V i- 7. ^' Kir ' V - r- ^ r <^ < .•■ r U^ t. PS''' ^. X ^ :.. mf^ K,- -,: . t-^'h^ *■■■■■■ i ^^1 '• • tf ' - . ^ > «',.'♦. 1^ ' / i- 1./ s V t '-3 ,\0 1 1,> TECHNICAL MECHANICS ""■• STATICS AND DYNAMICS BY EDWARD R. MAURER Professor of Mechanics in the University of Wisconsin FOURTH EDITION, REVISED AND ENLARGED TOTAL ISSUE TWENTY-TWO THOUSAND NEW YORK JOHN WILEY & SONS, Inc. London: CHAPMAN & HALL, Limited - J ,. 111? Engineering Library Copyright, igo3, 1914, 193^ BV Edward R. Maurer Stanbope ipress , GILSON COMP. BOSTON, U.S.A. H. GILSON COMPANY 19-01 PREFACE The following paragraph is an adaptation from the preface of the first edition of this work, published ten years ago; it applies to the present edition. This book might be described fairly as a theoretical mechanics for students of engineering. It is not comparable to books commonly called Theoretical Mechanics, generally intended for students of mathematics or physics; nor to books commonly titled AppUed IMechanics which generally include a treat- ment of strength of materials, hydraulics, etc., for students of engineering. The title Technical Mechanics seems fairly appropriate for this book; and inasmuch as it is not otherwise used in this country, it was so adopted. On the theoretical side, practically each subject discussed herein has a direct bearing on some engineering problem. The applications were selected and presented for the purpose of illustrating a principle of mechanics and for training students in the use of such principles, — not to furnish information, except incidentally, about the structure, machine, or what not to which the application was made. Ten years use of the book as a text in the author's classes has suggested many changes; and in recent years the need of a new collection of problems has become urgent. Accordingly, a revision was undertaken, and the effort has resulted in a practically rewritten book. Indeed the only portion of the former edition used again with little or no change is the present Appendix A. Though containing fewer pages than the old book, the new one — because of its (nearly one-third) larger printed page — contains more material than the old. Inasmuch as Mechanics deals mainly with subjects permanent in character, the revision consists principally of changes in arrangement and presentation. Both were determined upon to a large degree by a desire to furnish an ade- quate course of instruction for students in engineering in one semester, "five times per week." To this end, it was necessary to sacrifice logical order of arrangement more or less. As in former editions, Statics is presented first because relatively simpler than Dynamics. Kinematics, as such, is not given a place. The chapter on Attraction and Stress has not been retained. Dis- cussion of Friction and Efficiency has been amplified, and Dynamics has been extended to provide a quantitative explanation of simple gyroscopic action. Many solved numerical examples have been added to elucidate principles. The collection of problems to be solved by students has been completely changed. lU 4942^^5 IV All of Statics except Arts. 23, 25, 26, and 27 may be mastered with no knowl- edge of mathematics beyond trigonometry. Calculus methods are used in Dynamics, but a good knowledge of the elements only of that branch of mathematics is presupposed. Graphical methods are used freely, as much as the algebraic in Statics. The author is pleased to acknowledge with thanks the helpful suggestions and criticisms of the teaching staff in Mechanics at the University of Ilhnois; of his colleague, Professor M. O. Withey; and of Professor C. H. Burnside of Columbia University. He thanks also American Machinist, Engineering Record, and Engineering News for permission to copy and for gifts of cuts; and individuals and other journals named in the text for similar favors. Madison, Wisconsin. December, 1913. To the edition above described there has been added a second collection of problems, pages 354-377; and articles 38, 44, 49, 5^, 5^, 55, 5^, 58 have been modified. September, 191 7. TABLE OF CONTENTS CHAPTER I COMPOSITION AND RESOLUTION OF FORCES Article Page 1. Introduction i 2. Force; Definitions 4 3. Parallelogram and Triangle of Forces 7 4. Composition of Concurrent Forces 11 5. Moment of a Force; Couples 16 6. Graphical Composition of Coplanar Nonconcurrent Forces 20 7. Algebraic Composition of Coplanar Nonconcurrent Forces 23 8. Moment of a Force; Couples 27 9. Noncoplanar Nonconcurrent Forces 3° CHAPTER II FORCES IN EQUILIBRIUM 10. Principles of Equilibrium - 34 11. Coplanar Concurrent Forces 4° 12. Coplanar Parallel Forces 44 13. Coplanar Nonconcurrent Nonparallel Forces 46 14. Noncoplanar Forces 5° CHAPTER III SIMPLE STRUCTURES 15. Simple Frameworks (Truss Type) 54 16. Graphical Analysis of Trusses; Stress Diagrams 59 17. Simple Frameworks (Crane Type) 64 18. Cranes 69 CHAPTER IV FRICTION 19. Definitions and General Principles 74 20. Friction in Some Mechanical Devices 78 CHAPTER V CENTER OF GRAVITY Center of Gravity of Bodies 86 Centroids of Lines, Surfaces, and Solids 9° 23. Centroids Determined by Integration 93 24. Centroids of Some Lines, Surfaces, and Solids 98 V 21. 