l are given by *i = 2 L (^ + ^ - *i tan ), E, = bfa + 2^ tan fa + ~ ( Ftan ftl + H + 3 7). Equation (II) gives the bottom base b 2 for a properly designed low wall of trapezoid section. If the value of & 2 , as given by (II), substituted in (I), gives the limit h^ greater than the actual height, the wall is low. If less, the wall is high and equation (II) does not apply. TOP BASE. For economy the top base b^ should be assumed as small as possible con- sistent with practical and local considerations. We should not in any case assume b l greater than by If then we make b z = b v in (II), we have for the greatest allowable value of ^ max. b l = B+ V + E, (2) where the quantities B and E are given by B = ^(lf- 3 ^ tan & E = w, (v tan fr+ ff +* T )' Equation (2) gives then the largest allowable value of b l for low wall, that is, when A l is less than or equal to limit k^ as given by (i). Design of High Wall Trapezoid Section. As we have seen, if the value of 2 as given by (II), substituted in (I), gives the limit h v greater than the actual height, the wall is low. If less, the wall is high. In this case e is greater than - 2 , and/ = C. 43 ST/tTICS Of RIGID BODIES. [CnAi-. IV. LOWER BASE. From the third of equations (III), page 428, we have For / = C we have _ * A fiy.... - 3^ 43<5 STATICS OF RIGID BODIES. [CHAP. IV. as /?, decreases. So also does E 2 . Hence 2 in (III) has its least value when /?, = o, or the most economic section for high dam of trapezoid section is that for which the back is vertical. HIGH DAM ECONOMIC TRAPEZOID SECTION. If then we make fa = o in (III), we have for high dam, economic trapezoid section, _i_ c- and for the greatest allowable value of b l 6Th ............ (30 CONDITIONS FOR RECTANGULAR SECTION. If we make a = b l in (I), and fa = o, we have for limit of /^ for low dam of rectangular sectron Ts W and from (2^) for the value of b v From this, if we make h = h v and solve for Ji lt we have for water level with top of dam M!_^. 1 V r Y We see from this that if ^ is equal to \/-^~ we have h^ o, and if ^ is less than this, //, is imaginary. We have then mn. = and from (5), making /* = 7/j , and inserting the value of h^ from (4), Local and practical considerations must control the choice of top base b^ and we should take it as small as such considerations will allow. We should not take it greater than given by (7). If we take it greater than given by (6), the section can be rectangular. If we take it less than given by (6), the section cannot be rectangular. Examples. (i) Design a rectangular dam of granite ashlar 20 feet high, the water-level to be 2 feet below the top, -with and without ice-pressure. ANS. We have h\ = 20, h = 18, bi = b\ , ft\ o, P = 40000, and, from table page 424, C = 60000 d = 165. From equation (I), page 434, making ft, = o and 1 = 1, we have for the limit h\ for low dam 2el a feet below the top, with and without ice-pressure. ANS. We have hi = 20, h 18, fti = o, tan ft = , b% = bi + 2, T = 40000, and, from table page 424, 6 = 165, C= 60000. From equation (II), page 434, making bi = <$, + 2 and solving for , , we have for low dam without ice- pressure bi = 3 + ^115.45 = 7.74 ft., and hence a = 9.74 ft.; and with ice-pressure bi = 3 + ^ \ 424. 54 = 34.74 ft., and hence b t = 36.74 ft. From (I), page 434, inserting these values of b\ and b*, we have for the limit of h\ for low dam without ice-pressure limit h\ = 268.5 ft-> with ice- pressure limit h\ = 188.5 ft- Both these limits are greater than 20. The dam is then low in both cases, and the values of b\ and bi just found hold good. We have then for the area of cross-section without ice-pressure A = 174.8 sq. ft., with ice-pressure A = 714.8 sq. ft. This is the most economic section for the given proportions. We see, then, that there is a saving of material over the preceding examples by making the back vertical. We could save still more, however, by increasing the front batter angle or decreasing the top base, as in the next example. (5) Design a trapezoidal dam of granite ashlar 20 feet high, top base 2 feet, water-level 2 feet below top, for economic section. ANS. We. have h\ = 20, /*=i8, bi = 2, T= 40000, 5 = 165, C = 60000, and for economic section fti = o. From (IT), page 435, we have for low dam without ice-pressure = i + I/ 5 H -- ^ - -' -= 10.16 ft., 6 x 20 165 x 2 / 62.5x;8-?2 6x40^00x18 w.th .ce-pressure * = - i + |/ 5 + ~^^ + ,65 x 20 From (I), page 434, inserting these values of b*, we have for the limit of h\ for low dam b*C 10 1 6 x 60000 without ice-pressure limit hi = -r-r 2" = 33- 8 i 8(6 t + 6,) 165 x 1 2.16 with ice-pressure limit hi = = 344.8 ft. 105 x 30.74 Both these limits are greater than 20 ft. The dam is then low in both cases and the values of A, just found hold good. We have then for the area of cross-section without ice-pressure A = 121.6 sq. ft., with ice-pressure A = 387.4 sq. ft. We see that there is a large saving of material over the preceding cases. This is the most economic section for the given dimensions. We could save more only by reducing the top base still more. (6) Design the San Mateo dam (page 433). taking the top base at 20 feet and the back baiter i to 4 as actually built, and taking into account wave-pressure. ANS. We have h\ = 170, h = 165, b\ = 20, tan fti = - y 62.5, 5 = 150, C = 34000, T= 24000. CHAP. IV.] . DESIGN OF DAMS-EXAMPLES. 439 From equation (II), page 434, we have for low dam fa = 5.432 + ^29.51 + 15461.38 = 119.03 ft. From equation I), page 434, we have for the limit of hi for low dam limit hi = " x 34000 _ 4 x 150 x 139.03 159 x 139.03 62.5 x 165' This is greater than the height 170 ft. The dam is therefore low and the value of , just found holds good. We have then the area of cross-section A = 11818 sq. ft. The weight per foot of length is W = SA = 150 x 11818 = i 772 700 pounds per ft., the vertical pressure V (page 431) is Y IK* 62.5 x 170* V = '- tan fti = g = 22578 pounds per ft., and from the second of equations (III), page 428, the greatest unit pressure is V) p - -yj - = 33580 pounds per sq. ft., or less than the allowable stress C = 34000 pounds per sq. ft., the value of e being bi. The dam as actually built (see page 433) has a base b* 176 ft., an area A = 16660 sq. ft., the value of e is greater than hi, and the greatest unit pressure^ = 15930 pounds per sq. ft. We see, then, that by properly designing we have for e = bi, so that the entire base is still in action, a saving of over 29 per cent, and the dam is still safe against rotation and crushing. For sliding we have, as on page 427, for the coefficient of safety _n(W+ V) _ 0.6 x 1998480 _ H + T 874780 Since n is greater than unity, the resistance to sliding for through joint at base is greater than the thrust * H + T; but if friction only is relied upon, the coefficient is not large enough. By breaking joints, however, sliding is impossible. By making the back vertical we could make a still greater saving, as we see from the next example. (7) Design the San Mateo dam for economic section, taking into account wave-pressure and the top base 20 ft, as actually built. ANS. We have b\ = 20, hi = 170, h = 165, y = 62.5, 8 = 150, c 34000, T = 24000, and for economic section fi t ~ o. From equation (IT), page 435, we have for low dam ./ 62.1; x (165)' 6 x 24000 x 165 , b* = 10 + 4/ 500 + -^ : iL + 2 101.54 ft. 150 x 170 150 x 170 From equation (I), page 434, we have for the limit of hi for low dam limit hi = l '*> X 34000 =i8 9 ft. 150 x 121.54 This is greater than the height 170 ft. The dam is therefore low and the value of bi just found holds good. We have then the area of cross-section A = 10331 sq. ft., 44 ST/tTICS OF RIGID BODIES. [CHAP. IV the weight per foot of length W 8 A = \y>A = i 549650 pounds per ft., and from the second of equations (III), page 428, the greatest unit pressure 2/F p = -j- = 30523 pounds per sq. ft., f or less than the allowable unit stress C = 34000 pounds per sq. ft. The dam as actually built (see page 433) has a base , and the greatest unit pressure is only/ = 15930 pounds per sq. ft. By making the back vertical and e = -6,, so that the entire base is still in action, we save over 38 per cent on the section, and the dam is safe against rotation and crushing. For sliding we have, as on page 427, for the coefficient of safety I*W 0.6 x 1549650 -7TTT = 87-4780- Since ft is greater than unity, the resistance to sliding for through joint at base is greater than the thrust // + T, but the coefficient is not large enough if friction only is relied upon. By breaking joints, however, sliding is impossible. (8) Suppose the height in the last example to be taken at 200 ft. ; the -water level 5 ft. from the top. ANS! We have , = 20, //, = 200, h = 195, y = 62.5, S 150, C = 34000. T = 24000, and for economic section /Si = o. From equation (IT), page 435, we have for low dam = _ ,o + 4/ SOP + 62.5 x (195)* + 6 x 24000 x ,95 f 10 X 2OO 150 From equation (I), page 434, we have for the limit of hi for low dam limit h, = I2 = I9? ft . 150 x 140 This is less than the height 200 ft. The dam is therefore high and the value of bt just found does not hold good. We should then use equation (III'), page 436. We obtain, then, ./I5o*x 200 x (20)* + 62.5 x(i95) + 6 x 24000 x 195 * = \ - = 120.30 it. 34000 The area of cross-section is then A = 14038 sq. ft. Design of Dam Economic Section. We have seen, pages 435 and 436, that for low and high dam and trapezoid section the greatest economy is for back vertical. We have also seen, page 436, that/or rectangular section b^ must not be greater than * 67- + ~* nor less than | . , ....... (i) 6T Local and practical considerations must control the choice of top base , , and we should take it as small as such considerations will allow. CHAP. IV.] DESIGN OF DAM ECONOMIC SECTION. 441 FIRST SUB-SECTION. For given top base b lt then, less than max. b l and greater than min. b v as given by (i), the section should be a rectangle, A^B^DE. D &, We can run this rectangular section down to a distance /t l below the top, d until e shall be just equal to b^ so that the entire joint A l B l acts, 3 provided this joint is not overloaded. We have from equation (2'), page 435, for rectangular section and A __ Let d be the distance of the water-level below the top. Then h = h^ d. Substituting this value of h and solving for // 1 , we have X V - (2) Equation (2) will give the value of h^ for the first sub-section. From equation (I), page 434, we have for the limit of h^ for rectangular section limit h, = (3) We can then run the first rectangular sub-section A^B^ED down for the distance /^ given by equation (2) without overloading, provided that /^ as given by (2) is less than the limit of h^ given by (3)- If b is taken less than */$!, there can be no rectangular sub-section, but the first sub-section should be a trapezoid, as given in the next article. SECOND SUB-SECTION. If the height of dam is greater than the value of h^ as given by equation (2), we still continue the back vertical, but the breadth must increase so that e ^ shall be equal to -6 r We have then a second sub-section, A^B^B^A^ with front batter and vertical back. Since the vertical pressure remains unchanged for dam empty, we can run this second sub-section down to a distance h z below the top, until the back edge distance s 2 for dam empty is also equal to --b z , provided the greatest unit stress/ is less than or equal to C. Or we can run it down until p = C, provided s 2 is greater than or equal to -b r We have then two cases, s z = e 2 = -3* 2 , and / less than or equal to C, and e 2 = - 2 , p C, and s 2 greater than or equal to e r The height of the second sub-section is then (h z - k^. It is required to find //, and b v 442 ST4T1CS OF RIGID BODIES. [CHAP. IV. Since the resultant for full dam is to cut the base b 2 = AyB 9 in both cases at a distance ^ a = -, from 3 , we have in both cases From equation (5), page 427, We have also Inserting these values, making e 2 = 2 , and solving for b z , we have for the base b^ of the second sub-section in both cases where the quantities B 2 and E z are given by If we have b v less than / . ( there is no rectangular section (page 436), and we have A, = o in (4). V The weight W l per foot of length of the first sub-section acts at a distance b v from A 2 and \\\ at the distance (b 2 ^ 2 ) from A 2 . The resultant W l -f- W* ac ts then at the distance s t from Ay given by Inserting the values of W 7 , , W^ 2 and ^ 2 , we have -'tj When s 2 = & 2 we have, solving for // 2 for the limit of // 2t i 'when e n = S 9 = b 9 and 2//.(. ,) 2 2 7 2 > limit //, = 1 M i (6) ( /^, / less than or equal to C, ' and when e -b and/ = C, we have from the second of equations (III), page 428, CHAP. IV.] DESIGN OF DAM ECONOMIC SECTION. or, inserting the values of W l and W 2 and solving for // a , we have for the limit of A 2 when s 2 > b* e 2 and / equal to C, .. . . limit h = 443 (7) If we substitute the value of 2 given by (4) in equations (6) and (7), the least of the two values for limit h 2 thus obtained will be the limiting value of h z for the second sub-section. We can then find b z from (4) for any value of h z up to this limiting value. If we have b l less than \ / , there is no rectangular section, and we make h l = o in (6) and (7). The limit h 2 in this case is THIRD SUB-SECTION. If the height of the. dam is greater than the limiting value of /i 2 just found, we have a third sub-section. The limit h 2 for the second sub-section may be given by equation (6) or by equation (7) as we have seen, and we thus have two cases to consider. ist Case. If the limit h z is given by equation (6), that is, if e 2 = s 2 - b 2 and p less than or equal to C, we must batter both face and back, so that e 3 shall be greater than 6 S and / = C for dam full. We have then the sub-section A 2 B 2 3 A S with front and back batter. The back batter angle we denote by fl y We can run this sub-section down to the base of the dam, so that h^ is the given height of dam ; and since the vertical pressure is unchanged for dam empty, we should have the edge distance s 3 from A s for dam empty equal to the edge distance e^ from B 3 for dam full. It is required to find b^ and fi y The resultant W l + J^ is at the distance s 2 = -& 2 from A 2 for dam empty. The weight W 3 is at a distance -? 3 from B^ given from equation (5), page 427, by f, d &J + &7 V 1' "\ ' A y V i "T T-^A- tan (9) There is a vertical water-pressure V on the inclined back A 2 A 3 . For the sake of sim- plicity we neglect V. This omission, which is on the side of safety, involves no prac- tical error. Neglecting V, then, we have the edge distance s 3 from A 3 given by + (A, - tan (10) ST/1TICS OF RIGID BODIES. [CHAP. IV. For dam full, neglecting V t we have the edge distance e^ from B^ t * 8 - - (h, - kj tan A + Wf, - - T(h, - d) where W v and W^ are already known and ,_*(*! + W- A.) rr w - ~~ From the third of equations (III), page 428, we have when e s is greater than - 3 and p = C, neglecting F, Cbj 2 + wy If we equate (n) and (12), insert the values of W y H and ^ 3 , and solve for 3 , we have for the base 3 where the quantities B 3 and 3 are given by 2 ( yr t +;)-*(*,-*,) tan /?, ~" If we have b l less than * /-_., there is no rectangular section (page 436), and we make W l = o, //, = o. If we equate (10) and (12), insert the value of W^ and solve for /? 3 , we have for B^ _ 2(^. + ^(2b, - b^ + 6(h, - k&b* + V, - ^) - Cb} (*. - *,)[*(*, - ^ 2 )(^3 + 2^ 2 ) + 6(^ H- H/)] If we have b v less than A / , there is no rectangular section (page 436), and we make PF, = o, //, = o. From (13) and (14) we can find 3 and /? 3 . 2d Case. If the limit h^ is given by equation (7), that is, if s 2 is greater than e a = b z and / equal to C, and if the height of the dam is greater than this limit, we have also a third sub-section. In this case we still continue the back vertical, but the breadth must increase so that f 3 = ^ 3 shall be greater than # 3 and / = C. We have then the third sub-section A 2 B 2 B 3 A 3 with vertical back. CHAP. 1V.J DESIGN OF DAM-ECONOMIC Since the vertical pressure remains unchanged f section down with vertical back until the edge dist from the back for dam empty is equal to. the edge e 3 from B z for dam full. The height of this sub-s( SECTION. 445 or dam empty, we can run this sub- ance s^ D E distance d x n action is .A, / = S 2 I* 1 A* (15) 7/ 3 hy It is required to find 7/ 3 and b y Evidently we have only to make /J 3 = o and equations (13) and (14) and we have \ X \ b z = B + VB* + E . . . . where the quantities B 3 and E 3 are given by A, 6, B ^ C ' E * C C , C&* 2(W.+ limit k. = h-A . .. * . + C If we substitute the value of 3 given by (15) in (16) and solve for 7/ 3 , we shall have the limiting value of h z for the third sub-section in this case. We can then find b z from (15) for any value of h^ up to this limiting value. /' sr *r* If we have b l less than A I , there is no rectangular section (page 436), and we make W^=. o and h^ = o. , FOURTH SUB-SECTION. If the height of the dam is greater than the limiting value of h 3 , when we have the third sub-section with vertical back, we must batter both face and back so that e shall be greater than b and p = C for dam full. We have then a fourth sub-section, A^ByB^A^, with front and back batter. We denote the back batter angle by A- We can run this sub-section down to the base of the dam, so that // 4 is the given height of dam ; and since the vertical pressure is unchanged for dam empty, we should have the edge distance s 4 from A for dam empty equal to the edge distance e^ from B t for dam full. It is required to find b and /? 4 . In this case we have from (9) and (10), by making /3 3 = o, and from (13) and (14) 07) 446 ST/tTlCS OF RIGID BODIES. where the quantities B and t are given by 2( W, H- W % + **g - 6(A<- h, y tan ft ~ [CHAP. IV From (17) and (18) we can find b t and /? 4 . If we have b, less than A /- , there is no rectangular section (page 436) and we make V " ^ = o, A, = o. For dam full and empty/ = C, and in both cases e t = s 4 , or, from (12), (19) Examples, (t) The height of the proposed QuakerBridge Dam, New York, is 170 ft., top thickness 20 ft., density of masonry 150 pounds per sq. ft. , depth of water 163 ft., allowable compressive stress 20000 pounds per sq. ft. Find the economic section without ice-pressure or wave-thrust. ANS. We have d = 7, , = 20, C 20000, 8 = 150, y = 62.5, and without ice-pressure or wave-thrust T o. The top base b\ = 20 is less than maximum b\ and greater than minimum b\ as given by equations (i) page 440, viz., max. bi = i/ -r- s = 43 ft. and min. b\ = y = o. The first sub-section is then rectangular. First Sub-section. For the height of this rectangular sub-section we have, from equation (2), page 441, //i s - 2i//i a 813^1 = 343, or //, = 41.02 ft. This value of hi is less than the limit //i given by equation (3), page 441, viz., limit *, = =- 300 = 66* ft. The base is then not overloaded. The unit stress is then, from the second of equations (III), page 428, D 6, E d / B '\ A - X 2Wj bi ' We have for the wejght W^ and cross-section A, Wi Sbihi = 123060 pounds per ft. Ai 820 sq. ft., ^ and hence for the greatest unit stress at front edge B \ P = 12306 pounds per sq. ft., and for the greatest unit stress at back edge At , for dam empty, P = r- = 61530 pounds per sq. ft. V B, Second Sub-section. If in equation (4), page 442, we assume for a first approximation h* = 78, we have #* = 54-54, "1 = 5760.32 and A = - 54-54 + ^8734.92 = 38.91- CHAP. IV.] EXAMPLES. 447 Substituting this value of bi in equations (6) and (7), page 443, we find the least value of limit hi given by (6), and from (6) we have limit hi = 77.5, or less than we assumed it. If we assume again hi = 78.5 in equation (4), we have bi = 39.33, and from (6), hi = 79.2, or greater than we assumed it. The value of hi is then between 78 and 78.5. If we try again for h* = 78.1, we have, from (4), hi = 39.07. and, from (6), hi 78.187, or slightly larger than we assumed it. The value of hi is then between 78 and 78.1, It will then be quite accurate if we take hi = 78.1 ft. and bi = 39.07 ft. f The weight per ft. of the second sub-section and the area of cross-section are then = 164262 pounds per ft., Ai = 1095 sq. ft. Since the least value of limit Jii is given by equation (6), page 442, we have s-i = bi = bi, and hence from the second of equations (III), page 428, we have for the greatest unit stress for both dam full and empty 2 ( Wi + Wi} p - ^ = 14715 pounds per sq. ft. The tangent of the front batter angle is given by ~ = 0.51. Third Sub-section. Since the least value of limit hi is given by equation (6), page 442, we have the first case, given on page 443. The third sub-section then extends to the base of the dam and has a front and back batter. We have the base b a and back batter /3 S from equations (13) and (14), page 444. If in (13) we assume tan ft 3 0.2, we have a = 8.02, E t = 18288.38 and 3 = - 8.02 + 4/18352.7 = 127.45. If we substitute this value of b* in (14), we have tan /J 3 = 0.16, or less than we assumed it. If we try again for tan /3 t = 0.15, we have,^ = 9.61, E = 17643.69 and ft = 9.61 + 4/17736.04= 123.56. With this value of t>, in (14) we have tan fi a = 0.17, or greater than we assumed it. The value of tan yS s is then between 0.2 and 0.15. If we assume tan /3 3 = 0.17, we have B\ = 8.97 E = 17901.57 and b* = 8.97 + 4/17982.03 = 125.12. With this value of bs in (14) we have tan /$ 3 = 0.169, or almost exactly what we assumed it. We have then for the third section //a hi = 91.9, bt = 125.12, tan /3 3 = 0.17. The weight per foot of the third sub-section and the cross-section are then h ^ = i 131 670 pounds per ft., A, = 7544 sq. ft. The tangent of the front batter angle is given by = o 76. The unit pressure back and front is 20000 pounds per sq. ft. The front edge distance is, from (12), ;, = 46.64, and the back edge distance the same. We have then the following table: h b A tan P back. O O 0.17 tan /3 front. e / back. / front. 41.02 78.1 . 170 20 39-07 125.12 820 1095 7544 O 0.51 0.76 ~ 6.6 13.02 46.64 10 13.02 46.64 61530 ?47T3 2OOOO 125060 M7I3 20OOO 9459 44^ STATICS OF RIGID BODIES. [CHAP. IV. In this table the first column contains the distance in feet from the top of the dam to the bottom of each sub-section, the second the base in feet of each sub-section, the third the area in square feet of each sub- section, the fourth and fifth the tangent of the back and front batter angles, the sixth and seventh the back and front edge distances in feet, the last two the greatest unit stress back and front in pounds per square foot. The total area is 9459 sq. ft. We see, from example (3), page 433, that in the case of the San Mateo dam, which has the same height and density and an even greater allowable unit stress C, the area is 15810 sq. ft. There is a saving in this case of 40 per cent, due to economic section. (2) Design tht proposed Quaker Bridge Dam, Neu York, taking ice-pressure into account, dimensions and data as given in the preceding example. , ANS. We have T = 40000 and, as before, d = j. t* = 20, C = 20000, 8 = 150, y = 62.5. Th top base , = 20 ft. is less than y -r- ** t ft-, and therefore (page 436) there is no rectangular sub-section. We have then to make (f i = o and ^1=0 wherever they occur in our equations. First Sub-section. The first sub-section, Ihen, is given by equations (4) and (7), page 443, remembering to wake \V\ = o, h\ = o. If we assume h* = 102 ft. in (4), we hsve .#, = 10, E, = 5239.48 and , - - 10 -f- 4/5339.98 = 63.07 ft. Substituting this in (7). we fcav< /*, = 99.6 ft., or less than we assumed. If we assume h* = 101 ft. in (4), we have 2? = 10, E* = 5315.61 and fa = - 10+ 4/5415-61 = 63.59 ft. Substituting this in ft), we have h* = 101.43 ft., or greater than we assumed. The value of h t is then between 101 and 102 *t. Assuming // t = 101.5 ft. in (4), we have B t = 10, E t = 5353.97 and Substituting this in (7), we have //j = 101.47 i*-> or almost exactly what we assumed. We have then for the first sub-section a trapezoid with vertical back and height and lower base given by ht = IOI.5 ft" * = 63.7 ft. We have then the weight W 9 per foot of length and the area of cross-section At , W, = S ^ * W * = 633166 pounds per ft., A* = 4214 sq. ft. We have also for the dam full e 3 = t>, = 21.33 fc- an{ ^ tne greatest unit pressure p = C= 20000 pounds per sq. ft. For the dam empty we have from (5), page 428, making ^, = o, s* = 22.82 ft., and from the third of equations (III), page 442, the greatest unit pressure p = 18388 pounds per sq. ft. The tangent of the front batter angle is given by = 0.43. h* Second Sub-section. Since the limit of // is given by equation (7), page 443, we have the second case, given on page 445 by equations (15) and (16), page 445, after making //i = o, W\ = o. If we assume h% = 107 ft. in (15), we have = 31.66, E t = 8827.015, and bt = 31.66 + 4/9829.365 = 67.48. Substituting this in (16), we have h\ = 111.04, r greater than we assumed. If we assume A t = 110.5 ft- > n ^5) we nav e B* = 31.66, E* = 9315.26 and b t = 31.66 + 4/10317.61 = 69.91. Substituting this in (16), we have A s = 110.81, or greater than we assumed. If we assume h % = 1 10.8 in (15), we have B t = 31.66, E t = 9358.2 and b* = 31.66+ 4/10360.55 = 70.12. CHAP. IV.] , EXAMPLES. 449 Substituting this in (16), we have h* = 110.81, or almost exactly what we assumed. We have then for the second sub-section a trapezoid with vertical back and height and lower base given by 7i t h = 9.3 ft., b = 70. 12 ft. The weight W* per ft. of length and the area of cross-section A* are then We have for the dam full and empty/ = C = 20000 pounds per square foot, and in both cases e =s s t , or from equation (12), page 444, making W^ = o, , and P % be the equal normal pressures upon the equal por- tions a'a! ', a'a', and H the horizontal pressure at the crown. In Fig. 2, lay off H from O to o horizontally, and let Oo represent the magnitude of H. Then lay off o i and I 2 parallel and equal in magnitude to jP, and />,, and draw the rays Oi, O2. In Fig. I, let H act at a, and prolong its direction till it meets /^ at b. From b draw be parallel to O\ till it meets P 3 at c. From c draw ca parallel to O2. Then (page 413) abca, Fig. I, is the equilibrium polygon. We have by similar triangles P'.H::cb'.bCor cC\ cb cC The same holds true no matter how many equal portions a'a' we take. But as we increase the number of portions, the polygon approaches a curve. For an indefinitely great p number of portions we have for the curve of equilibrium cb = ds, also - = p = the unit pressure and cC = r = the radius of curvature. Hence or H rp X ds. But/ must be constant, and we see from the construction that H is constant. There- fore r is constant and the equilibrium curve is a circle. If, then, we make the dam circular in cross-section, as shown in Fig. i, the curve of equilibrium will coincide with the centre line, and the horizontal pressure H at the crown acts at the centre line and is given by ds. (0 Also, since in Fig. 2 the force polygon 012 becomes a .circle of radius // when the seg- ments of the arch are indefinitely great in number, and since any ray, as o i in Fig. 2, gives the stress in the corresponding segment, cb, Fig. I, of the equilibrium polygon (page 413), it is evident that the pressure at every point of the centre line is tangent to the centre line at that point and equal to H. If, then, C is the allowable stress per square foot, we have for the area A of cross- section of the arch one foot in depth at any distance below the surface X ds CHAP. IV.] ARCH DAM. 45' where/ is the unit pressure at any distance below the surface of the water, and C is given by our table page 424. FIRST SUB-SECTION. Let the water-level be at a distance d below the top (Fig. 3). Let the top base be b^ , and let the dam be rectangular for a distance /i l below the top. At this depth, then, we have \~f~ - - - *~ ds C' But the unit pressure of the water at the depth ti l d is y(h-i d}, where y is the mass of a cubic foot of water, or 62.5 Ibs. If T is the ice-thrust at the surface per unit of length, is the unit pressure due to ice-thrust if h is the depth of water. The total unit pressure is then We have then or the height k l of the first rectangular sub-section in order that the allowable unit stress C may be just attained is given by J, J+. "i d ~\ ~T. r yr yh (2) TOP THICKNESS. The choice of top thickness b l must in general be determined by local and practical considerations. We can scarcely take b^ = o, or have no rectangular portion, as it would not be allowable to bring the top to an edge. Some breadth must be assumed. We have, from (2), . ..../.,,. . . .'. . (3) Equation (3) will give b l for any value of h^ and 'd we choose. Thus if we take \ = -j-^,, we shall have h* = d. We ought not to take b less than this. If we take hi. we shall have h v equal to the entire height h 2 of the dam, and the entire cross-section will be rectangular. We cannot take b l greater than this. Between these two values of rT yr and * = ~ rT (4) we can assume x what we wish, and will have then /^ greater than d and less than There will then be a second sub-section not rectangular. 452 ST/tTKS OF RIGID BODIES. [CHAP. IV. SECOND SUB-SECTION. For any distance y below the top greater than //, and less than // 2 (Fig. 3) we have the thickness and the total unit pressure Hence we have for the thickness at any distance y from the top when y > (5) The arch dam requires far less masonry than the gravity dam, but the pressure on the arch stones increases with the depth and span, and so does the thickness. When the thick- ness becomes great we cannot be sure that each arch stone will take its own share of the pressure. The distribution of the pressure over the cross-section is then uncertain. For such reasons the arch dam is best restricted to short and low dams. It is also evidently unwise to make the stability of a dam depend wholly upon its action as an arch, except under the most favorable conditions as to rigid side hills for abutments, and the most unfavorable conditions as to cost of masonry. Although it is not, then, generally wise to make the stability of a dam depend wholly upon its action as an arch, it is well to make a gravity dam curved up-stream, so that the arch action may give additional security. There are but two examples of the pure arch type: the Zola dam in the city of Aix in France, and the Bear Valley dam in the San Bernardino Mountains, Southern California. The first is of rubble masonry, height 120 ft., radius 158 ft., thickness at top 19 ft., at base 42 ft. The Bear Valley dam is of granite, height 64 ft., radius 300 ft., thickness at top 3.16 ft., at base 20 ft. Examples. (i) The Bear Valley dam in\the San Bernardino Mountains, Southern California, is an arch dam of granite ashlar, radius r = 300 ft., top base b\. = j. /7 ft. , bottom base bi 20 ft., depth of water h = 60 ft., height hi = 64 ft., and therefore d = 4 ft., face vertical and other dimensions as shown in the following figure. Examine its stability, ANS. We have from the given dimensions and from equation (5), neglecting the ice- thrust T, for y = 12 24 36 48 64 ft. [/ = 448 5-79 7-1 8.42 20 " C= (y d) = 16.72 32.22 42.25 49. 