IN MEMORIAM FLORIAN CAJORl ^•yOriyp'i- a^^<-- WORKS OF PROF. W. WOOLSEY JOHNSON PUBLISHED BY JOHN WILEY & SONS. An Elementary Treatise on the Integral Ca!culus. Founded on the Method of Rates. Small 8vo, ix + 234 pages, 36 figures. Cloth, f 1.50. The Theory of Errors and the Method of Least Squares. i2mo, x-{- 172 pages. Cloth, $1.50. Curve Tracing in Cartesian Coordinates. i2mo, vi4- 86 pages, 54 figures. Cloth, $1.00. Differential Equations. A Treatise on Ordinary and Partial Differential Equations. Small 8vo, xii + sSS pages. Cloth, I3.50. Theoretical Mechanics. An Elementary Treatise. i2mo, xv4- 434 pages, 115 figures. Cloth, $3.00, net. An Elementary Treatise on the Differential Cal- culus. Founded on the Method of Rates. Small Svo, xiv -j- 404 pages, 70 figures. Cloth, $3.00. THEORETICAL MECHANICS AN ELEMENTARY TREATISE, W. WOOLSEY JOHNSON, pyofessoy of Mathematics, U. S. Naval Academy. FIRST EDITION, THIItD THOUSAND. NEW YORK: JOHN WILEY & SONS. London: CHAPMAN & HALL, Limited. X 1906 Copyright 1901, BY W. W. JOHNSON. ROBERT DRUMMOND, PRINTER, NEW YORK. PREFACE. In preparing the present work, which was designed to include in a single volume of moderate compass the elementary portions of Theoretical Mechanics, no formal division of the subject into Kinematics, Statics and Kinetics has been made. The topics often included under the first head it was thought best to intro- duce separately, each at the point where it is required for imme- diate application to the treatment of the motions produced by forces. For example, the expressions for radial and transverse accelerations are not introduced until required in the discussion of Central Forces. The subject of Statics is, to be sure, to a large extent separable from the idea of motion. But, on the one hand, as has been recognized in all recent treatises, the fundamental notions of force are best presented, and the Parallelogram of Forces is best estab- lished, on the basis of the Laws of Motion. This requires what may be called a dynamical introduction to Statics. On the other hand, the subject cannot be completed without the Method of Virtual Velocities, an application of the Principle of Work. This principle, which is dynamical as involving forces acting through spaces, advantageously precedes the study of kinetics into which time enters explicitly, and prepares the student for the notion of Kinetic Energy, or work embodied in motion. Accordingly, in the present volume, Chapter I consists of such a kinetical introduction to the whole subject as is referred to iV!304888 IV PREFA CE. above; Chapters 1 1- VI are purely statical ; Chapter VII treats of the dynamical Principle of Work with its application to Statics and to the notion of the Potential Function; and the remaining chapters treat of purely kinetical topics. The chapters are further subdivided into sections followed by copious lists of graded examples aggregating over 500 in number; many of these were taken from examination papers set at the Naval Academy, and not a few were prepared expressly for this work. In these examples, as well as in the numerical illustrations introduced in the text, gravitation units of force have for the most part been employed. These units in fact not only have the advantage of being rendered familiar to us by the common usages of every-day life, but they are actually more convenient than absolute units in mechanical problems, since in them the forces arise principally from the weights of bodies. Thus their use is forced upon even those writers who most deprecate the employ- ment of a variable unit of force. The conception of an absolute unit of force, dependent upon mass and motion and not upon weight, is indeed essential to the gaining of correct ideas of the nature of force. Hence the introduction by Prof. James Thomson of the poundal, which serves this purpose when the English system of weights and measures is used, has been of very great value. At the same time the employment of the pound as a unit of mass as well as a unit of force has been the cause of confusion, so that a student is sometimes in doubt whether the result of the use of a formula is the number of pounds or of poundals, or, as he may phrase it, whether the formula is expressed in gravitation or in absolute units. To prevent this confusion, care has been taken in the present volume, while using gravitation units, to avoid such expressions as, for example, " a mass of 6 pounds," and to speak instead of " a body whose weight is 6 pounds." There is no doubt that the same body would be intended in either expression, but the former would imply that, in the formula W^=: mg, 6 is the numerical value of m^ and the latter that 6 is the numerical value PREFA CE. of W. Inasmuch as the pound, though an absolute '* standard " of mass, is properly called and legally styled a " unit of weight," the latter is the more natural course. Accordingly the student is directed on page 13 to follow it and to remember that all the forces are thus expressed in local pounds. If the result is desired in poundals, neither is the formula changed nor is the result found in oi?e unit and then changed to the other, but the number of pounds is taken as the numerical value of m. For the same reason, we should not say that the " weight " of a body varies when it is taken to a place where g has a different value, because the number which legally expresses its weight remains the same. The force of its gravity has indeed changed, but it is (when we use gravitation units) the unit of this force, and not its numerical measure, which has changed. In the treatment of kinetics, the conception of the forces of inertia has been freely employed, and that without the apologies that some writers have thought necessary. It would seem that the resistance of a body in motion to acceleration in any direction is as much entitled to be regarded as a force as is the resistance of other bodies which, in the case of a body at rest, prevent motion. By including the latter as forces, we obtain the idea of a system of forces in equilibrium; so also, by including the former as forces, we extend this idea to the case of a body in motion, and D'Alembert's Principle presents itself in the form of " kinetic equilibrium," instead of requiring for its statement a set of hypothetical ' ' effective forces. ' ' The study of Mechanics is here supposed to follow an adequate course in the Differential and Integral Calculus, and to form a very important application of its principles. But, when these applications occur, the results are not merely presented in the shape of general formulae in the notation of the Calculus, leav- ing the student unaided in the process of evaluation. Instead of this, pains has been taken to instruct the student in the methods best adapted in various cases to obtaining numerical results. Particularly in the treatment of statical moments and of moment? VI PREFA CE, of inertia it is hoped that the book will be found a useful supple- ment to the course of instruction in the processes of integration. Throughout, the practice of relying upon substitution in general formulae is discouraged as far as possible, and the opposite practice inculcated — namely, that of applying general principles directly to the problem in hand. Special prominence is given to those results which it is the most important to make familiar to the student of Applied Mechanics, and to the readiest ways of recalling them when they have slipped the memory. Although preference is given to analytical processes, a not inconsiderable use is made of graphical methods. These have, however, been introduced rather as diagrammatic aids to the comprehension of general principles, and to the calculation of numerical results, than as methods of obtaining results by meas- urement from accurately constructed diagrams — the latter belong- ing rather to the province of Applied Mechanics. W. W. J. April, 1901. CONTENTS. CHAPTER I. DEFINITIONS AND LAWS OF MOTION. PACK I. Motion in a Straight Line I Velocity or Speed 2 Variable Speed ? Acceleration and Retardation t Variable Acceleration '^ The Laws of Motion Q Inertia 9 The Measure of Force 10 Mass II Equation of Force and Motion 11 The Units of Force and of Mass 12 Absolute Units of Force 14 Momentum and Impulse 14 Reaction ;[6 Transmission of Force 18 Examples I. 18 II. Composition of Motion 20 Composition of Velocities 20 Resolutions of Velocities 22 Motion in a Plane Curve 23 The Hodograph 24 Acceleration in Curvilinear Motion .^ 25 Component Accelerations 27 Application of the Second Law of Motion to Forces in Different Directions 29 VI U CONTENTS. PACK Momentum as a Vector Quantity 31 Examples IL 32 CHAPTER II. FORCES ACTING AT A SINGLE POINT. III. Statics 36 The Resultant of Two Forces 37 Statical Verification of the Parallelogram of Forces 37 Three Forces in Equilibrium 3g Resolution of Forces 40 Effective or Resolved Part of a Force in a Given Direction 40 The Resultant of Three or More Forces 41 The Resolved Part of the Resultant 43 Reference of Forces in a Plane to Coordinate Axes 44 Rectangular Components in Space 45 Another Method of Constructing the Resultant 46 Examples III. 48 IV. Conditions of Equilibrium for a Particle 51 Number of Independent Conditions 52 Solution by Means of a Triangle of Forces 54 The Condition of Equilibrium in a Plane Curve 55 Condition of Equilibrium on a Fixed Curve in Space 57 Conditions of Equilibrium on a Surface 58 Equilibrium of Interacting Particles 59 Examples IV 60 CHAPTER III. FORCES ACTING IN A SINGLE PLANE. V. Joint Action of Forces on a Rigid Body 64 Construction of the Resultant 65 The Resultant of Two Parallel Forces , 66 The Resultant of a Number of Parallel Forces in One Plane 68 Couples 70 Measure of Turning Moment 71 Moment of a Force about a Point 71 Varignon's Theorem of Moments 73 CONTENTS. PAGE Three Numerical Elements Determining a Force in a Given Plane 75 Resultant of a Force and a Couple 77 Moment of a Force Represented by an Area 77 Forces in a Plane Referred to Rectangular Axes 78 Examples V 79 VI. Conditions of Equilibrium for Forces in a Single Plane S2 Number of Independent Conditions 84 Choice of Conditions 85 Case of Three Forces 86 Equilibrium of Parallel Forces 88 Examples VI 89 VII. The Rigid Body regarded as a System of Particles 92 The Funicular Polygon for Parallel Forces 94 The Funicular Polygon for Forces not Parallel 96 The Suspension Bridge 98 Form of the Suspension Cable 99 The Catenary. 102 Approximate Formulae. 104 Equilibrium of a System of Solid Bodies 106 Examples VII 109 CHAPTER IV. PARALLEL FORCES AND CENTRES OF FORCE. VIII. Resultant of Three Parallel Forces not in One Plane 113 Forces in Opposite Directions 114 Couples in Parallel Planes 115 Case in which the Resultant is a Couple 116 Composition of Couples in Intersecting Planes 116 Resolution fo a Force into Three Parallel Components 117 Moment of a Force about an Axis 118 The Centre of Parallel Forces 119 Case in which R = o 121 Conditions of Equilibrium 122 The Centre of Gravity of n Particles 122 Examples VIII 124 IX. The Centre of Uniform Pressure on a Plane Surface 126 The Centre of Gravity of an Area 128 CONTENTS. PAGK The Centre of Gravity of a Uniform Curve 131 Employment of Polar Coordinates 132 The Theorems of Pappus 134 Examples IX 1 38 X. The Centre of Gravity of Particles not in One Plane < 141 Statical Moments with Respect to the Coordinate Planes 142 Centre of Gravity of a Volume or a Homogeneous Solid 144 Employment of Triple Integration. 146 Solids of Variable Density .v 148 Centre of Gravity of a Solid of Variable Density 149 Stable and Unstable Equilibrium 151 Equilibrium in Rolling Motion , 152 Limits of Stability 155 Examples X. 156 CHAPTER V. FRICTIONAL RESISTANCE. XL Laws of Friction 161 The Angle of Friction " 162 Limits of Equilibrium on a Rough Inclined Plane 163 The Cone of Friction 165 Frictional Equilibrium of a Rigid Body 166 The Moment of Friction 168 Friction of a Cord on a Rough Surface 169 Examples XI 172 CHAPTER VI. FORCES IN GENERAL. XII. Lines of Action neither Coplanar nor Parallel 178 The Moment of a Force about any Axis 178 Representation of a Couple by a Vector 180 Moment of a Couple about an Oblique Axis 181 Resolved Part of a Couple 182 Composition of Couples 182 Joint Action of a System of Forces 183 The Principal Moment of a System at a Point 184 Poinsot's Central Axis 184 CONTENTS, XI PACK Forces referred to Three Rectangular Axes 187 Six Independent Elements of a System of Forces 188 Conditions of Equilibrium igi Equilibrium of Constrained Bodies 192 Examples XII. 194 CHAPTER VII. THE PRINCIPLE OF WORK. XIII. Work done by or against a Force 197 Work done by the Components of a Force 198 Virtual Work of a Variable Force 200 The Principle of Virtual Work 200 Work done by Internal Forces 201 Virtual Work in Constrained Motion 202 Expression of the Virtual Work in the Displacement of a Solid 203 Stability of Equilibrium 205 Case of Several Degrees of Freedom 207 Determination of Unknown Forces by the Principle of Virtual Work 207 Equilibrium of Interacting Solids 209 Examples XIII 210 XIV. Total Work of a Force in an Actual Displacement 211 Graphical Representation of Work 212 Work done when the Path of Displacement is Oblique or Curved 213 Potential Energy 214 Work done by a Resultant 215 Work expressed in Rectangular Coordinates 217 The Work Function for Central Forces 218 The Potential Function 219 The Potential of Attractive Force Varying Directly as the Distance.. 220 The Work Function in General 221 Equipotential Surfaces 223 Exarnples XIV 225 Xll CONTENTS. CHAPTER VIII. MOTION PRODUCED BY CONSTANT FORCE. PAGE XV. Inertia regarded as a Force 227 The Centre of Inertia 227 Rectilinear Motion 228 Integration of the Equation of Motion when the Force is Constant.... 230 Kinetic Energy 231 Laws of Falling Bodies 232 Body Projected Upward 234 Motion on a Smooth Inclined Plane 236 Spaces Fallen through in Equal Times 236 Body Projected up an Inclined Plane 237 Motion on a Rough Plane 238 Examples XV 240 XVI. Kinetic Equilibrium 243 Acceleration of Interacting Bodies 245 Application of the Principle of Work 247 Resolution of Inertia Forces 248 Examples XVI 251 XVII. Motion Oblique to the Direction of the Force 254 Parabolic Motion 255 Kinetic Energy of the Projectile 256 The Trajectory referred to the Point of Projection 257 The Range and the Time of Flight 259 With a Given Initial Velocity to Hit a Given Point 261 Constant Value of the Total Energy 263 Examples X VII 263 CHAPTER IX. MOTION PRODUCED BY VARIABLE FORCE. XVIII. Rectilinear Motion 267 Attractive Force Varying Directly as the Distance 268 The Period of Harmonic Vibration 270 The Energy of Vibration 271 Motion Produced by a Component of the Force 272 Repulsive Force Proportional to the Distance * 273 CONTENTS. XI 11 PAGB Attraction Inversely Proportional to the Square of the Distance 276 The Gravitation Potential 280 Examples XVIII 282 XIX. Curvilinear Motion 286 Tangential and Normal Components of Acceleration 286 The Normal Component of Inertia 287 Centrifugal Force 288 The Conical Pendulum 290 The Centrifugal Force due to the Earth's Rotation 291 The General Expression for the Normal Acceleration 292 Examples XIX 294 XX. Constrained Motion under the Action of External Force 296 Motion of a Heavy Body on a Smooth Vertical Curve 2c,7 X he Cycloidal Pendulum 299 Motion in a Vertical Circle 3c 2 The Simple Pendulum 306 The Seconds Pendulum 307 Comparison of Small Changes in /, «, and g 309 Examples XX 311 CHAPTER X. CENTRAL ORBITS. XXI. Free Motion about a Fixed Centre of Force 313 Attraction Directly Proportional to the Distance 314 Elliptical Harmonic Motion 315 Acceleration Along and Perpendicular to the Radius Vector 317 Area described by the Radius Vector 319 The Differential Polar Equation of the Orbit 322 The Central Force under which a Given Orbit is described 324 The Equation of Energy 325 The Circle of Total Energy or of Zero Velocity 327 The First Integral Equation of the Orbit , 329 The Apsides of the Orbit 331 The Radius of Curvature at an Apse 333 Circular Orbits 334 XIV CONTENTS. FACT Attraction Inversely Proportional to the Square of the Distance 33t) Elliptical Motion 339 The Periodic Time 341 Kepler's Laws 341 Time of Describing a Given Arc of the Orbit 343 Examples XXI. 347 CHAPTER XL MOTION OF RIGID BODIES. XXIL Action of Inertia in Rotation. 351 Moments of Inertia 353 Moment of Inertia of a Continuous Body 355 The Radius of Gyration 355 Interaction of Inertia in Rotation and Translation 356 The Energy of Rotation 357 Work done in an Angular Displacement 358 Moment of Inertia of a Geometrical Magnitude 359 The Moment of Inertia of a Plane Area 361 The Polar Moment of Inertia of an Area 363 Employment of Polar Coordinates 365 The Moment of Inertia of a Solid 366 Separate Calculation of 2mx^, 2m_y^, and ^mz"^ 368 Selection of the Element of Integration 370 Examples XXII 372 XXIII. Relations between Moments of Inertia about Different Axes 374 Moments of Inertia about Parallel Axes 375 Application to the Moment of the Element 377 The Principal Axes for a Point in the Plane of a Lamina 378 The Momental Ellipse of a Lamina for a Given Point 380 Principal Axes of a Lamina at the Centre of Inertia 381 The Moments of Inertia of a Solid for Axes passing through a Given Point 382 Momental Ellipsoid 383 The Principal Axes of Symmetrical Bodies 384 The Equimomental Ellipsoid 386 The Compound Pendulum 387 CONTENTS. XV PAGB Foucault's Pendulum Experiment 389 Pressure on the Axis of a Uniformly Rotating Lamina 390 Pressure on the Axis of a Uniformly Rotating Solid 392 Condition under which the Centrifugal System is Equivalent to a Single Force 393 Rotation about a Centroidal Axis 395 Pressure on the Axis when the Rotation is not Uniform 396 Plane Motion of a Rigid Body 397 Rotation and Translation Combined 399 Examples XXIII. 402 CHAPTER XII. MOTION PRODUCED BY IMPULSIVE FORCE. XXIV. Effect of Impulsive Force 405 Impact upon a Fixed Plane 406 Direct Impact of Spheres 409 Loss of Kinetic Energy in Impact 413 Energy of Driving and of Forging 414 Oblique Impact of Spheres 416 The Moment of an Impulse 417 Impulsive Pressure upon a Fixed Axis 418 Motion produced in a Free Body by an Impulse in a Principal Plane. 419 Motion of a System of Bodies 420 Conservation of the Motion of the Centre of Inertia 421 The Inertia Forces of the System 422 The Hypothesis of Fixed Centres of Force 423 External and Internal Kinetic Energy of a System 425 Examples XXIV 426 Lndex. 431 THEORETICAL MECHANICS, CHAPTER I. DEFINITIONS AND LAWS OF MOTION. I. Motion in a Straight Line. I. Mechanics is the science which treats of the motions of material bodies, and the causes of these motions. K force is an action, applied to a material body or to any part of it, which when unresisted produces motion. A solid body is one which resists relative motion between its parts, so that it does not readily change its shape. When the forces under con- sideration can produce no change of shape, the body is said to be rigid, and it moves only as a whole. If the motion of a rigid body is such that every straight line drawn in its substance re- mains always parallel to its original position, the motion is said to be one of translation. When this is the case, it is obvious that the motion of a single point of the body (whether it be in a straight or in a curved line) is sufficient to determine completely the motion of the body. The whole amount of matter contained in a body is often imagined to be concentrated at a single point. In this case it is called a material particle. The motion of a body in translation is completely represented by the motion of a particle. 2 DEFINITIONS AND LAWS OF MOTION. [Art. 2. 2. We discuss in this book only the motions of rigid bodies, and at first consider motions of translation, so that the body may be regarded as a particle, and the forces as acting at a single point. In this first chapter, we consider the general relations between forces and the motions they produce, from which is derived the mode in which they are measured and subjected to mathematical analysis. Velocity or Speed. 3. When a body is in motion, we have to consider both the speed and the direction of the motion. The term velocity is often used to include both these notions ; in such c^se, the velocity of a body is not said to be constant unless the direction of the motion as well as its speed is unchanged ; that is, unless the motion is rectilinear as well as uniform. In the first section of this chapter, we shall suppose the motion to be in a single straight line, so that speed only will at present be considered. The speed is untfor??i when the spaces described in any inter- vals of time are always proportional to the intervals. When this is the case, its measure is the number of units of space described in a unit of time. Thus, if / denotes the number (integral or fractional) of units of time in any interval, and s denotes the number of units of space described or passed over in that inter- val, the velocity is uniform when the ratio of ^ to / is the same for all corresponding values of s and /. Now putting v for this constant ratio, we have ^=;- (0 In this equation v is the value of s corresponding to / = i, and we take this as the numerical measure of the velocity. It is neces- sary to specify the units of time and space employed ; thus we speak of a speed of 10 feet per second, of 15 miles per hour, of a mile in two minutes and ten seconds, and so forth. 4. By means of equation (1), we can obtain the numerical measure of a constant speed from any given corresponding values § I.] VELOCITY OR SPEED. 3 of the space and time, and thus pass from one set of units to another. For example, to express the velocity of 30 miles per hour in feet per second. Here 30 miles is given as the space corresponding to the time one hour ; expressing the values of s and / in equation (i) in feet and seconds respectively, we have 30 X 528of ,, The arithmetical work shows that 44 ft. is the space correspond- ing to one second, and the customary mode of expressing the unit of velocity,* namely in the fractional form %, is suggested by the mode in which the symbols for the units of space and time occur in the equation. This result may therefore be ex- pressed thus : 3o™A = 44Vs ; and it is one which it is useful to remember, as giving the ratio between the numerical measures of any velocity as expressisd in these units. We shall regard the foot and the second as the standard units of time and length, and therefore the foot per second as the standard unit of velocity. Variable Speed. 5. When the spaces passed over in equal intervals of time are not equal, the speed is variable, and the quotient arising from dividing the space by the time gives what may be called the average speed for the given time. But at any given instant of time the speed has a definite value of which the numerical meas- ure is the immber of units of space which would be described in a * Separate names fqr units of velocity have been proposed, but have not been generally accepted. It is in fact better to keep the funda- mental units of space and time in evidence. It is said that the " knot " is the only single term for a unit of velocity in general use : thus we speak of a speed of 12 knots, meaning 12 sea-miles per hour. But the term knot is also often used as synonymous with sea-mile. 4 DEFINITIONS AND LAWS OF MOTION. [Art. 5. unit of time if the body moved uniformly throughout that interval with the speed which it had at the instant considered. Hence, if s denote the distance of the body, at any time /, from some fixed origin of distances taken on the path of the particle (here supposed to be a straight line), we have, by the definition of the derivative, ds , . ' = dt (=) This expression may also be regarded as the limiting value (when At is indefinitely diminished) of the ratio At ' where At is the increment of /, the time reckoned from some fixed instant taken as the origin of time, and As is the corresponding increment of s, that is, the space passed over in the interval At* (See Art. 390, Diff. Calc.) Writing equation (2) in the form ds = vdt, we see that, when the value of v is known for every instant or value of t (in other words, when v is given as a function of /), s is given by the equation = ^vdt, (3) which involves a constant of integration depending upon the position of the body at some given time. Again, using limits, we may write for the space described in a given interval s— s,= \ vdt, (4) where s„ and s correspond respectively to /„ and /, the values of the time at the beginning and end of the interval in question. § L] VARIABLE SPEED. 5 It is to be noticed that the result of supposing v constant^ and making j, and t^ each equal to zero, is j = vt^ equivalent to equa- tion (i), Art. 3. 6. The simplest example of a variable velocity, expressed as a known function of the time, is that of a body falling freely from a position of rest. It has been shown by experiment that the velocity at the end of any time after the instant when the body was dropped is proportional to the time ; so that we may put where ^ is a constant. This equation implies that v ■= o when / = o (that is, the body was at rest at the instant from which t is reckoned), and that v =. g when / =1, so that ^ is the velocity of the body at the end of one unit of time. Using our standard units, it is found that this velocity is about 32% ; hence, suppos- ing^ = 32, the equation shows, for example, that the velocity at the end of the first half-second is 16%, at the end of 2 seconds it is 64ys, etc. 7. If now we use this expression for v in equation (3), Art. 5, and perform the integration, we shall have where C is the constant of integration. Now if we agree to measure the space s from the position of rest, having already assumed that / = o when v = o, we must have j = o when / = o ; therefore C = o, so that s = ^gt^ is the space fallen through in / seconds from rest. In particular, putting /= i, we find 16 feet for the space fallen through in the first second ; putting / = 2, 64 feet is the space fallen through in the first 2 seconds. The difference of these, or 48 feet, is the space fallen through during the 2^ second, as would be directly obtained by using the limits I and 2 in equation (4). Since this 48 feet is described during a single second it measures the average speedy Art. 5, for that 6 DEFINITIONS AND LAWS OF MOTION. [Art. 7. second. It will be noticed that, in this case, the average speed is midway between the least and the greatest speed which occur during the interval, namely, 32Va and 64% , which correspond respectively to the beginning and to the end of the interval. Acceleration and Retardation. 8. The motion of a body is said to be hastened or accelerated when the velocity is increasing, and it is said to be unifor^nly accelerated when the increments of velocity which take plcae in any two intervals of time are proportional to the intervals. Thus, the motion considered in Art. 6, namely, that of a freely falling body, is a case of uniformly accelerated motion ; for the ex- pression V ^ gt shows that in any one second the velocity changes from gt to^(/+ i), that is, it receives the increment g ; in any two seconds it receives the increment 2g ; in any half-second, the increment ^g ; and so on. Under these circumstances, the increment of velocity received in a unit of time is taken as the measure of the acceleration. Thus, in the case of the falling body, the acceleration is constant and equal to ^. Supposing the motion to start from rest at the be- ginning of the interval, so that v = o when / = o, the acceleration is the same as the velocity acquired in the first second, or the quotient arising from dividing the velocity acquired in any inter- val by the number of units of time in that interval. Suppose, for example, that a train getting under way ac- quires a velocity of 18 miles per hour during one minute ; assuming the acceleration to be constant, what is its measure in the standard units — that is, the foot and second ? The velocity igni/h = 18 X (see Art. 4) ; dividing this by the number of 30 seconds in which it is acquired, and denoting acceleration by «, we have 30 X 60s 100 § I.] ' A CCELERA TION. Thus the foot-second unit of acceleration is 2, gain of velocity at the rate of 07te foot per second per second^ and the process shows how this naturally gives rise to the symbol Ys*. 9. The motion of a body is said to be retarded when the velocity is decreasing, and the rate of loss of velocity is called the retardation. For example, suppose that a stone projected along the ground with the velocity of 20 feet per second is observed to come to rest in 4 seconds. Here the velocity 20% is lost in 4 seconds ; hence, if we suppose the rate of loss to be constant, there is a loss of ^^/^ per second, that is, the retardation is 5%'- Variable Acceleration. 10. The acceleration a is defined as the rate of the velocity, whether that rate be constant or variable. Hence, using the notation of Art. 5, we have ds dv d'^s When we are dealing with motion in both directions along a straight line, it is necessary to assume one direction as the posi- tive one for measuring s from the origin. Then v is positive when the body is moving in this direction, so as to increase a positive or numerically decrease a negative value of s. In like manner, a is positive when a positive value of v is increasing or a negative value of v is numerically decreasing ; on the other hand, a is negative when a positive value of v is decreasing or a negative value numerically increasing. For example, when a heavy body is projected vertically up- ward, if the space is measured upward, the velocity is at first positive and decreases ; hence there is a retardation. The value of a is therefore negative ; and, on account of this negative accel- eration, the positive velocity is lost in a certain time, and after that converted into a negative and numerically increasing velocity. As a particular case, suppose the upward velocity of 8 DEFINITIONS AND LAWS OF MOTION. [Art. lo. projection to be i28Vs, the negative acceleration being 32. Reckoning the time / from the instant of projection, the loss of velocity in / seconds is 32/ ; hence the velocity at any instant is given by z^ = 128 — 32/. Putting z^ = o, we find that the whole velocity is lost when 32/ = 128, that is, when / = 4. Again, if we put / = 5, we find V — 128 — 160 = — 32, showing that at the end of 5 seconds the body is descending, and has acquired a negative velocity of 3 2 Vs. II. An example of motion with variable acceleration is af- forded by any vibratory motion, like that of a pendulum. For, since the velocity changes sign alternately from + to — , and from — to + , the acceleration must also change sign. The simplest case of vibratory motion in a straight line is that which is called harmonic, in which the distance of the particle from the origin at the time / is given by the equation S — a %\Vi QDt. (i) When / = o, the particle is at the origin ; when oot = \tc or / = — , it is at the distance a from the origin on the positive 200 side ; when / has twice this value, it is again at the origin ; at the end of three times this interval, ^ = — ^, the particle is at its greatest distance on the negative side; and so on. By successive differentiation, equation (i) gives 1) z=. aoD cos Got. (2) £t = — aca)^ sin oot (3) A comparison of equations (3) and (i) shows that the accelera- tion is negative whenever s is positive, and positive whenever s is negative. It is zero at the origin ; and at that point v has its greatest positive or negative value. This is in accordance with the principles of maxima and minima, since a is the deriva- tive of V. § I.] NEWTON'S LAWS OF MOTION. 9 The Laws of Motion. 12. The science of Mechanics is based upon certain first principles which must be regarded as established by experience. These, having been first clearly formulated by Sir Isaac Newton in the Philosophiae Naturalis Principia Mathematical are known as Newton's Laws of Motion. We shall in the succeeding arti- cles give the three laws in literal translation from the Latin of the Principia^ each followed by the necessary explanations. Inertia. 13. Law I. — Every body keeps in its state of rest or of moving uniformly in a straight line^ except so far as it is compelled by forces acting on it to chatige its state. This law, which is sometimes called the Law of Inertia^ as- serts that, while some external cause which we call/.^r^^ is neces- sary to put a body in motion, no such external action is necessary to keep it in uniform rectilinear motion after it has acquired a velocity ; but, on the contrary, force is then required either to deflect it from a rectilinear path, or to alter its speed. This is contrary to the notion of the ancients, who regarded the earth as at rest, and attributed the observed tendency of bodies put in motion to come to rest to an inherent property of matter which they called inertia. On the other hand, we now hold that the earth itself is in motion, but that this does not in any way dis- turb the relative motion of bodies with respect to it. We regard inertia as opposed to any change of motion ; so that, when bodies already in motion come to rest relatively to the earth,* the fact must be attributed to external causes or forces. We cannot completely prove the first law of motion experi- mentally, because it is impossible to free the body on which we experiment entirely from the action of external forces ; but we can show that the nearer we approach to this condition the nearer we realize a state of uniform rectilinear motion. * It is noteworthy that Galileo, who was the first to hold correct views of the nature of fcice and motion, maintained also that the earth was in motion. 10 DEFINITIONS AND LOSS OF MOTION, [Art. 14, The Measure of Force. 14. Law II. — Change of motion is proportional to the moving force acting, and takes place in the straight line in which the force acts. This is the most important of the three laws. We defer to the next section its application to forces and motions in various directions, and here consider only the case of a single force acting upon a freely moving body. The direction of the force is of course that of the straight line in which the body, starting from rest, begins to move under the influence of the force acting freely — that is, when no other forces are acting. This line is called the line of action of the force. If, after the body has ac- quired motion in this line, the force continues to act in the same direction, the body will continue to move in the same straight line ; for there is no reason why it should deviate from it to one side rather than the other. In the case of the single body now under consideration, '' change of motion" means change of ve- locity. The second law therefore asserts that the change of velocity produced in any interval of time is proportional to the force acting during that time. 15. It follows that, if the changes of velocity in all equal inter- vals of time while the force is acting are equal (in other words, if the acceleration^ Art. 8, is constant), the intensity of the force is constant. Thus, because the motion of a freely falling body, Art. 6, is found to be a case of uniformly accelerated motion, we infer that the force which urges a body downward is a constant one. It is thus independent of the velocity with which the body is moving. On the other hand, the force of the wind upon a body moving before it is not constant, but depends in part upon the velocity of the moving body, for the acceleration in this last case is not constant ; in fact it disappears when the body has ac- quired the velocity of the wind. Again, the acceleration of a body sinking in water is not constant, because the constant force due to the body's weight in water is resisted by a force which de- pends upon the velocity. The acceleration in this case vanishes § I.] THE MEASURE OF FORCE. II when the resistance becomes equal to the weight in water, and the body then descends with a uniform velocity. So far as it relates to the motion of a single body, we may therefore express the second law as follows : Force is measured by the acceleration it produces in a freely moving body. Mass. 16. We come now to the consideration of the action of forces upon different bodies. When the bodies are regarded as particles, the only respect in which they differ is in quantity of matter, which is called mass. Moreover, so far as motion of translation is concerned, the size and shape of the body, and the mode of distribution of the matter within the volume, is of no conse- quence. The comparison of the masses of bodies \^ practically effected by means of their weights^ as indicated by jthe common balance. If two bodies are equal in weight, we assume that they are equal in mass. It will presently be shown why this assump- tion is correct ; but it is important to notice that the mass of a body is really measured by its resistance to change of motion. For, if two equal forces act in the same direction upon two equal bodies starting from rest, the bodies will acquire the same velocity, and will move side by side. They may therefore be con- sidered as forming a body of double mass acted upon by a double force. Thus a double force is required to produce a given acceleration in a double mass, and in like manner it can be shown in general that the force required to produce a given accelera- tion is proportional to the mass moved. In other words, the inertia of a body, which, as stated in Art. 13, is its resistance to change of motion (or the quality of matter by virtue of which it requires force to produce change of motion), is proportional to the mass of the body. Equation of Force and Motion. 17. The results of Arts. 15 and 16 may be combined in the statement that force \s jointly proportional to the mass upon which it acts freely and the acceleration it produces in that mass. This 12 DEFlNirjONS AND LAWS OF MOTION. [Art. 17. is the form in which the proposition was stated by the older writers, who always used proportion in comparing magnitudes of different kinds. But the modern practice is to adopt units for the various magnitudes, in accordance with which, the force is said to be proportional to the /r^^z/C, which last represents the velocity of the current. The geometrical subtraction effected in the process is equivalent to the addition to AC oi the vector CZ>, which is the nega- tive of the vector to be subtracted. 33. Graphic solutions may also be given for problems con- cerning velocities in which the data do not consist of completely given vectors. For example, in Fig. 2, while AB is still the direction of the required resultant, suppose that the rate of row- ing were given instead of the resultant speed. Then, drawing the vector CD from any point C of AB,wq find the point D. From this point as a centre, with a radius equal to the given speed of rowing, describe an arc cutting AB in the point A'] then A'D determines the proper direction of rowing. The construction shows that the least possible speed of rowing is represented by the perpendicular from Z>, and that in general there are two solutions giving different resultant speeds along AB. Resolution of Velocities. 34. A. given velocity is readily resolved into two components having the directions of any two straight lines lying in the same plane with the given line of motion. To do this, it is only neces- sary to draw parallels to the given lines through the two ex- tremities of the vector representing the given velocity. Thus, in Fig. 3, OX 2iVidi (9K being the given lines, the velocity AC is by drawing the parallels AE and CJS resolved into the component velocities AE and EC. ^ II.] RESOLUTION OF VELOCITIES. 23 When considering a number of velocities in one plane, we may thus, by adopting two intersecting straight lines as axes, replace each velocity by its two components in the directions of the axes. If we adopt a positive direction upon each axis, and take the component velocities along the /y axes themselves, it is apparent g"/ ^ tliat the algebraic sum of the com- ponents along either axis of two ' ^z""/"""";^D given velocities is the like com- 'y '" A; ponent of their resultant. Thus, c/- / -^^^N^ in Fisr. ^, AC and AB or its equal L ^__ / / y r-r, f • . • 1 '.' /O ^ ^^' D" LD bemg two given velocities, / ^'C'and CD' are the compo- Fig. 3. nents along OX, and their sum A' D' is the like component of the resultant velocity ^Z>. Again, along the axis (9Kthe com- ponents are A" B" and B" D" , of which the latter is negative ; accordingly their algebraic sum A" D" is the like component of the resultant AD. A velocity in a given plane is determined ]m?>\. as well by means of its two components along given axes as by means of its magnitude and direction, and the advantage of using this system consists in its simplification of the relation between given veloci- ties and their resultant. Motion in a Plane Curve. 35. When a particle moves in a curve, the direction of its velocity at any instant is that of the tangent to the curve ; and the vector representing the velocity is a portion of this tangent, measured in the direction of the motion, and equal in length to the speed or numerical measure of the velocity, which we shall denote by v. Thus, in Fig. 4, if a particle is moving in the curve AB, and AC = v be measured off upon the tangent, it will be the vector representing the velocity. In. this position, the vector AC is the space which the particle would describe in the next unit of time if during that interval its velocity remained the same both in amouiit and direction as it is at the point A. 24 DEFINITIONS AND LAWS OF MOTION. [Art. a?- Fig. 4. Now, since the particle moves in a curve, this vector is vari- able, because its direction varies, even if its magnitude remains constant. For example, when the par- ticle has arrived at B its velocity will be represented by a certain vector BD^ which will generally differ in direction from ACy whatever be the relative magnitudes of the lines. 36. In order to compare the veloci- ties at A and B^ we may draw a vector AE from A equal to the vector BD, that is, parallel to BD as well as equal to it in length. The vector CE will then represent the total change of velocity which the particle undergoes in passing from A to B (when direction as well as magnitude is taken into account), because it is the vector which must be geometrically added to AC in order to produce AE. Completing the parallelogram, we may also take the vector AE to represent the change of velocity ; it is in fact vectorially equal to AE + EE^ that is, to BD—AC. The Hodograph. 37. In order to compare the velocities at all points, in a case of plane curvilinear motion, the vectors representing the several velocities may all be laid off from a common point taken for con- venience in a separate diagram. For example, suppose a particle to describe the ellipse ABC^ in Fig. 5, with variable speed. From any point O let OA' be drawn parallel and equal to the vector representing the velocity at A. From the same point let OP' be drawn equal and parallel to the velocity at any point P. By sup- posing the point P to move continuously about the ellipse, the point P' describes a curve, which will be a closed curve if, as supposed in the figure, the particle arrives at A with the same velocity with which it started. This curve is called the hodograph of the given motion. The point O is called the pole ; and it must be remembered that, in order to represent a given motion of Py the hodograph must be taken in connection with a certain pole. §11.] THE HODOGRAPH. 25 Moreover, for any given motion of P^ the auxiliary point F' will have a certain corresponding laiu of motion. For instance, the hodograph of a motion at uniform speed in any curve whatever would be a circle referred to its centre as pole ; but the motion of F' in this circle would de- pend on the curvature of the path of P. 38. It follows from the construction of the hodograph that the vector A'F' repre- sents the change in velocity experienced by the particle in moving from A to P m Fig. 5, just as CE does in Fig. 4. So, in general, the change of velocity in any arc of the given motion is represented by the chord of the corresponding arc of the ^' hodograph. Thus the displacement or Fig. 5. change of position of P' indicates the continuous change in the velocity of P, Acceleration in Curvilinear Motion. 39. By an extension of its original meaning, the term accelera- tion is used to denote the rate of change of velocity when direction as well as speed is considered. Hence, when the hodograph is constructed, the acceleration of F is the rate of displacement of F'\ that is, the velocity of the auxiliary poijtt in the hodograph. Thus, acceleration in general is a vector quantity, that is, one having direction as well as magnitude ; and whenever the hodo- graph is a curve, it is variable in direction at least. Its magni- tude is the same as the speed of the auxiliary point P\ 40. As an example, let us consider the case of uniform circular motion. Let C, Fig. 6, be the centre of the circular path, and a its radius. Denote by Fthe constant value of the speed which is the length of the vectors AA\ FF\ The hodograph, con- structed as in Art. 37, is therefore a circle whose radius is V. Moreover, since the vectors AA\ FF^ are perpendicular to the radii CA^ CP,x\i& angles ACP, A' OP' are equal. It follows that 26 DEFINITIONS AND LAWS OF MOTION. [Art. 40. the point P' moves uniformly in the hodograph, completing a revolution in the same time that P does. The vector «, con- structed in the diagram to represent the velocity of P' <, represents also (Art. 39) the acceleration of P. Since this vector is per- pendicular to the radius OP' of the hodograph, it is parallel to -PC; hence the acceleration is con- stant in magnitude and is directed toward the centre of the circle in which the particle moves. Now the velocities are proportional to the radii, because the two circles are described in the same time, therefore a\ V =■ Via; whence Fig. 6. gives the magnitude of the acceleration. 41. Consider next the motion of a point in a circle rolling upon a straight line. For example, suppose the circle in Fig. 6 to be rolling, like a carriage-wheel, with uniform speed upon a horizontal tangent. The wheel then has a motion of translation toward the left with the speed F, and any point of the rim, as P, has in addition a uniform circular motion relatively to the carriage. The velocity of P at any point is now the resultant of this constant horizontal velocity and that which the point has by virtue of the rotation of the wheel. This is completely represented in the hodograph by removing the pole to the point 0\ at the ex- tremity of the horizontal radius ; for the vector O'P' is the resultant of O'O, representing the motion of translation of the carriage, and OP' representing the relative velocity of P. It thus appears that the hodograph and the velocity of P' in it are not affected by the constant velocity of translation, the only effect being to remove to a new position the pole of reference. §11.] ACCELERATION IN CURVILINEAR MOTION. 2/ Thus the hodograph of uniform cycloidal motion is a circle referred to a point on its circumference as pole, and the accelera- tion in rolling motion is directed toward the centre and has the same magnitude as in uniform circular motion. Component Accelerations. 42. Referring the motion, as in Art. 34, to coordinate axes, CD, in Fig. 3, is the change taking place when che velocity AC is changed to AD^ whence it is easily seen that the component along either axis of any change of velocity is the change in the like component of the given velocity. The coordinates of the moving particle being x and 7, these component velocities are denoted by ^ and ^'. dt dt' It follows that the rates of change of these component velocities, namely, ^ and ^, df df are the components along the axes of the acceleration. They are also called the component* accelerations^ and the actual acceleration which is their resultant, is sometimes called in distinction the total acceleration, 43. In the analytical treatment of questions of motion rect- angular coordinates are nearly always employed. Then, denot- ing by s the length of the path as measured from some fixed point of it, and by its inclination to the axis of x^ we have for the component velocities dx ds , ^ dy ds . , . . , . — - = — cos (p=-v cos 0, -r- = — sm = z' sm 0, . ( i) dt dt dt dt where v is the actual speed of the point ; whence •=s =/[©•+ (in (■) 28 DEMNITIONS ANV LAWS OF MOTION. [Art. 43. give the resultant velocity and its direction in terms of the com- ponent velocities. In like manner, if ^ denotes the inclination of the accelera- tion or, the component accelerations are df (4) and give the total acceleration and its inclination, in terms of the component accelerations. 44. As an illustration, we give the analytical treatment of the case of uniform circular motion which has been treated graphically in Art. 40. Taking the centre of the circle, O^ in Fig. 7 as origin of rectangular axes, denote by d the X angle POA made with the axis of x by the radius OP at the time /. Then, sup- posing A to be the position correspond- ing to ^ = o, ^ = csot, where gl? is a con- ^ ^ stant because the motion is uniform. It r IG. 7» is in fact the angular velocity of P, The coordinates of P are X ^^ a cos 9 =^ a cos Gl?/, y= a s\x\ = a sm cot^ . (i) whence, differentiating, the component velocities are dx ' ^ dy — = — aoo sm cw/, -f = aoo cos oot, dt di (^) Therefore, by equations (2) and (3), Art. 43, the linear velocity V and its direction are given by j/ = a'c»?% tan = — cot 0\ (3) g II.] COMPONENT ACCELERATIONS. 29 whence, measuring s from A^ so that v is positive when d in- creases, z; = — = «G7, , the opposite vertex of the parallelogram ABDC, in the unit of time ; therefore the joint action will give it the velocity AD in the direction of the diagonal. The impulse which acting . P ^ alone would produce this joint ^\~ effect is called the resultant im- \ ^^^^^..^ 1^ \ pulse. By Art. 22, impulses are ^\ ^^^"-^ \ measured by the momenta they Y^ ^^!>^ produce ; therefore the given and ^ resultant impulses, which have a Fig. 8. common mass factor, are propor- tional to, and in the direction of, the lines AB, AC and AD. It follows that, if two impulses are represented by proportional straight lines in the proper directions, their resultant will be represented by the diagonal of the parallelogram of which they are the sides.* *This is equivalent to Newton's proof of the "parallelogram of forces" {Corollaria I and // of the Axiomata sive Leges Motus); the forces being, as mentioned in the foot-note to Art. 22, what we now cail impulses. The result applied also to the " continuous forces" (vires acceleratrices) of the Frincipia, because these were measured by the actions produced in a given time. §11.] APPLICATION OF THE SECOND LAW. ^ 31 47. In the second mode of regarding the joint action of two given forces, we may suppose the particle of mass 7ti at A\ Fig. 8, to be already moving in any manner whatever, while AB and AC, drawn in the directions of the two forces, are vectors represent- ing the accelerations which the two forces each acting singly would produce in the mass m. Then, by the second Law of Motion, the joint action of the forces is to produce in the particle the acceleration represented in amount and direction by the diagonal AD of the parallelogram. But this acceleration would be produced by a single force of the proper amount in the direc- tion AD. The measures of the given forces and of the single force, which is called the resultant^ are, by Art. 17, niAB^ niAC and niAD^ respectively. Hence, if two forces acting upon a particle are represented by vectors having their directions and having lengths proportional to their magnitudes, the single equivalent force, or resultant^ will be represented in direction and magnitude by the diagonal of the parallelogram, or vectorial sum of the two vectors. The construction, when thus applied to forces, is known as the parallelogram of forces. It is to be noticed that, since the value of m is arbitrary, the scale in which forces are represented by lengths is purely arbitrary Momentum as a Vector Quantity. 48. Momentum being the product of mass and velocity has, like the latter, a definite direction : in other words, it is a vector quantity. The parallelogram of forces shows that, when forces act simultaneously upon a body, each may be regarded as producing a momentum in its proper direction, and that these momenta coexist subject to the law of composition of vectors. This is readily seen to extend to any number of forces. The momenta produced may exactly neutralize one another, and in that case no change of motion will take place, so that the body will either be at rest or moving uniformly in a straight line. In such a case, the forces are said to be in equilibrium. 32 DEFINITIONS AND LAWS OF MOTION. [Ex. IL EXAMPLES. II. 1. A body undergoes three displacements, of i, 2 and 3 units respectively, in the directions of a point describing the three sides of an equilateral triangle. What is the resulting displacement ? 1/3, in a direction perpendicular to the second side. 2. A ship is carried by the wind 3 miles due north, by the current 2 miles due west, and by her screw 6 miles E. 30° S. What is her actual displacement ? and if these displacements take place uniformly in half an hour, what is the velocity relative to the water <* (3 y'3 — 2) miles E.: 6 'f/3™/h. 3. The hood of a market van is 3^ feet above the floor : in driving through a shower the floor is wet to a distance of 11 inches behind the front edge of the hood. Assuming the rain- drops to fall vertically with a uniform velocity of 28%, what is the rate of driving ? 5 miles per hour. 4. A man jumps with a velocity of 8 feet per second from a car running ten miles an hour, in a direction making an angle of 60° with the direction of the car's motion. With what velocity does he strike the ground ? f 4/223 = 19.91^3. 5. Two ships, A and B, are approaching with uniform speeds the intersection of the straight lines in which they move. If the bearing of B from A is unchanging, show that the velocity of B relative to A is in the opposite direction, and that the ships will meet. If the speeds remain fixed and the courses vary, what is the locus of the point of meeting ? 6. A ship is steaming in a direction due north across a cur- rent running due west. At the end of an hour and a half it is found that the ship has made 24 miles in a direction 30° west of north. Find the velocity of the current, and the rate at which the ship is steaming. 8™/h ; 8 4/3™/h. 7. A street car is moving with the speed of 9 miles per hour. At what inclination to the line of motion must a package be projected from it, with a velocity of 24 feet per second, in § II.] EXAMPLES. 33 order that the resultant motion may be at right angles to the track ? cos -\— \\) = 123" 22'. 8. Assuming that the earth moves in a circular orbit about the sun, and that light travels from the sun to the earth in 8"' 20% find the apparent displacement of the sun due to the earth's motiorw 2o'^55. 9. A point is moving eastward with a velocity of 20^/5, and one hour afterwards it is found to be moving northeast with the same speed. Find the change of velocity, and the measure of the accel- eration, if the latter is assumed to be uniform. 20 i/(2 - V2)Vs N. N. W.; ^U V(2 - i/2). 10. Assuming the labor of rowing for a given time to be proportional to the square of the speed, and denoting the angle between AB (see Fig. 2) and the direction of the stream by (p, show that the labor of rowing from ^ to ^ is a minimum when the direction of rowing makes with AB the angle 90° — ^-0. 11. A carriage is travelling at the rate of six miles an hour. What is the velocity in feet per second of a point midway be- tween the centre and rim of the wheel: (ex) at its highest; and (/3) at its lowest point ? (a) 13.2 ; (/3) 4.4. 12. A point is describing a circle, of radius 7 yards, in 11 sec- onds with uniform speed. Find the change in its velocity after describing one sixth of a revolution from a given initial point. About i2Vs at an angle of 120° with the initial motion. 13. A train is travelling at the rate of 45 miles per hour, and rain is driven by the wind, which is in the same direction as the motion of the train, so that it falls with a velocity of 33 feet per second at an angle of 30° with the vertical. Show that the apparent direction of the rain to a person in the train is at right angles to its true direction. 14. A train moving at the rate of 30 miles per hour is struck by a stone moving horizontally and at right angles to the track with a velocity of 33 feet per second. Find the magnitude of the velocity with which the stone strikes the train, and the angle it makes with the motion of the train. SS'A; tan- (- i). 34 DEFINITIONS AND LAWS OF MOTION. [Ex. II. 15. A ship is sailing due east, and it is known that the wind is blowing from the northwest; the apparent direction of the wind as shown by the pennant is from N. N. E. Show that the velocity of the ship is equal to that of the wind. 16. A person walking eastward at the rate of 3'"/h finds that the wind seems to blow directly from the north, and on doubling his speed it seems to blow from the northeast. Find the velocity and direction of the wind. 3 4/2 '"/h from N. W. 17. What is the amount and direction of the momentum received by a body of mass m moving uniformly with velocity v in a circle: («) in a half-revolution ; (^) in a quarter-revolution ? {pc) 2mv opposite the original direction ; (ft) mv 4/2 at an angle of 135°. 18. If the speed of a carriage be represented by the radius of the wheel, show that the velocity of a point on the rim at any instant is represented in length by the chord joining it with the point in contact with the ground, and is perpendicular to this chord. 19. Derive the acceleration in the case of the uniformly roll- ing wheel (Art. 41) from the equations of the cycloid. 20. Draw the hodograph for a point of the wheel midway between the centre and the rim, and thence show that the greatest inclination of the velocity of this point to the horizontal is 30°. 21. Show that, if one of two component velocities of a point in fixed directions is constant, the hodograph of the motion is a straight line. 22. A bicycle is " geared to 2b inches," and the length of the crank is a inches. Determine the arc in which back-pedalling is effective on a down grade whose inclination to the horizon is u ; also the value of a for which