22. VI CHAPTER VI SUSPENDED CABLES, WIRES, CHAINS, ETC. Article Page 25. Parabolic Cable 102 26. Catenary Cable 107 27. Cable with Concentrated Loads 113 CHAPTER VII RECTILINEAR MOTION 28. Velocity and Acceleration 118 29. Motion Graphs 126 30. Simple Harmonic Motion 131 31. Motion and Force 138 CHAPTER VIII CURVILINEAR MOTION 32. Velocity and Acceleration 144 33. Components of Velocity and Acceleration 148 34. Motion of the Center of Gravity of a Body 155 CHAPTER IX TRANSLATION AND ROTATION 35. Translation 163 36. Moment of Inertia and Radius of Gyration 168 37. Rotation 176 38. Axle Reactions 180 39. Pendulums 183 CHAPTER X WORK, ENERGY, POWER 40. Work 189 41. Energy 193 42. Power 196 43. Principles of Work and Energy 203 44. Efficiency; Hoists 211 45. Kinetic Friction 221 CHAPTER XI MOMENTUM AND IMPULSE 46. Linear Momentum and Impulse 228 47. Impact or Collision 232 48. Angular Momentum and Impulse 237 49. Gyrostat 243 vu CHAPTER XII TWO DIMENSIONAL (PLANE) MOTION Article Page 50. Kinematics of Plane Motion 256 51. Kinetics of Plane Motion 261 52. Rolling Resistance 268 53. Relative Motion 273 CHAPTER XIII THREE DIMENSIONAL (SOLID) MOTION 54. Body With a Fixed Point, Kinematics of 280 55. Body With a Fixed Point, Kinetics of 284 56. Gyrostat 288 57. Principal Moments of Inertia, and Axes 292 58. Any Solid Motion; Summary of Dynamics 296 APPENDIX A. THEORY OF DIMENSIONS OF UNITS 302 APPENDIX B. MOMENT OF INERTIA OF PLANE AREAS 308 PROBLEMS 323 TECHNICAL MECHANICS I. Introduction Mechanics had its origin in the experience of ancient peoples with de- vices for lifting and moving heavy things. The devices included the so-called simple machines or mechanical powers; namely, the lever, the pulley, the wheel and axle, the inclined plane, the wedge and the screw. That experience probably afforded fairly definite and full knowledge of the practical advantages of these various devices, but the simple and precise mechanical principles involved in them were long unrecognized. The first recognition of such a principle marked the real beginning of the science of Mechanics. History records that the principle of the lever is the mechanical principle first discovered, and that Archimedes (287-212 B.C.), famous Greek mathe- matician, was the discoverer. He perceived the application of this prin- ciple to the wheel and axle (continuous lever), to the pulley (movable lever), and to certain combinations or systems of pulleys and cords, one of which still bears his name. The discovery of the principle of buoyant effort on a body floating on or immersed in a fluid is due to him. Apparently no additions to these achievements of Archimedes were made during the sixteen centuries following his time. The principle of the lever as understood by Archimedes covered only the special case of two heavy weights suspended from a horizontal bar sup- ported at a point (fulcrum) between them. For such case he stated that the weights are inversely as the distances from the fulcrum to the points of suspension. The principle was extended to include the case of forces ap- plied obliquely, by Leonardo da Vinci (145 2-1 5 19), famous Italian artist and engineer. He perceived that the efficacy of such a force depends on the distance from the fulcrum, not to the point of application of the force, but to its line of action. The principle next discovered was that of the inclined plane, first defi- nitely stated by Simon Stevin (i 548-1620), Dutch mathematician and en- gineer. His statement of the principle was somewhat as follows: The force (acting along the plane) required to support a (frictionless) body resting upon it is to the weight of the body as the height of the plane is to its length (measured along the slope). This principle afforded the explana- tion of the wedge (double inclined plane) and the screw (continuous inclined plane). Stevin deduced the parallelogram law for two forces at right 2 ' ■ ' Art. I •a'n^ies' 'from the principle of the inclined plane; and from his study of pulleys he noted that what is gained in power is lost in speed. Thus he caught the first glimpse of two important principles, — that of the parallelo- gram of forces, and that