28.12 tons per square foot. From page 424, the allowable unit stress fought not to exceed 30 tons per square foot. The dam as built then has, except at top and bottom, a much higher unit stress than ordi- nary practice would consider allowable. (2) Design an arch dam of the same height and radius as the Bear Valley dam, for same depth of water and for an allowable compressive stress of 30 tons per square foot. ANS. We have h = 60, hi = 64, , -f-/> 2 ) on the face ac is just the same, or NA = -(p l -j-/ 2 ), Fig. 3 (b], FIG. . laid off normal to ab. Also, by Case 2, the unit pressure on the face ab which balances -\ (p l p^ on the face be and ~(p l p^) on the face ac is just the same, or -(/, A)' but ^ makes an angle N'A Fon the other side of A F equal to NA V. If then we lay off, in Fig. 3 (b), NA normal to ab and equal to (A~t~A) an< ^ with N as a centre and NA as a radius describe the arc of a circle intersecting the vertical A V at the point S, then SN will give the direction of -(p l / 2 ) acting on the face ab. If then we lay off along this line SN the distance NR = -(p l / 2 ) and draw RA, the line RA will give the magnitude and direction of the resultant unit pressure p on the face ab when the nmoral unit pressures p^ and p^ on the faces be and ac are unequal. It is also evident that the angle RNF is equal to twice the angle VAN, or, if we bisect the angle RNF, we shall have the direction of AV or /,. Suppose, now, the faces ac and be, Fig. 3 (b], to remain invariable in direction and the normal unit pressures p l and/ 2 on these faces to remain unchanged, but let the third face,, vary its inclination. Then the magnitudes of NA = (p l -j-/ 2 ) and of RN (p l p^ in Fig. 3 (b} will be unchanged, but their directions will change according as the face ab changes its inclination. It is evident that the greatest possible value of the angle RAN which the resultant pressure/ = RA on the face ab makes with the normal NA to that face will be when RN is perpendicular to RA, or when the angle ARN = 90. Now for earth-pressure the greatest possible value of the angle RAN is the angle of friction or of repose for earth on earth. The angle RNF is then equal to 90 + 0, and, as we have just seen, if we bisect this we have the direction of /,. Hence the angle VAN of /> t with the normal NA is Y. CASE 4. In Fig. 4, let ab be any earth-surface, and let the resultant unit pressure RA =/ on this surface be given in direction and magnitude. It is required to find ~(A~I~A)' ~(P\ ~~ A) and tne direction of p r Draw AF normal to the surface ab, and AR' making the angle of friction with this normal. Then find by trial a point N in this normal AF such that if we take TV as a centre and NR as a radius, the arc ARR' will be just tangent to AR'. When this point N is thus found by FIG. 4. trial, then, by Case 3, the distance AN will be -(/>, -f~ and R'N = RN will be -(p l /,). Also, if we bisect the angle RNF by the line NS', we have the direction of />, . APPLICATION TO THE RETAINING WALL. Let AD, Fig. 5, be the back of the wall, and D^FI the earth-surface CHAP. V.] MAGNITUDE AND DIRECTION OF EARTH-PRESSUREGRAPHIC DETERMINATION. 457 Pass a plane AA l through the foot of the wall FIG. 5. making the angle OL with the horizontal. A parallel to the earth-surface, and draw AJ vertical and A^F normal to the earth-surface. The pressure upon every square foot of this plane AA^ is vertical and equal to the weight of a column of earth of vertical height AJ and cross-section I sq. ft. X cos a. If y\ is the mass of a cubic foot of earth, then we have Yl X A~JX I sq. ft. X cos a for the mass of this column. But /ij/cos a = A^F, hence the mass of this column is y t X A^F X I sq. ft. If then we draw A^F perpendicular to the earth-surface and revolve A^F about A l as centre to the vertical A^ , we have for the vertical unit pressure/ p = y v X A l R l pounds per sq. ft., where Y\ is the mass in Ibs. of a cubic foot of earth, a.ndA l R l is measured in feet. We have then, as in Case 4, / given in magnitude and direction, and we can find, as in Case 4, ~(A~t~A) "(A A)> and the direction of/ r Thus, as in Fig. 4, draw A^R' (Fig. 5) making with the normal A^the angle R'A^F , equal to the angle of friction or repose for earth on earth. Find by trial a point JV l on the normal A ^ such that the arc of a circle with N l as a centre passes through R l and is tangent to A^R' ' . Then, as in Case 4, ;( i -A) =n -^Pi- Bisect the angle R^N^F by the line N V S' . Then the line N^' gives the direction of /j , as in Fig. 4. Now lay off at the foot of the wall A (which may be considered as identical with A^) the distance NA N l A l in a direction perpendicular to the back of the wall at A. Draw the line AS parallel to TV^S', the direction of/j already found. Then, as in Fig. 3 (b}, with N as a centre and NA as a radius describe an arc of a circle intersecting AS at S, and lay off along NS the distance NR = N' l R r Then, as in Case 3, RA represents the magnitude and direction of the pressure per square foot at the foot of the wall. Thus, if ^ is the density of earth and we measure RA in feet, the pressure per square foot at the foot A of the wall is given in magnitude by and its direction is the direction of RA. STATICS OF RIGID BODIES. [CHAP. V. Since the pressure is zero at the top D v and greatest at the foot A, and varies for any point directly as the distance of that point below /?, , the average pressure per square foot is 2 ri X The total pressure P is then for a wall of one foot in length P -y l X RA X D^A, where y\ * s tne mass of a cubic foot of earth and RA, D^A are taken in feet, and Pis the pressure per foot of length. This pressure P acts (page 454) at a point A!" at a distance above the base of the wall equal to -//, where h is the distance D^O of the earth-surface above the base of the wall, and is parallel to the direction of RA already found. We can thus find by a simple graphic construction, in any given case, the magnitude, direction and point of application of the earth-pressure P on the back of the wall. The vertical and horizontal components Fand //"of Pare then easily found, and then the general principles and equations already given in the preceding chapter, apply at once, and we can investigate the stability of, or design, a retaining wall. Magnitude and Direction of Earth-pressure Analytic Determination. From the graphic construction just given we can easily derive the corresponding general formulas for the magnitude and direction of the earth-pressure P. NOTATION. Let h = D X O be the height of the earth-surface at D l above the base AB of the wall ; the angle of the earth-surface with the horizontal is ; the batter angle of the back of the wall with the vertical is /?, ; the earth-pressure /"makes the angle with the normal to the back of the wall ; the angle R'A 1 N 1 = is the angle of repose for earth on earth ; the angle R 1 N 1 F= 7;, and the angles R^N^ and S t N t F are each |; the angle RA S= e; the angle RSA = GO; y l is the mass A B of a cubic foot of earth. SOLUTION. Now by the graphic construction we have -(A + A) sin -(A -A)- We have also CHAP. V.] MAGNITUDE AND DIRECTION OF EARTH-PRESSURE ANALYTIC DETERMINATION. 459 and since A^R^ = A^F, we have by construction I V /I ^A - A) sin r? ^-p cos (ft a) sin a. ....... (2) We have also by construction [^ A + A) + j(A - A) cos V ] 2 + [J( A - A) sin T?] = \j- ft cos (ft - )] , (3) and also ^( A + A) + ^( A - A) cos 7/ = -^j-jjj cos (ft - a) cos or (4) From (i), (2) and (3), eliminating -(p l -\- / 2 ) and -(A A) we ODta in siir <* / . / sin* \ cos 77 = h \ / (i sin* ar)^l 2 j (1) We have also directly from the figure GO = angle NSA angle NAS, or fi , = 9 o-ft_^ + a (II) From (2) and (i) we have yJi cos (# ^- ft") sin (i -- 1 - sin 0) . , ^, = ^ : - : \S) ri cos ft sin sin ?/ _ ^// cos (a ft) sin Q'd sin 0) ,^ A ~ cos ft sin sin /; We have also from the figure RS sin co tan e = -- RS COS r3 But YI RS = A and y^ AS= (A ~r A^ cos Therefore Substituting the values of A and A from (5) and ( 6 )> i sin tan e = . -; I -\- sin We have also directly from the figure tan e = L^_ s , an = tan' 45 - tan (III) i -}- sin (IV) 460 STATICS OF RIGID BODIES. [CHAP. V. Also, __ __ r , . RA = V/ 2 8 sin 8 GJ + ( ri . ZS - / 2 cos a?) 8 = V/ 2 8 sin 8 a? -f/>, 8 cos 2 GO, or, substituting the values of p l and/ 2 from (5) and (6), we have for the earth-pressure P or Yl # cos (ft - ) sin a ^ ^ _ . ..... 2 cos 8 /?! sin sin 77 From (i) and (4) we obtain _ y^k cos (ft l a) cos a (i -f- sm 0) * "" cos ytfj (i + sin cos ^) Comparing this with (5), we have sin a cos a sin sin 77 i -\- sin cos ;/' (7) Making this substitution in (V), we obtain an equivalent expression for P which can be used when a = o, viz., v,// 2 cos (8, a) cos a ./. P -*-*- -. 4 r V(i -j- sm 0r 4 sin sin 2 GO. . . . (VI) 2 cos 2 /^(i +sm cos rj) We see from the figure page 458 that the earth-pressure P makes the angle 6 -\- ft l with the horizontal. We have then for the vertical and horizontal components of P V = P sin (0 + A)' ' (VII) The values of Fand // being thus known, the principles and general equations already given in the preceding chapter, apply at once. The case of water-pressure is a special case of the equations just deduced. Thus for water we have y in place of y { , since the surface is level we have a = O, and since there is no friction 0, = o. We have then, from (\), rj o; from (II), ca = 90 /?, ; from (III), 6=0?; from (IV), 6 o. The pressure of water is therefore normal to the surface. From (VI) we have 2 COS and from (VII) These are precisely the values of Fand H given on page 431 for water-pressure. We see, then, that water-pressure is but a special case. Surface of Rupture. If there were no wall and we were to disregard cohesion, a prism of earth AD^G would tend to slide off along a plane AG which would make with the hori- zontal the angle of repose for earth on earth. CHAP. V.] MAGNITUDE AND DIRECTION OF EARTH-PRESSURE ANALYTIC DETERMINATION. 461 But on account of the wall this plane AG makes with the horizontal an angle ip greater than 0. This angle fy we call the angle of riipture, the plane A G is the plane of rupture, and the prism AD^G which thus tends to separate along AG and force the wall is the prism of rupture. If, in the figure page 458, p v remains unchanged in direction and magnitude, while AA^ is resolved about A until the pressure upon AA l makes with the normal A l N l the angle 0, then this new position of AA l is the plane of rupture. But for this new position p l makes the angle 45 + "~ W1 *-h t ^ ie norm al. The normal A^N lt and hence the plane AA lt have then been revolved through the angle 45 -f - - . The angle which the plane of rupture A G makes with the horizontal is then In the case of water, a o, = o and, from (I), 77 = 0. We have then for water # = 45- General Metkod.Wz have in any case the following general method ; 1st. Find rf from (I). 2d. Find GD from (II). 3d. Find e from (III). 4th. Find from (IV). The angle ft gives the inclination of the earth-pressure with the normal to the back of the wall. 5th. Find P from (V) or (VI). 6th. Find V and H from (VII). If desired we can find the angle of rupture from (VIII). Knowing now J^and H, we can use the general equations pages 426 to 430, making in them T= o. Special Cases. The formulas (I) to (VIII) just given are general and admit cation for special cases. CASE i. EARTH-SURFACE HORIZONTAL If the earth-surface is horizontal, we have a = o, hence, from (I), r, = O and, from (II), = 9 o - & We have then, from ( [I), tan e = tan 2 (45 - f ) cotan ft ,....'...-.- (8) and then, from (IV), 0= 9 o -A - e; .......... from (VI), D_ri /z2 / r 4 sin . : ~7~V cos2 / ? i (i + sin 0) 2 ' and from (VII), = Psin (90 - e), H = P cos (90 - e) ...... ( l 462 STATICS OF RIGID BOOTES. [CHAP. V. From (VIII) the surfact of rupture makes with the horizontal the angle V> = 45 + f. . . . : . .' . ~. . (12) CASE 2. EARTH-SURFACE HORIZONTAL BACK VERTICAL. If in the preceding case we make fa = o, we have e = 90, = o and therefore the earth-pressure is normal to the back of the wall, or horizontal. From (10), then, and from (u) V= o. The surface of rupture makes as before the angle if) with the horizontal given by CASE 3. EARTH-SURFACE HORIZONTAL BACK BATTER ANGLE EQUAL TO 90^. In this case a = o, fa = 90 ^. We have, from (I), r/ = o and, from (VIII), $ = 45+ J- Hence #=45- ^, and, from (II), GO = *f> = 45-f ^; from (III), e = 45 ~ ^\ from (IV), 0=0, or the pressure makes the angle of 'friction with the normal to the back. From (VI) we have p= 2 cos ^45 |j But by Trigonometry \/(l + sin 0) 2 - 4 sin sin 2 (45 - |). Hence sin = I 2 cos 2 (45 -f- -J, I -f sin = 2 2 cos 2 (45' cos 2 = 4 cos 2 (45 + |) ~ 4 cos* (45 + j). Inserting these values and reducing, we have , cos cos ^45 -j From (VII), (15) CHAP. V.] MAGNITUDE AND DIRECTION OF EARTH-PRBSSURE ANALYTIC DETERMINATION. 463 CASE 4. EARTH-SURFACE INCLINED AT THE ANGLE OF REPOSE. In this case a = 0> or the surface of rupture is parallel to the earth-surface. CASE 5. EARTH-SURFACE INCLINED AT THE ANGLE OF REPOSE BACK VERTICAL. In this case we have only to make /3 1 = o'm the preceding case, and we then have r/ = 90 + 0* GO = 45 -f- -, and, from (16), tan e = or, since, by Trigonometry, 1 + S !" = tan 2 (45 4- ^j, ' ~ sin ^ = tan 2 (45' i sin V T 2 J> j _^ sin ^ ^3 i sin /I -|- sin /I sin / 0\ tan ^ = x + sin y rz 1HT0 = V i + sin = n ^ 45 ~^' Hence e = 45 ^, and, from (17), or the pressure makes the angle of friction with the normal to the back. From (i 8), P = p*(l + sin 0) 2 - 4 sin sin 2 4 5 or, reducing as in Case 3, cos __ r (20) STATICS OF RIGID BODIES RETAINING WALLS. [CHAP. V From (19), The surface of rupture makes, as before, the angle with the horizontal. Values of 0, ^ and y r We give in the following table the value of the angle of friction 0, of the coefficient of friction /* = tan 0, and of the density y\ f r earth, sand and gravel. Material. Angle of Repose *. Coeffcient of Friction P* Density in Ibs. per cu. ft. Yi 30 o 58 40 o 84 Sand dry si o. 70 0.84 o <;8 Earth dry 0.84 4C Examples. (i) At Northfield, Vt., on the line of the Central Vermont R.K., is a retaining wall 13 ft. high, top base 2 ft., bottom base b ft. The wall is composed of large blocks of limestone without cement. The density of the masonry is about 170 Ibs. Per cubic foot. The face of the wall has a batter of / inch horizontal for every foot of height. The wall is over 30 years old and in as good condition as when laid. Investigate the stability, taking the angle of repose 38 , the density of the earth 90 Ibs. per cubic foot, and the coefficient of friction for the masonry // = 0.66 (page 424). , or ANS. We have h = hi = 15, di = 2, b* = 6, 5 = 170, tan fii = 0.66. Take a section of the wall i foot in length. Then the weight of this section is = 10" 23', = 38, 90. IV = 10200 pounds per ft. From Case i, page 461, we have p _ 90 x 225 , = 2983 pounds per ft>> tan e = tan* 26 cot 10 23', or e = 52 26', V = P sin (90 ) = 1814 pounds per ft., H = P cos (90 e) = 2364 pounds per ft. Now that we know Kand H, we can proceed as on pages 426 to 430, making T = o in all equations in which it occurs. Thus for stability for sliding we have from (I), page 427, for the factor of safety for sliding n _ o.66(to2oo + 1814) _ 2364 There is therefore ample security against sliding even for through joints. If there are no through joints, there is in any case no possibility of sliding. For stability for rotation we have, from equation (5), page 427, 72 + 24 -4 - 15(4 + 6) 2.75 - 5 - = 2.646 ft., CHAP. V.] RETAINING WALLS EXAMPLES. 465 and from (II), page 428, since d\ = h = 5, 10200 x 2.646 + 1814^6 - 5 X ^' 75 V 2364 x 5 10200 + 1814 " = 2 ' 03 - Since e is positive, the resultant pressure on the base falls within the base and there is no rotation. For stability for pressure, since e is greater than b* = 2, we have, from the third of equations (III), page 428, for the greatest unit pressure 2(10200 + i8i4)/ 3 x 2.03\ p = g \ 2 ~ 6 J = 3944 P unds P er ^ ft From our table page 424 we see that the allowable unit stress is C = from 50000 to 60000 pounds per sq. ft. The wall is then abundantly secure against sliding, rotation and crushing. We also see that the bottom base might have been considerably less, thus saving much material, without being insecure. Check by Graphic Construction, page 457. (2) Design the preceding wall properly for the same top base, height and back batter, for an allowable stress of C = 40000 pounds per sq. ft. ANS. We have h = hi =15, b\ = 2, 8 = 170, tan /?i - L -^-, u = 0.66, and of course the same values for V and H as before, viz., V ' 1814, H = 2364. From equation (II), page 429, making T = o, we have for the value of bi for low wall , = 1.065 + i /I - I 34 + 5-95 * 6.16 ft. Substituting this value of <5 a in equation (I), page 429, we have for the limit of hi for low wafl . 6.16 x 40000- 3628 . ltmtt hl = -- -- = I75 t ' Since this is greater than the actual height, the wall is low and the value of , just found is the value ired. We have for this base e = We have then the weight per ft. required. We have for this base e == t>, and p less than C. I7Q x 816 x 15 From (I), page 427, we have the coefficient of safety for sliding / . _ 0.66(10494 + 1814) _ * I 2364 There is no danger of sliding even for through joints. Check by Graphic Construction, page 457. (3) Design the -wall of the preceding example for the same top base and height and vertical back, for an allowable stress C = 40000 pounds per square foot, and show that there is a saving of over 17 per cent due to vertical back. ANS. We have h = hi = 15, & = 2, S = 170, /?i = o, = 38, y, = 90, // = 0.66. From Case 2, page 462, we have 466 STATICS OF RIGID BODIES RETAINING WALLS. [CHAP. V. From equation (II), page 429, making T = o, we have fa = - i + */33/J3 = 4.77 ft. Substituting this value of b* in equation (I), page 429, we find the limit of h\ for low wall greater than the given height of dam. The wall is then low and the value of fa just found is the value required. For this base we have e = -fa and p less than C. We have then the weight per ft. <5(, + fa)t 170 x 6.77 x 15 --- 8632 pounds per ft., as against 10404 pounds per ft. in the preceding example, or a saving of 17.03 per cent. From (I), page 427, we have the coefficient of safety for sliding There is then no danger of sliding even for through joints. Check by Graphic Construction, page 457. (4) Find the bottom base of a trapezoidal retaining wall of granite ashlar with vertical back, 20 feet high, earth surface horizontal and level with the top, < = 33" 40' ', y 100 pounds per cubic foot. ANS. In this case 0, = o, V = o, // = //,. From Case (2), page 462, we have the pressure P normal to the back and given by P = H = - tan" 28 1 5' = 5774 pounds per ft. From equation (II), page 429, we have If we take the top base <5, = 2 ft. and S = 165 Ibs. per cubic ft. (page 424), we have fa = 7.66 ft. Check by Graphic Construction, page 457. (5) Same as preceding example, with back batter ft\ = 8. ANS. P = 6420 pounds per ft., = 18 9', H = 5758 pounds per ft., V = 2825 pounds per ft., fa = 7.9 ft. (6) Design a retaining wall 20 ft. high, back batter fti = 8, S = 770 Ibs. per cubic foot, earth-surface inclined to horizontal at angle of repose = 33 40' , y^ = 100 Ibs. per cubic ft., earth-surface at top of wall. ANS. From Case 4, page 463, we have e = 21 22', 6 = 32 28', P 21740 pounds per ft., H = 16522 pounds per ft., V = 13230 Ibs. per ft. If we take the top base fa = 2 ft., we have, from equation (II), page 429, fa = 9.6 ft. Check by Graphic Construction, page 457. (7) A wall 15 ft. high retains an earth filling which supports a double-track railway. The top base is 3.5 ft. Find the bottom base for y, = 100, = jj 40', fti = 8, S = 170. ANS. If we take the train-load at 6000 pounds per linear ft. and top of fill 15 ft., the pressure per square ft. on top is 400 pounds, which is equivalent to a column of earth 4 ft. high. We have then ^, = 15. h = 19, and from Case i, page 461, P = 5795 pounds per ft., 6 = 18, H = 5200 pounds per ft., V = 2540 pounds per ft. From equation (II), page 429, fa = 7 ft. Check by Graphic Construction, page 457. COHESION OF EARTH. Cohesion is that resistance to motion which occurs when two surfaces of the same kind are in contact. It is found by experiment that cohesion is directly proportional to the area of contact, varies with the nature of the surfaces, in contact, and is independent of the pressure. CHAP. V.] EQUILIBRIUM OF A MASS OF EARTH. For any given surfaces, then, the cohesion is 467 where A is the area of contact and c is the coefficient of cohesion. EQUILIBRIUM OF A MASS OF EARTH. Let ADG be a mass of earth, the batter angle of the face being fi. If there were no cohesion, a prism of earth ADG would tend to slide off along a plane AG which would make with the horizontal the angle of repose 0. But if there is cohesion, this plane, which is called the plane of rupture, will make an angle ^ with the horizontal greater than the angle of repose 0. We call rp the angle of rupture. Let the angle of the earth-surface DG with the horizontal be , and the weight of the prism ADG per unit of length be W. The weight W acting at the centre of mass C can be resolved into a force N normal to the surface of rupture AG and a force P parallel to the surface. We have then P= W sin ^, N= W cos $ (i) The force P tends to cause sliding. This force is resisted by the friction and the cohesion. The friction is pN, where p = tan is the coefficient of static sliding friction, and the cohesion is c . AG, where c is the coefficient of cohesion. We have then for equilibrium P tN c . AG = or P c. AG, Now, for any plane which makes an angle with the horizontal greater or less than ^ there will be no sliding, and for that plane P pN will be less than c .AG. For the plane of rupture, then, we must have AG = a maximum Let the vertical height of the prism be h. Then AD = ^r^> the angle DAG is _ ( _j_ ^ the area of the prism ADG is then AG 2cos/? > and the weight W per foot of length is, if y l is the density of the earth, '. co. (/+) 2 COS ft (4) 468 STATICS OF RIGID BODIES EARTH EQUILIBRIUM. [CHAP. V. Inserting this value of W in (i) and the corresponding values of P and N in (3), we have, since p = tan 0, yh cos (8 +0) . sin (tb 0) = c = a maximum ...... (5) 2 COS ft COS ANGLE O'F RUPTURE. Equation (5) is a maximum when cos(/? + ^) = sin fy - 0) = cos [90 - ty - 0)], or when * = 45 + ( --^ ..... (6) Equation (6) gives, then, the angle of rupture, or the angle which the plane of rupture AG makes with the horizontal. COEFFICIENT OF COHESION If we insert this value of tf> in (5), we have yh cos [45 + :k0 + /?)] sin [45 - ^(0 + /?)] = 2c cos ft cos 0, ^[i sin (0 + /?)] 4^ cos/? cos ........ (7) Let ft = o, and let // in this case be // . Then we have, from (7), c = 4 cos From (8) we can determine c by experiment. Thus if a trench with vertical sides, of R D considerable length as compared to its width, is dug in the earth, with a transverse trench at each end, so that lateral cohesion may not prevent rupture, it will be observed after a few days to have caved in along some plane, as AG. Let the observed depth AD be 7/ . Then from (8) we can compute the coefficient of cohesion c. HEIGHT OF SLOPE. If we substitute the value of c from (8) in (7), we obtain // (i - sin 0) cos/? i-sin(0 + /?) ' From (6) and (9) we see that the angle of rupture tf> and the height of slope h are independent of the inclination of the earth-surface. Equation (9) gives the limiting height // of slope when sliding is about to take place. Let be a factor of safety, so that if n is 2 or 3, the height can be taken two or three times the safe height before sliding begins. Then we have for the safe height * = %'"!?! ,*>!?/-.. -co) Equation (10) gives the safe height of slope for any given batter angle ft. CHAP. V.] ANGLE AND CURVE OF SLOPE. 469 ANGLE OF SLOPE. If h is given and the corresponding batter angle /3 is required, we have, from (10), i - sin (0 + ft) __ h Q (i -sin 0) = cos ft nh where the second member, being a known quantity, is denoted by a. If we develop the numerator in the first member and substitute for sin ft and cos ft their values in terms of tan ft, viz., I I 2 tan -ft i tan 2 -ft sin ft = - , cos ft = - , I + tan 2 -ft i + tan 2 -ft 2 2 we obtain a quadratic whose solution gives Equation (u) gives the batter angle for a factor of safety when the height is given. CURVE OF SLOPE. Let a be any point of the slope Da A, whose vertical distance below D is da = y, and let aG be the plane of rupture at the point a, making the angle ip with the horizontal. Then the prism DGa of weight W per foot of length tends to slide along aG and is prevented by friction and cohesion. Let ^V and Pbe the components of W normal and parallel to aG. Then if n is the factor of safety and jw is the coefficient of static sliding friction, we have (12) Let A be the area daD between the curve of the slope and any ordinate da y. Then Y^A is the weight of the prism daD per foot of length, if Y l is the mass of a unit of volume of earth. The area daG is ^ per foot of length. Hence the weight IV of the prism DaG is If we insert this value of W in the equations (i) for P and N, and then substitute in (12), - ^Vsin ^ p cos ^) *-j--7 = O, y v ' ' sin w V we obtain, since aG = ^ - 47 STATICS OF RIGID BODIES EARTH EOJJ1LIBR1UM. [CHAP. V. or, dividing by sin ^, cot = If <* makes an angle with the horizontal greater or less than 0, we have, from (12), n(P )*N) less than c . aG, or the left side of equation (13) less than zero. The value of ^> must then make (13) a maximum. If then we differentiate (13) with reference to cot ^ and put the first derivative equal to zero, we obtain o. . . . (14) Eliminating cot $ from (13) and (14), we obtain A = \- n ^^y + V - * * / 2c(nri}> + 2^(1 + ^)]. ... (15) Equation (15) gives the area A between the curve of the slope and any ordinate da =y. Examples. (i) A bank of earth without cohesion stands 3 ft. high with a slope of jo ft. Find the coef- ficient of friction and the angle of repose. "?O ANS. The horizontal projection of the slope is 40 ft. Hence u = tan = =0.75, and

f 1.25 to f horizontal. Find the slope after rupture. Also, if the mass of a cubic foot is 100 Ibs., find the coefficient of cohesion. ANS. We have ft = o and, from equation (6), page 468, ^> = 45 + . The tangent of the angle of repose is n = tan = 0.75. Hence is about 35 and il> is about 62. From equation (8), page 468, since // = 3 ft., y\ = 100 Ibs. per cubic ft., = 35, 100x3(1 sin 35) 128 c = - = 39 pounds per sq. ft. 4 cos 35 3.28 (3) A bank of earth the same as in the preceding example has a height of 30 ft. and a batter of 45' '. Find the limiting height for the same slope, also the factor of safety. ANS. From equation (9), page 468, since A = 3, ft = 45, = 35, the limiting height is h = 3('~ sin 35') cos 45 = I sin 80 or the factor of safety is 2. (4) A bank of earth the same as in example (2) is required to have a height of 30 ft. and a factor of safety of a. Find the batter of the face. ANS. From equation (i i), page 469, ft = 45. (5) A bank of earth -with vertical face caves for a distance of 5 ft. below the surface. The same earth loose and without cohesion takes a slofie of 1.25 to i horizontal. The mass of a cubic foot is too Ibs. Find the angle of rttpture and the coefficient of cohtsion. If the batter of the face is made 45 and the height jo ft , find the factor of safety. ANS. The angle of repose is about 35. The angle of rupture ^ is about 62. The coefficient of cohesion is c = 65 pounds per sq. ft. From equation (10), page 468, 5(1 - sin 35) cos 45 _ 30(1 - sin 80) CHAP. V.] EARTH EQUILIBRIUM- EXAMPLES. 47 1 (6) Find the batter angle of the slope in the preceding example for a height of 30 ft. and a factor of safety ANS. From equation (u), page 469, ft = 45. (7) Find the curve of the slope in example (5) for a factor of safety of 3 and a height of 40 feet. ANS. We have jj, = 0.75, c = 65, n = 3, yi = 100, and equation (i 5), page 470, becomes If we take we have ^=10 30 413 40 ft., 777 sq. ft. Considering the area between the slope and any ordinate as made up of trapezoids as shown in the figure, we have i . loxDa' = 33, or Da' 6.6ft.; DC' 6' c' 10 + 20 33 + 167+ = 167, or a'tf = 9 ft.; = 413, or b'c 1 = 9.8ft.; = 777, or c"d' = 10.4 ft. 40 We see from equation (15), page 470, that for small values of y, A is negative. The equation should not be used for y less than //o, and the upper part of the slope should be rounded off as shown in the figure. (8) // is desired to cut a bank 30 ft. high into three terraces as shown in the figure, with a factor of safety O f i j- The height of each terrace is to be 10 ft., and there are to be two steps, ab and cd, each 4ft. wide. The mass per cubic foot is y\ = I0 Ibs., and and h^ as found by experiment are