7 and in fact is not generally true, that there is a single force or resultant equivalent to the two forces. In the present chapter, the forces will be regarded as acting upon a single particle, or at a single point of a rigid body througn which all the lines of action pass. The Resultant of Two Forces. 50. Let AB and AC, Fig. 9, represent two given forces, P^ and P^y acting at the point A, and let denote the angle BAC between their directions. The direction of the re- sultant divides the angle into two parts, ^i^ 6^ and 0^, In the triangle ABD, the angle ABD is the supplement of 0, and BD = AC', hence ^'''- 9- R' = P^'-\.P: Jr2P,P, cos be taken upon CJ^ pro- ^ III.] THE PARALLELOGRAM OF FORCES. 39 duced, and the parallelogram DECF be completed, it will be found, on measurement, that the lines CE, CF and CD have the same ratios as the weights F, Q and F, which agrees with the principle of the parallelogram of forces. Three Forces in Equilibrium. 53. Three forces which, as in Fig. 11, act upon a point which remains at rest are said to be in equilibrium. It is obvious that the resultant of any two of the three forces is a force equal to the third, but opposite to it in direction. Such forces therefore have lines of action lying in one plane, and in magnitude they are proportional to the sides and diagonal of a parallelogram drawn as in Fig. 11 ; or what is the same thing, to the sides CF, ED DC oi a triangle, such as CED in Fig. ii, whose sides are in or parallel to the lines of action. Such a triangle is called a triangle of forces for the equilibrium of the particle on which the forces act. Thus, in Fig. 11, either DCE or DCF may be taken as the triangle of forces for the equilibrium of C. 54* The directions of the forces are those in which a point must move in describing the complete perimeter of the triangle in one direction; for example, in the direction DCE in the first of the triangles mentioned above, or DCF in the other. The angles between the directions of the forces are the supplements of the angles of the triangles, and therefore have the same sines. It follows that /* : <2 : ^ = sin BCR : sin RCA : sin ACF; that is to say, three forces in equilibrium are proportional each to the sine of the angle between the other two. When a rigid body is in equilibrium under the action of three forces, they may have different points of application, but their lines of action must lie in one plane and must meet in one point ^ Whenever a triangle of forces is drawn, it must be remembered that the forces do not act along the three sides of the triangle, but in lines parallel to them which, if there are no other forces acting meet in a single point. 40 FORCES ACTING AT A SINGLE POINT. [Art. 55- Resolution of Forces. 55. If, in any plane containing the line which represents a given force, lines be drawn through its extremities in any given Q directions, a triangle will be formed ^.'-''"X the sides of which represent in magni- ,^''' ''\ tude and direction tivo forces of which the given force is the resultant. These foroes are called components of the given force. Thus, in Fig. 12, if AB represents the given force, and AC Fig. 12. and CB drawn in the given directions intersect at C, the lengths AC and CB represent in magnitude the two components of AB. The given force is then said to be resolved into a pair of components in given directions. But, if AC is taken as the line of action of one component, the com- ponent represented in direction and magnitude by CB must act in the parallel line AD. If the given force is regarded as acting upon a rigid body, it may, of course, be resolved into similar components both acting at B^ and in the lines CB and DB respectively. Effective or Resolved Part of a Force in a Given Direction. 56. The component of a force in a given direction is not determined unless the direction of the other component is given. Thus, in Fig. 12, AC ma.y be drawn in the given direction; but its length (representing the magnitude of the component of AB in this direc- tion) is not determined unless we know the direction in which to draw BC. Now, in many cases, what we require is the effectiveness of the force to pro- duce motion in the direction in question when no other motion can take place. Suppose, for example, that a particle at Fig. 13. Af Fig. 13, is confined in a smooth tube AC, or constrained in §111.] RESOLVED PART OF A FORCE, 4I some other manner, so that it can move only along the line AC, If now the particle is acted upon by the force AB it will move along the tube in the direction AC. The force which prevents any other motion is now the resistance of the tube, which, since the tube is supposed to be perfectly smooth, acts at right angles with it. Taking this as the direction of the other component, as represented in the figure, we have AC^ one of two rectangular cotnponentSy as the measure of the action of the force in the given direction. 57. This rectangular resolution of a force is of such im- portance that the rectangular component in a given direction is generally referred to simply as the resolved part of the force in the given direction ; and this is always to be understood by the term ** resolved part " or " component," when no other direction is mentioned. The length AC^ in Fig. 13, is the projection of AB upon the line ACy and, denoting the angle BAC by ^, its value is AB cos B\ hence the resolved part of a force P in the direction of a line making the angle Q with its line of action is cos e. Equation (3), Art. 50, expresses that the sum of the resolved parts of two given forces in the direction of their resultant is the resultant itself. It is readily seen also from Fig. 9 that the resolved parts of the given forces in a direction at right angles to their resultant are equal and opposite forces. These components counterbalance one another, so that there is no force tending to move the body to either side of the diagonal AD. The Resultant of Three or more Forces. 58. Let /*, , P^ and P^^ Fig. 14, represent three forces acting on a particle at O, If from any point A we lay off AB equal and parallel to Z',, and then from B lay off -^C equal and parallel to /*,, we shall have determined (without completing the parallelo- gram) the point C, such that AC represents in direction and 42 FORCES ACTING JT A SINGLE POINT. [Art. 58. magnitude the resultant of P, and P^. Denoting this resultant by Qy we shall in like manner, by laying off from C CD equal and parallel to /*, , arrive at the point Z>, such that AD represents in direction and magnitude the resultant of Q and /*,. Now the joint action of the three forces is the same as that of Q and P^ acting at O, and therefore is equivalent to the action of a single Fig. 14. force R acting at O and represented in magnitude and direction by AD. This force is called the resultant of the three forces /*, , P, and P,: 59. The forces are here represented hyvectorSy and the process is an extension of the geometrical or vectorial addition mentioned in Art. 29. Since the order in which any two of the vectors are taken is immaterial, we arrive at the same final point Z>, whatever be the order of geometrically summing the three vectors. The three forces may or may not lie in one plane. When they do not, ^, , /*, and -P, at O may be regarded as three edges of a parallelopiped of which the diagonal from O represents the resultant. The different orders in which the vectors can be added then correspond to the different paths by which a point might move from O to the opposite vertex passing over three edges of the parallelopiped. 60. The process of Art. 58 is evidently applicable to any number of forces. When the final point arrived at in the geo- metrical addition coincides with the initial point, the resultant is zero, and the forces are said to be in equilibriu7n. For example, we shall have such a system of forces if, in Fig. 14, in addition to -P,, P^ and /*,, there were acting a.t O a. fourth force equal and §111. J RESULTANT OF THREE OR MORE FORCES. 43 opposite to the resultant R, which would be vectorially represented by AD. The closed perimeter such as A BCD A formed in this case is known as the polygon of forces^ and the theorem is that : If any number of forces aciifig at a point are represefited in direc- tion a?td magnitude by the sides of a closed polygon, each taken in the direction of the motion of a point describing the complete peri7neier, the forces are in equilibrium. If the lines of action of the forces are not all in one plane, the theorem still holds, the polygon of forces being, in that case, not a plane figure, but what is called a skew polygon. The Resolved Part of the Resultant. 61. If, through the extremities of a vector AB, planes perpen- dicular to a given line be passed, the length which they intercept on this line is called the projection of AB upon the given line. With this definition, the projec- D tion of AB upon any two parallel ^-^ lines is the same ; for the same y^><' I projecting planes are used, and ^^'"'^ ^v/ the projection is the perpendicu- f^f' \ C; lar distance between these planes. 1^ i • ' ' N The line MNy Fig. 15, upon ? B' C d' which the projection A' B' is Fig. 15. made, may not be in a plane with AB\ but the lines AA\ BB* will in all cases be perpendicular to MN. If now we define the inclination of two lines which do not intersect as the same as the angle B between intersecting lines parallel to them, we shall have, as in Art. 57, for the length of the projection of AB, A'B' = AB cos e. 62. Now if we take the broken line ABCD, formed in the vectorial addition of the forces /^, , P, and /*,, Fig. 15, and pass planes perpendicular to MN through A, B, C and D, we see that A' D' , the projection of the resultant AD^ is the algebraic sum of the projections of the vectors AB, BC and CD. Denoting by ^,, ^,, etc., the angles between the direction taken as positive along 44 FORCES ACTING AT A SINGLE POINT. [Art. 62. J/iVand that of the forces respectively, P^ cos ^,, etc., (Art. 57,) express the resolved parts along MN of the given forces (which are represented by the projections) with their proper signs (the projection being negative when B is obtuse). Hence, if is the inclination of the resultant R^ i? cos = P, cos e, + P^ cos (9, 4- P^ cos Q^. The result of course extends to any number of forces; that is to say, the resolved part of the resultant of a number of forces in any direction is the algebraic sum of the like resolved parts of the given forces. Reference of Forces in a Plane to Coordinate Axes. 63. In the systematic treatment of forces in a plane, coor- dinate axes are assumed, and the components of a force P along the axes of x and j/ respectively are denoted by X and F. The demonstration applied in Art. 34 to velocities shows that, for any quantities represented by vectors and which combine by the vector law, the component of the resultant in the direction of either axis is the sum of the like components of the given quan- tities. Thus, in Fig. 3, AB and AC may be taken to represent the forces /*, and /*„ and AD^ their resultant R, Hence the com- ponents of R are X, + X, and F, + F,. In like manner, for any number of forces, using the sign of summation ^, the components of the resultant are :2X and ^r, 64. When, as is usually most con- venient, rectangular axes are used, the components are " resolved parts " ^^°* ^^* (Art. 57). Denoting by 6^ the inclina- tion of the force P^ Fig. 16, to the positive direction of the axis of :r, the inclination to the positive direction of the axis. of >> is the complement of d and we have iiii] RECTANGULAR COMPONENTS. 45 = P cos ^, ) = P sin e, \ (i) From these we derive for the determination of P and ^, when X and Y are given, P» = X' + F', tan e = |. (2) It follows that, if denotes the inclination of the resultant P of any number of forces, P* = (:SXY + (2r)\ tan = -jj. Rectangular Components in Space. 65. When the forces under consideration do not all lie in one plane, a system of coordinate axes in space may be assumed, as in Fig. 17. Let the given force P be represented by a line OA drawn from the origin. Draw AB parallel to the axis of y to meet the plane of xz, and join OB. Then OB and BA represent in magnitude and direction a pair of components of the given force. It will be noticed that in general the component in a given plane is not determined either in mag- nitude or direction unless the direction of the other component is known. But, when this other component is perpendicular to the given plane, as AB in Fig. 17, where the axes are supposed rect- angular, OB is a definite line known as the projection of OA upon the plane y and the component it represents is called the resolved part of P in the plane. The magnitude of this resolved part is P cos AOB^ where AOB is the inclination of the force to the plane. 4^ FORCES ACTING AT A SINGLE POINT. [Art. 66. 66. Again, resolving this component in the direction of rectan- gular axes of X and z, by drawing BC parallel to the axis of z, we have P resolved into three components represented in magnitude and direction by OC^ CB and BA^ each of which is the resolved part in the direction of one of the axes. Denoting by oc^ fi and y the angles AOx^ AOy and AOz, or direction angles of OA, we have for the three components of P X=Pcosa, V=Pcos/3, Z = P cos y. . (i) We also have OA' = OB' + AB' = OC + CB'-{-AB'; that is, P' = X' -{- V -\- Z' (2) The factors cos a, cos /?, cos y in equations (i) are called M^7^\ numbers, m and «, the resultant of two /y^ / \ forces represented in direction and jt^ / -^ magnitude by mAB and nAC, will be V' represented m direction and magnitude Pj ^g by {m 4- 7i)AD, where D divides BC in the ratio 71 : w. On a line parallel to BC, through A^ lay off AE^ AP\ equal to and in the direction of BD, CD respectively ; then, by hy- pothesis, niAE = nAF, AEDB and AFDC are parallelograms ; hence the resultant of mAB and niAE is mAD^ and that of uAC and nAF IS nAD. Thus the resultant of the four forces niAB^ § III.] CONSTRUCTION OF THE RESULTANT. 47 mAEy nAC, tiAF is {m -\- n) AD\ and, since the forces w^^ and nAF are equal and opposite, they neutralize each other, so that {m + n)AD is the resultant of niAB and nAC^ which was to be proved. 68. The theorem proved above leads to another method of graphically determining the resultant of several forces, as fol- lows : Let n forces acting at O be represented by OA^, OA^, . . . OAn, Fig. 19. Bisect A^A^ in B ; then, by the theorem, 2 OB is the re- sultant of OA^ and OA^. Join B with ^,, and cut off BC = \BA^ ; then, by the same theorem, ^OC is the re- sultant of 2OB and OA^ ; that is, of OA^, OA^ and Fig. 19. OA^. In like manner, if we lay off CD = \CA^^^ OfOD is the re- sultant of the first four forces, and so on. We finally reach a point Z, such that nOL is the resultant of the n forces. 69. In this construction it is not necessary to suppose that the forces all lie in one plane. Whether the n points ^,, ^,, . . . -4„ do or do not lie in one plane, the final point L is calle'd their centre of position. This point has the property that its dis- tance from any plane is the average distance of the n points from that plane. To prove this, denote the perpendiculars from the given points to the selected plane by /,,/,, . . . /«, and those from B^C^ . . . L by/^,/>^, . . . p^\ then it is readily seen that the construction gives A=A + i(A-AX A=A + i(A-A), A=A + i(A-A)- «, give These equations, multiplied by 2, 3, 4, . 2A= A+A, 2d>c = 2/,4-A=A+A+A» 48 FORCES ACTING AT A SINGLE POINT. [Art. 69. «A= A+A+ . . . +A = ^/; hence p^ = ~^, which is the average or arithmetical mean of the perpendiculars. The theorem proved in the preceding article may now be stated thus : If n forces, OA^^ OA^, . . . 0A„, act at O, their re- sultant is n times the force represented by the line drawn from O to the centre of position of A^, A^, . . . A„. If the forces be in equilibrium, O will itself be the centre of position of ^,, ^,, . . . An. As an illustration, let the resultant be required of eight forces represented by lines joining a given point O to the eight vertices of a parallelopiped. The centre of position of the eight vertices is obviously the centre of the figure ; hence the resultant is rep- resented by eight times the line joining O to the centre of the parallelopiped. EXAMPLES. III. 1. Find the resultant of forces of 3 and 4 pounds, respec- tively, acting at right angles. 5 pounds. 2. At what angle must two equal forces act in order that the resultant shall equal either force ? 120°. 3. Show that, if the resultant of two forces is equal to one of them, the forces act at an obtuse angle ; also that, if the resultant is at right angles to one of the forces, it is less than the other. 4. Two forces when acting in opposite directions have a re- sultant of 7 pounds, and when acting at right angles they have a resultant of 13 pounds. What are the forces? 12 and 5 pounds. 5. Forces /* and 2P have a resultant at right angles to one of them. At what angle do they act? 120°. 6. If two of three forces in equilibrium are equal to I* and the angle between them is 0, what is the other force ? 2P cos ie. § III.] EXAMPLES. 49 7. Forces of 5 and 3 pounds have a resultant of 7 pounds. At what angle do they act ? - 60''. 8. Forces of 3, 4, 5 and 6 pounds, respectively, act along the straight lines drawn from the centre of a square to the angular points taken in order. Find their resultant. 2 ^2 pounds. 9. Three forces P^ 2/^, 3/* act at angles of 120° to each other. Determine the resultant. P Vs at right angles to 2P. 10. Lines parallel to the sides of a parallelogram intersect at a point O within it. Show that the resultant of four forces at O represented by the segments of these lines acts through the centre of the parallelogram. 11. Show that the resultant of three forces acting at the vertex A of & parallelopiped and represented by the diagonals of the three faces meeting at A is represented by twice the diagonal of the parallelopiped drawn from A. 12. ABODE is a regular hexagon ; at A forces act repre- sented in magnitude and direction by AB^ 2AC, 3^Z>, 4^^, and 5^i^. Show that the length of the line representing their resultant is AB 1^351. 13. The chords A OB and 00P> of a circle intersect at right angles at O. Show that the resultant of forces represented by OA, OB, 00, OD is represented by twice the line joining O to the centre of the circle. 14. If P is the orthocentre (point of intersection of the per- pendiculars) of the triangle ABC, show that the resultant of forces acting at a point and represented in magnitude and direction by AP, PB and PC is represented by the diameter from A of the circumscribing circle. 15. If O is the centre of the circumscribed circle of the tri- angle ABC, and P its orthocentre, show that the resultant of forces represented by OA, OB and OC is represented by OP. 16. If P is the total pressure produced by the wind normal to the sails, supposed flat and making the angle B with the keel, what is the effective force driving the ship ahead ? P sin B. 17. A weight W is sustained by a tripod of equal legs, so 50 FORCES ACTING AT A SINGLE POINT. [Ex. III. placed that the distance between each pair of feet is equal to a leg. Find the compression of each leg. W 76- i8. A and B, standing on opposite sides of a weight of loo pounds, pull upon ropes attached to it and making angles of 45° and 60°, respectively, with the horizontal. Find the ratio of their pulls if the resultant is vertical; also the value of .^'s if the weight is just raised? 1:4/2; ioo( //3 — i) = 73.2 pounds. 19. If, in example 18, ^'s rope be shifted to make an angle of 30° with the horizontal, show that his pull must be the same as before, but A's must be multiplied by 4/3. 20. A force acting at A is represented by the line AB. Show that the resolved part or action of the force in any direction AC is represented by the chord from A o( a. sphere whose diameter is AB ; also, that the action in any plane through A is represented by the diameter from A of the small circle in which the plane cuts this sphere. 21. Show that if four forces in given directions which are nof in one plane keep a particle in equilibrium, the ratios of the forces are determined ; but if four forces in one plane, or more than four in general, are in equilibrium, their ratios are not determined by their given directions. 22. Show that, by resolving the system of forces X, V, Z in Fig. 17 along a line whose direction angles are A, ju, v, we obtain cos ip = cos ex cos X -j- cos /? cos /< + cos y cos v, which is the expression for the cosine of the angle between two lines in terms of their direction cosines. 23. If the sides of the triangle ABC be produced, namely, BC to B>, CA to jS, AB to Bf so that the parts produced are propor- tional to the sides, show that forces acting at a point and repre- sented vectorially by AD, BE and CB' are in equilibrium ; also, if O be any point in the plane, the forces OD, OE and OF have a resultant independent of the ratio BD : BC §IV.] CONDITIONS OF EQUILIBRIUM. 5 1 IV. Conditions of Equilibrium for a Particle. 70. As stated in Art. 25, when a body acted on by a single active force is prevented from moving by the resistance of a fixed body with which it is in contact, this resistance is regarded as a force equal and opposite to the active force, and thus, with it, producing equilibrium. So also, when a body acted upon by several known forces is kept at rest by bodies with which it is in contact, the resistances or reactions of these bodies are regarded as forces ; and these, together with the known forces, constitute a system of forces in equilibrium. 71. In the case of a single particle, or of a body which maybe regarded as such, the forces all act at a single point, and it is im- portant to represent in the diagram all the forces, including the resistances. This is usually done by drawing lines from the point in the directions of the forces. For example, let it be required to find the force P which, acting horizontally, will sustain the weight ^upon a smooth plane inclined at the angle a to the horizontal. Let the weight act at the point A^ Fig. 20; then the force W is represented by a line drawn vertically downward from A, The only force except JV and P acting upon the body is the resistance of the plane, which, because the plane is smooth, acts in a normal to the plane. This is represented as in the diagram by a line R drawn from the plane, because it prevents motion in the opposite direction. Since there are but three forces in equi- librium, their lines of action will lie in one plane, Art. 53. Accordingly, P'% line of action must lie in the vertical plane which is per- pendicular to the inclined plane. The diagram is of course supposed to be in this plane. Now, since the forces represented in the diagram are in equilibrium, the resolved part of their resultant (which is zero) in any direction 52 FORCES ACTING AT A SINGLE POINT, [Art. 71. whatever must vanish. That is, by Art. 61, the algebraic sum of the resolved parts of the forces along any straight line is zero, or, what is the same thing, the sum of the resolved parts in one direc- tion along a given line is equal to the sum of those in the opposite direction. For the present purpose, let us take resolved parts along the inclined line AB. R, being perpendicular to this line, has no resolved part along it, and equating the resolved part of P up the plane to the resolved part of W down the plane, we have P cos = ^sin 0, (i) whence P = W tan 0. 72. An equation formed as above by resolving forces in equi- librium along a given line is called a condition of equilibrium ; and in equation (i) we chose the direction in such a way that the value of R did not enter the equation. In like manner, if we wish to obtain R directly in terms of W^ we may resolve verti- cally so as not to introduce P ; thus Rcos

— '~> ds in which the ratio dy : dx is to be derived from the equation of the curve. The result is an equation between x and y which, with the equation of the curve, determines the position of the point (x,y) of equilibrium. 78. For example, suppose that a particle restricted to a smooth ellipse is acted upon by two forces, one toward each focus, and each varying directly as the distance ; required the position of equilibrium. The equation of the ellipse referred to its axes, as in Fig. 22, is -Z lZ. (I) and the distance of either focus from the centre, OJ^^ or 01^^, is ae^ where ^ is the eccentricity of the ellipse. Let A be the particle in equilibrium and P^^P^ the forces directed along the lines AF^ and AF^. Draw the ordinate AB^ then the force P^ and its components, -ST, and F,, are proportional to the sides of the triangle F^AB. Since Fig. 22. the force P^ is proportional to AF^ or r^ , for different positions of A, we put P^ = /i, r, , yw, being the intensity of the force at a unit's distance. We have then >"i^, X^ = ^^{x + ag), F, = lAj ; §IV.] EQUILIBRIUM IN A PLANE CURVE. 57 and, in like manner, We have then for the sums of the forces in the directions of the axes ^ = (/^i + ^^oc + (//. - //,)«^, .... (2) F=(/., + ;.,);, (3) Differentiating equation (i), we obtain dy-^-^-^^d^ ... (4) Substituting in the condition of equilibrium Xdx . /)/ = o, we find «M(>"i + Z'.)^ + (>"i - /^«)^^] - ^'-^(/^i + /^> = o. One solution of this equation is jv = o, which shows that each extremity of the major axis is a position of equilibrium. The other solution gives When this is numerically less than a, it determines two interme- diate positions of equilibrium. Condition of Equilibrium on a Fixed Curve in Space. 79. If the curve upon which the particle is constrained to lie is not a plane curve, it may be referred to three rectangular axes in space. Then, if s be the length of the arc as measured from some fixed point to the point {x,y^ z)^ and A, //, v the direction- angles of the tangent at {x^y^ z)^ we shall have - dx dy dz cos A = — -, cos M = — -, cos r = — . ds ds ds Now, if Xy Y and Z denote the sums of the resolved forces in the directions of the axes, the particle is in equilibrium under the 58 FORCES ACTING AT A SINGLE POINT. [Art. 79- action of the forces X, F, Z and R the resistance of the curve. Since this resistance acts in some line perpendicular to the tan- gent, we shall obtain an equation independent of R by resolving along the tangent ; namely, X cos A + F cos /< + Z cos ^ = o, or, substituting the values of the direction cosines, Xdx -{- Ydy -\- Zdz =1 o (i) In this equation, the ratios of dx, dy and dz are those which result from the differentiation of the two relations between x,y and z, which define the line to which the particle is restricted. As in the case of the plane curve, we have thus a single condition of equilibrium. Conditions of Equilibrium on a Surface. 80. Let us next suppose that the point is only restricted to lie in a given surface, of which the equation is «=/(-^,J,^) = o (i) Then it is plain that, if the particle at {x, y, z) is in equilibrium, it would be in equilibrium if it were restricted to any line which could be drawn on the surface through the point {xy y, z). It must therefore satisfy a condition of the form Xdx -\- Ydy -\- Zdz = o, (2) for every direction in which the point can move on the surface. When a point moves upon the surface (i), the differentials dx, dy and dz must satisfy the differential equation (3) du du dy + du^ where du du du Tx' Ty^ dz are the partial derivatives of the function/". §iv.] EQUILIBRIUM ON A SURFACE, 59 Comparing equations (2) and (3), which must each hold for all values of the ratios of dx^ dy and dz^ we see that X^ Y and Z must be proportional to the partial derivatives, that is, ~ = — =~ du ~ du ~ du ^ dx dy dz (4) This expresses the two independent conditions of equilibrium which must hold for a particle subject only to the single restric- tion of lying upon a given surface.* Equilibrium of Interacting Particles. 81. When a mutual action exists between two particles in equilibrium, the intensity of this action (which, by the third Law of Motion, is the same for each particle) is one of the forces to be considered when the conditions of equilibrium are applied to the particles separately. For example, suppose two bodies whose weights are P and Q to rest at A and B^ Fig. 23, upon two planes perpendicular to one another, which intersect in a horizontal line (perpendicular to the plane of the diagram at C). The bodies are held apart by a rod AB of fixed length ; required the position of equilibrium. De- note the given inclination of the plane ^C by a^ and by Q the angle BAC, which, when found, will determine the position of equilibrium. The forces which act on either particle are, as C Fig. 23. * When the given surface is the boundary of a solid substance which the particle cannot penetrate, the resistance can only act out- ward ; but, in the general case, we suppose that the resistance may change sign ; as. for example, when a particle is constrained to a spher- ical surface by means of a rod connecting it with a fixed point, the rod b?ing capable of resisting either compression or tension. 6o FORCES ACTING AT A SINGLE POINT. [Art. 8i. represented in the diagram, its weight, the resistance of the plane on which it lies, and the thrust T of the rod AB (which is in compression), acting upon each in its proper direction. In the complete solution of the problem there are four equations (two for the equilibrium of each particle) and four unknown quantities, namely, the resistances R and -5", the thrust Z", and the angle 0, But, when required only to find the angle 6^, we may avoid the resistances and obtain two equations for the remaining quantities by resolving in each case along the plane only. Thus we obtain -P sin « = r cos e, and ^cos a ^T* sin ^ ; whence, eliminating T^ we have tan =z -p- cot o(^ which determines the value of 6, EXAMPLES. I\f. 1. Show that three forces represented in magnitude and direc- tion by the medial lines of a triangle, and acting at a point, are in equilibrium. 2. A weight of 25 pounds hangs by two strings, of which the lengths are 3 and 4 feet respectively, from two points in a hori- zontal line, distant 5 feet from each other. Find the tension of each string. 20 pounds ; 15 pounds. 3. A body is sustained upon an inclined plane by two forces, each equal to half the weight, one horizontal and the other act- ing along the plane. Determine the inclination of the plane. = 2 tan~'^. 4. A weight of 50 pounds, moving in smooth vertical guides is supported by a string, attached to a point at a horizontal dis- tance of 12 feet from the guides and 5 feet above the weight. Find the tension of the string and the pressure on the guides. 130 pounds ; 120 pounds. § IV.] EXAMPLES. 6 1 5. Two weights, P and Q^ are attached to the extremities of a string which passes over a smooth peg at a distance b vertically over the centre of a sphere of radius a^ on whose surface P rests while Q hangs freely. Find the distance s oi P from the peg when in equilibrium. _ Qb '~~P' 6. A picture, whose weight is W^ hangs from a nail by means of a cord, whose length is /, attached to two screw-eyes the hori- zontal projection of whose distance is a. What is the tension of the cord Wl 7. Weights P and Q are attached to the extremities of a cord of length /, which passes over a smooth peg at the distance h vertically above the centre of a smooth sphere of radius ^, upon whose surface P and Q rest in equilibrium. Show that the cord is divided at the peg into segments inversely proportional to the weights, and find the tension of the cord. PQl 8. An anchor weighing 4000 pounds is supported by two tackles from the fore and main yards of a vessel, making angles of 30° and 45° respectively with the vertical. Find the tension on each tackle. 2928 and 2070 lbs. 9. A weight W is supported by a tripod each leg of which is 3^ feet long, the feet making a triangle each side of which is 2\ feet long. Find the thrust in each leg. 14W 39 10. Two smooth rings of weights P and Q rest on the convex side of a circular wire in a vertical plane, and are connected by a string subtending the angle 2a at the centre. Determine the inclination 6 of the string to the vertical in the position of 11. Two smooth pegs are in the same horizontal line and 6 feet distant. The end of a string is made fast to one of them, and passing over the other sustains a weight of 10 62 FORCES ACTING AT A SINGLE POINT. [Ex. IV. pounds, while a smooth ring weighing 13 pounds is suspended on the bight. Find the length of string between the pegs. 120 . ^ feet. 4/231 12. A weight H^ is attached to a ring A which slides on a smooth circular hoop in a vertical plane. An equal weight is also attached to the ring by means of a string which passes over a smooth peg at the extremity B oi 2, horizontal diameter of the hoop. Find the position of A. 120° from B. 13. A barrel four feet long, weight 500 lbs., is hoisted from a ship's hold by means of a pair of can-hooks 52 inches long. Find the tension on each leg of the can-hook. 650 lbs. 14. A body is kept in equilibrium on a smooth plane of incli- nation or by a force P acting along the plane and a horizontal force Q. When the inclination is halved and the forces P and Q each halved, the body is observed to be still in equilibrium. Find the ratio oi P Xo Q, P .a -p: = 2 COS — . Q 4 15. A ring of weight W^ slides on a smooth rod fixed at an inclination of 30° to the horizontal. A weight W is attached to one end of a string which passes through the ring, the other end being attached to a fixed point not in the rod. Prove that there is no position of equilibrium unless W < W. 16. Two spheres of radii a and b and weights JV and W (which are supposed to act at the centres) are connected by a string of length / attached to points in their surfaces. They are in equi- librium when hung by the string over a smooth peg, their surfaces being in contact. Find the parts into which the peg divides the string. W'l + Wb - Wa Wl -^ Wa - Wb W -\- W W^ W 17. A bead is movable on a circular wire whose plane is vertical ; a string attached to it passes through a smooth ring at the highest point of the circle and supports a weight at its other end equal to that of the bead. Find the angle between the parts of the string in the position of equilibrium. 60°. IV.] EXAMPLES. 63 18. The ends of a string are attached to two heavy rings of weight W and W which are free to slide upon two smooth fixed rods making the angles a and /? with the horizontal and in the same vertical plane ; the string carries a third ring of weight M which slides upon it. Prove that, if is the angle which each part of the string makes with the vertical, cot : cot >5 : cot « = J/ : J/ + 2 ^' : J/ 4- 2 ^ CHAPTER III. FORCES ACTING IN A SINGLE PLANE. . V. Joint Action of Forces on a Rigid Body. 82. When forces act upon a solid body, the points at which the forces are applied are of importance in considering the motion produced. But in statics, the present division of our sub- ject, we are concerned only with the tendency to motion, or change of. motion, while the body is in a single definite position. The principle of transmission of force shows that no change of action is produced if the point of application is transferred to any point of the line of action, provided the new point of appli- cation is a point of the same rigid body — in other words, rigidly connected with the original point of application. It follows that it is the position of the line of action only, and not that of the point of application, which is at present of consequence. When however, we come to treat of the motion of the body the latter will be of consequence, because it affects the position of the line of action. We confine ourselves in this chapter to the action of forces whose lines of action lie in one plane, and we shall find that the joint action of such forces is in general the same as that of a cer- tain single force which is called their resultant. The body upon which the forces are supposed to act is frequently not represented in the figure at all ; but if there is a resultant force, it is of course assumed that some point of its line of action is rigidly §v.] CONSTRUCTION OF THE RESULTANT. 6S connected with the supposed body, so that it might serve as the point of application. Construction of the Resultant. 83. Let P^ , P^ and P^ , Fig. 24, represent in magnitudes and lines of action three forces acting in a plane. Let the lines of action of P^ and P^ intersect at A ; these forces may be trans- ferred to A and their resultant Q constructed at that point. The joint action of P^ , P^ and P, is evi- dently the same as that of Q and P^- Hence, constructing, in like manner, the resultant of Q and P^ at B^ the point of intersection of their lines of action, we have a force R whose ac- tion is the same as the joint action of P^ , P^ and P^, In like manner, we may construct the resultant of any number of forces in a plane, pro- vided that at no step the forces to be combined have parallel lines of action. It is evident that, in this construction, the magnitude and direction of the resultant are the same as if the forces, retaining their magnitudes and directions, had all acted at a single point. Thus the resultant considered only as a vector may be found from the given forces by simple vectorial addition as in Art. 58. But, in the present construction, we have in addition found the position of the resultant line of action. 84* The construction is simplified by separating completely the determination of the resultant vector from that of the line of action, as illustrated in Fig. 25. Let P, , P, , P^ and P ^ be four forces given in magnitude, direction and position, to find the resultant. Taking any point O as origin, we first construct the resultant vector by laying off from O successively OA equal and parallel to P^ , AB equal and par- allel to P^ , BC to P, , and CD to P ,. Then, as in Art. 58, OD is the vector representing the resultant. Now, to find the position 66 FORCES ACTING IN A SINGLE PLANE. [Art. 84. Fig. 25. of the line of action, join OB^ OC. Through, the intersections of the lines of action of P^ and P^^ whose vectors were used in find- ing B^ draw a line parallel to 0B\ the resultant of P^ and P^ is the force OB acting in this line. Next, through the intersection of this line with the line of action of P^ draw a parallel to OC ; the resultant of P^ , P^ and P^ is the force OC acting in this line. Proceeding in this way we finally arrive at the line in which acts the required resultant R^ which is equal and parallel to OD^ as represented in the diagram. If any change is made in the order of the forces, we shall arrive in the first part of the construction at the same point Z>, and in the second part, at some final point of intersection which will determine the same line of action for R. Such a new con- struction may be used to test the accuracy of the drawing. The Resultant of Two Parallel Forces. 85. The method given in Art. 83 fails when the two forces to be combined act in parallel lines, because there is no point of intersection to which we can transfer them. The difficulty is ob- viated by introducing two equal and opposite forces in any line of action which intersects the parallel lines. Thus, in Fig. 26, let the forces P and Q act in parallel lines at A and B. To find their resultant, let two forces, each equal to F^ acting in opposite directions in the line AB^ be intro- duced. Since these last forces coun- terbalance each other, it is plain that the system of four forces acting at A and B will have the same Fig. 26. § v.] RESULTANT OF TWO PARALLEL FORCES. 6/ resultant as F and Q. Constracting now the resultants of the pair of forces at A and the pair at B^ we have two forces whose resultant is the same as that of P and Q. Their lines of action meet at C, which is therefore on the line of action of the final resultant. If we now again resolve the two forces at C into com- ponents equal and parallel to their original ones, we shall have the system replaced by four forces acting at C; namely, two forces equal to Fy which counterbalance each other, and the forces P and Q^ of which the resultant is the force P -\- Q acting at C. 86. As before, the resultant, regarded as a vector, is the same as if the forces acted at a single point ; but we have also deter- mined the position of the line of action. Let this line intersect AB in M ; then the triangles CM A and CMB are similar to those used in the construction ; therefore .^=Z and ^-e MA F MB - F' Dividing, MB ^P MA Q (I) hence the point M divides the line AB inversely in the ratio of the forces. Since A and B are any points in the lines of action of P and Qy we see that the lines of action of two parallel forces P and Q and their resultant R — P -\- Q cut any transverse line in points A^ By and M such that P:Q:R = BM : MA : BA, .... (2) each force being proportional to the distance between the lines of action of the other two. We have seen in Art. 50 that, in the general case, each force is proportional to the sine of the angle between the lines of action of the other two. The present proposition is in fact the limiting case of the former. 87. Conversely, a force F may be resolved into components acting in any two lines parallel to its line of action and in one 68 FORCES ACTING IN A SINGLE PLANE. [Art. 87. plane with it. Thus if, in Fig. 26, the force R is given in posi- tion and magnitude, and the given lines of action of P and Q are on opposite sides of it, we have from equations (2) P=~^R and Q = -^R. ... (3) which determine the magnitudes of the components. 88. If the two given parallel forces act in opposite directions, the construction is the same, but the point C falls beyond the line „ of action of the greater force P. ^ yA ^ The demonstration and deduc- ^^'^ / [ tion of equations (i) and (2) are y' / I precisely the same as in the pre- x' / / ceding case; but the resultant is Q / / / in the direction of the larger force f// F a/ /m Z', and equal to Z* — ^. The line F AP is said to be divided ex/er nally in the inverse ratio of the forces. So also, in the converse P problem, where R is given to be ^^' ^'' resolved into two components in given parallel lines of action both on the same side of the line of action of i?, equations (3) of the preceding article hold for the magnitudes of the components ; but the component nearer to R exceeds it in magnitude, while the more distant is in the op- posite direction. The Resultant of a Number of Parallel Forces in One Plane. 89. It is an obvious consequence of the preceding articles that, for any number of parallel forces, the resultant considered as a vector, or resultant force ^ is the algebraic sum of the given forces. But, in finding the position of the line of action graphi- cally, it is more convenient, instead of combining the parallel forces two by two, to use a process similar to that of Art. 84. Thus, in Fig. 28, let forces P ^, P^,P,ar\d P^ act in the parallel §v.] RESULTANT OF PARALLEL FORCES. 69 11 . ! p. 'p. ■'?, 1 / .J,,^ \—- M '■■' Fig. 28. rines as represented. Assume also a force C acting in a line intersecting the line of action of /*,. Taking any point (7, we construct vectorially the resultant of Q^ Z',, -P,, /'a and P^ by lay- ing off OA equal and parallel to Q, and then ^^, BC, CD, DE equal and parallel to the given forces. Completing the figure, as in Art. 84, OE represents vectorially the re- sultant of Q and the given forces. To find the line of action of this resultant, draw through the intersection of the lines of action of Q and P^ a parallel to OB ; the resultant of these two forces is tlierefore OB acting in this line. Again, through the intersection of this line with the line of action of /*,, draw a line parallel to OC : the resultant of Q, jP, and P, is the force OC acting in this line. Continuing in this manner, we finally obtain the line of action of the resultant OE. Now the resultant of this force with one equal and opposite to Q in the same line of action, is the resultant of the parallel forces ; hence the resultant line of action passes through A/, the intersection of this last line with the line of action of Q. Thus the resultant of the parallel forces is the force P equal to their algebraic sum acting in a parallel line through M. 90. If we reverse the direction of the resultant force OE, and denote it (in the direction EO) by Q\ we have six forces, namely, Q, Pit P,f P,, P^a.nd Q\ in equilibrium, as represented vec- torially by the closed polygon OABCDEO in the left-hand figure, which is called \.\\q force diagram. The resultant of Q and Q' is therefore the reverse of the resultant of the parallel forces. Supposing the parallel forces to be the weights of given bodies acting in vertical lines, Q and Q' will be oblique forces which are together capable of sustaining the weights while acting in the given lines. The broken line formed in the construction of the right-hand figure may be regarded as a cord to which the given yo FORCES ACTING IN A SINGLE PLANE. [Art. 90. weights are knotted at the points of intersection, and of which the extremities are attached to fixed points in the lines of action of Q and Q\ The figure is hence called a funicular polygon for the given parallel forces. Taking into consideration the action and reaction at its two ends of the tension in each of the intermediate segments, we notice that the triangles in the force diagram represent the separate equilibrium of each of the knots. The term funicular polygon is sometimes extended to the more general case, in which the points of application of the forces are supposed con- nected by rods, some of which may be in compression instead of tension. Compare Arts. 122 and 125. Couples, 91. When the parallel and oppositely directed forces in Art. 88 are equal, the construction fails because the lines by which the point C was found are, in this case, parallel. Indeed the mag- nitude of the resulting force considered as a vector is now zero, and yet the forces are not in equilibrium. This combination of two equal opposite forces acting in parallel lines is called a couple p because it cannot be reduced to any simpler mechanical equivalent. Thus Fig. 29 repre- sents a couple acting upon a rigid body, which may here be regarded as a lamina or thin plate in the plane of the two parallel lines of action. The mechanical action of the couple is obviously a tendency to turn the lamina in its own plane. This tendency is called a turning moment or simply a moment. It cannot be counteracted by means of a single force, and if motion is pre- vented by the resistance of fixed bodies in contact with that on which the couple acts, the reaction of these bodies is equivalent to a turning moment in the opposite direction of rotation. § v.] MOMENT OF A FORCE, 7 1 Measure of Turning Moment. 92. We have seen in Art. 86 that the resultant of two parallel forces in the same direction acts in a line dividing a transverse line in the inverse ratio of the A P forces. A force opposite to this _ resultant will therefore produce -^ — g equilibrium. Thus, in Fig. 30, in M — ^Q which the transverse line AB is B^ taken perpendicular to the parallel Fig. 30. lines of action, the force P -^ Q acting at M is in equilibrium with P and Q^ if AM'.MB =Q;P^ (i) or Py. AM = (2 X MB (2) Distinguishing the direction of P and Q as the positive direction along the parallel lines, the forces in equilibrium may be regarded as forming two couples; namely, P acting at A ^ith — P acting at M^ and Q acting at B with — Q acting at M. These two couples are therefore in equilibrium; in other words, their turning moments, which are in opposite directions, are equal in magnitude. Equation (2) shows that the product of the magnitude of the force and the distance between the parallel lines of action is the same for each of these couples. This product is therefore taken as the meas- ure of the moment of the couple. For example, if ^ = 2 pounds and BM = 3 feet, the moment of the ^-couple is said to be 6 pounds-feet, and when algebraic signs are used this moment is taken as positive because it tends to produce positive rotation. Moment of a Force about a Point. 93. If the point B of the lamina in Fig. 29 is fixed, and the lamina, while free to turn about B^ is acted upon by a force P at the point A^ it will tend to turn about the point B. The measure of this tendency is called the moment of the force about the point B. The resistance at B which prevents the motion of 72 FORCES ACTING IN A SINGLE PLANE. [Art. 93. that point of the lamina is a force equal, parallel and opposite to Py and the turning effect is produced by the couple thus formed. Therefore its measure is taken to be the same as that of the couple; that is, the moment of a force about a given point is the product of the magnitude of the force and the perpendicular from ike point upon the line of action. Accordingly in Fig. 30 the forces P and Q are said to have equal and opposite moments about M \ and if AB is a rigid bar free to turn about a fixed point J/, it will be in equilibrium when parallel forces P and Q act at A and B^ provided J/ divides AB in the inverse ratio of the forces. In this arrangement, the bar is called a lever ^ the point M the fulcrum ^ and AM^ MB the arms. The proposition just stated, upon which is based the measure of moments, has been known from the early days of mechanical science as the principle of the lever. The perpendicular from the point is often called the arm of the moment so that the moment is said to be the product of the force and the arm. In like manner the distance between the parallel lines of action is called the arm of the couple. 94' It is sometimes convenient to employ, as the line factor in the expression for a moment, the dis- b tance of a definite point of application of the force from the point about which the moment is taken. This can be done by means of the following theorem: The moment about O of a force P acting at A is equal to the product of OA and the resolved part of P in the direction perpendicular to OA. To prove this, let AC, Fig. 31, be perpen- dicular to OA, and OB perpendicular to the line of action; then, denoting the angle Fig. 31- BAC by )/;, we have OB = OA cos i\ and the resolved part of P in the direction AC is P cos tp. Hence, denoting the moment by If, we have If z= P X OB = P X OA cos Jp = P cos ip . OA, / §v.] THEOREM OF MOMENTS. n that is, the product of OA by the resolved force. Since OA is the arm for the resolved force, we may state the theorem thus : The moment about O of a force acting at A is the same as the moment of its resolved part perpendicular to OA. Resolving forces in the direc- Varignon*s Theorem of Moments. 95. Varignon's theorem, that : The moment about atiy point of the resultant of two forces is the algebraic sum of the moments of the given forces, follows directly from the theorem of the preceding article. Thus, in Fig. 32, let O be the point about which the moment is taken, or origin of moments, and let /^ and ^ be the forces acting in lines which in- tersect at A. Construct the resultant R at A, join OA, and draw AB at right angles to OA. tion AB by projecting the lines representing P, Q and R upon AB, we have AJ\r = AZ + AM. Multiplying by OA, \ve derive the equation OA . AN = OA . AL-^rOA . AM, in which, by Art. 94, the several terms are the moments of R, P and Q about O. Hence, denoting the perpendiculars from O by r, p and q, rR=pP^ qQ, In the diagram, AB is taken as the positive direction for the resolved forces, because a force in that direction has a positive moment about O. Then the signs of the moments are the same as those of the resolved forces ; and, for all positions of O, we have the moment of the resultant equal to the algebraic sum of the moments of the given forces. 96. When the forces are parallel, let a perpendicular be 74 FORCES ACTING IN A SINGLE PLANE, [Art. 96. drawn, as in Fig. 33, from O^ cutting the parallel lines of action in A^ B and C, at which points we may regard A JP, Q and R as acting. Then we have seen in C Art. 86 that F .AC=Q.BC. . . . (1) Denote the distance OC by x ; then the sum of B the moments of P and Q about O is F(x + AC) -f Q{x - BC\ which by equation (i) reduces to {P^-Q)x. „ But this is the moment of R about O. since Fig. 33. R^= P -\- Q^ and x is the arm OC* Hence, as before, the moment of the resultant is the sum of the moments of the components. The proof is readily extended to cases in which O is between the lines of action, and to that in which the forces have opposite directions. Thus, in Fig. 34, we have as before P . AC= Q,BC, R~ and the algebraic sum of the moments is P{x - AC) - Q(x - BC), P which reduces to {P — Q)x or Rx, Thus Varignon's Theorem is true for any par- ^ allel and unequal forces. 97. Finally, when the two given forces are parallel, equal and opposite, their resultant is the couple which they form, and the theorem is that: Fig. 34. T^e algebraic sum of the moments of the forces forming a couple has for every point in the plane of the couple the same value as the moment of the couple. To prove this, draw through O a line OBA perpendicular to the parallel lines of action, as in Fig. 35. Supposing O to be beyond the lines of action as indicated, denote its distance OB from the nearer line by x, and the arm AB of the couple by a. Then the moment about O of P acting at A is P{a + x), and that of P acting at B is A p B a? a P § v.] MOMENT OF A COUPLE. 75 Px in the opposite direction ; hence the algebraic sum of the moment is P{a -\-x) - Px = Pa, which is independent of x, and has been already defined in Art. 92 as the moment of the couple. When the point O is between the lines of action the moments of the compo- nents have the same algebraic sign, and the moment of the couple is their numerical sum. 98. If H^ , H^ and H^ denote the mo- ments of the forces P^, P^ and P^ about a selected point O in the plane of the forces, and Q the resultaht of P^ and /*,, the mo- ment of Q about (9 is ^, + H^ by Varignon's Theorem. Again, if R is the resultant of Q and P^ , its moment about O is, by the same * '^^' theorem, ZT^ + ZT, + H^, But R is the resultant of P^, P^ and P^ ; hence the moment of the resultant of these forces is equal to the algebraic sum of the moments of the forces ; and, in like manner, for any number of forces, if K denote the moment of the resultant Ry we have This resultant moment of a system of forces in a plane is fre- quently called simply the moment of the system with respect to the given origin of moments. Three Numerical Elements Determining a Force in a Given Plane. 99. A force acting in a given plane and at a given point, or con- sidered merely as a vector in a given plane, requires two numerical values for its determination. These may be the values of P and B, the magnitude and one angle determining the direction of the force ; but we have seen in Art. (i2> ^^^ the most convenient determining elements (or coordinates, in the general sense of the term) are the values of X and F, the components of the force in two standard directions, because these are combined in the result- ant by simple algebraic addition. J^ FORCES ACTING IN A SINGLE PLANE. [Art 99. When the forces in the given plane are not confined to a single point of application, they are not completely represented by vec- tors, and we require in addition a third numerical element to fix the position of the line of action after its direction has been fixed. This third element might be taken as the perpendicular from a fixed point of reference upon the line of action ; but the theorem of moments shows that it is more convenient to take-for the third element the moment about the point of reference, because then the values of this e)ement also are combined in the resultant by simple algebraic addition. We therefore take X, F and ^(the resolved forces in two fixed directions and the moment about a fixed point) for the three determining elements of a force jP in a given plane ; then :2X, 2F and 21/ are the corresponding determining elements of the resultant of a system of forces in one plane. 100. Let J^ denote the resultant vector of the given system of forces, J^f as in Art. 98, the resultant moment, and p the perpen- dicular from O the origin of moments upon the line of action of the resultant. Then, by the theorem of moments, which determines the value of/. The direction of the vector ^ determines that of the perpendicular line upon which/ is to be laid off, and the sign of X determines in which of the two opposite directions it is to be laid off from O. It is to be noticed that, in the case of the single force, if we have X = o and K = o, we shall have I* = o, and therefore If = o for any position of the origin of moments O. But in the case of the resultant, we may have 2X =0 and 21^=^ o, (whence ^ = o,) without having K — o. In this last case, the resultant is not a force, but the couple whose moment is K\ and the value of K is, in this case, independent of the position of O. The system of forces will produce equilibrium only when all three of the elements vanish, that is, when 2X = o, -^F = o and 2H =0. § v.] RESULTANT OF FORCE AND COUPLE. yy Resultant of a Force and a Couple. lOI. Since a couple in a given plane has no element of magni- tude except that of moment, the direction and magnitude of the force P, employed in the graphical representation of a couple, are immaterial, provided only the arm a be so taken that aP = If, the given value of the moment of the couple. Hence, to find the resultant of a force P acting at A, Fig. 36, and the couple FT, we may put ^ B H = aP '^ (which determines a), and then represent H by the force P reversed at A and an equal force acting in a parallel line at u J- AD r u • • , Fig. 36. the distance AB = a from the origmal line of action. (In the diagram AB is laid off on the supposition that H '\% positive.) Then the two forces acting at A neutralize each other ; therefore, the resultant of /* at ^ and the couple ^is the force P acting at B. Thus the result of combining a couple with a force, is not to change it as a vector, but to shift its line of action. The effect is algebraically to increase its moment with respect to any point by a constant quantity. 102. Conversely, the force P acting at B, Fig. 36, may be re- solved into an equal and parallel force, acting at any selected point A, and a couple; and the moment H oi this couple is that of the force about the selected point. The process of combining forces into a resultant in Art. 100 may in fact be described thus: We first resolve each of the forces into a parallel force acting at the selected origin of moments O and a couple H\ we next combine the forces acting at O into a resultant /v?, and the couples H (by algebraic addition) into a couple K\ finally, unless ^ = o, we combine the couple K with R acting at O, so as to shift its line of action as in Art. 101. Mon\ent of a Force Represented by an Area. 103. When a force P is represented by a line of definite length and in a definite position, its moment about a point O is rep- resented by twice the area of the triangle whose base is the line 78 FORCES ACTING IN A SINGLE PLANE. [Art. 103. representing P and whose vertex is O ; for the altitude of this tri- angle is /, the arm of the moment, and the area is one half the product of the base and altitude. As an application, let us consider a system of forces acting in the sides of a polygon in one consecutive direction around the perimeter and proportional to the sides in which they act. Thus, in Fig. 37, let the forces be represented in magnitude and posi- tion by AB^ BC, CD, DE, EA. Take any point O ; then, joining Q it with A, By C, D and ^, the moments of the several forces about (9 are represented by the doubles of the triangles AOB, etc. Therefore K, the sum of the moments (which in the diagram have all the same sign), is twice the area of the polygon. Since the vectorial sum of the forces is by hypothesis zero, the resultant is not a force. It is therefore a couple K measured by twice the area of the poly- gon. Accordingly, we find here, as in Art. 97, that the resultant moment is the same for every position of the point O. Fig. 37. Forces in a Plane Referred to Rectangular Axes. 104. When the forces and their points of application are re- ferred to rectangular axes, the moments are usually taken about the origin. In Fig. 38, let X and Y be the rectangular components along the axes of the force P, and let x andjv be the coordinates of its point of application A. The moment of P about the origin is the algebraic sum of the moments about the same point of X and Y acting at A. The numerical values of these moments are ^A" and ^F. When ■ Xy y, X and Y are all positive, as in the figure, it will be noticed that the mo- Fig. 38. ment of Y is positive and that of X is negative ; hence the mo- ment of P about the origin is H = xY -yX, (i) § v.] REFERENCE TO RECTANGULAR AXES. 79 105. When the force is determined by given values of the ele- ments A', Y and H^ as in Art. 99, the point of application {x, y) is not fully determined; for, it has only to satisfy equation (i), which is therefore the equation of the line of action. We have seen, in Art. 64, how the inclination to the axis of ^and the value of P are determined; and, denoting by / the perpendicular from the origin upon the line of action, p is determined by Zr= Pp. In like manner, in the case of a system of forces, we have, for the rectangular components of the resultant and the resultant moment about the origin, X' = 2X, Y' = 2Y, K = :SxY - 2yX -, hence the equation of the line of action of the resultant is ^Y' - yX = K. (2) Accordingly, we find that, if X = o, this equation represents a line passing through the origin. If X' = o, it represents a line parallel to the axis of y; and if Y' = o, a line parallel to the axis of X. If X' = o and Y' = o, while X does not vanish, the equa- tion becomes the impossible one representing the line at infinity, the resultant being, in this case, the couple whose moment is X. EXAMPLES. V. 1. ABCD is a square; a force of one pound acts along AD^ a force of two pounds along AB, and a force of three pounds along CB. Determine the resultant and its line of action. 2. ABCD is a parallelogram; forces represented in magnitude and position by AB^ BC and CD act on a body. Determine the resultant. Reversing the resultant, so as to produce equilibrium, explain the result by the theory of couples. 3. Show that a force may be graphically resolved into two com- ponents, one acting along a given line of action coplanar with that of the force, and the other at a given point in the same plane. 4. Show how to resolve a force graphically into three com- ponents acting along three given lines coplanar with the line of 80 FORCES ACTING IN A SINGLE PLANE. [Ex. V. action. When is this impossible? and when does one component vanish ? 5. ABCD is a square. A force of 3 lbs. acts from A \q B^ d. force of 4 lbs. from ^ to C, a force of 6 lbs. from D to C, and a force of 5 lbs. from A to D. Show that the line of action of the resultant force is parallel to the diagonal AC, and find where it crosses AD. At a distance of \AD from A. 6. The distance between the lines of action of two parallel forces P and Q is a. What is the moment of either force about a point in the line of action of the resultant ? aPQ 7. A man supports two weights slung on the ends of a stick 40 inches long placed across his shoulder. If one weight be two thirds of the other, find the point of support, the weight of the stick being disregarded. 16 inches from the larger weight. 8. To a rod 10 feet long, carried by A and B, a weight of 100 pounds is slung, by means of two cords; one, 4 feet long, attached to a point 2 feet from A'% end, the other, 3 feet long to a point 3 feet from B's end. Determine the portions carried by A and B, 48 lbs.; 52 lbs. 9. If the side of the square in Ex. i is two feet in length, find the resultant moment about C and about D. 2 pounds-feet in each case, directions opposite. In the following proble7ns, the weight of a uniform beam or rod is regarded as acting at its middle point. 10. A horizontal uniform beam, 5 feet long and weighing 10 pounds, is supported at its ends on two props. How far from one prop must a weight of 30 pounds be placed on the beam, in order that the pressure on that prop may be 25 pounds ? 20 inches. 11. A rod weighing i pound per foot turns about a smooth hinge at one end, and is held by a string fastened to a point 10 inches from the other end. If the string can only sustain i^ pounds tension, find the limiting lengths of the rod, when held in a horizontal position. i and 5 feet. ^ v.] EXAMPLES. 8 1 12. A bar AB^ weighing \ oi z. pound per linear inch, rests on a prop at A and carries a weight of 125 pounds at a point 10 inches from A. Find the length of the bar, in order that the force P acting at B to produce equilibrium may be the least pos- sible. 8 feet 4 inches. 13. The forces P and Q act at A and B perpendicularly to the arms of a bent lever, or "bell-crank," -4 Ciff which turns about the fulcrum C. Show that in equilibrium the resultant of P and Q passes through C, and thence derive the measure of a turning moment. 14. Weights of 3 pounds and 5 pounds respectively hang from pegs in the rim of a vertical wheel, whose radius is 2 feet, at the extremity of a horizontal radius and at a point 120° distant. What is the resulting moment at the centre ? and if the wheel be blocked by a fixed peg touching a spoke at a point 3 inches from the centre, what is the pressure on the peg ? I pound-foot; 4 pounds. 15. Demonstrate the equivalence of two couples having the same moment but different forces and arms, by reversing one of them and showing that the four forces are in equilibrium. 16. ABC is a triangle. Show how to construct the line of action of a force whose moments about A^ B and C are in the ratios l\ vr. n. 17. Verify geometrically the value of / found from the value of H'vci Art. 104. 18. If H^ is the moment at the origin, and X, Kthe resolved parts of a force referred to rectangular axes, show that expresses the moment of the force about the point (x^y). 19. If four forces acting along the sides of a quadrilateral are in equilibrium, prove that the quadrilateral is plane ; and that, if it can be inscribed in a circle, the forces are proportional to the opposite sides. 20. Four forces act in, and are inversely proportional to, the sides AB^ BC^ CD and DA of a quadrilateral inscribed in a circle. 82 FORCES ACTING IN A SINGLE PLANE. [Ex. V. Show that the resultant moment about the intersection of AB and CD vanishes; and, thence, that the Ijine of action of the resultant passes through the intersections of pairs of opposite sides. 21. Four forces acting in the sides of a trapezoid are in equi- librium. Prove that the forces in the non-parallel sides may be represented by the sides themselves, and those in the parallel sides each by the opposite side. VI. Conditions of Equilibrium for Forces in a Single Plane. I06. The forces acting upon a rigid body at rest, including the resistances of other bodies with which it is in contact, form a system in equilibrium. We consider in this section cases in which the forces act in a single plane, but at different points of applica- tion. When such a system is in equilibrium, not only must the resultant force R^ considered as a vector, vanish; but A", the result- ant moment (Art. 98) of the forces about any selected point, must vanish. We therefore have, in addition to the general condition of equilibrium used in § IV (namely, that the resolved forces in the direction of any straight line must balance each other), a new general condition ; namely, that the algebraic sum of the moments of the forces about any point in the plane must vanish; or, what is the same thing, that the sum of the moments of the forces tending to turn the body in one direction about the point must be equal to the sum of those tending to turn it in the opposite direction. 107' As in Art. 71, it is important to represent, in the diagram constructed for a problem, d;// the forces, including the resistances, which act upon the single body whose equilibrium is considered, and only those forces. As an illustration, take the following problem : Let AB^ Fig. 39, represent a uniform heavy beam, §vi.] CONDITIONS OF EQUILIBRIUM. 83 6 feet long, resting at A upon a smooth horizontal plane, and at D upon the smooth top of a vertical post, 3 feet high, fixed in the plane. The end A is prevented from slipping by a cord ^C, 4 feet long, connecting it with the foot ot the post. Required the tension of this cord. The only force acting from a distance is the weight of the beam, which, because the beam is uniform, may be regarded as acting at its middle point M, The remaining forces are the resistances of fixed bodies in contact with the beam opposing its motion. Care must be taken to assign to them the directions, not of the action of the beam upon the obstacle, but of the reaction of the obstacle upon the beam. For the horizontal plane, this reaction is vertically upward as repre- sented by R^ because the plane is smooth. For the cord, it is the required tension T in the direction of the cord. Finally, the action at D is perpendicular to the beam for the same reason that the action at A is perpendicular to the ground, and the arrow represents the action of the fixed point Z>* upon the beam. 108. Since, in the right triangle ACD, AC — 4 and CD ~ 3, we have^Z> = 5, and denoting the angle DAC by or, sin « = I, cos <^ = i; a is thus a known angle, and the directions of all the forces are known. There are therefore in this problem only three unknown quantities, namely, the magnitudes of the three resistances R. T and S. * Although we speak of Z> as a fixed point, there are really two surfaces in contact, just as if D were a round peg; and when the con- tact is smooth, the direction of the mutual action is a common normal to the surfaces at their point of contact. 84 FORCES ACTING IN A SINGLE PLANE. [Art. io8. We have seen in § IV that equations of equilibrium may be found by resolving forces in various directions. But it is shown in Art. 73 that we can in this way obtain but two independent equations ; hence it is necessary in this case to obtain at least one new condition by the principle of moments. Resolving vertically and horizontally for two conditions, we have, from vertical forces, W =^ R -^ S cos «, from horizontal forces, 7" = 5 sin a. For the third condition, it is convenient to take the origin of moments at A where R and T have no moments ; thus, from moments at ^, fT X AM cos nr = 6' X AD. Substituting the numerical values of AM^ AD and the trigo- nometrical functions, we have W^R^>iS, (.) T=\S, (2) ^W=$S, (3) from which by elimination we obtain Number of Independent Conditions. 109. The solution given above shows that, in a problem involving the equilibrium of a solid body under the action of coplanar forces, three unknown quantities may be determined by ineans of three equations of condition, of which one must be derived from the principle of moments. It may furthermore be shown that, if three conditions thus found are satisfied, all other equations found by taking moments W//J/ be satisfied. For, using the notation of Art. 100, when the § VI.] INDEPENDENT CONDITIONS. 85 two equations derived by resolving are satisfied, R (the resultant of the system considered merely as a vector) vanishes, so that the system is either in equilibrium or else equivalent to a couple. But when, by the third condition, the moment about any one point vanishes, the resultant is not a couple; therefore the system is in equilibrium, and the moment about every point is zero. It follows that a problem of this kind is indeterminate if more than three independent quantities are unknown. 110. It is to be noticed that, although the equations derived by resolution are generally the most simple, yet two or even all three of the independent conditions may be found by taking moments. For, let us consider what follows when it is known that the moment of a given system of forces about a given point A vanishes. The resultant of the system, in this case, cannot be a couple, but may be a force whose line of action passes through A. If now the moment of the system about some other point B is also known to vanish, the line of action of the resultant, if there be one, must be the line AB. When these two conditions are given, the sum of the resolved forces in a direction perpen- dicular to AB necessarily vanishes, and so does the resultant moment about any point in the line AB. But an independent third condition is furnished by the vanishing of resolved forces in any other direction, or of moments about any point not collinear with A and B.'^ Choice of Conditions. III. When only one or two of the three unknown quanti- ties are required, it is desirable to use conditions which are inde- * The vanishing of the resolved force in a given direction is but the limiting form of the vanishing of the moment about a distant point. Thus, the vanishing of horizontal forces may be said to express the vanishing of the moment about an infinitely distant point in the verti- cal direction. Hence, the conditions are always the vanishing of the moments about three points, and the conditions are independent when the three points do not lie in a straight line. 86 FORCES ACTING IN A SINGLE PLANE, [Art. iii. pendent of one or more of the unknown quantities whose value is not required, because we shall then be able to employ a smaller number of equations. In the case of a condition ob- tained by resolving forces, we have seen in Art. 72 that this is done by resolving perpendicularly to the line of action of the force which is to be avoided. In the case of a condition obtained by taking moments, it is done by choosing a point on the line of action for the origin of moments. Thus, in the example solved in Art. 108, supposing 7" only to be required, we may avoid introducing R and write only equations (2) and (3). Again, if S only were required, equation (3) would suffice, since by taking moments about A we have avoided both R and T, Case of Three Forces. 112. When the number of forces acting in a plane upon a solid body in equilibrium is but three, the lines of action must either meet in a point or be parallel ; for, if two of the lines in- tersect, the forces in these lines have no moment about the point of intersection; hence the third force can have no moment about the same point. Thus the principle that: The lines of action of three forces in equilibrium must meet in a point is equivalent to a condition of equilibrium derived from the principle of moments. 113. This form of the condition is of frequent application when one of the unknown quantities determines the direction of one of the B forces acting at a given point. For example, let the uniform heavy rod AB^ Fig. 40, rest in a vertical plane with its upper end B against a smooth vertical wall to which it is inclined at an angle of 45°, and the lower end held at a fixed point A in such a manner that there is no resistance to turning exerted at the point. This is frequently expressed by supposing the end of the rod fastened § VI.] CASE OF THREE FORCES. 