of virtual velocity or work. The first discoveries of laws of motion were made by Galileo (i 564-1 642), Italian astronomer and physicist. For 2000 years it had been believed that heavy bodies fall more rapidly than light ones. This Galileo disproved by actual trial at the leaning tower of Pisa. Next he was led to inquire about the manner in which a body falls, or how the speed changes. He made several guesses at this law, and finally verified one of them by indirect experiment and deduction. Up to Galileo's time, it was believed that rest was the natural condition for a body; and that motion was unnatural, requiring some outside cause (force) to maintain it, and ceasing only when the force ceases. Galileo perceived that motion is just as natural as rest; that motions cease not because they are unnatural, but because of some influence (force) from the outside operating to reduce the motion and eventually to destroy it. In short, he discovered the so-called first law of motion, usually credited to Newton. He invented the telescope. Huygens (1629-1695), Dutch physicist, made some important contribu- tions to this science. He developed the theory of the pendulum, determined the acceleration due to gravity from pendulum obser\^ations, and deduced certain theorems regarding centrifugal force. He invented the clock pen- dulimi and escapement. Newton (1642-1727), English mathematician and physicist, is generally regarded as the founder of Mechanics. At an early age he began an at- tempt to explain the motions of the planets, whose orbits and speeds were then well known, in terms of experience with more familiar motions. He succeeded in thus explaining many features of the planetary motions, and established that there are certain principles common to the motion of all bodies, celestial and terrestial. These principles are generally known as Newton's laws of motions (see index). His study of planetary motion led to other great achievements, among which may be mentioned the discovery of the law of universal gravitation, and the invention of the calculus (also invented independently by Leibnitz, German mathematician). Since Newton, "no essentially new principle [of Mechanics] has been stated. All that has been accomplished since his day has been a deductive, formal, and mathematical development on the basis of Newton's laws."* Such development consritutes the body of knowledge which we call Me- chanics, or sometimes Rational and Theoretical Mechanics, to distinguish it from Applied Mechanics. It may be defined as the science of motion, but it includes the science of rest as a relatively minor part. * For a full and critical account of that development, see Mach's " Science of Me- chanics," from which the quotation was taken, or Co.x's " Mechanics " for a good but less critical account. Art. I « Adaptations of rational mechanics have played an important part in the development of the science of engineering, particularly in the depart- ments of structures and machines. Such adaptations, together with our knowledge of friction, strength of materials, and certain properties of fluids, constitute Applied Mechanics. Among the pioneer workers in this field should be mentioned the following: Coulomb (1736-1806), Navier (1785- 1836), Poncelet (1788-1867), Morin (1795-1880), Saint-Venant (1797-1886), Weisbach (1806-71), Rankine (1820-72), Grashof (1826-93) and Bauschinger (1834-93).* Under Technical Mechanics, the present author includes those prin- ciples of rational mechanics which are especially applicable in various fields of engineering, and some of our knowledge of friction. The book is divided into two parts called Statics and Dynamics. The first deals with certain of the circumstances of bodies at rest, and the second with those of bodies in motion. The certain circumstances dealt with will become ap- parent to the student as he progresses in the subject. * See Keek's " Mechanik " for an account of their work and fuller list. STATICS CHAPTER I COMPOSITION AND RESOLUTION OF FORCES 2. Force; Definitions Bodies act upon each other in various ways, producing different kinds of results. Any action of one body upon another which, when exerted alone, would result in motion of the body acted upon, or in change of motion if the body is already moving, is called force; the word is a general term for push and pull. Our earliest notions about forces are based on our experience with forces exerted by or upon ourselves. Through this experience we have learned that a force has magnitude, place of application, and direction, sometimes called the characteristics of a force. To express the magnitude of a force, we must of course compare it to some other force regarded as a unit. Many units of force are in use; the most convenient are the so-called gravitation units. They are the earth- pulls on our standards for measuring quantity of material (as iron, coal, grain, sugar, etc.), commonly called standards of weight.