ooo pounds per linear ft. of trad;. ANS. The weight of locomotive is 6000 pounds on 12 square feet, or 500 pounds per sq. ft. This is equivalent to a mass of earth 5 feet high. 472 STATICS OF RIGID BODIES -EARTH EQUILIBRIUM. [CHAP. V. We take, then, h 35 feet in equation (u), page 469, and have tan -ft = ]-Tg 4 [o- 82 9 + ^0.0286] = 0.416, or ft 45". The embankment with this batter contains 47 cubic yards per lineal foot, while with the natural slope of 34 it would contain 62. There will then be a saving in cost of construction if the expense of protecting the slope to preserve the cohesion is not greater than the saving in embankment. (10) A railway cut is made in material for which yi = too, $=.34, At =.5/1. The depth of cut is h = 40 ft., and the road-bed is '6/t. Find the batter angle for a factor of safety of j. ANS. We find ft = 47. The cut with this batter is 87 cubic yards per linear foot. For the natural slope it would be in. There is then a saving in cost if the expense of protecting the slope to preserve the cohesion is not greater than the saving in excavation. STATICS OF ELASTIC SOLIDS. CHAPTER I. ELASTICITY AND STRENGTH FOR TENSION, COMPRESSION AND SHEAR. Elasticity. A body is said to be "elastic" when force is necessary to change either its volume or shape and when the continued application of that force is necessary to maintain such change, so that the body recovers more or less completely its initial volume and shape when the force is removed. If the body recovers completely its initial volume, it has perfect elasticity of volume. If it recovers completely its original shape, it has perfect elasticity of shape. If the recovery is imperfect, the elasticity is imperfect. All bodies have some elasticity of volume. If a body possesses no elasticity of shape, it is called a fluid, as air, water, etc. If a body possesses elasticity of shape as well as volume, it is called an elastic solid. We deal here only with elastic solids. Prismatic Body. We shall treat also only of prismatic solids. A prismatic body is one whose volume can be generated by the motion of a plane area, moving so that its centre of mass describes any curve, the generating plane being always at right angles to that curve. This curve is the AXIS of the body, and the generating plane at right angles to the axis is the CROSS-SECTION. The area and form of this cross-section may be constant or variable. If either the area or form of cross-section varies, the cross-section is said to be VARIABLE. If neither the area nor form varies, the cross-section is said to be uniform. Stress and Force. As we have seen, page 397, we distinguish three kinds of force and stress. Tensile force, tending to pull the particles of a body apart in parallel straight lines in the direction of the force, and tensile stress, resisting such separation. Compressive force, tending to push the particles of a body together in parallel straight lines in the direction of the body, and compressive stress, resisting such approach. Shearing force, tending to make adjacent particles move past one another at right angles to the line joining the particles, and shearing stress, resisting such motion. Other stresses with which we have to do are combinations of these. We measure force and stress, then, in pounds, and unit force and unit stress in pounds per square inch. STRAIN The change of distance between two particles of a body, in a direction contrary to coexisting stress between those particles, is called STRAIN. 473 474 ST/IT1CS OF ELASTIC SOLIDS. [CHAP. I. Thus if a tensile force F is applied to a straight homogeneous bar of uniform cross- section A in the axis of the bar, and this force F is uni- FIG. i. formly distributed over the end area A, so that the unit force is , we have a corresponding equal and opposite unit-stress S at every point of the cross-section, and the bar is extended a small amount A in a direction opposite to coexisting stress. The strain A between end particles is one of extension, or tensile strain. If a compressive force F in the axis is uniformly distributed Fie. 2. over the end area A, so that the unit-force is , we have a A _A_c corresponding equal and opposite unit-stress S, and the bar is compressed a small amount A in a direction opposite to co- existing stress. The strain A between end particles is one of compression, or compressive strain. If a shearing unit-force is applied, we have a Xjil sponding equal and opposite unit-stress S, and the cross- section slides on the adjacent one a small amount A in a direction opposite to coexisting stress. The strain A is one of shear, or shearing strain. Strain, then, being a distance, is measured in units of length, as feet or inches, and the term " strain " is general and signifies always a change of distance between points of a body, always in opposition to coexisting stress between those points, without specifying whether that change of distance is due to extension, shortening, or sliding, or whether the form of the body is changed or not.* It should be noted that when there is no coexisting opposite stress, 'there is no strain. Thus when the bar in Fig. 2 is compressed from n to c, let the force F be removed. The bar would expand from c to , and during such expansion we have unbalanced stress in the direction of expansion. Such expansion is not, then, strain. It is not opposite to coexisting stress, and is simply displacement. The bar is not strained by such expansion, but, on the contrary, the original strain A diminishes to zero at the neutral point ;/. But after the end of the bar reaches the neutral point , if it still continues to expand, as in Fig. I, the stress will be opposite to the direction of expansion, and such expansion is strain. The bar is strained by such expansion. Law of Elasticity. When a force F is applied to a homogeneous elastic bar of uniform cross-section A, so that the unit-force is and acts to elongate, compress, or shear the bar, A as in Figs. I, 2, 3, a corresponding equal and opposite unit-stress S must always exist when there is equilibrium, and a corresponding strain A is always observed in the direction of the force F, and therefore opposite to the coexisting unit-stress. * It will be policed, therefore, that strain does not necessarily imply distortion. In the two first cases above there is strain without distortion or change of shape. In the third there is strain with distortion. If a homogeneous sphere is equally compressed radially in all directions, it remains a sphere. It has strain, but no distortion. The word " distortion" is not, therefore, equivalent to shear, and in many cases would be mis- applied if used as equivalent. The term "strain" in common language is used indifferently either for change of distance between points of a body, or for the force which causes this change. Our definition simply restricts its use to the first meaning. CHAP. I.] SET AND SHOCK. 475 So long as this unit-stress S = does not exceed a certain magnitude, all experiments ^1 prove that the unit-stress 6" and the corresponding strain A are approximately pioportional, so that I or = a constant. This is known as the law of elasticity. The point where this law begins to fail is called the ELASTIC LIMIT, and the correspond- ing unit-stress at the elastic limit is therefore the elastic limit unit-stress. We denote it by S e . The law of elasticity is then expressed by saying that for homogeneous bodies and within the elastic limit the strain and unit-stress are proportional. Beyond the elastic limit the strain increases more rapidly than the unit-stress, until finally rupture occurs. Set and Shock. As we have seen, page 473, if when the force F is removed the strain A entirely disappears, the bar is perfectly elastic. If the strain A does not entirely disappear, the bar is imperfectly elastic, and the residual strain is called the SET. For very small stresses the set, if any, is too small to be detected by measurement, and the bar may be regarded as practically perfectly elastic. As, however, no solid body is perfectly elastic, the point at which set is first observed depends upon accuracy of measure- ment. Theoretically there is a set for any stress. The point at which set is first observed is often taken as marking the limit of elasticity. But in view of the preceding it should not be so taken, and experiments generally show a slight set before the limit of elasticity is reached. A .suddenly applied stress or shock is found by experiment to be more injurious than a steady stress or a stress gradually applied. Determination of Elastic Limit and Ultimate Strength. Let a straight homogeneous bar of given material have a length / and uniform cross-section A. Let a tensile" force F be applied in the axis and be uniformly distributed over the end cross- section so that the unit-stress is 5 = -r opposite in direction to F. Let the observed strain of elongation be A. Now, according to the law of elasticity, provided the elastic limit is not exceeded, if we double F we shall observe a strain 2A. If we apply $F, we shall observe a strain 3A, and so on. If then we lay off the successive unit stresses _F _2F ^ l ~ A 1 2 ~ A' " 3 ~ A' A = ~T, etc., to scale along a horizontal line from O, and lay off to scale as ordinates the corresponding observed strains A 1 , A 2 = 2 A t , A 3 = 3AP etc., it is evident that within the S, S, S, S^ elastic limit we have, by the law of elasticity, a straight line OL. The point Z, where the straight line begins to curve, marks the elastic limit, and the corresponding unit-stress S, is the elastic limit unit-stress. As the straight line passes into a curve gradually, it is evident that the point L is net 476 STATICS OF ELASTIC SOLIDS. [CHAP. I very definite, and its location, even with very accurate experimental results, will vary moie or less within certain limits. When the elastic limit is passed the elongation increases more rapidly than the unit- stress, and the cross-section decreases, until the bar ruptures. The unit-stress at which rupture occurs is called the ULTIMATE STRENGTH for tension. We denote it by U t . We determine the elastic limit unit-stress S t for compression, and the ultimate strength for compression U e , in precisely the same way by experiment. The force F is axial and uniformly distributed over the end cross-section. The length of the test specimen should be small, not over five times its least diameter, in order to avoid bending. After the elastic limit is reached the strain increases more rapidly than the unit-stress, and the cross-section enlarges. For shear the strain A is as shown in the figure. The elastic limit unit-stress S t and ultimate strength U t are then theoretically determined in the same way as for tension and compression, by measuring A for different values of ?. A It is, however, difficult to measure A for shear, and hence the experimental determina- tion of S e is more uncertain than for tension or compression, and reliable experiments are wanting. The values of S e , U t , U c and U t vary considerably for the same material, according to grade and quality. Thus, for instance, for timber there are different kinds, and each kind varies according to soil, climate, season when cut, method and duration of seasoning, direction of fibres with reference to stress, etc. So also iron and steel vary according to quality, process of manufacture, whether in bars, plates or wire, etc. We give in the following table such average values as may be used for ordinary preliminary computations and estimates. For actual design, the necessary values for the material used should be accurately known by special tests. TABLE OF ELASTIC LIMIT AND ULTIMATE STRENGTH AVERAGE VALUES. Material. Elastic Limit Unit-stress in pounds per square inch. Ultimate Strength in pounds per square inch. Tension. S, Compression. s e- Tension. V* Compression. Shear. U,. Wrought iron Steel (structural) 25000 40000 6000 3000 25OOO 4OOOO 55000 lOOOOO 20OOO lOOOO 55000 150000 90000 8OOO 6OOO 25OO 5OOOO 700OO 20OOO < 600 longitudinal. ) 3000 transverse. Timber 3000 Brick Coefficient of Elasticity. In the experi- ment described on page 475, we find for a homogeneous bar of given material, cross- section A and original length /, extended by an axial force F uniformly distributed over the end area, that within the elastic limit, accord- ing to the law of elasticity, F_ A\ CHAP. I.] COEFFICIENT OF ELASTICITY. 477 is constant. We see from the figure that this constant is the tangent of the angle which the straight line OL makes with the vertical. But if we had taken a different length / x , everything else being the same, we should have for the same loads a different constant or tangent, F AX,' Now if the bar is homogeneous, the stress and hence the strain between two consecutive particles in the line of strain will be the same along the whole length. The total strain will then be the sum of all the strains between consecutive particles, or proportional to the length. We have then for different lengths V- K ::/ t :/, or A, =^. Substituting this, we have a = a constant. If then A, is known for any length / by experiment, this expression will give the constant or tangent of the angle of OL with the vertical for any other length l r Let the length / x be unity. Then we shall have for the constant in this case _ ATC This latter constant is called the coefficient of elasticity, and we denote it by E. We have then Now -T is the unit-stress, and 7 is the strain per unit of length, or the unit-strain. A I We can then define the coefficient of elasticity as the ratio of the unit-stress to the unit- strain, within the limit of elasticity. We have also, from (I), and from the preceding figure we see at once that we may also define E as that unit-stress which would cause an elongation equal to the original length of the bar, if the law of elasticity held good without limit. The value of E thus determined by experiments well within the elastic li^nit is then a measure of the elasticity of the body. We determine the coefficient of elasticity E c for 47 STATICS Oh ELASTIC SOLIDS. [CHAP. I. compression in precisely the same way, having regard to the precautions mentioned on page 476. For shear we have the same expression for E, as given by (I), if we put for / the distance ab, as shown in the figure, and for A. the distance ac. The direct measurement of \ is difficult and uncertain, but E, can be determined indirectly by experiments on torsion and bending, as will be explained later (pages 511 and 555). The values of E tt E, and E, vary for the same material according to grade and quality, just as the values of S t , for the reasons given on page 476. We give in the following table average values which may be used for ordinary preliminary computations. For actual design the values should be accurately known by special test. Equation (I) then holds for tension, compression or shear, within the elastic limit, if we put E t , E c , E, in place of E. In using (I), since E is always given in pounds per square inch, we should take / and A in inches and A in square inches. TABLE OF COEFFICIENT OF ELASTICITY AVERAGE VALUES. Material. Coefficient of Elasticity in pounds per square inch. Tension. *t Compression. E c Shear. E t 25 oooooo 30000000 15 oooooo I 500000 25 oooooo 30 ooo ooo 15 oooooo I 500000 6000000 15 oooooo 7 oooooo 6 ooo ooo 400000 longitudinal. Timber . Brick As an example, we give the following record of an actual series of experiments made upon a wrought-iron bar | inches diameter and 12 inches long. The first four columns give the experimental record. In the fifth column we give the corresponding values of. o Stress Elongation A. T 1 C pounds per 1 olal btress in Pounds. F. square i^ch Load on A Load off A A / S^ ~E- A' in inches. in inches. 2245 5OOO O.OOI 0.000 o.ooo ooo 20 449 IOOOO O.OO4 o.ooo o . ooo ooo 40 6375 15000 O.OC5 o.ooo o.ooo ooo 33 8980 20OOO O.OO8 o.ooo o . ooo ooo 40 9878 22000 0.009 o.ooo o.ooo 00041 101776 24000 O.OIO .000 o.ooo 00042 11674 260OO 0.0105 .000 o.ooo 00040 12572 28000 O.OII .000 o.ooo 00039 13470 30000 0.013 .000 o.ooo 00043 14368 32000 0.014 .000 o.ooo 00044 15266 34000 0.015 .002 0.00000044 16164 37000 O.022 .007 0.000 OOQ 6l 17062 380OO 0.416 3995 o.ooo ooi 09 17960 40OOO 0-5445 523 0.00001361 25450 5OOOO 1.740 1.707 0.000034 80 *?1I *7C 5 1600 2.468 ''J 1 /!> Specimen broke at 51600 pounds per square inch. CHAP - !] STRAIN DUE TO WEIGHT. 479 We see that the ultimate tensile strength is U t 51600 pounds per square inch; also that the elastic limit unit stress S e lies between 34000 and 36000 pounds per square inch. If we should judge by the beginning of noticeable set, the elastic limit would be between 32000 and 34000 pounds per square inch, and up to this point the elasticity appears to be perfect. The elasticity, however, is not perfect, and more accurate measurement would undoubtedly show the set beginning earlier. We cannot, then, take the beginning of observed set as the limit of elasticity. Since the elastic limit lies,' then, between 34000 and 36000 pounds per square inch, we have for the average -^ up to this point o = O. ooo ooo 387 3, and hence the value of E t given by this experiment is t = = OQO QOO 3 87 3 = 3 984 P unds P er s q uare inch- Examples. (i) A steel rod jo ft. long and 4 square inches cross-section is subjected to a tensile force of 40000 pounds. The elongation is observed to be of an inch. Find the coefficient of elasticity. ANS. The unit stress is loooo pounds per square inch. From our table page 476 we see that this is well within the elastic limit unit-stress S e , and equation (I) can therefore be applied and we have Et = = = 30000 ooo pounds per square inch. 4X 100 (2) A rectangular timber strut is 40 feet long and i3\ inches deep. If E c is i 200 ooo pounds per square inch, find the -width so that the strain under a stress of 270000 pounds may be one inch; all lateral bending being prevented. ANS. We have, from equation (I), Fl 270000 x 40 x 12 A = -rrr = = 108 square inches. -C.A i 200 ooo x i This gives 2500 pounds per square inch, which we see from our table page 476 is within the elastic limit. Equation (I) therefore applies, and we have then a width of , = 8.1 inches. Strain Due to Weight. If a homogeneous straight prismatic body of uniform cross- section A has a considerable length /, the strain due to its own weight may be considerable. Let <5 be the density or weight of a cubic foot, For any length x the weight is then 6 Ax, and the corresponding strain in an element of length dx is then, by equation (I), _ $Ax . dx _ d_ Integrating between the limits x / and x o, we have for the entire strain dl* _ 6A/ Z A ~ 2E ~ 2AE' or, since dAl is the entire weight W, Wl or, as we see from (I), the strain is one-half as much as that due to the same weight at the end. 480 STATICS OF ELASTIC SOLIDS. [CHAP. I. Example. How long must a homogeneous bar of wrought iron of uniform cross-section A be in order that when suspended vertically the greatest unit-stress may be equal to the elastic limit unit-stress, and what is the extension f ANS. The weight 8AI must equal S,A, or dl = S,, or / = ^. From our table page 476, S, = 25000 pounds per square inch; and since a bar of wrought iron one square inch in cross-section and 3 feet long weighs jo pounds (page 19), <5 = -7 Ibs. per cubic inch. We have then 2 5000 x 36 /= ,ox-iT- =750ofeet - The extension is, taking "=25 oooooo pounds per square inch (page 478), St/ 25000x7500 _ ~ 2E ~ 2x25 oooooo ~ 3-75 Stress Due to Change of Temperature. We have, from equation (I), where A is the strain corresponding to the unit-stress 5 = -j for a bar of length / and uniform cross-section, the coefficient of elasticity being E. If the bar is constrained so that it cannot change its length and then is exposed to change of temperature, there will be a unit-stress S equal to that due to a strain equal to the change of length for the same unconstrained bar under the same change of temperature. Thus if e is the coefficient of linear expansion for one degree of temperature, / the number of degrees change of temperature, and / the original length, the change of length of the unconstrained bar would be A = e//. The coefficient of expansion e = r is then the strain per unit of length per degree. If the bar is constrained so that its length cannot change, we have then a unit-stress 5 = ~ = Eet. which is independent of the length /. The total stress, if the area is A t is then AS=AEet. We give the following average values of the coefficient of linear expansion e for one degree Fahrenheit : Wrought iron e = 0.0000067 Steel e = 0.0000065 Cast iron = 0.0000062 Stone and brick e = 0.0000050 Example. A wrought-iron rod of length I and 2 square inches cross-section has its ends fired rigidly when the temperature is bo* F. Taking the contraction at 0.00000671 for one degree, what tension will be exerted when the temperature is 20" f.f ANS. The shortening due to cooling' is A = 0.0000067 x 4 We have then, from equation (I), EA\ E x 2 x 0.0000067 x 4' F = = ~ - *= 0.000536^. CHAP. I.] WORKING STRESS FACTOR OF SAFETY. 481 If E = 30 ooo ocx) pounds per square inch, F = 16080 pounds. This gives the unit-stress 8040 pounds per square inch, which we see, from our table page 476, is well within the elastic limit, and equation (i) therefore applies. Working Stress Factor of Safety. The greatest unit-stress to which a material can safely be subjected in practice is called the WORKING STRESS. We denote it by S u . The working stress should never equal the elastic limit unit-stress S e , since if it exceeds that limit the elasticity is impaired. For security, then, we take a fraction of S e . The number /by which S e is thus divided is called the factor of safety. We have then the working stress '' where / is the factor of safety. For steady stress it is customary to take f 1.5, and for repeated stress or suddenly applied stress /= 3. As we have seen, page 476, the exact determination of S e is difficult. It is therefore also customary to take S v a certain fraction of the ultimate strength [7 t , U c , U s for tension, com- pression or shear. The following table gives the factors of safety usually adopted when this method is used: FACTORS OF SAFETY FOR $ ^ & E ^ Steady Stress (Buildings). Varying Repeated Stress (Bridges). Shocks (Machines). 6 Steel (structural). 5 j IO 6 IO ie 3 Brick je 25 In order to find the area of cross-section A for simple tension, compression or shear, we have then simply to divide the given total stress by the working stress S w , We have then, when flexure is not to be apprehended, for steady or varying stress or shocks A = total stress Sometimes we have alternating stress, i.e. tension and compression alternating, as in the connecting-rod or piston-rod of an engine. In such case we find the area of cross-section for each stress and take the greatest. Thus, if flexure is not to be apprehended, total tensile stress total compressive stress A = or whichever is the greatest. When flexure of long columns is to be guarded against we must proceed as on page Variable Working Stress. The fact that the working stress S w , as determined in the preceding article, is constant in any given case, is by many engineers considered objec- tionable. The total stress can in general be divided into two portions. The one portion is a steady stress always existing, such as that due to weight or dead load. The other portion is a repeated stress, such as that due to loads recurring at intervals of time. 4 82 ST /tTlCS OF ELASTIC SOLIDS. [CHAP. 2. Evidently when the ratio of the steady stress to the total stress is great, we should !>< able to use a greater working stress than when this ratio is small. Thus when the steady stress is equal to the total stress there is no repeated stress at all, and the working unit- stress should have its greatest value. On the other hand, when the steady stress is zero we have repeated stress only, and the working unit-stress should have its least value. It is therefore customary to take for the working unit-stress, when flexure is not to be apprehended, for repeated stress of any kind U steady stress\ total stress /' From (i) we see that when the steady stress is equal to the total stress, that is, when there is no repeated stress, we have S u = -^, where U is accordingly the ultimate strength and /the factor of safety, just as in the preceding article. But when the steady stress is zero, we have only repeated stress, and (i) gives us S v = -. Hence Up is the ultimate strength for repeated stress only, or the "repetition strength " as it may be called. In like manner, when flexure of long columns is not to be apprehended, we have for the working unit-stress for alternating stress least of the two stresses greatest of the two stresse (2) From (2) we see that when the least of the two stresses is zero, we have S w = ', as in the previous case, for steady stress zero. But when the two alternating stresses are equal, we have S w = ~ '. Hence U, is the ultimate strength for equal alternating stresses, or the "vibration strength " as it may be called. The difficulty and uncertainty of determining U f and U v by experiment, and the few experiments thus far available, make the method of the preceding article the most generally accepted. The present method is, however, extensively used with assumed average values for U, U f and U v as given in the following table: w / U- Up UP Up- U, Up Wood 2 i 2 2 2 Steel (structural) 17800 3 5 7 15 These values are for direct tension or compression. For shear we take four-fifths of S v as determined above. CHAP ' *] WORKING STRESS- EXAMPLES. 483 In order to determine the area of cross-section A, we have, then, in all cases _ total maximum stress " ^ -. When flexure of long columns is to be provided against we must proceed as on page 569. Examples (i) A wrought iron tie-rod in a roof -truss is under a stress of 20000 Ibs. Find its area of cross-section. ANS. Taking the ultimate tensile strength (page 476) U t = 55000 pounds per square inch, and factor of safety 4 (page 481), we have e _ Ut 55000 > -j- - - = 13750 pounds per square inch. and hence 1375 Or again, from (i), page 482. if steady stress is equal to total stress, we have the same result. (2) Suppose the rod is a bridge member and the dead-load stress is 5000 pounds, the total stress being 20000 pounds as before. ANS. We have, as before, U t = 55o. and (page 481) the factor of safety 6. Hence W = Also by (i), page 482, we have . = 7500(1 + -^=9375 and A = = 2.13 sq. in. \ 20000/ 9375 (3) 7^, and f S/iftJ "+S a do>' -S.dfo, -S ( d( let t be the unit stress of direct tension and s the unit stress of direct shear. We have then the two equal and opposite tensile stresses -f- thw and ////, and the two equal and opposite shearing stresses -\-shw and shw. These last two stresses form a couple an d can only be held in equilibrium by an equal and opposite couple -f- sbw and sbw. Let d be the diagonal and a the angle of the diagonal with the direction of /. Let S t be the unit shear along the diagonal, and S t the unit tension normal to the diagonal. Then we have along the diagonal the components -f S t dw and S^w, and normal to the diago- nal the components -|- S,dw and S,dw. We have then for equilibrium S t dw thw cos a sbw cos a -f- shw sin a = O, Sdu' thw sin a sbw sin a shw cos a = O. Now we have h b sm a = -; and cos a = -. ; d d hence, dividing these equations by dw t we have S, = / sin a cos a -{- s cos 2 a s sin 2 a sin 2a -f- s cos 2a, S t = f sin 2 a + 2s cos a sin a = cos 2a + s sin 2a. dS t dS t If we differentiate these equations and put the first differentials -7-* = o and -;- = O, we da da have, when S t is a maximum, tan 2a = -, or sm 2a = and when S t is a maximum, tan 2a = , or sin 2a = and cos 2 a = and cos 2 a = (0 - (2) Inserting these values, we have for the maximum combined unit shear wax. S t = and for the maximum combined unit tension . S t = - -f \l / + - (3) (4) CHAP. I.] COMBINED TENSION AND SHEAR. 485 If then the direct tensile and shearing unit stresses t and s are given, we have from (3) the maximum combined unit shear S s , and from( i) its direction, or the angle a which it makes with the direction of /. We also have from (4) the maximum combined unit tension S t , and from (2) its direction, or the angle a which it makes with the normal to t. For compression and shear combined we have to substitute for t the direct compressive unit stress c, and then the same equations hold. Example. A rivet- inch in diameter is subjected to a tension 0/2000 pounds, and at the same time to a shear 0/3000 pounds. Find the combined maximum tensile and shearing unit stresses and the angles they make with the axis of the rivet. ANS. We have the area , where d is the diameter of the rivet. Hence the direct tensile unit stress is 4 > an( j t fr e direct shear is s - - . . From (3) we have S s = 7155 pounds per square inch, Ttu making, from (i), an angle tan la = , or a = 9 13'. with the axis of the rivet, and from (4), S t = 9420 pounds per square inch, making an angle tan 2 = -, or 54 23', with the normal to the axis, or 35 47' with the axis of the rivet. CHAPTER II. STRENGTH OF PIPES AND CYLINDERS. RIVETING. Strength of Pipes and Cylinders. A practical application of the principles of the preceding chapter is the determination of the size of pipes and cylinders, subjected to internal pressure. Let / be the pressure per square inch on the interior surface of a pipe or cylinder, due to the pressure of water or steam or air, or other fluid. It is a well-known principle of physics that the pressure of a fluid in any direction is equal to its pressure on a plane at right angles to that direction. Hence, in the figure, the pressure P, say in a vertical direction, is equal to the pressure on a horizontal plane of area Id, where / is the length and d the interior diameter. We have then P pld. If 5 W is the working stress for the material, and / is the thickness, we have then Pipes come in commercial sizes, and the preceding formula enables us to select the nearest commercial size for given unit-pressure, diameter and working stress. If we consider the preceding figure as a closed cylinder, then the pressure on the head Ttlf is/ X --- a "d the area of cross-section is ntd. We have then p X = ntdS^ or / = ^- . 4 4-S. Hence the thickness to resist longitudinal rupture is twice that necessary to resist end rupture. For water-pressure, if the head h is taken in feet, the pressure in pounds per square inch is/ = O.434//. Examples. (l) A cast-iron water-pipe 12 inches diameter and - inch thick is under a head of 300 f,'et. Find the factor of safety. ANS. The unit-pressure is 300 x 0.434 = 130.2 pounds per square inch. Hence the working stress is Pd 130.2 X 12 5 = = ~ - = 1230 pounds per square inch. 2 X 2 X g 486 CHAP. II.] THEORY AND PRACTICE OF RIVETING. 