8/ to yi by a smooth hinge. Thus the action at A is simply a force acting on the rod at that point in a direction as yet unknown. The only other forces are the weight acting at the middle point Z>, and the action of the wall, which is horizontal because the wall is smooth. Producing the known lines of action to meet in O, AO '\% the line of action of the resistance at A. The geometry of the figure now shows that the inclination to the horizontal of the force acting at A is tan "'2. Thus one of the three unknown quantities has been determined by a single condition ; the other two may now if required be found by the other conditions of equilibrium, or by a triangle of forces. It is obvious that in this problem we might have employed as in the problem of Art.'ioy, the horizontal and vertical com- ponents of the force at A^ for two of the unknown quantities, instead of its direction and magnitude. 114. The principle is particularly useful in problems where one of the unknown quantities determines the position of equilib^ Hum of a movable rigid body under given circumstances. For example, suppose the uniform heavy rod AB^ Fig. 41, of length 2^, to be in equilibrium in a vertical plane, with its lower B end A against a smooth vertical ^\^ wall (perpendicular to the plane of ^^v the diagram), and resting at some, point of its length upon a smooth , horizontal rail (piercing the plane of the diagram at D) parallel to the wall and at a distance b from it. It is required to determine the inclination 6 of the rod to the hori- zon. The forces acting on the rod are its weight W, acting vertically at the middle point C, the resistance I^ of the wall, acting horizontally at A because the wall is smooth, and the resistance F of the rail, acting at right angles to the rod because the rail is smooth. Hence, to be in equilibrium, the position of the rail must be such that the lines of action of these three 88 FORCES ACTING IX A SINGLE PLANE. [Art. 114. forces meet in a point O as represented. We have therefore OA = a cos 6, AD = a cos' ^, and hence b = a cos' 0^ which determines B. The values of P and J^ may now be found by resolving vertically and along the rod, namely: P = IVsec e, R= IV tan 6, Equilibrium of Parallel Forces. 115. When all the forces are parallel and in a single plane, the resolved forces in that direction in the plane which is at right angles to the lines of action vanish. In this case, then, we have to consider but two conditions of equilibrium, one at least of which must be obtained by taking moments. The most familiar examples are those in which the forces are the weights of bodies applied at different points of a rigid body, such as a beam in a horizontal position, the two unknown quantities being the magni- tudes of the upward supporting forces, or the position and magnitude of a single force. Thus, let the beam AB, weighing 280 pounds and 20 feet long, be supported at its ends, and loaded with a weiglit of 200 pounds at a point 4 feet from A, a weight of 320 ■^ pounds at a point 5 feet from jB, in addition to its own weight acting at the middle point; required the reac- ^ tions, F and Q, of the supports at A and B. Taking moments about A, 200 00 T 1 el 5.5 Fig. 42. we obtain a condition of equilibrium independent of F, namely, 20^ = 4 X 200 4- 10 X 280 + 15 X 320 = 8400, whence Q = 420 pounds. F may be found in like manner, or § VI.] EQUILIBRIUM OF PARALLEL FORCES. 89 more simply (having found Q) from the equation of vertical forces, p J^ Q=z 800, whence P = 380 pounds. 116. We may also proceed as follows, which is sometimes more convenient. Resolving each of the downward forces into components acting at A and B^ P is evidently the sum of the components acting at A^ and Q is the sum of those acting at B. Thus, the 200 pounds is divided into components inversely pro- portional to 4 and 16, the distances of its point of application from A and B\ that is to say, -J of it, or 160 pounds, is the com- ponent at A^ and \ of it, or 40 pounds, the component at B. In like manner, \ of the 280 pounds, or 140 pounds, is the compo- nent of this force at A^ and 140 pounds acts at B. Finally, \ of the 320 pounds, or 80 pounds, acts at A^ and | of it, or 240 pounds, acts at B. Hence, adding the like components, we have P = 160 + 140 + 80 = 380, ^ = 40 + 140 + 240 = 420. EXAMPLES. VI. 1. A uniform beam of weight W rests against a smooth ver- tical wall, and a smooth horizontal plane with which it makes the angle a. Its lower end is attached by a string to the foot of the wall. Find the tension of the string. ^ W cot a. 2. The ends A and B oi a. uniform rod, of length 2/^ and weight PF^ are fastened by strings, whose lengths are 2a and «, respec- tively, to the point C in a smooth vertical wall, the rod and strings lying in a verti- cal plane perpendicular to the wall, against which B rests below C, as in Fig. 43. De- noting by the inclination of the longer string to the wall, find the tensions S and T of the strings. 90 FORCES ACTING I A' A SINGLE PLANE. [Ex. VI. 3. A uniform beam of weight W, and length 3 feet, rests in equilibrium with its upper end A against a sn.ooth vertical wall, while its lower end £ is supported by a string, 5 feet long, whose other end is attached to a point C in the wall. Find AC and the tension of the string. AC = -^ feet • T=^^-^W 4/3 ' 8 * 4. A uniform lower boom AB = /, whose weight is IV, is sup- ported at C in a horizontal position by a topping-lift, inclined at 45°. Given AC = |/, find the tension of the lift, and the thrust at the goose-neck A. 2 4/2 4/5 3 ' 3 ' 5. Prove that three forces acting at the middle points of the sides of a triangle perpendicularly inward, and proportional to the sides, are in equilibrium, and extend the theorem to a plane polygon of any number of sides. 6. A uniform beam of length / rests upon the horizontal rim of a hemispherical bowl of radius a, with its lower end upon the smooth concave surface. Det*?rmine its inclination 6 to the horizon when in equilibrium. n I , !/(/'+ 128^-) cos iJ = — - H — . \oa 16a 7. A bar, whose length is /, rests upon a smooth peg at the focus of a parabola whose axis is vertical and whose parameter is 4a, with its lower end on the smooth concave arc. Find its inclination to the axis. . , ^ a cos if^ = — 8. A uniform rod AB^ of length 4^, has the end A in contact with a smooth vertical wall, and one end of a string is fastened to the rod at the point C, such that AC = «, while the other end is fastened to the wall. Show that, in order that equilibrium may exist, the string must have a definite length, but that its inclina- tion is indeterminate. 9. A uniform yard-stick weighing 10 ounces is supported in a horizontal position by the thumb at one end, and the forefinger at a point 3 inches from the end. What is the pressure on the thumb and on the finger 1 " 50 oz. ; 60 oz. § VI.] EXAMPLES. 91 10. If, in Fig. 39, the cord is attached to the post at a point I foot from the ground, find the values of R and 7", supposing ^=250 pounds. r= 18 4/17 lbs.; 7? = 136 lbs. 11. A uniform beam ^j9, 3 feet long, weighing 10 pounds, rests in equilibrium with 4 pounds hanging from A and 13 pounds from B. What is the position of the point of support ? 2 feet from A. 12. The scales A and B of a false balance are at unequal distances, a and ^, from the fulcrum or point of support, but they balance when empty. Find the true weight W of a body whose weight appears to be P when placed in the scale A^ and Q when placed in the scale B\ find also the ratio of b to a. b 13. If, in example 12, the scales do not balance when empty, A tending downward with the moment If, find the ratio b to a and the value of If, supposing the true weight IV known. b _ IV- Q W - PQ a ^ P-W' P-W • 14. A beam AB weighing -J ton per running foot and 18 feet long is loaded with 4 tons at A and 5 tons at B\ it is supported at points 4 feet from A and 6 feet from B. Find the supporting forces P and Q. P = 5I tons; Q = 12^ tons. 15. A topgallant yard 40 feet long, weighing 1680 pounds, is supported in a horizontal position by lifts attached to the ends, which if produced would meet in a point 21 feet above the yard. Find the tension on either lift. 1160 pounds. 16. A uniform plank 20 feet long, weighing 42 pounds, is placed over a rail; two boys weighing 75 and 99 pounds, respec- tively, stand each at a distance of a foot from one end. Find the position of the rail for equilibrium. I foot from the middle point. 17. A uniform rod 22 feet long, weighing 80 pounds, rests with its upper end against a smooth vertical wall and its lower end supported by a cord, 26 feet long, attached to a point in the wall. Find the tension of tlie cord. 130 pounds. 92 FORCES ACTING IN A SINGLE PLANE. [Ex. Vi. 1 8. An iron rod weighing 4 lbs. per linear foot projects from a cask 3 feet in diameter and 4 feet high, the part within the cask forming a diagonal of the vertical section through the axis. The weight of the cask is 60 pounds and is assumed to act at its centre. Find the length of the rod if the cask is on the point of overturning. 15 feet. VII. The Rigid Body regarded as a System of Particles. II7« When a limited number of forces act upon a body at rest, each of the several points of application may be regarded as a particle in equilibrium. In this point of view, the forces which act upon the body (including of course the resistances of other bodies) are called the external forces. The forces which act at any one of the points of application, and there produce equilib- rium consist of some of the external forces together with the resistances of other parts of the body. These resistances are called the internal forces. The whole set of internal forces consists of actions and reac- tions between the several parts of the body, forming stresses; of these the simplest kind are the tensions and compressions con- sidered in Arts. 26 and 27. The nature of the stresses thus produced in a body by external forces depends partly on the material and form of the body ; and their study, together with that of the changes of form the body may undergo before reach- ing a state of equilibrium, forms a special branch of the Science of Mechanics, namely, the Mechanics of Structures and Materials. 118. But when we are, as at present, considering only the equilibrium of the external forces, we may imagine the simplest possible rigid connection to exist between the several points of application. For example, in Fig. 24, p. 65, the forces /*, , /*,, P^ and the force — R (which is equal and opposite to the resultant R^ and in the same line of action) form a system of four forces in equilibrium. Now, if we take A as the point of application for § VII.] RIGID BODY AS A SYSTEM OF PAKTICLES. 93 F^ and /*,, and B as the point of application of F^ and — R, a single rod joining A and B will be sufficient to form the rigid connection, and this is the simplest body upon which the four external forces could act so as to produce the equilibrium. This rod will be in a state of compression, and the two phases of this stress, which are equal and opposite by the law of reaction, namely, Q acting at B and — Q acting at A^ are the internal forces. Altogether then we have, six forces, forming two sets of three each, which are in equilibrium at A and at B respectively. 119. An equally simple mode of practically illustrating the equilibrium of the four external forces in Fig. 24 would be to suppose jP, and — -/? to act at the intersection of their lines of action, and /*, and F^ to act at the intersection of their lines of action. Then if we connect these points by a rod, it will be in a state of tension, and there will as before be three forces in equi- librium at each end. But if the four external forces acted at any other points of their respective lines of action, taken for instance upon a thin plate in the plane of the forces, the stresses produced in the plate, would be of a much more complicated nature. 120. Conversely, a system of particles having a mutual action, as, for example, the weights F and Q in the example of Art. 81 p. 59, may be replaced by a solid body. Thus, in Fig. 23, we may consider the rod AB as a solid in equilibrium, acted upon by the four external forces /*, Q, R and S^ excluding the two forces equal to T, which are now regarded as internal forces. The problem thus becomes one of three unknown quantities, R, S and ^, to be determined by the three conditions of equilibrium for forces in a plane (acting on a solid); whereas it was before treated as one of four unknown quantities, R, S, 6 and 7", to be deter- mined by four conditions, namely, two for the equilibrium of each particle. When a single unknown quantity, for example, in this prob- lem, is required, we can frequently use a smaller number of equations. Thus, in the solution given in Art. 81 we obtained two equations independent of R and -S", and so had only to 94 FORCES AC7VNG IN A SINGLE PLANE. [Art. 120. eliminate T, On the other hand, treating the bar as a solid, we can obtain two equations independent of R by resolving along CA and taking moments about A. Denoting the length of th"^ rod by ^, these equations are S-(P -\- Q) sin a, Sa sin 6 = Qa cos (<^ — 0); and these will be found, on eliminating ^S", to give the result already found in Art. 81. The Funicular Polygon for Parallel Forces. 121. We have seen in Art. 90 that the graphic construction employed in Fig. 28, p. 69, to find the resultant of a number of parallel forces in a plane (supposed in the figure to be the weights of given bodies acting in given lines), gives rise to a broken line or polygon, which is called funicular because it is the form of a cord to which the weights may be knotted and sustained in equilibrium by means of supporting forces at the ends of the cord. Regarding the cord as a rigid body, the external forces in Fig. 28 are the weights and the oblique forces, Q and Q which are applied to the ends of the cord by means of two fixed points to which they are attached. 122. In Fig. 44 we give a modification in which the funicular polygon is closed by a bar connecting the ends of the cord. This permits the supporting forces to be vertical, and to act, like the weights, in given lines. The data taken are the same as those of the problem solved in Art. 115. (See Fig. 42.) The lines of action, both of the weights and the upward forces, are drawn at their proper distances in the right-hand figure. The left-hand figure, ox force diagram, is constructed by laying off AB, BC, CD in a vertical line, to represent on a selected linear scale the weights taken in order; and then joining A, B, C and Z> to a point (9, called the pole, taken at random. Then, starting from any point M in the line of action of P, the funicular polygon ^VII.] FUNICULAR POLYGON FOR PARALLEL PORCES. 95 M i2T^N 1% constructed as in Art. 89; that is to say, its sides are drawn successively parallel to AOy BO^ CO and DO. Finally, the polygon is closed by the line MN, and OX is drawn parallel to MN in the force diagram. 123. Suppose now the closed polygon to be composed of bars con- nected by smooth joints or hinges, and let us de- termine conversely what 200 Fig. 44. external forces' applied at the joints are necessary in order that the several links of the chain from M \o N may be subjected to tensions proportional to the lines to which they are parallel in the left-hand diagram. Since two sides of the triangle ABO are parallel and propor- tional to the two internal forces we have supposed to act at the joint I, it is a triangle of forces for the equilibrium of that joint; therefore AB represents the required force there acting. In like manner, BCO is a triangle of forces for the equilibrium of the joint 2; the forces acting in the directions AB^ BO, OA in the first case, and in the directions BC, CO, OB in the second. Moreover, since the line BO represents in these two triangles the action of the joint 2 upon i, and that of i upon 2, which are equal by the law of reaction^ these triangles represent the several forces upon the same scale. It follows that the forces at the joints I and 2 necessary to complete the equilibrium must act ver- tically downward and be represented on the same scale by AB and BC. In like manner, at all the joints the necessary external forces must be weights proportional to the segments of the vertical line. Now, in constructing the figure these segments were drawn to represent the given weights; hence the polygon is capable of supporting the given weights in their given lines of action. 96 FOI^CES ACTING IN A SINGLE PLANE. [Art. lU- 124. Moreover, if we suppose MN subject to a compression represented on the same scale by OX, the triangles AOX^ XOD will be triangles of forces for the equilibrium of the points M and N. Hence vertical supporting forces represented on the same scale by XA and DX, acting at M and iV, will complete the equilibrium of the whole polygon. Thus the process is a graphic method of determining the supporting forces P and Q, which were (for the data of this problem) found in Art. 116, by the method of moments, to be 380 and 420 pounds respectively. If the lines from the; pole O had been drawn to represent any other stresses, the sides of the closed polygon ABCDX would still have represented vectorially the external forces necessary to produce these stresses. The Funicular Polygon for Forces not Parallel. 125. In the preceding artioV^s the system of external forces, which is itself in equilibrium, is a system o: parallel forces, and we have seen that it is possible to draw a funicular polygon (having its vertices upon the given lines of action), which may be regarded as a chain of jointed bars in equilibrium under the action of the given forces acting at the joints. In precisely the same way a funicular polygon may be drawn for any coplanar system of forces in equilibrium. For this purpose, the given forces are taken in a certain order, as P ^y P^, /*,, P^^ P^ in Fig. 45, and the polygon of forces ABCDE is drawn by laying off the corre- sponding vectors in the selected order from any point A. This will be a closed polygon if the forces are in equilibrium. A pole O is then taken at random in the plane of the vectorial or force polygon, and joined to the several vertices. Then, starting from any point in the line of action of P ^ a parallel to OB is drawn to intersect the line of action of P^. This is a side of the funicu- lar polygon. In like manner the other sides are drawn suc- cessively, each corresponding to a vertex of the force polygon. Then, if the forces are in equilibrium, the parallel last drawn will § VII.] FUNICULAR POLYGON FOR CO PLANAR FORCES. 97 pass through the point on the line of action of P^ from which we started ; that is, the funicular polygon will close as represented in the diagram. The proof is pre- cisely the same as in Art. 123. The condition that the vectorial polygon shall close is the graphical equivalent of the condition ^ = o of Art. 100, which is equivalent to two analytical conditions. Hence the condition that the funicular polygon shall close is the graphical equivalent of the third ana- lytical condition K = o,* which was derived from the principle of moments. 126. The funicular polygon is practically employed in deter- mining three unknown elements in a system of forces in equilib- rium. For example, if only the point of application M of P^ were known (its direction and magnitude being two unknown elements), and only the line of action of P ^ (its magnitude being a third unknown quantity), we could complete the two diagrams as follows: Starting from A, the data enable us to draw ABCD^ and through D the line DF parallel to P ^ upon which E must lie. Then, selecting the pole, we join AO^ BO^ CO and DO, We can now, in the funicular polygon, starting from M^ draw all the sides but one in the usual manner, arriving at a point N on the given line of action of P ^. Joining MN^ we have the final side of the funicular polygon. Finally, returning to the vectorial * The closing of the vectorial polygon is equivalent to two condi- tions because the final side must have a given direction and a given length; but the closing of the funicular polygon is but a single condi- tion, because the final side is only required to have a certain direction. 98 FORCES ACTING IN A SINGLE PLANE. [Art. 126. diagram, we draw OE parallel to this final side ; thus determin- ing the point E and completing the force diagram. The length of DE and the length and direction of EA give, respectively, the magnitude of P ^^ , and the magnitude and direction of P ^ which were to be determined. The Suspension Bridge. 127. In the suspension bridge the weight of a uniform plat- form is carried by vertical rods to a cable consisting of jointed bars. Supposing the vertical rods to be spaced at equal hori- zontal distances and attached to the joints, the cable becomes, when the weight of the bars is neglected, a funicular polygon for the case of equal parallel forces acting in equidistant lines, as represented in Fig. 46. The extremities M and N are fixed Fig. 46. at the same level at the tops of solid piers symmetrically situated with respect to the weights. If the slopes of the extreme bars are given, the pole O is found by the intersection oi AO and EO parallel to these bars; then OB, OC, etc., give the direction and stress in the intermediate bars. The perpendicular OC will represent the stress of the middle bar (if there be an odd number of bars), which will be horizontal. This stress, which we shall denote by If, is the horizontal component of the stress in every bar. Instead of a closing line MJV, of which the compression would be If, the horizontal component of the pull of the end § VII.] THE SUSPENSION BRIDGE. 99 links at J/" and N is balanced by the like components of the ten- sions of ties connecting M and N with fixed points. Now, drawing OK and OL in the force diagram parallel to these ties, and supposing the piers to take only vertical compres- sion, the equilibrium of the point M is represented by the triangle OKA, and that of N by the triangle OEL, Thus OK and LO give the tension of the ties, and KA, EL give the supporting forces supplied by the piers. Each of these exceeds the half sum of the weights suspended from the cable by the vertical component of the tension of the tie. Form of the Suspension Cable. 128. If the chain be replaced by a perfectly flexible cord, and the number of points at which the weight is applied be increased indefinitely, the limiting form to which the cord ap- proaches will be a continuous curve. To find this limiting form, let the lowest point O, Fig. 47, be taken as the origin of rectangular coordinates, and the horizontal tangent at that point as the axis of x. The portion of the cord OF between the origin and any point {x, y) of the curve would remain in equilibrium if it became rigid. Now, considered as a rigid body, it is acted upon by three external forces, namely, the resultant of the entire weight sus- pended from it, and the tensions at O and /*, which are the actions of the other parts of the cord upon it. The weight sus- pended from each unit of length in the horizontal projection of the cable is by hypothesis constant. Denoting it by w, the whole weight suspended from OP is w . OR, or wx', and, because it is uniformly distributed, the line of action of the resultant bisects OR in M. Since the lines of action of the three forces must meet in one point, it follows that the line of action of 7*, the tension at /*, must pass through My the intersection of the ICX) FORCES ACTING IN A SINGLE PLANE. [Art. 128. lines of action of the weight wx and the horizontal tension H at O. Then, because the ordinate PR is vertical, the triangle PMR is a triangle of forces; and since MR = \xy we have H_^\x wx y* Hence, the equation Hy = iwx\ (i) which is true for every position of the point Py is the required equation of the curve. The curve is therefore a parabola with its vertex at O and its axis vertical. 129. The constant If in this equation is readily determined if the coordinates of any one point as referred to O are known. Thus, if / denotes the total length MJV, or spartj and /i denotes the depth of the parabolic arc (or distance of O below the line MJV)y the coordinates of the extremity of the cable, correspond- ing to JV of Fig. 46, are i/ and ^. Substituting in equation (i) these values of x and y, we find „_wr _wi . where, in the third member, W is the whole weight suspended upon the cable. Denoting by a the inclination to the horizontal of the cable at this point, we have, from Fig. 47, and the tension at this point is §VII.] FORM OF THF SUSPENSION CABLE. lOI which is the greatest tension of the cable. For example, if the span is loo feet and the depth lo feet, we find If=ilVsind T^ = ^-^ W, for the least and the great- est tension of the cable. 130. It is obvious that the cable will still be in equilibrium if we regard the portion of it between any two points as a solid connected with the portion on either side by smooth joints. Thus the cable, which is flexible at every point, may be replaced by a chain consisting of jointed bars provided the joints are situ- ated on the parabolic arc. The weight is here still supposed to be distributed uniformly along the horizontal projection and to be attached at all points of the bars, exactly as if they were uniform heavy bars (but of weights proportional to their horizontal pro- jections, not to their lengths). Under these circumstances there will be at each joint only two forces acting, namely, the equal action and reaction of the two bars, which will be in the direc- tion of the tangent to the parabola. 131. Suppose now that the weight suspended from any one bar be divided into two equal parts and applied directly to the pins which constitute the joints at its two ends. The resultant of the vertical forces will not be changed, and therefore the equilibrium will not be disturbed. When all the weight is thus concentrated at the joints we have the funicular polygon of Fig. 46, which is therefore a polygon inscribed in a parabola. When the number of bars is considerable the depth of the poly- gon will not differ sensibly from that of the parabolic arc, and the tensions (which are now in the directions of the bars them- selves) do not differ sensibly from the mutual actions mentioned in the preceding article. Therefore the formulae of Art. 129 may be used in this case also. I02 FORCES ACTING IN A SINGLE PLANE. [Art. 132, The Catenary. 132. The form assumed by a uniform heavy and perfectly flexible cord hanging from two fixed points is called a catenary. We refer the curve to rectangular co- ordinates as in Art. 128, the vertex, or lowest point, O' being the origin, Fig. 48; and, as before, regard the portion OP as acted upon by three forces. These are the horizontal tension H dX 0\ the tension T 2X P va the direction of the tangent to the curve, and the resultant of the forces of gravity act- ing upon the portion O' P. This last force must as before act in a vertical line passing through the intersection M of the tangents. Denoting by w the weight of the cord per unit length, and by s the length of the arc measured from 0\ the weight of the portion O' P is ws. The horizontal component of T is equal to the constant H^ and its vertical component is 'W5\ therefore Fig. 48. 133. The triangle PMR is, as in Fig. 47, a triangle of forces; but, in this case, we do not know the position of M. Employing, however, the differential triangle whose sides dx^ dy' ^ and ds are parallel to the three forces, we have dx\d/\ds=. H: ws : |/(wV + If'). (i) The differential relation between y and s is here the most simple, giving §VII.J THE CATENARY. IO3 whence, by integration, wy' = ^{w's' -h H^) 4- C To determine C the constant of integration, we notice that J = o when^' = o; whence C ^=^ — H^ and wy' = J^iw's^ + H^) - H. .... (2) It is convenient to introduce, in place of H^ the constant c such that wc = H, (3) and then to take a new origin O at the distance c below O'. Equation (2) then becomes w{y' + ' referring to the right-hand support, I = 2x and h—y — c. Since z* z^ ^= I +^r + --j + - +..., we have, by expansion, from equation (6), ^=^-, = ,(^. + ^-£1 + ...). When h is very small compared with x^ c is very large, and the first term gives a close approximation, namely, . x^ X* /i = — ; whence ^ = — r . . . . (i) 2C 2/1 may be taken as the value of c for a flat catenary (that is, one which is nearly a straight line) whose length is 2x and whose depth is h. Since H =■ wc, this gives for the tension of the cord, which in this case does not differ sensibly throughout the curve, This agrees with the formula of Art. 129 for the suspension bridge, as we should expect, since the weight is now distributed nearly uniformly along the axis of x. 136. An approximate formula for ^, when s and x are given and their difference is very small, is obtained by expanding equa- tion (7) of Art. 134 ; thus, I06 FORCES ACTING IN A SINGLE PLANE, [Art. 136. whence, taking two terms of the expansion, ' = \/w^)-- • ■ • ■ ■ • (3) Again, substituting this in the approximate expression for A, we obtain /, = i i^[6x{s -x)] = i 4/[6/(5 - /)], ... (4) where in the third member S denotes the whole length, and / the span. For example, let the length of a wire suspended between two points 100 feet apart in a horizontal line be 100 feet i inch ; to find the depth A of the middle point below this line. Here / = 100, S — / = Y^, and, substituting in equation (4), ^ = J 4/2, or about i foot 9.2 inches. The tension of this wire may now be found from equation (i), or directly from equation (3) because If = cw. It is ^ = 500 \/2Wy that is, the weight of about 707 feet of the wire. 137. When h and x are given and h is small, the difference J — ^ is very small, and equations (3) and (1) give the very close approximation j-^ = — ; whence S-l^—. . . (5) Equilibrium of a System of Solid Bodies. 138. A number of movable solid bodies, between which such mutual actions exist that they can only occupy certain positions relatively to one another and to fixed bodies, form a system of solids ; and the possible positions are called the configurations of the system. When some or all of the bodies are acted upon by forces external to the system^ it may happen that equilibrium can exist only when the system is in a particular configuration. The mutual actions which exist between the bodies of the system are the internal forces. The external forces of the system will them- § VII.] EQUILIBRIUM OF A SYSTEM OF SOLID BODIES. lo; selves be in equilibrium, exactly as if the system formed a single body. But it is only by considering the separate bodies that we can determine the configuration of equilibrium. 139. For example, suppose four uniform bars, each of length za and weight W, jointed together so as to form a rhombus, to be suspended as in Fig. 49 from two smooth pegs fixed at the same height and at the distance 2c in a vertical wall ; and let it be required to determine a position of equilibrium in which the diagonal AC '\^ vertical. The angle B between the side AB and the vertical will serve to determine the required configuration. Since the forces act symmetrically upon the pair of upper bars and the pair of lower bars respectively, it is evident that the mutual action of the bars at A and at C is horizontal, and that it is only necessary to consider the equilibrium of the bars AB and BC. These bars are drawn separately in the diagram to show more clearly the forces acting upon them. In the case of AB, these are four in number ; namely, its weight W acting at its middle point, the horizontal action S at A^ which we shall assume Fig. -49. to act to the left, the resistance R of the peg at E, which acts perpendicularly to the bar (and therefore in a direction depend- ing upon ^), and the action at B of the lower rod, of which the direction as well as the amount is unknown. We may therefore take V and ZT, the vertical and horizontal components of this force, for two of the unknown quantities. In the case of the Io8 FOJ^CES ACTING IN A SINGLE PLANE. [Art. 139. lower bar we have these forces Fand ZT acting in the opposite directions at B^ the weight W acting at tHe middle point, and tlie horizontal force T acting at C. Hence there are in all six unknown quantities, ^, R^ S, T, V and H, and the conditions of equilibrium, three for each bar, are sufficient in number to determine them. The equilibrium of the rod BC gives, by taking vertical and horizontal forces and moments about B^ H = T, 2 Ta cos B = Wa sin 6; and the equilibrium of AB gives RsinO = V-\- W i? cos (9 = S ^ H, Re cosec d = Wa sin ^ -|- 2 Va sin 6 -\- 2Ha cos B, The first five of these six equations determine the five un- known forces in terms of ^, namely : R = 2W cosec d, S = W{2 coiO - i tan ^); and then the sixth equation gives which determines 6. If the value of is found to exceed tan~*2, the value of S will be negative and its direction will be opposite to that which we assumed it to be in the diagram. § VII.] EXAMPLES. 109 EXAMPLES. VII. 1. Solve Ex. IV. 10 by assuming a rigid triangle turning about the centre instead of a circular wire and string. 2. Two uniform and equal smooth cylinders rest in contact, and each is in contact with one of two smooth planes inclined at angles a and (i to the horizon. Determine the inclination B to the horizon of the plane passing through their axes. tan Q = i(cot ex — cot /?). 3. A polygon composed of bars without weight has the form of five sides of a regular dodecagon with the middle side horizon- tal. Show that three equal weights may be hung one from the middle of this bar and one from each of the two upper joints. 4. A funicular polygon is inscribed in a semicircle with horizontal diameter, the sides taken in order being chords of 60°, 30°, 30° and 60°. Find the ratio of the weights at the joints. [1/3 — 1:2— i/s : i/3 — I or] 2 : 1/3 — 1:2. 5. Determine the ratio of the weights when the sides of the funicular polygon in example 4 taken in order are chords of 60°, 60°, 30° and 30°. 1/3:1:1+ 1/3. 6. Forces acting in lines bisecting the angles of a plane polygon are proportional to the cosines of the half angles, and all act out- ward. Prove that they form a system in equilibrium, and that the polygon of forces can, in this case, be inscribed in a circle. 7. Show that, if one of the vertices of the polygon of external forces be taken as the pole, two sides of the funicular polygon coincide with the lines of action of two of the forces, the result being the funicular polygon for the resultant of these two forces and the remaining forces. Compare Fig. 25. 8. In the suspension bridge, the horizontal projections of the bars being equal, prove that the tangents of the inclinations of successive bars are in arithmetical progression ; also, if the number of bars is even, the heights of the joints taken in order above the middle joint are proportional to the squares of the integers in natural order. 9. Show, by mechanical considerations, that the vertical no FORCES ACTING IN A SINGLE PLANE. [Ex. VII. through the intersection of tangents at the extremities of any arc of the parabola in Fig. 47 bisects the chord of the arc; and thence that the tangent at the point where this vertical cuts the arc is parallel to the chord and bisects the distance between the inter' section of the tangents and the middle point of the chord. 10. Derive the equation for the suspension cable by integration. 11. A bridge of 360 feet span is supported by two suspension cables. The weight of the bridge is \ ton per foot run, and the dip of each cable is 37^ feet. Find the least and the greatest tension of the cable; also, if the stays and the cable are equally inclined to the vertical at the top of a pier, what is the whole pressure on the pier. 108 tons; 117 tons; 180 tons. 12. Derive the value of H^ Art. 129, directly from the extreme case in which the suspension cable is reduced to two bars. 13. Prove that the perpendicular from the foot of the ordinate in Fig. 48 has the constant value c ; and thence that the involute of the catenary is the iractrix (the curve in which the part of the tangent between the point of contact and the axis of x is constant). 14. Prove the following values of x^y and c in terms of s and h in the catenary where h = y — <: as in Art. 134 :^ / + /^' s' — h' s^ - h\ s -{- h y — ; — » ^ = ; — > ^ = 7~~ log 7. •^ 2h ' 2h ' 2h ^ s — h The expanded form of the last expression is r 2 h' 2 h' 2 h' "I X = S\ I -a i 1 — , . .. L 1.3 s 3-5 -f 5-7 -f J 15. A cord weighing 2 oz. per linear inch hangs over two smooth pegs at the same level; the tension of the cord at its lowest point, which is 6 inches below the pegs, is 20 oz. Find the whole length of the cord. 56.98 inches. 16. The wire for a line of telegraph cannot sustain more than the weight of 4000 feet of its own length. If there are 22 poles to the mile, what is the least sag allowable ? 21.6 inches. §VII.] EXAMPLES. Ill 17. How much per mile does the actual lengfh of the stretched wire in example 16 exceed the straight line ? 9504 inches. 18. Two rods, AC and BC^ of uniform weight per linear inch, are jointed together at C and to two fixed points in the same vertical at A and B, Show that the direction of the action at C bisects the angle ACB. 19. Two uniform heavy rods have their ends connected by weightless strings and are supported at the middle point of one of them. Prove that in equilibrium either the rods or the strings are parallel. 20. Two equal rods without weight are hinged together at their common middle point, C, and placed in a vertical plane on a smooth horizontal table. The upper ends, A and B, are connected by a light string, ADB, upon which a heavy ring can slide freely. Show that in equilibrium the height of D above the table will be three-fourths that of A or B. 21. A BCD ... is a closed polygon formed of any number of bars jointed together, and is in equilibrium under the action of forces acting at right angles to the bars at their middle points. Show that the actions at the joints are all equal. If perpendiculars to the actions at B and C meet in O^ OBC may be taken as a triangle of forces for the rod BC (turned through 90°). Thence show that, if the forces are proportional to the sides on which they act, the polygon can be inscribed in a circle. 22. Two equal uniform spheres of weight W and radius a rest in a spherical cup of radius r. Find the resistance R between either sphere and the cup, and the pressure P between the spheres. * ^ r - a _ __± w |/(r^ -- 2ar) ' J^{r' - 2ar) 23. A string, 21 inches long, is fastened to two nails 14 inches distant in a horizontal line, and weights P and Q are knotted to it at points § and 6 inches, respectively, from the two ends. Deter- mine the ratio of P to Q, so that in equilibrium the intermediate portion of the string shall be horizontal. /^ : ^ = 3 : 11. 24. Two equal, uniform rods, each of length 2b, are jointed 112 FORCES ACTING IN A SINGLE PLANE. [Ex. VII. together at one end of each, and rest in equilibrium on a smooth cylinder with horizontal axis and radius a. Show that the angle d which each rod makes with the horizontal is determined by a s\x\ S ^=z b cos^ B. 25. A uniform cylindrical shell of radius c without a bottom stands on a horizontal plane, and two smooth spheres with radii a and ^, such that a ■\- b > c, are placed within it. Show that the cylinder will not upset if the ratio of its weight to that of the upper sphere exceeds 2c — a — b \ c. 26. Two equal uniform planks, of length b and weight /*, are attached together and to two fixed points in a horizontal line by smooth hinges so that the angle each makes with the vertical is 0. A sphere, of weight W and radius a, rests between them. Find the tension on the lower hinge. X= iY b-a^o\.B P tan B 2b sin B cos B 2 27. A crane is formed of a vertical post fixed in the ground, with the part AB 15 feet long above the ground at A^ a horizontal bar BD 12 feet long, jointed to the post at B^ and a strut jointed to the post at A and to BD at a point C 8 feet distant from B. At the end Z> hangs a weight of 10 tons. Find the action at the joints Cand B. I'j tons; 9.434 tons. CHAPTER IV. PARALLEL FORCES AND CENTRES OF FORCE. VIII. Resultant of Three Parallel Forces. 140. We shall in this chapter consider systems of forces whose lines of action are all parallel but not in a single plane, the most familiar instances of which are afforded by the action of gravity upon different bodies or the parts of the same body. It is convenient, in this case, to assume in the diagrams that the lines of action are perpendicular to the plane of the paper, and to suppose them to act at the points where the lines of action pierce the plane of the paper. We have seen that the resultant of two parallel forces is, in general, a force equal to their algebraic sum and acting in a line parallel to their lines of action. It evidently follows that the resultant of three parallel forces acting in lines perpendicular to the plane of the diagram is also a force equal to their algebraic sum, acting in a line per- pendicular to the same plane. The point in which this resultant line of action pierces the plane is called the centre of the parallel forces. \\\. The position of this centre depends upon the ratios of the given forces. In Fig. 50, let the forces P, , P^ and P^y which are in the ratios I : m : n^ act at the points A, B and C respectively, in lines perpendicular to the plane of the diagram. By Art. 85 the resultant of P^ and /*, is the force P^ -\- P^ acting at the point F in which AB is cut inversely in the ratio / : /«, so that AF\FB = m',l, Fig. 50. 114 PARALLEL FORCES AND CENTRES OF FORCE, [Art. 141. as represented in the diagram. Join CF^ then the resultant of this force and P^ is the force R = P^-\- P^-\- P^^ acting at the point O where CF is cut inversely in the ratio P^\ P^-\- P^^ orn : m -\- 1, that is to say, so that CO:OF= m~j- /:n, 142. In this construction for the centre of the three parallel forces P^ , P, and P^ , we shall arrive at the same point O if we take the forces in any other order. Thus, if PC be cut at D inversely in the ratio of the forces P^ and P^, that is, in the ratio n : m, O will be a point of the \mQ AD. Hence O may be found as the intersection of CF and AD. When this is done and O joined to B, it is readily seen that the areas of the triangles AOC, COB are in the ratio m : /, and that the areas AOCy BOA are in the ratio m : n. Thus the triangle ABC is divided into three parts bearing the ratios /: m:n; and, if BO be produced to F, AC is cut in the ratio « : /, as indicated in the diagram. The point O, thus determined by means of the ratios /: m : n, is also called t^e centre of gravity of three particles having these ratios and situated at A^ B and C respectively: for, supposing the plane of the diagram horizontal, the forces may be taken as the weights of these particles, and O is the point at which the resultant acts; or, as it is sometimes expressed, the point at which the total weight may be regarded as concentrated. Forces in Opposite Directions. 143. Parallel forces in opposite directions are called unlike parallel forces. If P^ and P^ are unlike, the numbers / and m have opposite signs, and we have seen in Art. 88 that the resultant cuts the transverse line produced in the inverse ratio of the weights; that is, on the side of the greater weight, and so that the whole line is to the part produced as the greater force is to the less. The construction for three forces is similar to that of Fig. 50, but the point F falls on AB produced, and O is §VIII.] COUPLES IN PARALLEL PLANES. 115 found outside of the triangle ABC. When two of the three given forces have opposite directions, the third will be in the direction of one of them, and therefore one of the three points Z>, E and F will be on a side of the triangle. We may, in this case, take two of the numbers /, in and n as positive and the other as- negative. In Fig. 51, we consider the special case in which 7n and n ape positive, and I = — m, so that the forces P^ and P^ form a couple in the plane passing through the line AB, and perpendicular to the plane of the diagram. The point D will now be upon CB^ E will be upon AC produced, and F will be at an infinite distance on AB^ so that CO is parallel to AB. The resultant acting at 6> is the algebraic sum of the forces, Ej-'" Q,'-' which is in this case P^ , because Fig. 51. Couples in Parallel Planes. 144. If we reverse the direction of this resultant, we shall have four forces in equilibrium; namely, — P^ acting at ^, P, acting at B, P^ acting at C, and — P^ at O. These constitute two couples in parallel planes, cutting the plane of the diagram in AB and CO respectively. By similar triangles, we have therefore CO'.AB = m:n = P^:P^\ P^XCO=^ P^X AB; that is to say, the moments of these couples are equal. Since the forces at B and C have the same direction, inspection of the figure shows that these couples tend to turn a rigid body about a perpendicular to AB and CO in opposite directions. Thus we Il6 PARALLEL FORCES AND CENl^RES OF FORCE. [Art. 144. have proved that couples having equal moments in parallel planes are equivalent. The construction in Fig. 51 is in fact, like that of Art. loi, the composition of a force and a couple, and the effect is, as in Fig. 36, not to change the magnitude or direction of the force, but to shift the line of action. Case in which the Resultant is a Couple. 145. A special case arises when the algebraic sum of the three forces is zero, so that I ■■\- m -\- n = o. Let us suppose m and n to have the same sign ; then, in Fig. 52, the point D is upon the side BC of the triangle ABC. The resultant of P, and P^ is the force P^ + P^ acting at D. Now P,- - {P^ + P^) by hypothesis ; hence this force forms with P^ a couple. Thus / the resultant is in this case a couple 9<' in the plane passing through AD and / \N^ t^6 li^^ o^ action of P^ , and having y' \ ^v''* for its moment P, X AD. y M_il__\'^ •'^^ ^^^^^ case, the point j^ will lie F y' \ yB upon AB produced, and ^ upon CA V'' produced, as in the figure, and we might equally well take for the re- FiG. 52. sultant the couple P^ X CP, or the couple P^ X BP. In fact the three couples have equal mo- ments and act in parallel planes, and are therefore equivalent by Art 144. Composition of Couples in Intersecting Planes. 146. The construction given above may be regarded as the composition of couples in planes which are not parallel. For, since P^ = — (P^ -{- P^), this force may be separated into parts equal and opposite respectively to P, and P^. Therefore the given forces are equivalent to four forces forming two couples ; namely, P^ acting at B with — P^ at A, and P, acting at C v.-ith §VIII.] PARALLEL COMPONENTS OF A FORCE. WJ ~ /*, at A. These two couples act in planes perpendicular to that of the diagram and intersecting it in AB and ACy and their moments are P^ X AB and P^ X AC^ respectively. The con- struction therefore shows that the resultant of these couples is the couple (/*, + P^AD in a plane passing through AD and the intersection of the planes of the given couples. The given couples and the resultant couple are here proportional to mAB^ uAC and {ni -f- n)AD ; hence, comparing the construction with that of Art. 67 for the resultant of two forces, it appears that the magnitude of the resultant and the parts into which the diedral angle between the planes is divided are precisely the same as in the case of two forces and the parts into which the plane angle between their lines of action is divided. Resolution of a Force into Parallel Components. 147. A given force R can be resolved into components acting in any three given lines parallel to its line of action. Let the force act at (9, and let the given lines intersect a plane through O perpendicular to the line of action in A^ B and C. If these points are in a straight line, the problem is not determinate; but if they form the vertices of a triangle, the components have definite values. First, suppose O to lie within this triangle as in Fig. 50. Join OA, OB and OC^ and let /, m and n be three positive numbers prov portional to the areas of the triangle BOC^ CO A, AOB\ then, by Art. 142, R acting at O is the resultant of three forces pro-, portional to /, m and //, and having R for their sum. Hence the required components are IR mR nR I •\' m -\- fC / -\- m -j- n I -^ m -\- n The ratios of /, m and n may of course also be determined by means of the segments of the sides of the triangle as indicated in Fig. 50. I 18 PARALLEL FORCES AND CENTRES OF FORCE. [Art. 148. 148. If O falls upon one side of the triangle, the component at the opposite vertex vanishes, and the problem reduces to that of Art. 87 for determining two components coplanar with the given force. If O falls outside of the triangle, one or two of the triangles whose vertices are at O must be taken as negative ; for example, in Fig. 51 the value of / is negative and the component at A has a direction opposite to that of R. As an illustration, we may suppose A^ B and C to be the points in a three-legged table directly over the three feet. Then, when a weight W is placed at O upon the table, the components are the pressures produced by W at the three feet and resisted by the floor. If O lies beyond the line BC^ as in Fig. 51, the di- rection of the component at A is reversed, and, since it now acts upward, the foot must be held down to produce equilibrium. Moment of a Force about an Axis. 149. In Art. 91, the action of a couple upon a lamina in its plane is explained as a tendency to turn it in its own plane, that is to say, about an axis perpendicular to the plane of the couple. Furthermore, in Art. 93, we have seen that if the axis, say through B^ Fig. 29, is fixed, while a single force acts on the lamina at A, the resistance of the axis, together with the given force, constitutes a couple, and the moment of this couple is called the moment of the force about the point B^ or more properly about the axis through B perpendicular to the plane of the diagram. In Fig. 53 we represent the axis MN passing through B as in the plane of the diagram, while the force at A is supposed to act perpendicularly to the plane of the diagram. The arm AB of the moment is — N perpendicular both to the line of ^, , action and to the axis. If the riG. 53. force P be transferred to the point A' equally distant with A from the axis MN^ it will have §VIII.] THE CENTRE OF PARALLEL FORCES. 1 I9 the same moment; and since we have now seen, in Art. 144, that couples of the same moment in parallel planes are equivalent, it is not necessary to specify the point B at which the arm meets the axis. We can therefore compare directly, and employ in the theorem of moments, the moments of forces about a given axis, although the lines of action are not in one plane. It is only necessary for our present definition of a moment that the line of action and the axis (though not intersecting) should be in direc- tions at right angles to one another. 150. In the case of parallel forces, the diagrams being drawn, as in the present section, in a plane perpendicular to the lines of action, the forces have moments about all straight lines in the plane of the diagram, and we can apply the theorem of moments with respect to any such line as axis. Thus, in Fig. 50, the moments of /'j and of P^ about the axis AB are both zero; hence, by the theorem of moments, the moment of P^ about AB is the moment of the resultant at O about the same axis. Ac- cordingly, the perpendiculars from C and O are inversely as P^ is to *Ry which agrees with the result in Art. 142, since the tri- angles ABC and A OB are proportional to their altitudes. The Centre of Parallel Forces. 151. The graphical determination of the resultant of any number of parallel forces acting at given points in a plane per- pendicular to the direction of the forces is an extension of the process of Art. 141, involving successive applications of the operation of cutting a given line in a given ratio. At each step we combine two forces of the system into one, and the order in which the forces are taken is arbitrary. Thus, if four equal forces of magnitude P act at A^ B, C and Z>, Fig. 54, we may determine the resultant of the forces at A and B by bisecting AB\ and then (instead of combining 2P acting at this point with one of the other forces, which would require a trisection) combine the forces at C and D in like manner at the middle point of CD. The system is thus reduced to two forces, each y D c 7" fG A • X, r" ' \20 PARALLEL FORCES AND CENTRES OF FORCE. [Art. 151. equal to 2F, and we have finally to combine them by a third bisection. If we change the order in which the points are taken, we sliall arrive at the same final position; thus it appears that the lines bisecting pairs of opposite sides or diagonal's of a plane quadri- lateral bisect each other. 152. But when the positions of the points of application are given by means of their distances from given lines, it is generally more convenient to determine the centre of the parallel forces by means of their moments with re- spect to the given lines. The al- gebraic sum of the forces gives the magnitude of the resultant force R\ •^ and, supposing this not to be zero (in which case the resultant is a couple), the resultant moment of the system about any line is the moment of the resultant, and therefore determines the distance from that line at which it acts. Thus, supposing the forces referred to rectangular axes, the moment about the axis of y of the force /*, acting at (^, , ^,), Fig. 54, is /*,a:, , that of Z', acting at (x^^y^ is ^,^, , and so on, for all the forces of the system. The algebraic sum of these moments is equal to the moment of R acting at the centre of force (^, y), which is Rx^ and therefore determines the distance of that point from the axis of ^. Using, in like manner, moments about the axis of x^ we have for the determination of R and the coordinates of its point of action: i? = p, + /', + + />,. =2^,. . . (.) Rx= P,x, + P,x, + + I',x„ = SPx, . . (2) J(J = I',y, + P,y,+.... + -Pnyn=:^2/'y. . . (3) §VIII.] CASE IN WHICH J^ = o. 21 Case in which R = o. 153. If the given forces have not all the same direction, some of the terms in 2P are negative and the sign of ^ determines the direction of the resultant, and together with that of the total moments 2xP and 2yP the signs of x and J in equations (2) and (3). When i? = o, and 2Px or 2Py is not zero, at least one of the quantities ^ or y is infinite. The resultant is now not a force, but a couple which may be determined as follows : Assume two equal and opposite forces Q and — Q acting at the origin in a line perpendicular to the plane of the diagram. The addition of these forces to the system will not change the result- ant. Now, since Q acting at the origin has no moment about either axis, the moments of the system consisting of the given forces and the force Q are still 2Px and ^-Py, and the resultant of this system (which includes the force Q but 7iof the force — Q) is Qf because 2P is by hypothesis zero. Hence the resultant of this system is the force Q acting at {x\ /) when Qx' = 2Px and Q/ = :^Py. Therefore the resultant of the given system is the couple formed by the force Q acting at (x\ j') and the force — Q act- ing at the origin. This couple acts in a plane perpendicular to that of the paper intersecting it in the line joining (^',jf') to the origin, and its arm is the distance of {x' y y') from the origin. Hence, denoting its moment by H and the inclination of its plane to the axis of x by 6^, we have and _/ _ ^^y tan <9 _ , _ . X ^Fx 122 PARALLEL FORCES AND CENTRES OF FORCE. [Art. 154. Conditions of Equilibrium. 154. The system of parallel forces is in equilibrium if the resultant force and the resultant moments about each axis all vanish, that is, the conditions of equilibrium are :2P = o, 2xP = o, :SyP = o. Thus, for forces whose lines of action are all parallel, as well as for forces whose lines of action are all in one plane, there are Mr', we have dM=px j/(a* — x*)dx. The figure shows the limits of integration to be o and tf; hence M^p f V - x')^x^x = - ^(a' - a:')*T = ^'. Jo 3 Jo 3 128 PARALLEL FORCES AND CENTRES OF FORCE. [Art. i6o. Denoting the abscissa of the centre of pressure by "x^ the moment of the resultant, or total pressure /*, is Fx^ and this must equal the total moment about the axis of 7. Hence Px=^ M; and, substituting the values of J^ and M in this example, _ Ttpa^ pa^ , _ Aa x.-^ — ^=^~ — ; whence jc = -^— . 4 3 3^ 161. The value of p, the ordinate of the centre of pressure, may be found in the same way, using the element of area xdy^ and in this example the value will be the same as that of Ic^ be- cause X and y are, in the equation of the circle, interchangeable. But it may also be found by employing the element of area used in Fig. 56, so that dP=pydx still denotes the element of pressure. It must be noticed, however, that the points of this element are not all at the same distance from the axis of x\ hence, denoting the moment about the axis of x by M^y the arm of the elementary moment dMx is the average value of the distances of the several points. Since the element has a uniform breadth, this average value is obviously \y. Therefore dM^ = \pfdx\ substituting the value of y and integrating, M, = i/> [ %• - x'')dx = ip(a'x - ix'ij = ipa\ Finally, putting Py for M^ , we find y = — . The Centre of Gravity of an Area 162. A uniform heavy lamina is a thin plate of uniform thick- ness and material, so that the weight of a unit of area taken in § IX.] THE CENTRE OF GRAVITY OF AN AREA. I29 any part of it is equal to a constant w. When such a lamina is in a horizontal plane, the resultant of its weight is evidently the same as that of a uniform pressure w upon the area of the lamina. Hence the centre of uniform pressure is also the centre of gravity of the lamina; that is, the point at which, if the whole weight were concentrated, it would produce the effect of the resultant weight. The fac4ror w does not affect the position of this point, which is therefore called the centre of gravity of the areuy and also sometimes the centroid of the area. In finding its position, as in Arts. 160 and 161, w, which takes the place of /, is generally put equal to unity, and the moments about the axes are then called the moments of the areay or the. statical moments of the area. 163. The centroid of a plane area corresponds to the centre of position of all its points, so to speak; for equal weight is given to all its points, that is, to all equal small areas. Accordingly, its distance from any straight line is regarded as the average distance of all its points. The position of the centroid is often obvious from considerations of symmetry. For example, the centroid of a circle, of an ellipse, or of a regular polygon is the geo- metric centre; that of a parallelogram is the intersection of the diagonals. Again, whenever there is an axis which bisects a set of ele- ments of uniform width which make up the area, the centroid must be upon this axis. For the centre of gravity of each element is upon this axis, so that the weight of the whole area has the same resultant as that of a number of particles situated upon the axis, and there is no reason why this resultant should be on one side of the axis rather than ^^' ^^" the other. In particular, the centroid of a triangle, Fig. 57, is thus seen to be upon a medial line; and, since for the same reason it is on each of the other medial lines, it is at the point where the medial lines meet. Comparing with Art. 141, it is the position of O when I := m =^ n. Hence the centre of gravity of a triangle I30 PARALLEL FORCES AND CENTRES OF FORCE. [Art. 163. is the same as that of three equal particles at its vertices. That is to say, it is on a medial line at a distance of two thirds of the medial li?ie from the vertex, and its perpendicular distance from the base is one third of the altitude. 164. When the centre of gravity of a figure is known as well as its area, its statical moment about any axis in the plane may be found without resorting to integration*. The centre of gravity of an area made up of parts which are £— ^ J ? thus known is then readily deter- y\i ^^^i,^^^ \ mined. For example, given the / \ I ^^"">>^\ trapezoid ABCD, Fig. 58, whose A 6 B parallel sides are a and b and whose Pjq -g altitude is h. By Art. 162, the centre of gravity is on the line join- ing the middle points of AB and CD. To find its distance from the base ^, we divide the area into two triangles by the diagonal BD, and take statical moments about AB, The moment of ABD is its area \bh multiplied by the distance of its centre of gravity from AB.^ which, by Art. 163, is \h. In like manner, the moment of DCB is \ah X \h. Denoting the required distance by jf, the moment of the whole figure is the total area multiplied by ^. Hence \h{a + b)li = \bh . \h + \ah.\h ; therefore _ b ■\- 2a X = -7 r— Tx/'' 3(^ + b) 165. In like manner, the given area may be the difference of areas whose centres of gravity are known. For example, from a square whose side is 2^, Fig. 59, an equilateral triangle con- structed on the side AB is removed, and this triangle is turned about AB into the position shown in . the diagram ; it is re- quired to find the centre of gravity of the figure thus formed. The centre of gravity is upon the axis of symmetry CD bisecting AB^ and we have only to find its distance x from AB. The § IX.J CENTRE OF GRAVITY OF A UNIFORM CURVE. I3I moment of the square about BA is 4^' . a = 4^;'. The area of the triangle is ^V3> since the alti- tude is a^2> i ^^^^ centre of gravity is at a distance \^2>^, therefore the moment of the triangle is a^. To find the moment of the figure in its present position, we must subtract from the moment of the square the moment of the triangle removed, and also algebraically add the moment of the triangle in its new position, which is negative because it is on the other side of AB. This total moment is to be put equal to the total area multiplied by the arm x. Thus ^a X 4a ~ a —a whence X = ^a. The Centre of Gravity of a Uniform Curve. 166. The weight of a thin uniform rod, or wire, having the form of a plane curve may be regarded as uniformly distributed along the length of a mathematical line. Supposing the plane horizontal, the resultant weight pierces the plane in a point, which is called M ^ i^'^^^ ^ '^^' 168. For example, let it be required to find the centre of gravity of the area of the half cardioid, Fig. 61, whose polar equation is r = a{i -{- cos 6), § IX.] EMPLOYMENT OF POLAR COORDINATES. 133 Substituting this value of r, we have for the elements in terms of 6^ dA = \a\i -f cos eyds, dMy = ia'(i + cos ey cos (^dO, dM, = ia\i + cos ey sin Odd. The figure shows that the limits of integration foi* d are o and zr. Hence, A = -T(i + cos ey dd = - V{i + 2 cos 6^ -h cos' e)de 2 J o 2 Jo My = -f (i + COS ey cos 6*^6^ 3 Jo = — I (cos ^ + 3 cos' ^ -h 3 cos' ^ + cos* e)d0 3 L2 4-2. 8 ;i^ ^M^ I m' • /9^AV «'(i-f-cos^n''_4^\ J/, — — I (i + cos e) sin edu = -^^ '- I = — ; 3 Jo 3 4 _io 3 whence, since mA = My and J' A = M^^ we find ■x:- 5 i6 6 ()Tt for the coordinates of the centre of gravity. 169. If we employ the ultimate polar element of area, r dr de^ which is situated anywhere within the area {r and e being now independent variables), the expressions for the area and the mo- ments are the double integrals A =1 \rdrde, xA =f fr'cosl?^^, yA =f fr'sin^^^. 134 PARALLEL FORCES AND CENTRES OF I'ORCE. [Art. 169. Performing the r-integration first, as is usually most con- venient, the limits are zero and the r of the curve ; and the results under the single integral sign are the elements given in Art 167, in which r represents a given function of 0. In like manner, the elements of moment for single integration given in Arts. 158 and 160 may be derived from the moments of the ultimate element of area dy dx^ namely, X dy dx and y dy dx, by performing the ^'-integration between the limits zero and the ordinate of the given curve. The Theorems of Pappus. 170. The centre of gravity regarded as the point at the aver- age distance has a useful application to surfaces and solids of revolution. Suppose an arc of a plane curve to revolve about an axis in its plane, but not crossing the arc, thus generating a sur- face of revolution. Taking the axis of revolution as the axis of /, every element ds of the arc describes in the revolution a circle whose radius is x. The circumference of this circle, or path of dsy is 27tx^ hence it generates the element of surface inxds. Therefore the whole surface generated is the integral of this ex- pression taken between the limits which define j. Denoting it by Sy we have then S= 2n\xds\ but the integral in this ex- pression is the statical moment of S about the axis of 7, which is the value of 'xs. Hence the surface of revolution is ^ = 2 7tX J, in which 2iix is the circumference described by the centre of gravity. Hence the surface generated is equal to the product of the length of the arc and the path of its centre of gravity. 171. Again, let a plane area revolve about an axis in its plane but not crossing its surface, thus generating a volume of revolu- ^ IX.] THE THEOREMS OE PAPPUS. 135 lion. Taking as before this axis as axis of y\ every element of area dA describes a circle whose radius is x and circumference mx. Hence it generates the element of volume 27txdA, and the whole volume generated is the integral of this expression taken with the same limits which define the area A. Denoting the volume by F, we have then F'= 2n\xdA^ where the integral is the statical moment of A with respect to the axis of j. The value of this moment is xAy where x is the distance from the axis of y of the centre of gravity of the area A. Hence V = 27rxAt in which 27tx is the circumference described by the centre of gravity. Therefore ^/te volume generated by the revolution of a iylane area about an axis^ in its plane and not crossing its surface^ is the product of the area and the path of its centre of gravity. rhese theorems, sometimes called Guldin's Theorems, are ])roperly the Theorems of Pappus, having been first given in the *' Collection " of Pappus, a mathematician of Alexandria who flourished probably about 300 a.d. 172. It will be noticed that, in each of the theorems, " the path of the centre of gravity " is the average length of the path of the elements of the generating line or area as the case may be, just as the arm of the total moment is the average arm of the ele- ments.* The useful applications are not to cases in which it would be necessary to find the centre of gravity by integration, f * Pappus's theorems evidently apply to any part of a revolution, or to any motion in which the elements are always moving in a direction perpendicular to the plane of the generating figure, and in like directions. Compare with the description of an area by a straight line, Int. Calc, Art. 163, where the motion considered is the resolved part perpendicular to the generating line, as recorded by the wheel in Amsler's Planimeter. f The method for a volume of revolution given in Int. Calc, Art. 136, involves precisely the integral we should employ in finding the statical moment about the axis of revolution, with the addition of the factor 2n, 136 PARALLEL FORCES AND CENTRES OF FORCE. [Art. but to those in which the position of the centre of gravity and the area are known. For example, if the circle of radius a, Fig. 62, revolve about the axis AB at a distance b from the centre, it generates a solid gen- erally called an "anchor - ring." The centre of gravity being the <:entre of the circle, its path is 271b. Therefore the volume of the an- chor-ring is B r D \N A as. a/^ \ I J ( i — " / Fig. 62. V = TTa* . 