* The earth-pull on any of these standards is called by the name of the standard; thus the earth-pull on the pound standard (also any equal force) is called a pound; the earth-pull on the kilogram standard (also any equal force) is called a kilogram, etc. Since the earth-pull on any given thing varies in amount as the thing is transported from place to place, gravitation units of force are not constant with regard to place. But this variation need not be regarded in most engineering calculations because any error due to such disregard is generally smaller than errors due to other approximations in the calculations. The extreme variation in any gravitation unit is that between its magnitudes at the highest elevation on the equator and at the poles; this difference is but 0.6 per cent. For points within the United States the extreme variation equals about 0.3 per cent. For any two * In common parlance the word weight is used in at least two senses. Thus, suppose that a dealer sells coal to a consumer by weight, and engages a teamster to deliver it by weight; to the consumer, the weight of each wagon load represents a certain amount of useful material, but to the teamster it represents a certain burden on his team due to the action of gravity on the coal. That is, weight suggests material to the one man and earth-pull to the other. Art. 2 5 points on the surface of the earth, the variation equals that in the values of g in the formula g = 32.0894 (i + 0.0052375 sin^O (i ~ 0.0000000957 e) computed for the two places; I denotes latitude, and e elevation above sea level, in feet. The place of application of most forces with which we shall deal is a portion of the surface of the body to which the force is applied. A notable exception is earth-pull, or gravity, which is applied not to the surface of a body but throughout the same. All such are called distributed forces. The places of application of some forces are very small compared to the sur- faces of the bodies to which they are applied, and for many purposes these places may be regarded as points of application; any such force is called a concentrated force. The line of action of a concentrated force is a line indefinite in length, parallel to the direction of the force, and containing its point of application. A concentrated force may act along its line of action in one of two ways, — to the right or left, up or down, etc. We say that the sense of a force is toward the right, toward the left, up, or down as the case may be. That is, sense refers to " arrow-headedness " (see next paragraph). Since a force is a vector quantity,* it can be represented in part by a vector (a straight line of definite length and direction), the length of the a vector representing the magnitude of the force according to some scale, and the direction of the vector giving the direction of the force. Thus, if the pressures of the driving wheels of the locomotive on the rails (Fig. i) is 12 tons, then the vector Aa (0.4 inch long) represents the magnitude and direction of the pressures, the scale being one inch " equals " 30 tons. If the force to be represented is a concentrated one, as in the illustration, then the line of action also can be represented by the same vector which repre- sents the force magnitude by drawing it through the point of application of the force. Thus the vector Bb represents magnitude, line of action, and direction of the pressure of the first driving wheel. We might extend this scheme further so as to indicate also point of application of the force by the head, say, of the vector as Cc; but we will not plan to do that because the point of application is not of importance in this subject, — Statics. * A vector quantity is one having magnitude and direction, as, for example, a definite displacement of a moving point. A quantity having magnitude only, as the volume of a thing for example, is a scalar quantity. Chap, i f 1 \ B Fig. 2 This unimportance of the point of application is definitely expressed in the principle of transmissibility of force, which for the present purpose may be stated as follows: The effect of any force applied to a rigid body at rest is the same, no matter where in its own line of action the force is applied. The principle may be roughly verified by experiment, when the body on which the force acts is at rest, with the apparatus represented in Fig. 2; it con- sists of a rigid body