487 sq From our table page 476, the ultimate strength of cast iron for tension is 20000 pounds per square inch. The factor of safety is then 2OOO __ a b out jg 1230 (2) Find the thickness of a cast-iron pipe 18 inches in diameter under a head of water of 300 feet, taking a factor of safety of 10. ANS. From our table page 476, we have the ultimate strength for cast iron in tension 20000 pounds per uare inch. Our working stress is then S w = 2000 pounds per square inch. Hence /= Pd = 300x0.434x18 inch 25 ro 2 X 2000 (3) A wrought-iron pipe 4.3 inches internal diameter weighs 12.3 pounds per linear foot. What pressure can it carry with a factor of safety of 8 ? ANS. A bar of wrought iron i square inch in cross-section and 3 feet long weighs 10 pounds. Hence the area of cross-section of the pipe is 12.5 x -- = 3.75 square inches. The thickness is then / = 7 -^ = - 10 -zitr 4 inch. We have, from our table page 481, the working stress S w = ^= . Hence , 2tS, u 2 x 55000^ . ^ = T- = TTT-. = 763 pounds per square inch. Theory and Practice of Riveting. Another important practical application of preceding principles is the determination of the size and number of the rivets with which plates are fastened together. KINDS OF RIVETED JOINTS. We may distinguish the following FIG. i. joints : ist. Simple "Lap-joint, Single -riveted. Fig. I shows this joint front and side. The two plates overlie each other by an amount equal to the " lap, " and are united by a single row of rivets. The distance p from centre to centre of a rivet is called the pitch. We denote the diameter of rivet by d, and the thickness of plate by /. 2d. " Lap "Joint, Double-riveted. This joint is similar to the preceding, except that two rows of rivets are used. In both cases the rivets are in single shear. In all cases where more than one row of rivets is used the rivets are " staggered," or so spaced that those in one row come midway between those in the next, as shown in Fig. 2. Lap joints are used in tension only. o o FIG. 3. FIG. 2. O O O ooo o o o o o o o 3d. "Butt" Joint, Single-riveted, Two Cover-plates. Here the two plates are set end to end, making a " butt "joint, and a pair of "cover-plates" are placed on the back and front and riveted through by a single row of rivets on each side of the joint (Fig. 3). The plates in such a joint are in general not allowed to actually touch, and the entire stress, whether tensile or compressive, is therefore transmitted by the rivets. The thickness of the cover- 4 88 STATICS OF ELASTIC SOLIDS. [CHAP. II. FIG. 4. o o o o o o o o o o o 000 plates should not be less than half the thickness of the plates joined, except when this rule would give a thickness less than inch. Owing to deterioration of the metal by the action of the weather, no plate is used in construction less than inch in thickness. Hence if the plates joined arc less than inch, the cover-plates should be ^ inch. 4th. "Butt" Joint, One C for double shear I d\ 2.A p - d\\ + 1.57087] = ^+ -y- The plate section is reduced by punching from // between two rivets to (p - d)t, so that in the case of a tension joint the strength is reduced in the ratio or We see at once that for a given thickness / a large rivet gives a large pitch and less reduction in strength than a small rivet. Small rivets allow a less pitch at a sacrifice strength But the less the pitch the tighter the joint. When strength rather than tightness is desired, as in bridges and parts of buildings and machines, we should then use a large When tightness is essential, as in steam-boilers, we should use a small rivet witl rivet. sacrifice of strength. Practical Restrictions.-^'^ to risk of injury to the material in punching and liability to tear out, rivets are not allowed a pitch of less than 3 diameters, or if this distance 490 ST/tTICS OF ELASTIC SOLIDS. [CHAP. II. than 3 inches, as it usually is, less than j inches. Rivets should not be spaced farther apart than 6 inches in any case, or, when the plate is in compression, 16 times the thickness of the thinnest outside plate. This last is to guard against buckling of the outside plate between rivets. With these restrictions we may apply (5). Number of Rivets. Guided by the preceding restrictions and rules, we can select in any case a suitable size of rivet. This done, we can easily determine the number required. A rivet' is considered as failing either by shearing across or by crushing. In any case, then, the diameter being chosen, we must take such a number as shall give security against these two methods of failure, choosing the greater number. In general the number to resist crushing will be more than enough to resist shear. Still we should try for both. The bear- ing area of a rivet is the projection of the hole upon the "diameter, or dt. The allowable compressive stress is about 12500 Ibs. per square inch. The allowable shear is taken at 7500 Ibs. per square inch for single shear. In the following table we have given the safe shearing and bearing resistance for rivets of different sizes and for different thicknesses of plate. Having chosen, then, the size of rivet, -an inspection of the table will give its resistance. The stress to be resisted being known, the number to resist this stress either by bearing or shearing is easily determined. The greatest of these two numbers is taken, with enough over in any case to complete the row or rows. As most practical cases are double shear, the greatest number will usually be determined by the bearing resistance. Distance from End to Edge. The distance between the end and edge of any plate and the centre of rivet-hole, or between rows, is fixed by practice at never less than i inches, and when practicable it should be at least 2 diameters for rivets over f inch diameter. Joints in Compression The size and number of rivets are determined for joints in com- pression precisely as for joints in tension, because the joints are not considered as in contact, and hence the rivets must transmit the stress in either case. RIVET TABLE. SHEARING AND BEARING RESISTANCE OF RIVETS. Diameter of Rivcl in inches. Area of Rivet in square inches. Single Shear at 7500 Ibs. per square inch. Hearing Resistance in pounds for Different Thicknesses of Plate at 17500 Ibs. per square inch = 12500 x at. Frac- tion. Deci- mal. i" iV t" TV" i" TV r H" 1" 1 3" T 1" t 0-375 0.1IO4 828 1170 1465 1760 V 0-4375 0.1503 II3O 1370 1710 2050 2390 * 0.5 0.1963 14/0 1560 1950 2340 2730 3125 rV 0.5685 0.2485 1860 1760 220O 2640 3080 3520 3955 1 0.625 0.3068 2300 1950 2440 2930 3420 3900 4300 4 88o H o 6875 0.3712 2780 2I5O 2680 3220 3760 4290 4830 5370 5908 I 0.75 0.4418 33io 2340 2930 3520 4100 4690 527" 5%o 6440 7030 it 0.8125 0.5185 3890 2540 3170 3800 4440 5080 57o 6350 980 762O 8250 I 0.875 o 6013 45io 2730 3420 4100 4780 5470 6150 68*0 7520 8200 8890 9570 il 0.9375 0.6903 5180 2930 3660 4390 5130 5860 6590 7320 8050 8790 9520 10250 i I 0.7854 5890 3125 3900 4690 5470 6250 7^30 7810 8590 9370 10160 10940 IA 1.062? 0.8866 6650 3320 4150 4980 5810 6640 74/0 8300 9130 9900 10790 11620 n 1.125 9.0940 7460 3520 4390 5270 6150 7030 7910 8790 9667 10550 11420 12300 IA 1.1875 1.1075 8310 3710 4640 5570 7420 8350 9280 I02fX> HI30 12060 12990 CHAP. II.] WETTING EXAMPLES. 49 I Examples. (i) A boiler is to be made of ivr ought-iron plates f inch thick, united by single lap-joints. Find the size and pitch of rivets. If the boiler is 30 inches in diameter and carries a pressure of 100 Ibs. per square inch above the atmosphere, find the factor of safety, taking the ultimate strength at 55000 Ids. per square inch. ANS. From (4), page 489, we have -in. rivets. But from (3), page 488, we have f-in. This size would be chosen for ordinary construction work. In this case we wish a tight joint, and therefore use a small rivet at sacrifice of strength. Let us take, then, f-in. rivets. Then from (5), page 489, we find the pitch f in. But this violates the practical restriction that rivets should not have a less pitch than three diameters. We take the pitch, then, 2 inches. The pressure on a length equal to the pitch is 30 x 2 x 100 = 6000 Ibs. If S is the unit stress, the resisting stress is 5(2 |-]/ = p5. Hence S = = 11640 Ibs. per square inch. \ / 4 .33 The factor of safety is then about 5. If this is considered too small, we should use a less pitch or a larger rivet. A larger rivet would not be tight enough. For a less pitch the holes must be drilled and not punched. (2) Required to unite two \-inch plates by a butt joint with two cover-plates ; the stress to be transmitted being 40000 Ibs. and the unit working stress 10000 Ibs. per square inch. ANS. The area of the plates must then be 4 square inches net if the joint is in tension, gross if in com- pression. The cover-plates can be each \ inch thick. Our rule (4), page 489, gives for diameter of rivet d= \l inch. This is greater than given by (3). page 488, therefore we take it. From our table page 490 we have for the resistance to shear of a ff-inch rivet 3890 Ibs. The rivets are in double shear in a butt joint, hence we require ~ = about 5 rivets. The bearing resistance from our table is 5080 Ibs. We require, 40000 then, for bearing ^ = about 8 rivets. This, then, is the number we should use. For the pitch we have from (5), page 489, 2.887 inches. This is less than 3 inches. We therefore take the pitch 3 inches. We must have at least ij inches for distance from end and ed(! _ k]l = o, or R, = (i - k)P. The bending moment for any point C^ between the left cud and the load is then h i-*)z ^ M = = - (i - k}Px. (4) STATICS OF ELASTIC SOLIDS. [CHAP. III. For any point <7 a between the load and the right end the bending moment is x). . .... . (5) In both cases the bending moment is a maximum at the load and equal to k(i k)Pl, and the bending moment at any point is given to scale by the ordinates to two straight lines as shown in Fig. 8. I r pi If the load is at the centre, k = -, and we have R, -P, M '= -- . 2 ' 4 CASE 4. If the beam has a uniformly distributed load of w per unit of length, Fig. 9, the FIG. o. wl reaction at the left end R l = , and the bending mo- ment at any point C is wlx . wx* M m . 2 2 The bending moment is a maximum at the centre wl* and equal to ^-, and the bending moment at any point is given to scale by the ordinates to a parabola whose vertex is at the centre as shown in Fig. 10. (See page 420.) Resisting Moment. The resultant moment of all the fibre stresses at any cross-section, relative to the neutral axis of that cross-section, resists bending. It must therefore be equal and opposite to the bend- ing moment. We call it there- fore the resisting moment. Since we have taken for the bending moment the resultant moment of all external forces on the left, the resisting moment must be taken ( C for all fibre stresses on the right of the cross-section. Let 5" be the unit stress in any fibre of the cross-section, and a the area of cross-section of the fibre. The stress in the fibre is then Sa. If v' is the dis- tance of the fibre from the neutral axis cc of the cross-section, we have for the moment of the fibre stress - Sai/. The resisting moment for the cross-section is then - 2Sav'. But from equation (i), page 494, within the limit of elasticity, if any cross-section plane upon bending remains plane after, CHAP. III.] RESISTING MOMENT. EXAMPLES. 497 where S f is the unit stress in the most remote fibre at a distance v. Hence we have for the resisting moment But 2av'* is the moment of inertia / of the reaction relative to its neutral axis cc. Hence the resisting moment is S f l resisting moment = * . Now if M is the resultant moment of all the external forces either on right or-left of section relative to the neutral axis of the section, we have in general for equilibrium, within the limit of elasticity, if any cross-section plane before bending remains plane after, M * - = o. or = M. (II) v v If in equation (II) we always take M on the left of the section, it is evident that Sf will be positive when the stress is tension and negative when the stress is compression. It is for this reason that we always take the bending moment M on the left of the section. In applying (II), since S f is always given in pounds per square inch, all distances should be taken in inches. In finding M the change of lever-arm due to bending is neglected. That is, tJie deflection is assumed as very small compared to the length. ( By inserting for M its value as found on page 495, we can find in any case the load which will give any desired outer fibre stress S f within the limit of elasticity. For values of / for different cross-sections see page 42. We give here some of the most common. I d Rectangular cross-section / = bd* t v = . Circular " / = , v = r. 4 Triangular " / = v = A, -h. 3 6 33 Examples. (i) A timber beam of length I = 10 feet and constant rectangular cross-section of breadth b = 4 inches and depth d = 6 inches, rests horizontally upon two supports and sustains a load of P = 3000 pounds at the centre. Find'the maximum fibre stresses. P P \ ' ANS. We have R\ = . The maximum bending moment is at L . 2 the centre and given by (page 496) L p 2 4 The maximum fibre stress Sf, if the elastic limit is not exceeded, is then given, from (II), by c _ Mv V- / Now / = bd*\ and since v = -\ for the top fibre, we have for the top fibre \Pl T, x 2000 x 1 20 Sf = ~ M = ~ 2 x 4 x 3 6~ = - 2 5 The minus sign shows that the top fibre is in compression \Pl T, x 2000 x 1 20 Sf = ~ M = ~ 2 x 4 x 3 6~ = - 2 5 Pounds per square inch. 49^ STATICS OF ELASTIC SOLIDS. [CHAP. TIL For the bottom fibre we have v = , and hence S/ = + 2500 pounds per square inch tension. We see from our table page 476 that this is within the elastic limit. (2) Let the cross-section be triangular with the point up, the base of the triangle being S inches and the height to inches. Find the maximum fibre stresses. ANS. We have as before M . But / = g, and for the top fibre v = + -h. Hence for the top fibre 6/Y 6 X 2000 X 120 6/ = j-rr = = = looo pounds per sq. in. bh l 8 x loo For the bottom fibre, since v = h, s f = + ^Ta = + 9 pounds per sq. in. The maximum fibre stresses are then 1800 pounds per square inch compression in the top fibre and 900 pounds per square inch tension in the bottom fibre, at the centre of the span. We see from our table page 476 that the elastic limit is not exceeded. (3) Required the depth of a rectangular beam supported at the ends and carrying a load P at the middle in order that the elongation of the lou>est fibre shall equal of its original length. ANS. We have from (i), page 477, the elongation A given by S f l I E A = -FT = . Hence Ss = E 1400 1400 Also from (II), page 497, s _Mv E_ Now from (4), page 495, J/ = . Also in the present case v = - -, I =~M*. Hence nr rr or d =' (4) A cast-iron rectangular beam rests horizontally upon supports 12 feet apart, and carries a load of 2000 pounds at the centre. If the breadth is one half the depth, find the area of cross-section, so that the unit stress stress in the loiuer fibre may nowhere exceed 4000 pounds per square inch. ANS. The given unit stress 4000 is less than the elastic limit (page 476), so (II) applies. The greatest will be at the centre, where the bending moment is greatest. We have then for tension in lower fibre Mv S f = 4000 = . In the present case, from (4), page 495, M = ,v = -,/ = _. Hence r ^ = 3x2000x13x1^ or _- ' " 61 ' Taking S' = 3000 as before, we have for the cfippling load 3000 x 8 x loo P. = - ^ - = 3333* pounds. Taking R = 9000 as before, we have the breaking load P = loooo pounds, or for a factor of safety of 8 (page 481) a working load of 1250 pounds. If R were unknown, we should have, taking the ultimate strength for tension and compression Ut and Uc, in the place of R, from our table page 476, and v -h for the lower or tension fibre and -h for the upper 3 3 or compression fibre, _ \UtI 12 x loooo/ " = ~~ 4*7,7 6 x 8ooo7 8000 x 8 x 100 or P = - -- = -- -.-. - = -- ? - = 88884 pounds. 2. hi 6 x 120 hi 3 The second gives P the least value and should be taken. (3) A beam of yellow pine, 14 inches wide, 15 inches deep, resting horizontally upon supports 10 feet 6 inches apart, broke under a uniformly distributed load of 12 1940 pounds, rupture beginning in the lowest fibre at the centre. Find the coefficient of rupture. ANS. We have, from (8), Pvl 3/Y 3 x 121940 x 126 R = -%-= = -2;-= = - - -- = 3658.2 pounds per sq. in. 87 4W. 4 x 14 x 15 x 15 (4) A wroitght-iron beam 4 inches deep and i\ inches wide, fixed horizontally at one end and loaded with 7000 pounds at the free end, ruptured at a point 2 feet 8 inches from the load. Find the coefficient of rupture R. ANS. From (5), Pvl 6PI 6 x 7000 x 32 R = =_=: - L 2- = 56000 pounds per sq. in. | x 4 x 4 (5) A ivr ought-iron beam 2 inches wide and 4 inches deep rests horizontally upon supports 12 feet apart. Find the uniformly distributed load it will carry in addition to its own weight for a factor of safety of 4, ANS. From our table page 500, R = 55000 pounds per square inch. 502 STATICS OF ELASTIC SOLIDS. [CHAP. III. A bar of iron 3 feet long and one square inch in cross-section weighs 10 pounds. The weight of the beam is then 320 pounds. We have then, from (8), for the breaking load Hence for a factor of 4 P + 32 = vl 8AY Rbd* 55000 x 2 x 16 3 x .44 = 4 74 P " and P = 3754 pounds. (6) Find the length of a beam of ash 6 inches square which would break of its ovvn weight when supported horizontally at the ends, the weight of the timber being jo pounds per cubic foot and R = 7000 pounds per square inch. ANS. The weight per cubic inch is -|jg; and since the cross-section is 36 square inches, the weight per inch in length is / = 3 ' * g 3 = g-. We have then, from (8), Hence V 32 x 7000 x 6 x 36 1796 inches = 149! feet. (7) A wrought-iron beam 12 ft. long, 2 inches wide and 4 inches deep is supported at the ends. The beam weighs \ pound per cubic inch. Taking R at 54000 pounds per square inch, find the uniform load it can carry to rupture. ANS. Without the weight of beam, 16000 pounds. Besides the weight of beam, 15712 pounds. Comparative Strength. We see at once from (5), (6), (7) and (8), page 500, that, taking the semi-beam of Case I as unity, the same beam with unifoim load will carry twice as much, the beam on two supports with load in centre will carry four times as much, and the same beam uniformly loaded eight times as much. Examples. (i) A round and a square beam are equal in length and have the same loading. Find the ratio of the diameter to side of square so that the two may be of equal strength, R being the same for both. ANS. Since P, R, I are the same in both cases, we see from our equations page 500 that must be Diameter \/ 2 equal in both cases. Hence gr^ =f 2|/ (2) Compare the relative strengths of a cylindrical beam and the strongest rectangular beam that can be cut from it. ANS. Let D be the diameter of the cylinder, and b and d the breadth and depth of the rectangle. Tlie / bd* strength of the rectangle is proportional to;- = /-, or to btP. But we have Hence or t = j = W - b*. If we differentiate for l> and put the differential coefficient equal to zero, we have for maximum strength D /~2 D* 3// free end equal to PI For any value of x greater than X Q the breadth is given by (3). If the breadth is constant, we have, in (2), b = b l and hence (4) The height then varies as the ordinates to a parabola from b l at the fixed end to zero, theoretically, at the free end. Here, again, we must have the height at the free end equal to Substituting this for h in (4), we find that the cross-section must be constant and equal to p ^j// = r; for a distance x from the free and equal to For any value of x greater than x^ the height is given by (4). If both b and // vary, but the cross-section at every point is rectangular, we have b l : //j : : b : //, or b = -, h ' . Substituting these in (2), we have ,=*, (5) The height and breadth vary, then, as the ordinates to a cubic parabola from //, and , at the fixed end to zero, theoretically, at the free end. The area at -any point is then A = The area at the free end should be, then, P i ' The cross-section should therefore be constant and equal to or 9 = QQQQ x t = 7 degrees. (5) Compare the strength of a square shaft with that of a circular shaft of equal area. ANS. From (2), page 510, we see that the strength is proportional to. For a square shaft = = . v v 34/2 For a circular shaft = . The ratio is then 3 . For equal areas we have icr* = d*. Hence the ratio is V 2 * . (6) A circular shaft 2 feet long -is twisted through an angle of 7 degrees by a couple of 200 pounds with a lever-arm of 6 inches. Find the angle for the same shaft 4 feet long when twisted by a couple of 500 pounds with a lever-arm of 18 inches. ANS. 105 degrees. (7) Find the combined unit stresses for a wrought-iron shaft 3 inches diameter and 12 feet long, resting on bearings at each end, which transmits 4 horse-power while making 120 revolutions per minute, upon which a load of 800 pounds is brought by a belt and pulley at the centre. ANS. The unit stress for flexure is Mr Plr PI 800 x 12 x 12 x 8 Sf = 7- = -r = 3 = = 10800 pounds per square inch. 1 4-* 1t r it X. 2y The unit stress for torsion is, from (6), page 512, 198000 x 40 x r 198000 x 40 x 2 x 8 ,S S = - 1 a = 5 = 4000 pounds per square inch. it x I20/, TT" x 120 x 27 The combined stresses, then, from (7) and (8), are, for shear, S, = 6700 pounds per square inch, and for tension or compression, S t or S c = 12100 pounds per square inch. 514 STATICS OF ELASTIC SOLIDS. [CHAi-. III. (8) A vertical shaft weighing with its loads 6000 pounds is subjected to a twisting moment bv >i f.'ne couple of joo pounds acting with a leverage of 4 feet. If the shaft is of wrought iron 4 feet long and 2 inches diameter, find its unit stress, provided flexure is prevented. ANS. The unit stress of compression is , = 1910 pounds per square inch. The unit shearing stress for torsion is _ Mflf 300 x 48 x 2 S $ = j- =9172 pounds per square inch. From equations (i) and (2), page 484, for combined compression and shear, we have for the shearing unit stress S, = 4/9172* + = 9215 pounds per square inch, and for the compressive unit-stress S e = - + -1/9172* + - = 10170 pounds per square inch. (9) Find the diameter of a short vertical steel shaft to carry a load of 6000 pounds when twisted by a force of 300 pounds with a leverage of 4 feet, taking unit stress for shear at 7000 and for compression at 10000 pounds per square inch. ANS. About 2.5 inches. CHAPTER IV. WORK OF STRAINING. DEFLECTION OF FRAMED STRUCTURES. PRINCIPLE OF LEAST WORK. REDUNDANT MEMBERS. BEAMS FIXED HORIZONTALLY AT ENDS. Work of Straining. If the force F is gradually applied, increasing from zero up to F t F the average force is ; and if A, is the strain, the work done by I 7 is p work = A. Since the stress is equal and opposite to F, the work of straining, within the elastic limit, is one half the product of the stress and strain. Now, from (I), Fl Hence we have for the work done by the force or against the stress, within the elastic limit, FH Since E is always taken in pounds per square inch, if we take A in square inches, .Fin pounds and / in feet, the result will be foot-pounds. Work and Coefficient of Resilience. If the unit force is equal to the elastic limit unit A stress S e , so that we have, from (II), for the work done in straining up to the limit of elasticity S?AI work = If we regard the body as practically perfectly elastic up to the elastic limit, this is the work which the body can do in returning to its original dimensions when the force is removed. It is therefore called the WORK OF RESILIENCE. Since, for uniform A, Al is equal to the volume V, we can write S? S? work of resilience = ^ . Al = ^ . V. STATICS OF ELASTIC SOLIDS. [CHAP. IV. s? The coefficient jr, or the work per unit of volume, is called the coefficient of resilience. The work of resilience measures the ability of the body to withstand shock or suddenly applied force due to the impact of another body. To bring such a body to rest requires work. If this work is not greater than the work of resilience, the elastic limit is not exceeded. From our tables, pages 476 and 478, for average values of S, and , we can compute the following average values of the coefficient of resilience : Coefficient of Resilience. Wrought iron .................... 12.5 inch-pounds per cubic foot Steel (structural) ................. 26.6 Cast iron ....................... 1.2 " " " " Timber ...... ................... 3 " " " " Deflection of a Framed Structure. Equations (I) and (III) find direct application in the computation of the deflection of a framed structure. Thus let S be the stress in any member due to the actual loading, and / and a the length and area of cross-section of the member. Then, from (I), the strain of the member is Now let s be the stress in the same member due to any arbitrary load /, supposed to rest at the point where the deflection is required. The work due to this load,/ is s\ 2 Sst zaE' The total work in all the members due to this load / is then Ssl Now if A is the deflection at the point where/ acts, the work done by/ is. . Hence we have 1 -j~ pE Sst - a Example. Suppose a truss (see figure) composed of two inclined rafters of length bo inches, two vertical ties of length 48 inches, an upper chord of length 60 inches and a lower tie of length / J-? inches, the two end panels 36 inches and the centre 60 inches. Lft there be a diagonal strut cf whose length is 76.84 inches. Suppose a load of j tons atf and 10 tons at e. Required the deflection at e, the areas of cross-section being known. ANS. Let the coefficient of elasticity E= 12500 tons per square inch, and the areas of cross-section of the members as given in the following table. f 5 tons 10 tons CHAP. IV.] REMARKS ON THE PRECEDING EXAMPLE. 517 Member. af, Length / in inches. 60 in tons per sq. in. I25OO 12500 I25OO 12500 12500 12500 12500 12500 12500 5 in tons. 7-9545 4-7727 - 10.7954 + 6.4772 + 6.4772 + 4-7727 + 6.3636 -f- IO.OOOO + + + -f + n tons. 3.4091 2.0454 9.0909 5-4545 5-4545 2-0454 2.7272 IO.OOOO Cross-section in sq. ins. 1.8 5 I.OO 1.85 1-5 1-5 i-5 2.0 2.0 / 3 9 + 14-6582 + 9.7621 + 53.0436 + 23.5532 + 23.5532 + 6.5080 + 8.6777 + 50.0000 + 12.7067 A in inches. \ 0.0372 ! 0.0199 J V 0.0308 be 60 c d 60 de 16 ef 60 31250 3 9 l6 bf. . . 48 31250 6 48 15265 6 ,f 76 84 15265 76.84 125000 A = 0.0879 i- at e We take for the value of/ the load of 10 tons at e and find the stresses s in every member due to this single load. We also find the stresses S in every member due to the actual loading. In the product Ss these stresses must be taken with their proper sign. Thus if .yis compression or minus and S'ls also compression or minus, the product Ss is positive. If one is tension or positive and the other compression or negative, the product is negative. If the signs of S and s are carefully observed, the signs of the products Ss will thus take care of themselves. If we take E in pounds or tons per square inch, S, s and p must be taken in pounds or tons, and / in inches and a in square inches. We have taken / at e equal to 10 tons, or the load actually acting there. But if there were no load acting there, we could still assume/ = 10 tons or i ton or any convenient amount, and proceed as before. The stresses S due to actual loading are, strictly speaking, affected by the change of shape. This can, however, be disregarded without perceptible error, as the deflection in all practical cases is very small com- pared to the span. Remarks on the Preceding Example. In our example we assume E as constant for all -members. We also assume that every member has its exact length and area of cross-section, that all pins fit perfectly tight, and all adjustable members, if any, are accurately adjusted. A truss after erection may then be tested by calculating the deflection at the centre for a given loading and comparing with the actual observed deflection due to this loading. A good agreement is thus a test of the close fit of all pins, of the proper adjustment of all adjustable members, of the agreement of the lengths and cross-sections of members with those called for by the design, of the constant value of E and its proper assumption as to magnitude, and finally of the fact that the elastic limit is not exceeded. It is evident, however, that when so many conditions must concur, a discrepancy between the observed and the calculated deflection has little practical significance. The last-mentioned fact, that the elastic limit is not exceeded, is the most important, and this is proved, not by any close agreement between actual and calculated deflections, but by observing whether the deflection is constant under repeated applications of the same loading after the structure has attained its permanent set. Computations of deflection are then of little value as a means of testing framed structures, and the calculated result cannot be expected to agree very closely with the actual deflection. Principle of Least Work. We have seen, page 311, that for stable equilibrium of a material system the conditions of static equilibrium must be fulfilled, and, also that the STATICS OF ELASTIC SOLIDS. [CHAP. IV. potential energy is a minimum. When an elastic body is strained by external forces within the elastic limit, the work done by the external forces is the work the body can do when released, or the potential energy is equal to the work of the external forces. Hence for stable equilibrium of an elastic body the work of the external forces must be the least possible consistent with the conditions of static equilibrium. This is called the prin- ciple of least work. As an illustration of the application of this principle, suppose a rectangular table of L length L and breadth b to have four legs, all of equal length / and uniform cross-section A, one at each corner. Let a load Prest on the table, and let x and y be the co- ordinates of its point of application. Let P l , P 2 , P 3 , P t be the loads carried by the legs. We have for the conditions of static equilibrium P 3 L + P 2 L - Px = o, P,b + PJ> - Py = o. From these equations we obtain where P^ , P 2 and P 3 are given in terms of P 4 . But P^ is unknown. We have four unknown quantities and only three equations of condition. We need another equation of condition. This is furnished by the principle of least work. Thus from equation (III) we have for the work of compressing the legs, assuming the floor and table-top to be rigid, = [/> + P* + P* + P*]. If in this we substitute the values of P l , P 2 , P 3 , we have - zT- '4 - z k + We thus have the work given in terms of P t . Now P t must have such a value that the work shall be a minimum. Therefore putting the differential of the work relative to P 4 equal to zero, we have CHAP. IV.] PRINCIPLE OF LEAST WORK. 519 Hence P Px Py P Px p y ^ = 4~ 2 -Z + 2* and P * = -4 + 2Z+ 2 -f P-^_^0.^ 2 ~ 4 2b ~^~ 2U = $P_r^_Py 1 4 2 2b' If P is at the centre, x = , y = - and P P = P P -P f\ -*a ~ * a *4. v* 4 If P is at the middle of a side /, x , y = o and If P is over leg three, x = Z, ^ = < and ?!