2 7rb = 2 rr^a'b. Again, the centre is also the centre of gravity of the circum- ference which generates the surface of the anchor-ring. Hence we have for the surface S = 27ra . 27rb = ^n^ab, 173. The segment of the anchor-ring cut off by a plane per- pendicular to the axis is generated in the revolution by a seg- ment such as MDN of the generating circle. The centre of gravity of this segment, and also that of the arc MN^ is at the same distance b from the axis of revolution. Denoting the half of the angle subtended by the chord MN by a, the length of the arc is 2aa. The area of the circular segment is the area of the sector OMDN^ which is a^a^ diminished by that of the triangle OMN^ which is a^ sin a cos a. Hence we have, for the volume and surface of the segment of the ring, V =. 27tba^(a — sin a cos a) and »S = ^naab. 174. Another useful application of Pappus's theorems is the determination of the centre of gravity of the generating area, when the area itself and the volume generated are both supposed known. For example, if the semi-circle CND in Fig. 62 revolves about the diameter CD it will generate a sphere. Denoting the § IX.] APPLICATIONS OF PAPPUS'S THEOREMS. 17 distance of the centre of gravity of the semi-circle from the diameter by 'x, we have by the second theorem, since the known volume of the sphere is f tt^j', \na'' . 27rx = ^Tra'; Aa whence x = — , agreeing with the result found by integration* in Art. i6o. In like manner, because the semi-circumference DNC gener- ates the surface of the sphere, of which the value is ^nd"^ we have, when 'x denotes the distance of the centre of gravity of the semi-circumference, 7ta , 27rx = 47ra'' _ 2a . . whence x = — , agreeing with Art. 166. 175. The reason for the restriction that the axis of revolution must not cross the generating area is that the distance x, in the expression xdA of the demonstration in Art. 171, has opposite signs for the portions of the area on opposite sides of the axis. Hence, in the result, the volumes generated by the two portions have to be taken with opposite algebraic signs, just as the statical moments of these parts have opposite signs. For example, let the circle in Fig. 63 revolve about the chord ABj which cuts off 120° of the circumference, and therefore bisects the radius CD to which it * If the volume had to be found by integration, it might be found directly, as mentioned in the preceding foot-note, by the same integration which is performed in Art. 160 to find the statical moment. But, in the process above, we are free to employ other methods of integrating for the volume. 13^ PARALLEL FORCES AND CENTRES OF FORCE. [Art. 175. is perpendicular. The centre of gravity of the circle is at a dis- tance \a from AB^ and the theorem gives V = na^ . na = 7t^a\ which is M(? difference between the volumes generated by the segments AEB and ADB respectively. Now since this is the same as the volume generated by the lune-shaped area ACBE^ it will enable us to find the centre of gravity of that area. For the area of the lune, being the dif- ference between the segments AEB and ACB or A£>B, is Therefore, regarding the volume /^as generated by this area, and denoting the distance of its centre of gravity from AB by 'x, we have ^^-^^' .27rx = 7t*a\ whence x = -, — = .82 1<^, so that the centre of gravity is 27r + 3i/3 ^ ^ at a distance .321^? from the centre of the circle. EXAMPLES. IX. 1. Find by integration the distance from the base of the centre of gravity of a triangle and of a trapezoid. 2. Find the centre of gravity of the area between the parabola y' = —X and the double ordinate 2b. x = ^/i. 3. Find the ordinate of the centre of gravity of the upper half of the area in Ex. 2. J> = |^. 4. Find the centre of gravity of the area between the parabola in Ex. 2, the axis o( y, and the perpendicular to it from the point § IX.] EXAMPLES. 139 5. Find the centre of gravity of the area between the semi- cubical parabola ay^ = x^ and the double ordinate which cor- responds to the abscissa a. x = ^a. 6. Find the distance of the centre of gravity of a sector of a circle from the centre of the circle, the radius being a and the angle of the sector 2a, _ _ 2^ sin a ^~ 3« * 7. Determine the centre of gravity of the area included between the parabola jv* = 4ax and the straight line^ = mx. _ Sa ^ _ 2a ^ ~ 5^" ^ ~ ~m' 8. A uniform wire is bent into the form of a circular arc and its two bounding radii. Determine the angle between them if the centre of gravity of the whole wire is at the centre. tan-"^. 3 9. Determine the centre of gravity of a loop of the curve r ^= a cos 20. _ 128 i/2 X = a. 10. Determine the centre of gravity of that part of the area of the cardioid in Fig. 61 which is on the right of the axis of _>'. j6 -\- ^TT 10^ ■^' ~ 16 + a^r"^' ^~84-3;r- 11. Find the centre of gravity of the arc of the same half cardioid. _ _ 4 X = y = —a. 5 12. Find the centre of gravity of the area above the axis of X contained between the curves^" = ax andy = 2ax — x^, X = a — —; y =■ -. i57r — 40 ^ in -% 13. From a circle whose radius is a two segments are cut off by chords drawn from the same point, each subtending 90° at the centre. Find the distance from the centre of the centre of gravity of the remaining area. 2a 3(^+2)* 140 PARALLEL FORCES AND CENTRES OF FORCE. [Ex. IX. 14. The middle points of two adjacent sides of a square are joined, and the triangle formed by this straight line and the edges is cut off. Find the distance of the centre of gravity of the remainder from the centre of the square. ^^ of the diagonal. 15. If the small triangle in Ex. 14 is folded over instead of cut off, find fhe distance. -^ of the diagonal. 16. Find the centre of gravity of the arc of the cycloid X = a{^ — sin ?/.), J = ^(i — cos ^). — - 4 X = na^ y =. — a. 3 17. Find the distance from the base of the centre of gravity of the area of the cycloid in Ex. 16. _ _ 5 18. Find the centroid of the arc in the first quadrant of the four-cusped hypocycloid x^ _|_ jl _ ^1. - _ - _ _^ 19. Find the centre of gravity of the area enclosed between the arc of Ex. 18 and the coordinate axes. _ _ 256^ X =y =-^ . 20. Find the centre of gravity of the arc of the catenary, Fig. 48, Art. 132. f . _f ^ = ^-^~--~~r-'y = 7s + 7- e' — e ^ 21. An area is composed of a semi-ellipse and a semi-circle having the minor axis for its diameter. Find the distance of the centre of gravity from the common centre. ^{a — b^ 3^~ * 22. Prove that the area between the parabolas j'* =^ px and x^ = qyis^pgy and find the coordinates of its centroid. X = ^phA ' y = 3^f,\q\ 20^ ^ 20^ ^ 23. Find the volume of a cone by Pappus's Theorem. 24 Show that the outer part of the anchor-ring (generated by the semi-circle DNC, Fig. 62) exceeds the inner part by twice the volume of the sphere whose radius is a. § X.] EXAMPLES. 141 25. An ellipse whose semi-axes are a and b revolves about a tangent at the extremity of the major axis. Find the volume generated. 271^ a^b. 26. Find the volume enclosed between the surface of the solid of Ex. 25 and a tangent plane. 10 — 37r The Centre of Gravity of Particles not in One Plane. 176. We have seen in Art. 156 that the point through which the resultant of the weights of particles in a single plane passes when the plane is horizontal, is such that the resultant passes through it when the plane is inclined in any way; in other words, when gravity has any direction with respect to the plane. This of course applies to a heavy lamina, and is the basis of an experi- mental method of determining the centre of gravity. For, if the lamina be suspended from any point, the centre of gravity will, in equilibrium, be vertically beneath the point of suspension, because the resultant weight and the supporting force must have the same line of action. This will enable us to draw a line on the lamina upon which the centre of gravity must lie. By sus- pending the lamina from a point not in this line, we can determine another line containing the centre of gravity ; this point must therefore be the intersection of the two lines. If the lamina were suspended from a third point, the vertical through the point of suspension would be found to pass through the point so found, thus giving an experimental verification of the existence of a centre of gravity which is independent of the direction of gravity relatively to the body. We have now to show that such a point, independent of the direction of gravity, exists for particles not all in one plane, and hence also for any solid body. This point is the centre of gravity, and, if the body were suspended from any point, would always be found in the vertical through the point of suspension. 142 PARALLEL FORCES AND CENTRES OF FORCE. [Art 177. 177- I" the case of particles, this point may be found by an extension of the graphic process in Art. 141, which it will be noticed is independent of the direction of gravity. For example^ if there be a fourth particle at a point Z>, not in the plane of ABC, Fig. 50, and if /, m, n, p be the weights of the particles, the centre of gravity G of the four particles will be on the line OD, and will divide it inversely in the ratio of the weight of the particle/ at D to the combined weights of the other three particles at O. That is to say, OG: GD=p:i-{- m-\-n. The same point would be found by grouping the particles in any other way. In fact, the construction shows that G is situated in the plane passing through two of the particles and the centre of gravity of the other two. The six planes of this character meet in a point, and any three of them which pass through edges of the tetrahedron ABCD not meeting in a point would serve to determine the centre of gravity. In like manner, for any number of particles, the centre of gravity may be found, as in Art. 151, by substituting at each step for two of the particles a particle equal to their sum at their centre of gravity, the process applying as well when the particles are not as when they are in one plane. In particular, when the particles are equal, the centre of gravity is the centre of position of the points (see Art. 69), of which it was shown that the distance from any plane is the average of the distances of the particles from that plane. Statical Moments with Respect' to the Coordinate Planes. 178. In the analytical treatment of the centre of gravity, the positions of the particles are referred to three rectangular axes, and we shall suppose at present that the plane of xy is horizontal. Let the particles whose weights are F^, F^, . , . P^ be situated at the points (^,,J^',, ^,), {x^, y^y ^J, . . . (^«,J^'«, Zn). Since the weights of the particles act in lines parallel to the axis of -s, their resultant is (by the principle of the transmission of force) the same g X.J Sl^A TJCAL MOMENT WITH RESPECT TO A PLANE. I43 as if they were situated at their projections in the plane of xy\ that is to say, at the points {x^,y^), {x^,y^, . . . (;c„, j«). It fol- lows that the resultant weight of the system of particles acts in a line which pierces the plane of xy at the point (^ j), defined by the equations of Art. 152; that is, x:2P = :2Px, j:2p = :2Py. The terms P^x^^ -^-^a, etc., which make up ^Px, are now the products of the particles, each multiplied by its distance from the plane oi yz. Such a product is called the statical moment of the particle with respect to the plane. Accordingly, '^Px is called the total or resultant statical moment of the system of particles with respect to the plane oi yz. 179. In considering the various positions of a body, or of a system of particles supposed to be rigidly connected so that they maintain their relative positions, it is convenient to retain the coordinate axes in their positions relative to the system, and to imagine the relative direction of the force of gravity to be changed. If then we suppose gravity to act in the direction of the axis of x^ the resultant of the weights of the particles may be shown in like manner to act in a line piercing the plane of yz at the point (y, z) defined by the equations -y^P = :2Py, -z:2P = :2Pz. The value of y is the same as before, and the lines of action in the two cases intersect in the point (a;, y, z), all of whose coor- dinates are defined by means of the statical moments of the par- ticles* with respect to the coordinate planes. These coordi- * We do not speak of the moment of ^ force or of a system of forces^ in general, with respect to a plane ; but the notion of the statical moment of a particle with respect to a plane arises from that of the moment of a force about an axis. The forces involved in the idea of the statical moment of a system of particles with respect to a plane are parallel forces acting in some direction parallel to the plane, and the axis iS any line in the plane perpendicular to that direction. 144 PARALLEL FORCES AXD CENTRES OF FORCE. [Art 179. nates are the weighted means of the like coordinates of the particles. 180. To show that the resultant of the weights passes through {x^y^ z) for all directions of the force of gravity relative to the coordinate planes, let /, rfi and 71 be the direction cosines (see Art. 66) of the direction of gravity, and resolve each of the forces into components parallel to the three axes. Then X^ — //*,, Y^ = mP,, Z, = tiF^, X^ = IP^, etc., and :2X = 12P, 2y= m2P, 2Z = n2P. The system of forces is now resolved into three systems of parallel forces. The resultant of the Z-system acts in a line parallel to the axis of z and piercing the plane of xy in the point {xy y) ; for we have seen that the resultant of the original system acts in this line when the ^'s have this direction, and the Z-system consists of the same forces each multiplied by the constant factor n. In like manner the resultant of the F-system is w-S"^, acting in a line parallel to the axis of j, which pierces the plane of xz in the point (x, z), and that of the X-system is /^P^ acting parallel to the axis of x at the point {y, z) in the plane of yz. Thus the whole system is reduced to three forces acting in lines which meet in the point (x, J, z). Therefore the resultant is the force 2P acting, as was to be proved, in a line which always passes through this point, which is for that reason called the centre of gravity. Centre of Gravity of a Volume or a Homogeneous Solid. 181. The position of the centre of gravity of a solid depends not only upon its size and shape, but upon the distribution of the matter within the volume. When the weights of equal volumes taken from any part whatever of the given volume are equal, the body is said to be homogeneous, and the weight of a unit volume is taken as the measure of the density. Denoting this weight by w, we have, for the weight of the homogeneous body of volume F^ JV= wV, § X.] CENTRE OF GRAVITY OF A VOLUME. 1 45 The distance of the centre of gravity from any plane is now found by the condition that the statical moment of W at that point, with respect to the plane, is equal to the total moment of the elements. The weight of the element of volume, dV^ is wdVx hence we have, for the moment with respect to the plane oiyz^ xlV= \wxdVf or, dividing by w, since W ■= wV^ xV={xdK (i) The value of each member of this equation is called the statical moment of the volume. 182. For example, let it be required to find the centre of gravity of a right cone of radius ^ and height a. Fig. 64 represents a section through the geometrical axis upon which the centre of gravity obviously lies. Let this axis be taken as the axis of Xj and the vertex O, as the origin. The section made by a plane perpendicular to tlie axis at the dis- ^^g* ^4. tance x from the vertex is ttj'^, and from Fig. 64 we have i^x TTd^'x* r=— ; whence dF= — ^-dxr ard equation (i) gives -xV^^Vx^dx^ a Jo Hence, knowing the volume of the cone to be V — \nb'^a^ we have 7\; = J^ ; that is, the centre of gravity of a cone is at a dis- tance of one-fourth of the altitude from the centre of the base. 146 PARALLEL FORCES AND CENTRES OF FORCE. [Art. 183. Employment of Triple Integration. 183. In the preceding article we supposed the area of the element for which the arm of the moment is constant to be known, and also the volume of the body. But the complete problem of finding the moment when only the equations of the bounding surfaces are known is, like that of finding the volume itself, one of triple integration. Suppose, in the first place, that the three independent variables used are the rectangular coordinates x^ y and z ; then the ultimate element of volume is dx dy dz, and that of the moment with respect to the plane of ys is x dx dy dz. Hence the moment is the triple integral of this expression taken with the same limits that would be used in finding the volume by the triple integration of dx dy dz. 184. For example, let us determine the centre of gravity of the solid represented in Fig. 65, which is common to the cylinder ^' +/ = <»', and the half, on the right of the plane oi yz, of the cylin- der x' +/ = «•. By symmetry, this centre of gravity is on the axis of x. Fig. 65. The element of moment with respect to the plane of yz is x dx dy dz ; and, if the integra- tions are performed in the order x, z, y, we have y "^ \ \ xdxdz dy^ § X.] EMPLOYMENT OF TRIPLE INTEGRATION. I47 in which x^ and s, are limiting values determined by the equations of the bounding surfaces- The fact that x occurs only in the equation of the second cylinder shows that the whole volume can be covered by one integration, in which the limit x^ is taken from that equation.* Hence, performing the integration for x, and substituting the value of ^j, we have xV=i\ [\a'-f)dzdy (2) J -a J — 2j Next, performing the ;2-integration, *^ = JV-/)V>'. (3) Finally, putting _y = a sin 6 in this integral, IT xV=: a'V cos* ddO = 2a'l^ - = ^7ta\ . . (4) J IT 4 • ^ 2 8 (Int. Calc, formula (P), p. 120.) Following the same order of integration, we have, for the value of r, V= [" p V'cixdzdy^V \ {a'-/ydzdy J -a] -ZiJo J -at —z., = 2 {a - y)dy = -— -. J -a 3 * If this were not the case, it would be necessary to find the volume in two parts. For instance, in the present example, if the jj/-integration were performed first, the limiting value oi y would in a part of the volume be determined by one of the cylinders, and in the other part by the other. 148 PARALLEL FORCES AND CENTRES OF FORCE. [Art. 184, Substituting in the value found for ^ F,* we find 64 Solids of Variable Density. 185. For a solid which is not homogeneous, let the variable 7V denote the density, at any point; that is to say, the weight f of a homogeneous unit of volume having the density of the body at that point. If now the law of the distribution of the matter within a solid of given volume is known — in other words, if w is given in the form of a function of the coordinates of the point — the weight of the body, as well as its statical moments, will have to be found * In finding a volume by triple integration, the first two integrations are equivalent to finding the area of a section parallel to one of the co- ordinate planes; and if we can employ a single integral, it is because this area is already known. So also in the example of Art. 182, in find- ing the value of i \\xdxdydz, we were able to begin with the form \xdV, because the result of the first two integrations would be the area of the section perpendicular to the axis of x, which in that case was known to be tt;/', ^ being a given function of x. But, in the example of Art. 184, it was not convenient to find the area of a section parallel to the plane oi yz. We might, however, have used the section parallel to the plane of xz, which is the double square 2^", so that the element is ix'^dy; because we know that its centre of gravity is at its geometric centre, and there- fore at a distance |;f from the plane oi yz. Therefore the element of moment is x^dy, giving at once the above expression for xF as a. simple integral, equation (3). f The density is often defined as the mass of a unit of volume, so that the weight of the unit is w = gp, but in using gravitation units it is more convenient to use the weight of a unit volume, which is here de- noted by w and called density. This measure of density would be properly called specific weight had not the term specific gravity h^^w applied to the ratio of density to that of water, for which w is 62^ pounds. § X.] SOLIDS OF VARIABLE DENSITY. I49 by integration. In this case, dV being an element of volume, wdVwiW be the element of weight, and the total weight will be JV= [wdV taken between the limits which define the volume. 186. For example, let us determine the weight of a sphere whose radius is ^, when the density varies inversely as the square of the distance from the centre, and is w^ at the surface. The conditions give w\w^^= a^\ r^^ whence w = -^^, where r is the dis- r tance of the point from the centre. Since p is in this case given in terms of r, it is convenient to use an element of volume such that r has the same value for all its points. The area of the spherical surface at the distance r is \nr'^\ hence, taking as the element the spherical shell of thickness dr, we have dV = ^vtr'^dr. Therefore W = IwdF — 47rw^a^\ dr = ^Ttwa^, Since the volume of the sphere is |7r^', this sphere has three times the weight of a homogeneous sphere of density equal to that at the surface. Hence its average density is 3t£/. Centre of Gravity of a Solid of Variable Density. 187. In finding the statical moment of a solid which is not homogeneous, it will generally be necessary to use the ultimate element of volume as in Art. 184, because the lamina parallel to the plane of reference, used as an element in Art. 182, will not have the same value of w for all points of its area ; and therefore, although we may know its area, we cannot write the expression for its weight. For example, let the centre of gravity of one half of the sphere considered in the preceding article be required. 150 PARALLEL FORCES AND CENTRES OF FORCE. [Art. 187. The centre of gravity is in the radius perpendicular to the base of the hemisphere. Let JBAC, Fig. 66, be a section through this radius OA^ intersecting the base in the diameter BC, and F any point within the hemisphere. The fact that Te/ is a function of r makes it advisable to take r as one of the three independent variables ; for the other two, let us take 6 the angle BOP^ and the angle between the planes BOP and BOA. When these quantities vary separately, the differentials of the motion of P are dr^ rdO and r sin Od(p ; and, since these differentials are mutually rectangular, the element of vol- ume is r^ sin 6 dd dcpdr, and that of weight is Fig. 66. dW^r'w sin Odddcpdr, The distance of P from the base of the hemisphere or arm of the moment, is r sin cos

we have xW = U'Tta'X rdr = — ^ ; Jo 2 and since the weight of the hemisphere as found in Art. 186 is W = 27Tw^a'', we derive x = \a. §X.] STABLE AND UNSTABLE EQUILIBRIUM. I5I Stable and Unstable Equilibrium. 188. We have seen that, when the conditions of a problem define the forces which act upon a body for all positions of the body, or at least for a series of positions which the body is free to take, there are positions of equilibrium. If now the body in a position of equilibrium suffer any of its possible displacements, the lines of action, and sometimes also the magnitudes, of the forces will be so modified that equilibrium will, in general, no longer exist. If the action of the forces in the new position, assumed to be indefinitely near to that of equilibrium, is such as to cause the body to return to the position of equilibrium, that position is said to be one of stable equilibrium. If, on the other hand, the action is such as to urge the body away from the posi- tion of equilibrium, it is said to be one of unstable equilibrium. 189. For example, we supposed the weights in Fig. 11, page 2i^y to be allowed to adjust themselves into a position of equilib- rium. This would not be possible if it were not a position of stable equilibrium, but a little consideration will show that this is the case. For instance, if C be displaced downward, the result- ant of P and Q will become greater than -/?, and the total action on the knot will be upward. Again, in Fig. 21, page 54, the position is one of stable equi- librium, because, if A be brought nearer to C, the repulsive force is increased. In Fig. 20, page 51, -^ was so determined as to produce equilibrium, and we cannot pronounce it as stable or unstable unless P is defined for all positions of A regarded as movable along the line AB. If now we suppose P to remain constant in magnitude and direction, equilibrium will still exist when A is displaced. The position is therefore called one of neutral or astatic equilibrium. 190. When a rigid body is displaced in any manner involving rotation, the forces of gravity upon the several parts retain their directions and magnitudes, while their lines of action are shifted into new relative positions. In proving the existence of a centre 152 PARALLEL FORCES AND CENTRES OF FORCE, [Art. 190 of gravity, we have shown that there is a point at which if the body be supported it will remain in equilibrium for all possible displacements, that is, it will be in astatic equilibrium. Accord- ingly a system of parallel forces having definite points of appli- cation in a body is said to have an astatic ce?ttre. In the case of a system of coplanar, but not parallel, forces having definite points of application in a rigid body, and invaria- ble in direction and magnitude when the body is turned in the plane, it can also be shown that an astatic centre exists. See examples 23 and 24 below. 191. If the heavy rigid body be supported at any other point than the centre of gravity, the reaction of the support will be equal to the weight, and with it will form a couple which will cause the body to turn, if free to do so, unless the centres of gravity and of suspension are in a vertical line. In the latter case, equilibrium will exist, and it will plainly be unstable when the centre of gravity is above the point of support, and stable when it is below it. We may, in this case, regard the centre of gravity as a heavy particle which is constrained to lie in a spherical surface, and therefore rests in stable equilibrium only at the lowest point of the surface. Again, if the body be supported upon an axis about which it is free to turn, the centre of gravity describes a circle (unless the axis passes through it), and will seek the lowest point of the circle if it lies in a vertical or oblique plane; but, if the plane of the circle is horizontal, that is, if the axis is vertical, the body will be in neutral equilibrium. Equilibrium in Rolling Motion. 192. When a body with a curved surface rolls upon a fixed surface, equilibrium can exist only when the point of contact is in a vertical line with the centre of gravity; otherwise there will be, as in the preceding article, a couple which will cause the body to roll. In some cases, the stability of the equilibrium is readily deter- §x.] EQUILIBRIUM IN ROLLING MOTION. 153 mined by considering the path of the centre of gravity. For example, when a cylinder rolls on a horizontal plane, if the centre of gravity is not on the geometrical axis it will obviously be at the lowest point of its path when between the axis and line of contact, and at the highest point when vertically above the axis. The former is therefore a case of stable, and the latter one of unstable, equilibrium. 193. In general, the stability of the equilibrium is more con- veniently determined by means of the couple formed when dis- placement takes place. For example, suppose the heavy body to rest with its convex surface in contact with the convex surface of a fixed body, the common tangent plane being horizon- tal and the centre of gravity G vertically above the point of contact, so that equi- librium exists. Let Fig. 67 represent a vertical section through these points, and let the sections of the surfaces at first be supposed circles whose centres are C and B and whose radii are r and R, If the surfaces are smooth, the equilibrium is unstable, because as soon as displacement takes place the body will slide down the inclined surface. But suppose them to be rough, so that rolling takes place, and let Ca be the new position of the radius CA upon which G lies, while A' is the new point of contact. Then the arcs A' a and AA' are equal; and, denoting the angles subtended at C and B by and ^, we have r(f> = Rdj or (p : 6 =: B : r. . (0 Let a vertical line through A' intersect C'a in M; then, if G' is between C' and M^ the couple formed by the vertical forces, namely, the weight acting at G^ and the upward reaction of the fixed surface at ^', will tend to roll the body still further from its original position, and the equilibrium is unstable. If, on the other hand, 6^' is between a and M, the body will tend to return 154 PARALLEL FORCES AND CENTRES OF FORCE. [Art. 193. and the equilibrium is stable. Now, from the triangle CM A' we find CM _ sin e r ~sin(^+0) ^^) The limiting value, when B is small, of -: — 777— — tt is tt", — :, or, sm (c/ -|- 0) £7 -|- by equation (i), — ■ — — ; hence at the limit r -^ K C'^=7T^. and aM=-^. . . (3) It follows that, if AG^ the height of the centre of gravity above the point of contact in the position of equilibrium, is greater than rR ... ... ... • — j — ^, the position is one of unstable equilibrium, but if it is ^ r R less" than ^r, , the equilibrium is stable. 194. When the sections of the surfaces are not circles, the condition for stability is the same, R and r now standing for the radii of curvature of the sections. If the body rests in a concav- ity of the fixed body, R is negative, and putting R = — R\ we have, for the value which AG must not exceed if the equilibrium is to be stable, rR' R' - r In like manner, if the curvature of the section of the moving body be reversed, putting r = — r\ the expression for the limit- ing height becomes r'R r' - R' If the body rests upon a plane, R is infinite, and we have r for the limiting value, as obviously should be expected. Again, if §X.] LIMITS OF STABILITY. 1^5 r is infinite, so that a plane surface rests upon a curved one, we have R for the limiting value. If the curvature of the sections made by vertical planes passing through the line BC is variable, it is necessary for com- plete stability that AG should be less than the least value of rR r-\-R' Limits of Stability. 195. In cases of stable equilibrium, if the displacement be carried beyond certain limits, the body will not return to its original position. For example, in Fig. 67, though for small dis- placements G is found on the left of the vertical through A' , it will, if the angle of rolling be increased, reach that line, and the body will then be in a position of unstable equilibrium. Hence if it be still further displaced, it will not tend to return to its first position. In like manner, there is a position of unstable equi- librium on the other side, and these determine an interval within which displacements may take place without causing the body to leave the position of stable equilibrium. The equilibrium is said to be more or less stable according to the size of this interval. In the example, this interval, which is large when AG '\^ small, decreases as we increase AG\ and finally disappears when AG equals the limiting value given in equation (3), so that the posi- tion then becomes one of unstable equilibrium. 196. The notion of limits of stability is sometimes applied also to cases in which the body itself is not displaced with refer- ence to other bodies which are in contact with it and react upon it, but in which external forces can undergo changes within cer- tain limits before equilibrium is destroyed. For example, a three-legged table stands upon a horizontal plane. If the centre of gravity be moved, by changing the position of heavy bodies upon the table, the resistances at the three feet adapt themselves as explained in Art. 148. But, if it be moved until its projection upon the horizontal plane crosses one of the sides of the triangle formed by the feet, the equilibrium will no I 5^ PARALLEL FORCES AND CENTRES OF FORCE. [Art. 196. longer exist, unless the resistance at the opposite foot can change sign ; that is to say, unless this foot is held down, the table will topple over. Thus, the condition of equilibrium, when the feet are not held down, is that the projection of the centre of gravity- shall fall within this triangle, which is called the base. In like manner, for a body of any form resting upon a hori- zontal plane, the smallest convex polygon which encloses all the points of contact with the plane is called the base, and the con- dition of stability is that a perpendicular from the centre of gravity shall fall within the base. EXAMPLES. X. 1. Determine the centre of gravity of seven equal particles situated at the vertices of a cube. 2. Show that the centre of gravity of a tetrahedron is the same as that of four equal particles at its vertices, and cuts off one-fourth of the line joining the centre of gravity of either face with the opposite vertex. 3. Extend the result of Ex. 2 to any pyramid and thence to any cone. 4. Show that if a and b be any homologous lines in the bases of a frustum, and h the distance between the bases, the distance of the centre of gravity from the base in which a lies is a'^ 2ab + 3^% ^ 4(a' -]- ab + b') ' 5. A cone of height A is cut out of a cylinder of the same base and height. Find the distance of the centre of gravity of the remainder from the vertex. 3 . 6. Find the centre of gravity of the solid formed by the revo- lution of the sector of a circle about one of its extreme radii. The height of the cone being denoted by /i, and the •? radius of the circle by a, we have ^ = -^{a -\- h), o 7. A solid is formed of a hemisphere whose radius is a and a paraboloid with the same base. What must be the height of the g X . ] EX A MPLES. I 5 7 paraboloid, in order that the solid may rest with any point of the spherical surface upon a horizontal plane ? 4/6 2 8. Find the distance between the centre and the centre of gravity of one-half an anchor-ring generated by a circle whose radius is a and whose centre describes the circle whose radius is h. 271 b 9. A paraboloid whose parameter is \a stands on a plane whose inclination is a and is prevented from sliding. Find its height if just on the point of toppling over. h = ^6a cot' ex. 10. A paraboloid and a cone have a common base and vertices at the same point. Find the centre of gravity of the solid enclosed between their surfaces. The centre of gravity is the middle point of the axis. 11. A cone of height /i and radius a is hung up by a string over a smooth peg, one end being attached to the vertex and the other to the rim. Find the length of the string if equilibrium exists when the axis is horizontal. V{4^^ + ^^^) 12. Determine the centre of gravity of the surface formed by revolving the cardioid r = a{i + cos 0) about the initial line. - 50 X = —a. .63 13. A frustum is cut from a right cone by a plane bisecting the axis. If the frustum rests in equilibrium with its slant height upon a horizontal plane, find the greatest ratio the radius of the lower base can bear to the height. ' /17 N/y 14. Find the average density of a sphere whose density varies inversely as the distance from the centre, yu being the density at the surface. •? 2 15. The corners of a tetrahedron are cut off by planes par- allel to the opposite faces. Prove that if the parts cut off are equal, the centre of gravity of the remainder will coincide with that of the tetrahedron. 158 PARALLEL FORCES AND CENTRES OF FORCE. [Ex. X. 16. If a uniform lamina, whose form is that of the area be- tween the sinusoid jv — sin x and the axis of x, be suspended from one extremity of its base, show that the base will make the angle tan" ^4 with the horizontal. 17. The density of a sphere of radius a varies uniformly from p, at the centre to p, at the surface. Determine the centre of gravity of one hemisphere. _ 3(p -f- 4p )^ ^ ~ io(p, + 3P,)' 18. A cone of height h rests with its base upon the vertex of a paraboloid whose parameter is 4^2. Find the greatest value of h for stable equilibrium. Za. 19. A plank rests upon a rough cylinder of radius -^ in a horizontal position of stable equilibrium, h being the height of the centre of gravity above the point of contact. Show that the position of unstable equilibrium occurs when the plank is rolled through the angle d determined by li 20. A paraboloid, whose height is ^, and the radius of whose base is 3, rests with its convex surface on a horizontal plane. Determine the inclination a of the axis to the horizon, and thence determine the greatest value of h for which the equi- librium at the vertex is stable. . . 3^^ sm» a — 21. Find the distance between the centre of the sphere and the centre of gravity of the volume cut from the sphere of Art. 186 by a cone whose vertex is at the centre and whose semi-verti- cal angle is «. ^/ , \ ^ —{1 + cos a). 4 22. Determine the centre of gravity of a segment of the same sphere cut off by the plane x = /i, knowing that the centre of gravity of a spherical cap bisects its altitude. _ _ 1 a^ — /i^ — 2/i'^{]oga — log h) 4 a — h — -^(log a — log K) § X.] EXAMPLES. 159 23. If a rigid body, upon which two coplanar forces act at definite points of application, be turned in the plane of the forces, the forces retaining their magnitudes and directions, show that the resultant will always pass through a fixed point upon the cir- cumference which is the locus of the intersection of the lines of action. This point is the astatic centre of the forces (see Ar^. 190), and when they are parallel it becomes the "centre of parallel forces." 24. Show, hence, that any system of coplanar forces has an astatic centre ; and, if the forces are referred to rectangular axes, putting ^{xY-yX) = K and :2{xX -\- yY) = K, prove that the astatic centre is the intersection of the lines x:2r - y2X = X and x:SX -{-yl:Y= V. CHAPTER V. FRICTIONAL RESISTANCE. XI. Laws of Friction. 197. When a body is so constrained, by material bodies with which it is in contact, that motion can take place only along a certain line, the resistance of the line is normal to it when the surfaces in contact are smooth ; but when they are rough, the line offers a resistance which has a component along as well as one normal to it. The component of resistance along the line is called statical friction. This frictional resistance, like the normal resistance, is a passive force which adapts itself so as to produce equilibrium if possible; but, unlike the normal resistance, it cannot exceed a certain limit. Thus, if a brick rest upon a horizontal table, and a small horizontal force applied to it be gradually increased, this force will be resisted until it reaches a certain value which is called the limiting statical frictiofi. If the force exceed this value, the brick will move, but with the acceleration due to a force less than that actually applied, and the diminution thus suffered by the force is called the dynamical friction. 198. The following " laws of friction " were enunciated in 1 781 by Coulomb as the results of his experiments : I. The litniting statical friction is, for a gi7>e?t pair of surf aces in contact, proportional to the nortnal pressure. Thus, if a second brick of the same weight be placed upon the brick in the illus- tration above it is found that the limiting statical friction is doubled. § XL] LAWS OF FRICTION. l6r 2. The limiting statical friction is independent of the area of suiface in contact. Thus, if the brick be placed upon its side, the horizontal force required to move it is the same as when it rests upon its face. This law is easily seen to be a consequence of tlie first law. For, if the second brick be placed on the table and connected by a string to the first, the statical friction is doubled, and is therefore by the first law equal to the value which it has when the normal force is doubled, without change of the area of contact, by placing the second brick upon the first. 3. The dynamical frictio7i is independent of the velocity. Accord- ingly, after the body is in motion with a given velocity, the same force will suffice to keep it in uniform motion, no matter what the given velocity may be ; and, if the force applied exceed this, the acceleration will be constant. We should, therefore, expect the dynamical friction to be equal to the limiting statical friction and to obey the same laws, but it is found that the dynamical friction is somewhat the smaller. The third law cannot therefore be true for very low velocities, and it has also been found that the laws require modification in certain other extreme cases. But we are here concerned only with the limiting statical friction, and shall assume that, in accordance with the first two laws, it bears a fixed ratio to the normal resistance R ; so that it may be WTitten fxR^ in which /i is called the coefficient of friction. 199. The cause of friction is the roughness of surfaces con- sisting of small projections which, fitting into one another, must either be broken off, or cause the surfaces to separate when they move on one another. The coefficient of friction differs greatly for different substances, and is diminished by grinding and polishing the surfaces, and also by introducing lubricating sub- stances. The following table will give an idea of the general range of values of yw for unlubricated surfaces : For iron on stone, ^ varies from .3 to .7 ; For timber on timber, fx " .2 to .5 ; For timber on metals, jx " .2 to .6 ; For metals on metals, }x ** ,15 to .25. 1 62 FRICTIONAL RESISTANCE. [Art. 20a 200. The direction of frictional resistance in the case of a body resting upon a surface is opposite that in which motion would take place if the surface were smooth. The resultant of the normal and the frictional re- sistances is called the total resist- ance of the surface. Thus, if a particle at A^ Fig. 68, acted upon by forces, be kept at rest by the normal resistance AR of the plane AB and the friction AF^ the total YiQ, 68. resistance of the plane is repre- sented by A 7", the resultant of AR and AF. This force must, of course, be directly opposite to the resultant of the active forces which would otherwise produce motion. Hence, if AB is an inclined plane and there is no active force except the weight of the particle, ^7" will be directed verti- cally upward. 201. If we furthermore suppose the friction acting to be the maximum or limiting statical friction, we have, denoting the normal resistance by R and the friction by F^ F= }aR, and, if we denote by a the angle between the normal and the total resistance when limiting friction is acting, F tan a = -= M- This angle «', which is therefore the greatest possible inclina- tion of the total resistance to the normal, is called t/ie angle of friction. Fig. 68 shows that the angle of friction is the same as the inclination of the plane when the body is about to slide. It may be found by gradually increasing the inclination until motion takes place ;* the total resistance is, in this case, equal * Owing to the fact that the limiting statical exceeds the dynamical friction, the body will, on starting, move with a uniform acceleration. If the inclination be now decreased until its tangent equals the coeffi- cient of dynamical friction, the body will move with a uniform velocity. § XL] LIMITS OF^EQUILIBKIUM ON INCLINED PLANE. 163 to the weight W^ hence the normal resistance produced by the weight is ^cos a^ and the friction is W sin a. Limits of Equilibrium on a Rough Inclined Plane. 202. Suppose the weight Cresting at A upon a rough plane, whose inclination ^ is greater than the angle of friction, to be supported by the force P^ whose line of action lies in the vertical plane which contains the normal, and makes the angle with the in- clined plane, as represented in Fig. 69; let us find the limiting values of P consistent with equilibrium. The greatest value of P will occur when the body is on the Fig. 6g. point of moving up the plane, and the least value is that which is just sufficient to prevent the body sliding down the plane. In the latter case, the friction acts up the plane, as represented in the figure (that is, it assists in holding the body up), and, being the limiting friction, its value is /i^?, where R denotes the normal resistance. Resolving the forces perpendicularly to and along the plane, we* have R =. W Q.o^ S — P sm (f)^ fxR -(- P cos (p = JV sin 0. Eliminating R, P(cos — /^ sin (p) ='lV{s\n (9 — /i cos 6*) ; and, since pi = tan ^, where a is the angle of friction, we have for the least value of P, or the force which just sustains the weighty „ rrrSin S cos a — cos S sin a ,„ sin {d — a) JJ -m^ yy -rzz yy • cos cos «: — sin sin a cos (0 + «) (0 164 FKICTIONAL RESISTANCE. [Art. 203. 203. To find the greatest value of F, the body being on the point of moving up the plane, we have only to change the direc- tion of the frictional force p.R in the figure, since the limiting friction now acts down the plane. Hence, replacing /i by — yw, or o' by — «', in equation (i), we have for the greatest value of Fi or the force ivhich will just fail t-o move the body up the plane y P = iv ''"" \^ + "\ (.) cos [cp— a) ^ ' Equilibrium will exist for any value of F between the limits ^ivcii in equations (i) and (2). 204. When = a, F in equation (i) vanishes, irrespective of the value of (p, as should be expected, since friction alone will just sustain the body at this inclination of the plane. When <. a^ F in equation (1) becomes negative, and put- ting F' for its numerical value, F^ is the force which, acting in a direction opposite to AF in Fig. 69, will just fail to move the body down the plane. This force acts down the plane, and with a component pushing the body against the plane. Replacing F by — F' in equation (i), we have cos (a -\- (/)) The angle between the direction of the force F^ and the direction down the plane now lies below the plane ; hence, if we put 0' = — in the last equation, we shall have the value of F' when pulling down the plane *at the inclination 0'. That is to say, for the "greatest force which fails to move the body down the planCy we have ^.^^^ sin(^-^) cos {a — 0') ^^^ This formula may of course be derived directly from a diagram properly constructed. In each case, the value of F might have § XL] THE CONE OF FRICTION. 1 6$ been obtained without elimination by resolving in a direction perpendicular to that of the total resistance. 205- The values of the force in equations (2) and (3) may be regarded as the least values of the force which will start the body up or down the plane, as the case may be, when acting at the given inclination 0. Hence, when is arbitrary, the value which makes, the force a minimum is the most advantageous when the body is to be moved along the plane. Thus, the most advan- tageous value of for hauling the body up the plane is = ty, which makes P in equation (2) a minimum. As we increase the value of from zero, the component of P along the plane which must overcome the friction (as well as a component of W) is diminished, but this loss is compensated by the diminution of friction produced by the considerable component of P normal to the plane. The corresponding value of P is ^sin ifi -\- ^), which is the same as if the plane were smooth and its inclination were d -\- a. The most advantageous value of for keeping the body from sliding down the plane when ^ > a: is = — «', which makes P in equation (i) a minimum. This implies that, if the force be applied from above the plane, it should be a pushing force up the plane, a component of which increases the friction, which is now advantageous. The corresponding value of jP is W sin {B — «'), the same that would be required on a smooth plane inclined at the angle 6 — a. Again, the most advantageous value of for hauling the body down the plane, when 6 < a, is, from equation (3), (f)' = a \ thus, for hauling in either direction the best "angle of draught" is the angle of friction, the direction of draught being in each case perpendicular to the direction of total resistance. The Cone of Friction. 206. We have, in the foregoing articles, considered the force P in Fig. 69 as acting in the plane containing the normal at A and the vertical. When this restriction is removed, the total 1 66 FRICriONAL RESISTANCE. [Art. 206 resistance of the rough plane will not necessarily lie in the vertical plane. But equilibrium will exist whenever the angle which its direction makes with the normal does not exceed a. The limiting positions of the total resistance will, therefore, lie in the surface of a cone of which A is the vertex, the normal is the axis and ol is the semi-vertical angle. This cone is called the cone of friction, 207* As an application to the general case of a body resting upon a rough inclined plane, we notice that P is in equilibrium with the weight IV B.nd the total resistance, which we shall denote by T. Therefore P is the resultant of JV and T both reversed. When the limiting total resistance is reversed, it acts in the sur- face of the cone of friction produced downward below the in- clined plane. Hence, if T were known, we could construct P at ^ by first laying off /'^reversed (that is, upward), and then from its extremity a line representing T parallel to an element of the cone just mentioned. It follows that, if we lift this cone, without any change of direction, until its vertex is at the extremity of IV laid off upward, the end of the line representing P will lie in the surface of the cone. In other words, let a right circular cone be constructed with its vertex at a distance JV directly above A, its axis perpen- dicular to the inclined plane, and its semi-vertical angle equal to a. Then the body at A will be in equilibrium when acted upon by a force P represented by a line drawn from A to any point within this cone ; but, if the line representing P terminates out- side of the cone, the body will move. The point A will itself be outside of the cone when > a, sls supposed in Art. 202, and within it when 6 < a, sls supposed in Art. 204. Frictional Equilibrium of a Rigid Body. 208. In the case of a body resting upon a curved surface or upon several surfaces, the friction at different points must be considered separately. When friction is called into action to %Xl.]FJ^JC7^I0NAL EQUILIBRIUM OF A RIGID BODY. 1 67 produce equilibrium, the resistances are, in general, indeterminate. Thus, if a heavy rod AB rests with its ends upon a rough hori- zontal plane and a rough vertical wall, the total resistances at A and B will meet the vertical line through the centre of gravity in the same point. This point must be within each of the cones of friction constructed, as explained in Art. 206, at A and B\ otherwise, one at least of the frictional resistances would have to exceed its limiting value. Supposing the vertical line through the centre of gravity to pass through the space common to the two cones, the point in question may be any point of the seg- ment of the line within this space, and the values of the resist- ances at A and B are to a certain extent indeterminate. 209. In the limiting position of equilibrium, however, we must assume that, at each of the points where motion must take place if the equilibrium is broken, the maximum friction in the direction opposite to that motion is called into action. For example, let us find the greatest angle which the rod AB can make with the wall, supposing it situated in a vertical plane perpendicular to the wall, as represented in Fig. 70. The coefficients of friction at A and at B are assumed to be the same. In this case, if the equi- librium is broken, the motion which must take place at A is outward from A""-^,'p^ the wall, and that at B is downward. T^ • ^ ^^ ■ r 1 i F^G. 70. Drawmg the directions of the total re- sistances accordingly, so as to make the angle oc with the normal in each case, they will meet in C, and the triangle ABC is right- angled at C. Moreover, the point C is vertically above the cen- tre of gravity G ; hence ACG = a. The circumference of a circle described on the diameter AB will pass through C, and if CG produced meets it in Z>, the arc A£> which subtends the angle a a.t C will subtend the angle 2a at the centre. Hence Z> is a fixed point * relatively to AB. We * The total resistances are W cos a and PV sin a, which are con- c ^, "T^- .~-^^ y /j / / A ;«! / i \ / \^ ^ 1 1 68 FRICTIONAL RESISTANCE. [Art. 209. have thus for a given position of 6^ a graphical construction for the angle AGD, which is 6. If the rod is uniform, so that G is at its middle point, we shall have =: 2a. Moment of Friction. 210. When a definite portion of the surface of the solid body whose equilibrium is under consideration is in contact with the fixed body, the distribution of the normal pressure over the area of contact depends, as mentioned in Art. 158, upon the geometrical exactness of the surfaces in contact, and the rigidity of the materials. In accordance with the laws of friction. Art. 198, this distri- bution of pressure makes no difference in the total amount of friction when the friction at different points acts in parallel lines ; that is, when the motion resisted is one of translation. When, on the other hand, the motion resisted is one of rotation about an axis perpendicular to the surface of contact (which we shall suppose to be a plane area), this is no longer true. Thus, a heavy cylinder resting with its base upon a rough horizontal plane will, through friction, resist a force tending to turn it about its axis; but the maximum moment of this resistance depends upon the distribution of the pressure produced by the weight. If, owing to slight inaccuracies in the fitting of the surfaces, the weight rests chiefly upon the area near the centre of the base, the limiting moment of the friction will be small. If, on the other hand, it rests chiefly upon the rim, the moment will be comparatively large. 211. If, in the illustration of the cylinder, given in the pre- ceding article, we assume the pressure caused by the weight W slants; hence, if we suppose the limiting resistance to remain in action while the body ismoved, D will be their astatic centre (see Ex. X, 23). The forces are therefore equivalent to W downward at 6", and W upward at D. This indicates unstable equilibrium. Practically the equilibrium on one side is neutral; for friction, being a passive force, cannot act so as to produce motion; in other words, when S is dimin- ished, the friction no longer has its limiting value. § XL] MOMENT OF FRICTION. 1 69 to be uniformly distributed over the circular base whose radius W is ^, we shall have / = — 5-, while pdA is the pressure upon an element of area dA^ and }ApdA is the limiting friction. Hence, if r be the distance of the element from the centre of rotation, jAprdA is the element of the moment of the friction about the axis of rotation, which we assume to be the geometrical axis of the cylinder. Taking, for the element of area, the ring which is at the distance r from the centre, we have dA = 27trur\ whence we find, for the element of moment, and integrating. dM — T-r dr. M = -S — I ^ V^ = iapi W, a Jo Since the limiting friction is /^ ^, the moment is the same as if the whole friction acted with an arm |^, so that it is f of what it would be if the weight rested entirely upon the rim. Supposing the axis to be fixed, a horizontal force whose moment does not exceed f^yuW^can be applied to the cylinder without producing motion. Also, if the axis is not fixed, a hori- zontal couple not exceeding the same limit can be applied with- out producing motion, for it is easily seen that the moment of friction about any other point is greater than that about the centre. 212. The resistance of a body to rolling when in neutral equilibrium, as, for example, of a homogeneous cylinder upon a horizontal plane, is called rolling friction. Like true friction, its limiting value is proportional to the normal resistance, but its coefficient is, in general, very small, particularly when the sub- stances in contact are hard. Friction of a Cord on a Rough Surface. 213. We have seen, in Art. 51, that the tension of a cord is not altered when the cord passes over a smooth curved surface, I/O FRIC TIONA L RE SIS TA NCE. [Art. 213. because the resistance is always normal to the direction of the string. But, if the surface is rough, an inequality of tension may exist, the equilibrium being maintained by the friction of the cord upon the surface. In estimating the effect of this friction, it is necessary first to obtain an expression for the normal resist- ance of the surface at any point. Suppose the curve of contact of the cord and surface to be the circular aic AB^ Fig. 71. Denote the radius by «, the angle Q subtended at the centre by 6^, the /\ length of arc by s^ and the ten- sions at A and B^ which we shall at first suppose equal, by T. Producing the tangents at A and,^ to meet in C, we see that the resultant resistance of the whole arc AB acts in the direc- tion CO^ which bisects the angle Fig. 71. AOB^ and its value is 2 T sin ^e. The tension being in this case uniform, it is obvious that the normal resistance is uniformly distributed over the arc. The intensity of the resistance at any point of the arc of contact is the resistance which would be offered by a unit's length, if at every point the resistance were the same in intensity and direction. Denoting this resistaiice per linear unit by R^ the resistance offered by a length s under the same circumstances would be Rs. It follows that the value of R is the limit of the value of the re- sultant resistance divided by s (that is, by aB)^ when B is dimin- ished without limit. We have, then, 7? = 2rsin a ' (i) 214. Next suppose that the tensions T^ at A and T^ at B are not equal, and that 7", > T^. The component of the resistance in the direction OC is now {I\ -{- T^) sin \0. Proceeding to the i^Xl.] FRICTION OF A CORD OX A ROUGH SURFACE. IJl limit, we have for the normal resistance R the same value as be- fore, since 7\ = T^ at the limit. But the resistance has now a component perpendicular to OC^ namely, (r, — rj cos 16/, which balances the resolved parts of T^ and 7", in this direction. This is the resultant effect of the friction on the arc AB or aB. It follows that the friction for any element of arc, As = aAd^ is AT cos\Ad. Hence, denoting the intensity oi friction at any point by F^ we have, proceeding to the limit, _ ATcos^Ad -\ _ dT aAe A^^^radd ^^^ Now, if the excess of T^ over T^ is such that the cord is on the point of slipping from B toward A^ the limiting amount of fric- tion is acting at every point, that is, F = }xR. Hence, from equations (i) and (2), dT _ T adO a * or In this expression is measured from B toward A; hence integrating between limits, we have log r. - log T, = }xd, whence T, = T,e<^' (3) This formula shows that the ratio of the tensions depends Only upon the coefficient of friction and the angular measure of the arc of contact, and is therefore independent of the size of the cylinder. 172 FRICTIONAL RESISTANCE. [Art. 215. 215. When a large tension T^ is to be sustained and the fo»ce T^ available is small, friction is taken advantage of by taking several turns about a rough cylindrical post. If n is the number of turns, we have, putting i^ = 2«;t, and, taking common logarithms, iog,„r, = iog,„r, + 2.7288«//, .... (4) in which the constant is the value of 2 tt log,o e. Thus, for ex- ample, if three turns of a rope under the tension T^ be taken around the post, and a force T^ = 100 pounds be applied to the other end of the rope, it will not surge, or slip upon the post, unless T^ is greater than the value determined by equation (4). Supposing /^ = 4, this equation becomes logj„ r, = 2 4- 2.0466, whence we find T^ = 11,133 pounds. 216. When the arc of contact of the cord and surface is not circular, equation (i) becomes where p is the radius of curvature. In equation (2), pdcp, which is the value of ds, takes the place of add. Thus and the final result is T = T e*^'^ Since the variable p has disappeared before integration, it appears that the ratio of the tensions is independent of the shape of the rough surface, depending only upon the coefficient of friction and the angle 0, which is the total change of direction which the rope undergoes. §XL] EXAMPLES. 173 EXAMPLES. XI. 1. On a rough plane of inclination 6 the greatest value of the force acting along the plane and producing equilibrium is double the least. What is the coefhcient of friction ? /^ = i tan B.. 2. Two unequal weights, W^ and W^^ on a rough inclined plane are connected by a string which passes through a smooth pulley in the plane. Find the greatest inclination of the plane consistent with equilibrium. ,, W, + W, '^" ^ = w-^:"- 3. Two rough bodies, W^ and ^, , rest upon an inclined plane and are connected by a string parallel to the plane. If the coefficient of friction is not the same for both, determine the greatest inclination consistent with equilibrium, and the tension of the string. 4. If the angle of friction is 30^, what is the least force which will sustain a weight of 100 pounds on a plane whose inclination is 60° ? 50 pounds. 5. A uniform pole leans against a smooth vertical wall at an angle of 45° with it, the lower end being on a rough horizontal ])lane and about to slide. What is the value of yu ? yU = |. 6. Two equal uniform beams, connected at one end of each by a smooth hinge, rest in a vertical plane with their other ends on a rough horizontal plane. If ft is the greatest possible angle at the hinge, what is the coefficient of friction ? /i = i tan \ft. 7. A heavy uniform rod, whose length is 2^, is supported on a rough peg, a string of length / being attached to one end of the rod and fastened to a point in the same horizontal plane with the peg. If, when the rod is on the point of slipping, the string is perpendicular to it, show that / = }xa. 8. A weight JV on a. rough horizontal plane is attached to a string which passes over a smooth pulley at the height a above 174 FRICTIONAL RESISTANCE. [Ex. XL the plane and carries a weight F hanging freely. It is found that / is the least length of string between W and the pulley con- sistent with equilibrium. What is the coefficient of friction? 9. Two equal rings of weight W rest on a rough horizontal rod; a string of length /passes through them and has both ends attached to a weight W, If ^ is the coefficient of friction for the rod and rings and there is no friction between the string and rings, what is the greatest possible distance between the rings ? // W'^ 2\- M\2lV-{- W')% 10. A uniform plank of weight W^ length /, and whose thick- ness may be neglected, rests horizontally on a rough cylinder whose radius*is a. Find the weight W which can be suspended from one end without causing the plank to slide, oc being the angle of friction. ^ , _ 2aa I — 2aa 11. A hemisphere is supported by friction against a vertical wall and a horizontal plane of equal roughness. Find 6, the greatest possible inclination of the plane base to the horizon. sm fi = —. — ■ ~. 3(1 + >" ) 12. Three equal hemispheres rest with their circular bases upon a rough horizontal plane and tangent to one another. They support a smooth sphere of the same material and radius. What is the least possible value of /i ? }^ — \ 4/2. 13. Show that, on a rough inclined plane, the locus of the extremities of lines representing forces which can be applied to a heavy particle along the plane without producing or permitting motion is a circle ; and that, when motion begins to take place, it will be in a direction parallel to the corresponding radius of this circle. 14. On a rough plane inclined at the angle it was found that the least angle which a force acting along the plane and § XL] EXAMPLES. I75 sustaining a weight could make with the horizontal line in the plane was 60°. What was the coefficient of friction ? /^ = J tan b. 15. Find the greatest horizontal force along the inclined plane, when 6 <, a, which can be applied to a weight ^without produc- ing motion. W ,, . ^ . „ zix ^ -i/(sin' a — sin' 6). cos a ^ ^ ' 16. A weight ^, resting upon a rough plane inclined at an angle of 30°, is attached to a string which passes in a horizontal direction parallel to the plane over a pulley, and supports a weight \V/ ^/2 hanging freely. If W\s on the point of moving, determine the coefficient of friction, and 0, the angle between the string and the direction of motion. // = I ; sin = J 1/3. 17. A uniform rod rests wholly within a hemispherical bowl in a vertical plane through its centre, and there subtends the angle 2/?. a being the angle of friction, determine ^, the inclina- tion of the rod to the horizon in limiting equilibrium. n _ sin 2a ~ 2 cos (P -h a) cos (P — a)' 18. Two weights, P and Q, of the same material rest on a double inclined plane and are connected by a string passing over a smooth pulley at the common vertex, ^ and tp being the in- clinations of the planes, and a the angle of friction ; Q is on the point of motion down the plane. Show that the weight which may be added to P without producing motion is _ sin 2a sin {ip + 6) sin {6 — a) sin (tp — a)' 19. Denoting AG in Fig. 70 by a, and GB by ^, what is the least coefficient of friction that will allow the rod to rest in all positions ? ^ = /:- 2o.' If, in example 19, /< is the coefficient of friction between the rod and the ground, and yu' that between the rod and the 17^ FRICTIONAL RESISTANCE. [Ex. XI. wall, show that the rod will rest in all positions if }xp! is not less a thp.n -. 21. If one cord of a balanced window-sash, whose height is a and breadth b, is broken, what is the least coefficient of friction in order that the other weight may support the window ? a 22. A cubical block stands upon a rough inclined plane and is attached to a fixed point by a cord passing from the middle of the upper edge, which is horizontal, in a direction perpendic- ular to it and parallel to the plane. Determine the greatest in- clination for which the block will stand. tan ^ = i + 2/^. 23. A uniform heavy plank AB rests with the end A on a. rough horizontal plane, and a point C of its length touching a rough heavy sphere whose point of contact with the plane is Z>. Prove that the magnitude of the friction is the same at each of the points A, C and D. If the coefficient of friction is the same at each point, and is diminished until slipping takes place, show that it will occur at A or at C, according as A and D lie on the same or on opposite sides of the vertical through B. 24. A uniform beam AB of weight ^ lies horizontally upon two transverse horizontal beams at A and C; a horizontal force P at right angles with AB is then applied at B and is gradually increased until motion takes place. Putting AB = 2a and AC = b, show that, if 3*^ > 4a, slipping will take place at C when B = \iaW\ and if 3^ < 4^, slipping will take place at A when r, rrr ^ — a B = ^W -. 