suspended from two spring balances. The springs are elongated on account of the weight of the body, and if a force, as F, be applied at A, the springs will suffer additional elonga- tions which in a way are a measure of the effect of the applied force. If the point of application of F be changed to B or C, the spring readings will not change; hence the effect of F will not have changed. Generally, when many forces are to be represented graphically and dis- cussed, it would be well to represent each force by a line and a vector, the first to represent the line of action of the force and the second to represent the magnitude and the direction of the force. Of course the line must be drawn through the point of application of the force, but the vector may be drawn where convenient. For example, consider the forces act- ing on the upper end of the boom (Fig. 3) of a derrick. There are three forces; namely, a down- ward force at pin i, one toward the left at pin 2, and one downward at pin 3. The lines marked ah, cd, and ef are the lines of action of the forces respectively; the vectors AB, CD, and EF (drawn where convenient but of proper length and direc- tion) represent the magnitudes and directions of the forces. The scheme of notation here used — two lower-case letters on opposite sides of the line of action of a force, and the same capital letters at the ends of the vector representing its value — is in common use. Any force so marked is re- ferred to in written statement by the two capitals used; thus the first force mentioned above would be called the force AB. The part of the drawing in which the lines of action of the forces and the body (here, a derrick-boom) are represented is called a space diagram; the part in which the vectors are drawn is called a vector diagram. The scales of these diagrams are of course different; the lengths of lines in the first represent distances, and those in the second, force magnitudes. Fig. Art. 3 7 Any number of forces collectively considered is called a system or a set of forces. The forces of a set are called coplanar if their lines of action are in the same plane, and noncoplanar if not in the same plane; they are called concurrent if their lines of action intersect in a point, and noncon- current when they do not so intersect; they are called parallel if their lines of action are parallel, and nonpar allel if the lines of action are not parallel. Force-sets are also described in accordance with the foregoing definitions; thus, a concurrent set, a noncoplanar parallel set, etc., according as the forces of the set are concurrent, noncoplanar and parallel, etc. Force- sets can be classified in various ways, as below for example^ — Coplanar , fcolinear i concurrent < ,, • \ nonparallel .... 2 nonconcurrent -! ^^^^ ^ W i ' " ' ^ \nonparallel .... 4 concurrent 5 Noncoplanar between them be 60 degrees; required, their resultant R. Rouglily, ABC is the triangle for the forces, AC representing the magnitude and direction of R, and the angle ABC = 180° — 60° = 120°. Then from the trigonometry of the triangle, R^ = 100^ + 150^ — 2 X 100 X 150 cos 120° = 47,500, or R = 218.3; 2.1so sin CAB/sin 120° = 150/i?, or CAB (the angle a between R and P) = 36° 35'. Employing the foregoing method, the following general form- ulas may be worked out for determining the magnitude and direction of the resultant, — Ri = p2-\-Qi^ 2 PQ cos <^; sin a = sin <^ • Q/R, and sin /3 = sin • P/R, where , a, and are the angles marked in Fig. 9. When the two forces P and Q are at right angles to each other (0 = 90 degrees) , then ^2 = p2_|_Q2^ and tarn a-=Q/ P. § 2. Resolution of a Force into Concurrent Components can be accomplished by applying the triangle or parallelogram law inversely. Thus, let it be required to resolve the force F (Fig. 10) into two components. We draw AB anywhere equal (by some scale) and parallel to F; join any point C with A and B, and draw lines through any point in ab parallel to ^C and BC; then AC and CB represent the magnitudes and directions, ac and cb the lines of action of two forces equivalent to F, that is, components of F. For the resultant of these two component forces is F, as shown by the tri- * For convenience and clearness of figure, a subdivided square (or rectangle) will herein- after represent a machine, or structure (derrick -boom, bridge, etc.), on which the forces under discussion act. If he prefers, the student might regard the subdivided square as representing a drawing board or some other definite object suggested by the square. It is important that he should have in mind the fact that forces act only on material things (bodies), and that the lines of action of the