=--, />, = +-, />=+^/>, /> -+ ' 4 ^4 ^4 4 The minus sign for /^ shows that the stress is reversed and leg one must be fastened down. If not, it is lifted off the floor and we have the table supported on three legs only. Example Take t/ie same table with a fifth leg at the centre, and show that the load carried by this leg is always -/' no matter where the load P may be placed. Remarks on the Preceding. The preceding problem of the four-leg table illustrates the principle of least work. It also illustrates much more it furnishes an example of theory misapplied. The theory is sound, and the results are therefore correct provided the assumptions are realized in fact. But these assumptions are not realized by any actual table. For instance, we have assumed the floor and table-top rigid, every leg of precisely equal length and the same uniform cross-section. Such a table is an ideal which has no physical existence. The theory is then misapplied, since the assumptions do not correspond to fact. A very small discrepancy in the length of the legs alone would entirely change the results. The reader, then, must regard the problem simply as an illustration of the principle of least work, and should note that even sound principles need care in application, and that for proper application the assumptions should accord with reality, otherwise the results are worthless. Thus in the present case the legs should be designed for three only. Then if any others are desired, they can be added of the same dimensions. This practical solution is not only simpler, but it is actually more accurate and scientific. FIG. i. c B 520 STATICS OF ELASTIC SOLIDS. [CHAP. IV. I Redundant Members. The principle of least work is also illustrated by the calculation of the stresses in a framed structure with redundant members. The method of procedure is the same as for the table with four legs. Thus take the simple truss shown in Fig. I, resting on supports at a and b with a load P at the centre. Let the inclined braces make the angle 6 = 45 with the vertical, so that the lengths of all horizontal and vertical members are equal to the height h of the truss. The length of the inclined members is then h ^2, and tan = I, sec d = ^2. Let the stress and area of cross-section of aA, aC, ac, Ac, AC, Cc be s l and a l , s t and "*> S * and ' CtC " aS lndi - FIG.,. FIG. 3. cated in Fig. i. This truss A consists of two statically de- terminate trusses, as shown in Fig. 2 and Fig. 3, super- posed upon each other. Let the truss of Fig. 2 a carry a certain fraction . CHOP Then we have the stresses (page 409) = -* 2 2 ' 2 = -(i-0) /'tan 6 = - (i - 0)/>, s 6 = />0. The value of must be that which makes the work a minimum. We have for the work, from equation (II), since the stresses and areas in the left portion of Fig. i are the same as the right, or, substituting the values of the stresses as found, work = , ,_ 2^ L 2, a 2 za 3 a 4 a, If we differentiate with reference to and put the value of - = o, we have for the value of which makes the work a minimum 2JK2_, I ~^~ a s 21/24 a. a. CHAP. IV.] N Q ECONOMY DUE TO REDUNDANT MEMBERS. 521 We can therefore find s lt s 2 , etc., by inserting this value of < in the equations for the stresses already given. It will be noted that the cross-section a lt a 2 , etc., must be known for each member in advance. If the cross-sections are all equal, we have Evidently the same remarks apply here as in the case of the table with four legs. Every member must be of absolutely true length and of the exact cross-section assigned. Any variation from ideal conditions invalidates the result. As such ideal conditions do not and cannot exist, the actual stresses in any given case will not agree with the computed stresses. We see, then, that the use of redundant members in a structure not only makes the calculation of stresses very involved and laborious, but also that the results obtained hold only for an ideal structure under ideal conditions and are by no means the actual stresses. No Economy due to Redundant Members. It remains to inquire whether there can be any compensating advantages in economy in the use of redundant members to offset the objections already noted. This inquiry is directly answered by the preceding article. We see that Fig. I is composed of two statically determinate trusses, Fig. 2 and Fig. 3, superposed upon each other. Each of these carries its own proportion of the loading. It is also evident that one of these trusses is more economical than the other. The combined truss of Fig. I must therefore have an economy intermediate between the two, and therefore must necessarily be less economical than one of the two. The same holds for any structure with redundant members. We may consider it as formed by the superposition cf a series of statically determinate trusses. The economy of the combination must be less than the economy of some one of this series. Hence any structure with redundant members is less economical than some statically determinate struc- ture included in the redundant structure. There is, then, no gain of economy by the use of redundant members. Work of Bending. As before, let 5^ be the unit stress in the most remote fibre of any cross-section at a distance v from the neutral axis of that cross-section ; then the unit stress 5 in any fibre at a distance v' is, from equation (i), page 494, S=^-S / . hence 5 / = 7 -,S ......... (i) Also, if is the bending moment, or moment of all the external forces on the left of the cross-section (page 494), we have, from (II), page 497, v Inserting the value of S^from (i), we have for the unit stress 52 STATICS OF ELASTIC SOLIDS. [CHAP. IV. If ds is the distance measured along the neutral axis between two consecutive cross- sections, and a the area of the cross-section of the fibre, we have, from (I), page 477, putting Sa tor F, a lor A, and ds for /, for the strain of any fibre SJs *~ ' or, inserting the value of 5" from (2), _ Mv'ds A ~ El ' (3) Now the work on the fibre is half the product of the stress and strain (page 515), or, from (2) and (3), A. *" ' 2 and since 2ai/* = /, the work on all the fibres of the cross-section is 2EI' For the total work, then, for all the cross-sections we have "=/ (IV) Equation (IV) is general whatever the shape of the beam before flexure. If the beam is straight before flexure, we have dx in place of ds t and / in place of s. Hence w= / -^ Beams Fixed Horizontally at the Ends. By means of equation (IV) and the principle of least work, we can solve all cases of beams fixed horizontally at the ends. CASE i. BEAM OF UNIFORM CROSS-SECTION FIXED HORIZONTALLY AT ONE END AND SUPPORTED AT THE OTHER CONCENTRATED LOAD Reaction. Let R l be the reaction at the supported end A on left, and let the load R, | R * P be at a distance kl from the supported^ end and (i k)l 4 f >| || from the fixed end B. Then for any point of the neutral axis distant x from the supported end we have the bending moment, when x < kl, M x R^x ; R( when x > kl, M x = - R,x + P(x kl}. The beam may be fixed at the end B either by letting it into a wall or by continuing it over the support at B and applying a load P . In cither case the tangent to the deflected beam must be Jiorizontal at the fixed end />', and the value of R^ must be that which makes the work of bending from A to B a minimum. The work on the remainder of the beam, if any, can be neglected. CHAP - IV ] BEAMS FIXED HORIZONTALLY AT THE ENDS. From (IV), then, we have for the work of bending 523 If we differentiate with respect to R l and put i wor ') _ Q>we havg f or t h e value of dR l which makes the work of bending from A to B a minimum, since E and /are constant, = o. Performing the integrations, we obtain If the load is at the centre, we have k = and R, = P. 2 16 Moment. The moment M 2 at the fixed end is and is positive. Hence RJ is less than P(i k)L The moment at the load is RJsl and' is negative. If we subtract this from M 2 , we have which is positive since Pis greater than R r The moment at the fixed end is therefore the greatest, and the moment at any point is given to scale by the ordinates to two straight lines as shown in the figure. Point of Inflection. We see that at the point C in the figure the bending moment is zero. This point, as we shall see (page 547), is the point at which the curvature of the deflected beam changes from concave to convex. We can easily find its position. Thus, from the figure, if x l is the distance of C from the supported end A, we have or x, = Substituting the values of M 2 and R^ already found, 2/ If the load is at the centre, k = and x^=. /. Breaking Load. Since, as we have proved, the bending moment M 2 at the fixed end is greatest, we have, from equation (4), page 499, 524 ST/1T1CS OF ELASTIC SOLIDS. [CHAP. IV. or, inserting the value of M t , we have for the breaking weight 2RI P = vlk(\ - For the load at the centre k == - and P = - -, or - as much as for the same beam 2 ->>vl 3 supported at both ends (page 500). The moment M 2 is a maximum for k = A /- = 0.5774, that is when the load -Pis distant O.5774/ from the supported end. For this position we have the least breaking load vl CASE 2. BEAM OF UNIFORM CROSS-SECTION FIXED HORIZONTALLY AT ONE END AND SUPPORTED AT THE OTHER UNIFORM LOAD. Reaction. In the preceding case we have found for concentrated load If for P we put wdx, and for k we put -j, we have in the present case Performing the integration, we have *,=!/. Moment. The moment M 2 at the fixed end is then wl* wl* "~~ '' The moment at any point distant x from the supported end is then This is the equation of a parabola whose vertex is at -/ from the supported end as shown in the figure o the moment at this point being z 5 . 1 2o Point of Inflection. Here again the moment is zero and we have a point of inflection 6 at a distance x 1 from the supported end given by CHAP. IV.] BEAMS FIXED HORIZONTALLY AT THE ENDS. 525 Breaking Load. Since the moment M 2 at the fixed end is the greatest, we have, from equation (4), page 499, RI => r wl* RI Hence the breaking load is 8RI or -as much as for the same load at the centre and just the same as for the same beam supported at the ends (page 500). CASE 3. BEAM OF UNIFORM CROSS-SECTION FIXED HORIZONTALLY AT BOTH ENDS CONCENTRATED LOAD. In this case we have the bending moment for * x Performing the integration, we have The moment M t on the left of the right end is the same. The moment at any point distant x from the left end is .. M x = M, - R,x + = -^-^V~ wx* wl* vox = -^-^ This is the equation of a parabola whose vertex is at '/ as shown in the figure, the moment at this point being - ~. The moment at the fixed ,?nds is the reatest. 528 STATICS OF ELASTIC SOLIDS. [CHAP. IV. Breaking Load. Since the moment is greatest at the fixed end, we have, from equation (4), page 499, wl* RI \2RI = , or wl -- . 12 V V Bending and Tension Combined. A beam may sometimes be subjected to bending and at the same time to tension. Thus, for instance, a lower chord panel of a bridge truss may be in tension and at the same time sustain loads applied by means of cross-ties between the panel points. In such a case let M be the maximum bending moment, y the deflection, Tthe tensile stress and A the area of cross-section, so that is the tensile unit stress. Then we have the A maximum moment and from equation (II), page 497, the unit stress in the outer tensile fibre is T (M+ Ty]v A^ I If S w is the working unit stress adopted, we have then T , (M+Ty)v ^ - ^ -t f The neutral axis is now no longer at the centre of mass of the cross-section, and a strict solution leads to results of great complexity. If, however, we disregard the small deflection y, we have in all practical cases Sv = -: -\ p , hence M = If we put for /its value Ax 2 , where K is the radius of gyration of the cross-section (page 32), we have From (i) we can find in any case the load for a given S w , T and cross-section A, and from (2) the cross-section for a given load. Bending and Compression Combined. This case is just the same as the preceding, except that we must put the compressive stress C in place of T and take S w the working stress for compression. If flexure is to be apprehended, we must take S u , as given on page 569. Examples. (i) Find the area of cross-section for a square beam of ra ft. span -which sustains a load of joo pounds at the centre and has at the same time a direct longitudinal tension of 2000 pounds, the working stress being taken at 1000 pounds per square inch. . ANS. We have S w = 1000, T= 2000, v = , K* = , A = d*, M = I5ox6x 12. Hence, from (2), A=d* = ^ + 2, or d = 4. 1 8 inches. CHAP. IV.] EXAMPLES. 529 (2) 'Find the area of cross-section for a square beam of 12 ft. span which sustains a load of 50 pounds per foot uniformly distributed and has at the same time a direct longitudinal tension of 2000 pounds, the working unit stress being taken at 1000 pounds per square inch. ANS. We have 5 W = 1000, T = 2000, v = d -, * = . A = d\ M = ^ = 5<>x 12x12x12 x Hence> from (2), A = l6x^ Hence, from (2), ^4 = 2^=^ + 2, or d = 7.86 inches. CHAPTER V. DEFLECTION OF BEAMS. Deflection of a Beam. Let ds be the distance measured along the neutral axis between any two consecutive cross-sections, and v' the distance of any fibre from the neutral axis ; then the strain A of the fibre is, from equation (3), page 522, Mv'ds where M is the bending moment of all the^external forces on the left. Take the origin at the left end A, and let x, y be the co-ordinates of any point P r At this point let a force F be supposed to act, and let its moment relative to any point P z on the right of P l given by the co-ordinates ~x, ~y be m. Then from equation (2), page 521, the stress due to I this force is mad The work of this force on any fibre of the cross-section at P 2 is then half the product of the stress and strain (page 515), or mMa&ds On all the fibres of the cross-section, since 2aiP = o, we have then mMds work = E j , and for all the cross-sections between the right end B and />,, if AB = s and AP l = s lt 'mMds work =f */ t t This equation is general whatever the shape of the beam, whether straight or curved, or the direction of F. Let F be vertical. Then its moment m at P t is m = F(x x), 530 CHAP. V.] DEFLECTION OF A BEAM. S3 1 and, from (i), its work is C s F(x - x)Mds work = / -^ =-t . ./,, 2EI But if A y is the vertical deflection at P lt the work of F is ^, and this must be equal and opposite in sign to the work done against the fibre stresses, or FA^ _ C*F(x-x)Mds 2 ~V,, 2EI Hence the vertical deflection at any point P l is given by C*M(x x)ds C'Mxds A > = J,-'-ET- -XTF Again, let Fbe horizontal. Then its moment m at P^ is andj from (i), its work is work -: If A x is the horizontal deflection at P lt we have then F(y-y)Mds 2EI FA, _ r* 2 / J *i Hence the horizontal deflection at any point P l is given by Mds C'Myds , ' r* *=- -ir+y j j, j ^ , rr Equations (V) and (VI) are general whatever the shape of the beam, whether straight or curved. For a straight beam y = o and the horizontal deflection A x = o. The vertical deflection is given by (V) if we put ds = dx, s l = x and s = /. Hence for a straight beam, calling the vertical deflection y, we have , = r%* f^f-- .'...'... (vn) If we differentiate this, we obtain MX dx MX dx C l M ^x or vjftfr dy __ C ~fc~ I El . j * If we differentiate again, we have (VIII) 53* STATICS OF ELASTIC SOLIDS. [CHAP. V. Equation (VIII) is the differential equation of the curve of deflection for a stra ight beam By integrating it we obtain (2), which gives the inclination at any point. Another integra- tion gives (VII), the deflection y at any point. When the deflection is very small compared to the length, we have, from the Calculus, d*y i ~dx*~~~p' where p is the radius of curvature. Hence we can write El = - M (IX) Equations (VIII) and (IX) can also be directly deduced as follows: Let ad, a l d l and a 2 6 2 be three cross-sections all parallel before bending, the distance between the first two being x, and the last two being consecutive at the distance dx. After bending let these cross-sections be still plane, and intersect at some point C. Let ccfa be the neutral axis, so that cC c^C c t C = p. Let the angle aCa^ = a be very small. Then the angle a^Ca 2 = da, and the distance cc l = s will be equal to x, and c^c t = ds, to dx, approximately. Through c l draw a'b' parallel to a 2 b z . Then the angle a^CjOt' = da. For any fibre at a distance v from the neutral axis, then, the strain is A = vda. Since the original length is dx, if the area of cross-section is a, we have from I), page 477, the stress of the fibre Eavda Ea\ dx dx The moment relative to c l is then EaiPda dx For all the fibres of the cross-section, then, since 2av* = /, we have the moment Elda dx ' This moment must be equal and opposite to the bending moment M. Hence . dx Now for very small deflection a = ~ and hence da ~. Substituting, we have dx dx which is equation (VIII). CHAP. V.] BEAMS FIXED HORIZONTALLY AT ONE END AND LOADED AT THE OTHER. Again, from similar triangles, we have vda : v : : as : p, or, since ds = dx 9 533 da dx Substituting this, we have which is equation (IX). Beam Fixed Horizontally at One End and Loaded at the Other. _(*) UNIFORM CROSS- SECTION. For uniform cross-section / is constant. Take the ori- gin at the free end. Then for any point of the neutral axis at a distance x the bending moment is We have then, from (VIII), Integrating, where C l is the constant of integration. Since the beam is fixed horizontally at the right end, the tangent at the right end to dy PP the curve of deflection is horizontal, and hence - = o when x = /. Hence C l = + . We have then Integrating again, PPx where C 2 is the constant of integration. Since the deflection at the fixed end is zero, we have y = o when x = /, and hence We have then PPx Px* - 6~ This equation gives the deflection at any point. The deflection is evidently greatest at the free end. Making, then, x = o, we have the maximum deflection ' ' ^-ff r ' ; : : The minus sign shows that the deflection is downwards. 534 STATICS OF ELASTIC SOLIDS. [CHAP. V. If the cross-section is rectangular, / = bd* and we have 4/Y (b) UNIFORM STRENGTH. For uniform strength the cross-section varies and hence / is variable. We have then in general for rectangular cross-section and uniform strength d*y _ _Px_ i2Px d**~ El Ebd*' where b and d are variable. Constant Depth. If the depth is constant and always equal to d l , then, as we have seen (page 503), b = jb l , where b l is the breadth at the fixed end. Hence for rectangular cross-section d* 12.PI As we have seen (page 532), -^ = i where p is the radius of curvature. Hence for constant depth and uniform strength the radius of curvature p is constant and the curve of deflection is a circle. Integrating, since ^- = o for x = /, dy \2Plx I2/V 2 Tz- ~'Eb^' Integrating again, since y = o for x /, _ . Ebji? EbJ* For the deflection J at the free end x = o and 6/V 8 or as much as for same beam of uniform cross-section. Constant Breadth. For constant breadth b l we have (page 504) d* = jd*. Hence d*y _ I2/Y4/7 Integrating, since -r- = o for x = /, ~dx Integrating again, since 7=0 for x = /, y= CHAP. V.] BEAMS FIXED HORIZONTALLY AT ONE END AND LOADED AT THE OTHER. 535 For the deflection A at the free end x = o and or twice as much as for same beam of uniform cross-section. For similar cross-sections we have (page 504) d* = *d*. Hence bd* = ^\** and Integrating, since ~ = o for x = /, Integrating again, since ^ = o for x = /, _ S*^ 1 ** -pV^ + For the deflection J at the free end * = o and A- , or as much as for same beam of uniform cross-section. The maximum deflections are then as -, 2 and -, or as 15, 20 and 18. If we call the volume of the beam of constant cross-section V, then in the first case the volume is V t in the second case V, in the third case V, or the volumes are as 15, 20 and 1 8. The maximum deflections, then, for a beam of uniform strength in three cases are as the volumes. Beam Fixed Horizontally at One End and Uniformly Loaded. (a) UNIFORM CROSS- SECTION. If w is the load per unit of length, we have for any point of the neutral axis at a distance x from the free end, the bending moment . wx* J/ = + . Hence, from (VIII), A dy Integrating, since = o for x = /, dy wx* wt s EI dx = -~6 + ~^' S3 6 ST/tTlCS OF ELASTIC SOLIDS. [CHAP. V. Integrating again, since y = o for x /, wx* wl*x wl* "iT H ~6 r- The deflection A at the end is for x = o or only as much as for an equal load at the end. o () UNIFORM STRENGTH. We have then in general for uniform strength and rectangular cross-section Constant Depth. For constant depth and uniform strength, as we have seen (page 505), X* b j^b l and d d r Hence if we make the breadth constant and equal to ^, and the depth variable, so that d = di\/ T- "P to the next bar and then uniform, p will be constant still. Such a spring is shown in the figure. The A crippling load and flexibility are as before, the length / being measured from the ends BD, BD, and not from the centre. 542 STATICS OF ELASTIC SOLIDS. [CHAP. V. Beam Supported at Both Ends Uniform Cross-section Concentrated Load. Let the load P be distant from the left end a distance //, where k is any given fraction. Then the R/ Ri distance from the right end is (i >&)/, and the reaction at the left end is P( i K). Take the origin at the left end. Then for any value of x less than kl we have when xkl M = - P(\ k)x + P(x - kl). Hence from (VIII) we have when xkl ^fj~5 = ^ l ~ ) x ~~ P ^ x ~ &fy Integrating, we have dy P(i k}x* _ dy P(i k\x* Px* EI-r pC, and EI-j- = h Pklx dx 2 dx 2 2 dy For x = kl these two values of -7- are equal, and hence dx 2i 2 ' Inserting this value of C 2 and integrating again, we have k** Px* Pklx* Ely = - - + C v x + C, and Ely = In the first of these equations, when x = o, y = o, and hence C 3 = o. For x = kl these two equations are equal, hence " 4 = -= . For x = o, y in the second equation is zero, hence - k)(2 - k) and therefore Pl*k(2 + *) C *~ ~6~ Substituting these constants, we have when x^/ y=-' ^ ~ ' ( 2 lx - &P - x*). brLl CHAP. V.] BEAM SUPPORTED AT BOTH ENDS. 543 If we make x kl, we have th6 deflection at the load If we insert the values of C^ and C 2 in the equations for ~ and put -^ = o, we have for the value of x which makes the deflection a maximum when xkl x=l When x = kl in these equations the maximum deflection will be at the load and will be the greatest possible. Placing, therefore, x = kl, we obtain from both these equations the condition That is, the greatest maximum deflection is at the load when the load is at the centre. For any other position of the load the maximum deflection is on the right of the load when /< , and on the left of the load when k> . That is, the maximum deflection is always between the load P and the farthest end. Inserting, then, these values of x in the values for y when x>kl, or when x we ^ ave f r tne point at which the moment is zero, or ^ changes sign, that is for the point P of inflection, = o, or x = * 2 4 Beam Supported at One End and Fixed Horizontally at the Other Uniform Cross section Concentrated Load. Let the load P be distant IR, j from the supported end a distance kl, where k is any given W -f- fraction. Then the distance from the fixed end is (i k)l. M/ Take the origin at the supported end. Then for any value of x less than kl we have for xkl M=-R l x+P(x- kl). Hence, from (VIII), for xkl EIj^ CIX l*X> Integrating we have EI-j- = ^ -|- C^ and Efj- = - dy For x = kl these two values of ~ are equal, and hence STATICS OF ELASTIC SOLIDS. [CHAP. V. Inserting this value of * 2 and integrating again, we have 6 6622 4 * In the first of these equations, when x = o, y = o and hence C = O. For x = kl these two equations are equal, hence " 4 == . For x = l,y in the second equation is zero, hence 6 Also for 4r = /, = o, and hence ax Hence d ^4=fl, and * 1 = f( I - Wz +*). This is the same value for R^ that we have already found (page 526) by the principle of least work. -f- &} We have then (7, = , and all the constants are determined. 4 If we insert the values of C\ and C 2 in the equations for and put -j- = o, we have for the value of x which makes the deflection a maximum when x < kl x I ' when x > kl When x = kl in these equations the maximum deflection will be at the load and will be the greatest possible. Putting, therefore, x = kl, we obtain from both these equations the condition k= 4/2 i = 0.414213. That is, the greatest maximum deflection is at the load when the load is at a distance of ^2 i =0.414213 of the span from the supported end. For any other position of the load the maximum deflection is between the load and the supported end when k> 4/2 1, and between the load and the fixed end when k < 4/2 I . Inserting, then, these values of x in the values of j>, we have for the maximum deflection in general P(\-KfkP I k * P(\ - when k < 4/2 i A = - 3 7(3 - CHAP. V.] BEAM FIXED HORIZONTALLY AT BOTH ENDS. 547 Both of these are equal and have their greatest value when k = i/J I. Inserting this value of k, we have for the greatest maximum deflection at the load 3^7 or only about - as much as for beam supported at the ends. 100 If the load is at the centre of the span, k and , - _ and since k > 4/2 I, we have the maximum deflection in this case when k = and this maximum deflection is at a point between the load and the supported end given by For the point of inflection, if we put ~ = o, we have the distance of the point of inflection from the supported end 2/ If the load is at the centre of the span this becomes /. Beam Fixed Horizontally at Both Ends Uniform Cross-section Uniform Load. wl For a load wj)er unit of length the reaction is at each end, and the bending moment at any point of the neutral axis distant x from the left end is wlx wx* + . Hence, from (VIII), wlx wx* Integrating, since f = o for x = o, we have 548 STATICS OF ELASTIC SOLIDS. Integrating again, since for x = o, y = o, we have wlx* wx* '~~+-' [CHAP. V Since for x = - we have also -j- = o, we have , , -- L 4- - - ^ = o, or 2 " 16 48 Inserting this value of M v , we have 12 wPx wlx* _ dx ~ 12 " 4 ~~6~' a/;*" 4 -. 24 24 12 If we put the first of these equations equal to zero, we find for the point at which the deflection is a maximum x ' -. The maximum deflection is then at the centre and given by A - - or only one fifth as much as for beam supported at the ends. If we put -7-3 = o, we have for the points of inflection x = --=. and x = - + 21/3 21/3 = o.2ii3/ and x = o. 78877. Beam Fixed Horizontally at Both Ends Uniform Cross-section Concentrated Load. Let the load P be distant from the left end a distance kl, where k is any given fraction. Then the distance from the right end is (i )/. Take the origin at the left end. Then for any value of x less than kl we have for x kl M = MI - R^x + P(x kl). Hence, from (VIII), for xkl lX - P(x - < V ' ] BEAM FIXED HORIZONTALLY AT BOTH ENDS. 549 Integrating, we have, since ~- = o for x = o ax For x = M these two values of -_ are equal, and hence Inserting this value of C 2 and integrating again, we have, since y o for x = o, For x = / these two equations are equal, hence For x = /, y in the second equation is zero, hence Also for x = /, ~ = o. Hence From these two equations we obtain M l = Plk(i - ) 2 , and R^ = P( I - ) 2 (i -f 2k\ This is the same value for R l that we have already found (page 526) by the principle of least work. dy If we put the values of ~- o, we have for the value of x which makes the deflection a maximum when x < kl x = when x > kl 3 -2k' When x = kl in these equations the maximum deflection will be at the load and will be tne greatest possible. Putting, therefore, x^kl, we obtain from both these equations the condition *.i. 55 STATICS OF ELASTIC SOLIDS. [CHAP. V. That is, the greatest maximum deflection is at the load when the load is at the centre of the span. For any other position of the load the maximum deflection is between the load and the farthest end from the load. Inserting, then, these values of x in the values of y, we have for the maximum deflection in general i 2/W(i - Kf >- J= - I 2 3 / (3 _ 2 ^- Both of these are equal and have their greatest value when k = . Inserting this value of k, we have for the greatest maximum deflection at the load I 7^/8 when k = - A = - or only one fourth as much as for beam supported at the ends. For the load at the centre PI P M, = +^, and R^= -. For the points of inflection, if we put ~ = o, we have for the distance of the points tix ' of inflection from the left end kl (2 - K)l and x l+2k ' 3-2/1' ' If the load is at the centre of the span, k = -- and these values of x become - / and - /. 