2a — b 25. A man, by taking 2J turns around a post with a rope and holding back with a force of 200 pounds, just keeps the rope from surging. Supposing fx = 0.168, find the tension at the other end of the rope. 2800 pounds. 26. A hawser is subjected to a stress of 1 0,000 pounds. How many turns must be taken around the bitts, in order that a man who cannot pull more than 250 pounds may keep it from surg- ing, supposing // = 0.168 ? 3^. ^XL] EXAMPLES. 1/7 27. A weight of 5 tons is to be raised from the hold of a steamer by means of a rope which takes 3^ turns around the drum of a steam-windlass. If /^ = 0'234, what force must a man exert on the other end of the rope ? 65 pounds. 28. A weight of 2000 pounds is to be lowered into the hold of a ship by means of a rope which passes over and around a spar lashed across the hatch-coamings so as to have an arc of contact of \\ circumferences. If yw ^g'V, what force must a man exert at the end of the rope to control the weight ? 164 pounds. 29. A weight is supported on an inclined plane by a cord parallel to the plane. If the cord can just sustain the weight when the plane is smooth and the inclination 45°, what is the greatest possible inclination if /^ = i ? 75°. 30. A uniform ladder weighing ico pounds, and 52 feet long, rests against a rough vertical wall and a rough horizontal plane, making an angle of 45° with each. If the coefficient of friction is at each end f, how far up the ladder can a man weighing 200 pounds ascend before the ladder begins to slip? 47 feet. 3i^l^heavy homogeneous hemisphere rests with its convex surfao^Bfekrough inclined plane. If the inclination be gradu- ally inc^^Hli, the hemisphere will roll until it either slides or tumbles Vk- ^i M = h will it tumble or slide ? CHAPTER VI. FORCES IN GENERAL. XII. Lines of Action neither Coplanar nor Parallel, 217. When the lines of action of two forces acting on a rigid body lie in a single plane, they either intersect or are parallel; and we have seen that, in either case, we can find a single force whose action is equivalent to the joint action of the two given forces. Hence, in the case of a coplanar system of forces, and also in the case of a system of parallel forces, we were able, by combining the forces two by two, to reduce the joint action of the system to that of a single force, called the resultant^ except when the final pair of forces happen to form a couple. But, when the lines of action of two given forces do not lie in one plane (that is to say, neither intersect nor are parallel), there is no single force whose action is equivalent to the joint action of the two forces. We proceed, in this section, to analyze the joint action of a system of forces in general, and shall find that the simplest mechanical equivalent of such a system consists of a force together with a couple. The Moment of a Force about any Axis. 218. In the preceding chapters, the moment of a force about an axis has been defined only in the case of an axis perpendicu- lar to (though not intersecting) the line of action. In other § XII. J MOMENT OF A FORCE ABOUT ANY AXIS. 179 words, supposing the solid upon which the forces act to be free lo turn about a fixed axis, we have considered the turning moment, or tendency to turn about the axis, produced by a force whose line of action lies in some plane perpendicular to the axis. We have now to consider the turning effect of a force whose line of action is oblique to the axis. Let CO^ Fig. 72, be the axis, and F the force acting at A in the line AB^ which does not lie in the plane MN passing through A and perpendicular to the axis. Draw AE par- allel to the axis, and let AD be the intersection of the plane EAB with the plane MN which cuts the axis in O. AD is then the projec- tion of AB upon this plane, the projecting plane BAD being perpendicular to MN. Now let P be resolved into rectangular components act- ing in the lines AE and AD. Fig. 72. The first of these components, being parallel to the axis, obviously has no tendency to turn the body about the axis. Hence the turning effect of the force F is entirely due to the component AF along AD. We therefore define the moment of F about the axis CO to be the same thing as the moment of the component AF in a plane perpendicular to the axis. Denoting this moment by H^ the inclination of the line of action to the plane just mentioned (or angle BAD in the figure) by 0, and the distance from O to the projected line of action by Uy we have AF = F cos 0, and therefore H = aF cos 0. 219. It will be noticed that and a will have the same values wherever the point of application A be taken on the line l8o FORCES IN GENERAL. [Art. 219. of action of P. The distance a may be defined as the distance between the axis and a plane parallel to it through the line of action, or as the distance between two parallel planes passing each through one of these lines and parallel to the other y or finally as the common perpendicular to (or shortest distance between) the axis and the line of action. If d denotes the angle EAB^ in Fig. 72, which is the comple- ment of 0, we have H = aF sin ^, where 6 is the inclination of the line of action to the axis (or angle between any intersecting lines parallel to them), and a the shortest distance between them. 220. The combined turning effect of two forces about a given axis is the sum or difference of the moments of the forces according as they tend to turn the body in the same or in op- posite directions. In like manner, adopting one direction of rotation as positive and the opposite as negative, it is readily seen that the joint turning effect of a system of forces, or result- ant moment of the system about a given axis, is the algebraic sum of the moments of the several forces. In the case of a system of forces in equilibrium, the resultant moment must vanish for every axis. Representation of a Couple by a Vector. 221. We have seen in Art. 97 that the moment of a couple is the same as the algebraic sum of the moments of the two forces which constitute the couple about any point whatever in the plane of the couple, that is to say, about any axis perpendicular to the plane of the couple. Such an axis is called the axis of the couple', and, since we have seen in Art. 144 that couples of the same moment in parallel planes are equivalent, it appears that the direction of the axis and the magnitude of the moment are the only essential features of a couple. Provided these are § XII.] MOMENT OF A COUPLE ABOUT ANY AXIS. I8 given, the position of the plane and that of the axis are im- material. It follows that, denoting the moment of the couple by H^ a length representing H on any given scale laid off upon an axis of tke couple 7villy by its magnitude and direction^ completely represent the couple. In doing this, one direction upon the axis will be chosen to represent a particular direction of rotation about the axis. For example, the couple in the plane MN shown in Fig. 73 is usually represented by a length AE measured from A in the plane of the couple toward O, where O is on that side of the plane from which the direction of rotation produced by the couple appears as positive or counter-clockwise. Moment of a Couple about an Oblique Axis. 222. Consider now the turning effect of the couple H in the plane J/jY about an axis oblique to its plane. Let ACy Fig. 73, be the oblique axis cutting the plane MN 2X A, and take AO, the perpendicular to the plane at this point, as the axis of the couple. Denote by the angle OAC between the axes, and put H = aP. Draw AD the projection of'^C upon the plane MN^ and AB = a per- pendicular to it in the plane MN, Then the force P act- ing at B^ parallel to AD and in the proper direction, will have a moment about AO equal to that of the given couple H\ and this force, together with P acting in the opposite direction at A^ will represent the couple. Now, since AB is the common perpendicular to AC a.r\d the line of action of P, and is the complement of their inclina- tion, the moment of P about AC is aP cos Fig. 73. 1 82 FORCES IN GENERAL. [Art. 222. (in fact P cos is the resolved part of /* in a direction perpen- dicular to AC), Now P acting at A has no moment about AC^ hence aP cos is the entire moment of the couple about AC, That is, since H = aP^ the moment of the couple H about an axis inclined to its axis at the angle (p is H cos 0. Resolved Part of a Couple. 223. The axis of the couple is sometimes called its principal axis, in contradistinction to an oblique axis about which its moment may be considered. When a line AE representing ZT, as explained in Art. 221, is measured off on the principal axis (see Fig. 73), the projection AF of this line on the oblique axis is If cos 0, which we have seen is the moment about the oblique axis. Thus lAe effective part of the couple ZTin producing rota- tion about the oblique axis is equivalent to a couple whose mo- ment is H cos 0, and is represented by the projection of the line representing H upon the new axis, exactly as the effective part of a force P m 3, given direction is represented by the resolved part of the line representing P. It should be noticed that is the angle between the direc- tions taken as positive along the two axes; if the opposite direction were regarded as positive on the oblique axis, the angle would be obtuse, and H cos 0, the resolved part of the couple, would be negative. Composition of Couples. 224. It follows from the preceding articles that, if we have any number of couples about axes in different directions, their joint moment about any axis is represented by the sum of the projections of the lines representing the couples. These lines, having direction and magnitude only, are simple vectors, and if they be added vectorially, as is done for forces in Fig. 15, p. 43, the sum of the projections will be the projection of the vectorial sum. Hence the joint effect of a system of couples in producing rotation about a given axis is the same as that of the couple represented by the vectorial sum of the vectors representing tlie ^XIL] JOINT ACTION OF A SYSTEM OF FORCES. 1 83 given couples. Since this is true for every direction of the given axis, the couple just mentioned is the exact equivalent of the system of couples and is therefore called their resultant. In the case of two couples, this agrees with what is proved in Art. 146, for the axes of the planes of the given couples and their resultant evidently make plane angles equal to the diedral angles made by the corresponding planes. Joint Action of a System of Forces. 225. We have seen in Art. 102 that, given a force P and a selected point A not in the line of action, we may, by assuming two equal and opposite forces acting in a parallel line at A (see Fig. 36), replace the force P by an equal parallel force at A together with a couple. The plane of this couple is that con- taining A and the line of action of -P, and its moment is the moment of P about A. Suppose now that this is done for each of the forces of a given system. We shall then have replaced the whole system of forces by a system of equal and parallel forces acting at A^ together with a system of couples in different planes. The forces at A may be combined, by vectorial addition, into a result- ant force acting at A^ and the couples may, in accordance with the preceding article, be combined in like manner into a single resultant couple K. Thus the whole system of forces has been replaced by a force R acting at a selected point A together with a couple K, The plane of the couple K will not in general be parallel to the line of action of R^ so that the force and couple cannot be replaced, as in Art. loi, by a single force. 226. This combination of a force and a couple which cannot be reduced to a single force is called a dy?iame. We have thus found that the resultant of a system of forces is, in general, not a single force, but a dyname. Since the magnitude and direction of R are vectorially deter- mined from the given forces, they are independent of the posi- tion of the selected point A^ so that the resultant force-vector is 1 84 FORCES IN GENERAL. [Art. 226. constant. But the moment of the couple K and the direction of its axis depend upon the position of A. For, if A be moved to a point B not on the line of action of R, the effect is to combine with R acting at A a. couple (as in Art. loi); then the reverse couple, which must be combined with X, will make an alteration in the value of R'. The dyname consisting of R acting at A and the couple R may be denoted by (R ^ , A'). The Principal Moment of a System at a Point. 227. Since the given system of forces is equivalent to the dyname (Ra, R^), t^e moment of the system about any axis pass- ing through A is the moment of X about that axis, because R acting at A has no moment about any axis passing through A. It fol- lows that R^ is the greatest value of the moment of the system about any axis passing through A. It is therefore called the principal moment of the system at^, and its axis, or principal axis (see Art. 223), is called the principal axis of mo7ne?it at A. It follows that the moment of the system about an axis making the angle with the principal axis is K cos 0, exactly as in the case of a single force or of a couple. Poinsot's Central Axis. 228. In Fig. 74, let AB represent the resultant force R of the system acting at A^ and let AC represent, as in Art. 221, the axis and magnitude of the resultant couple K, Draw AE perpendicular to AB in the plane BAC, and let the couple K be resolved into rectangular components whose axes are AD, in the direction of the line of action of -/?, and AE perpen- dicular to it. Denote by G the first of these couples, and by tp the angle BAC, Then, by Art. 223, Fig. 74. G — R cos if), . . . (i) Draw ^4(9 perpendicular to the plane BAC; then the plane of § XII.] POINSOr-S CENTRAL AXIS. 185 the other component couple whose axis is AE is the plane -^^(^, containing the line AB. This couple, whose value is K sin ^, can therefore be combined with the force R by the method of Art. 10 1. For this purpose determine a so that aR = K.sxxi tpy (2) and lay o^ AO =^ a on that side of A which makes the moment of R acting at O about the axis AE agree in direction with the couple K sin tp. Then R acting at O is equivalent to R acting at A together with the couple AE. Hence the whole system of forces is equivalent to the dyname consisting of the forc^ R act- ing at O and the couple G whose axis is in the directio7i of the line of action of R. 229. The line of action of R when the system is thus reduced to the dyname (i?, (?), in which the line of action is also the axis of the couple, is known as Poinsofs Central Axis j and the dyname of this character has been called a wrench. The central axis of a system of forces has a definite position independent of the position of the initial point A. For, suppose it possible that the system could be reduced to another wrench {R\ G')y where i?' has a different line of action from R. Since its direction is the same as that of R, it has a parallel line of action, and the axis of G' is the same as that of G, that is, G and G' are couples in the same plane. Now by combining 7(" and G^ each reversed with the system {R, G), we shall have a system in equilibrium. Therefore the couple formed by R and R' reversed is in equilibrium with the couple G — G'; but this is impossible unless they both vanish, because these couples are in different planes. Hence R and R' act in the same line, and G' = G. 230. Equation (i). Art. 228, shows that G is less than any other value of JC ; so that the central axis is the locus of the points for which the principal moment is a minimum. Supposing R, G and the central axis to be known, to deter- mine the principal moment R and the direction of the principal 1 86 FORCES IN GENERAL. [Art. 230. axis at any point A, let a be the distance of A from the central axis; then from equations (i)and (2) we derive K= ^{a'R''-\-G\ (3) and , aR "" tan ^ = -— - (4) 231. If, for a given system of forces, we find the force-vector R reduces to zero, the dyname reduces to the couple K^ which in this case will be independent of the position of A. If, on the other hand, we find K ^ o, the dyname reduces to a single force R acting at A. But the general condition that the dyname, or resultant of the system of forces, should reduce to a single force is that G shall vanish. This occurs, according to equation (i), Art. 228, not only when ^ = o, but when ^ = 90°; in other words, when for any selected point A the axis of the couple K is perpendicular to the line of action of R. The pro- * Referring to Fig. 74, it follows that, for any point A on the cylin- drical surface whose axis is the central axis an4 whose radius is a, the axis of the couple K or principal axis is tangent to a spiral described on the surface, making with the elements the constant angle ^ deter- mined by equation (4). The portion of an element intercepted between two whorls of this spiral is 27ta cot tp = —y^, A which is independent of a ; therefore every such spiral is the inter- section of a cylinder with a helical or screw surface whose pitch is 27r6' §XII.] FORCES REFERRED TO RECTANGULAR AXES. 1 87 cess in Art. 228 then gives the line of action of the single force which is the resultant of the system. -a? Forces Referred to Three Rectangular Axes. 232. In referring a system of forces to three rectangular axes, we shall take them in such a manner that positive rotation about the axis of z (that is, positive rotation in the plane of xy as viewed from the side on "U which z is positive) shall be rotation from the positive direc- tion of the axis of x to that of the axis of _y, as in Fig. 75. It follows that positive rotation about the axis of x is rotation from y to ^, and that about the axis of ^ is rotation from z to x.^ Let {x,y^ z) be the point of application, and X, F, Z the • ^^^- 75- resolved parts, in the direction of the axes, of a force P. The moment of P about the axis of x is the algebraic sum of the moments of the resolved parts Y and Z, since X which is parallel to the axis of x^ has no moment about it. The moment of Z about the axis of x^ to which it is perpendicular, isj^Z, since 7 is the common perpendicular to the axis and the line of action. This moment is positive, because, when y and Z are positive as in the figure, Z tends to turn the ordinate y toward the positive direction of the axis of z. In like manner, the moment of Y about the axis of :r is sF, but this moment is found to be nega- tive. Hence, denoting the moment of P about the axis of x by Z, we have L^yZ- zY. (i) * The diagrams being drawn as if the observer were situated in the first octant, the letters x, y, z appear to follow one another in positive rotation about the origin. 1 88 FORCES IN GENERAL, [Art. 232. Similarly, denoting the moments about the axis of y and z by M and N respectively, we have M=zX- xZ, (2) N=xY-yX. (3) 233. The six quantities X^ Y, Z , L ^ M and N may be taken as the determining elements or coordinates of a given force, and each of these quantities has a definite value for a given force; but it can be shown that they are not six independent elements. For, suppose these six quantities to be given in equations (i), (2) and (3); if x,y and z (regarded now as unknown quanti- ties) admit of any actual values, we shall have, by multiplying the equations by X^ V and Z respectively and adding, ZX-\-MV-}-JVZ=o (4) This is therefore a necessary relation which must exist be- tween the six elements of a force. If it does not hold true, it is impossible to find values of x, y and z, the equations being, in that case, inconsistent. But, if it does hold true, the equations will not determine definite values of x^y and z\ they are, in that case, the equations of three planes which intersect in one line, and this line is the line of action of the force. Thus, as we should expect, the point of application is not determined, but only the line of action. Any two of the equations (i), (2) and (3) may be taken as the equations of the line of action. Six Independent Elements of a System of Forces. 234. Thfe advantage of employing the six elements A", K, Z, Z, J/", N arises from the fact that the joint effect of a system of forces is found by simply adding the like elements of the several forces. Thus, in the case of a system of forces P^^ . , . P^^ put x' = :^x, v' = :sv, z' = :sz, ) [ . . (0 z' = :sz, M' = :em, n' = :sjv; ) §XII.] SIX INDEPENDENT ELEMENTS. 1 89 then X' , V\ Z', Z', M\ N' constitute the like elements of the total resultant of the system. Moreover, these are six independ- ent elements which may have any values whatever, not generally satisfying a relation like equation (4) of the preceding article. It is only when they happen to satisfy such an equation that there exists a single force equivalent to the system. 235. In the general case, suppose the origin to be taken as the selected point of Art. 225; then we have seen that the system is equivalent to a force R acting at the origin together with a couple K. Since ^ at the origin has no moment about either axis, Z, M and N are the moments of K about the three axes respectively. Therefore, by Art. 223, they are the resolved parts of the couple K about the axes, and the axial representations of them are the resolved parts or projections of the vector K, just as Xy Y and Z are the projections of the vector -^. Let «', p, y be the direction angles of R^ so that i?=|/(Ar"+F" + Z"), _(2) and X' Y' Z' cos 0^= -^y cos /? = -^, cos K = ^- • (s) These equations determine the magnitude and direction of R, Similarly if A, /i, r are the direction angles of K^ we have X=y(Z" + ^" + ^"), (4) , Z' M' JV' , ^ cos ^= ^y cos /^ = — , cos ^=-Y* • (5) which determine the magnitude and direction of K. 236. To find the value of G, the minimum couple, which is associated with R acting in the central axis, let tp denote the angle between the directions of R and ZT, as in Fig. 74; then G, the projection of K upon R, is the sum of the projections of Z', M' and N' (compare Art. 6\). Hence, G •=^ K cos tp = V cos a -\- M' cos /? -{- N' cos y\ 19C> FORCES IN GENERAL. [Art. 23! or, substituting the values in equations (3) and (2), ^ VX' ^ M'Y' -\- N'Z' 4/(^'"^4- Y" ■\- Z") (6) The condition that the resultant of the system may be a single force is that G shall vanish; hence it is rX' + M'Y' + N'Z' = o, which agrees with the result found in Arts. 233 and 234. Con- versely, if this condition is satisfied, the resultant must be either a single force or a couple. 237- To determine the central axis, we observe that, because J? in the central axis and the couple G are together equivalent to the given system of forces, the sum of their moments about the axis of X must be Z'. The moment of G about that axis is G cos a; therefore that of i? in the central axis is V — G cos o', and, in like manner, the moments about the axes of y and z are M'—Gcos/3 and N' — G cosy. Hence, by Art. 233, any two of the equations yZ' - zY' = Z' - 6^ cos a, zX' - xZ' ^ M'- G cos /?, xY' - yX' = N'- GcosyJ (7) determine the line of action of this force; that is to say, these are the equations of the central axis. §XII.] CONDITIONS OF EQUILIBRIUM. IQI Conditions of Equilibrium. 238. The system of forces is in equilibrium when the six elements of the resultant X' , Y\ Z\ L\ M' ,N' all vanish; that is, when ^X = o, 21^ = o, 2'Z = o, JS'Z = o, :SM = o, 2JV = o. Thus, when the forces of a system in equilibrium are unre- stricted, there are six independent conditions of equilibrium which must be fulfilled; and from these it is possible to deter- mine six, and not more than six, unknown quantities. The equations above are the simplest form of the conditions of equi- librium when the forces are referred to coordinate axes; but a condition of equilibrium can, of course, be found by resolving forces in any direction, or by taking moments about any axis. The conditions obtained by resolving forces are precisely the same as if the forces all acted at a single point; hence, as in Art. 74, only three independent conditions can be found in this way, and in order to be independent the three directions of resolving must not lie in one plane. It follows that three at least of the six independent conditions must be derived by taking moments.* 239. It is possible, however, to obtain all the conditions of equilibrium by taking moments about different axes; but, in order that they should be independent, there are some restric- tions upon the choice of these axes. For example, if two of the axes intersect in a point A, the vanishing of the moment about a third axis passing through A and in the plane of the two axes will nof give an independent condition. If the third axis passes through A but is not coplanar with the other two, it gives an independent condition. Again, in this last case, the moment *In like manner we have seen in Art. 109 that at least one of the three conditions, when the forces are restricted to a given plane, must be derived from the principle of moments. 192 FORCES IN GENERAL. [Art. 239. necessarily vanishes about any other axis passing through A^ so that a fourth axis passing through A would not give an independ- ent condition. Equilibrium of Constrained Bodies. 240. Suppose a rigid body to have its possible motions limited or constrained by means of fixed bodies with which it is in con- tact. This may be done, for example, by having one or more of its points fixed or confined to fixed surfaces or lines, or by having its surface in contact with a fixed surface. If such a body be acted upon by external forces, it will in general move subject to the constraints; and, if it is at rest, the external forces together with the resistances of the fixed bodies must form a system of forces in equilibrium. Let n denote the smallest number of numerical elements which will serve to determine the unknown resistances or forces producing the constraint. Since the whole number of unknown quantities is six, n must be less than six; therefore, if these // unknown quantities were eliminated from the six equations of equilibrium, there would remain 6 — n equations independent of the forces of constraint^ which are therefore conditions imposed upon the external forces in order that equilibrium may exist. These equations, whether found by elimination or directly by a method which will be given in the following section, are called the condi- tions of equilibriu7n for the constrained body. 241. If we put 6 — « = w, the body thus constrained in its motion is said to be subject to n degrees of constraint, and to possess m degrees of freedom. Thus, the perfectly free rigid body has 6 degrees of freedom. The constrained body with ni degrees of freedom requires the knowledge of the values of tn numerical determining quantities or elements to fix its position ; and, if the external forces are either given quantities or known functions of these m elements, the latter will be the unknown quantities to be determined by means of the m conditions of equilibrium. In the case of a ix^^ particle, there are but three degrees of freedom and accordingly three determining elements or coordi- §XII.] EQUILIBRIUM OF CONSTRAINED BODIES. I93 nates fix its position. But, in the case of a rigid body, after a point A of the body is fixed, the body still has three degrees of freedom. This may be clearly seen as follows : When A is fixed, a point B of the body at a given distance from A is thereby re- stricted to the surface of a given sphere. Two determining ele- ments or coordinates are therefore necessary to determine the position of B, But, after B is fixed ^s well as A^ the body is still free to turn about the axis AB. A point C, not in the line AB^ is now restricted to a given circle ; and therefore one more de- termining element will fix it, and thus completely determine the position of the rigid body. 242. It will be noticed that, in the illustration above, the fixing of the point A is equivalent to three degrees of constraint. The body retains three degrees of freedom, and three equations of equilibrium are necessary. The remaining three of the six conditions of equilibrium of the general case would serve to de- termine the resistance at A^ which, being unknown in magni- tude and direction, involves three unknown elements. Again, when two points are fixed, so that the body rotates about a fixed axis, the body retains but one degree of freedom, and but one condition of equilibrium is necessary. The remain- ing five conditions of the general case would serve to determine the reactions of the axis. 243. As a further illustration, if three points (not in a straight line) of a rigid body are constrained to remain in a given plane, we shall have n = 3, because (see Art. 240) three unknown quan- tities, namely, the values of the normal resistances or reactions of the plane at these three points, are sufficient to determine a set of forces capable of producing the constraint. Therefore the body will be subject to three degrees of constraint. Hence we have also m ^=- t^: the body has three degrees of freedom, and three conditions of equilibrium are required. The case is that of a body capable of plane motion only, just as if it were a lamina subject only to forces acting in its plane. If this plane is taken as that of xy^ the conditions of equilib- rium, in the standard form of Art. 238, reduce to three, namely, 194 FORCES IN- GENERAL. [Art. 243. ^X = o, -S'y = o and ^iV= o, which are independent of forces in the direction of the axis of z. These are identical with the conditions given in Art. 100 for coplanar forces. EXAMPLES. XII. 1. When a force is represented by a line AB^ show that its moment about any axis through O is represented by double the projection of the area OAB on a plane perpendicular to the axis; also that, when a couple is represented by an area, the resolved part of the couple in any plane is represented by the projection of the area. 2. Show that four forces acting in the sides of a quadrilateral which is not plane, and represented by them taken in one con- tinuous direction about the perimeter, are equivalent to a couple in a plane parallel to the two diagonals and represented by double the area enclosed by the projections of the sides on this plane. 3. Show directly that the forces constituting a couple in a plane and those constituting the reverse couple in a parallel plane are in equilibrium. 4. Show that lines laid off from O representing the moment of a force, or of a system of forces, about different axes passing through O form chords of a sphere which passes through O and of which the diameter represents the principal moment. 5. If P be the value of each of two equal forces, 2a the short- est distance between the lines of action, and 2a the angle between their inclinations, show that the central axis bisects the distance and the angle, and determine R and G. R — 2P cos a\ G =^ 2aP sin a. 6. Prove that, if the moment of a system of forces about each side of a triangle vanishes, the resultant is either a force or a couple in the plane of the triangle. 7. Six equal forces act in consecutive directions along those edges of a cube which do not meet a given diagonal. Find their resultant. The couple 2Pa\^ 3. §XII.] EXAMPLES. 195 8. A force 3 acts parallel to the axis of z at the point (4, 3), and a furce 4 acts in the negative direction in the axis of x. Determine the central axis and the values of R and G. 4^ + Sx= 12; ) ^' 5 9. OABC is a tetrahedron, of which the edges meeting at O are mutually at right angles. Forces are represented in magni- tude, direction and line of action by OA, OB^ OC, AB, BC, CA, Taking the first three edges as axes, and equal a, b, c, show that the resultant is a force represented by the line joining the origin with the point («, by ^), and the couple ^{b'^c^ + ^^<^ + ^'^O in the plane ABC. Determine also the value of G. G= 3^^^ 10. If P and Q are two forces whose directions are at right angles, show that the central axis divides the distance a between their lines of action inversely in the ratio F"^ : Q^ and that 11. An upper half port, whose weight, 48 pounds, acts at its middle point, is $6 inches long and 15 inches broad, and is held in a horizontal position by a laniard, the single part of which passes through a hole in the bulwark 8 inches above the hinges. Find the tension on the bridle, which is 39 inches long and secured to the corners of the port, and the total action on each hinge. - 66.3 lbs.; 25.5 lbs. 12. A pair of "sheer legs" is formed of two equal spars lashed together at the tops, so as to form an inverted V. They stand with their " heels " 20 feet apart on the ground, and would be 40 feet high if vertical. They are supported in a position 12 feet out of the vertical by a guy made fast to a point in the ground 60 feet to the rear. Find the tension on the guy and the thrust on each leg when lifting a 30-ton gun. T = 12.8 tons, B = 19.5 tons. 13. The legs of a pair of sheers are at an angle of 60° with each other, and the plane of the sheers is inclined 60° to the 196 FORCES IN GENERAL. [Ex. XT I. horizontal; the supporting guy is inclined 30° to the horizontal. Find the thrust on each leg when a weight of 20 tons is lifted. 20 tons. 14. The line of hinges of a door is inclined at an angle oc to the vertical. Show that the couple necessary to keep it in a posi- tion inclined at an angle ^ to that of equilibrium is proportional to sin OL sin /?. 15. A load of 10 tons is suspended from a tripod whose legs are inclined 60° to the horizontal. A horizontal force of 7 tons is applied at the top in such a manner as to produce the greatest possible thrust in one leg. Find in tons that thrust and the stress on each of the other legs. 13- 18; — 0.82. 16. A square is formed of uniform rods of length a and weight W^ freely joined together. One rod being fixed in a horizontal position, find the couple required to turn the opposite rod through the horizontal angle B. aW ?\.vi \Q. 17. Each of two strings of the same length has one end fastened to each of two points, whose distance is horizontal and equal to a. A smooth sphere of radius r and weight W is sup- ported upon them, the plane of each string making the angle a with the vertical. Find the tension of either string. Wa Sr cos a 18. A rod of length a can turn about one end in a horizontal plane. A string tied to the other end passes over a smooth peg at a distance /^ vertically above it, and is then attached to a given weight. The rod is then turned through an angle 6, and is kept in position by a horizontal force J^, applied at the end of the rod perpendicularly to it. Prove that i^ is a maximum when tan — 2 ~ ^' + 4a" CHAPTER VII. THE PRINCIPLE OF WORK. XIII. Work done by or against a Force. 244. When the point of application of a constant force P is displaced in the direction of the force through a space J, the force is said to do worky and the product Ps is taken as the measure of the work done. When the displacement is in the di- rection opposite to that of the force, work is said to be done against the force. The unit of work is a compound unit involv- ing the unit of space or length and the unit of force ; thus the ordinary unit of work is the foot-pound^ which may be defined as the work done by the gravity of a pound descending through one foot, or the work done against gravity in lifting one pound through the space of one foot. That part of Mechanical Science which deals with forces as overcoming resistances through definite spaces is known as Dynamics^ in distinction from Statics, in which the points of ap- plication of the forces are regarded as fixed. Compare Art. 49. 245. If the displacement of the point of application takes place in aline oblique to the line of action of the force, while the force remains constant in direction as well as in magnitude, the work done by the force is defined as the product of the force and the projection of the displacement upon the line of action. Thus, in Fig. 76, let AP = P represent the force, and let AB — i be the displacement, making the angle with the direction of P. 19^ THE PRINCIPLE OF WORK. [Art. 245. Denote the projection AC of s upon the line of action by /. Then Pp is the work done by the force during the displacement; and, since/=i- cos 0, we may also take Ps cos (fy as the expression for the work done. It will be noticed that, when the angle is obtuse, p has a direction opposite to that of the force, so that work is done against the force. In this case, the expression for the work becomes negative. Thus, work done against a force is regarded as negative. 246. The work done by a force in a given displacement oblique to the line of action is the same thing as the work done by the resolved part of the force in the direction of the dis- placement ; for, in Fig. 76, this resolved part is AD = P cos 0; and, multiplying this by the displacement s, we have the work of the resolved part equal to Ps cos 0. Thus the "resolved pait" of the force (see Art. 56) is the only " effective " part of the force in respect to work done. It will be noticed that the ex- pression for the work done by the other component of the force in this case vanishes, because the angle between that compo- nent and the displacement is a right angle. Work done by the Components of a Force. 247. Let P be the resultant of any number /*,,/',... /'^ of forces acting at a single point of application which undergoes the displacement s in any direction. Then, denoting the incli- nations of P, P^j P, . , . P^ to the direction of s by 0, 0, , 0, , • • . 0„» we have, by resolving the forces in the direction of displacement, i? cos = /*, cos 01 + P.^ cos 0, -h . . . + ^„ cos 0^. (l) § XIII.] WORK OF THE COMPONENTS OF A FORCE. 199 Multiplying by j, we have Rs cos

,,/, . . ./ are, as in Art. 245, the projected displace- ments taken in the directions oiP,P^^P^.,.P^. 248. Accordingly, when several forces act upon a particle, the total work of the forces is the algebraic sum of the works of the several forces, each reckoned independently of the existence of the others. Thus, if a weight W, Fig. 77, be displaced through the space i- up a plane inclined at the angle <^, the total work is the sum of those of the several forces ; namely, the weight W acting vertically, the normal resistance P and the frictional resistance F (if the plane is rough). If h is the vertical height through which the body is raised (so that h =^ s sm 6), — Wh is the work of Wy which is negative because h is measured upward or against the force. No work is done either by or against the normal resistance P. The work of the frictional resistance Fis — /\y, because work is done against friction. Thus the total work is — (Wh -\- Fs) ; that C b Fig. 77. is, the work Wh -\- Fs must be done against the forces in pro- ducing this displacement. If the body were displaced down the plane, the work would be Wh — Fs ; work would now be done by gravity, but against friction, because the latter force would now act up the plane. 200 THE PRINCIPLE OF WORK. [Art. 249. Virtual Work of a Variable Force. 249. When the force P is variable either in magnitude or direction, if ds is an element of displacement in the direction of the force, Pds is the corresponding element of work. Again, if the element of displacement makes the angle with the direc- tion of /*, the element of work is Pds cos 0. In either case, the total work is an integral of this element. A small displacement denoted by ^s and treated as an ele- ment is sometimes called a virtual displacement^ and the corre- sponding expression, PSs cos 0, is called the virtual work of P in this displacement. Thus, in the expression for virtual work, the magnitude of the force and the direction, both of the force and of the displacement, are regarded as constant. It follows (see Art. 247) that, for any forces acting at a single point of application, the virtual work of the resultant is equal to the algebraic sum of the virtual works of the components. In particular, // the forces are in equilibrium, the resultant vanishes, and therefore the algebraic sum of the works of the forces in any virtual displacement is zero. The Principle of Virtual Work. 250. The principle stated above, for the case of forces in equilibrium acting at a single point, is called the principle of vir- tual work. The conditions of equilibrium obtained in this case are identical with those obtained from the resolution of forces; for the work in any virtual displacement is merely the product of Ss and the sum of the resolved parts of the forces in the direc- tion of displacement. But we have seen (Art. 117) that a solid body acted upon by forces having several points of application may be regarded as a §XIII.] WORK DONE BY INTERNAL FORCES. 20 1 system of interacting particles, each of which is in equilibrium under the action of certain forces. These forces taken together for all the points constitute the external and internal forces. Now, when the solid begins to move in any way, the several points of application will have certain initial velocities, which are sometimes called their virtual velocities. If we determine these virtual velocities, we can write the expression for the rate at which work begins to be done in the displacement ; and, equating this to zero, we have a method of obtaining conditions of equilibrium distinct from the methods of resolving forces and of taking moments. B' Work done by Internal Forces. 251. The internal forces mentioned above are stresses between pairs of particles between which a mutual action exists. Let A and B^ Fig. 78, be two such points between which a stress P exists, and let P tend to increase the distance AB. Assume rectangular coordi- dates such that the axis of x is parallel to AB^ and let x^ and x^ be the abscissas of A — and B, so that AB = x^ — x^. Then the stress P acts paral- lel to the axis of x in the positive direction at B^ and in the negative direction at A. Sup- pose now a displacement takes place in which AB assumes the position A' B' , The virtual velocities of A and B are the pro- jected velocities of these points at the beginning of the motion namely, dx^ ~di' Fig. 78. dx^ -H and dt dx^ Hence the rate at which P does work at ^ is ^ -^^ and that at at 202 THE PRINCIPLE OF WORK. [Art. 251. dx which work is done against P dX Ax^ P -r^. Thus the total work- rate of the two phases of the stress is that is to say, it is the product of the stress P and the rate of change in the length of AB. Denoting this length by p, the virtual work is Pdp. This expression shows that the internal work of a sys- tem of interacting particles depends solely upon their relative motions^ and not upon their absolute motions. 252. In the case of a rigid body the distance between any two points is invariable. Therefore the work of the stresses between the parts of a rigid body vanishes, so that the internal forces do not appear in the expression for the virtual work in any displacement . of the solid. Thus, in the case of any forces acting in equilibrium upon a rigid body (but not at a single point of application), the algebraic sum of the virtual works of the forces in any displacement is zero. Virtual Work in Constrained Motion. 253. When a solid is free, a condition of equilibrium might be obtained from any virtual displacement which is possible to the body as a whole. The simplest displacements are translations in fixed directions, in which every point of application of an external force has the same displacement Ss^ and the condition of equilibrium is (as we have seen in Art. 250, in the case of a single point of application,) identical with that obtained by the resolution of forces. 254. The displacements next in point of simplicity are rota- tions about fixed axes. If we draw a perpendicular AR from the point of application A of a. force to the axis of rotation, and resolve the force into three rectangular components, two of which are paralled to the axis and to ^^i? respectively, it is evi- dent that only the third component, which we may denote by P, § XIII.] VIRTUAL WORK FOR A DISPLACED SOLID, 203 does work, when the body undergoes a virtual angular displace- ment dS about the axis. Moreover, the virtual displacement of the point A is AR^B, hence the virtual work of the force is F.ARdB. Now, by Art. 218, Z'. AR is the moment of the given force about the axis, since F is the only component which has a moment about the axis and AR is its arm. Therefore, in rotation, the work done by a force is the product of the moment of the force and the angular displacement. Now, in the expression for the total work, the angular displacement dd will occur as a common factor; hence the total virtual work in the case of a system of forces is the product of the resultant moment of the system and the virtual angular displacement. It follows that the condition of equilibrium obtained by the principle of virtual work, in the case of rotation, is identical with that obtained by taking moments about the axis. 255. But, when the body is constrained in its possible motions, we can, by considering only such displacements as are possible under the constraint, obtain equations which are free from the forces of constraint. For, at each point of contact of the body with fixed bodies, the reaction which constitutes the constraining force is perpendicular to the displacement at that point; that is, it does no work and therefore does not appear in the equation.* It is in this way that we can, as mentioned in Art. 240, obtain directly the conditions of equilibrium for constrained bodies. Expression of the Total Virtual Work in the Displacement of a Solid. 256. As stated in Art. 241, the number of these equations is w, the number of the body's degrees of freedom; and, when the question is one of finding the position of equilibrium, the un- known quantities are in fact the m numerical elements determin- * The constraints are here supposed smooth, for if there were fric- tional resistances, as in the illustration of Art. 248. their limiting values would involve the values of the normal resistances. 204 THE PRINCIPLE OF WORK, [Art. 256. ing the position, sometimes called the coordinates of position of the constrained body. In any displacement of the solid, it is necessary, in order to obtain an expression for the work done, to express the linear virtual displacements of the points of application of the several forces in terms of these coordinates of position. 257* For example, the end A o( a. uniform rod of length 2a, and weight IV, Fig. 79, is con- strained to move in the horizon- tal line AC, while the end B is constrained to move in a vertical line intersecting AC in C; the constraints being smooth. By means of a string attached to A and passing over a pulley at C, the weight of a given body P acts at A in the direction AC; to find the position of equilib- FiG. 79- rium. The rod has, in this case, but one degree of freedom, and accordingly its position is fixed by a single coordinate. Taking ^, the inclination to the horizontal, for this coordinate, and denoting AC by x, we have 2b cos 0. (i) The linear virtual displacement of the point of application A in the direction of the force /'is — dx, because I* acts in the direc- tion of X. decreasing. Hence the virtual work o( P is — Pdx. Differentiating equation (i), we have dx 2b sin d dd therefore the virtual work of P, in the displacement of the solid indicated by dO, is 2bP sin 6 dd. g XIIL] STABILITY OF EQUILIBRIUM. 205 In like manner, denoting by ^ the height of the point of appli- cation of j^ above AC^ the work of W '\% — Wdh^ and h = bsinO, dh = b cos 6 dO, Therefore the virtual work of ^ is — bW cos Odd^ and the total virtual work is (2bF sine - blV cos B)dd. Putting this equal to zero, we obtain for the position of equi- librium Stability of Equilibrium. 258. When, as in the preceding example, the solid has but one degree of freedom, any point of it during its possible motion is describing a definite path. In a position which is not one of equilibrium, the point tends to move in one of the two opposite directions along this path. By applying at the point an additional force of proper magnitude in the direction opposite to that in which the point tends to move, the motion may be prevented, and the body put in equilibrium. If now the body be displaced in the direction of the previous tendency to motion, the whole vir- tual work of the forces (including the additional force) is zero ; but the work of the additional force is negative, therefore the total work of the original forces is positive. It follows that, when a solid has one degree of freedom, // tends to move in such a man- ner that the virtual work of the forces is positive, 259* The principle proved above enables us to give a more complete discussion of the stability of equilibrium considered in Art. 188. The rate at which the displacement or motion takes place is measured by the rate of change in the coordinate of position ^, dQ that is, -J- ; and, if we divide the expression for the virtual work dt 2o6 THE PRINCIPLE OF WORK. [Art. 259. by dty we shall have the rate at which work is done by the forces, or the work-rate of the forces. The expression for" the work-rate will then be of the form f^and the side AB, If there are no external forces acting, the configuration of equilibrium (but not the absolute position of the system) may be determined by an F^g. 80. equation of work involving internal forces which are capable of doing work. Thus, suppose the points A and C to be connected by an elastic string of natural length / and strength K\ that is to say, such that a force K will produce an elongation of one unit of length in excess of its natural length /. By Hooke's Law the tension of such a string is proportional to its extension in length; therefore, when the string has the lengthy its tension is P = K{y — /). Again, if B and D are connected by an elastic string of natural length /' and the same strength K^ the tension of BD when its length is ;s: is ^ = K{z—l'). Now, in any dis- placement the total virtual work of the forces is — Pdy —Qdz, or - K\{y - l)dy + (^ - ndz\ Denoting the length of each bar by tf, the values ofj^and^ in terms of d are y = 2a sin B^ 2a cos when< dy = 2a cos B dS^ dz =■ ^ 2a sin B dB. Substituting, and equating the virtual work to zero, 210 THE PRINCIPLE OF WORK, [Art. 264. 2a COS s\n 6 — I cos 6— 2a sin 6 cos 6 -\- I' %m(f ^= o: from which we have Xditi 6 z= - to determine the configuration of equilibrium. It is readily seen that the configuration is one of stable equilibrium. EXAMPLES. XIII. 1. A continuous source of energy which can do 33,000 foot- pounds of work in one minute, or 550 in one second, is said to have one horse-power. What is the horse-power developed by a locomotive which keeps a train weighing 50 tons moving uni- formly at the rate of 30 miles an hour on a level track, the resist- ance being 16 pounds per ton ? 64. 2. What should be the horse-power of a locomotive to move a train of 60 tons at the rate of 20 miles an hour up an incline of i foot in 100, the resistance from friction being 12 pounds per ton ? 110.08. 3. Assuming a cubic foot of water to weigh 62^ pounds, how many cubic feet will an engine of 25 horse-power raise per minute from a depth of 600 feet ? 22. 4. A train, whose weight is 100 tons, is ascending uniformly a grade of i in 150, and the resistance from friction is 12 pounds per ton. If the locomotive is developing 200 horse-power, what is the rate in miles per hour? 27.85. 5. Determine the value of B in the problem of Art. 81 by the principle of virtual work, and show that the position is one of unstable equilibrium. What may be inferred with respect to the locus of the centre of gravity of the weights F and Q ? 6. Solve Ex. VII, 2, by the principle of virtual work, and show that the position is one of unstable equilibrium. 7. The ends A^ B oi a uniform heavy rod lie in a smooth ellipse whose major axis is vertical ; referring the ellipse to the focus and directrix, prove that the rod is in equilibrium if it § XIV.] EXAMPLES. 211 passes through the lower focus. Show geometrically that this is a position of stable equilibrium. 8. Solve the problem of Art. 114 by virtual work, and prove that the equilibrium is unstable. What may be inferred with respect to the locus of the point Ci XIV. Total Work of a Force in an Actual Displacement. 265. When the displacement is in the direction of the force, the element of work is Pds\ and, if this direction remains con- stant while the force P is variable, the whole work done in a dis- placement is the integral \pds taken between proper limits, where P is supposed to be ex- pressed as a function of s. As an illustration, consider the force of an elastic string or wire AB which is stretched beyond its natural length /to a length / + J,- Suppose the end A to be fixed and the force which pro- duces the extension s in the length to act at B in the direction AB, By Hooke's Law, the tension of the stretched string, when it has any extension s beyond its natural length, is directly pro- portional to the fxtension, and may be denoted, as in Art. 264, by Ks, where -^ is a constant which is called the strength of the spring, because it is the value of the force when the extension is unity. Treating i' as a variable, P = Ks is the expression for the variable force producing the extension in terms of j, which is measured from B, so that the lower limit of the integral express- ing the work is zero. Thus the total work done in producing the extension s^ is Pds = k\ sds = ^Ks*. THE PRINCIPLE OF WORK. [Art. 265. Since Ks^ is the final value of the variable force, it appears that, when Hooke's Law applies, the work done is one-half as much as would have been done had the force been constant and equal to its final or greatest value. Graphical Representation of Work. 266. If, at all points of the line in which the point of appli- cation of a variable force travels, perpendiculars be erected, representing on some selected scale the corresponding values of the force, the curve marked out by their extremities will give a graphic representation of the mode in which the force varies with the space. This is, for example, mechanically done in the formation of the "indicator diagram " of a steam-engine. Thus, in Fig. 81, suppose the horizontal motion of the pencil describing the curve A^A^ to represent (generally on a re- duced scale) the motion of the piston, while the distance AB of the pencil from the hori- zontal line Ox is by proper mechanism caused to be always pro- portional to the force acting on the piston. Taking as origin the point from which s is measured, we thus have a curve in which the abscissa and ordinate of any point represent corresponding values of s and P. 267. Let A^A^, Fig. 81, be any curve of force thus con- structed: the integral I Pds, which represents the work done JSi when the point of application passes over the space B^B^ = J, — J,, is also the value of the area A^A^B^B^ inclosed between the curve, the axis of abscissas and the ordinutes corresponding to the limits. Hence, by the construction of the curve of force, we are able to represent the work done by an area. When the §XIV.] GRAPHICAL REPRESENTATION OF WORK. 21 3 A' curve is mechanically constructed as supposed in Art. 266, the area is measured either by one of the approximate methods or by the Planimeter. 268. The graphic representation of force by an area is some- times useful when the law of the force is known. In this case, the curve of force will be a known curve, and the areas representing the work may be obtained directly from geometrical principles. For example, in the illustration of Art. 265 let AB, Fig. 82, be the nat- ural length of the string, and BC = s^ the final extension. Hooke's Law, B = Xs, gives for the curve of force the straight line B£> passing through B, the origin from which s is measured. The value of CD is the final value of the force, Ks^, and the area of the triangle BCD^ which represents the work done, is one-half the product of the base and altitude, that is, \K5^. In like manner the work done in any displacement not starting from the point of zero force B would be represented by a trapezoid, and its value found from the known expression for the trapezoid. Fig. 82. of area Work done when the Path of Displacement is Oblique or Curved. 269. When the elementary displacement ds makes the angle with the direction of the force, the element of work is Pds cos 0. In this expression, may be variable either on account of a change in the direction of the force or of that of the path de- scribed by the point of application. The total work is now the value between proper limits of the integral I P cos (t>dsy in which P^ and s are supposed to be expressed in terms of a single variable. 214 THE PRINCIPLE OF WORK. [Art. 270. 270. Let US consider first the special case in which the force P is the weight JV oi a. particle, which is constant in amount and direction, while thepath described is curved so that is variable. Referring the curved path. Fig. 83, to rectangular coordinates, the axis of x being horizontal, (p is the inclination of the curve to the axis of y so that Fig. 83. dy cos

y^, is said to have a greater potential energy when at the point A than when at the point B. The difference of potential energy at these two points, that is, the potential energy expended in the displacement of the body, is taken as equivalent to the work done, W{y^ — y^. In the reverse displacement, the work done against the force is, in §XIV.] POTENTIAL ENERGY. 21$ like manner, represented by an equivalent gain in potential energy. When it is desired to assign an absolute value to the poten- tial energy at a point, it is necessary to assume the position of zero-potential. For example, in the present case, it is convenient to assume the potential to be zero on the axis of x^ so that Wy^ is the potential at A^ and Wy^ that at B. The value of the po- tential when the body is below the axis of x is negative, but there is still a positive expenditure of potential energy when it passes to a position for which the value of the potential is algebraically smaller. 272. It is not necessary to suppose the curve in Fig. 83 to be a plane curve. The locus in space of the points of zero-poten- tial is, of course, a horizontal plane. Accordingly the potential has a common value for all points upon any other given hori- zontal plane. For this reason, such planes are called, with refer- ence to gravity, equipotential surfaces. In passing from one such surface to another, the loss of potential is equal to the work done by the force, or the gain of potential is the work done against the force. If the path begins and ends in the same potential surface, that is, at the same levels there is as much work done by gravity as against it, and the total work is zero. The force of gravity is called a conservative force^ because the work done against it is stored up, as it were, in the form of po- tential energy, or difference of potential, and can be reconverted into an equivalent amount of work. On the other hand, a force like that of friction is non-conservative y because work done against it does not produce any potential energy or power to do work. Work done by a Resultant. 273. We have seen in Art. 247 that, when constant forces are acting at a single point of application which undergoes displace- ment, the algebraic sum of the works done by the several forces has the same value as the single expression for the work done by the resultant. The same thing is obviously true of the ele- 2l6 THE PRINCIPLE OF WORK. [Art. 273. ments of work done in the elementary displacement ds^ when the forces are variable. Thus, using the same notation as in Art. 247, R cos ^=~r> .... (2) dx dy dz ^ ' the partial derivatives of Fwith respect to .r,_y and 2; respect- ively. It must be remembered that in accordance with the no- tation of the Differential Calculus ^Fhas a meaning in each of the fractions different from that of dF in equation (i). For this reason we shall here use the notation of the virtual displacements of the preceding section, and write equation (i) in the form 6 V = X^x -i- YSy -{- ZSz, ..... (3) in which ^x, Sy and Sz are the projections in the direction of the axes of the virtual displacement Ss, and accordingly SFis the virtual work done in the displacement Ss. 284. Dividing this equation by dsy we have rr=4+4+4: ('^) Now, denoting the direction angles of the displacement Ss by A, //, v^ this becomes —-= X cosX-\- Y COS ix-\- Zcos V, ... (5) Since 6V =^ F cos (pds, the second member of this equation is an expression for the resolved force in the direction of the displace- ment.* * This is readily seen geometrically, for if P, X, Fand Z are con- structed as in Fig. 17, p. 45, X cos A is the projection of Xupon a straight line in the direction of <5j, F cos /< that of F, and Z cos v that of Z. Their sum is therefore the projection of P on the same line, that is, P cos . g XI V. ] EQ UlPO TENTIA L S URFA CES. 223 Equipotential Surfaces. 285. The potential function U is defined as in Art, 279, so that U = C — V, and its derivates are simply those of the work- function with their signs changed. If we put U" = C^ where C is any constant we have the equation in x, y, 2 o( sl surface. If the displacement (^j takes place in any direction along this surface, we have SU = Oy that is to say, no work is done. Such a surface is called an equipotential surface. We have already seen that, for a constant force such as gravity, these surfaces are-parallel planes; also that, for any central force, they are concentric spherical sur- faces. In general, they form a system of surfaces which do not intersect one another; for U =^ C and U =^ C are contradictory equations. The direction of the force is at every point of an equipotential surface normal to it.* A line (in the general case a curved one) whicTi is normal to every surface of the system is called a line of force. 286. By integration of the element of work, equation (i), * If a, P, y are the direction-angles of the force P^ X—P cos a, Y—P cos (5, Z=Pcosy. Accordingly the partial derivatives of F, see Art. 283, are proportional to the direction-cosines of the normal to the surface V ^ C. When these values are substituted in the expression for the deriva- tive of V in the direction of 8s (of which the direction-angles are A, yU, v), equation (3), Art. 283, we have 8V -^ = P{cos a cos A -[- cos (i cos n + cos y cos r) = P cos 0. This derivative may be called the space-rate of energy expended ; it is zero for any direction A, fx, v which satisfies cos a cos A -\- cos (i cos u -\- cos y cos r = o, that is, for any tangent line to the surface, and it is a maximum when cos a cos A -{- cos (i cos i^i -j- cos y cos r = i, which is satisfied only by A = a, // = /J, v = y. 224 THE PRINCIPLE OF WORK. [Art. 286. Art. 283, we find that the work done in any displacement from the point A to the point B is V ~ V where F, is the value of the work-function at the point B^ and Kj that at A. If we use instead the potential function, it is that is, the work done is equal to the loss of potential, and nega- tive work is represented by gain of potential, as in the special cases already considered. 287. If the particle is so constrained that it can move only in a path which crosses the equipotential surfaces, it will tend to move from the surface of higher to that of lower potential; that is, in accordance with Art. 258, in that direction in which the virtual work of the forces is positive. At a point where the path is tangent to an equipotential surface no virtual work is done, and we have a position of equilibrium. Supposing the path to suffer no sudden changes of direction, a position of maximum or minimum potential is such a point, the equilibrium being unstable in the first case and stable in the second. If the path, were tangent to an equipotential surface but also crossed it, the equilibrium would be stable on one side, and unstable on the other side, of the position of equilibrium. In like manner, when the particle is restricted only to remain in a given surface, a position of maximum potential is one of unstable equilibrium, and one of minimum potential is one of stable equilibrium.* * The intersections of the surface of a mountain with horizontal planes at different altitudes, which are the equipotential surfaces in the case of gravity are called contour lines. These lines form a good illus- tration in two-dimensional space of the equipotential surface in three- dimensional space ; the variable force to which they correspond being the component of gravity along the sloping surface. As we pass from one contour line to another, the work done is the difference of corre- §XIV.] EXAMPLES. 22$ EXAMPLES. XIV. 1. What is the work done in raising to the surface the water in a cistern lo feet square and 6 feet deep? 112,500 ft.-lbs. 2. A well is to be made 20 feet deep and 4 feet in diameter. Find the work in raising the material, supposing that a cubic foot of it weighs 140 lbs. 351,900 ft.-lbs. 3. What part of the work of emptying a conical cistern is done when the depth is reduced one-half? \\. 4. Find how many units of work are stored up in a mill-pond which is 100 feet long, 50 feet broad, and 3 feet deep, the point at which the water is'discharged being 11 feet below the surface of the pond. 8,906,000. 5. Show that, in accordance with Hooke's Law, the work done io stretching the string through any space is the product of the space and the arithmetical mean of the initial and final tensions. 6. The wire for moving a distant signal is, when the signal is down, stretched 16 inches beyond its natural length, and has a tension of 240 pounds, which is produced by a back weight of 270 pounds resting with a portion (30 pounds) of its weight upon its bed. If the signal end of the wire is to move through 2 inches in raising the signal, show that the end which is attached to the hand-lever must have a motion of 4 inches. Find the work done when the hand-lever is suddenly pulled back and locked before the signal begins to move, and find how much less work is neces- sary if it be pulled back slowly. 90 ft.-lbs.; 2\ ft.-lbs. 7. Assuming the earth to be a sphere of radius a^ and the at- sponding potentials. The lines of greatest slope are those which at every point give the direction of the greatest force, thus corresponding to the " lines of force " of Art. 285. Accordingly, they cross the contour lines at right angles, just as the lines of force cross the equipotential surfaces at right angles. In passing between consecutive contour lines along a path oblique to the line of greatest slope, the distance is in- creased in the same ratio as that in which the effective force or force along the path is diminished, so that the product, or work done, is un- changed. 226 THE PRINCIPLE OF WORK. [Ex. XIV. traction of a body to the earth to be inversely proportional to the square of its distance from the centre, show that the work done in removing a body whose weight is W from the surface to an infinite distance is Wa. 8. Show that the work done against friction in dragging a body along a rough curve in a vertical plane, by a force which is always tangent to the path, is independent of the form of the curve. 9. Show that, if the force in Ex. 8 makes a constant angle ^ with the tangent to the path, but never becomes vertical, the whole work done is still independent of the form of the path, and find its ratio to that done when ^ = o^ a being the angle of friction. cos ft cos (x cos (y6f — a)' 10. A weight Amoving in vertical guides rests upon a bar which turns upon a horizontal axis at the distance a from the guides. The weight is raised by turning the rod through the angle from the horizontal position. Show that, if // and f^i\ the coefficients of friction between the weight and guides and the weight and rod respectively, are small, the work done against friction is approximately i^«(yu + /i') tan' 6^. 11. A weight W h drawn up a rough conical hill of height // and slope «', and the path cuts all the lines of greatest slope at the constant angle /?. Find the work done in attaining the summit. Wh{i + /f cot a sec /?). 12. In the example of Art. 78, p. 56, show that the equi- potential lines are circles, and that at positions of equilibrium not on the axis (when they exist) the equilibrium is stable. 13. A weight H^ attached by a string to a ring moving on a smooth horizontal rod hangs vertically, the string passing through a fixed smooth ring at a distance b below the rod. Verify, by direct integration of the work done in removing the ring through a distance s^ that it is the same as that of raising the weight. Find also the work done against friction if the rod be rough. CHAPTER VIII. MOTION PRODUCED BY CONSTANT FORCE. XV. Inertia regarded as a Force. 288. We have seen in Art. 13 that the property of matter through which it resists any change of motion, in accordance with the First Law of Motion, is called Inertia. The change of motion which is resisted is measured by the product of the mass and acceleration, that is, by ma, which, in accordance with the Second Law, is taken as the measure of the force which, acting freely, produces the motion. Now, just as the resistance of a fixed body in contact with that upon which the force acts, and preventing its motion, is regarded as a force equal and opposite to the force which would otherwise produce motion, so the resistance to motion in the body when free is regarded as a force equal and opposite to the active force which produces the motion. Thus the force of inertia acts upon a particle of mass m only when there is an acceleration ^, and its value is ma, while its direction is opposite to that of the acceleration. The Centre of Inertia. 289. When a rigid body has a motion of translation (see Art. i), all its points have at every instant a common velocity, and therefore a common acceleration; so that the forces of inertia acting on its several parts form a system of parallel forces pro- 228 MOTION PRODUCED BY CONSTANT FORCE. [Art. 289. portional to the masses of the parts, exactly as the forces of gravity do. It follows that, in this case, the resultant of the inertia forces is their sum acting at the same point as the result- ant of the gravity forces regarded as a system of parallel forces. This point, usually known as the Centre of Gravity, is in fact more ])roperly called the Centre of Inertia. Thus, for motions of trans- Lition, a rigid body may be regarded as a particle situated at the Centre of Inertia; and the weight of the body, when that is in question, is a force acting at the same point.* Rectilinear Motion. 290. We now resume that part of Dynamics to which Chap- ter I is introductory, namely, that which treats of the action of forces in overcoming the resistances of inertia. It is known as Kinetics,f because it is concerned with the production of motion. We consider in this chapter the motion produced in a particle (or a solid regarded as a particle of mass m situated at its Centre of Inertia) by a force constant in direction and magnitude; and, in the present section, we further suppose the particle to have no motion except in the line of action of the force. In Art. 17, it is pointed out that the units of force, mass and acceleration are so taken that F =^ mf^ where /" stands for the acceleration produced by F acting freely; but, in virtue of this equation, /may also be taken as the force acting upon a unit of * The position of the Centre of Inertia of a body depends only upon its volume and the distribution of its mass. The identity of the Centre of Gravity with this point is due to the fact that we regard the forces of gravity, near the earth's surface, as constant forces proportional to the masses and acting in parallel lines. f From Ki rr/ai^, movement. The term Dynamics, from dvvauiS, force or power, is sometimes used to cover the whole range of Theoretical Mechanics. It has in this book been employed, in accordance with common usage (in such phrases, for example, as "dynamical friction "), to imply the action of a force through a space. Compare Art. 244. § XV.] RECTILINEAR MOTION. 22Q mass. When only a single body is in consideration, we may onvit the factor ;«, and equate the force/ acting upon the unit mass to the acceleration produced. For rectilinear motion in the line of action of the force, the acceleration is dv d's ^ = ^ = ^"- where v denotes the speed, and s the distance of the particle from some fixed origin taken on the line of motion. The differ- ential equation d^__ r dt'~^' where /is the ** accelerating force," or force acting on each unit of mass, is called the equation of motion for a particle moving in a straight line. 291. The solution of this equation of the second order is the relation between the variables s and / found by integration, and involving two constants of integration. But, since in Mechanics the velocity defined by the equation ds ""^It is a variable of equal importance with j, we may with advantage regard this equation together with the equation of motion in the form dv _ ^ Jt~^ as two simultaneous differential equations of the first order between the three variables v^ s and /. For integration, these equations are written in the form dv=/dt, (i) ^s = vdt (2) 230 MOTION PRODUCED BY CONSTANT FORCE. [Art. 2qi. Eliminating dt between them, we also have vdv = fds, (3) a differential relation between v and s. Integration of the Equation of Motion when the Force is Constant. 292. When/is a constant, equation (i) contains only two vari- ables and can be directly integrated. The result may be written v-Vo-\-ft, (i) in which the constant of integration is expressed by the symbol z'o, because it is the value of v when / = o. Using this value of Vy equation (2) becomes ds — vjt -{- ft dt. Hence a second integration gives s = s,-^vJ-\-yt\ * (2) in which the constant of integration is denoted by s^ because it is the value of s when / = o. This last equation is the complete solution of the differential equation of*the second order, dt' ^* when /is constant, and equation (i) is called Si first integral of that equation. 293. Equation (3), Art. 291, contains only the variables 7> and s; it therefore also admits of direct integration, giving §XV.] KINETIC ENERGY, 23 1 This is also a first integral of the differential equation of the second order. If we determine the constant of integration by- means of the condition that v = Vq when j = ^o as in the preced- ing article, we shall have \{v'-v:)=f(s-s:) (3) This relation between v and s might have been found by elimina- tion of / from equations (i) and (2); in other words, by elimina- ting / after y instead of before, integrating. Kinetic Energy. 294. The equation vdv = fds is integrable, not only when/ is constant, but when it is a vari- able depending for its value only upon s\ that is to say, when/ is a function of s. Multiplying by /«, the mass of the body, and integrating, we have, since F = w/. J ^o where Vo is the velocity corresponding to the lower limit So. The second member is, by Art. 265, the work done by the force F in the displacement of its point of application (which is the particle, or the centre of inertia of the body) through the space s — Sq. The quantity ^mv^ is known as f/ie kinetic energy of the mass m moving with the velocity v. If Vo = o, the equation expresses that the kinetic energy is equal to the work done by the force F in imparting to the body the velocity v. Thus kinetic energy is the measure of the work done by a force against inertia; and the general equation asserts that the work done by a force upon a body moving in the line of action is equal to the gain in kinetic energy. If the force is opposite in direction to the displacement, the work is negative, and there is a loss of 232 MOTION PRODUCED BY CONSTANT FORCE. [Art. 294. kinetic energy, which is thus equal to the work done by inertia against the force. The equation of this article is called the equation of energy. Laws of Falling Bodies. 295. In the particular case of a body falling freely from rest, the position of rest is usually taken as the origin of Sy and the instant of falling as the origin of time, or instant when / = o; tims the " initial circumstances " or known corresponding values of the variables are / = o, ^ = o, z; = o. The space being measured downward, that is, in the direction of the force, the acceleration is positive and its value is g. Hence ^_ Integrating successively with respect to /, and determining the constants by the initial circumstances, we have i>=g^ (0 s^W (2) Eliminating / between these equations, we have also V" =^2gS (3) These three equations, expressing the relations between each pair of the variables /, v and s^ are sometimes said to express the laws of freely falling bodies. It must be remembered that in accordance with the initial circumstances / denotes the time in which the velocity v is acquired, and in which the space s is described, from rest. 296. The second equation shows that the space fallen through in the first second is ^g, which is one-half the space that repre- § XV.] LA tVS OF FALLING BODIES. 233 sents the velocity acquired. Again, the space described in the interval between the instants /, and /, is k('.'-O = /-^H/,-0. Defining the average velocity in a given interval as that with which, as a constant velocity, the body would describe in the interval a space equal to that which actually is described, and denoting the average velocity in the interval /, — /, by z; , the space described is v^(t^ — /,). Comparing this with the expres- sion written above, we see that the average velocity in any inter- val is where v^ and v^ are the velocities at the beginning and end of the interval. That is, the average velocity, in the case of constant acceleration, is the arithmetical mean of the extreme velocities: it is also the same as the velocity at the middle instant. The average velocity during the «th second, found by putting /, = « and /, = « — !, is accordingly the same as the space described in that second, namely, i(2« - i)^. Thus the spaces described in successive seconds are proportional to the successive odd numbers. 297. The velocity acquired by falling from rest through the height h is, by equation (3), V = \/(2gh) This velocity is often called the velocity due to the height h. Conversely, the height 234 MOTION PRODUCED BY CONSTANT FORCE. [Art. 297. is called the height due to the velocity v. It is the distance through which gravity must work upon a body originally at rest to give it the velocity v or the kinetic energy \mv^. Accordingly, multi- plying by W^ we have Wh = Jwz;'. Body Projected Upward. 298. in the case of a body projected upward, it is convenient to measure the space positively upward : therefore gravity pro- duces a retardation of g feet. Taking the point and the instant of projection as the origins of space and time, and Vo as the ve- locity of projection, the initial circumstances are V = Vo, s = o when / = o. Integrating ^- _ dt* ~ ^ successively, and determining the constants accordingly, we find v^Vo-gt, (i) S =zVot- \gt\ (2) and, eliminating /, V" z=Vo — 2gS (3) 299. Equation (i) shows that the velocity, originally positive, V is decreasing; it vanishes when t — —, which is the time required o for gravity to overcome the initial velocity. Equation (2), being a quadratic for /when s is given, shows that there are two instants at which s has a given value. For example, if the velocity of projection is 64%, to find the instant when the body is 48 feet \ above the point of projection, we have the quadratic 48 = 64/— 16/'. The roots of this are / = i and / = 3; the first indicates the I XV.] BODY PROJECTED UPWARD. 235 instant at which the body reaches the height of 48 feet while rising, and the second that at which it returns to the same point while falling. 30c. Equation (3) shows that the two values of v which correspond to the same value of s are numerically equal and of opposite signs, that is, the body passes a given point with the same speed in ascending and descending. The greatest height H from the ground to which the body will rise is found by putting z^ = o in the same equation to be which is the height due to the initial velocity Vo (see Art. 297). The height due to a given velocity may therefore be defined as that to which a body will rise if projected directly upward with that velocity. 301. Multiplying equation (3) by \in^ and introducing H in place of z'o, we derive the equation ^mv" = mg{H -s)=^ W{H - s), which shows that the kinetic energy at any point is equal to the work of gravity corresponding to the distance of the point below the highest point reached. If we take the ground, that is, the level of the point of projec- tion, as that of zero-potential (see Art- 271), Ws is the potential energy of the body when at the height s. Then, writing the equation in the form ^mv" -\- JVs= WI£^imVo\ it asserts that the sum of the kinetic energy and the potential energy at any point is constant. This is the simplest example of the principle of the Conservation of Energy in its two mechanical forms of potential and kinetic energy. When the body leaves the ground the whole energy is in the kinetic form, and when it reaches the highest point it is all in the potential form. 236 MOTION PRODUCED BY CON STAN 7' FORCE, [Art. 302. Motion on a Smooth Inclined Plane. 302. For a body moving on a smooth inclined plane, as in Fig. 84, the only force acting in the direction of the motion is the resolved part of the weight W which acts down the plane. De- noting the inclination by ^, this is \V sin 6^, hence the acceleration or force acting on a unit mass is ^ sin d. Hence, for a body falling from rest, the equations, found as in Art. 295. are \'\i^. 64. z; = ^ sin 6^ . /, . s = i^ sin 6^ . /', v^ = 2g sin d . s, (i) {2) (3) Let AC t= ^ be the height of the starting-point A above the bottom of the plane, and AB — c the length of the plane ; then ^ = ^ sin B. Putting s = c m equation (3), we have then V* — 2gh\ hence, comparing with Art. 295, we see that the ve- locity acquired by falling through the length of the plane is equal to that acquired by a body falling freely through the same height. Multiplying by ^m^ we have or the kinetic energy acquired is, as before, equal to the work done by gravity. Spaces fallen through in Equal Times. 303. To compare the spaces described in the same time by the freely falling body and that on the inclined plane, let / be the time occupied by the freely falling body in describing the space ^ XV.] MOTION ON A SMOOTH INCLINED PLANE. ■37 h. By equation (2), Art. 295, h — \gf. Substituting in equa- tion (2) above, we have s =^ h sin 6. This value of s is AD in Fig. 84, constructed by drawing CD perpendicular to AB. Thus, if the bodies start from rest at the same instant, they will reach C and D respec- tively in the same time. Suppose now that while A and C are fixed points, the inclination ^of the plane is varied. Because ADC is a right angle, the locus of D is a circle described on AC as a diameter. Hence the time of falling through any smooth chord drawn from the highest point of a vertical circle is the same as the time of falling through the vertical diameter. The same thing is obviously true of chords drawn to the lowest l)oint of the circle. The proposition may be used in the graphical solution of certain problems involving the straight line of quickest descent. For example, to construct the straight line of quickest descent from a given point v4 to a given curve we have only to draw the smallest circle of which A is the highest point and which meets the given curve. This circle is evidently tangent to the given curve. Body Projected up an Inclined Plane. 304. For a body projected up a smooth inclined plane, the initial circumstances being taken as in Art. 298, and the space measured up the plane, the equations become v= Vo — g SVCid .t, (i) s = Vot — ig sin 6.t*f (2) v^= 7>o' — 2g sin 6 , s (3) 238 MOTION PRODUCED BY CONSTANT FORCE. [Art. 304, Denoting by ^ the greatest vertical height to which the body will ascend, we have, putting z; = o in equation (3), the same result as in Art. 300. Thus the body will rise to the same height as if it were projected vertically upward. 305. Multiplying equation (3) by \fn^ we have \mv^ = \'invQ — W svcvB .s, or putting s .^\ViO ■= h (so that // is the vertical height correspond- ing to the velocity z/), and introducing ZT, \mv' = W{^H - h). Hence, although the velocity is in a direction oblique to the force, the kinetic energy at any vertical height is the equivalent of the work done by gravity in the vertical distance of the body below the highest point. Again, if the level of the point of projection be taken as that of zero-potential, we have ^mi' + Wh =z Wlf = imVo\ which expresses that the sum of the potential and kinetic energies is constant. There is a continual transferrence of energy from the kinetic to the potential form and 7>ue versa^ but no loss of total energy. Motion on a Rough Plane, 306. Let us next suppose the plane to be rough, then when the body is moving down the plane with the same initial circum- stances as in Art. 302, the friction acts up the plane, and its value is — i^R, where // is the coefficient of dynamical friction and i? = JVcosO. See Fig. 84, p. 236. Therefore the acceleration g XV.] MOTION ON A ROUGH PLANE. 239 down the plane is / = ^(sin d —}x cos ^). The equations now become V — ^(sin 6 — }x cos 6)f^ (i) s = i^(sin 9 — fj. cos 6^)/', (2) v*-= 2^(sin 6 — fi. cos 6)s (3) These equations, of course, presuppose that the expression for/ is positive, so that motion actually takes place ; that is, tan ^ > /^, or > a^ the angle of friction. This being the case, suppose the body to fall from A, Fig. 84, to B ; then, putting s = c^ equation (3) gives, for the kinetic energy at the bottom of the plane, \mv' = Wc sin 6 - /x^ccose = Wh - fxWb, . (4) where b is the base BC of the plane. Therefore the kinetic energy acquired in the fall is less than Wh, the potential energy expended, by the amount fjiWb ; this is therefore the energy ex- pended in overcoming the non-conservative force of friction. 307. If the body is projected up the plane, the initial circum- stances being as in Art. 304, the friction as well as the resolved part of the weight will act down the plane and / = — ^(sin 6 -\- fx cos 0), The relations now become V = Vo — g{s\n ^ + yw cos 6)t^ . . . (i) s = Vot — ig{sm 6 -\- }x cos 6)t''f ... (2) ?;' = Vo' — 2^(sin -{• M cos 6)s. ... (3) If B, Fig. 84, is the point of projection and A the highest point reached, we find, by putting z; = o in equation (3), z/o' = 2^r(sin ^ + /^ cos 6'), and multiplying by Jw, 240 MOTION PRODUCED BY CONSTANT FORCE. [Art. 307. Taking the potential energy as zero at the bottom of the plane, the first member expresses the total energy at the instant of pro- jection. At the highest point A^ this energy has been expended ; the part Wh has been converted into potential energy, and the remaining part, p. Wd, has been used in doing work against friction. 308. The equations above apply only up to the time when the body reaches its highest point, because if the body descends friction will act up the plane, thus changing its direction. If 6 > a, there will be a downward motion in accordance with the equations of Art. 306 ; but, if ^ < cy, the motion will cease. In particular, if ^ = o, we have the case of a body projected along a rough horizontal plane. Such a body is subject to a retarda- tion yu^, and the space s, which will be described before the body comes to rest, is given by the equation which expresses that the initial energy is all expended in work against friction. EXAMPLES. XV. 1. If a body start with a velocity of 4 feet per second and move with one foot-second unit of acceleration, in what time will it acquire a velocity of 30 miles per hour? 40 seconds. 2. A stone skimming on ice passes a certain point with a velocity of 20 feet per second and suffers a retardation of one unit. Find the space described in the next 10 seconds, and the whole space described when the stone has come to rest. 150 ft.; 200 ft. 3. A body whose velocity is uniformly accelerated has at a certain instant a velocity of 22^5. In the following minute it travels 10,320 feet. Find the acceleration. S-/s». 4. A uniformly accelerated body passes two points 30 feet apart with velocities of 7 and 13 feet respectively. What is the acceleration? 2 ft.-sec. units. 5. A body whose motion is uniformly retarded changes its velocity from 24% to 6% while describing 12 feet. In what time does it describe the 12 feet ? | sec. § XV.] EXAMPLES. 24 1 6. A steamer approaching a dock with engines reversed so as to produce a uniform retardation is observed to make 500 feet during the first 30 seconds of the retarded motion and 200 feet during the next 30 seconds. In how many more seconds will the headway be completely stopped ? 5. In the following examples take g =32 when numerical results are required : 7. A body is let fall from a point 576 feet above the ground. With what velocity should another body be projected vertically upward from the same point and at the same instant, in order that it may strike the ground 4 seconds after the first body ? io2.4ys« 8. A body is dropped from a height AB = h^ and at the same moment a body is projected vertically upward from B. What must be the initial velocity if they are to meet half way ? 9. To what height will a body projected upward with a velocity of 40 feet per second rise ; and at the end of what times will it be 9 feet from the ground ? 25 ft. ; i and 2J sec. 10. Two bodies are let fall from the same point at an interval of one second. How many feet apart will they be at the end of four more seconds ? \g. 11. A body projected vertically upward remained for 4 seconds above the 960-foot level. What was the velocity of projection ? 256 ft. per sec. 12. A balloon ascends with the uniform acceleration \g: At the end of half a minute a stone is dropped from it; how long will it take to reach the ground ? 15 sec. 13. A ball is projected vertically upward with a velocity of 128 feet per second: when it has reached f of its greatest height, another is projected from the same point with the same velocity. At what height will they meet ? 240 feet. 14. A stone is dropped into a well, and the sound of the splash is heard 7.7 "seconds afterward. Find the depth of the well, supposing the velocity of sound to be 11 20 feet per second. 784 feet. 242 MOTION PRODUCED BY CONSTANT FORCE. [Ex. XV. 15. With what velocity in feet per second must a body be pro- jected upward to reach the top of a tower 210 feet high in 3 sec- onds ; and with what velocity will it reach the top ? 118 ; 22. 16. A body projected upward from the top of a tower a feet high reaches the ground 4 seconds later than a body dropped at the same time. What was its initial velocity ? * ^°-^^ 4/^ + 2 1/(2^) • 17. Show that the distance between two falling bodies in the same vertical line is a uniformly varying quantity. Thence find the velocity with which a body must be projected down- ward, to overtake in / seconds a body which has fallen from rest at the same point through a feet. o. y , i \ 18. A body is projected vertically downward from the top of a tower with the velocity F. One second afterwards another body is dropped from a window a feet below the top. Determine in how many more seconds it will be overtaken by the first body, and explain the result when it becomes negative. 2^ — 2 V—g 19. A body is projected down a smooth inclined plane whose height is j^-^ of its length with a velocity of 7J miles per hour Find the space passed over in two minutes. 3240 feet. 20. Show that the times of falling down smooth planes of the same height are proportional to the lengths of the planes. 21. A body weighing 30 pounds falls down a rough inclined plane of height 30 feet and base 100 feet. If /< = ^, what is the kinetic energy acquired ? 300 foot-pounds. 22. A weight of 40 pounds is projected along a rough hori- zontal plane with a velocity of 150 feet per second. The coeffi- cient of dynamical friction being ^, what is the work done against friction in the first five seconds, and in the five seconds imme- diately preceding rest ? 35oo ; 250 foot-pounds. 23. A train weighing 60 tons has a velocity of 40 miles an hour when the steam is shut off. If the resistance to motion is § XVI.] EXAMPLES, 243 10 pounds per ton, and no brakes are applied, how far will it travel before the velocity is reduced to 10 miles an hour ? 11,2934 feet. 24. Show that the straight line of quickest descent from a point to a curve in the same vertical plane makes equal angles with the vertical and the normal at its extremity ; and that the line of quickest descent between two curves makes the same angles with the two normals at its extremities. 25. Show how to construct graphically the straight line of quickest descent from a given point to a given circle. 26. What is the angular distance between the highest point of a vertical circle and the point from which the time down the radius is the same as the time down the chord to the lowest point ? 60 o . 27. Show that, if the plane is rough, the locus of the point corresponding to Z>, Fig. 84, p. 236, is the arc of a circle, and that the locus of the point corresponding to B (where the velocity of a body starting from rest at A is the same as that of the freely falling body at C) is a straight line. 28. A heavy body projected up a rough plane whose inclina- tion is 15" came to rest in 5 seconds after sliding 200 feet along the plane. Find the coefficient of friction. /f = •2497. 29. A body with constant acceleration acquires a velocity of 45™A i^ 4 "oxAq from rest. In what time is the \ mile described ? 80 sec. XVI. Kinetic Equilibrium. 309. We have seen that the inertia of a body undergoing acceleration may be regarded as a force balancing that which produces the acceleration. So also, when more than one force beside the inertia acts, we have, by including the inertia-force, a system of forces in equilibrium. In employing this principle, 244 MOTION PRODUCED BY CONSTANT FORCE. [Art. 309. which may be called that of kinetic equilibrium^'^ the unknown quantity derived from the condition of equilibrium may be a force instead of an acceleration. For example, suppose a man whose weight is W to be standing on the floor of an elevator which begins to descend with the known acceleration a. The forces acting on the man are his weight, W = mg^ acting down- ward, his inertia, ma^ acting upward because the acceleration is downward, and the resistance R of the floor of the elevator act- ing upward. Since the forces are all vertical, there is but one condition of equilibrium, namely, JV = R -{■ ma. W Substituting — for /«, we find - ="-[-.-} For example, taking ^=32, if « = 8, we find R — \W\ in other words, three-fourths of the man's weight is sustained by the floor, the other one-fourth going to produce the acceleration without which the man would not follow the elevator in its motion. When the elevator has assumed a uniform velocity, a vanishes and R = Wf exactly as if there were no motion. When the ele- vator is coming to rest, ex changes sign in the equation as written above, because the acceleration has changed its direction. Hence, during the retardation, the pressure upon the floor is greater than the weight. 310. As a further illustration, suppose a brick of mass m to be dragged over a rough horizontal table by means of a string parallel to the table. If the velocity is constant, the force exerted, which is the tension of the string, is equal to the dynam- ical friction. But, if the brick is to receive the acceleration/. *This principle in its application to the general equations of motion is known as D'Alembert's Principle. § XVI.] ACCELERATION OF INTERACTIJVG BODIES. 245 the tension must be increased by the amount w/to overcome the resistance to acceleration, that is, the inertia. Again, suppose the tension to fall below the friction, the brick will be retarded, and until it comes to rest the inertia will act in the direction of motion and assist the tension in overcoming friction. Acceleration of Interacting Bodies. 311. When the mutual action of two bodies is such as to furnish a relation between their motions, the kinetic equilibrium of the two bodies may be used to deter- mine at once their accelerations and their mutual action. For example, if the weights W^ and W^ are connected by an inextensible string passing over two smooth pegs or pulleys, as in Fig. 86, the downward ac- celeration of the greater weight is evi- dently equal to the upward acceleration of the less. Their mutual action is the tension T of the string, which acts up- ward in each case. Then, denoting the common acceleration by or, we have to determine the two unknown quantities, T and ot^ by means of two equations, one derived from each of the bodies and expressing the kinetic equilibrium of vertical forces. When a single acceleration is involved, it is convenient to place upon one side of the equation the algebraic sum of all the external forces, regarding the direction of the acceleration as positive. The second member will then be the product of the mass and acceleration, which is in fact the inertia force acting in the oppo- site direction. Thus, in the present case, if W^ > W^ , the Wx m Fig. 85. acceleration of W^ is downward; hence we write W (0 246 MOTION PRODUCED BY CONSTANT FORCE. [Art. 311. equating the excess of downward force to the inertia it over- comes. In like manner for the other body we have W T-lV, = ~^a (2) Eliminating J", we derive g whence ^ U/ -I. Ur ^' '3) Again, eliminating a from equations (i) and (2), whence This value of T is intermediate in value between W^ and W^, It is in fact the so-called " harmonic mean " of these quantities. 312. Since the bodies W^ and IV^ have the same speed, they may in a sense be regarded as a single mass which has the acceleration a. The force producing this acceleration is then W^ — W^, and equating this to the product of the total mass into the acceleration we have W -\- W o giving the equation (3) at once. The arrangement shown in Fig. 86 is the essential part of Attwood's machine, by which the acceleration of gravity may be diminished in any chosen ratio, so that the velocity produced can be, conveniently measured. Thus, if the two weights were each 15^ ounces and an extra weight of one ounce were added to one § XVI.] APFLICA TION OF THE PRINCIPLE OF WORK. 247 of them, we should have a total mass of two pounds moved by a force of one ounce, hence by equation (3) the acceleration will be i^ of g. Application of the Principle of Work, 313. The principle of virtual work, or of work-rate, is some- times employed when one of the forces in question is that of iii- ertia. For example, a train weighing 160 tons is hauled up a grade of i in 140, the resistances from friction, etc., being 12 pounds per ton. Required to find the acceleration at the instant the speed is 15 miles an hour, if the engine is then developing 200 horse-power, that is to say, doing work at the rate of 200 X 550 foot-pounds per second. This work is done against the resistance Ry the component of the weight along the inclined plane, which is y|-Q^ W^ and the iner- tia ma. Since the speed is 22 '/s, the space through which the sum of these forces works in one second (or rather the rate per second at which they are at the instant working) is 22. Hence / ^ ^ , 160 X 2240 , 160 X 2240 y 200 X 550 = 22 12 X 160 H -^ A OL\y whence we find ex = -^^-^. It is obvious that the process is equivalent to equating the forces which act at the two ends of the draw-bar, since the foBce is the result of dividing the work by the space through which the force acts. 314. In Section XIV. we have employed the total work done in a displacement in solving questions involving forces and spaces only. Such questions usually imply the transferrence of a mass from one position to another, which generally brings into action the force of inertia. Thus, if the initial position is one of rest, some motion, and therefore some acceleration, must take place. We have seen in Art. 294 that the work done against inertia takes the form of kinetic energy, and that during retarda- 248 MOTION PRODUCED BY CONSTANT FORCE. [Art. 314. tion an amount of work is done by inertia equivalent to the loss of kinetic energy. Hence, if the final position is also one of rest, the force of inertia does not appear in the equation derived from the total work. 315. But, in applying the principle to the more general case where the initial and final circumstances involve velocities, the changes in kinetic energy must be reck- oned as part of the work done. As an illustration, take the following example : A body A^ weighing I pound, is connect- ed with a body B^ Fig. 87. weighing 2 pounds, by means of a string which passes over a smooth pulley at the edge of a rough horizontal table on which B rests, while A hangs at a distance of 18 inches from the floor. Supposing /< = 4, if ^ is allowed to drop, find the distance s which B will travel after A strikes the floor. The whole work done by gravity upon the system consisting of the two weights is here i^ foot-pounds. The friction F \m^)^ ^^m.h which are connected m,/ ^m^g by a string which passes over a smooth pulley at the top of the wedge; it is re- quired to find the acceleration of the wedge, and the acceleration of the masses relatively to the wedge. Let/ denote this last acceleration, that is to say, the rate of change of the speed with which the string passes over the pulley. Let h denote the horizontal acceleration of the wedge toward the left, which is also shared by the masses Wj and m^\ and let 7", R^ i?j and J^^ be the tension of the string, the resistance of the horizontal plane, and the actions between the wedge and the masses m^ and m^. > To determine these six unknown quantities we have two conditions of equilibrium for each of the bodies w,, ;«, and M. The forces acting on w, and w, respectively are shown in separate diagrams for clearness. Taking, in each of these cases, forces along, and perpendicular to, the face of the wedge, we have .... (I) R^ = m^g cos OL — tnji sin or, . . T = m^g sin a -f- Wj(// cos a — /), R^ = m^g cos a -f- ^^h sin ^, . . , T = tn^g sin oc — mj^h cos a — /). (2) (3) (4) The forces acting upon M are its weight, its inertia Mh acting as before to the right, the reactions of the resistances and 250 MOTION PRODUCED BY CONSTANT FORCE. [Art. 316. tensions which act on w, and ///, , and the resultant upward resist- ance R of the fixed plane. Hence, taking horizontal and vertical forces acting on J/, Mh={R-R:)^xx.a, (5) R~ Mg-\r {R, + R,) cos a-}- 2 T sin a. . . (6) 317. Eliminating R, and R^ by means of equations (i) and (3), equation (5) gives (w, — fn^)g sin ex cos ex ~ M -\- {m^ -\- w,) sin'' a ' and the substitution of this value gives M -\- 2m. sin' a -^1 = ^iJT cos a ; 1 r r-=— , „ M -\- 2m, sin' a R^ = m^g cos a- M + (w, + m^ sin' a* Again; eliminating h and /from equations (2) and (4), we find T— ^^i^^^-g" sin a ^ w, + w, ' and, this being substituted in equation (2), we obtain . , , . 2w, p- sin a f=zh cos a -\- g sm a ^ — ^ ■. m^ + /«, which, when the value of h found above is substituted, becomes (m, — m.,)(M -f w, 4- m,)g sin a / = (m, + m,)[M + (^^ + m,) sin' o']' Finally, the value of R is most conveniently found by first substituting in equation (6) the values of R^ and R^ directly ^ XV I .] EXAMPLES. 2 $ I from equations (i) and (3), and that of 2 7" which results from adding equations (2) and (4); thus R — (M + w, -f w,)^ — (w, — m^)/ sin a, in which / has the value given above. This equation shows that the pressure on the plane is less than it would be if motion were prevented, the diminution being the excess of the upward com- ponent of the inertia of m^ over the downward component of the inertia of w,. EXAMPLES. XVI. 1. If the weight of a balloon and its appendages is 4500 pounds, and that of the air displaced (which is the upward force) is 4800 pounds, with what acceleration does it begin to ascend ? -f^g. 2. Two bodies weighing 3 pounds each are connected by a light string passing over a smooth peg. If a third body of the same weight is added to one of them, how much is the pressure on the peg increased ? 2 pounds. 3. Two weights of 5 and 4 pounds respectively are attached to one end of a string which passes over a smooth pulley and has a weight of 7 pounds on the other end. The two weights descend through a distance j, and the 4-pound weight is then detached. How much farther will the 5-pound weight descend. f j. 4. In an Attwood's machine a 40-gramme weight is drawn up by a 50-gramme weight 2.18 metres in 2 seconds. What is the value of ^ in centimetres per second per second ? 981. 5. If a 3-pound weight hanging over the edge of a smooth hori- zontal table drags a 45-pound weight along it, determine the acceleration and the tension of the string. « = tV.^; T= 2 lbs. 13 oz. 6. Two equal weights are connected by a string 7 feet in length, one of them resting upon a smooth horizontal table, 3 feet high, at a point 6 feet from the edge, where the string passes over a 252 MOTION PRODUCED BY CONSTANT FORCE. [Ex. XVI smooth pulley to the other weight hanging freely. In what time from rest will the first weight reach the edge of the table ? I second. 7. If, in the preceding example, the table is so rough that the body just reaches the edge, in what time will it do so ? J 4/ 5 sec. 8. If the string in Fig. 86, p. 245, can only sustain a tension of \ of the sum of the weights, show that the least possible value of the acceleration is 1^4/2. 9. A train weighing 100 tons is drawn on a level track by a locomotive developing 150 horse-power, and the resistance is 14 pounds per ton. What is the acceleration when the train is moving 15 miles an hour?- 47^ 4480 10. A bicyclist and his machine weigh 180 pounds. What horse- power does he exert in riding on a level track, whose resistance is one per cent of the weight, at the rate of 20 miles per hour ? .096. 11. Weights of II and 5 pounds are suspended from the ex- tremities of a cord which passes over a smooth fixed pulley. What is the velocity of either weight at the end of 5 seconds from rest, and the pressure on the supports of the pulley ? 6oVs; isflbs. 12. Two unequal weights, W^ > W^,, on a rough inclined plane are connected by a string passing through a smooth pulley fixed to the plane so that the parts of the string are parallel to the plane. Determine the acceleration /, the inclination being ^, and the coefficient of friction p.. 13. A mass^ draws a mass -^ up a smooth inclined plane by means of a string passing over the vertex. Determine the inclina- tion of the plane so that A may draw ^ up a given vertical height in the shortest possible time. ^ _ • x :^ 2B 14. The height 01 an inclined plane is 5 feet and its length 13 §XVI.] EXAMPLES. 253 feet. A weight of 10 pounds is suspended from one end of a cord which passes over a smooth pulley at the top of the plane and is attached to a weight of 3 pounds resting on the plane. Find the tension of the string during motion: ist, if the plane is smooth; 2d, if /i = ^. 3.20 lbs. ; 3.90 lbs. 15. A string passing over a fixed pulley carries a weight of 2 pounds on one end and a pulley on the other, over which passes a string carrying a weight of one pound at each end. The sys- tem being at rest in equilibrium, a force is applied to one of the one-pound weights. Prove that when it has moved the weight down three inches each of the other weights has risen one inch. 16. A train weighing 100 tons is ascending a i-per-cent grade, and the frictional resistance is 12 lbs. per ton. What is its greatest speed, if the engine can develop 200 horse-power at that speed? 21.8 Vh. 17. A weight of 10 pounds rests on a rough horizontal table, /x = "i; a String attached to it passes over a smooth pulley at the edge of the table to an equal weight at the distance of 2\ feet from the floor. If this second weight is let fall, find the tension of the string, the time to reach the floor, and how much farther the weight on the table will go. 6 lbs.; \ sec; 5 feet. 18. Two weights P and Q are connected by a string which passes over a pulley at the top of a smooth plane inclined 30° to the horizon. Q^ hanging freely, can draw P up the length of the plane in half the time that P would take to 'draw Q up. Find the ratio of ^ to Z'. 3 : 2. 254 MOTION PRODUCED BY CONSTANT FORCE. [Art. 318. XVII. Motion Oblique to the Direction of the Force. 318. In this section, we suppose the particle, acted upon by a force constant in direction as well as magnitude, to have an initial motion oblique to the direction of the force. The most impor- tant application is to the case in which the force is the weight of a particle, or the weight of a body regarded as acting at the centre of inertia. We therefore take as the representative case the mo- tion of a projectile or body projected from a point in a direction not vertical. The plane of projection is the vertical plane through the line of projection, which is a tangent to the path of the pro- jectile. Since there is no component of force tending to move the body out of this plane, the path, which is called the trajectory y will be a plane curve lying in this plane. 319. Let us first suppose the body to be moving through the point Oy Fig. 89, with the velocity Fin the direction of the hori- zontal line Ox. Then, if gravity did not act, the particle would at the end of one second arrive at the point A^y where OA^ = V; at the end of two seconds at A^ , where 0A^ = 2 V; and at the end of any time / at A^ where OA = Vt. Resolving the actual velocity of the particle P at any instant into its hori- zontal and vertical components, there is no force tending to disturb the horizontal velocity; hence, at the instants mentioned above, the particle P will actually be found at points vertically below the points ^,, A^ and A respectively. Consider now the vertical velocity. This is zero at the point O, at which point, therefore, the trajectory is tangent to Ox. Now, since the vertical force is not affected by the horizontal motion, the vertical ve- locity is at every instant the same as that of a particle falling freely from O, in the vertical line Oy. It follows that, if ^,, B^ D Ai A2 A ^ B2 B ^ \ y \? Fig. 89. §XVII.] VARABOLIC MOTION. 255 and B are the positions of the freely falling body at the instants I, 2 and /, OB^, OB^ and OB will be the actual distances of the particle at these instants from the horizontal line Ox\ hence the actual positions of the particle will be as indicated in the diagram. 320. Referring the position of P to Ox and Oy as axes of co- ordinates, we have at the time / =c=Vt, (.) and, by Art. 295, y = W- • • (») These equations connect the position of the body with the time, and suffice to solve such questions as the following : A body is projected horizontally with a velocity of 2oy8 from a tower standing 128 feet above a horizontal plane; when and where will it strike the ground ? Putting jf = 128 in equation (2), and taking^ = 32, we find / = 2 4/2; whence equation (i) gives jc = 40 4/2; that is, the body hits the ground about 56.57 feet from the foot of the tower. To find the equation of the trajectory, or direct relation be- tween X and J, we have, by eliminating / from equations (i) and y=-^,x\ ....... (3) which is the equation of a parabola with its axis vertical. Parabolic Motion. 321. Before proceeding to the equation of the trajectory re- ferred to the point of projection, we shall use the symmetrical form of the equation found above in deriving some properties of the motion. Let H denote the height due to the velocity F(see Art. 297), then >-=2^jy, and Zr=— . 256 MOTION PRODUCED BY CONSTANT FORCE. [Art. 321 Substituting in equation (3), the equation of the curve takes the form x" = 4By, Comparing this with the usual form of the equation of the parab- ola, we see that 4!^ is the parameter or latus-rectum, and H is the distance of the vertex from the focus or from the directrix. Hence, measuring (7Z)=^ vertically upward in Fig. 89, the hori- zontal line through D is the directrix. 322. The horizontal and vertical velocities of the particle are denoted by — and J- respectively, and, by Art. 43, the actual speed in the curve is given by '■ = (f)' + m (.) In the present case, ~t7 — ^y and, by Art. 295, i-j\ = 2gy. Substituting, e,' =. V -V 2gy= 2g{II-\- y) (2) Now, in Fig. 89, !!-{-)/ is the distance of the particle P below the directrix ; hence, by Art. 297, the velocity at any point is equal to that due to the distance of the point below the directrix. Kinetic Energy of the Projectile. 323. If we multiply the general equation (1) of the preceding article by \m^ we derive \mv^ — \mvx + \mvy^ Vx and Vy denoting any pair of rectangular components of the §XVII.] KINETIC ENERGY OF THE PROJECTILE. 257 velocity. This equation shows that the actual kinetic energy of a body is the sum of the kinetic energies it would have if moving with one and then the other of its resolved velocities. The kinetic energy can thus, as it were, be resolved into two component parts, but only when the component velocities are rectangular. Treating the particular equation (2) in like manner, we have ^mv" = W{H -^-yY that is, the kinetic energy in the trajectory is the sum of the con- stant term WH^ due to the constant horizontal velocity, and the variable part Wy^ due to the vertical velocity. The Trajectory referred to the Point of Projection. 324. When a body is projected obliquely upward from a point on the ground taken as origin, it is convenient to measure^ up- ward. The equations of motion for the two component motions, which, as we have seen, may be considered separately, are d'^x d^y , X ^ = o, and ^=-S (i) Let O, Fig. 90, be the point of projection, Fthe velocity of projection, or initial velocity, and a the inclination of the line of projection to the horizontal. The initial horizontal and vertical velocities are then Fcos a and Fsin a. The first integration of equations (i) gives ~z=Vcosa, -£ = Fsma-gt, ... (2) 258 MOTION PRODUCED BY CONSTANT FORCE. [Art. 324. in which the constants of integration are the initial values which correspond to /= o. A second . integration gives, since x = o andjv = o when / = o, X — Fcos a .t^ \ i \ y =V%\Tia.t-\gt\ \ -^^ The value of y shows that the ^iG- 90- vertical distance of the particle P below the tangent at O is that through which it would have fallen freely in the time of describing the arc OP. 325. Eliminating / between equations (3), we have, for the equation of the trajectory y=x\2iXia -^ 5— (4) If we put, as in Art. 321, V' = 2gH, H^—, so that H is the height due to the initial velocity, the equation may also be written y = X tsina — 5— , . . • (5) in which -^is now, by Art. 322, the distance of the point O (not of the vertex) below the directrix; that is, the height of the directrix above the ground. 326. Equating to zero the vertical velocity, equation (2), we have F sin or / = £: for the time in which the initial vertical velocity is destroyed by gravity; and using this value of f in equations (3), we find §XVII.] EQUATION OF THE TRAJECTORY. 259 X— , y = for the coordinates of the point A^ Fig. 90, which is the highest point reached by the projectile, that is, the vertex. of the parabola. Substituting the value of V"^ in the preceding article, these co- ordinates may be written a; = -^sin 2^. When written in the form it corresponds to the general equation CI -{- V= C, Art. 279, and expresses the conservation of energy in its two mechanical forms. The total energy of vibration Jwyw^' varies as the square of the amplitude. It is all in the potential form at A and at B, Fig. 91, and it is all in the kinetic form when the body is passing through O. 2/2 MOTION PRODUCED BY VARIABLE FORCE. [Art. 340. Motion Produced by a Component of the Force. 340. It is a peculiarity of the law of force we are discussing that a resolved part of the force follows the same law, and there- fore produces a motion of the same kind. Thus, suppose the particle My Fig. 92, constrained to move in the smooth line AB^ to be acted upon by a force, directed toward the fixed . point C (not in AB)^ and propor- 7M ' tional to its distance from C, which we shall denote by r. Let CO = b be the perpendicular from C upon the line AB^ and take O as the origin of distance for the line AB. YiG^ <)2. Then, /i denoting, as in Art. 335, the force acting on the unit mass at a unit's distance, the force acting upon M in the direc- tion MC is mjir. The component in the direction MO is there- fore m}^r cos OMC\ that is to say, nt^xs acting toward the point O. Hence the force in the line of motion is precisely the same as that considered in Art. 335, and the motion is the same as if the particle were free, and O were the centre of an attractive force proportional to the distance ; that is to say, it is harmonic motion. 341. Let C be the point symmetrically situated to C on the other side of the line AB\ then, if the body were attracted to each of the points C and C by forces each equal to \mjAry the force in AB would be as before nijxs^ but there would be no force transverse to AB^ so that the constraint of the line might be dispensed with, and the same harmonic motion would still exist. If the force attracting the body to each of the points C and C' were constant and equal to \m}J.by the force in AB would be mub- ; and, since the fraction — , for small values of s, differs very r r § XVIII.] APPROXIMATE HARMONIC MOTION. 2/3 little from its maximum value unity, we shall have, for small oscillations, very nearly the same harmonic motion as for the force considered in the preceding articles. Again, whatever function of r the force may be, if P is its value at O^ the attrac- tion to C and C will, for small oscillations, differ very little from P\ hence the motion will still be very nearly harmonic. If P is the tension of a light elastic string of length CC = / = 2^, and m is the mass of a bead at the middle point (7, we shall have, by giving to P the value assigned to the constant force above, P = \m^b — \m^l^ whence /i = ^ • Substituting this value of p. in the equation of Art. ZZ^* ^c have for the period of small complete oscillations.* Repulsive Force Proportional to the Distance. 342. Let us now suppose the force to be repulsive and pro- portional to the distance from a fixed point in the line of motion, * In the case of musical strings, the mass is uniformly distributed throughout the length; each particle has harmonic motion, but the period differs from that, when the mass is concentrated at the centre in the ratio of 2 to IT. Thus, \ml and if n is tne number of vibrations per second and w the weight of the string per linear unit (so that mg ■=. wt)^ we have Hence, the vibration number is inversely proportional to the length, and to the square root of the mass per linear unit, and directly propor- tional to the square root of the tension. 274 MOTION PROD UCED B V VARIABLE FORCE. [Art. 342. SO that we have only to change the sign of ^ in the differential equation of Art. 335 ; thus, vdv = pLsds ; (i) whence, putting pi z= n* a.s before, v' = n's' + C. The second integration for this force takes different forms, according as this first constant is positive or negative, the cases corresponding to radically different kinds of motion. Thus, i/ there is a position of rest^ we may take for initial values J = d5, » = o, / = o, and the first integral equation becomes v^ = n\s^-a') (2) t the constant of integration being negative. Substituting in ds = vdty and separating the variables, '^' =«dt (3) Integrating, log[^ + |/(x' - a')] = nt+ log a, . . . (4) the value of the constant being determined by the initial circum- stances. 343. To express s directly in terms of /, we have t + j/Cx- - a') _ <", §XLVIII.] REPULSIVE FORCE, 275 of which the reciprocal is a '^' * whence s — ^a{e*^ 4- ^-«') = a cosh «/. .... (5) Differentiating with respect to /, V = ^na{e*'^ — e'"*^) = na sinh nt, .... (6) In this case, the body recedes indefinitely from the centre of force, and when moving toward the centre a is its least distance. 344. In the second case, the body may pass through the centre of force, and it is convenient to take for initial circum- stances J = o, V = na^ / = o. The first integral then takes the form v' = n\s' + a'); whence ds = nt. Integrating, and determining the constant, log[^ + V{s' + a')-] = ni + log a, from which, proceeding as before, s = ^aie*"^ — (-*"■ ) = a sinh «/. Differentiating with respect to /, f; = ^na(g''^ -j- s|t^'' ...... (4) i^s ds __ 7c g XVIII.] PROPORTIONAL TO THE INVERSE SQUARE. 279 for which purpose assume 4/j = y<2 tan ^, that is j = a tan" ^, whence 4/(j + tf ) = |/a sec ^, and «- ^j= 2tf tan ^sec' B dB, Thus equation (4') becomes 2a tan* (9 sec B dB ^ l^ dt. Integrating, and determining the constant so that, when / = o, 6 = o and therefore j = o. si 2/i ^ ^ I -}- sin ^ — . I = a tan c/ sec u ^ a log tf ^ cos C7 whence This is therefore the time of describing the distance s from the centre, when the velocity is approaching the limiting value \2JX SJa' 349. Finally, the integration takes still another form in the intermediate case when the constant in equation (2), Art. 346, is zero. We now have ^ = 'i (3") so that the velocity never vanishes, but has zero for its limit when s = 00, If we suppose the body to be receding from the centre of force, i/sds=^(2M)di^ (4") 28o MOTION PRODUCED BY VARIABLE FORCE. [Art. 349. Integrating, and supposing s =■ o when / = o, we have Sy»=4/(2yu)/; therefore /= ^-s\ (5") is the time of describing s from the centre when the velocity does not vanish for any finite distance but has zero for its limit. This value of / is in fact the limiting value of that given in equa- tion (5') when a is made infinite. The Gravitation Potential. 350. The law of force considered above is that of the attrac- tion of gravitation when the attracting body is regarded as a fixed centre of force. The force acting on the mass ni is The potential function for this force is in which is the value of the work-function V i^hxi. 277). The constant C may be so taken as to make the absolute value of the potential zero for any particular point. But, since there is no inconvenience in negative values of the potential, it is in general best to take C = o, so that the potential (for positive values of s only, see Art. 346) is defined by which is zero at infinity and negative for all finite values of s. § XVI 1 1.] THE GRAVITATION POTENTIAL. 28 1 351. Now, multiplying equation (3), Art. 346, by Jw, it may be written a Hence, in the first of the above cases, the total energy is a nega- tive constant equal to the value of the potential when the body is at rest. In the second case, that of Art. 348, equation (3') gives, in like manner, a Hence the total energy is in this case a positive quantity : The kinetic energy at every point exceeds the value of the negative potential, and the body may recede to infinity with a limiting velocity In the third case, the total energy is zero, the kinetic energy having at every point a value numerically equal to the potential energy. 352. The same law of inverse squares holds for the attraction to the centre of a sphere composed of matter either of uniform density, or of variable density which has the same value at points equally distant from the centre. Assuming the earth to be such a sphere of radius Ry it is convenient to take the potential so as to vanish at the surface. We therefore add to the expression for U the positive constant —^^ equal to the negative value assigned in Art. 350 to U at the surface, and thus obtain R{R + h) for the potential at the distance from the centre s =^ R -\- h\ that is, at the height h above the surface. 2^2 MOTION PRODUCED BY VARIABLE FORCE. [Art. 35^. The acceleration at the surface is ^ = -7^3: hence, substituting }x = gK^y we have When h is small compared with R^ this is nearly equivalent to Wh^ which is the potential energy at the height h when the force W is regarded as constant, and the potential is so taken as to vanish at the surface. Putting // = 00, the expression above gives mgR for the poten- tial at an infinite distance. Accordingly, this is also the value of the kinetic energy of a body falling with no initial velocity from an infinite distance to the surface of the earth. Putting mgR = \niv'^y we have for the corresponding velocity V = ^/{2gR\ The numerical value of this velocity is about 7 miles per sec- ond; hence, if a body were projected upward with such a velocity and free from any other resistance, it would escape from the sphere of the earth's attraction. EXAMPLES. XVIII. 1. A particle moves in a straight line subject to an attraction proportional to s-^. Show that the velocity acquired in falling from an infinite distance to the distance a is equal to that acquired in falling from rest at ^ to a distance \a. 2. If a particle be attracted to two centres of force propor- tional to the distance with intensities yu^ and /<, , show that the resultant at all points is directed toward the weighted centre of gravity of the given centres, is proportional to the distance and has an intensity /^, + Z^, at a unit's distance ; and hence that the motion will be harmonic. Show also that the property extends to any number of centres of force proportional to the distance. (See Art. 67.) ^* XVIII.] EXAMPLES. . 283 3. Let a weight W \izxig from an elastic string without weight, stretching the string to a length exceeding its natural length by e. Assuming Hooke's Law, show that, if the weight be now drawn down a further distance less than e and then released, its vertical motion will be harmonic, and find the time of a complete vibration. I e 4. Let a light elastic string be stretched to an additional length ^, the tension being /*, and let it carry a bead of mass m at its middle point. Show that the motion of the bead, when displaced in the direction of the string and released, is harmonic, and find the time of a complete vibration. ) em 5. A heavy body, attached to a fixed point by an elastic string whose natural length is ^, hangs freely, stretching the string to an additional length e. It is drawn down through a further dis- tance r > ^ac -\- ^', show that the body will rise until the string is stretched vertically upward to the length a ■\- Xy where x is determined by the equation (^ + ^)« = r' - 4.^^. 7. Find the time it takes the back-weight in Ex. 6, XIV., to rise, when the hand-lever is suddenly pulled back and locked. Find also its final velocity, and verify that its final kinetic energy is equivalent to the extra work done in the sudden motion. 3 8. A particle of unit mass is attached by a straight elastic string to a centre of repulsive force equal to }x times the dis- tance; the string is at first of its natural length a^ and its tension when stretched one unit is k. Supposing// < ^, find the greatest = 5 /Jl. 1 I: 284 MO TION PROD UCED B V VARIABLE FORCE. [Ex. XVIIL distance from the centre of force which the body will reach, and the time it will take to return to its first position. k-\- pi 27T a ' 9. A perfectly flexible rope whose weight is le/ per linear unit and length 2/ rests in equilibrium on a smooth peg. If now one end be raised a distance a and then released, find the time in which this end will rise to the height x above its original posi- tion, and the tension at that instant of the rope at the point where it passes over the peg. 4 / , x+ i/(x' - a') t -x' _,Og ; „,_^ 10. If, when in equilibrium, the rope in the preceding example had been given an initial velocity 7'o, how long would it take to drop from the peg ? I / j/j/g) + Vfe + ^^q) 11. In the case of a force inversely proportional to the square of the distance, if /o, ^o, ^'o denote the acceleration, distance and velocity at any point, show that the motion belongs to the case considered in Art. 347, Art. 348 or Art. 349 according as 2'1 is less than, greater than or equal to 2/0^0- 12. Show that the time of descent from rest through the first half of the distance to a centre of attraction varying as (distance) ""* is to that through the last half as ;r -J- 2 : ;r •— 2. 13. If h be the height due to a given velocity at the earth's surface, supposing the attraction constant (see Art. 297), and If the corresponding height, when the variation of gravity is taken into account, prove that I _ I __ I 14. Assuming the attraction of a sphere upon a particle to be the same as that of the entire mass supposed concentrated at the centre, show that, if a sphere of the same density as the earth attract a free particle placed at a distance from its surface bearing § XVIIL] EXAMPLES. 285 a given ratio to the radius, the time of falling to the surface will be the same as that of a particle falling to the earth's surface from a distance bearing the same ratio to the earth's radius. 15. If the intensity of an attractive force be -;j, show that, when « > I, the velocity acquired by falling from an infinite dis- tance to the distance a is \n — I and that, if the potential is so taken as to vanish at infinity, its value at a is the negative of the corresponding kinetic energy. 16. In the preceding example, if n < i,show that the velocity acquired in falling from rest at the distance a to the centre of force is \ I — « and that the potential maybe so taken as to vanish at the centre; its value at the distance a being then the kinetic energy corre- sponding to this value of v. 17. If, in Ex. 15, n = i, show that the potential is infinite both at infinity and at the centre, and, if so taken as to vanish at J = «, is s U = mfJi log a Also, being given that I e *V^ = ii^^> fii^, and also PV putting — for m, we have the following expression for centrifugal force : _ ^^' _ ^<*^' _ 47r*aW _ 4n'7r*a W These expressions show that, while the centrifugal force is in- vgrsely proportional to the radius for a given linear velocity, it is for a given angular velocity, or for a given time of revolution, directly proportional to the radius. 290 MOTJON PRODUCED BY VARIABLE FORCE. [Art. 360. 360. c ^ . \ 7i \ ^- """aAM. u W^ The Conical Pendulum. In the case of the heavy particle moving in a horizontal circle, the resultant of the weight (which in Art. 357 we supposed neutralized by the resistance of a smooth horizontal plane) and the centrifugal force will lie in the vertical plane which passes through the radius of the horizontal circle. In Fig. 93 let this resultant meet the vertical line through the centre in C, and denote CO by //. The triangle OCM will serve as a triangle of forces, and gives Fig. 93. whence h^ ^■n a JT'' ^7t (i) Thus, for a given time of revolution, the height h is independent not only of the weight, but of the radius of the circle described. 361. The forces of constraint may now be completely replaced by the tension of a string or rod connecting the particle J/ with C. This string will, in the revolution, describe the sur- face of a right cone. For this reason, the arrangement is called the conical pendulum. Denoting the length CM by /, and the angle OCM by 6^ we have h -==■ I cos 6. Also, in the case of rapid motion, let n be the number of revolutions per second, so that nT = 1. Then, from equation, (i) we have cos^ = ^'=-^, .... 4/7r' 4«V;r' ' (2) which determines for a given rapidity of motion. The tension of the string is then W stc 0. The principle of the conical pendulum is employed for the §XIX.] THE CONICAL PENDULUM. 291 regulation of the speed of a shaft in the "governor" of the steam-engine, in which the increase of the angle B beyond the desired limit is made to operate a valve cutting off steam. It is also used in the clockwork for driving a telescope equatorially mounted, in which case the increase of d causes sufficient friction of the body M against a metal ring (whose inner surface is the desired circle of revolution) to produce the necessary retardation. The Centrifugal Force due to the Earth^s Rotation. 362. Let NQS, Fig. 94, be a section of the earth supposed a sphere, and Q a. point on the equator. By Art. 359, the cen- trifugal force acting on a unit mass at Q by virtue of the earth's rotation is -'^ (0 /='- T' where T is the number of seconds in the sidereal day, and ^ the number of feet in the radius of the earth. The value of /is thus found to be 0.1113, which is about ^^^ of the observed value, namely 32.09, of g at the equator. This force tends directly to diminish the weight of the body. Hence, denoting by G the value which g would have if there were no rotation, G = g -^ f = 32.20, and the centrifugal force diminishes the weight of a body at the equator by about -^\-^. 363. Let 7^ be a body at a place whose latitude is A, and draw FD perpendicular to the axis ; then F describes a circle whose radius is FD = F cos A. Hence, by Art. 359, the centrifugal force on a unit of mass ^n^F cos A at F is ^, (i) reduces to , which by equation D — y<^ ; / cos A. This force acts in the direction DF as represented in the figure. Resolving it into rectangular com- S Fig. 94. 292 MOTION PRODUCED BY VARIABLE FORCE. [Art. 363. ponents along, and perpendicular to, the earth's radius, they arc / cos" A, and / cos A. sin A. The first of these, which tends directly to diminish the weight, decreases with the increase of latitude, and causes an increase in the value of g as we approach the poles. The second produces a deflection in the direction of gravity, which is in fact the resultant of the earth's attraction and the centrifugal force. Since the sea- level is everywhere perpendicular to this resultant, the centrifugal force causes the earth to assume a form of equilibrium which has been proved to be a spheroid, the polar diameter being smaller than the equatorial. This, in accordance with the law of gravita- tion, still further increases the difference between the values of g at the equator and the poles. The General Expression for the Normal Acceleration. 