2 44 Examples. (i) A rectangular beam of wrought iron 5 feet long , 3 inches wide and 3 inches deep is deflected of an inch by a load of 3000 pounds applied at the centre. Find E. ANS. We have (page 543) PI* _ /Y />/ , hence E = Inserting numerical values ^ = 3000 x 60 x 6o_x_6oj x 60 x 3 x .2 ]Qooo / 4 / 4x2x3x27 Since the elastic limit (page 476) is 25000. the elastic limit is not exceeded. (2) An iron rectangular beam whose length is 12 ft., breadth l\ inches, coefficient of elasticity 24 ooo ooo, has a load of 10000 pounds at the centre. Find the depth in order that the deflection may be - of the length. PHAP v n EXAMPLES-FLEXURE. ANS. We have d* = -^L = I000 X 144 X 144 X 144 X 2 X 480 tfEA 4 x 3 x 24000000 x 144 =691.2 inches, hence d = 8.8 inches, provided the limit of elasticity is not exceeded. In order to find whether this is the case, we have as before C, _ Plv _ lOOOOX 144 X 4.4 X 12 X 2 ~ 4T - ~T^7^8^8T8^8- = l8595 P unds P Since the elastic limit is 25000 pounds per square inch, the elastic limit is not exceeded. (3) Find the depth of a rectangular beam so loaded at the centre that the elongation of the lowest fibre shall equal ~^ o of its original length. ANS. We have A = _L, and (page 477) A = ^. We also have (page 497) S f = ^. Hence Since v ,f= bd* and M , we have '2IOO/Y "T Eb ' provided that S f is less than the elastic limit. (4) Find the radius of curvature at the middle point of a wooden beam when the load at middle is 3000 pounds, the length to feet, breadth 4 inches, depth 8 inches and E is i ooo ooo pounds per square inch. ANS. We have (page 532) El 4Y 4x1000000x4x8x8x8 P = -irr = ^r = - -= 1896 inches, M PI 1 2 x 3000 x 1 20 provided the elastic limit is not exceeded. To find whether this is the case we have c Plv 3000x120x4x12 5 ' = 77 =! ' 4x4x8x8x8 = 2I 9 pounds per sc * uare inch - Since the elastic limit is 3000 pounds per square inch (page 476), the elastic limit is not exceeded. (5) A wrought-iron 13 inch I beam, whose moment of inertia is 6os in inches, has a length of 30 feet and E = 24 ooo ooo pounds per square inch. If supported at the ends with a uniform load of 7 5 pounds per inch of length over the first to feet, find the deflection at the end of the load. ANS. Deflection =0.23444 inch. Find the deflection at the centre. ANS. Deflection = 0.24421 inch. Find the deflection 10 feet from the unloaded end. ANS. Deflection = 0.19537 inch. Where is the point of greatest deflection and what is the greatest deflection ? ANS. At 13.1676 feet. Greatest deflection = 0.24847 inch. // the weight of the beam itself is 5.57 3 pounds per inch of length, find the deflection at the centre. ANS. Deflection = 0.07349 inch. If the same io-foot load is moved to the centre, find the deflection at the centre. ANS. Deflection = 0.50063 inch. If the uniform load of 73 pounds per inch covers the whole span, find the deflection at the centre. ANS. Deflection =0.98905 inch. If the same beam is half loaded with 75 pounds per inch, find the deflection at the centre, the maximum deflection and the point at which the deflection is a maximum. ANS. Deflection = 0.494525 inch. Max. deflection = 0.49855 inch within the loaded portion at 14.48 inches from centre. If the same beam has three weights of 4500 pounds each placed at intervals of bo inches beginning at one end, find the deflection at centre. ANS. Deflection = 0.6154 inch. 55 2 S7V/7YCS OF ELASTIC SOLIDS. [CHAP. V. If tht beam is fixed horizontally at both ends and loaded uniformly with 75 pounds per inch, find the deflection at 10 feet from either end and at the centre. ANS. Deflection = 0.1563 inch ; at centre = o. 19781 inch. If only one end is fixed, the other supported, find the dfflection at 10 feet; at centre; at 20 feet. Find ///< maximum deflection. IV here is it? ANS. At 10 feet = 0.39074 inch ; at centre 0.39563 inch ; at 20 feet = 0.27352 inch. Maximum deflec- tion = 0.41018 inch at 151.7524 inches from supported end. If the beam is fixed horizontally at both ends, with a load of 27000 pounds at the centre, find the deflection at the quarter points and at the centre. Where are the points of inflection ? ANS. At quarter points = 0.19781 inch; at centre = 0.39562 inch. Point of inflection at 90 inches froqa each end. If only the right end is fixed and the other supported, and the load of 27000 pounds is at the centre, finil the deflections d at the quarter points and at the centre, and the maximum deflection. ANS. At the quarter points 0.5316 and 0.3091 inch; at centre 0.69234 inch; maximum deflection = 0.70732 inch at /y from supported end.. CHAPTER VI. SHEARING STRESS. Shearing Stress in Beams. Let A& and A 2 B 2 be two consecutive cross-sections, so that the distance C^C^ , between them is ds, and let M l be the bending moment at C l and M 2 the bending moment at C r Then by equation (II), page 497, the unit stress S/ in any fibre at a distance v' is M?' and the unit stress S, in the same fibre at the other cross- C, section is The difference v-. is the horizontal unit shear at this fibre. But J/ 2 M l is */J/, and if Fis the shear, dM = Vds. Hence the horizontal unit shear at this fibre is Vv'ds and if the area of cross-section of the fibre is a, then the horizontal shear for this fibre is Vds . av' For the total horizontal shear at the neutral axis we have then Vds . 2av' H = ~7 the summation extending to all the fibres above C. Let be the breadth of cross-section at the neutral axis. Then b^ds is the area sheared, and the horizontal unit shear at the neutral axis is 553 554 STATICS OF ELASTIC SOLIDS. [CHAP. VI. Thus for a rectangular beam of breadth b and depth d we have 7 = , 2ai/ = x - = -g- and hence For a cylindrical beam of radius r, I and 2av = x 4 - = ^-, = ir, and hence Take the elementary fibre at the neutral axis of length ds and depth dv' and area of cross-section a b dv' '. Let S v be the vertical unit shear at the neutral axis. Then the vertical shearing force is _ v S v a = S v b dv' and the horizontal shearing force is S h b ds. * We have then for equilibrium S k d ds X dv' = S v b dv' X ds. Hence V2av' That is, the horizontal and vertical unit shear at the neutral axis are equal. Work of the Shearing Force in Beams. We have from equation (I), page 477, for the strain due to shear _ Sjts _ = - = where E, is the coefficient of elasticity for shear. If then V is the shear, we have the work of the shear between two consecutive sections V\ 2 = The total work of shear is then For straight beams we have s = /, and dx in place of ds. Influence of the Shearing Force upon Deflection. If we divide (X) by -, we have for the deflection due to the shear, for straight beam, deflection = and this d flection should be added to the deflection due to bending, as already found in the preceding pages. CHAP. VI] DETERMINATION OF COEFFICIENT OF ELASTICITY FOR SHEAR. 555 Thus for a beam of constant cross-section, fixed at one end, with a load P at the free end, we have already found the deflection due to bending (page 533) where E is the coefficient of elasticity for tension or compression. For the deflection due to shear we have, since V=P A = where E s is the coefficient of elasticity for shear. The total deflection is then For rectangular cross-section b n = b, I = bd z , 2av = , and hence 12 8 A AJP/ while for bending only we should have A = -75 If we assume the ratio = 3, we have . Ebd*_ 8/ 2 / A.P' 3 We have then for- = 10, ^ i.oi i ^r-. We see at once tixati. for depth small compared to the length, the effect of the shear can be disregarded. In all practical cases where / is greater than io A = cc-i is, by equation (I), page 477, Pdx *-~AE' If now the column deflects towards the left, the unit stress on the inner compressed fibre at a distance v from the axis is, from equation (II), page 497, P . Mv and hence the compression ad = A/ of that fibre is _Pdx , Mvdx -zr H m - 559 560 STATICS. OF ELASTIC SOLIDS. [CHAP. VII. The distance a^d is then Mvdx Ol d = \ -&- Tip. If now p = OC is the radius of curvature of the axis, we have at once, from the figure, p i dx A : : v : a^d t or "^--JE] " P / But, from (i), i -j-g = j. Hence I _ d*y Now, from Calculus, ;; ~ --% Hence (3) Comparing with equation (VIII), page 531, we see that when we take / x = / we have bending only without compression. If we multiply both sides of (2) by 2dy and integrate, we obtain, since M= Py, When y~A = the maximum deflection, ~- o. Hence C . , and we have dx dy Integrating again, we have When y = o, x = o. Hence C = o, and we have CHAP. V'll.] THEORY OF THE IDEAL COLUMN. 5 6i (a) COLUMN FIXED AT ONE END, FREE AT THE OTHER. For a column fixed at one end and free at the other we have from (3), when x = i lt y = J, and hence IP or, since Ia= ,/4/c 2 , where K is the radius gyration of the cross-section, P (b) COLUMN WITH Two PIN ENDS (Fig. i). In this FIG. i. FIG. a . FIG. 3. case we have only to make, in (3), y = A when x = We thus obtain (c) COLUMN FIXED AT ONE END, PIN AT THE OTHER (Fig. 2). In this case we have, in (3), y = A when x = - / r We thus have COLUMN FIXED AT BOTH ENDS (Fig. 3). In this case we have, in (3), y when x - / lt and hence GENERAL EQUATION. All these equations are of the form P (4) where we have for n the values^, \n, |TT, 4r for one fixed and one free end, two pin ends, 2 & one fixed and one pin end, and two fixed ends. p EQUATION OF THE ELASTIC CURVE. Substituting the value of -^ from (4) in (3), we have for the equation of the elastic curve and hence A ' "* = A sin -j- , dy nA -r = ~r cos dx / (5) (6) 5&* STATICS OF ELASTIC SOLIDS. [CHAP. Vlf. WORK OF P DURING BENDING. During the direct compression of the column from P P*l the length / to /, , the work done is (/ /J, or, from (i), A c . 2. J . / /: If now, when the column is pushed slightly to one side, it continues to bend and assumes a deflection A, we have, from equation (6), for the moment M at any point of the neutral axis M= Py = PA sin^. '\ From equation (IV'), page 522, we have then for the work of P during this bending C' l M*dx C l * I ^ = / J 2hl J DJ . . nx work of P during bending = I ^ = / - sin 2 -=- dx = 2EI / WORK OF P NECESSARY TO PRODUCE THE DEFLECTION A. The horizontal compo- dy Pdy dy nent of P at any point of the neutral axis is P~r- Its work is -y*- , and hence, from (6), we have work of P necessary to produce the deflection == / f- = cos* . where, as always, K Is the least radius of gyration of the cross-section, and c is a constant to be determined by experiment, depending upon the material and the end conditions. Since the column bends easiest in the direction of its least dimension, ive take for K the least radius of gyration. Equation (R) is Rankine's formula for long columns. It holds for all values of . P I We see that Rankine's formula gives = S t for- = o. The tangent at A (figure, page A K p i 556) is horizontal, and we have -7 = for - = oo . It therefore complies with the conditions A K for the average actual curve given on page 564. It is not so simple or easily applied as the straight-line or the parabola formula, and the experimental constant c must be determined before it can bo used in any case. Gordon's Formula. Since K is a function of the least dimension d of the cross-section, we may also write for the crippling unit stress where c is again a constant, to be determined by experiment. Equation (G) is known as Gordon's formula for long struts. It also holds for all values of -, and the same remarks apply as for Rankine's formula. Merriman's Formula. The equation of the curve AB (figure, page 564) has been assumed by Prof. Merriman (Engineering News, July 19, 1894) as identical in form with Rankine's formula. We have, then, S, Instead, however, of regarding b as an experimental constant, Prof. Merriman deter- mines b precisely as in the case of the straight-line and parabola formulas, by the condition of tangency. CHAP. VII.] ALLOWABLE UNIT STRESS FACTOR OF SAFETY. 5 6 9 We thus obtain where, as always, K is the least radius of gyration of the cross-section. Equation (M) is Merriman's formula for long columns. Like Rankine's fornvila, it complies with the conditions of the average actual curve given on page 564. It is preferable to Rankine's in that it contains no experimental constant. It is therefore probably nearer the true curve for an average actual column than any of the formulas thus far given. ' Allowable Unit Stress Factor of Safety. The preceding formulas will enable us to find tne crippling unit stress. In practice only a portion of this is taken as the allowable working unit stress. This portion is called the factor of safety (page 481). For quiescent loads (buildings, etc.y this factor is taken at/= 4 for wrought iron and steel, and/= 6 for cast iron and wood, For variable loads a variable factor of safety is used equal to f = 4 + - , for wrought iron and steel, f = 7 -I -- for cast iron, 2Oa f = 6 -I- - for wood, ^ 2Qd where /is the length in inches and d the least dimension in inches of the rectangle which encloses the given cross-section. We have then in general for the working unit stress where is found by any one of the preceding formulas, and the value of / is taken as just given. ^1 Examples. (i ) Let the ratio of the length of a steel column to the least radius of gyration of its cross- section A be - = 100, and to the least dimension of the enclosing rectangle be - = 25. Let S e = 40000 and K E jo oooooo pounds per square inch. Find the crippling and working unit stress by the straight-line formula. - ANS. We have, using the practical values of n on page 565, ^r = IO less than " V ^ fn a11 Cases ' Hence by the straight-line formula the crippling unit stress is 7i* Hence for Two pin ends ^ = 0.655, = 26000 pounds per square inch ; P M One pin end and one flat end -. = 0.695, = 27600 75 Two flat ends ^ = 0.725, = 28800 The factor of safety is / = 4 + ^ = 5-25- Hence the working stress in these three cases is S w = 4952, 5257, 5486 pounds per square inch. STATICS OF EL/tSTIC SOLIDS. [CHAP. VII. (2) Find the crippling and working unit stress by the parabola formula. ANS. We have by this formula 570 - j.- Hence for p Two pin ends ...................... . = 0.805", = 32000 pounds per square inch A p One pin end and one flat end ....... - = 0.845, = 33600 A Two flat ends ..................... ^ = o.865, = 34400 " " " " A The factor of safety is as before/"= 5.25. Hence the working stress in these three cases is 5 w = 6ioo, 6400, 6550 pounds per square inch. It will be seen that the values given by the parabola formula are greater than those given by the straight-line formula. For very perfect columns and load very accurately centred the oarabola formula results are preferable; for poorer columns, the straight-line. (3) Find tht crippling and working unit stress by the Merriman formula. ANS. We have by this formula P = S. _ A i + < 3' Hence for p Two pin eods ................... = 0.555, = 22000 pounds per square inch; fl p One pin end and one flat end ....... = o.6o5, = 24000 " " " * V '* p Two flat ends ................. ... -= 0.655, = 26000 " " " " A The factor of safety is as before/ = 5.25. Hence the working stresses in these three cases are 5, =* 4190, 4570, 4950 pounds per square inch. CHAPTER VIII. THE PIVOT OR SWING BRIDGE. Pivot or Swing Bridge. The pivot or swing bridge is a girder continuous or partially continuous over three or four supports. If over three supports, it is a pivot span. If over four supports, the length of the short centre span is the width of the turn-table, and loads in this span come directly on the turn-table and hence cause no stresses in the truss members. If the bracing is carried through the centre span as shown in the figure, it is evident that a load in one end span, as AB, tends to lift the span from the support C. It will be found in general difficult to hold the span down at C. For this reason the bracing in the centre span is omitted. The continuity in such case is partial, but the span can be held down at C. Centre Span without Bracing. By means of the principle of least work we can find the reactions at the supports for a load P placed anywhere, and then the stresses in the truss members for this load are easily found. ist. LOAD P IN SPAN AB = l v . Let the span be AB =-. /,, BC= / 2 , CD = / 3 . Let the load P be at the distance k^ from the left end, where k^ is any given fraction. Let J/ 2 be the moment on the left of the second S, I, t k T l r S 2 It s s b VD , 3 support. Let the pressure on the right of A be S lt on the left of B be S 2 ', on the right of C, S 3 , and on the left of D, S s '. < There are no braces in the span BC and hence no pressure on right of B or left of C, and the moment on left of C will also be M y the same as on left of B. Taking moments about B, we have Taking moments about D we have From these equations we have >, = ^=_.S,'. . (i) 571 STATICS OF ELASTIC SOLIDS. [CHAP. VIII. Equations (i) give the pressures at the supports in terms of M v If then we can determine J/ 2 , we can find these pressures. For any point distant x from A we have the moment, between A and P, M = - S,x = - P(i - > ; between P and B, M= - S,x + P(x - / ) = - - P( i - k,)x + P(x - /,). l i For any point between B and C we have For any point distant x from D we have, between C and D, M= S^x = 3, Let v be the lever-arm for any truss member, and M the moment at the centre of moments for that member. Then the stress in that member is . Let a be the cross- v section of the member, and s its length. Then, from equation (III), page 515, the work of straining that member is the total work of straining all the members is then work and this work must be a minimum by the principle of least work, page 517. We have then, in the present case, inserting the value of M just found, work = - /Xl - * + 2 - /M - + ^ - ~ ^ 2 Suppose the span AB to be fixed horizontally at B and the support at A removed, and let be the stress in any member due to a unit load at A, and / the stress in any member due to a unit load at P. Then we have uv = I X x, or - = K 2 CHAP - Vin -] SMNG BRIDGE. J ' If, then, we insert these values, and put ^^ = o, we have for which makes the work a minimum . -A/TI - *,)> 573 Q{ or, solving for M v . ..... ' ' '" Equation (2) gives the value of M v or the moment at the support B, for a load P at any distance k^ from the left end A in the first span. If M 2 is known, then equations (i) give the pressures at the supports. When these are known the stresses can be easily calcu- lated. Note that in equation (2) u is the stress in any member due to a unit load at A, and / the stress in any member due to a unit load at P, considering the span AB as fixed hori- zontally at B and without anv support at A . 2d. LOAD P IN SPAN CD = / 3 . Let the load P be in the third span / at the distance 3 /3 fr m support A where 3 is any given fraction. Then we have c ~ X-X 4 A JA 5 1+ AA 4 /\ Z\[ M . B M - c ^-H^H and proceeding now precisely as before, we find in the same way ^, = *y, /.,_/. ^ 11777- L TR (4) Equation (4) gives the value of 7J/ 2 , or the moment at the support B, for a load /* at any distance - 3 / 3 from the right end D in the third span. If M 2 is known, then equations (3) give the pressures at the supports. When these are known the stresses can be easily calcu- lated. Note that in equation (4) u is the stress in any member due to a unit load at D, and / the stress in any member due to a unit load at P, considering the span CD as fixed hori- zontally at C and without any support at D. 574 STstTICS OF ELASTIC SOLIDS, [CHAP. VIII. Special Cases. The preceding equations are general and include all cases. Thus if the two end spans are equal, we have only to make /, = / 3 = / and k l = k^ = k. If we have only two spans, we make / 2 = o. Hence we have THREE SPANS, END SPANS EQUAL. Making | S ' t* _ j ' /, = / 3 = / and , = 3 = k, we have the following A -- ' -- B * C * ec l uations: 1st. Load in span AB : - 2d. Load in span CD : And in both cases M t = 2d. Load in span BC : tfs . l*'*u*s (5) (6) M PI o a */ a (7) l *+t^'JL \ Two SPANS ONLY. Making / 2 WC AS/ *H t^* l s; have the following equations : 1 // II?* 1 1st. Load in span AB : A, B C 5 i = -y 2 + ^!-^i). ^=T 2+/ ^' 5,= -^'=^, . (8) (9) ), . . (10) M 2 = / 2 a _^^ *./. | s Two EQUAL SPANS. Making / 2 = o and I /, = / s = /, we have the following equations : C 1st. Load in span AB : 2d. in span BC : CHAP. VIII.] And in both cases SWING BRIDGE. 575 (14) Values Of a Indeterminate. It will be noted that our equations for M 2 require that the area of cross-section a for each member shall be known, while it is our object to determine these areas by first finding the stress in each member and then dividing this stress by the allowable unit stress. It is necessary, then, to make a first approximation by supposing a the same for each member. It will then cancel out of our equations for M y We can then find M 2 approxi- mately, and determine the stresses and corresponding areas a. With these values of a we can again determine J/ . A short example will thoroughly illustrate the application of our equations. Example. Find the stresses for a swing bridge of two equal spans /= So ft., each span divided into four panels of 20 ft. Let the spans be symmetrical, the centre height Be = 10 ft. and the height at ends bi = 7 //. I --- ...... ............. I *-- -k-l- V M, B For these dimensions the lever-arm v for any upper-chord member, as de, will not differ appreciably from the height ^3, and the length s for any upper-chord member, as cd, will not differ appreciably from the panel length. We have then the following lever-arms A\ 1-2 2-3 3-Z? be cd de Ab b\ e\ t2 d2 d$ e^ lever-arms 7 8 9 10 7 8 9 39.64 140 52 160 65.66 180 80.5 We can now calculate the stresses u in every member due to a unit load at A, and the stresses p in every member due to a unit load at P, considering the span AB as fixed horizontally at B and without sup- port at A . We can then form the following table : (0 (2) (3) (4) /i (5) /a (6) A (7) J A) /.* (9) tea (.0) / 163 25 20 250.00 6 666 883 88 CQ2 CO 2-3 3-Jf 20 8.000 _|_ o 8e7 6.000 4.000 2. COO 1280.00 163 35 " 960.00 64O.OO 320.OO 500 oo 250.00 _i_ 6 5^5 888 88 592 59 206 to 5. 14 7 1 2 6o2 114 64 *33 77 21.54 5 2 5 f-2 -4-2 di6 72 87 85 42 Q7 60 z-d 21.93 o 888 4 66 ^3 3-f eB 9 22.36 10 + 1.490 0.600 + 1-739 0.700 + 1-987 0.800 + 2.236 O.gOO 49.64 3.60 57-94 4.20 66.20 4.80 74-49 5.40 4833.90 2936.42 1406.53 399 39 576 STATICS OF ELASTIC SOLIDS. [CHAP. VIII. In the first column we place the designation of each member ; in column (2) the length s of each member ; in column (3) the stress u in each member for a load of unity at the end A, considering the span AB as a semi-girder fixed horizontally at fi\ in columns (4), (5), (6) the stress^ in each member for a load of unity at apex i, 2 and 3, considering the span AB as a semi-girder fixed horizontally at B. In column (7) we have the values of u*s for each member; in columns (8), (9), (10), the values oi PUS for each member. At the bottom of columns (7), (8), (9), (10) we give the summation of the values in each column. We have then, from equation (14), where k has the values , , at apex i, 2, 3, and the summation 2pus is given by columns (8), (9), (10). We have then for load P at apex i, from equations (12), M, = + 0.0712 PI, Si = + 0.67 S8P. For P at apex 2, Mi = + 0.1045 PI, & = + 0.395 5/>. For P at apex 3, M t = + 0.0836 PI, Si = + o.i664/>. For P at apex 4, S t will be the same as S for P at 3, or Si = 0.0836 P. For P at apex 5, 5i will be the same as S 3 ' for P at 2, or Si = O.IO45/*. For P at apex 6, Si will be the same as S* for P at i , or 5, = - 0.07 12P. Negative values denote that 5, acts downwards. We can now find the stresses in each member for each load, and then by tabulation can find the loading which gives the maximum stress and the maximum stress itself in each member. Solid Beam Uniform Cross-section. We can easily find, from the equations already deduced, the value of M 2 for a solid beam of uniform cross-section. Thus, let x be the distance of the point of moments for any member from the left end A, and v its lever-arm. Insert these values in equation (2) and we have = PL Now if the girder is a solid beam of uniform cross-section, we have s = dx, and at? constant. Hence we have - /o /*' JQ _ __ __ rw^+A 8 rv.*+f> r*** Jo Jo 's /o Af t = PI, If we perform the integrations, we obtain for load in span AB ff^L ( I5 ) CHAP. VIII.] SWING BRIDGE. 577 The pressures of the supports are then given by equations (l). In the same way we have, from equation (4), for load in span CD M 2 = Pl * k *( * ~_jyl (16) 2 (/i ~T 3fz ~r ^3) The pressures of the supports are then given by equations (3). For three spans, end spans equal, the pressures on the supports are given by equations (5) for load in span AB, and by equations (6) for load in span CD, and in both cases For two spans only we have for load in span AB M 2 = ^~~^~T~r\* 0&) 2 (A ~T~ ^2) The pressures of the supports are given by equations (8). -^ for load in span BC M 2 = The pressures of the supports are given by equations (10). For two equal spans the pressures on the supports are given by equations (12) for load in span AB, and by equations (13) for load in span BC, and in both cases (20) Example. In the example given on page 575, consider the girder as a solid beam of uniform cross-section. In this case we have, from equation (20), M* = zoPk(l -/&'), where k has the values , , 444 We have then for load P at / from left end, from equations (12), 4-687 $P, For load P at -/ from left end, For load P at / from left end, M a = + 6.5625P, 5, = + o.i68o/>. For P at f rom right end, from equations (13), M = + 6.5625^, 5, = - o.o820/>. For P at -/ from right end, 4 M t = + 7- $P, & = 0.0938^. For P at / from right < 1 , M = + 4.6875/ ) , Si = - o.o$86P. CHAPTER IX. THE METAL ARCH. Three Kinds of Metal Arch. We may distinguish three kinds of metal arch, viz., arch hinged at crown and ends; arch hinged at ends only; arch without hinges. If the arch is a framed structure, the stresses in the members can be found in any case, if for a given load we can find the horizontal thrust and vertical reactions at the ends and the moments, if any, which exist at the ends. Framed Arch Hinged at Crown and Ends. This form of construction is an arch only in form, but in principle is simply two braced rafters the thrust of which is taken by the abutments instead of by a tie-rod. It is therefore a very simple matter to find the end reactions for a given load. Let the span AB be /, and P the load at the distance /from the left end, where k is any given fraction. Let the rise, measured always from the chord AB to the hinge C at the crown, be denoted by r. Then taking moments about the right-hand hinge at B t we have for the reaction V^ at the left end for any position of P - k) = o, or V l =P(i ). (0 Taking moments about the hinge Cat the crown, we have for the horizontal thrust H at the left end, when kl is less than -, that is when P is on the left of the centre, )=o, or = V 2r PI 2r - 2k) = Pkl 2r ' When kl is greater than -, that is when P is on the right of the centre, VI 7V(i -*) or H =it = Tr (3) These values of V^ and H are independent of the shape of the arch. Change of temperature causes no stresses in the arch hinged at crown and ends. Each half is free to turn about the hinges and accommodate itself to any change of shape due to change of temperature. Equations (i), (2) and (3). hold for a solid as well as for a braced arch. We can then determine the stress in each member for each load, and then by tabulation can find the loads which give the maximum stress and the maximum stress itself in each member. 578 CHAP. IX.] FRAMED ARCH HINGED AT ENDS ONLY. 579 Framed Arch Hinged at Ends Only. In this case we have just as before, taking moments about B, for any position of P It remains to find the horizontal thrust H. Let v be the lever-arm for any member, as determined by the method of sections, page 401, and M the moment at the centre of moments for that member. Then the stress in that member is . Let a be the cross-section of the member, and s its length. Then from equation (III), page 515, the work of straining that member is The total work of straining all the members is then work = * and by the principle of least work, page 517, this work must be a minimum. Let x and y be the co-ordinates of the point of moments for any member, as deter- mined by the method of sections, page 401. Then for any member on the left of P we have the moment M= Hy V^x = Hy P(i k}x, . . and for any member on the right of P we have M = Hy Vjc + P(x /) = Hy P(i k)x -f P(x kl). We have then for the work of straining all the members work = ^\Hy- P(i - ^XP^^ + 2E {_Hy - P(\ - K)x + P(x -kl^J^p If we differentiate with reference to H and put the differential coefficient equal to zero, we have for least work d, de subtend an angle of 30 at the centre, while Aa, eB subtend an angle of 15. We can now find the stress h in each member for a nega- tive thrust of unity at the end A, and also the stress p in each member for a unit load at a, b d, e, considering the arch as simply supported at the ends. We can then draw up the following table. In the first column we have the members; in column two the CHAP. IX.] TEMPERATURE STRESS-FRAMED ARCH HINGED AT ENDS. 5 8r lengths s of the members ; in the third column the stresses h ; in the fourth column the quantities h*s, and in the following columns the quantities /,/fo, p^hs, p*hs, p^hs. The minus sign for a stress denotes compression, and the plus sign tension. h A*s jM* >** P*hs p _ is made as in the preceding example, / is the number of degrees o a rise or fall of temperature above or below the mean temperature of erection, e is the coefficient of expansion or the change of length per unit of length for one degree, and E is the coefficient of elasticity (page 478). The plus (+) sign is taken for rise of temperature, and the minus ( ) sign for fall of temperature. We have for one degree Fahrenheit : For cast-iron. ... e = 0.00000617, wrought-iron. . e = 0.00000686, steel e = 0.00000599. Values of E are given on page 478. Equation (3) requires that the areas of the members should be known in advance, whereas these are what we wish to find. We must in general, then, first assume a constant value fora in (3) equal, say, to the section at the crown. Denote this assumed constant section by . We have then for our first calculation CHAP. IX.] SOLID ARCH HINGED AT ENDS ONLY. 583 Example, In the preceding example, page 580, let the cross-section at crown be a a = 2 square inches Let the arch be of steel, and let us take = 30000000 pounds per square inch. Let the change of temper- ature be / = 40. Then from the table page 581 w.e have 2 o Ws = 635.297.; and since /= 76.21024 ft., we have for 0.00000599 Ht = 1728 pounds. The stresses due to this thrust can now be found. Solid Arch Hinged at Ends Only. For any load P we have, just as for the framed arch page 579, the left vertical reaction We have found on page 579, equation (2), for the horizontal thrust of a framed arch If the arch is a solid beam, we can put ds for s, and ai? = I = moment of inertia of the cross- section. Hence if x and 7 are the co-ordinates of any point of the neutral axis, we have This equation is general. If the moment of inertia / of the cross-section is constant, /cancels out. Instead of performing the integrations indicated in (2) we can in any case divide the neutral axis into a number of equal arcs of length s. We have then, since s cancels out, . T *. T 0/ 2 - . .... (3) If 7 is constant, it cancels out. From (i) and (3), then, we can find V l and H for any given load, and can then find the moment M at any point of the neutral axis for a load anywhere. Then by tabulation we can find the loading which gives the maximum moment at any point of the neutral axis, and this maximum moment M m . A *. itself. We have then, from equation (II), page 497, (4). where ^T max is in pound-feet, S f is the allowable unit stress in pounds per square inch in the most remotS fibre at a distance v infect, and / is given for dimensions in inches. 584 STATICS OF ELASTIC SOLIDS. [CHAP. IX. 