364. The hodograph may be used to find the expression for the normal component of acceleration in the general case as well as in that of uniform circular motion. In Fig. 95, the right-hand diagram represents the curve in which the particle P moves, denoting the in- clination of the tangent to a fixed line. The left-hand diagram is the hodograph, F"iG. 95. which we refer to polar co- ordinates, the initial line being in the direction = o. Then, by the construction of the hodograph, the polar coordinates of P\ the point corresponding to Py are r — Vy 6 = (f). We have seen that the acceleration a oi P \s the same as the velocity of -/"; hence the tangential and normal components of the acceleration are the resolved parts of the velocity of P' in § XIX.] EXPRESSION FOR THE NORMAL A CCELERA TION 293 the direction of, and perpendicular to, the radius-vector r. Hence (Diff. Calc, Art. 317) they are dr dv rdB vd(f> The first of these expressions is the value of the tangential ac- celeration already given in Art. 355. The normal acceleration is more conveniently expressed in terms of the radius of curvature at the point P, which is (Diff Calc, Art. 332) ds , defy ds V /o = —7 ; whence —r- = — - = — . d(p dt pdt p Substituting in the expression above, we have Normal acceleration = — . 9 The result found in Art. 40, for the special case in which v and a are constants, agrees with this general expression. 365. If the body is constrained to move in a smooth fixed curve, and there is no external force acting upon it except the reaction of the curve, there will be no tangential acceleration, and therefore v will remain constant. The pressure on the curve caused by the normal inertia, or centrifugal force, will now be -i or . 9 g9 If the curve is horizontal and the weight of the body is also regarded as acting, this is, of course, only the horizontal com- ponent of the action between the curve and the body. If the curve is rough, the friction caused by this pressure will be a tangential force causing a retardation. The equation of the motion will therefore be dv V* which is directly integrable when p is constant. See Ex. 1$. 294 MOTION PRODUCED BY VARIABLE FORCE. [Ex. XIX. EXAMPLES. XIX. 1. A cord two feet long passes at its middle point through a hole in a smooth horizontal table. It carries at its lower end a weight of two pounds, and at the other a weight of one pound. With what velocity must the latter weight revolve in a circle to prevent the lower weight from descending? V = i/(2g) = 8 Vs. 2. If, in the preceding example, only i of the cord lies on the table, how many revolutions must be made per minute to sustain t*he weight ? io8. 3. With what number of turns per minute must a weight of 10 grammes revolve on a smooth horizontal table, at the end of a string half a meter in length, to cause the same tension that would be caused by a weight of one gramme hanging vertically at a place where the value of ^ in meters is 9.81 ? About 13.4. 4. A weight of IV pounds is connected by a string of length a to a. fixed point of a smooth horizontal table; the string can only support a weight of ^j pounds. What is the greatest number of revolutions per second which W can make without breaking the string ? _ ^ I ^i^ "^^nJ- IVa 5. A string can just carry one pound. What is the shortest length of this string which can connect a bullet weighing one ounce and moving with a velocity of 40 feet per second to a fixed point ? 3^ feet. 6. If the masses of the bodies in Ex. II. 24 are m and ;//, the length of the string /, and the angular velocity go, show that the tension of the string is m -\- m! 00 and find its value, supposing the bodies to weigh respectively i and 5 pounds, the string to be 3 feet long, and 200 revolutions to be made per minute. 34-32 pounds. 7. A stone weighing one pound is whirled round by means of a string so as to describe a horizontal circle in a plane 2 feet be- § XIX.] EXAMPLES. 295 low the point of suspension. Find the time of revolution and also the tension, / being the number of feet in the length of the string. I 2 sec; 4/ pounds. 8. A railway curve has a radius of a quarter of a mile, and trains are to run over it at the rate of 20 miles an hour, the gauge being 4 ft. 8 in. How much should the outer rail be raised above the level of the inner one to prevent lateral pressure on the rails ? About \\ in. 9. A particle rests in equilibrium at any point of a bowl in the form of a solid of revolution rotating once in 7" seconds about its axis, which is vertical. Show that the form is that of the gT paraboloid whose latus rectum is ^. 10. The length, weight and period of a conical pendulum being given, show that the tension of the string is independent of the value of ^. 11. A weight attached to a fixed point by a string describes a horizontal circle, the string being inclined 60° to the vertical. Show that the velocity is equal to that due to a height equal t(? three-fourths of the length of the string. 12. A plummet is suspended from the roof of a railway car. How much will it be deflected from the vertical when, the train is running 45 miles per hour over a curve of 300 yards radius ? 8-^ 36'. 13. Assuming g = 32, and the earth's radius 4000 miles, in what time could a body revolve freely round the earth close to its surface ? i^ 25"" 4'. 14. Supposing the earth a sphere of 4000 miles radius, find approximately the greatest value of the deviation of gravity from the direction of the radius. 6'. 15. A particle without weight is projected tangentially with the velocity Vo into a rough circular tube of radius a, yu being the coefficient of friction. Show that the space described in the time /is s — — log -. 296 MOTION PRODUCED BY VARIABLE FORCE. [Ex. XIX. Show also that the times in which successive revolutions are made are in geometrical progression, and that, when the particle has the velocity v^ it cannot have been moving more than 7— seconds. 16. Show that the hodograph of the motion in the preceding example is the logarithmic spiral r = v^e M» 17. A smooth tube rotates with uniform angular velocity co about a vertical axis intersecting it at right angles. A particle in the tube at the distance a from the axis is released Show that its distance r at the end of the time / is r — \a{e^* + ^ - "O —a cosh a?/, so that the polar equation of its path is r = ^ cosh B, XX. Constrained Motion under the Action of External Force. 366. .When a body acted upon by a force is constrained to move in a smooth curvilinear path, the tangential component of the force is resisted only by the corresponding component of the body's inertia. The change of velocity is therefore determined solely by this component of the force, exactly as in the case of rectilinear motion. That is to say, v is determined by the inte- gration of vdv = fds, where / is the tangential force acting on a unit mass expressed in terms of the distance measured along the arc from some fixed point. The position of the body at any time / may then be de- termined by the integration of vdt — ds. 367. Resolving forces normally to the curve, we have a con- { §XX.] HEAVY BODY ON SMOOTH VERTICAL CURVE. 297 dition of kinetic equilibrium which involves three forces, namely, the resistance of the curve, the normal component of the external force, and the centrifugal force, or normal component of inertia. This last is given by the expression P (Art. 365) after v has been determined, as explained in the pre- ceding article, by means of the tangential force. It follows that the resistance necessary to keep the body in the given path may reverse its direction, and if the body moves on the surface of a fixed solid, so that it is free to leave the curve on one side, it will do so at the point where the resistance of the surface vanishes; that is, where the normal component of the force is equal and opposite to that of inertia. It is obvious that at such a point the curve of constraint has the same curvature as the free path in which the body subse- quently moves. Motion of a Heavy Body on a Smooth Vertical Curve. 368. In the case of a body sliding down a smooth curve in a vertical plane, let us refer the curve to rectangular axes, that of X being horizontal. Let My Fig. 96, be the posi- tion of the body whose mass is m and weight W ] then, de- noting the inclination of the tangent to the axis of x^ W sin is the tangential force and / = ^ sin 0, V y ./.-. K/ ^^^ '" h \ - X Fig. 96. the acceleration down the curve. Hence, if s is measured as usual in the direction determined by 298 MOTION PRODUCED BY VARIABLE FORCE. [Art. 368. the angle (that is, up the curve in the diagram), the equation of motion is or, since sm = -f , as vdv = ~ g sin (f>ds vdv = — gdy, (i) Let A be the point at which the body is at rest, and let h be the ordinate of A ; then, integrating, and determining the constant in such a way that v =■ o when 7 = ^, we have v^ — 2g{h—y) (2) 369. Introducing the factor w, equation (2) may be written ^mv" -}-Wy= Wh. Regarding the potential energy of a body as zero upon the axis of AT, this expresses that the sum of the kinetic and potential ener- gies is constantly equal to the initial energy, which is all in potential form at A. The motion will be continuous until the body reaches a point on the same level with A, in which case it will come to rest, and then, unless the tangent at that point is horizontal, it will return and again come to rest at the point A. The velocity at any point is, by equation (2), that due to the distance of the point below the level of A. Hence, if the velocity at any point is known, this level may be constructed even when the curve lies entirely below it. It is sometimes called the level of zero-velocity^ 2LT\d corresponds to the directrix in the case of free parabolic motion. (See Art. 321.) 370. Supposing the body to move on the surface of a solid, the centrifugal force will, in any position of the curve which \s convex as viewed from above, diminish tlie pressure upon the surface, and the body will leave the curve when the centrifugal force be- § XX. ] THE C YCL 01 D A L PEND UL UM. 2 99 comes equal to W co% 0, the normal component of the weight; that is, when yv cos = — - . 9 Substituting the value of v* found in Art. 368, this equation be- comes \p cos (p = h — y (i) If we draw the circle of curvature, which in this case will lie below the curve, it is easily seen that 2/9 cos is the vertical chord through M oi this circle. Hence, by equation (i), the point at which the body will leave the curve is that at which the vertical chord of curvature is four times the distance of the point below the level of zero-velocity; or, what is the same thing, the point for which the centre of curvature is three times as far as the point itself is from this level. The parabola in which the body subsequently moves is that of which the level of zero-velocity is the directrix. It follows that the centre of curvature for any point of the parabola is three times as far as the point itself is from the directrix. The Cycloidal Pendulum. 371' To determine the position of the body moving in a smooth vertical curve at a given time, it is necessary (see Art. 366) to integrate the equation ds = vdt, which, by equation (2), Art. 368, is in this case 7(^) = ^<^->^'- To integrate this^ must be expressed as a function of s (or y and ds in term^ of some other variable) by means of the equation of the curve. 300 MOTION PRODUCED BY VARIABLE FORCE. [Art. 371. For example, suppose the curve to be an inverted cycloid. The equations of the curve, the vertex being at the origin, are (Diff. Calc, Art. 290) X = a(tp + sin tp)^ y = a{i — cos ip) = 2a sin^ itp; whence dx = a(i -\- cos tp)dip dy ■= a sin ^ ^^, and, from ds =^ ^ (dx^ + ^y) [• (2) ^90) c \-^ -f B\ m/^ ^^>^^_ ^J^^ X Fig. 97. ds = 2a cos ^tp dtp (3) Hence, if s is measured from the origin, we have, by integration, s = 4a sin ^ipf . (4) and therefore, since _>' = 2a sin' ^^, (5) y = Sa Substituting this value of y in terms of s, the differential equation above becomes or 4 d^ ds (6) in which h stands for the ordinate of the point A where the body is assumed to start from rest. This equation is of the same form or * as equation (3), Art. 336, the value of /i being — . Hence the 4a ^ XX.] THE CYCLOIDAL PENDULUM. 301 motion as measured along the arc is harmonic; and, by Art. -ifZ'^i the time occupied in passing from A to jB and returning to A is = 4- J J. Since this is independent of the position of A, the time of vibra- tion is the same for all arcs of the cycloid ; the curve is for this reason sometimes called f/i€ Tautochrone. 372. In the equations above, a stands for the radius of the generating circle of the cycloid. The evolute of the cycloid is an equal cycloid, having its vertices at the cusps of the given cycloid as represented in Fig. 97 (Diff. Calc, Art. 357). Hence, if a heavy particle be suspended from Cby a string of length 4a, and in its vibration be made to wrap upon solid pieces having the form of the cycloidal arcs CZ>, CE^ it will describe the cycloid AOB. Such an arrangement is called a Cycloidal Pendulum. Putting / = 4a for the length of the string, and r = J7', we have --1 for the time of passing from A to B^ which is the time of a single swing, or beat, of the cycloidal pendulum. If the cycloidal pieces be removed we have the simple pendulum ^ the particle describing the circle of curvature of the cycloid at the vertex. It is hence evident that for small oscillations t is very nearly the period of the beats of the simple pendulum. 373. We found in Art. 337 that harmonic motion resulted from an attractive force proportional to the distance measured from a fixed point of the path. Accordingly, the harmonic mo- tion, in this case, results from the fact that the tangential force (which alone produces the motion) is proportional to j, the dis- tance measured along the path, and acts toward O. For this force is/= —g sin 0, and by equations (2), (3) and (4), Art. 371, dy s P's sin = ^ = sini?^ = — : hence /= — — • ds "'^ 4a -" 4a 302 MOTION PRODUCED BY VARIABLE FORCE. [Art. 374. Motion in a Vertical Circle. 374' Let C, Fig. 98, be the centre and a the radius of a circle in a vertical plane. Take (9, the lowest point, as the origin; let F'be the velocity at 6? of a heavy particle moving smoothly in the circle, and B the angle OCM which defines the position of the particle M at the time / reckoned from the instant when the particle was at O. The acceleration down the curve is g sin B acting in the opposite direction to that in which s is measured. Hence the equation of motion is df = ~ g sin 6. This is equivalent to vdv = — ^ sin ^ ds, and the first integra- tion gives, as in Art. 368 (since = ^), v' = 2g(/i - y). (i) The constant of integration /i is the height due to the velocity when y = o; that is, Let Olf, Fig. 98, be this height, and draw the level of "zero-velocity." In the diagram we have assumed /i > 2a the diameter of the circle, so that M will, in this case, move continuously around the circle. 375. Since s = «^, ds = adB^ and equation (i) gives for the second integration v=^— = 4/(2^) ^{h -y)\ V{2g) whence 'dt = dO («) §XX.] MOTION IN A VERTICAL CIRCLE. 303 in which y is to be expressed in terms of B. Draw MD perpen- dicular to CO^ then V = OD z=z a — a cos 6^, or y — 2a sin' \i) (3) Making this substitution, and putting ^ for \B^ equation (2) may- be written a 4/(1- jsin'^fc) 2a or, putting ~r = k^^ and integrating, V __ j- dtp . . 2a ], |/(i-/c«sin'^')' ' ' ' W the lower limits being the value of tp which corresponds to / = o. The integral is Legendre's Elliptic Integral of the first kind, which he denoted by F{ip, k). tp is called the amplitude, and K the modulus of Eit/.^ k). Legendre considered the integral to be in its standard form when k is less than unity (as it is in the present case), and for this form he published tables of its nu- merical values. (Legendre's Fonctions Elliptique^ Vol. II.) 376. As mentioned in Art. 374 the particle in this case makes complete revolutions in the circle to which we may imagine its motion to be constrained by means of a rod of length a connect- ing it with the fixed point C. Denoting by T the time of a com- plete revolution, \T \s, the value of / when y = 2a; that is, by equation (3), when = Jtt. Therefore i:^_ r # ^a 04/(1 - /c'sin'V)"'^* .... (5) This definite integral, which is Fi^n^ k), is called tAe complete elliptic integral^ and is usually denoted by K. Separate tables of 304 MOTION PKODUCED BY VARIABLE FORCE. [Art. 376. the values of K for different values of k are given by Legendre, and also in Bertrand's Calcul Integral^ p. 714, Greenhill's Elliptic functions, p. 10, etc. When K = o, the value of X, see equation (5), reduces to ^tt. Hence, when /i increases indefinitely so that the limit of k is zero, equation (5) gives VT — 2a7T, as it should, since in the limit the velocity is evidently constant. 377* Let us next suppose h < 2a, then v in equation (i) vanishes whenj^' = /i. The level of zero-velocity, cutting off 0/f = /i on the vertical diameter, will now cut the circle in A and B, Fig. 99. The particle will come to rest at A and the motion will be one of oscillation between A and B. The relation between / and or 2tp is still expressed by equation (4), but the value of K will be greater than unity, so that the integral will not be of the standard form used by Legendre. If r denote the time occupied by a single oscillation, that is, the time of moving from B to A, the maximum value of (9, corresponding to / = -^r, will be a, the angle ACO or BCO in the diagram. 378. But, returning to equation (2), Art. 375, if we first elim- inate instead of y, the use of another variable is suggested, which will reduce the expression for / to a complete elliptic in- tegral of the standard form. From equation (3), we derive Fig. 99. whence e=z dS^ 2 sm . Vy il{2a)* dy Vy Vi^a ^y) Substitution in equation (2) gives § XX.] THE SIMPLE PENDULUM. 305 ^(^^ . sin 0, jy^{/i—y)= |/y^.COS0, and dy = 2/1 sin

&» sin' + "^k' sin* + _1i1jl| ^. sij^. _|_ . . . 2.4 2.4.6 Integrating each term in the definite integral by the formula f' sin- = '•3-5.--(^«-i) rt_ Jo 2.4.6... 2« 2 §XX.] THE SECONDS PENDULUM. 307 (Int. Calc , Art- 86), we have Therefore, putting / in place of ^, equation (i) gives where / is the length of the pendulum, and \(x. the quarter angle of swing. The Seconds Pendulum. 381. If we put e for the sum of all the terms but the first of the series, equation (2) becomes --■J> + ^) (3) which, when e = o, reduces to the expression for the cycloidal pendulum, the motion of which is shown in Art. 371 to be har- monic, and is said to be isochronous because the time is indepen- dent of the amplitude of swing. This is not true of the motion of the simple pendulum, but that motion is said to be approxi- mately isochronous because e involves, not the first power, but the square and higher powers of the small quantity sin ^a* 382. Putting r = I, and e = o, the first approximation to the length of the pendulum which beats seconds when a is small is ^=1^' .• • . (4) which is usually given as the length of the seconds pendulum. It * As remarked in Art. 373, the tangential force, in the case of the cycloidal pendulum, is proportional to the arc, measured from the lowest point. In the present case, it is proportional to sin 6, and there- fore very nearly proportional to the arc when 6 is small. 308 MOTION PRODUCED BY VARIABLE FORCE. [Art. 382. is really the limiting length of the pendulum which beats seconds, when the arc of swing is indefinitely decreased. Now, denoting by / the length of the pendulum which beats seconds when swinging through the arc 2a^ we find, by piittng T = I in equation (3), ^ = ^VtI.V =" "^(^ - 2e + 36' - . . .), in which (see Art. 381) e = - sin' \a ~\- ~- sin* \a ■\- , , , 4 04 Substituting, we have /= zfi — -sin' Jo- —^ sin* !«:—.. .) . . (c) V 2 32 / for the corrected length of the seconds pendulum designed to swing through the angle 2a, 383. By equation (3), the actual time of the beats of a "sec- onds pendulum," whose length is Z, when swinging through the arc 2^, is I + e; hence, if N is the number of seconds in any defi- nite interval, for example, in a day, the actual number of beats will be ■ N = iV^(i - e 4- e' - . . .). i + e We may therefore take eN or \N sin" \a approximately for the number of beats lost in TV" seconds when the amplitude of swing is considerable. It follows that, if the length is already adjusted to the swing 2a^^ the number of beats lost when the swing is 2a^ is (e^ -- e^N or iA^(sin' \a^ — sin' \a^^ § XX.] COMPARISON OF SMALL CHANGES IN l^ n AND g, 309 which may also be written in either of the forms ^iV(cos a^ — cos a,) or ^N sin \{a^ -\- a^) sin ^{a^ — a^ Comparison of Small Changes in /, n and g. 384. If n denotes the number of beats of a pendulum whose length is / in iV seconds, we have nr = JV^ whence, from equa- tion (2), Art. 381, Supposing the angle of swing, and therefore e, to be unchanged, we have, by logarithmic differentiation, when g and / vary, * dn _dg __ dl^^ ^ ^ ^ . ^ ^ ^ n 2P' 2/ Hence, if Al is a small error in the length of a pendulum intended to beat n times in iV^seconds, and Ati the consequent change in n (g remaining unchanged), we derive ^« = - -^ T' (^) the negative sign showing that n decreases as / increases. If An is known by observation, the error in / is given by ^/=-^ (3) For example, if the pendulum is intended to beat seconds, and n is the number of seconds in a day, An is the number of beats gained in a day ; then equation (3) gives the approximate error in /, which is too short if An is positive. 3IO MOTION PRODUCED BY VARIABLE FORCE. [Art. 385. 385. Again, for a small variation in g while / remains un- altered, equation (i) gives ^^ = ^ (4) a formula used in determining differences in the values of g by comparing the number of beats in a given time of the same pen- dulum in different localities. Experiments to determine An to be used in this formula are called " pendulum experiments," and are usually made with a seconds pendulum. But this is not necessary, it is only essential that the length should be unaltered, and that n and n -\- An should be the number of beats made in precisely the same interval of time. 386. Since, as mentioned in Art. 352, the attraction of the earth upon a particle is inversely proportional to the square of the distance from the centre, denoting this distance by r, we have ^' S — §• ^» where ^o and R are the values of g and r at the sea-level. By logarithmic differentiation, ^ ~ _ 2— . hence, if Ag is the variation of gravity from ^o for the height Ar = // above sea-level, we have approximately 2go ' but it is found that, when h is the altitude of a place above sea- level, the local attraction of the mountain or table land modi- fies this result very considerably. Hence the result, namely, § XX.] EXAMPLES. 3 1 1 ^ = , of eliminating Ag from this equation by means of n equation (4) is not trustworthy as a means of determining geo- graphical altitudes by pendulum experiments. EXAMPLES. XX. 1. A particle is allowed to slide from any point of a smooth hemisphere. Show that it will leave the hemisphere after describ- ing one-third of its vertical height above the centre. 2. If the particle in the preceding example rests at the top of the hemisphere of radius «, what is the least horizontal velocity that must be given to it in order that it may leave the hemisphere at once? I^(^^)- 3. If a particle starts from the cusp of a smooth inverted cycloid, prove that the pressure at any point is double what it would be if the particle started from that point. 4. Show that in the motion of the cycloidal pendulum the vertical velocity is greatest when one-half the vertical distance has been described. 5. A particle slides off a cycloid in erect position. Show that it will leave the curve when half the vertical height above the base has been described. 6. A heavy body is attached to a fixed point by means of a string 10 feet long. What is the least velocity it can have at its lowest point in order to describe a vertical circle keeping the string taut ? 40 Vs. 7. A particle of weight W attached to a fixed point by means of a string moves in a vertical circle. Determine the tension P of the string at any point, using the notation of Art. 374. _= 2 + 3COS 6^ = — ^ ^. W a a 8. Find the point at which the string becomes slack, and show that the result agrees with the construction given in Art. 370 for the centre of curvature of the parabola. 312 MOTION PRODUCED BY VARIABLE FORCE. [Ex. XX. 9. Show that the time of revolution of the conical pendulum is the same as that of a complete vibration through a small arc of the simple pendulum whose length is h. 10. Show that the motion of E in Fig. 99 tends at the limit when h is small to become uniform circular motion. 11. Prove that the time down the chord to the lowest point of a circle is to the time down the arc, when the arc is small, in the ratio 4 : tt. 12. A seconds pendulum in a railway car moving at the rate of 60 miles an hour on a circular track is observed to make 121 beats in two minutes. What is the radius of the circle? J mile. 13. Find the length of the seconds pendulum, at a place where ^ = 32.2. 39-i5 inches. 14. A clock which should beat seconds was found to lose 2 minutes a day at a place where \jO- = 32.2. How many turns to the right should be given to a nut raising the pendulum-bob, the screw having 50 threads to the inch? 5'4375' 15. Taking the earth's radius to be 6366 kilometers, how many beats a day will a seconds pendulum lose at the top of the Eiffel Tower, whicli is 300 meters in height ? 4-07« 16. A pendulum beats seconds when swinging through an angle of 6°. How many seconds will it lose a day when swinging through 8°, and through 10° ? 11.56 sec. ; 26.35 sec. 17. How much shorter than the "seconds pendulum " is that which beats seconds when swinging through an arc of 20° ? 0.149 i"- CHAPTER X. CENTRAL ORBITS. XXI. Free Motion about a Fixed Centre of Force. 387. We shall next consider the motion of a free particle sub- ject to the action of a force directed always to or from a fixed point, or centre of force, and having an initial motion oblique to the direction of the force. As in g XVIII, we shall in this chapter suppose the intensity of the force to be a function solely of the distance of the particle from the centre of force. The line of the initial motion, together with the centre of force, determines a plane to which it is evident that the motion of the particle will be restricted, because the force has no component tending to move it out of the plane. The path of the particle is therefore a plane curve. It is called the orbit of the particle under the given force, and may or may not be a closed curve according to the law of the variation of the force. 388. In referring the particle to rectangular and to polar co- ordinates in this plane, we shall take the centre of force as the origin and pole, and the initial line coincident with the axis of Xy so that we have the usual relations, X ■=■ r cos B. y =^ r sin 6, Also, if F is the force along r, since d is its inclination to the axis of Xy its components along the axes are X=^FcosO, Y=Fsmd, 314 CENTRAL ORBITS. [Art. 388. The acceleration /, or force acting upon a unit of mass, is by hypothesis a given function of r, the distance of the particle from the centre of force, and may be written /(r). If ni is the mass of the particle, the force is hence we have m in J ^ ' J. •> and the equations of the two components of the motion are Attraction Directly Proportional to the Distance. 389. We shall first consider the case of an attractive force whose intensity is proportional to the distance from the centre, and which, as we have seen in Art. 337, produces harmonic mo- tion when the initial velocity has no component transverse to its direction. Putting f{r) = — ywr, the two equations of motion become d'x d^y each of which is of the same form as that of Art. 335. Hence putting yw = «' as in that article, and denoting by a the value of dx X which corresponds to the component velocity , = o (that is, dt the maximum value of x), we have X =^ a sin {nt -\- C). ^ XXL] ELLIPTICAL HARMONIC MOTION. 315 Similarly, if b is the maximum value of y^ we have J = /^ sin («/ + C"). If we take for the origin of time the instant when a: = ^, cor- responding to i- = ^^ we have C = \n\ and, denoting the corresponding value of C" by a, the equations become X — a cos «/, (i) y = b ^\Tv {nt -\- a) (2) 390. It follows that the motion of the particle is a combina- tion of two harmonic motions having the same period, namely, T = - ^ = — B See Art. 338. To values of / differing by any multiple of this period correspond the same values of x and the same values of y\ hence the particle returns to the same position periodically; therefore its path or orbit is a closed curve, which, as repre- sented in Fig. 100, is inscribed in the rectangle whose sides are 2a and 2^, parallel to the axes, and whose centre is the origin. Elimination of / between equations (i) and (2) obviously gives an equation of the second degree between x and j, hence the orbit is an ellipse. The amplitudes a and b determine the size of the circumscribing rectangle, and the constant a^ depending on the difference of the phases of the harmonic motions, determines the positions of A and B^ the points of contact with its sides. Fig. 100. 31^ CENTRAL ORBITS. [Art. 391. 391. In particular, if d^X , V which is an exact differential equation, giving the first integral dy dx The constant h in this equation has the same meaning as in the preceding article ; for the equation may be written ^ dy dx\ds , %XXl.]y4/i:£A DESCRIBED BY THE RADIUS VECTOR. 321 in which the quantity enclosed in parentheses is the expression for the perpendicular/ from the origin upon the tangent (Diff. Calc, Art. 316), so that the equation is pv = h (3) Now if we join the extremities of the line representing the velocity to the origin,/?' is double the area of the triangle thus formed; hence h is double the area which would be described by the radius vector in a unit of time if the velocity were uni- form; that is to say, double the rate at which area is being swept over by the radius vector. 397. This important theorem was proved geometrically by Newton substantially as follows: Let AB^ Fig. loi, be the line which a moving particle describes by virtue of its velocity in a certain interval of time. Then, if there were no force acting, it would in the next equal interval of time describe the equal line ^^ in AB produced, and the areas swept over by the line joining the particle to any fixed point ^^ S would be the equal triangles ^''>' / A SB and BSc. Now, when the body is at B^ let a force ^„^^'' ^!^ ^'B act, with a single impulse toward 6", giving the body a velocity which would cause its"--.^ /^ to move over the space BD in """--^^ / the interval of time considered. i^"""-^^^ / Then, by the second law of ^"**^ , , ,,, ,, Fig. ioi. motion, the body will actually describe the line BC, the diagonal of the parallelogram BDCc, and the radius vector connecting the body with 5 will describe the triangle ^.SC But since (4) where7> is the perpendicular upon the tangent. For instance, at the nearest vertex where/ = r = a(i — e), we find v= l'w/", the potential is, as in Art. 281, 1/ = ijur^ (taking /// = i), which increases without limit as r in- creases. The circle of total energy, therefore, always exists in this case. For motion in a line with the centre of force, which is simple harmonic motion, it is the circle whose radius is a^ the amplitude, so tliat the body just reaches it; but for the body de- ^XXl.]F7/^Sr INTEGRAL EQUATION OF THE ORBIT, 329 scribing an orbit, as in Art. 390 or in Art. 391, it is the circle whose radius is ^{a^ + F)^ which encloses without touching the elliptical orbit. 408. When the law of variation of the force is such that the potential at infinity is finite, (of which the gravitation potential, Art. 350, affords an example,) the total energy in the orbit may be equal to, or it may exceed, the potential at infinity. In the first of these cases, the circle of total energy or of zero velocity is at an infinite distance, and the velocity at every point of the orbit is that due to infinity. A body projected with such a velocity directly away from the centre of force would never cease to recede from it, and it is pos- sible, although not necessarily the case, that a body describing an orbit with the velocity due to infinity may also so recede, the ve- locity in that case approaching zero as a limit. The relation pv = h shows that, under these circumstances,/ increases with- out limit; that is, the infinite branch of the orbit is of parabolic character. 409. So too, when the total energy exceeds the potential at infinity, or, what is the same thing, when the velocity in the orbit exceeds that due to infinity, the body may, but does not neces- sarily, recede without limit. If it does so recede, its kinetic energy, and therefore its velocity, will approach a finite limit ; and the relation /z/ = /? shows that the perpendicular from the centre of force upon the tangent will also approach a finite limit; that is to say, the orbit will have an asymptote whose distance from the centre is the limiting value of />. The First Integral of the Equation of the Orbit. 410. We have seen in Art. 400 that, supposing m = i,the dif- ferential equation of the orbit may be written in the form d'u P 330 CENTRAL ORBITS. [Art. 410. The first integral is found, as in similar cases, by direct inte- gration after multiplying by 2 -7-7; thus we have Now, since r = — , ^^ =' 7, whence (Art. 404), and therefore equation (2) may be written The form of this equation shows that it is identical with equa- tion (3) of Art. 404, hence the first term is an expression for \v\* the kinetic energy when m ~ i. The constant C is therefore the total energy of a unit mass in the orbit. 411. If the radius of the circle of zero velocity be given, the constant implied in equation (2) may be determined by simply using the corresponding value of u as the lower limit in the indefi- nite integral. For this makes the integral the expression for the potential, so taken as to vanish on the given circle ; and, when this is done, C vanishes for the given orbit. Thus, for example, the equation of the orbit in which the velocity is that due to infinity is (duV , , 2 [Fdu * Accordingly, z/' = — and — = ( —^\ -f- «', as shown in Diflf. Calc, P P \ d^ I Art. 321. n § XXI.] THE APSIDES OF THE ORBIT. 331 The Apsides of the Orbit. 412. A point at which the radius vector is normal to the orbit is called an apse. The corresponding value of r is called an ap- sidal distance^ and is either a maximum or a minimum value. When r is a maximum, « is a minimum, and vice versa ; hence the du apsidal distances may be found by putting — = o in equation (3), Art. 410. That equation may be written in the form B]=^w (0 where f{u):=Uc-U)-u^ (2) Since we are concerned only with positive * values of «, it follows that an apsidal value u^ is a positive root of the equation ^(«o) =0. ....... (3) Again, equation (i) shows that there can be no orbit having values of C and h which make "^^iu) negative for all positive values of u. If ^{u) is positive for all such values, the orbit will in one direction recede to infinity, and in the other direction pass to the centre of force. But, when there is an apsidal value «o» ^(«) will * Unless P is an odd function of r, so as to change its sign, but not its numerical value, when r is changed to — r, negative values correspond to a different law of force. For example, if P is an even function, nega- tive values of r imply a repulsive force with the same law of variation. Compare the laws of force treated in Art. 335 and in Art 346; in the former case s can change sign, in the latter it cannot. 332 CENTRAL ORBITS. [Art. 412 generally change sign as u passes through this value ; hence the circle whose radius is the apsidal distance is the boundary of a region which the orbit cannot enter. The orbit will now pass in both directions to infinity, or to the centre of force as the case may be, unless it reaches another circle on which ^(u) vanishes. In this last case, the orbit is confined to the annular space be- tween these two circles, and their radii are the maximum and minimum values of r. Thus, there cannot be more than two apsidal distances in a given orbit, although there may be any number of apsides. 413. If p is the value of B for an apsidal value u^y it is ob- vious that, for neighboring values of z/, on that side for which ^{u) is positive, there are two real values of B which become equal when u = Uo. Now, by equation (i), we have M = ± '^" Integrating, we have for the equation of the orbit du e = p± f o4/^(«) (4) This equation expresses the two values of S^ which become equal* when u —- Uo. Thus, the third constant of integration, /?, now introduced into the equation of the orbit, determines simply the direction of an apsidal radius vector, and has no connection with the shape of the orbit, which depends solely upon the constants h and C. * If u^ in the integral were not an apsidal value of u, the values of the constant to be used with the upper and lower sign, in a given orbit, would be different. Owing, however, to the multiple values of the in- tegral, the orbit is completely represented when a single sign is em- ployed, whether the lower limit is an apsidal value or not. gXXI.] RADIUS OF CURVATURE AT AN APSE. 333 414. A central orbit is symmetrical to every apsidal radius vector. For, taking /? = o, that is to say, reckoning B from the direction of the apsi'dal radius vector, equation (4) shows that to every point {uy 0) on the orbit there corresponds a point («,— B) also on the orbit, but this is the point symmetrically situated to (u, 6). The form of equation (2), Art. 398, shows that the fourth con- stant of integration, referred to in Art. 399, determines simply the epoch or time of passing a given point of the orbit ; and, if we reckon the time from the instant when the body passes the apse, the times / and — /correspond to the symmetrically situated points. It is obvious, also, that the orbit may be described in either direction, so that the epoch should include the direction of motion as well as the time of passing the given point. 415. In the case of an orbit having two apsidal distances, let u^ be the other apsidal value of u, then, by equation (4), the angle between two consecutive apsidal radii is du ^Ku) This is called the apsidal angle. As we have already seen, it is necessary not only that u^ and «, should be roots of the equation ^'(«) = o, but that the value of the functions should be positive for intermediate values of u. The Radius of Curvature at an Apse. 416. At an apse the radius vector coincides with the perpen- dicular upon the tangent, that is/ = r, and the centrifugal force is directly opposed to, and therefore in equilibrium with, the attractive force F. Let v^ be the velocity, and p^ the radius of curvature at the apse whose distance is r^ , P^ being the corre- sponding value of F ; then, from h = pv, we have v^ = - (0 334 CENTRAL ORBITS. [Art. 416. and, from the expression for centrifugal force, ~ — -^ o » /?0 whence Po = ;^. ....... {^) If the value of p^ thus found exceeds r^ , the centre of curva- ture lies beyond the centre of force, and the orbit lies outside of the circle whose radius is r°; that is, r^ is a minimum apsidal dis- tance. This corresponds to the case in which ^(u) has positive values for greater values of ;', that is, for values of u less than u^. On the other hand, if f>^ is less than r^ , the latter is a maximum apsidal distance. 417. If a second apsidal distance r^ be possible, but none whose value lies between r^ and r^., r^ will be found to be a maxi- mum or a minimum, according as r^ is a minimum or a maximum. If the maximum be greater than the minimum, ?/'(//) will be posi- tive in the annular space between the apsidal circles, and the case is that of an orbit with two apsidal values. If the contrary be the case, ^'(«) will be negative in the annular space and we infer that, with the same values of h and C (namely, those employed in forming the function ?/'), two orbits are possible — one situated beyond the annular space between the apsidal circles and passing to infinity or to another apsidal distance, the other within the smaller circle, and passing to the centre of force or to another apsidal distance. Circular Orbits. 418. For any central attraction depending solely upon the dis- tance, a circular orbit with a given radius is possible if the veloc- ity be properly determined. For this purpose it is only necessary to equate the centrifugal force to the attraction at the given distance. Denoting the required velocity by V^ we have § XXL] CIRCULAR ORBITS, 335 — ^P\ whence V = ^/{rF) is the circular velocity at the distance r. 419. When a circular orbit is regarded as a special case of the orbit described under the given law of force, the given value of r corresponds to the apsidal distance r^ of Art. 416, and /Oq = r^. We must therefore suppose C and h to have been so taken that, from equation (2), Art. 416 (or from h = Vp — Vr)^ V = rlF^, . . . . . . . . (l) while, at the same time, the reciprocal of the given value of r^ is a root of the equation ip{u) = o. The value of /i is therefore determined by the equation just written, and then substituting in the equation i^iu^) = o, we have (see Art. 412) c-=iA.4-K^„ (2) With these values of C and /i, u^ becomes a double root of tp{u) = o, and the function does not change sign as u passes through the value u^. If its value is negative for values of u on each side of u^ , the orbit is the limiting form of orbits having two nearly equal apsidal distances, and lying in an annular space where ^{u) is positive. These orbits, as the space narrows, ap- proximate more and more nearly to the circle. The circle is in this case said to be described with kinetic stability. In the oppo- site case, that is, when the values of ^{u) are positive for values of u on each side of u^ , we have the limiting form of the second case mentioned in Art. 417, in which the annular space (which vanishes at the limit) is one in which ^(u) is negative, so that orbits approximating to a circle are not possible. The circular orbit is in this case described with kinetic instability; that is to say, the slightest change in the direction or velocity of the body will cause the orbit to assume a totally different form; namely, one which has only one apsidal distance equal to r^ , so that (unless "^(u) = o has other roots besides those which have become equal) it will become one which passes either to infinity or to the centre. 33^ CENTRAL ORBITS. [Art. 420. Attraction Inversely Proportional to the Square of the Distance. 420. The most important case of central orbits is that in which the force is an attraction varying inversely as the square of the distance, which is the actual law of gravity. Putting, in equation (i), Art. 410, p = ^ = .«' (so that /i is the attraction acting upon the unit mass at the unit of distance), the differential equation of the orbit is or, putting du' Multiplying by ^-js; and integrating, we may write &T+«"=^' (^) since the constant of integration is necessarily positive. We have, then, du' de = and, integrating again, 6-{-p = sin-' - § XXL] /^C^i^Ci? VARYING AS THE INVERSE SQUARE, 337 or « - ^a = '^ sin (6> 4- /?), (3) which is the equation of the orbit involving three arbitrary con- stants //, c and /?. A maximum value of «, and hence a mini- mum value of r, occurs when 6 -\- fi = ^tt. Therefore, if we take the prime vector (corresponding to ^ = o) in the direction of such an apsidal value of r, we shall have^ = i;r, and equation (3) may be written in the form « = J(i+s Laws. 427. The following laws with respect to the planetary mo- tions were deduced by Kepler from a great mass of observations made by Tycho Brahe, combined with his own conjecture, re- garding the variable distances of the planets. 1. The straight line joining a planet with the sun describes equal areas in equal times. 2. The planets describe ellipses having the sun at a focus. 3. The squares of the periodic times are proportional to the cubes of the mean distances. Kepler was not possessed of correct notions regarding the nature of motion and force, but we have seen in Art. 397 how Newton, upon the basis of the true laws of motion, derived from the first of Kepler's laws the fact that the force acting upon the planets is directed toward the sun. From the second he showed (compare Art. 401) that the force acting upon any one planet varies inversely as the square of the distance. Finally, he showed that it follows from the third law, by means of the result expressed in equation (2) of the preceding article, that, regarding the sun as a fixed centre of force, the same law of variation with the distance governs its action upon the several planets. But, as we shall see hereafter, a slight modification of Kepler's third law is due to the fact that in each case the sun is not a fixed centre of force, but, like the planet itself, is free to move under the mutual attraction of the two bodies. f XXL] TIME OF DESCRIBING A GIVEN ARC. 343 Time of Describing a Given Arc of the Orbit. 428. The relation between r and ^, or equation of the orbit, given in equation (i), Art. 423, involves the constants of integra- tion h and I + / 2. Show also that the circular orbit is stable when « = 2, and unstable when « > 3. I71 the following examples the law of attraction is that of gravity^ namely, F = //«'. 18. Express the function ^(«) in terms of e and h, and thence obtain the apsidal distances. h'tp {u) = J^' (e' -i) + 2Mh'u - h'u\ 19. Determine the radii of curvature at the apsides and at the extremity of the minor axis by means of the normal acceleration. 20. Show that at the mean distance the kinetic energy of the body is a mean proportional between its extreme values, and that at the point where S = 90° it is an arithmetical .mean between the same values. 21. A body is projected with the velocity F in a direction making the angle/? with the prime vector, upon which the point of projection is situated at the distance J^ from the centre of force. Prove that, a being the vectorial angle of the'perihelion and ^^ + sin'/?[_-^ - ij , e cos a = — I. 22. From the relation between the mean and eccentric anom- alies, Art. 429, show that the time of falling from rest on the 350 CENTRAL ORBITS. [Ex. XXI. circle of total energy, that is from the distance 2^z, to the distance r from the centre of force, is ^ ^ _ft r , r -a _^ ^ (2ar - r')' VH' [cos- -F+^^-^-.-=^]- Compare equation (5), Art. 347. 23. Show that, for a central orbit, the hodograph is the curve inverse to the pedal from the centre of force turned through 90". Thence show that the hodograph of the planetary motion is a circle. CHAPTER XL MOTION OF RIGID BODIES. XXII. Action of Inertia in. Rotation. 432. We have seen in Art. 289 that, in motions of translation of a rigid body, the resultant of the inertia forces acts at the centre of inertia; so that, when there are no external forces acting except those of gravity (of which the resultant acts at the same point), the body may be treated as a particle. In this chapter we shall consider the action of inertia in other kinds of motion, and of external forces applied at points other than the centre of inertia. Let us first suppose the rigid body to admit of no motion except rotation about a fixed axis. A perpendicular of indefinite length drawn from a point of the axis in the substance of the body generates in the rotation a plane. Let be the angle which this perpendicular makes at the time / with a fixed direc- tion in the plane; then GO = —- dt is called the angular velocity of the rotation. The unit of angular velocity is of course the angle whose arcual measure is unity, 352 MOTION OF RIGID BODIES. [Art. 432. sometimes called the radian. The linear velocity of a point at a distance r from the axis is ds rdQ '^ ~Jt~ ~~dt '~ ^^' When the angular velocity is constant, the rotation is said to be uniform; every particle has uniform circular motion, and, denoting its mass by m and its distance from the axis by r, the inertia of the particle is simply its centrifugal force mod'r (Art. 359). Since the centrifugal force of each particle acts in a line passing through the axis, the resultant of the whole inertia will be balanced by the resistance of the axis. Hence, if the axis be smooth, there will be no resistance to uniform rotation. 433. If the angular velocity is not constant, every particle at a distance r from the axis will have the tangential acceleration dv^_ d_^_ d'd dt ~^ dt~^ di^' Hence, m being the mass of the particle, it exerts a tangential force of inertia equal to d'^d "^'-df' This component of inertia resists change in the angular speed of rotation, hence its efficiency must be estimated (like that of a force in producing rotation, Art. 93) by means of its moment about the axis. The arm with which the tangential inertia acts is r, hence its moment is and, since the normal inertia has no moment about the axis, this is the whole moment of the inertia of m. § XXII. J MOMENTS OF INERTIA. 353 The whole moment resisting the rotation of the body is found by summing the expressions of this form for all the par- ticles of the body; that js to say, it is d'^Q . . . . since the factor — -, which is the angular accelerattony is common to all the expressions. Moments of Inertia. 434. The moment of the impressed force (or, if more than one force is acting, the resultant moment of the impressed forces) which produces an angular acceleration is equal to the moment of the inertia which resists it. Denoting the former by K^ and putting / for the factor ^mr^ in the expression found above for the moment of inertia, we have, therefore, the equation -"^ (■) which enables us, when / has been found for the given body and axis, to determine the angular acceleration which will be pro- duced by a given force or system of forces. It has become customary to call the factor / the moment of inertia^ although properly the second member of equation (i) is the moment of inertia, or rotational inertia. The factor /, which is analogous to the mass in the formula for linear acceleration, is only the mass-factor of the rotational inertia, just as M is that of the inertia of, translation, the other factor being in each case an acceleration. 354 MOTION OF RIGID BODIES. Art. 435. 435. Let M = ^m be the whole mass of the body whose mo- ment of inertia is /= 2mr^. If the particles were all at a com- mon distance, r = a, from the axis, we should have /= a^M, For example, in the case of a heavy fly-wheel revolving about an axis passing through its centre and perpendicular to its plane, the mass may, with very little error, be assumed to be concentrated into the circumference of a circle whose radius a is the mean radius of the fly-wheel. Now suppose that a force P is applied at the circumference of an axle whose radius is b. The moment of the applied force is ^ = bP^ and the mo- ment of inertia of the fly-wheel is Ma^, iG. 104. Hence, substituting in equation (i) of the preceding article, we have bP = d'M For instance, if the radius of the fly-wheel is 3 feet, its weight 100 pounds, the radius of the axle 6 inches, and the applied force 40 pounds, we have ^ = i- X 40 = 20 pounds-feet and ... 100 , . / = 3 X ; therefore o - Qoo d'^S 20 = - 32 dt' d*6 72 whence —7-7- = — . Thus, the given moment produces in this d/' 45 wheel an angular acceleration of Jf radians. That is to say, if it acted /or one second upon the fly-wheel, originally at rest, it would produce an angular velocity of Jf (which is about .113 of one revolution) per second. The linear velocity acquired by each point of the rim is in this case 2^j Vs. gXXIL] THE RADIUS OF GYRATION. 355 Moment of Inertia of a Continuous Body. 436. In the expression / = ^mr^ the mass is regarded as made up of separate parts treated as particles, each particle having its special value of r. For a continuous mass we must (as in the case of statical moments) replace m by dMy an element of mass, and the sign of summation by that of integration. Thus we write -\- dM, in which dM is an element of mass at the distance r from the axis of rotation. If we can express the entire element of mass at the distance r in terms of r, we can find / by a single integration. Suppose, Tor example, we have to find the moment of inertia of a homogeneous cylinder of length / and radius a about its geomet- rical axis. The entire element at the distance r from the axis is the mass of an element of volume of thickness dr and having a cylindrical surface of radius r and length /. Then, denoting the uniform density by p, we have dM = p, 27tr/dr. Substituting in the expression for /, we find 7tf)la* I = 27tpl \?'- The Radius of Gyration. 437- The moment of inertia of a particle of mass M zi 2l. dis- tance k from the axis is k''M\ hence, if we put /= k'M, (i) k is the radius of a circumference upon which if the whole mass were concentrated (as in the illustration of the fly-wheel, Art. 435), it would have the same moment of inertia that it actually has. 356 MOTION OF RIGID BODIES. [Art. 437. Thus k may be regarded as the radius of the equivalent fly-wheel; it is called the radius of gyration of the body for the given axis. Equation (i) when written in the form shows that ^' is the average value of the squared distance of the particles from the axis. When, as is usually the case, the value of M\% known, we need only, in questions involving the moment of inertia, to know the value of k^^ which, being simpler than that of /, is more easily remembered. Thus, in the example of the pre- ceding article, M is known from the known volume of a cylinder, namely, M = npla". ■* Hence, from the value of / found above, we have for the squared radius of gyration of a homogeneous cylinder about its geometrical axis. Interaction of Inertia in Rotation and Translation. 438. When a mutual action exists between two bodies, one having a motion of translation and the other one of rotation, their accelerations and mutual action may be found by a method similar to that employed in Art. 311. As an illustration, suppose a homogeneous cylinder of weight ^and radius a, Fig. 105, mounted on a smooth horizontal axis, to have a fine string wound about it, to the free end of which a weight W is attached. Let us find the acceleration, and the tension, 7", of the string. Denoting the space through which W falls by ^, and the angle through which the cylinder turns by ^, we have s — aO^ whence the Fig. 105. §XXII.] THE ENERGY OF ROTATION. 357 linear and angular accelerations are connected by the relation df ~ ^ dt" ' The moment of inertia of a cylinder whose mass is Af is found above to be / = i^'^, and the impressed moment is K = aT \ hence, by equation (i), Art. 434, the kinetic equi- librium of the cylinder gives «7' = i«W^, or 2g dt' In like manner, that of W gives T^—-jr^' (0 -----f $ <■) W -\- 2W' d^s Adding, to eliminate Ty W 2g dr' whence ^ _ 2g^' dr~w-\-2W'' and, substituting in equation (i), WW' T = W-^2W" The Energy of Rotation. 439. The kinetic energy of a body rotating about a fixed axis is the sum of the kinetic energies of its particles. The linear velocity of a particle of mass m at the distance r from the axis is 358 MOTION OF RIGID BODIES. [Art. 439. rOD^ and therefore \mr'^QD^ is its kinetic energy. Hence, the whole kinetic energy of rotation is \:2mr'c^ - \IoD\ in which the quantity / again appears as analogous to M in the corresponding expression involving the velocity of translation, namely, \Mi^. Using this expression, we may apply the principle of work and energy directly to questions involving spaces and velocities. For example, in the illustration of the preceding article, to find the velocity acquired when W falls from rest through the space s : Denoting this velocity by v, the angular velocity of the cylinder is go, where v = aao. The work done by gravity is W's, and this work produces kinetic energy in each of the bodies. That of W is ^Igd\ where / = ^a'M; hence, W W kinetic energy oi W — — «'cw' = — z;», 4g 4g W kinetic energy of ^' = z''. o Therefore or ,»- 4^V^ - IV+ 2W' ' This velocity is, of course, the same that would be found by treating ?F' as a body moving with the constant acceleration found in the preceding article. Work done in an Angular Displacement. 440. Let a constant force P act upon a body free to rotate about a fixed axis, as in Fig. 104, p. 354, the force acting with a constant arm b so as to have a constant moment K = Pb. In ^5 XXI I.] WORK IN ANGULAR DISPLACEMENT. 359 an angular displacement through the angle B the force works through the space bd, equal to the arc of the axle from which the string is unwound^ Hence the work done is PbO^ or Kd. That is to say, the work done in an angular displacement is the product of the turning moment acting and the angular displacement. It will be noticed that the latter factor is an abstract number or ratio, and accordingly the units of work and of moment are the same, namely, the foot-pound or pound-foot. 441. The equation of rotary motion about a fixed axis is, by Art. 434, d^ Jdw ^ K df dt r where de 00 = -— . dt Eliminating dt^ after the analogy of Art. 291, we have codoo — —dd. The integral of this between limits is the equation of energy, The second member (in which K may be a function of &) expresses the work done by K^ while Q varies from 6^ to ^, ; hence the equation shows that this work is equal to the kinetic energy gained, as in the corresponding equation of Art. 294. Moment of Inertia of a Geometrical Magnitude. 442. In the case of a homogeneous solid of density p and vol- ume V, the mass is J/ = ftV, and the moment of inertia of this 360 MOTION OF RIGID BODIES. [Art 442. mass is k^M = k^p V. Omitting the constant factor p, the quan- tity /= k'V is called the moment of inertia of the volume with respect to a given axis, just as, in Art. 181, the product xF is called the statical moment of the volume with respect to a certain plane. In like manner, if k is the radius of gyration of a mass regarded as concentrated with uniform density into a given sur- face of area A^ /= k'A is called the moment of inertia of the area. Again, if k is the radius of gyration of a mass regarded as concentrated with uniform density into a line of length s^ /= k'^s is called the moment of inertia of the line. 443. In the case of the line, the expression for / as an integral is = [r'ds, where r is the distance of the element ds from the axis. This expression involves but a single integration, of which the limits are the values of s at the two extremities of the line. For example, let us find the moment of inertia of a line of length a about an axis perpendicular to it passing through one end. Taking this end as origin, the element is dx^ and its distance from the axis is x ; hence Jo z z In the last member we have written / in the form /^*j, so that k^ = \a^ is the squared radius of gyration. § XXII.] THE MOMENT OF INERTIA OF A PLANE AREA. 36 1 The displacement in a direction parallel to the axis of any portion of the mass which is supposed concentrated in the line evidently cannot change the moment of inertia ; therefore \c^ is also the squared radius of gyration of a rectangle whose sides are a and b about a side of length b. 444. As another example, let us find the radius of gyration of the arc in Fig. 60, p. 13 r, about the axis of x. The element ds is at the distance 7 from the axis; and, expressing ds zxid. y in terms of ^, we have ds = adQ,y = a sin Q. Hence ■f Q dd = a'(a — sin a cos a). Dividing by 5, which is 2aay , _ «'/ sin a cos a \ ~~ 2 \ a r When flf = o, we have k^ = o ; when a: = ^zr, we have /^' = — 2 for the radius of gyration of a semicircumference about the bisecting diameter. When a = tv^ the same value is found for the complete circumference about a diameter. This obviously should be the case, because both the moment of inertia and the length have now double the values which correspond to the semi- circumference. The Moment of Inertia of a Plane Area. 445. In finding the moment of inertia of a plane area about an axis in its plane, we shall suppose its curved boundaries to be referred to rectangular coordinate axes. For example, let us find the radii of gyration of an ellipse about its axes. The equation of the ellipse referred to its axes is ^ + ^' = x. 362 MOTION OF RIGID BODIES. CArt. 445. To find the moment of inertia about the axis of ^ by a single integration, we take for the element of area 2ydxy as in the dia- gram. Since all points of this element are at the same dis- tance X from the axis of ^, the moment of inertia about that axis is Fig. 106. -'\ yx^dx. Substituting the value of y from the equation of the curve, /= 2-^ i/(a' - x')x'dx; a J - a and, putting x = a sin 6^ this becomes i Aa'b I ' cos' e sin' d dO = ^a'b'-^ - = — . ' 4224 Since the area of the ellipse is ^ == nab, we have k^ = ^a^. 446. To illustrate the employment of double integration, we shall apply it to find the moment of inertia of this ellipse about the axis of x. The element of area is now the point-element d^A = dxdy situated at the point (^, >'), and its moment of inertia about the axis of X is d^'I^^/dxdy. The two integrations required may be performed in either order. If we perform the jc-integration first, we shall be summing up the elements along a line parallel to the axis of x^ and the remaining part of the process will be the same as that of the preceding article, with an interchange of the coordinates x and y and the § XXII.] POLAR MOMENT OF INERTIA OF AN AREA, l^l constants a and b. But if we perform the ^'-integration first, we obtain an integral of different form. Thus dl = dx V' y^dy = %y^dx; where 7i is the ordinate of the ellipse. Hence, substituting and integrating, rr ^Fcos*^^^=4^3j-i Jo 3 4-2 ^Pcos*^ ^^ = 4^ i:-i ^ = '^ 3 Jo 3 4-22 4 Hence, for this axis, k^ = J^^, agreeing with the result previously found. The Polar Moment of Inertia of an Area. 447* The general expressions for the moment of inertia of an area about the axes of x and y respectively are I^ = ^my'f Jy = 2mx^. Consider now the moment of inertia of the area about an axis passing through the origin and perpendicular to the plane. If r denote the distance of the particle m situated at the point (x^y) from the origin, 2mr^ is the required moment of inertia, which may be denoted by /z. Now, since r' = x' +y, * In this process we have summed up t.he point elements along a line parallel to the axis of y. Accordingly we have obtained the mo- ment of inertia of the element drawn in Fig. io6. Since the radius of gyration of this element about the axis of x is obviously the same as that of the ordinate jj', if we take the result of Art. 443 as known, this value of a'/ can be obtained by multiplying dA, which is 2y(Ix, by its squared radius of gyration. 364 MOTION OF RIGID BODIES. [Art. 447. we have, by summation, Hence we have, for any area, ^, = ^.+-^, (i) The moment of inertia Iz about an axis perpendicular to the plane is often called 3. polar moment of inertia. Equation (i) then shows that ihe. polar motneni of inertia of an area about a given axis is equal to the sum of the moments of inertia about any pair of axes in the plane which intersect the polar axis, and are at right angles to each other. 448. Dividing equation (i) by the mass J/, we derive K = kl-\-k] (2) For example, from the results found in Arts. 445, 446 we find, for the squared radius of gyration of an ellipse about an axis through its centre and perpendicular to its plane, 4 The theorem expressed by equation (i) or equation (2) gives usually the best method of finding a polar radius of gyration; but if this radius is known we may sometimes use the theorem to find the radius of gyration for an axis in the plane. For example, the process of Arts. 436, 437 is equivalent to showing that the value of J^ for a circle about its geometrical axis is k\ = ^a^. Taking this as known, and noticing that for the circle h^ and hy are equal by symmetry, we have hl = ia' = 2hl, giving hi = Jtf', for the circle about a diameter. 449. Again, we found in Art. 443 that, for a square of side a, the squared radius of gyration about a side is ^a^. Hence, by equa- gXXIL] EMPLOYMENT OF POLAR COORDINATES. 365 tion (2), we have, for an axis through a vertex and perpendicular to the plane, J^ = \a^. Now passing to a square of side 2a and a polar axis through its centre, we have for this also k^ = f«^, be- cause we have multiplied by four both the area and the moment of inertia. Furthermore, take any two axes in the plane, the moments of inertia about them are equal by symmetry. Hence the squared radius of gyration for one of them is one-half that found above, namely ia^. This is therefore the squared radius of gyration, for the square whose side is 2a, about any axis in its plane passing through its centre. Employment of Polar Coordinates. 450. When the boundary of the area is given by its polar equation, the ultimate or point ele- y ment of area d'A = rdrdO should be employed. Its distance from the initial line or axis of x is then r sin 0, that from the axis of y is r cos 0, and that from the axis of z is r. For example, we may thus find the moment of inertia of the cir- cle in Fig. 107 about the tangent Oy. The polar equation of the circle re- Fig. 107. ferred to the pole O upon its circumference, the diameter being the initial line, is Tj = 2a cos 0, The moment of the element d^A about Oy is d*Iy^=.r* cos' ddrdd. Whence It ir /j, = 2 f ' r* r'dr cos' edd = l\\\ cos" 6 dd. 366 MOTION OF RIGID BODIES. [Art. 450. Substituting the value of r, from the equation of the circle, /,= 8«« f^ cos' d de = 8.«|^^ !^ = SZ^' = 5^ ^. ^ J^ 6.4.22 4 4 The Moment of Inertia of a Solid. 451. The moment of inertia of any solid of revolution about its geometrical axis can be found, as in Art. 436, by means of a single integration. In general, the length of the cylindrical element is variable, and must be expressed in terms of its radius. For ex- ample, in the case of the cone of which Fig. 64, p. 145, is a sec- tion through the geometrical axis,j>^ is the radius, and a — x the length of the cylindrical element. Hence the volume of the element is dV = 27ty{a — x)dy ; whence dl — 27ty*{a — x)dy, a Substituting the value x =-t/, and integrating, ^ Jo 10 10 452. We can express the amount of inertia by a single in- tegral also when the polar moment of inertia of the section of the solid perpendicular to the axis is known. For example, the equation of the ellipsoid referred to its rectangular axes is ^ 4. ->:! 4. f! _ required to find the moment of inertia about the axis of z. The § XXI I.J THE MOMENT OF INERTIA OF A SOLID. 367 section parallel to the plane of xy at the distance z from that plane is the ellipse X ^ y _ c —z of which the semi-axes are « = ^ |/(^« _ z') and /? = - |/(^' - z*). Now the area of this ellipse is na^; and by Art. 448, its polar squared radius of gyration is ^{0^ + /?'). Hence we have, for the element of volume, dV~ 7t^{c'' - z^)dz, and for the square of its radius of gyration about the axis of z Therefore and 2C Jo = 2£^(fl+i!) (, - f + ^y = If^^ (a- + *'). 