'P. P, Example. Let us take a circular arch of radius of neutral axis 40 ft. and central angle 120, so that 69.282 ft. and rise r = 20 ft. We first suppose / constant so that it cancels out. Divide the neutral axis into a sufficiently large number of equal segments, say six, so that the length of a segment is s = 13.963 ft., and let the end segments Aa and Bf be each one half of s. Let the ordinates to the middle points a', V, r> p> T) 54-337 0.216 A 64.817 0.065 0.0657*.. [n order to find the summations 2 kl y(x kl} for each load we can draw up the following table. Pi Pi P* P* kl = 4.465 14-945 27.8 41.481 i k = 0.935 0.784 0.599 0.401 Hence from equation (i). page 583, the values of F t for each load are V\ = o.93$Pi, O.784/V O.599/V 0.40 i/V o.2\6P t , PI- fk ft A fk Ft X - kl. X-r -*/) X - kl. X-r - */) x- kl. y(*-kt). x-kl. y(- *0. x-kl. **-*/). x-kl. \Ax-U). ab be ed de . IX. at < 4 2 /.( y ) A,a - 3.2065 + 0.8801 118.0680 8.8968 ab - 3-2938 + 0.9041 247-0949 18.6170 bd 1-2053 - 0.3417 2-6593 de + 3 . 2947 + 0.9041 247. 2350 18.6170 5.3654 + I36.6IIO i -- J.874 + 3-2947 + 0.8801 124.6811 8.8 9 6S - 2.5669 + 1.0322 + 67.1200 26.9902 + 17-9295 7.2098 A if + 4.8012 - 1.3158 216.6358 16.2710 fg + -4733 + 1.3017 40.4494 31.5760 - -4741 + 1-3017 40.4930 31.5760 hi .4741 + 1-3017 40.4930 31.5760 iBt -8073 0.8115 217.6870 6.1888 - 3-9885 - 1-0437 + 152.1792 + 39.8248 + 25.6887 + 6.7227 Aif .0064 + 1.3116 13.2183 22.5580 f + -2735 - 0-3495 12.9744 .0-9772 fr + -4768 1.4300 135.5730 45-1920 tf - -2758 + 0.6747 0.6085 3-6418 g* + .8600 o 180.7691 dh - .7660 + 0.6747 4.6940 3-6418 di - .4768 - 1.4300 135-5730 45.1920 o + 135-5755 + 78-2755 n - . 2694 - 0.3495 12.8910 0-9772 0.3085 + 7.1350 + o. 1964 iB\ + .0064 + 1.3116 13.2183 22.5580 o + 13.2811 -fI3.28lI + I7-30S7 1-17.30*7 1835.5346 319.6127 I 12.-J 1M,; 0.0115 + 5II.90I8 + 26.1157 + 176.8662 - r6.8sx6 We have then, from equations (5), page 588 and the tables, 2". 8882 CHAP. IX.] TEMPERATURE STRESS FRAMED ARCH FIXED AT ENDS. 59' and hence for the apex loads A , A at apices d, e apex d M = 4. 2342 A We have also apex * - 0.004 A I835.5346 and hence for the apex loads A and A at apices d, e apex d Fx = +0.2788 A We have also apex* + 0.01427*4 2 U (* -J apex b + 0.55347*,, apex d + 0.5 534 A apex e + o.o 5 26 A apex b + 0.721 27*i apex d + 0.2788 A apex e + o.o 1 42 A apex b 26.7172 apex*/ apex e + 6.0801 7*9 + o.6ioi7* t O.O I 22 A 319-0127 and hence for the apex loads A and A at apices 'd, e apex d apex e H = + 0.5 5 34 A + 0.05 26 A We have the same values of H lor similarly placed loads at apices a, b. Hence apex a H = + 0.0526A Also, from (8) we have apex a F, = + 0.98 58 A From (6) and (7) we now have apex a (/-*/)= 6.9925 Mi = + 5.89827*. We can now find the stresses in each member for each apex load, and then by tabulation find the maximum stress in each member. Temperature Stress Framed Arch Fixed at Ends. The effect of a change of tempera- ture is to cause a horizontal thrust H t at the point O, or a horizontal thrust HI and moment M t at the left end A, where For the moment at the point '* of moments for any member we have We have then for the work of straining all the members work = If e is the coefficient of expansion and / the number of degrees rise or fall of tempera- ture, the change of length of the span is elt. The work is then H t elt work = Hence H t elt 592 STATICS OF ELASTIC SOLIDS. [CHAP. IX. or Eelt or, just as on page 588, Eelt Where considering the arch fixed at the right end and free at the left, h is the stress in any member for a negative unit horizontal force at the left end A, and m the stress in any member due to a negative unit moment at the point of moments for that member. 1 s The summation^ (h ymf- is made as in the preceding example. The plus(-}-) sign o is taken for rise of temperature, and the minus ( ) sign for fall of temperature. The values of e are given on page 582, and of E on page 478. The value of M t at the left end A is given by (i) when H t is known. Equation (2) requires that the areas a of the members should be known in advance. We must then in general assume a constant value for a in (2) equal, say, to the section at the crown. Denote this assumed constant cross-section by a . We have then for our first cal- culation Eeajt H t = (3) (h - ynifs Example. In the preceding example, page 589, let the cross-section at crown be a = 2 square inches. Let the arch be of steel, and let us take E = 30000000 pounds per square inch. Let the change of temper- ature be / = 40. Then from the table page 590 we have ^> (h ~ym?s = 319.6127, and since /= 76.21 ft., we have for i e = o.ooooo 599, from (3), Ht = 3428 pounds, and from (2), since jT ;= 10.4433 feet, Mt = T 35800 pound-feet. Solid Arch Fixed at Ends. If the arch is a solid beam, we can put in equations ^4), page 588, ds for s and / for av*, where 7 is the moment of inertia of the cross-section. Hence if x and y are the co-ordinates of the neutral axis, we have for a load P anywhere on a sym- metrical arch A*-*)* ~ /o y ds yds ? (0 CHAP. IX.] SOLID ARCH FIXED AT ENDS. 593 These equations are general. If the moment of inertia / is constant, it cancels out. Instead of performing the integrations indicated inequations (i), we can in any case divide the neutral axis into a number n of equal segments of length s. We have then, since s is constant, or, if / is constant, _ P where n is the number of segments; = -P If /is constant, it cancels out in these two equations. or, if / is constant, y = = 2 / where n is the number of segments. From (i) or (2), then, we can find M and V l and H at the left end A of the neutral axis for a load P anywhere on the right-hand half of a symmetrical arch. We have then for the moment M^ at the left end for a load on the right-hand half, just as on page 589, (3) and for the moment M 2 at the right end, or for a similarly placed load on the left-hand half, the moment M,' at the left end, (4) 594 STATICS OF ELASTIC SOLIDS. [CHAP. IX. The reaction l\' at the left end for a similarly placed load on the left-hand half is (5) V P V y \ * v r The value of H is the same in both cases, whether the similarly placed load is on the right- or left-hand half. From equations (i) or (2) and (3) we can find, then, J/,, V l and //at the left end for a load anywhere on the right-hand half of a symmetrical solid arch. From (i) or (2) and (4) and (5) we can find M t ' t V{ and //at the left end for a load anywhere on the left-hand half. We can now find the moment M at any point of the neutral axis for a load anywhere. Then by tabulation we can find the loading which gives the maximum moment at any point of the neutral axis, and this maximum moment M m ^ itself. We have then, from equation (II), page 497, r (5) where ^/ max . is in pound-feet, S/is the allowable unit stress in pounds per sqiiare inch in the most remote fibre at a distance v in feet, and /is given for dimensions in inches. Example. Let us take a circular arch of radius of neutral axis 40 ft. and central angle 120, so that the span / = 69.282 ft. and rise r = 20 ft. We first suppose /constant, so that it cancels out. Divide the neutral axis into a sufficiently large number of equal segments, say six. so that the length of a segment is s= 13.963 ft., and let the end segments Aa and fit be each one half of s. Let the ordinates to the middle points a', b', tf, etc., of each segment for origin at A be x, y, and take thejloads A, P\, etc., acting half way between A and a', a' and b', etc. We can now draw up the following table. X y / jr 2 ('-;)' y -y OK -7)" Aa 1.875 2.943 - 32-766 1073.611 10.291 105.005 01 be fd de (y-y)(x-kl) x - kl H<~"> (y-J)(.x-kl) x - kl H<- (y --&*-*!) de (y - yJ S Example, In the example page 594 let the moment of inertia at the crown be 2000 in. 4 . Let the arch be of steel, and let us take E = 30000000 pounds per square inch. Let the change of temperature be / = 40. Then from the preceding example we have ^ (y yf s = 4714.18, and since /= 69.28 ft., we have, from (4), for e= 0.00000599 H t = 1467 pounds. Since y = 10.38 ft., we have, from (3), Mt = T 1 5227 pound-ft., the top signs being taken for expansion and the bottom signs for contraction. CHAPTER X. THE STO NE ARCH. Definitions. The stone arch consists of a number of arch-stones or voussoirs which press upon each other. The central one of these is the keystone. The extrados is the exterior outline of the arch proper. The intrudes is the interior line, and the corresponding surface of the arch is the soffit. The sides of the arch are the haunches, and the spaces above are the spandrels. The ends of the arch or the area between in trades and extrados are \\\Q faces. The inclined surfaces or joints upon which the arch rests at the ends are the skewbacks. The permanent load supported by the arch in addition to its own weight is the surcharge. The masonry or other material which supports two successive arches is the pier; at the extreme ends this is the abutment. Thus in the figure cdd ' c' is a voussoir or arch-stone, and k is the keystone. The line abed, etc., is the extrados, and a'b'c'd' ', etc., the intrados, and the interior surface corre- sponding the soffit. The line rrr marks the upper limit of the surcharge. Between this and the haunches on either side is the spandrel space, and the material with which this space is filled is the spandrel filling or surcharge. The area between the extrados abed, etc., and the intrados a'b'c'd' , etc., is the face, and aa' is the skewback. At A and B \ve have the abutments or piers. Arch and abutments or piers are stone. The surcharge may be stoue or filled in with rubble or lighter material. The upper limit of surcharge may be level or on any desired grade. The extrados and intrados may be circular, elliptic or any desired curve, and may or may not be parallel. Often the depth at key is less than at ends. In all investigations and calculations we suppose the width to be one foot and the effect of mortar between the joints is disregarded. The neutral axis A CB or centre line is the curve passing through the centre of the voussoirs. The rise r of this axis is the rise of the arch and the span / of this axis is the span of the arch. 597 598 ST/1TICS OF EL/1 STIC SOLIDS. [CHAP. X. Reduced Surcharge. The arch proper is constructed of cut stone. The material above it may also be of stone or of some lighter material. The density or weight per cubic foot of the surcharge is then in general less than the density of the arch proper. Thus in the figure let, abed, etc., be the extrados, and rrr the roadway. Between the roadway and the extrados the surcharge may have a less density than for the arch proper. Say, for instance, that the density of the surcharge is of that of the arch. r. Then if we jflf^"*^ 2 "* draw verticals aa', bb' , cc' , etc., and lay off aa" equal to of aa 1 ', bb" equal to of bb' and so on, we obtain the line a"b"c", etc., which marks the limit of the reduced surcharge. We can then treat and discuss all the area below this line as if it were homogeneous and of the same density as the arch itself. Pressure Curve. Let Fig. (a) represent one half an arcli with its surcharge rr, and a'f the line of reduced surcharge, so that all the area below can be considered t as homogeneous and of the same den- sity ai the arch. Let us take the width as one foot, and divide the area into a suit- able number of slices by vertical lines. In this case we have five. The weight and centre of mass of each slice can be found. Let I, 2, 3, 4, 5 represent the weights acting at the centres of mass. Let H be the hori- zontal thrust at the crown due to the pressure of the other half of the arch. Let the magnitude and point of action a of // be known. In Fig. (b) lay off the weights I, 2, 3, 4, 5 to scale, let Oil be the known thrust to scale, Then O\ is the resultant of H and weight i ; (92 is the is the resultant of (?2 and weight 3. and so on. and draw Oi, 02, #3, 04, O$. resultant of O\ and weight 2 ; In Fig. (rf) produce //acting at a till it meets weight I. From I draw 1-2 parallel to Ol till it meets weight 2 ; from 2 draw 2-3 parallel to 02 till it meets weight 3, and so on CHAP. X.] CONDITIONS OF STABILITY OF THE ARCH. 599 We thus have a polygon a I2$4.$f, each segment of which is in the direction of the resultant of the forces acting at its end. Thus the resultant of H and weight I is in the direction 1-2, the resultant of Oi and weight 2 is in the direction 2-3 and so on. As we increase the number of divisions this polygon approaches a curve tangent at the points of division a, b, c, d, e, f. This curve is the curve of pressures in the arch. Conditions of Stability of the Arch. The conditions for stability of the arch are the same as for walls, page 425. Thus: ist. The joints of the voussoirs should be so arranged that the tangent to the curve of pressure at each joint shall make an angle less than the angle of friction with the normal to the joint, otherwise there is danger of sliding. 2d. The curve of pressures must lie entirely within the arch,' otherwise there is danger of rotation. 3d. The curve of pressure must not approach too near the edge of a joint, otherwise there is danger of crushing. Let d be the depth of any joint, N the normal pressure on the joint, and e the least edge distance of N. Then for a width of one foot we have, as already shown page 426, for the maximum unit pressure/, when e is greater than -d, p = 7- (2 -j j ; I 2N when e = -a, p = -j- I 2N when e is less than -d, p . o In any case the value of / must not exceed the allowable compressive unit stress C, which for stone may be taken at the average value of 25 tons per square foot or 50000 pounds per square foot. (See table page 424.) VALUE OF N. If we make the joints nearly at right angles to the curve of pressure, the first condition of stability is complied with, and we have with sufficient accuracy, if we denote by P n the sum of all the loads between the crown and any joint on the right, VALUE OF e. It Mis the moment at any point of the neutral axis, then ^ is the distance of N from that point. If we subtract this from ^ we have for the edge distance c' from the intrados, with sufficient accuracy, If ,' at any joint is negative or is positive and greater than 4, the equilibrium cun e runs outside of the arch. If M is negative, / is greater than - and the least edge distance is II Mi* positive, / is less than j and the least cd-e distance is e . V, = + 0.98477', +o866 9 /> 3 + 0.6377 K + o. 3 62 3 /> 4 + o.i3 3 i7>. +0.01537'. H = + 0.06587 3 , +o. 4 o8i7>, +0.7513^ + o. 75 r 3 7\ + 0.4081 P. + 0.06 5 87>. 602 STATICS OF ELASTIC SOLIDS. [CHAP. X. Let us first assume the constant depth at 2 feet. If we take one foot width, we have for the loads J\ = P. = 25966, P* =7' = 16253, Pt = P t = 7169 pounds. Hence 2Afi = + 23087 pound-feet, 2 V\ = + 49388 pounds, 2H = + 27455 pounds. For the values of 2 f ( x ~ M) we have for the points of division A, a', b' t c f , d'. e 1 ', B 4.4647*1 16.4957*1 + 6.0157*, 30.1767*, + 19.6967*, + 115912 526071 1152712 + 33-377A + 27. 3637*, 1877432 55.881 A + 51.422(7', + P) 2655412 2 o P(*-W = 69.282(7', + P, + 7 1 ,) 3421700 We have then from equation (8), page 601, for the moment at any point of the axis M = + 23087 - 49388-r + 27455^ + ^ 9 P( X ~ ft). For the points ^ ', d' , /, B we have then x = 34.641 = 20 d' 48.322 17.588 60.3525 10.642 B 69.282 ft. o ft. M = + 14050 3130 10014 + 23087 pound-feet. From equation (2), page 599, we have then d' 28375 B 56506 pounds. N = 27455 For the edge distance of N from the tntrados we have now, from equation (3), page 599, e 1 0.488 1. 1 1 1.27 and hence the least edtfe distance of N at each point is t = 0.488 d' 0.99 0.73 B 0.592 ft,, B 0.592 ft. CHAP. X.] THE STONE ARCH EXAMPLES. 603 We have then, from equations (i), page 599, for the maximum unit pressure p . = 8.93 x 400 = 3572 pounds, 7*3 = 7*6 = 12.03 x 4 = 4812 pounds, P, 7*4 = 13.681 x 400 = 5472 pounds. Let us find the moment at c 1 and B due to each of these loads. For the moment at c' we have for any load on left of c f , if r is the rise of the neutral axis, and for any load on right of c 1 M = Mi - + Hr. For the moment at B we have for any load M - Mi - VJ + P(l - kl). We have then at the crown c' M = + 2.8467*1 - 0.98477*. x 34.641 + 0.06587*. x 20 + 7*. x 30.176 = + 810 pound-feet, M = + 2.8427*, - 0.86697*, x 34.641 + 0.40817*, x 20 + 7*, x 19.696 = + 3220 M = - 1.1387*. - 0:63777*, x 34-641 + 0.7513^3 x 20 + 7*. x 6.841 = -- 7450 M = - 3.8387*4 - 0.36237*4 x 34.641 + 0.75137*4 x 20 = - 745 M 2.882 A -0.13317*5 x 34.641 + 0.40817*5 x 20 M- - 0.5587*8 - 0.01537*. x 34.641 + 0.06587*. x 20 = + 8l 6o 4 STATICS OF ELASTIC SOLIDS. [CHAP. X. at the springing 7? M = + 2.8467% 0.98477*, x 69.282 + Pi x 64.817 = 1996 pound-feet, M = + 2.8427% - 0.86697% x 69.282 + 7% x 54.337 = - 13866 M = 1.1387% - 0.63777% x 69.282 + 7% x 41.482 = 20997 " M = - 3.8387% - 0.36237% x 69.282 + 7% x 27.8 = - 6232 J/= 2.8827% - 0.13317% x 69.282 + Pi x 14.945 = + 13674 M = 0.5587% 0.01537% x 69.282 + 7% x 4.465 = + 10169 Since for the surcharge alone we have at c", M = + 14050, we see that there can never be a negative moment at c', and the maximum moment at c 1 is when the live loads Pi, 7%, 7%, 7% only act together with the surcharge, and is A/mm*, at Aa 2.n 2.70 68.89 4746.0321 20.69 428.0761 at 9.04 10.29 - 61.96 3839.0416 13.10 r7I.6TOO cd de 19.56 44.00 57-34 19.07 26.13 31.29 34-44 ~ 51-44 27.OO - 13.66 2646.0736 1575.2961 729.0000 186.5956 - 4-32 h 2.74 + 7-90 + 11-05 18.6624 7.5076 62.4100 I22.IO25 JS 71.00 35-50 + 12. II 146 6521 hi ik kl Im 84.66 98.00 110.69 122.44 132-96 34-44 31.29 26.13 19.07 10.29 + 13.66 + 27.00 + 39-69 + 51.44 + 61.96 186.5956 729.0000 1575.2961 2646.0736 3839.0416 + H.05 + 7.90 + 2.74 - 4-32 13.10 122.1025 62.4100 7.5076 18.6624 I7I.6IOO mB 139.89 2.70 + 68.89 4746.0321 2O.69 428.0761 280.64 22698.0459 I339-3032 We have then 2^y = 280.64. Note that in taking this summation and the other summations of the table, since the end segments Aa and mB are only half length, we take in the summations one half the values for Aa and mB. We have then 280.64 y = ^ = 23-39 ft., and can now fill out the last two columns. For the values of kl, (i K) and /(i - K) for each load we have Pi P* P* P t P* P, Pi kl= 4.52 14.30 25.435 37.655 50.67 64.17 77.83 /(i ) = 137.48 127.70 116.565 104.345 91.33 77.83 64.17 i k = 0.968 0.899 -82 1 0.734 0.643 0.548 0.452 We can now draw up the following table for the loads P-, to Pi, P, 91-33 50.67 -357 P Pi. P.i P tt 104.345 116.565 127.70 137.48 37.655 25.435 14.30 4.52 0.266 0.179 - I0 i 0.032 />, f, f x-kl H<~> (y -yte - *0 X - kl H)*-> (>-(*-*/ X-kl (*-;) fcr-*0 O - >x* - *f) gh hi ik kl Im mB 6.83 20.17 32.86 44.61 55-13 93.2978 544.5900 1304.2134 2294.7384 3415.8548 + 75.4712 + 159.3430 + 90.0364 - 192.7152 722.2030 6.67 19.36 31.11 41.63 180.0900 768.3984 1600.2984 2580.3948 + 52-6930 + 53-0464 I34-3952 - 545-3530 6-35 18.10 28.62 252.0315 93L0640 W-2952 + I7.399C - 73.1920 - 374-9220 62.06 4275.3136 1300.7776 48.54 3343.9206 1017.3984 35-54 2449.039^, - 745-1280 + 190.63 +9790.3512 1240.6564 +123.04 +6801.1419 1080.7080 + 70.84 +4180.9105! 808.2700 /> **n A x - kl H<-*> (y-yte -W x-ti H<-o (,-yXx-M X - kl H)<~, (r-JX-r- */) kl Im mB 5-87 16.39 302.0702 1015-5244 - 25.3584 2I4.7090 5.26 325.9096 - 68.9006 23-32 1606.5148 - 488.7872 12. 2O 839.7690 - 255.5024 2.40 1 166.0248 - 50.5136 + 33-32 +2120.8520 484.4610 + 11-36 + 745-7941 - 196.6608 + I.2o| + 83.0124 1 25.2568 6o6 STATICS OF ELASTIC SOLIDS. [CHAP. X. Note that in taking the summations, since tnB Is of half length, we take one half the values for mB in summing up. We have then, from these tables and equations (4), page 600, Af t = 15.8867%, 10.2537*. , 5.9037*., 2.8277*1., Vi = + 0.4317*7, + 0.2997%, +0.1847*., H + 0.9267*1, + o.8oi/%, +0.6037*., and from equations (5), (6) and (7), page 601, Mi = 6.9447%. 7-759^. 6.9437%. Mi'= - 10.927*., -7.3067%. -2.357%. V\ = + 0.5697*., + 0.7017%, + 0.8167*4, Since the depth is 2.5 ft. and density 160 Ibs. per cubic foot, we have for one foot width the loads 7% = 7%, = 48200, 7*, =7%i = 40000, 7% 7*,o = 29400, 7*4 = 7*. = 19100, 7% = 7*. =i looo, 7*. = 7*i = 6600. Hence 2Af t = + 272750 pound-feet, ^F, = 154300 pounds, ~2.H = 87750 pounds. 9870685 2.8277%. , 0.9477%,, o.ioo7%,; + 0.0937%. , + 0.0337*. , . + 0.0047*,,; + 0.3627%., + 0.1477%,, + 0.0197*1, ; 4.6917%., 2.0427% ,. o.26o/',,; + 2.8427%, + 5-53A . + 3432/>, 1 + 0.9077%, + 0.9677',, + 0.9967V For the values of 2* P (x /) we have for the points of division A, a', &, etc., A a' V (f d' / /' 2?p(x kr) = o 217864 935328 2144550 .3758000 5654936 7717664 h' i k' I' B = 12090728 14392982 16809712 19338720 21910600 We have then, /rom equation (8), page 601, for the moment at any point of the axis M = + 272750 - 154300* + 877507 + 2^ P(x kl). For the points/',^', K ', etc.. we have then 1 10.69 y = 35-50 34.44 31-29 26.13 M = + 150500 +102500 12500 From equation (2), page 599, we have then g 1 84.66 h 1 98 122.44 132.96 142 19.07 10.29 1 21000 -136650 -1300 +272750. g 88010 h' 89550 f 955 k' 109860 /' 137685 B 177510 N- 87750 For the edge distance of N from the intrados we have now, from equation (3), page 599, = - 0.47 + 0.08 k' + 1.39 + 2.52 t' + 2.50 /' B 1.26 0.29 We see that the curve of equilibrium passes outside of the arch and below the axis at the crown /' and springing 77, and outside of the arch and above the axis at /'. The arch is then unstable and will fall, the joints at the crown and springing opening at the extrados, and at /" opening at the intrados. In other words, the haunches sinking and the crown rising. This is precisely what happened when the arch was erected. In order to make the arch stable we should malte one of two changes in the design. We can either increase the depth or make the surcharge lighter over the haunches, by building up the surcharge with hollow spaces at the haunches or lightening the surcharge there l>y filling in with gravel instead of stone. Tlie arcli was actually rebuilt with hollow spaces in the surcharge over the haunches. CHAP. X.] THE STRAIGHT ARCH. 607 The Straight Arch The straight arch is a stone beam fixed at the ends, subjected to compression and bending. The beam is composed of voussoirs and therefore will not resist tension at any joint. Let the loading be uniform and equal to w pounds per foot of length, and the length of the span be /. We have at the left end the reaction wl Pi = and the thrust H acting at the distance y^ below A. The moment at any point of the axis is then wx 1 2 ' The work is work r' =: / -/O C*- r-^-"::---/ \ -2EA+ V \ r i r wx J L-^-T / If we differentiate with respect to H and put ^ ' = o, we have for the value of H which makes the work a minimum ..dx If we substitute / = Ax 2 , where /c 2 is the square of the radius of gyration of the cross- section A, and integrate, we have 1 TT ~* ./I / \ * = o, or -" = , , . av (0 Let the angle of the ends with the vertical be , and let the ends be at right angles to the curve of equilibrium. Then the normal pressure on the ends is _ H cos of and the end areas are A, = ^-^. Hence = -^-, and if we take the breadth unity We have then from equations (i), page 599, for the maximum unit pressure p at the end where e is the edge distance of H, or, disregard signs, e = - - y 6o8 STATICS OF ELASTIC SOLIDS. [CHAP. X. Hence The work will be a minimum when/ is a minimum. If we differentiate and put y = o, we have for the value of _y t which makes/ a, mini- mum, since /c 3 = , Substituting in equation (i), we have, since yc 2 = , *=--. ... ; Let the maximum allowable unit stress be S f , then we have w/ 2 (^3~-0 __._^ or rf= Equation (c) gives the depth pounds per foot will -be given by wu v The corresponding strain A. is then (page 477) A = a where s is the length of truss member, a its area of cross-section in square inches and E its coefficient of elasticity in pounds per square inch. If s is taken in feet or inches, A will be given in feet or inches. Let / be the stress in pounds in the member due to one pound placed at the centre of the truss. That is, / is the stress in pounds per pound of load at centre. Then the work on the member due to this load is /A. i pound X = p X i pound. The total work on all the members is then Work = -SE X T P und - CHAP. XL] OLD THEORY OF SUSPENSION SYSTEM. 613 But if A is the deflection at the centre the work is also i pound X - Hence we have A w ^ n i pound X - = ^Sr- X I pound, or A = ^>~f-, .... (7) where E and a are as already specified. If s is taken in feet or inches, A will be given in feet or inches. Temperature Load for Truss. When the cable expands or contracts the centre falls or rises a distance d t given by (6), and the centre of the truss falls or rises with the cable a dis- tance given by (7). Let w t be the uniformly distributed load in pounds per foot which would cause this deflection. Then, equating (6) and (7), we have Old Theory of Suspension System. The theory of the suspension system heretofore in use is due to Rankine, and is based upon the assumption that the cable carries the entire load, dead and live, the office of the truss being simply to distribute FIG. i. a partial loading over the cable, and thus prevent change of Slla P e - A A A A A A A* A jjt^TJ^t MAXIMUM SHEAR IN TRUSS OLD METHOD. Let the A y T TTTTTT I r / "'^B uniform live load w for unit of length extend over the distance jt~~~ s from the right end (see Fig. i). Then the load is wz, and, since by assumption the cable carries all this load, the upward load on the truss due to the cable is wz or, -^ for unit of length. Let R A be the reaction at the left end A of the truss. We have, taking moments about the right end B, c wz* wz(cz} -R A c- w *.- + = 0, or R A = ^ Since this is negative, the truss should be tied down at the ends. If the load ' extends over the distance z from the left end (Fig. 2), we have FIG. 2. p-4 1 tmttftj je:: ,^^. . . . . w In the first case (Fig. i), when the load comes on from the right, we have for the shear at any point distant x from the left end : wz when x is less than c z Shear = R A -}- ~~^~ x '* when x is greater than c z Shear = R A + -~x w\x~ (c z~)~] ; 614 STATICS OF ELASTIC SOLIDS. [CHA1-. XI. or, inserting the value of R A from (i), when x < c z Shear = ~ \2x (c ~) ; when x > c z Shear = 2x (c z) j iv\x (c *)]. From the last of these equations we see that the shear is a positive maximum when the last term is zero or when z = c x. That is, the shear for the unloaded portion is a positive maximum at the head of the load. From the first of these equations we have the shear a negative maximum when z = x. That is, the shear for the unloaded portion is a negative maximum at any point when the distance covered by the load is equal to the distance of the point from the centre. Inserting these values of z = x unloaded portion distant x from the left end : Inserting these values of z = x and z = c x, we have for any point of the w (c x \x maximum positive Shear = -\- - , unloaded portion -j c \ 2 j- (3) [ maximum negative Shear = . In the second case (Fig. 2), when the load comes on from the left, we have for the shear at any point distant x from the left end : when x < z Shear = R A -j " x wx, or. inserting the value of R A from (2), when x < z Shear = \2x 4- (c zV\ wx. 2C l This is a negative maximum for z = x. That is, the shear for the loaded portion is negative maximum at the head of the load. It is a positive maximum when z (- x. That is, the shear for the loaded portion is a positive maximum at any point when the distance between the point and the end of the load is equal to the half span. Inserting the values of z = x and z = -{- x, we have for any point of the loaded portion distant x from the left end : (4) loaded portion j w(c maximum negative Shear = CHAP. XI.] OLD THEORY OF SUSPENSION SYSTEM. 615 We see from equations (3) and (4) that we have for the maximum shears for I I x = o c -c 4 2 c , , we ywc we Shear== y i_ _ That is, the maximum shear is practically constant and varies but litle from -5-. o It is therefore customary by the old method to design every brace for the maximum shears due to live load. we Shear = y (i) MAXIMUM MOMENT IN TRUSS OLD METHOD. For the moment M at any point of the unloaded portion (Fig. i) distant ' x from the left end, if w is the uniform live load for unit of length, we have A * " 2C ' or, substituting the value of R A from (i), M = -^[_x - (c - z}} (5) For any point of the loaded portion (Fig. 2) we have wzx* wx* M R A x -1 -, 2C 2 or, substituting the value of R A from (2), In (5) M o for x = c z, and in (6) M= o for x = z. That is, the moment at the head of the load is zero. Also, if* is less than c z in (5), the moment is positive, and if greater than c - z, the moment is negative. The head of the load is then a point of inflection, and the loaded and unloaded portions may be considered as simple trusses uniformly loaded. The greatest moment for each portion will then be at the centre of each portion. Making, then, * - in (5) and * = - in (6), we have for the moment at the centre of each portion wz(c , x and --(<-*) _ These are a maximum, respectively, for z = -c and z = - 616 STATICS OF ELASTIC SOLIDS. [CHAP. XI. Hence the maximum positive moment is at the middle of the unloaded portion when tlie load extends over one tliird tlie span, and tlie maximum negative moment is at the middle of the load when // eovers two thirds the span. We have then for the maximum positive moment at any point of the unloaded left half span, by putting z = -c in (5), (7) and for the maximum negative moment at any point of the loaded left half span, by putting = -cin (6), (8) From equations (7) and (8) we have the maximum moments for * = \ c r ? we 1 we* we* M = T* 54 ^ I 1VC* That is, the maximum moment beyond -c is practically constant and varies but little from . It is therefore customary by the old method to design every chord panel for the maximum moments due to live load Temperature Load for Truss. From (I) and (II) we can then easily find the area a of each truss member due to live load. Thus for straight truss of height /r, if tr is the working stress, we have for the area a of the chords we* = 54*? ;-.' ......... (9) and for the area of a brace which makes the angle # with the vertical We have then from (8), page 613, the temperature load per unit of length, (in) Stress in Truss. This temperature load should be taken into account together with the live load in finding the maximum truss stresses. We have then to add to the shear and moment given by (I) and (III) the shear and moment due to the temperature load w t . The actual stresses in the truss members, then, are greater than those due to the live load only, and hence the areas assumed in (9) and (10) are too small i:ml the corresponding value of w t CHAP. XL] SUSPENSION SYSTEM. CABLE STRESS AND AREA. 617 given by (III) is too small. We should therefore assume w t somewhat larger than given by (III), and then find the stresses for this assumed w t and the live load. The corresponding areas should, when inserted in (III), give us pretty closely the value for w t we assumed. If not, we can make another approximation. Cable Stress and Area. We can now find the stress and area in any cable segment. Thus let the dead load per unit of length be w , the live load w, and the temperature load w t , assumed as above. Then the total unit load for the cable is (w -\- W Q -}- w t }, and from (5) we have for the stress in any segment whose length is s c n(w -f- Stress in segment ~s or (IV) If the working stress for cable is -f- w^). (V) If aa\ : 13 9 + 5.6*5 -16.875 + 0.625 -0-375 + 26.367 + 56.953 f 52.734 + 56.953 aia t 9 27 -0.75 + 182.25 + 182.25 a**, 9 -30.375 1. 125 + 307-547 + 307.547 AAi 9 o o o o AiA* AiA* 9 9 + 16.875 + 27 + 0.375 + 0.75 + 56.953 + 182.25 + 56.953 + 182.25 + "85.257 For each member in the half truss we give in the table the area a in square inches already found, the length s in feet, the stress in pounds due to a uniform load of one pound per foot of length or 9 pounds at each lower apex, and the stress p in pounds due to one pound at the centre-line apex. In the last column we have then the quantities fu t s and <^L for each member of the half span. CHAP. XL] SUSPENSION SYSTEM. EXAMPLE. 619 The table gives us for the half span ^-^- = + 1185.257, and hence for the whole span Inserting this and the value of 2sy one truss and one cable. CABLE STRESS AND AREA. We have found already for the lengths of the cable segments A'a' = Vto6, a'b' = Vyo, V -^, ^ where a e is the area of cross-section of cable segment, and E e is the coefficient of elasticity for the cable. Work on Truss. The truss is subjected to a uniformly distributed upward load due to 0/> the cable of pounds per foot of length, and also supports a load P at any distance z from the left end. Let be the stress in pounds in any member for a uniformly distributed load of one fhP pound per foot of length. Then the stress for pounds per foot will be Let / be the stress in pounds in any member due to one pound at the distance z from the left end. Then the stress due to P will be Pp. The stress, then, in any member can be written Stress =P P - <^ =P ( f -^}. . . . ..... (4) where the stresses u and / are to be inserted with their proper signs (-[-) for tension and ( ) for compression. Now the work of the member is, from page 515, (Stress) 2 .? Work = = - F^ , 2Ea where s is the length, a the area of cross-section and E the coefficient of elasticity. Insert- ing the value of the stress just found, we have for the work of all the members, or Work of truss = P~^\p ) ^-. ... (5) ^r c i 2Ea Value of 0. We can now find the value of or that fraction of the load P carried by the cable. Thus, adding the works given by (i), (3) and (5), we have Total work = + s c * / \ 2 s + ^*2(/ ~ TV 25' The value of is that which makes this work a minimum. P carried by the cable which acts as a uniformly distributed upward load on the truss. This gives an upward P apex load at every apex of -- . We can now find and tabulate the stress in every truss member due to each apex live 4>P load P and the upward apex load -- at every apex, and thus obtain the maximum live-load stresses. To these must be added the stresses due to temperature load w t as given by equation (III). The cable stress is then given from (IV) by substituting for w\ Stress in cable segment = . h Sr . . . . (VII) The suspender stress is in the same way, from (V), Stress on suspender = \- IV Q -j- w t -. (VIII) From these stresses the corresponding areas a, a c and a r can be determined, and if not sufficiently close to those assumed, another approximation can be made. Example. NEW METHOD. We take the same example as before. Data.c = 54 ft., r = 9 ft., h = 12 ft., = 6, E = E c = E, = 30000000 Ibs. per sq. in., tr e = 30000, and = 492.375 + 670.55 624 STATICS OF ELASTIC SOLIDS. [CHAI-. XI. s a o Ut's a A i A A A A aA 12 2.93 -22.5 + 2080 0.833 -0.666 -0.5 -0.333 -0.166 OiAt 13 2.29 13-5 + 955 + 0.166 - 0.666 -0.5 -0.333 -0.166 Ot.'lt 12 .66 - 4-5 + 146 + 0.166 + 0.333 -0.5 -0.333 -0.166 at A i 13 35 o o o o o fl//4 12 .66 - 4-5 + 146 -0.166 -0.333 -0.5 +0.333 + 0.166 atAt 12 .29 -13-5 + 955 -0.166 -0.333- -0.5 -0.666 + 0.166 bB 12 .92 -22.5 + 2080 -0.166 0.333 0.5 - 0.666 - 0.833 aA, 15 .66 + 28.125 + 3241 + 1.041 + 0.833 + 0.625 + 0.416 +0.208 aiAt 15 .87 + 16.875 + 1488 o. 208 + 0.833 +0.625 + 0.416 + o. 208 a,A* 15 .08 + 5-625 + 228 o. 208 0.416 + 0.625 + 0.416 + 0.208 a 4 A t 15 .oS + 5-625 + 228 + o. 208 + 0.416 + 0.625 0.416 0.208 a t A< 15 .87 + 16.875 + 1488 +0.208 + 0.416 + 0.625 + 0.833 -0.208 bAt 15 .66 + 28.125 + 3241 + 0.208 + 0.416 + 0.625 + 0.833 + 1.041 aai 9 .08 - 16.875 + 1232 0.625 -0.5 -0.375 0.25 0.125 Oirti 9 79 -27 + 2352 -0.5 I.O -0.75 -0.5 -025 a t ^ s f Or one pound at A, , At , At , A 4 , and A t . a a a a a aA + 77-05 + 61.64 +' 46.23 + 30.82 + 15-41 a,Ai "-79 + 47.16 + 35-37 + 23-58 + n-79 a,A t - 5-4 - 10.8 + 16.2 + 10.8 + 5-4 atAt o a 4 A 4 + 5-4 + 10.8 + 16.2 - 10.8 5-4 atAt + 11-79 + 23-58 + 35.37 + 47.16 - "-79 bB + I5-4I + 30.82 + 46.23 + 61.64 + 77-05 aA, + M9-87 + 95-90 h 71-92 + 47-95 + 23.97 a t A t - 18.34 + 73.36 + 55-02 + 36.68 - 18.34 a t At - 8.44 + 16.87 + 25 3i + 16.87 + 8.44 a 4 A t + 8.44 - 16.87 + 25.31 - 16.87 - 8.44 a t A t + 18.34 + 36.68 + 55-02 + 73.36 - 18.34 bAt + 23.97 + 47-95 + 7I-92 + 95.90 + 119-87 aa, + 45-65 + 36-52 + 27-39 + 18.26 + 9-13 a,a t I- 43-54 + 86.08 + 65.31 + 43-54 h 21.77 a ta t + 33-94 + 67.88 + 101.82 + 67.88 h 33-94 o t at H 33-94 + 67.88 + 101.82 + 67.88 r 33-94 a 4 a t + 21.77 + 43-54 + 65.31 + 86.08 f- 43-54 atb + 9-13 + 18.26 + 27.39 + 36.52 + 45-65 AA, o o o O A,At 1- 45-65 + 36.52 -f 27.39 + 18.26 + 9-13 AtAt + 43-54 + 86.08 + 65.31 + 43-54 + 21.77 AtA 4 + 21.77 + 43-54 f- 65.31 + 86.08 i- O-54 A 4 At + 9-13 + 18.26 + 27.39 + 36.52 + 45-65 AtB o o O o \ + 544-36 + 921-65 , + 1074.54 1 + 921.65 + 544-36 i i SUSPENSION SYSTEM. EXAMPLE. We have then from (7) for the apex loads A , A , 7% , A , /% , f or cable of links, 625 />, = 0.49 0.83 A 0.96 Pt 0.83 0.49 We have P = ,8000 pounds in the present case and the upward load on truss * = 3 ooo0 pounds at every apex of the loaded chord. We also have the temperature load ,, = 700 pound's per foot We can then draw up the following table of stresses for the truss. aA a t A t a?A* a 3 A 3 aA, aiAt a,A P, P^ P^ P. P* - "325 - 5775 1800 o - 8265 - 4680 2265 - 795 - 5203 - 7560 - 4755 2265 - 3735 - 7245 o 7245 3735 + 14156 + 7219 + 2250 - 281 - 844 - 6506 + 10331 + 5850 + 2831 + 994 - 4668 - 9056 + 9450 1- 5944 + 2831 * \ + 19687 - 19687 + 11812 11812 + 3937 - 3937 15750 - 15750 - 9450 - 3150 Max. Stresses. .. \ + 433" 20812 + 31818 - 18318 + 22162 -17661 - 34650 31755 - 29235 25110 aai aia* a?a 3 AA* AiA, A*A, Pi P* P\ P* - 8494 - 4331 - 1350 o o 8494 10530 - 4860 - 1530 - 90 - 4590 - 10530 - 10530 - 5096 - 1789 o o o 168 - 506 + 4590 + 4331 + 1350 - 168 - 506 + 1789 + 5096 + 4860 + 1530 + 90 ., { + 11812 18900 + 18900 21262 -f- 11812- 18900 + 18900 - 21262 11812 11812 i + 11812 - 44404 + 18900 - 53797 -f 22083 19574 + 3226^ Max. Stresses...] - 25987 12486 21262 From this table we find the maximum stresses in the truss members. We have now from (VII), since iv t = 700, tv = 1000, 2 = 3.6, for the stresses in the cable segments, A'a' I344I5 123714 118102 The suspender stress is, from (VIII), 26100 pounds. We can now find the corresponding areas of cross-section of the members, and if these areas differ too much from those assumed, should make a new calculation with the new areas. 626 STATICS OF ELASTIC SOLIDS. [CHAI-. XI. COMPARISON OF THE Two METHODS. In the present case we have then the following results: Old Method. New Method. Old Method. New Method. oA 29250 22950 - 34650 - 3755 a t a t 30262 j + 18900 1 ~ 53797 atAt 16050 - 29230 AA, 2O8I2 - 12486 a t A t - 13500 - 25II5 A A 20812 j + 22083 aAl 36562 ( + 28687 + 433" 20812 AtAt 27900 1 ~ 19574 j + 32265 ( 21262 a\A\ a t At J - 27000 ' - 18318 + 22162 17661 A'a' /if + I7I495 + I58I75 + 149850 + I344I5 + 123714 + 118102 aa, 2O8I2 - 25987 Susp. + 30300 + 26100 ai a t 27900 I +II8I2 I 444O4 We see that the cable and suspender stresses are much less by the new method. For the truss members by the new method tke direct stresses are greater and the counter-stresses less. INDEX. ABUTTING joints 487 Acceleration, angular 82 axial and normal 83 couple 144 " in terms of linear 85 " moment of 90 rectangular components of 83 resolution and composition of 83 " resultant 83, 159 axial and normal 83, 160 " central m, 160 directly as distance 125 inversely as square of dis- tance . 112 of centre of mass 299 deflecting and deviating 80, 159 of gravity 100 " " experimental determina- tion of 340 instantaneous linear 75 axis of 151, 161 linear, resolution and composition of 76 linear and angular combined 150 " mean linear 75 " moment of 89, 90, 160 of moment of momentum 301 radial and axial 80 rectangular components of 77 resultant 159 analytic determination of 77 spontaneous axis of 152 " tangential 158 " and central 77 uniform 100 " " and variable 78 " " motion on an inclined plane 136 " " motion in a curve 103 variable 109 Adhesion 220 Advantage, mechanical 267 Alphabet, Greek i Analytic determination of resultant acceleration 77 " " " " velocity .... 67 ' " * " angular ve- locity 60 Angle, conical 7 of friction 263 of rupture for earth ' 468 unit of 6 Angular and linear acceleration combined 150 rate of change of speed 73 velocity combined. .' 144 displacement 52 resolution and composition of 1 58 in terms of linear velocity 70 speed and velocity 61 speed, rate of change of 96 velocity, composition and resolution of. 68 " moment of 90 rectangular components of. ... 69 " resultant 155 " analytic determina- tion of 69 " uniform and variable 65 Arch, conditions of stability of 599 ' ' dam 450 " framed, fixed at ends 586 " hiriged at ends 579 " three hinges 578 " metal 578 " pressure curve of 598 " reduced surcharge for 598 " solid, fixed at ends 592 hinged at ends 592 stone 597 " straight . 607 Area, material 23 " moment of 22 Astronomical unit of mass 206 Attraction of homogeneous shell or sphere 203 Axes, principal . / 35 Axial acceleration 160 " moment 194 " and normal angular acceleration 83 " " radial acceleration 80 Axis of acceleration, instantaneous 161 " instantaneous 145, 157, 371 " invariable 302, 378, 391 " moment about 180 " neutral 493, 509 " of rotation, instantaneous 145, 157 627 ($28 INDEX. Axis of symmetry 23 Axle friction 227 " " work of 266 BALANCE, determination of mass by 212 Ballistic pendulum 362 Beams, deflection of 530 " fixed horizontally at ends 522 " impact of 358 " shearing stress in 553 " strength of 499 " of uniform strength 503 " work of shearing force in 554 Bending and compression combined 528 " " tension " 528 moment 494 stress 493 " work of 521 Blackburn's pendulum 135 Butt-joint 487 Brae histoch rone 142 Brake, friction 265 CARTESIAN co-ordinates 12 Central acceleration ill, 160 directly as the distance 125 inversely as square of dis- tance 112 " and tangential acceleration 77 impact 342, 344, 347 Centre of gravity 22, 190, 207 " mass 20, 23, 189 acceleration of 299 conservation of 299 motion of 299 " " " resultant force at 193 " velocity of 298 " oscillation 337 " parallel forces 188, 414 " " percussion 339 Centrifugal force 245 Change of motion of a point of a rigid body 157 Coefficient of cohesion for earth 468 " elasticity 346, 476 " for earth 464 " friction 221, 224, 263, 267 for shear 511, 555 " resilience 515 " rupture 499, 510 Cohesion of earth 466 Column, the ideal 559 Columns, factor of safety for long 569 formulas for long 565 Euler's formula for long 563 Gordon's " " " 568 Merriman's " " " 568 parabolic " " " 566 " Rankine's " " " 567 the straight-line formula for long 565 " strength of 559 Combined bending and tension 528 , CAOt Combined bending and compression 528 " flexure and torsion 512 tension and compression or shear 484 Components of change of motion, invariant for. iM " " motion, invariant for 157 " rectangular 55 of acceleration 77 " angular acceleration. 83 " " velocity 69 " velocity 67 Composition and resolution of angular accelera- tions 83 " and resolution of angular displace- ments 58 and resolution of angular veloci ties 68 and resolution of forces 176 " " linear accelera- tions 76 and resolution of linear displace- ments 54 and resolution of linear velocities. 66 " moments. ... 88, 180 " of harmonic motions 129 Compound harmonic motion 129 " pendulum 337 Compression and bending combined 528 ' tension 484 Compressive stress and force 397 Concurring coplanar forces 404 forces 1 79 " resultant for 179 Cone of friction 222 Conical angle, unit of 7 " pendulum, simple 244 Conservation of areas 303 " centre of mass 299 " energy 276, 304 " momentum 300 " " moment of momentum, 302, 320, 369, 377, 390 Consolidation of earth 350 Constant of gravitation 205 Constrained motion curved path, uniform accel- eration 137 " in a circle, uniform accelera- tion 138 " " of a point 136 Co-ordinates, Cartesian 12 polar ii Cords and chains, friction of 231 Cosines, direction 13 Couple, angular acceleration 144 force 184 " resultant of 185 " or wrench 194 Curvature, unit of 7 Cycloid, motion in, uniform acceleration 138 Cylinders, strength of 486 D'ALEMBERT'S principle, 242, 297 INDEX. 629 Dams, masonry 424 design of 434 " economic section 440 Deflecting force 243 ' at earth's surface 246 " and deviating acceleration 80, 159 Deflection of beams 530 " framed structures 516 influence of shearing force on 554 Degree 6 Density 15 Derived unit 2 Determination of mass by balance 212 " centre of mass 23 Deviating and deflecting acceleration 80, 159 Deviation of falling body 252 Dimensions of a derived unit 2 Direction cosines 13 Displacements, angular, resolution and compo- sition of 58 line representative of 52 linear and angular 52 " resolution and composi- tion of 54 moment of 89 " relative 55 " triangle and polygon of 56 Dynamic components of motion 192 " equilibrium 210 Dynamics i, 169 Dyne 170 EARTH, angle of rupture for 468 " coefficient of cohesion for 468 " " " friction for 466 ' ' cohesion of 466 " consolidation 35 " ellipticity of 247 " . mass equilibrium of 467 " pressure 454 " slope 468 " surface, motion on 251 Eccentric impact 3^3 Effective forces....' 242, 297 " " fixed axis, moment of 316 " " rotation and translation, moment of 389 " " rotation about a fixed point, mo- ment of 375 " " translating axis, moment of 367 Efficiency 267 Elastic limit 475 " solids, statics of 473 Elasticity 473 " coefficient of 34 6 . 47& " for shear 5 11 . 555 law of 474 " modulus of 347 Elementary mass or particle 20 Ellipsoid of inertia 3 6 h.lipticity of earth 2 47 Energy, change of potential., PAGE 2 7 8 " conservation of 276, 304 " kinetic t 271 " " fixed axis 320 of a system 303 rotation about a fixed point 378 " rotation and translation 391 " translating axis 370 " law of 275,304 ' ' potential 274, 304 Epoch 128 Equations, homogeneous 3, 86 Equilibrium of earth mass 467 " " forces 209 " " material system 310 " a particle 277 " a system of bodies 398 limiting 221 " of non-concurring forces 211 polygon 413 stable 311 " static, molar, dynamic, molecular .. 210 Equipotential surface 287 Erg 261 Euler's formula for long columns 563 " geometric equations 167 External forces - 297 Eyebars and pins, theory of 506 FACTOR of safety " " " for long columns. Falling body, deviation of Fixed axis, rotation ' ' point, rotation about Flexure " and torsion combined Force " centrifugal " couple " " resultant of 185, " deflecting " " at earth surface " gravitation unit of " impressed and effective 169, " line representative of " lines of " moment " " unit of " resultant at centre of mass " and stress " system, invariant for " tangential Forces, centre of parallel " equilibrium of 209, " impressed and effective " non-concurring " " resultant for " revolution and composition ot " triangle and polygon of " work of non-concurring Formulas for long columns 481 569 252 315 375 493 512 170 245 184 194 243 246 171 242 176 287 1 80 iBi 193 473 197 255 188 211 297 IQI jSf> 630 1NDI-X. Framed arch 578 44 " three hinges 578 " hinged at ends 579 4 " fixed at ends 586 " structures 397,400 " " deflection of 516 Friction, angle of 263 41 axle 227 " brake 265 44 coefficient of 221,224,263,265,267 " " for earth 464 44 cone of 222 " kinds of 220 44 kinetic 263 laws of 222 " of cords and chains 231 " " masonry 424 44 44 oblique central impact 353 " 4< pivots 224 44 rolling 235 * 4 static 220 " wheels 230 44 work of axle 266 GEOMETRIC equations, Euler's 167 Gordon's formula for long struts 568 Graphical statics 404 Gravitation, constant of 205 force of 203 " law of 203 44 potential 286 Gravity 14, 173 " acceleration of 22 " centre of 22, 190, 207 41 experimental determination of accelera- tion of 340 " specific 16 Greek alphabet I Gyration, radius of 32, 338 HARMONIC motion 125 " simple 126 " " compound 129 Hemp ropes 235 High dam 434 44 wall 428 Hodograph 79 Homogeneous body A 15 " equations 3, 86 shell or sphere, attraction of 203 ICE- and wave-pressure 432 Ideal column 559 Impact 342 " of beams 358 1 ' direct central 342, 344, 347 ' ' eccentric 363 imperfectly elastic 347 " non-elastic 342 " oblique central. . 352 PAGE Impact oscillating body 361 perfectly elastic 344 14 rotating bodies 360 " and strength 356 Impressed force 169 " and effective forces 242, 297 Impulse 172, 257 and momentum 172,257 Inclined plane, motion on 136 Indeterminate stresses 407 Inertia 169 " moment of 31, 317 " * 4 determination of 38 " " experimental determination of 340 44 " ellipsoid of 36 " any axis in general 34 44 " relative to an axis 33 " a point 34 44 " " reduction of 32 "polar 33 Instantaneous angular acceleration 82 axis 371 " of acceleration 151, 161 14 of rotation 145,156,392 linear acceleration 75 and angular speed and ve- locity 63 rate of change of speed 75 Internal and external forces 297 " stresses in a beam 557 Invariable axis 378 14 and plane 302 " rotation and translation 391 Invariant for components of motion 157 " change of motion. . 161 " 4 ' force system 197 JOINT, abutting 487 " lap 487 " single-riveted 487 44 stability of masonry 425 Joule 261 KEPLER'S laws 120 Kinematics I, 51 44 of a point 91 14 of a rigid body 143 Kinetic energy 271 " fixed axis 320 14 rotation and translation 391 41 about a fixed point 378 " of a system 303 " translating axis 370 friction. 263 Kinetics I, 242 " of a material system 297 LAP-JOINT 487 Law of elasticity - 474 " "energy 275.304 " " gravitation 203 44 " the lever 183 INDEX. 63* Laws of friction 222 Least work, principle of 3H ( 517 Length, unit of 4^ Lever, law of ^3 Limiting equilibrium 221 Line, material , 20, 23 " " moment of 22 " representative of angular acceleration 83 " displacement.... 52 " velocity 66 " " force 176 ' moment 181 " linear acceleration 76 displacement 52 " " " velocity 66 " " " moment 83 Lines of force , 287 Linear acceleration, mean 75 " instantaneous...... 75 resolution and composition of 76 " and angular acceleration " 150 " " " rate of change of speed 73 " " " speed and velocity 61 " velocity combined 144 " density 15 " displacement 52 " resolution and composition of 54 " in terms of angular acceleration 84 " " " " " velocity 70 " velocity, resolution and composition of . . 66 Long struts, Euler's formula 563 " " factor of safety for 569 " " formulas for 565 ' " Gordon's formula 568 " " Merriman's " 568 " ' parabolic " 566 Rankine's " 567 " " straight-line" 5^5 strength of 559 Low dam 434 " wall 429 MASONRY dams " joint, stability of " weight and friction of Mass, acceleration of centre of 424 425 424 299 astronomical unit of 206 centre of 2O conservation of centre of 299 determination of, by balance 212 elementary ...... 20 independent of gravity 14, 173 measurement of r 4. *73 moment of 22 motion of *74 " " centre of 299 notation for *5 properties of centre of l8 9 reduction of 3 22 and space, measurable relations of 14 Mass, specific ................................. 16 " standard unit of ..................... 4, 5, 14 velocity of centre of ..................... 298 Material line ................................. 20, 23 " particle .................... .......... 169 " surface ...................... ....... 20 " system, equilibrium of ................ 310 " kinetics of .................. . 297 Matter, states of ............................... i Mean linear acceleration ....................... 75 " " and angular velocity ............... 61 Measurable relations of mass and space ......... 14 Measurement .................................. 2 of mass ....................... 14. 173 Measures, table of ............................. 8 Mechanical advantage ......................... 267 Mechanics ..................................... I Merriman's formula for long struts ............. 568 Metal arch .................................... 578 " springs ................................. 537 Middle-third rule .............................. 426 Modulus of elasticity .......................... 347 Molar equilibrium ............................ 210 Molecular " ............................. 210 Moment about an axis ....................... 88, 180 axial .................................. 194 " bending ................... ........... 494 " normal ................................ 194 " of acceleration .................... 89, 160 "aline .............................. 22 " angular acceleration ................ 90 " " " velocity ................... 90 "area ............................. 22 " " a vector quantity ................... 88 " " displacement .................. .... 89 " " effective forces, fixed axis .......... 316 " " " " rotation about fixed point ............ 375 " " " rotation and trans- lation ............ 389 " " " " , translating axis ..... 367 " force .............................. 181 " inertia ...................... 31.34.3*7 " " " determination of ............. 38 " " experimental 340 " " " polar ........................ 33 " " " reduction of ................. 32 " ' relative to an axis ........... 33 " ' " " " a point ............ 34 "mass ............................... 22 " " momentum ........................ 3 < " acceleration of .......... 301 " " conservation of.. 302, 320. 369 ' " rotation about fixed point 377 " rotation and translation. 390 " rotation, fixed axes ..... 318 translating axis ....... . 368 " velocity ......................... 8 9. *55 " "volume ............................ 22 resistng ....... resolution and composition of 180 63* /v/vr.v. Momentum 1 72, 255 " and impulse 172, 257 " conservation of 300 " ol a system 298 " rotation, fixed axis 318 about fixed point 376 translating axis 368 " translation and rotation 390 Motion, change of 157 " dynamic components of 192 " in a cycloid, uniform acceleration 136 " Newton's first law 169 " " second law 172 third law 174 " of a point of a rigid body 153 " of centre of mass 174, 299 of particle on earth surface 251 on an inclined plane 136 NEUTRAL axis 493. 509 Newton's laws of motion 169, 172, 174 Non-concurring forces 179, 191, 412 " equilibrium of 211 " resultant 186 " " work of 215 Normal and axial angular acceleration 83 moment 194 Notation for mass 15 OSCILLATING body, impact of 361 Oscillation, centre of 337 PARABOLA formula for long struts 566 Parallel forces, centre of 188 " graphic construction. . 414 Particle , 20 equilibrium of 277 " kinetics of 242 material 169 moving on earth surface 251 " stable equilibrium of 277 " unstable " " 277 Path of a point 51 Pendulum, ballistic 362 Blackburn's 135 " compound 337 " simple 336 " conical 244 " " time of vibration 138 Per, meaning of 3 Percussion, centre of 339 Perpetual motion 304 Phase 128 Physical science i Pile-driving 351 Pins and eye-bars, theory of 506 Pipes, strength of 486 Pivot- or swing-bridge 571 Pivots, friction of. 224 Plane, and axis of symmetry 23 " invariable 302 Planetary motion 120 Point, position of n " of reference n Polar co-ordinates n " moment of inertia 33 Polygon, equilibrium 413 offerees 176 " linear displacement 54 " " relative " 56 Position of centre of mass 21 Potential energy 274 change of 278 " of a system 304 " gravitation 286 Poundal 170 Power 262 41 transmission of, by shafts 512 Pressure curve for arch 598 " on fixed axis 319 Principal a*es 35 1 properties of 35 Principle of least work 311, 517 Properties of centre of mass 189 RADIAL and axial acceleration 80 Radian 6 " square 7 Radius of gyration 32, 338 Rankine's formula for long struts 567 Rate of change of speed 91 " ' " angular 96 " linear and angular 73 ' work 262 Reaction, influence of shearing force on 556 Rectangular components 55 of acceleration 77,83 " angular acceleration 83 " " velocity.... 69 " velocity 67 Reduction of mass -. 322 " moment of inertia 32 Redundant members 520 Reference, point of u Relations, measurable, of mass and space 14 Relative displacement 55 ." triangle and polygon of. . 56 Resilience, work of 515 Resisting moment 496 Resolution and composition of angular accelera- tion 83 Resolution and composition of angular displace- ment 58 Resolution and composition of angular velocity.. 68 " forces 176 " linear accelera- tion 76 Resolution and composition of linear displace- ment 54 Resolution and composition of linear velocity. .. 66 " moments 88, iSo Resultant acceleration 159 INDEX. 633 Resultant acceleration, analytic determination of. angular acceleration analytic determi- nation of velocity analytic determina- tion of concurring forces couple or wrench force at centre of mass for non-concurring forces moment 44 of a force couple velocity analytic determination of .... " work of Ropes, hemp " rigidity of " wire Rigid body, change of motion of a point of kinematics of motion of a point of Rigidity of ropes Rivet table Riveting, theory of Rolling contact, stability in " friction . Rotating bodies, impact of momentum of moment of momentum pressure on fixed axis Rotation, and translation 153, fixed axis fixed point 154, instantaneous axis of 145, 157, invariable axis of 302,378, spontaneous axis of " translating axis " " " momentum. . . 77 159 33 155 69 179 194 193 1 86 89 185 154 67 215 235 234 235 157 143 153 234 490 487 312 235 360 319 389 315 375 37i 39i 145 367 390 SAFETY, factor of 481 Science, physical i Screw spin 145 " wrench 192,197 Set 475 Shear, coefficient of elasticity for 511, 555 " and tension combined 484 Shearing force in beams, work of 534 influence of, upon deflection 554 " " " reaction 556 Shearing stress 553 " " and force 397 Shell or sphere, attraction of homogeneous 203 Simple conical pendulum 244 " harmonic motion 126 " pendulum '. 336 " time of vibration 138 Single-riveted joint 487 Solid arch fixed at ends 592 44 hinged at ends 583 Space and mass, measurable relations of. ....... 14 PACK Specific mass and gravity j6 Speed, and velocity ( Jl rate of change of 73 gi S / ln : .'.. -'i45 Spontaneous ax-s of acceleration 152 " " rotation 145.392 Springs, metal 537 Square radian 7 Stability, conditions of, for arch 599 of masonry joint 425 in rolling contact 3I2 of a wall 426 Stable equilibrium 3II of a particle 277 Standard units ^ " of lengch 5 41 " mass 5 Statement of a quantity 2 States of matter r Static equilibrium 2 ro friction 22O Statics 209 of elastic solids 473 graphical 412 Stone arch 597 Straight arch 607 " -line formula for long struts 565 Strain... 473 ' due to weight 479 Straining, work of 515 Strength and impact 356 " of beams 499 41 " long struts 559 " " pipes and cylinders 486 " ultimate 475 Stress 174- 397 " and force 473 " bending 493 " in framed structures 400 44 internal, in a beam 557 44 shearing 553 " working 481 Struts, Euler's formula for 563 " factor of safety for 569 41 formulas for long 565 41 Rankine's formula for 567 " strength of long 559 Superfluous members 402, 520 Surface, material 20 Surcharge, reduced for arch 598 Surface density 15 Suspension system 609 4< new theory 620 " o'd theory 613 44 stiffening truss 612 41 " temperature load 613 Swing-bridge ... 571 Symmetry, plane and axis of 23 TABLE of measures. 634 INDEX. Tangent acceleration 158 Tangential and central acceleration 77 force 255 Temperature stress, framed arch fixed at ends.. 591 " " framed arch hinged at ends 581 " solid arch fixed at ends.. 585 " " solid arch hinged at ends 595 Tensile stress and force 397 Tension and bending combined 528 " " compression " 484 " " shear " 484 Time, standard unit of 4 " of vibration, simple pendulum 138 Torsion 59 " and flexure combined 512 work of 512 Translating axis, conservation of moment of mo- mentum 369 ' " kinetic energy 370 " " moments of effective forces... . 367 " " moment of momentum 368 " momentum 368 Translation and rotation 153, 389 Transmission of power by shafts 512 Triangle and polygon of displacements 54. 56 ' " " " forces 176 Twisting moment 510 ULTIMATE strength 475 Uniform acceleration 100 motion in curve 103 " " on inclined plane. .. 136 " strength, beams of 503 and variable acceleration 78 " " " velocity 65 Unit of angle 6 " " conical angle 7 | " " density 16 " derived 2 " offeree 170 " " " moment 181 " " length 4 " " mass 4.14 " " " astronomical 206 " " time 4 " " work 261 Unstable equilibrium of a particle 277 VALUES of g ico Variable and uniform acceleration 78, 109 velocity 65 Vector quantity, moment of 88 Velocity along axis of rotation 155 " angular, composition and resolution of 68 " rectangular components of. ... 69 linear and angular 61 linear in terms of angular 70 " moment of. 89 " normal to axis of rotation 155 " of centre of mass 298 rectangular components of 67 resolution and composition of 66 resultant 67,69 Virtual work 215 Volume, material. . . 23 WALI 424 " design of high 429 "low 429 " retaining 454 " stability of 426 Water-pressure 430 Wave- " 432 Weight of a body 170 " and friction of masonry 424 Weights and measures 8 Wheels, friction 230 Wire ropes 235 Work 215, 260 " of axle-friction 266 " " bending 521 ' principle of least 311, 517 ' rate of 262 " of resultant 215 " " resilience 516 " " shearing force in beams 554 " " straining 515 " " torsion -. 512 " unit of 261 " virtual 217 Working stress 481 Wrench 192 " resultant 194 " screw 197 TA v.l APR APR KAY; MOV DE ft INT l\ University of California SOUTHERN REGIONAL LIBRARY FACILITY 405 Hilgard Avenue, Los Angeles, CA 90024-1388 Return this material to the library from which it was borrowed. 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