2^ 15 Since the volume is F = ^nabCy we have, denoting the radius of gyration about the axis of ^ by ^z, 368 MOTION OF RIGID BODIES. [Art. 452. and in like manner, In particular, when the semi-axes are equal, we have, for the radius of gyration of the homogeneous sphere whose radius is a about a diameter, k' = K. Separate Calculation of ^mx", '2my* and ^mz*, 453* We have seen in Art. 447 that, in the case of an area or a lamina referred to three rectangular axes, the axes of x and^ being in the plane of the lamina, / = ^mx^ + 2my*. Now the moment of inertia of any particle m about the axis of z does not depend in any way upon the value of z. Hence this equation is also true for a solid of any form. But the moments of inertia about the other axes are now I^ = ^my^ -f- ^mz^^ I^ = 2mz^ -\- 2mx^ ; so that we no longer have, as in the case t f a lamina, one of the three moments about rectangular axes equal to the sum of the other two. On the contrary, in the general case of a solid, each of these moments is less than the sum of the other two. 454. We may often, with advantage, use these expressions in calculating the moment of inertia of a solid. For example, to express 2mz^ for the ellipsoid of Art. 452 as a single integral, we need only to know the volume of the element at the distance z gXXIL] CALCULATION OF ^mx\ ^^my^, AND 2mz\ 369 from the plane of xy. Thus, using the value of ^F employed in that article, we have J —c ^ Jo _ 271 ab J . _ 47rabc^ _ ^ V In like manner, we have 5 5 whence, by the equations of the preceding article, 5 ^ 5 ^ . 5 agreeing with the result found in Art. 452. 455" The moment of inertia of a spherical shell about a diameter can be found without integration by means of the equations of Art. 453. For the equation of the spherical surface referred to rectangular planes passing through the centre is Hence, if M is the mass of the spherical shell, or mass supposed to be uniformly concentrated on the spherical surface, 2mx' 4~ ^^y + ^mz"^ = ^tna^ = a^M. But, by symmetry, ^mx'^ = 2my^ = 2mz^; therefore the value of each of these quantities is ia'M, and /^ = ia^M, whence k' = K is the squared radius of gyration about a diameter. 370 MOTION OF RIGID BODIES. [Art. 456. 456. This result furnishes a convenient method of finding the moment of inertia of a sphere when the density is a function of the distance from the centre. For example, to find the moment of inertia of the sphere considered in Art. 186, of which the weight per unit volume is w.c^ , , r 1 • , • '^^^'^ — ^, and therefore the mass per unit volume is /? = . r gr' Taking for element of volume the spherical shell of radius r and thickness dr, we have dF= ^Ttr'dr, dM=47rp r' dr = 1^!^ dr. S Multiplying by the value of k^ for the shell, which is f r', we find dl = ^— r dr. Integrating from o to a, and using the mass as found in Art. 186, we obtain 2>7tw^d^ a^ AfTtw^c^ 2c^ ~~ Zg Z~ S 9 ' whence for this sphere 1^ =■ ^a*. Selection of the Element of Integration. 457* The examples already given show that the mode of selecting the element depends chiefly upon the character of the bounding curve or surface, which determines the limits of inte- gration. As a further illustration, consider the solid generated by revolving the circle in Fig. 107, Art. 450, about the axis of y. The most convenient element of volume is that generated in this rotation by the element of area d^A. The path described by this element of area is the circumference of the circle whose radius is r cos 6; hence d'V= 27rr cos e . rdrdd. § XXII.] SELECTION OF ELEMENT OF INTEGRA TION. 37 1 Now, to obtain the moment of inertia about the axis of y, we multiply this circular element* by the square of its radius of gyration, which is its own radius, r cos 6. This gives ^V= 27tr'dr co%' Odd; whence ir rr I = 4,7tV V'r^dr COS* edd= ^[\\cos* Odd. J o J o 5 J o Substituting r, = 2a cos 6, from the equation of the circle, 5 Jo 5 8.6.4.22 2 By Pappus's Theorem the volume is 27r'a'; hence for this solid we have 458. For the anchor-ring in general, it is simpler to refer the generating circle to its centre, because the limits of each variable will then be independent of the other. Thus, referring to Fig. 62, p. 136, let us find the moment of inertia of the anchor-ring about its axis AB. The radius of the circle described by the element of area d^A = rdrdd is b -\- r cos By and this is also the radius of gyration of the element of volume generated by the rotation of the element of area. Hence ^ V = 27r(^ + r cos By rdrdB. * It is the shape of the solid, and not the position of the axis about which the moment of inertia is required, which determines the element to be used. We shall see, in the next section, that the element of vol-' ume thus determined can be used in finding the moment of inertia about any axis, because we have means of finding the radii of gyration of elements of simple form about any axis. 372 MOTION OF RIGID BODIES, [Art. 45S The limits for r are o and a, and the limits for Q are o and 2/T. Hence, expanding and performing the r-integration, we have I=27t f " {b' -- + s^'' - cos ^ + 3/^ - cos^ ^ + - cos' d)de Jo 2 3 4 5 = 27r --.27r + ^^ — .TT =-^(4^'+3^'). By Pappus's Theorem the volume is F'= 27t^ba^\ hence, for the anchor-ring, 4 This gives the excess of the radius of gyration of a fly-wheel whose rim has a circular section over the mean radius 3, which was, in Art. 435, taken as its approximate value when a is small relatively to b, EXAMPLES. XXII. 1. Find the moment of inertia of a triangle of base b and altitude h about an axis passing through the vertex and parallel to the base. . ^ h^ 2 ' 2. Find the moment of inertia of the same triangle about the base. ^ h^ ' 6' 3. If the altitude /^ of a triangle divides the base into the segments a and b, find the radius of gyration about the altitude. a' - ab-\- ^' k = ^ -. 4. If IV in Art. 438 is a drum or hollow cylinder whose thick- ness may be neglected, show that the acceleration and tension are the same as if ^K' dragged W^ along a smooth horizontal table as in Fig. 87, p. 248. 5. Find the radius of gyration for the arc of the cycloid X = a (tp — sin tp)j y = a {1 — cos ^), about the axis of x, ,0^2, ^ = a . 15 1 ■^XXIL] EXAMPLES. ■ 373 6. Find the radius of gyration for the area of the cycloid in Ex. 5 about its base. ,1 35 , ~ 36" ■ 7. Determine k"^ about the same axis for the surface of revo- lution generated by revolving the cycloid about its base. 35 8. Determine the radius of gyration of the area of the lemnis- cata r' — a^ cos 2S about a tangent at the origin. ^ _ nc^ 9. Find the radius of gyration of the area of the lemniscata about the axis of the curve. a"^ , . 10. Determine the radius of gyration of a fly-wheel of mean radius a when the rim has a rectangular section, / being its thick- ness, k^ =a^ -\. lt\ 1 1. A uniform door, 3 feet wide, weighing 80 pounds, is swing- ing on its hinges, and the edge has a velocity of 8 feet per second. How many foot-pounds of energy must be expended in stop- ping it ? 26f . 12. Find the moment of inertia of a lamina in the form of a regular hexagon whose side is a about one of its central diago- nals. 5_Vi_l 16 • 13. Show that, if a solid of revolution is referred to rectang,ular axes, that of x being the geometrical axis, '!2mz'^ = '^.viy^ = \I^ (the density of the solid being assumed unity). Hence, using the value found for the cone in Art. 451 and determining ^mx* independently, find the radius of gyration for a perpendicular to the geometrical axis passing through the vertex. k' = Mb' + Ah'). 20 14. Determine the radius of gyration of a paraboloid about its axis, the radius of the base being b and the height h. k' = ib\ 15. Determine by the method suggested in Ex. 13 the radius 374 MOTION OF RIGID BODIES. [Ex. XXII. of gyration of the paraboloid of the preceding example about a perpendicular to the axis passing through the vertex. 6 i6. A solid homogeneous 8-inch shot consists of a cylinder two calibres in length and an ellipsoidal head one calibre in length ; the rifling gives it a rotation about its axis of one turn in 25 feet. What ratio does the energy of rotation bear to that of translation ? i : 300. 17. A drum whose diameter is 6 feet, and whose moment of inertia is that of 40 pounds at a distance of 10 feet from the axis, is employed to wind up a load of 500 pounds from a vertical shaft. It is rotating at the rate of 120 turns a minute when the steam is shut off. How far should the load be from the shaft's mouth that the kinetic energy of the load and drum may just suffice to carry the load to the surface ? 41.9 ft. 18. The ogival head of a projectile is formed by the revo- lution of a semi-parabola about the ordinate ^, so that the height h is the radius of the bore or one-half the calibre d. Determine k"^ for the axis of revolution. 7,2 _ 2 , 21 19. Show that the moment of inertia of a uniform right prism of any cross-section about an axis in the plane of any right section is equal to the moment of inertia, about the same axis, which it would have if its mass were concentrated into the sec- tion as a lamina, increased by that which it would have if its mass were concentrated into its length as a rod passing through the axis. XXIII. Relations between Moments of Inertia about Different Axes. 459. The Statical Moment of a body with respect to a given plane is defined in Art. 178 as '^mp^ where/ is the distance of the particle m from the plane. The values of three such statical moments, for example, with respect to three coordinate planes of § XXIII.] MOMENTS ABOUT PARALLEL AXES. 375 reference, serve to determine the value of the statical moment with reference to any given plane; for they determine the cen- tre of inertia, and the statical moment is Mp^ where p is the per- pendicular distance of the centre of inertia from the given plane. In the case of moments of inertia about different axes, the re- lations are not so simple; but we shall find that, supposing the cen- tre of inertia, or Centroid, already found, relations exist by virtue of which the values of the moment of inertia about three particu- lar axes will serve to determine that with respect to any given axis. Moments of Inertia about Parallel Axes. 460. The first of these relations is that which exists between the moments of inertia of a body about parallel axes, one of which passes through the centre of inertia. Let Fig. 108 represent a sec- tion of the body made by a plane passing through the centre of inertia (9, and perpendicular to the axes, one of which pierces the plane of the diagram at O^ and the other at A^ at a distance OA = h. Assume rectangular coordinate axes, OA being the axis of x^ and the cen- troidal axis of moments that of z. Denote by I^ the moment of inertia about this axis, and by /, that about the parallel axis through A, Let P be the projection upon the plane of xy of the point at which the particle m is situated; then FO is equal to the distance of the particle from the cen- troidal axis, and PA is equal to its distance from the parallel axis through A. From the figure we have AP"" = / -f (^ + hY =/ + x' -\- 2hx -f >^'; OP^ ^f-\- x\ Fig. 108. 37^ MOTION OF RIGID BODIES. [Art. 460. Hence = /o + 2h^mx -\- h'^^m. Now ^mx = o, because it is the statical moment of the body with respect to the plane oi yz which passes through the centroid. Hence the equation reduces to /,=/, + A'Af, (i) where M is the total mass of the body. Introducing the radii of gyration this equation becomes ^ = ^^ + /^' (2) It follows that for all parallel axes the moment of inertia {and radius of gyration^ is lesat when the axis passes through the centre of inertia. 461. If the moment of inertia about all axes through the centre of inertia is known, this theorem determines the moment about every axis. For example, we found in Art. 452 that the mo- ment of inertia of a sphere about a diameter was |^' V. Putting h = a, \vt have therefore for the moment of inertia about a tan- gent / = la' V. 462. In many cases, the moment of inertia about an axis not passing through the centroid is more easily found by integra- tion than that for the centroidal axis. For example, we read- ily find, for the triangle about an axis through its vertex and par- allel to its base, k"" — ^i", where h is the altitude. Now the dis- tance from the vertex to the centre of inertia is ^h. Hence, by the theorem, we have, for the centroidal radius of gyration, ki = \h: - i-h' = ^sh\ Again, to find the radius of gyration about the base, that is to pass to the distance ^h from the centre of inertia, we have i' = ^l + ¥'' = Ws + iV'' = i^'. § XXIII.] MOMENT OF THE ELEMENT. 117 Application to the Moment of the Element. 463. The application of the theorem of Art. 460 to the ele- ment of moment is often useful in enabling us to express a mo- ment of inertia as a simple integral. For example, let it be required to find the moment of inertia of the cone represented in Fig. 109 about the axis of z^ that is to say a perpendicular through the vertex to the geometrical axis. The only convenient element of volume for simple integration, in ^ this case, is the circular section perpendicular to the axis of x. De- noting the height of the cone by h and the radius of the base by ^, the radius of this element is Fig. 109. ^=r (i) The element of volume is then ny^dx. To find its squared radius of gyration about the axis of ^, we notice that this axis is at a distance x from the parallel diameter of the element which is a centroidal axis. The squared radius of gyration about the latter is, by Art. 448, J/; hence, by the theorem of parallel axes, that about the axis of z is ^^ + i/. It follows that the moment of inertia of the element is dl^ ny\x'-\-\f)dx. Substituting the value oi y in equation (i), we have Therefore 378 MOTION OF RIGID BODIES. [Art. 463. 20 and since r= \7tb''h, k^ = ^^(4/^' + /^^). The Principal Axes for a Point in the Plane of a Lamina. 464. We have next to consider the relations which exist be- tween moments of inertia about axes passing through a given point. We begin with the case of a plane lamina and axes in its plane, and shall prove that, for any given point in the lamina, there are two such axes about which the moments of in- ertia are respectively greater and less than that about any other axis in the plane and through the point ; except in the case when the moments of inertia about all such axes are equal. 465. Let O, Fig. no, be the given point, and let rectangular axes through O be assumed. Let another pair of rectangular axes Ox\ 0/ make the angle a with those of X andji'. Then, if P be the position of a particle whose mass is m^ we readily obtain from the figure for its distances from the new axes x' = OC-\- AD — X cos a -\- y sm a, p y c. \ \-^ "^ ^^ \ .'^^D T ^^ \. i K. Fig. iio. y PD — AC =y cos a (i) X sm a. * It will be noticed that the first term of the integral, in this process, is 'Emx^. Therefore, since Iz = '2mx^ -\- Smy^, the second term is the value of Smy"^. (Compare the method employed in Ex. XXII, 13.) The triple integral expression for Sf/i}''^ would, in this case, be U'X: y'''dzdy dx. in which the limits for z are values in terms of y and x obtained from the equation of the conical surface, and those for;j/are taken from equa- tion (i) above. § XXIII.] PRINCIPAL AXES IN THE PLANE OF A LAMINA. 379 From these equations, we have x'"^ — x^ cos'^oc -\- 2xy sin a cos ^ + / sin' «, y = y cos* a — 2xy sin a cos a -f- x:^ sin' a, x'y' = xy (cos' a — sin' a) — (ji;' — y) sin a cos ^. . Multiplying by m^ and summing for all the particles of the body, we obtain ^mx'"^ = 2mx^ . cos' a -j- ^mxy -sin 2 a -\- 2my' . sin' a^ ) 2my" = 2my^ .cos' « — 2mxy . sin 2« + 2mx'' . sin' a, ) ^tnx'y' = 2mxy . cos 20: — ^2m(x^ — y) . sin 2a. . . . (3) Now, supposing ^mx^^ ^my"^ and 2mxy to have been found, « may be so taken that ^rnx'y' = o; for, in equation (3), this gives 2^mxy . . tan 2a — ^ — 5 ^^ — 2, (4) which is always possible, since the tangent of an angle may have any value, positive or negative. If a^ is a value which satisfies the equation, a^ + 90° also satisfies the equation, but this change in the value of a only interchanges the new axes of x and y. There is therefore, in general, but one pair of rectangular axes for which ^mxy = o. 466. The axes thus determined are called the axes of principal moment, ov principal axes oi the lamina for the point O. Suppose now that Ox and Oy in Fig. no are the principal axes, so that 2mxy = o ; and put /x for ^^y, ly for 2mx\ as in Art. 447. Then, putting la for 2my' ', the moment of inertia about Ox\ which makes the angle ex with the principal axis Ox, the second of equations (2) gives fa = fx cos' a -{- /y sin" a, .... (5) reducing to Ix when a = o, and to ly when a = 90°. If /x > /y, /a is less than /^ and greater than ly , so that /x is 38o MOTION OF RIGID BODIES. [Art. 466. its maximum and ly its minimum value. But if 1^= ly , la is constant ; that is to say the moment of inertia is the same for all axes in the plane of the lamina passing through the given point. 467. As an example, let us find the principal axes of the right triangle OAB^ Fig. iii, for the right angle O, Taking OA = a and OB = b for axes of x and y re- spectively, we have, as in Art. 462, ^^mx" = ia' . M, :2mf = \b\M. For 2mxy, we have (for unit density) fa fyi Fa I xy dy dx = i\ xyi dx^ in which, from the figure, the upper limit for jj' is y^--{a- X), Hence ^2 -a ' ^a ta 2mxy = — - 1 x{a — xYdx = — I (a^x — 2ax^ + x*)dx 2a Jo 2a Jo Substituting in equation (4), Art. 465, ab tan 2a = — The principal axis of least moment here corresponds to the value o'o in the first quadrant ; its direction lies between the medial line through O and the greatest side and admits of an easy graphical construction. The Momental Ellipse of a Lamina for a Given Point. 468. Let p be a length such that — , is proportional to /« , so that p represents M^ reciprocal of the radius of gyration^ and let a § XXIII.] PRINCIPAL CENTROIDAL AXES OF A LAMINA. 38 1 and b be its values for « = o and ol = 90°, corresponding to A and /y. Equation (5) of Art. 466 then gives I cos" a sin' a p' a^ ' b^ ' Let p be laid off from O on the axis to which it belongs, which makes the angle a with the axis oi x \ so that f> and a are the polar coordinates of a point which, as a varies, describes the curve of which the above is the polar equation. Multiplying by p\ a^^ b' is the rectangular equation of this curve, which is therefore an ellipse. Thus the radius of gyration of the lamina about an axis in its plane passing through O in any direction is represented by the reciprocal of the radius vector in that direction of this ellipse which is called the tnomental ellipse * of the lamina with respect to the point O, Principal Axes of a Lamina at the Centre of Inertia. 469. The most important principal axes of a lamina are the centroidal ones. When the lamina is symmetrical with respect to each of two rectangular axes it is easy to see that, taking them as coordinate axes, 2mxy = o, and therefore these axes are the principal axes. Thus, the axes of an ellipse are principal axes; the lines bisecting opposite pairs of sides of a rectangle are principal axes ; the diagonals of a rhombus are principal axes. If the principal moments are equal, the momental ellipse be- * If we had made p directly proportional to the radius of gyration, we should have obtained the curve inverse to the ellipse with respect to its centre. The reciprocal is taken because it leads to a simpler and more familiar curve. 3i^2 . MOTION OF RIGID BODIES. [Art. 469. comes a circle and the moments are equal about all centroidal axes. This is the case with the square, as we have already seen in Art. 449. Again, if three moments of inertia about axes through O (whether O is or is not the centroid) are equal, the momental ellipse becomes a circle. For example, this is the case at the centre of any regular polygon. The theorem of parallel axes shows that if the momental ellipse is a circle for the centroid, it is not a circle for any other point. But if it is not a circle for the centroid, two points can be found for which it is a circle. The Moments of Inertia of a Solid for Axes passing through a Given Point. 470. In discussing the moments of inertia of a solid about axes which pass through a given point (7, we shall at first suppose the plane of xy to be a plane passing through O taken at random. The value of z for any particle m of the solid will not affect the values of the quantities ^mx^^ 'Smy^, and ^mxy. Hence we can show, exactly as in Art. 465, that new axes of x' and / in the plane can be found such that ^mx'y' = o. Now taking these new axes for those of xy, so that 2mxy = o, we shall have, as before, from the second of equations (2), Art. 465, when the axes are turned through any angle ar, ^my'^ = 2my^ . cos' a -\- ^mx^ . sin' a. . . . (i) But the terms of this equation are not now moments of inertia. In fact, for the solid, /r = ^my'^ + 2mz^^ and /y = 2mx^ -j- 2mz^; and, if /« denotes the moment of inertia about the axis of x^ which is in the plane of xy and makes the angle a with the axis of X, /a = :Smy" + 2mz\ § XXIII.] THE MOMENTAL ELLIPSOID. 383 Hence, adding to equation (i) the identity ^niz^ = 2mz'^ . cos' a -\- ^mz^ . sin' a-, we have /a =Ix cos' a -\- ly sin' a (2) which is the same relation for the solid as that found in Art. 466 for the lamina. The Momental Ellipsoid. 471. It follows that, for any plane passing through a given point, we have a momenfal ellipse^ as in Art. 468, of which the radius-vector drawn from the centre in any direction is the reciprocal of the radius of gyration of the body about the cor- responding axis. Consider now the locus in space of the extremities of radii- vectores laid off in the same way for all axes passing through the given point. This locus is a surface of which we have just seen that the section by the plane through O is an ellipse. Since this is true for any plane passing through 6>, the surface is such that all its plane sections through the point O are ellipses. The sur- face is therefore that of an ellipsoid. This ellipsoid is called the momental ellipsoid of the solid with respect to the point (9, which is its centre. 472. The principal axes of the momental ellipsoid are called the principal axes of moment of the solid for the given point, and the moments about them are the principal moments of inertia. If «, b and c are the reciprocals of the principal radii of gyration, the equation of the momental ellipsoid referred to the principal axes is ,-r + ^.+^ = . (I) The greatest and the least moment of inertia for axes passing through the given point O correspond respectively to the least and the greatest axis of this ellipsoid. 384 MOTION OF RIGID BODIES. [Art. 472. Now let p be the length and a^ ^, y the direction-angles of the radius-vector of the ellipsoid, so that X = p cos a, y — P cos ^, z = p cos y'y then the equation of the ellipsoid may be written in the polar form, I cos' or cos'^ cos'r , . 7'^~^^~r-+-7- (') Denoting the moment of inertia about the axis whose direction- angles are ^, fi, y by Ia,p,y, we have, on multiplying equation (2) by the mass J/, /a, ^, y = /;c cos' a .-{- ly COS* /3 -j- I^ cos'' ;/. . . (3) By means of this theorem and that expressed by equation (i), Art. 460, we can find the moment of inertia of a body about any axis, when we know the principal axes and moments for the centre of inertia. The Principal Axes of Symmetrical Bodies. 473. The principal axes of the solid tn a plane, for which 2mxy = o, as found in Art. 470, are of course the axes of the ellipse in which the given plane cuts the ellipsoid. Hence, when the principal moments of the solid are all unequal, it follows from the nature of the ellipsoid that the principal planes are the only set of rectangular coordinate planes for which we have at once '2mxy = o, 2myz = o, 2mzx = 0.* * By using the general equations of transformation for passing to a new set of rectangular coordinate planes with the same origin, we might have obtained expressions for '2mx'y' , '2,ttiy'z' and ^mz'x' \ and then, by equating these to zero, we might have determined the position of the principal axes in terms of ^mx"^, 2mxy, etc., supposed to have been calculated for the assumed coordinate planes. In the process given in the text, we have confined ourselves to the proof of the existence of the principal axes and the expression of the moment of inertia about any axis in terms of the principal moments. § XXIII.] PRINCIPAL AXES OF SYMMETRICAL BODIES. 385 But if two of these equations are true, the axis in which the corresponding planes intersect is a principal axis. Thus, if 2mxs = o and 2myz = o, the axis of z is an axis of each of the ellipses in which the planes of xz and o( yz cut the momental ellipsoid. It follows that a plane tangent to the ellipsoid at the point where the axis of z cuts the surface is parallel to the plane of xy; hence the axis of z is an axis of the ellipsoid. The other principal axes are now principal axes for the plane of xy, and can therefore be determined by the method illustrated in Art. 467. 474. The position of one or more of the principal axes of a solid is sometimes obvious from considerations of symmetry. For example, suppose the body to be homogeneous and sym- metrical to a given plane, so that, taking this as the plane of xy^ to any particle situated at a point (x,y, z) there corresponds an equal particle at {x, y, — z); then it is evident that, no matter where the origin and axis of x be taken in the plane, we shall have ^mxz = o. In like manner we have 2myz = o. Therefore, at every point of the plane of symmetry, the line perpendicular to it is a principal axis of the body for that point. Since the centre of inertia is in the plane of symmetry, we can therefore readily find the principal centroidal axes. In accordance with this principle, a plate of uniform thickness, or any body in the form of a right prism, has at any point of its central plane the line perpen- dicular to it for a principal axis of inertia. It obviously follows, from Art. 447, that, for a thin plate, this axis is the axis of greatest moment. 475* I^ there be two planes of symmetry, we shall thus have, for any point of their line of intersection, the position of two principal axes; and, these being in the plane perpendicular to the line of intersection, that line will itself be the third principal axis. For instance, a right pyramid whose base is a rectangle has two planes of symmetry, each passing through the geometrical axis and the middle points of a pair of opposite sides of the base. Therefore, for any point of the geometrical axis, this axis itself and the lines joining the middle points of the right section of the pyramid are the axes of principal moment of inertia. 386 MOTION OF RIGID BODIES. [Art. 476. 476. The two planes of symmetry are usually at right angles, as in the illustration just given, and the corresponding principal moments will generally be unequal. But if the planes cut obliquely, we have the case in which each of two oblique axes fulfils the condition for a principal axis, and therefore all the axes in the plane give equal moments of inertia (see Art. 469). An instance is afforded by a right pyramid having an equilateral triangle for its base. The moments of inertia for all axes passing through a point of the geometrical axis and in a plane parallel to the base are equal. In such a case, the momental ellipsoid be- comes a spheroid. In like manner, the moments of inertia for all axes passing through the centre of a regular tetrahedron, or of any regular solid, can be shown to be equal, the momental ellipsoid for that point becoming in these cases a sphere. The Equimomental Ellipsoid. 477. For any given rigid body, there may be found a homo- geneous ellipsoid having the same mass and the same principal moments at the centre of inertia as the given body, and therefore, from what precedes, the same moment of inertia and the same radius of gyration as the given body for every axis. This ellipsoid is called the equimomental ellipsoid of the body. Let the centre of inertia of the given body be taken as the origin, and the principal axes as coordinate axes, and let a, b^ c be the semi-axes of an ellipsoid lying respectively in the axes of x^y and z. The principal axes of inertia of this ellipsoid at the origin, which is its centre of inertia, are, by Art. 475, the coordi- nate axes, for which the moments of inertia were found in Art. 452 to be ^1 = *(«'+*'), kl = K^' + ^). kr = i(^' +'»')• Now, if in these equations kx, ky, k^ are the principal radii of § XXIII.] THE EQUIMOMENTAL ELLIPSOID, 387 gyration of the given body, we have, by solving for a, b and r, the semi-axes of the body's equimomental ellipsoid; namely, c' = i{kl + k'y - kl). This ellipsoid, which, as stated above, must also have the same mass as the given body, may be substituted for that body in any question involving either the inertia of rotation or that of translation. The Compound Pendulum. 478. A heavy body of any form free to turn upon a horizon- tal axis not passing through its centre of gravity is called a compound pendulum^ in distinction from the simple pendulum, in which the mass is regarded as concentrated into a single particle. Let G^ Fig. 112, be the centre of gravity, and C the point where the axis is cut by a vertical plane perpendicular to it, passing through G. This point is called the point of suspension. The forces acting upon the body- are its weight, acting vertically downward at 6^, and the resistance of the axis. Since rotation about the axis is the only motion possible, we obtain the single equation of motion required, by taking moments about C Denoting CG by ^, the angle it makes with the vertical by 6^, and the radius of gyration for the given axis by k^ equation (i), Art. 434, gives Mgh sin e = Mk^-^^, at or, putting (i) y6> ^sm in Q, 388 MOTION OF RIGID BODIES. [Art. 478- This is identical with the equation of motion of a simple pendu- lum, Art. 374, if its length is /, since this makes s = l6. Hence the motion is the same as that of a simple pendulum of length /. Denoting the radius of gyration about a parallel to the given axis through G by k^^ equation (2), Art. 460, gives hence, putting h! for —^ so that J^. = hh\ (2) equation (i) gives /=| + >5=>5 + >5' (3) It follows that, measuring from C, Fig. no, the distance CL = /, which is the length of the equivalent simple pendulum^ the point L will lie on the other side of G at the distance GL — h' , The point L is sometimes called the centre of oscillation. 479. Equation (2) shows that h and h' may be interchanged, the value of / being by equation (3) unchanged. Hence the remarkable result that if the body be suspended from the centre of oscillation the time of vibration remains unchanged.* In the ordinary pendulum in which most of the mass is con- tained in a small bob, k^ is small relatively to h^ and therefore the centre of oscillation is but a short distance below the centre of gravity. But, making h small relatively to k^^ h' can be made as large as we please, and the centre of oscillation placed far be- low and outside of the body. Thus a body of limited size can be so mounted as to be the equivalent of a very long simple pen- *This principle is used in determining experimentally the exact position of the centre of oscillation of the pendulum which beats sec- onds, and thence the value of L. This is done in *' pendulum experi- ments " to determine the absolute value of gy by the formula g — n'^L, Art. 382. The experiments mentioned in Art. 385 are only for varia* tions in the value of ^. § XX 1 1 1 . ] THE COMPO UND FEND UL UM, 3 89 dulum. Advantage is taken of this principle in the pendulum of the metronome^ in which the centre of gravity is adjustable, so that h may be shortened, and the length / (and consequently the time of vibration) increased at pleasure. The observed time of vibration of a body mounted as a pendu- lum is often used to determine k^^ h being measured, and h' derived from the calculated value of /. Foucault's Pendulum Experiment. 480. A body at rest relatively to the earth, partakes of its rotary motion. Thus, if a body is mounted on an axis through its centre of gravity, parallel to that of the earth, it may be re- garded as rotating about that axis at the angular rate of 360° a day, or 15° an hour. Even if the axis were perfectly smooth, the body would continue to rotate at this rate, and thus have no rotation relatively to the earth. But, if an ideal body without mass could be thus mounted, in such a way as not to share the rotation of the earth, it would have a rotation relatively to the earth exactly equal and opposite to the real rotation of the earth. This apparent rotation would then afford an experimental proof of the rotation of the earth. Such an ideal body, mounted upon a vertical axis, is furnished by the plane of vibration of a pendulum so suspended as to be free to vibrate in any vertical plane. If this experiment, which was devised by Foucault, were performed at the pole of the earth, the rate of the apparent rotation of the plane of vibration v'ould be 15° an hour. 481. At any other place, the rotation thus put in evidence will be only'a resolved part of the earth's rotation. To find its auiount, let A be the latitude, and suppose the line tangent to the meridian to meet the earth's axis produced in C \ this line will describe the surface of a cone with vertex at C touching the earth in the parallel of latitude. The tangent line may at any instant be regarded as rotating about C, and it is readily seen that its angular rate is to that of the earth inversely as the length of 390 MOTION OF RIGID BODIES. [Art. 481. the tangent is to the radius of the parallel of latitude. Thus, if 00 is the angular rate of the earth, 00 sin A is the rate of rotation of the plane of the horizon about a vertical axis, and this is the apparent rate of the plane of vibration exhibited by the experi- ment.* Pressure on the Axis of a Uniformly Rotating Lamina. 482. When a body mounted upon a fixed axis is at rest under the action of external forces, these forces are subject to one con- dition of equilibrium, namely, that their resultant moment about the axis of rotation shall vanish. As mentioned in Art. 242, the other five conditions of equilibrium serve to determine the reac- tions of the fixed axis. If this axis is, as usual, supported at two points, the pressures resisted by the supports may be reduced to three — one in the direction of the axis, and one at each support in some direction perpendicular to the axis. Since each of those last mentioned involve two unknown quantities, we have thus five quantities in all to be determined by the five conditions. If now the body be in rotation, the rate will remain uniform because there is no moment about the axis. The inertia of any particle m at the distance r from the axis will now, as stated in Art. 432, consist solely of its centrifugal force, mroa'^ which acts in a line passing through the axis. We have now to consider the resultant of these centrifugal forces for all the particles of a body, and the additional pressure upon the axis thus produced. We begin with the case of a lamina rotating about an axis perpendicular to its plane. 483. Take any rectangular axes in the plane of the lamina passing through the point in which it is pierced by the axis of rotation, and fixed with reference to the substance of the lamina. * The experiment must be executed with great care to prevent lateral motion. Otherwise, the rotation of the longer axis of the orbit men- tioned in the foot-note, p. 317, will completely disguise the motion to be exhibited. § XXIII.] F/^ESSURE ON A FIXED AXIS. 39I Let X and^ be the rectangular, and r and Q the polar, coordinates of the particle m^ as referred to these axes. Then, g? being the angular velocity, the centrifugal force iiioo^r acts at the origin, or centre of rotation, in the direction making the angle 6 with the axis of X. Hence the resolved part of this force along the axis of x is mGo^r cos B = moo^x (1) It follows that the resolved part, along this axis, of the resultant of the whole system of centrifugal forces is X = G0*'2mx. . (2) Now 2mx is the statical moment of the whole mass, M == ^m^ with respect to the plane of yz ; so that 2mx = Mx, where x is the abscissa of the centre of inertia. Thus the resolved part of the resultant of the centrifugal forces has the value X = Go'Afx, which, in accordance with expression (i), is the same that it would have if all the mass were concentrated at the centre of inertia. In like manner, the resolved part of the resultant along the axis of J is GO^Myy which is the same as if all the mass were concen- trated at the centre of inertia. Hence the resultant centrifugal force is R^o='Mr, (3) where r is the distance of the centre of inertia from the centre of rotation, and this force acts in a line directed toward the centre of inertia. In other words, yi^r a lamina rotating about an axis perpendicular to its plane ^ the resultant centrifugal force is the same as if the whole mass were concentrated at the centre of inertia, 484. This centrifugal force McD^r acts upon the axis, at the point where it pierces the lamina, in a line which rotates with the body. If the axis is supported at two points or pivots, say one on each side of the lamina, the pressures upon these supports will 392 MOTION OF RIGID BODIES. [Art. 484. be parallel components of the centrifugal force (3), acting in lines M'hich, in like manner, rotate with the body, and their magnitudes will be found as in Art. 87. In particular, if the axis of rotation passes through the centre of inertia of the lamina^ and is perpendicular to its plane, there will be no pressure upon the supports to the axis resulting from the rotation. Pressure on the Axis of a Uniformly Rotating Solid. 485. Passing now to the general case, let the rigid body of any form be referred to any three rectangular axes, of which that of z is the axis of rotation. Suppose the body separated into laminae by planes perpendicular to the axis of rotation, each lamina being characterized by a particular value of z. The centrifugal force due to a particular lamina acts upon the axis at the point (o, o, s), and has no component in the direction of the axis of z. Its components in the direction of the other axes are, by Art. 483, X = Gj'^^^ymx, V = GD':2^^ymy, where ^x,y indicates summation extended to particles having all values of x and _>', but only the given particular value of z. Now, substituting in the equations of Art. 232, we have for the moments of this force about the axes of x and y £ = — zV = — GD^z^^^ymy, M = zX = co^z^^^mx^ and JV =^ o. 486. The six elements (Art. 234) of the system consisting of the centrifugal forces of all the laminae are found by summing the expressions above for all values of z. Thus they are 2X = Gj'^mx, 2V= GO^^rny, 2Z = o, . (i) 2Z = — oo'^^myz, 2M= co^^mzx, ^JV = o, . (2) where the summations in the second members now extend to all the particles of the body. § XXIII.] PRESSURE DUE TO CENTRIFUGAL FORCE. 393 It follows that the system of centrifugal forces is equivalent to a dyname (i?, K) in which the force R is the resultant of ^X and -^'i^' acting at the origin, and the couple K is the re- sultant of the couples 2L and 2M. Comparing with Art. 483, we see that-, he value of J^y vectorially considered, has the same expression as in the case of the lamina, namely, R = GD'^Mr, so that it has the same value and the same direction as if the whole mass were concentrated at the centre of inertia. But R is now regarded as acting at the origin, which may be any point upon th'? axis, and the value of K depends, as explained in Art. 226, upon the position chosen for the origin. 487. The force R and the axis of the couple K both lie in the plane of xy ; but they will not generally be at right angles ; so that the system cannot generally be reduced to a si?igle force. If the axis is supported at two points, as in Art. 484, the pressures upon the supports will now consist not only of a component of R at each point parallel to the direction of R ; but, in addition to these, of two equal and opposite forces, one at each point in a direction perpendicular to the axis of K. If a is the distance between the supports, the value of either of these forces is Qy where K = aQ. Condition under which the Centrifgual System is Equivalent to a Single Force. 488. Substituting the values found in Art. 486, the condition under which the system is reducible to a single force (see Art. 236) becomes 2mx . 2myz — 2 my . 2mxz = o (i) This is satisfied when 2myz = o and 2mxz = o, which, as we have seen in Art. 473, is the condition that the axis of z shall be a principal axis for the origin. In this case .AT = o, and the sys- 394 MOTION OF RIGID BODIES. [Art. 488. tem of centrifugal forces reduces to the force R acting at the origin. The condition is also fulfilled by any axis which lies in a plane of symmetry. For suppose this plane to be taken as the plane of xz^ and let b denote a special value of y ; then, by hypothesis, the laminae parallel to the plane of xz corresponding \.o y=-b and to^ = — b are precisely alike. Now, for the first of these laminae, the first member of equation (i) becomes b^mx . ^mz — b^m . 2mxz. For the lamina^ =^ — b, this expression changes sign; since, by the identity of the laminae ^m, ^mx, 2mz and 2mxz are un- changed. Hence, for the two laminae taken together, equation (i) is satisfied ; and, since the body consists of such pairs of laminae, it is satisfied for the whole body. It follows that, for an axis in a plane of symmetry, the system of centrifugal forces is equivalent to a single force equal to the centrifugal force of the whole mass supposed concentrated at the centre of inertia; but it must be remembered that this force does nof generally act at that point. 489. In the case of a lamina, the foregoing applies to any axis in the plane of the lamina. As an illustration, suppose the triangular lamina, Fig. iii, Art. 467, to be rotating about the side OB, which (to agree with the notation of the preceding articles) we now take as the axis of z. The value of R is the centrifugal force of the whole mass M rotating at the distance of its centre of gravity from the axis, which is ia. Thus R =:z^Go'^Ma. Since in this case ^L = o, the couple K is the same as 2M of Art. 486, and its plane is the plane of xz, which is the plane of the lamina. Hence it can be combined with jR acting at the origin, as in Art. loi, the resultant being R acting at the point (o, z^), where Now, as found in Art. 467, we have, in this case, 2mzx — ^abM^ hence z^ = \b\ that is, the resulting centrifugal force of a homo- §XXIII.] ROTATION ABOUT A CENTROIDAL AXIS. 39$ geneous right triangle rotating about a side acts upon the axis at a distance from the right angle equal to one-fourth of that side.* Rotation about a Centroidal Axis. 490. If, in the equations of Art. 486, ^tnx = o and 2my = o, we have J^ = o, — that is to say, if the axis of rotation passes through the centroid, the resultant of the centrifugal forces is the couple JC. The pressures produced upon two fixed supports to the axis at the distance a apart are, in this case, simply equal and opposite parallel forces acting in lines perpendicular to the axis of X, and of magnitude Q^ where aQ = K, If, in addition to these conditions, we have ^myz = o and 2mxz = o, the couple A' also vanishes, and the system of cen- trifugal forces is in complete equilibrium. In this case, the axis of z is, by Art. 473, a principal axis for the origin. It is readily shown that it is also a principal axis for the centroid, so that the three centroidal principal axes are the only ones about which the centrifugal forces are in equilibrium, except in the special cases mentioned in Art. 476, where two or all three of the principal moments of inertia are equal, that is, when the centroidal mo- mental ellipsoid becomes a spheroid or a sphere. f 491. If there are no external forces acting upon a body rotating about a centroidal principal axis, there will be no pressure what- * It is beyond the scope of the present volume to go further into the discussion of the principal axes of a body. It may, however, be here stated that every axis about which the centrifugal forces reduce to a single force is a principal axis of inertia of the body, but in general it is a principal axis only for the point at which the force acts. Thus, in the illustration above, OB is a principal axis for the point (o, i^), (the other two principal axes for this point being a perpendicular to the plane and a parallel to OA). For a principal axis passing through the centroid, there is no force R to define the point for which it is a princi- pal axis, and accordingly such an axis is a principal axis for every one of its points. t In the first of these cases the body is said to have kinetic sym- metry with respect to an axis, and in the second to have complete kinetic symmetry. 39^ MOTION OF RIGID BODIES. [Art. 491. ever upon the axis and it may remain unsupported. The body is then said to rotate freely about the axis. But it can be shown that it is only in the case of the centroidal principal axis of great- est moment that the rotation is in kinetic stability. Thus, in ac- cordance with Art. 474, the rotation of a thin plate about an axis through its centre of inertia and perpendicular to its plane is stable. Pressure on the Axis when the Rotation is not Uniform. 492. When a body mounted on a fixed axis is subject to ex- ternal forces which have a resultant moment about the axis, the body will not only be in rotation but will have an angular accel- eration. The pressures upon the supports to the axis will now consist not only of those due to the external forces and to the centrifugal forces, but also of those due to another set of inertia forces, namely, the tangential components of the inertia of the various particles. Referring to rectangular axes as in Art. 483, this component of the inertia of the particle m is doo ~di' tnr-rr (l) and it acts in a direction at right angles to the direction of the radius vector r. This force acting at the particle is equivalent to a parallel force acting at the origin together with a couple. The resultant of these couples for all particles of the body is the mo- ment of inertia which we have already discussed. The forces con- stitute a system of the form (i) acting at the same points as the centrifugal forces, considered in the preceding articles, of which the form is mrCsD^. Each force of this system, in fact, differs from the corresponding force of the centrifugal system only in contain- doD . 2 . . . ing the common factor — in place of oo , and in acting in a § XXIII.] PLANE MOTION OF A RIGID BODY. 397 direction which is 90° behind or in advance of its line of action according as the angular acceleration is positive or negative. It follows that the resulting pressures upon the supports to the axes bea^ these relations, as to magnitude and direction, to the pressures which have been discussed in the preceding articles. Plane Motion of a Rigid Body. 493. A lamina moving in its own plane, or a rigid body so moving that a certain plane section of it remains always in the same plane, is said to have plane motion. The forces acting upon a body in plane motion, including those of inertia, are assumed to act in the plane ; for, if the body is constrained to have plane motion, we need only consider the resolved parts of the forces which lie in the plane. In the case of a solid, the particles actually move in the direction of lines parallel to the plane of reference. Thus the inertia forces con- sidered are equal and parallel to the actual ones, and are the same as if the particles were all projected on the plane ; so that the body may be replaced by a lamina, which will, however, in gen- eral be one of varying density. 494. Any plane motion can be resolved into two component motions ; one being a rotation about any selected point O^ and the other the motion of translation represented by the motion of the point O. We have had an illustration in the rolling wheel of Art. 41. It was there shown that, for any point of the rim, the velocity was at any instant the resultant of that due to the rota- tion about the centre, and the velocity of the centre itself, which was, in that case, a uniform velocity in a straight line. The same thing is true of any other point connected with the wheel. 495. The total momentum of a solid in motion is the sum of the momenta of all its particles. Now, remembering that mo- mentum is a vector quantity, it follows from the preceding article that the total momentum, at any instant, of a solid in plane motion is the sum of that due to the motion of translation rep- 39^ MOTION OF RIGID BODIES. [Art. 495. resented by the motion of 6>, and that due to the rotation about (9 as a fixed point. The first of these parts is obviously the same as if the whole mass were concentrated at O. 496. To find the part due to the rotation, let rectangular axes passing through O be assumed, and let x^y be the rectangu- lar and r, d the polar coordinates of a particle ni. Then, denot- ing the angular velocity of rotation by oa, its linear velocity is rce?, and its direction, supposing 00 positive, makes the angle B -\- 90° with the axis of x. It follows that the resolved velocity of m in the direction of the axis of ^ is — roo sin d\ hence, because y = r sin ^, the resolved momentum of m in this direc- tion is dx , . /«— = — myoo (i j Summing, we have, for the total resolved momentum along the axis of Xy '2m— = — Go2my = — Maoy, .... (2) where JK is the ordinate of the centre of inertia. Comparing with equation (i), we see that this component of the total momentum is the same as that of a particle of mass M situated at the centre of inertia. The same thing may be proved, in like manner, of the resolved part of the momentum along the axis of y. Therefore the total momentum due to rotation is the same as if, during the rotation of the body about O as a fixed point, the whole mass were concentrated at the centre of inertia. 497. Combining the results proved in the preceding articles, we find that ^/le total momentum of a body in plane motion is the same as if the whole mass were concentrated at the centre of inertia. When the centre of inertia is itself taken as the point of refer- ence, the momentum due to the rotation vanishes.* * That is to say, the linear momentum vanishes. The body possesses in virtue of its rotation an analogous property called angular momentum, which will be considered in the next chapter, Art. 524. § XXII I. J ROTATION AND TRANSLATION COMBINED. 399 Rotation and Translation Combined. 498. Let us suppose a lamina in plane motion to be acted upon by no external forces, and let us take the centre of inertia, 6r, as the point of reference. The lamina is thus regarded as rotating at any given instant about G^ while G is moving at the instant in a certain direction. Suppose now that G were so con- strained by proper guides that it could move only in the straight line having this direction. No pressure upon the guides will be produced by the motion of the mass regarded as concentrated at G^ because the motion is in a straight line ; at the same time, by Art. 490, no pressure will be produced by the centrifugal forces due to the rotation, since G is the centre of inertia. It follows that the constraint may be dispensed with, that is to say, the lamina will continue to rotate uniformly about the centre of inertia while that point describes a straight line with uniform speed. It is obvious that the same reasoning applies to the case of a rigid body rotating about a principal axis through the centre of inertia, Gy because the centrifugal forces are, by Art. 490, in equi- librium. In this motion, the axis of rotation remains fixed in the body and retains its direction in space, and the straight line in which G moves may make any angle with it. For stability of motion it is, however, necessary that the axis shall be that of greatest moment of inertia. 499. If an external force act upon the body at G^ or a system of forces such as those of gravity whose resultant acts at G^ the rate of rotation will remain uniform and the motion of G will be the same as if the entire mass were concentrated at G. Now suppose a force to act in one of the principal planes, but not at the centre of inertia. By Art. 102, this force is equivalent to an equal force acting at the centre of inertia, and a couple whose moment is the moment of the force about the centre of inertia. The force will therefore produce the same motion of translation as if it acted at the centre of inertia. In addition, the couple will produce the same angular acceleration (determined by 400 MOTION OF RIGID BODIES, [Art. 499. the equation of Art. 434) which it would produce if the axis per- pendicular to the plane and passing through the centre of inertia were fixed. 500. For example, suppose the fly-wheel in Fig. 104, p. 354, instead of having its axis fixed, were resting with its plane hori- zontal upon a smooth plane (so that the resistance of the plane neutralized the weight). The effect of the force P will now be a linear acceleration of the motion of the centre determined by as well as an angular acceleration of rotation determined as in Art. 435 by In explanation of the double effect of the force, it is to be no- ticed that, in this case, the force works through a greater space than before, and so produces an additional amount of kinetic energy which takes the form of energy of translation. 501. In the following example, two similar conditions of kinetic equilibrium serve to determine an unknown force, as well as the two accelerations which, by rea- son of a known relation which exists between them, constitute but one un- known quantity : Let a cylinder whose centre of gravity is on its axis be placed with its axis horizontal upon an inclined plane rough enough to compel it to roll with- ^^°- ^^3- out slipping. Let C, Fig. 113 (which represents a section made by a plane perpendicular to the axis), be the centre of gravity, at which acts the weight W =^ mg, and let A be the point of contact. Leaving out the component of W normal to the plane and the normal resistance at A (which, having the same line of action, is in equilibrium with it), the impressed § XXIII.] ROTATION AND TRANSLATION COMBINED. 4OI forces are the component ^sin ^, acting at C parallel to and down the plane, and the friction F acting at A up the plane. When the body is rolling down the plane, the forces of inertia which are in kinetic equilibrium with these forces are equivalent to the force w/, where /is the linear acceleration, together with the couple which resists angular acceleration. The force mf is represented in the diagram as acting at C Equilibrium of the forces parallel to the plane gives nig sin 6 — F -\- mf. (1) The moment of the couple to which the forces represented are equivalent is Fa*; hence, by equation (i), Art. 434, Fa = mk'i (2) a ^ ' where k is the radius of gyration about C, and the angular accel- eration is -, because, when rolling takes place, the linear and angular velocities are connected by the relation v = aoo. Eliminating i^ between equations (i) and (2), we find gd' sin ^ = (^' + «')/; . ^ . . . . (3) whence and, substituting in equation (2), k" sin e F=W a" -\- k^' 502. If the cylinder is homogeneous, k^ = \a}^ and we find f z=i\g sin ^; that is, the constant linear acceleration is two-thirds *The forces in the diagram do not represent complete kinetic equi- librium because we have not represented in it the inertia couple which balances this couple. 402 MOTION OF\l^IGID BODIES. [Art. 502, of what it would be if the plane were smooth. Accordingly, when the cylinder has descended through a given vertical height, two- thirds of the potential energy expended appears in the form of energy of translation and one-third as energy of rotation. There is in this example no work done against friction, and therefore no energy lost. If we regard one-third of the work of gravity as done against the force F^ we must also regard F as doing the same work against the rotational inertia. EXAMPLES. XXIII. ^ I. Determine the radius of gyration of the paraboloid, height h and radius of base b^ about a diameter of the base., 2. Find the radius of gyration of a right triangle whose-sides are a and b about an axis perpendicular to its plane and bisecting its hypothenuse. _ (^ -{■ b"^ 12 3. Determine the radius of gyration of a thin spherical shell of radius a about a tangent. k"^ =^ ^ «"• 4. Determine the radius of gyration of a thick shell about a tangent, the exterior and interior radii being a and b. ,, ^ 7 (^* + ^'^ + a'b-') + 2[ab* + b') ^[a'-\-ab-\-b-^) 5. Show that the values of k"^ for a homogeneous rectangulaf prism about its edges whose lengths are a, b and c are w-\-^% w-^^") and w-^n and that these are also the principal values at the centre for the prism whose sides are 2a^ 2b and 2c. 6. Show that the squared radius of gyration of the second prism in Ex. 5 about its diagonal, is k^ — .. ^ , .^ , — rr— , and there- 3(a -\-b ~\- c ) fore that of the prism whose edges are a, b and c about a diagonal is _ a'b' + b\' + c^a' ^ - 6{a^ + b'-^c') ' § XXIII.] EXAMPLES. 403 7. A body consists of a hemisphere and a cone of the same base and of height equal to the radius. Determine the radius of gyration about an axis through the vertex and parallel to the com- mon base. ,., 10 r , /^^ = -7- ^ . 60 8. Find the radius of gyration of a cone about a diameter of its base, the radius being b and the altitude h, _ 3/;^ + 2//" 20 9. Show that for a point on the circumference of a circular hoop the principal axes are a perpendicular to the plane, a tangent and a diameter; and, a being the radius, determine the principal moments of inertia. 20" M\ \a''M\ \a^xM. 10. If the centre of inertia of a lamina referred to rectangular axes is at the point [h^ k\ and x^ ^y^ denote the coordinates of the point (^, JF), when referred to parallel axes and the centre of inertia as origin, prove that 2mxy = "Smx^^ + Mhk. 11. By means of the theorem of the preceding example, show, from the results in Art. 467, that the direction of the principal axes at the centre of inertia of the triangle in Fig. iii is deter- mined by - ab tan 2a = -, and that those at the middle point of the hypothenuse are paral- lel to the sides. Verify the last result by considerations of sym- metry. 12. Prove that a centroidal principal axis of any solid is a principal axis for every one of its points. 13. Find the moment of inertia of a cone, radius of base b and height /i, about an element or slant height. K^if^h'^ -X- P\M 2Q{b'' + h") 14. A perfectly flexible cord is wrapped round a homogeneous cylinder of radius a. The cord is hauled in as the cylinder falls 404 MOTION OF RIGID BODIES. [Ex. XXIII. with its axis horizontal. With what acceleration is it hauled in if the cylinder falls with the acceleration \g ? \g. 15. A rod of length b^ bent into the form of a cycloid, oscil- lates about a horizontal line joining its extremities. Find the length of the equivalent simple pendulum. \b. 16. What must be the ratio of the radius to the height of a cone in order that the centre of oscillation may be in the base when that of suspension is the vertex ? Equality. 17. The height of the eaves and the ridge of a roof are 37^ and 46^- feet, respectively, and the slope of the roof is 30°. Find the time in which a homogeneous sphere rolling from the ridge will strike the ground. 3 seconds. 18. What is the angle at the vertex of the isosceles triangle of given area which oscillates in the least time about an axis through its vertex and perpendicular to its plane ? 90°. 19. What is the ratio of the times of vibration of a homo- geneous thin circular plate about a tangent and about a line through the point of contact perpendicular to the plane ? 4/5: +^6. 20. A circular arc oscillates about an axis through its middle point and perpendicular to its plane. Show that the length of the equivalent simple pendulum is independent of the extent of the arc. 21. What is the least value of the coefficient of friction for which rolling will take place in the case of the cylinder of Fig. 113, p. 400, supposed homogeneous ? \ tan 6. 22. If an inclined plane is just rough enough to insure the complete rolling of a homogeneous cylinder, show that a thin hol- low drum will roll and slip, the rate of slipping at any instant being one-half the linear velocity. CHAPTER XII. MOTION PRODUCED BY IMPULSIVE FORCE. XXIV. Effect of Impulsive Force. . 503. A force which acts for so short a time that neither the intensity of the force nor the time of action can be directly measured is called an imptilsive force. Such a force is, for exam- ple, called into action when a body receives a blow from a moving body, or, while moving, comes into contact with a body at rest. Let T denote the short interval of time during which the action takes place ; and let F denote the intensity of the force, which may undergo variation during the interval r. The acceleration which takes place, during the interval, in the body which receives the action, is proportional to the force, and like it cannot be meas- ured ; but the whole change of velocity produced is measurable. Now we have seen, in Art. 22, that the whole change of mo- mentum produced is the measure of the impulse. Hence, if m is the mass acted upon, we have I Fdt = niv — mv^f (i) in which 7o and v are the velocities before and after the impulse. 504. Accordingly, in treating of the effects of impulsive forces, our equations deal, not with the force F itself, but with the impulse, which may be denoted by {F), This im- pulse has, of course, a definite direction, which, in equation (i), 406 'motion produced by impulsive force. [Art. 504. was assumed to be the direction of the velocities v and v^. But, by the Second Law of Motion, a velocity transverse to the direc- tion of {F) may exist without modifying this equation. Hence the equation (7^) = mv — mv^ applies to a body having any motion, provided v^ and v denote the resolved velocities in the direction of {F) before and after the impulse respectively. 505. The shorter we suppose the interval r to be, in a given impulse, the greater must we suppose F to be. In default of any knowledge of t, except that it is small, we are compelled to assume it so small, and F so large, that the effect of any ordinary force, and the change of position due to any existing velocity, during the interval r may be neglected. For example, when a body moving in a straight line strikes a fixed wall, it receives an impulse at the moment of contact which changes its direction; it then moves off in another straight line. We are obliged to assume that the two straight parts of the path meet at an angle, as ABC^ Fig. 114, whereas in reality they must be connected by a very sharp curve. We may indeed define an impulsive force as a force which is assumed to produce a sudden chaiige of motion, and the impulse as the total action of the force; and this, in the case of a freely moving body, is measured by the momentum produced in the (which has destroyed t\\Q original momentum in the opposite direction), the body has received a further impulse, which has produced the momentum mv cos fi in the direction of the impulse. We infer, therefore, that the body has a power of regaining its form (like an elastic spring) after being compressed, so that a portion of the work done upon it by the first impulse, or impulse of compression^ is for tlie instant converted into 408 MOTION PRODUCED BY IMPULSIVE FORCE. [Art. '508. potential energy, and that the second impulse is due to the expenditure of this potential energy and its reconversion into kinetic energy. This property of certain bodies is called elasticity^ and the impulsive force exerted in the second impulse is called the force of restitution. 509. It is obvious that the kinetic energy restored by the second impulse cannot exceed that lost in the first ; therefore the second impulse cannot exceed the first. It was experiment- ally found by Newton that its ratio to the first impulse is inde- pendent of the magnitude of the impulse, depending only upon the material of the bodies in impact. We shall assume, therefore, that the second impulse is e times the first impulse, where ^ is a proper fraction which is called the coefficient of restitution. A body for which ^r = i is said to be perfectly elastic^ and one for which 7/', so that, if u' is positive m overtakes m\ and if u' is negative the bodies meet. In either case, the action of m upon m' is an impulse in the positive direction ; and, by the Law of Reaction, the action of m' upon 7n will be an impulse in the negative direction, which is equal to the other in magnitude, be- cause the time of action as well as the intensity of the impulsive force is the same for each. 513. In the first place, let the bodies be inelastic^ then the im- pulse will just suffice to prevent the further approach of the bodies, but not to separate them after contact; therefore after im- pact they will move with a common velocity V. The momentum of m before impact is mu^ and after impact it is m F, therefore m\ loss of momentum is m {u ~ F). The impulse is equal to this loss, and is also equal to the gain of momentum in the positive direction by m'; hence, denoting it by (^), we have (J?) = m{u-V) = m\V- u'); . . . (i) whence, eliminating (^), (m -1- ;// ) V = mu + ^'^^ (2) This last equation expresses that t/ie whole momentum after impact is the same as that before impact. §XXIV.] IMPACT OF ELASTIC SPHERES. 4II Solving equations (i) and (2) for Fand (^), we have ./../ _ mu-\-mu . ^- m-^m' ' • ^^^ mrn'^u-u') m -\- m Thus the common velocity F'is the weighted mean between the velocities u and «', and it is to be noticed that the impulse (^) is proportional to u — u\ that is, to the relative velocity of m with respect to m\ or the velocity of impact. It is, in fact, the momen- tum of a mass equal to one-half the harmonic mean between the given masses and moving with this velocity. 514. Next let the bodies be elastic, the coefficient of restitution being e. Then, after they have been reduced by the first impulse to the common velocity F, another impulse ei^R) is received, act- ing as before in the negative direction upon w, and in the posi- tive direction upon m! . Denote by v and v' the final velocities of m and m\ then m{y — v^ is the momentum lost by tn, and m'{v' — F) that gained by m! through this impulse. Thus e{R) =m(V-v) = m'{v' - V), ... (5) and these equations, solved as in the preceding article for V and the impulse e{R)y give _ mv -f m'v' ' .V m -f- m n,m'(v'-v) m -\- m Comparison of equations (3) and (6) shows that mv -f- ^'^'' = ^u + m'u' ; (8) hence, as before, the whole momentu?n is the same after impact as before impact. Again, comparison of equations (4) and (7) gives y — >d' — — e{u — «'), (9) 412 MOTION PRODUCED BY IMPULSIVE FORCE. [Art. 514. which shows that the relative velocity after impact is numerically e times that before impact^ and in the opposite direction, 515' Equations (8) and (9), which express principles readily remembered, should be used in solving all problems involving the four velocities, the value of e and the ratio of the masses, any two of which may be the unknown quantities. 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