forces represented in any given figure are definitely related to th« body on which the forces act. Art. 4 II angle law applied directly. Since C was taken at random, it is plain that a given force can be resolved into many different pairs of components. If conditions be imposed on the components, the resolution is more or less definite. Thus, let it be required to resolve F (Fig. n), equal to 350 pounds, into two components, one of which must act along the left-hand edge of the board and the other through the lower right-hand corner. Since the three forces must be concurrent, the second component must act through point i; so we make AB equal and parallel to F and draw from A and B lines parallel to the two components; then AC and CB represent the values (200 and 320 pounds respectively) and the directions of the components. An important case of resolution is that in which the components are at right angles to each other. Each is called a rectangular component or re- solved part of the force. Rectangular components can generally be com- puted more easily than by geometrical construction. Let F (Fig. 12) be the given force to be resolved into horizontal and vertical components, the 9^ s F • /hk. b C=i^ A Fig. 10 B c!a I > i jJ^- r 1 -i--~. b c^ Fig. II angles between F and the components being a and /5 respectively. From ABC, a sketch of the triangle of resolution, not necessarily a scale drawing, it is plain that the desired components equal F cos a and F cos /3 respectively. And always the rectangular com- ponent of a force along any line Jthe magnitude ^ ■ tof the force J ^ " the cosine of the acute angle between Ithe force and that line. The components of a force along two rectangular coordinate axes x and y are called the x and y components of the force respectively; they will be denoted by Fj, and Fy. 4. Composition of Concurrent Forces In the preceding article we showed how to determine the resultant of two concurrent forces; in this article we show how to determine the resultant of any number of such forces. § I. CoPLANAR Forces. — Graphical method. By means of the paral- lelogram or triangle of forces (Art. 3) find the resultant R' of any two of 12 Chap, i the forces of the given set; then find the resultant R" of any other given force and R'; then the resultant of another given force and R"\ and so on until the resultant of all is found. Thus, suppose that the resultant of Fi, F2, Fs, and Fi (Fig. 13) is required: Taking the given forces in the order in which they are numbered, say, we first draw AB parallel to Fi and equal to Fi by some convenient scale, then BC in the direction of and equal to F2; then AC gives the magnitude and direction of R', the line of action of R' passing through O parallel to AC. Next we draw CD in the direction of Fig. 13 F3 and equal to F3; then AD gives the magnitude and direction of R", the line of action of R" passing through O parallel to AD. Next we draw DE in the direction of and equal to Fi; then AE gives the magnitude and direction of R'", the line action of R'" passing through parallel to AE. Of course the lines AC, AD, R', and R" are not really essential to the solu- tion; they were drawn here and referred to only for explanatory purposes. The force polygon for a set of forces is the figure formed by drawing in succession and continuously lines which represent the magnitudes and directions of those forces. A force polygon is not necessarily a closed Fig. 14 figure; thus ABCDE, not including EA, is a force polygon for Fi, F2, F3, and Fi. Many force polygons can be drawn for a given set of forces, as many as there are orders of taking the forces; if there are n forces in the set, then I • 2 • 3 • • • '11 different force polygons can be drawn. In Fig. 14 ad- ditional polygons ABCDE are shown for Fi, F2, F3, and Fi of Fig. 13; the lines AE represent the magnitude and direction of R. The bare con- struction for determining the resultant of a set of concurrent forces can now be stated thus: Draw a polygon for the forces; join the beginning and the end of the polygon, and draw a line through the point of concurrence of the given forces parallel to the joining line; the joining line, with arrow- head pointing from the beginning to end of the force polygon, represents Aet. 4 13 the magnitude and direction of the resultant, and the other line its line of action. Algebraic Method. — Choose a pair of rectangular axes of resolution, which let us call X and y axes, with origin at the point of concurrence of the forces to be compounded; then resolve each force into its x and y components at the origin, and imagine it replaced by them; the resulting system consists of forces in ihex and in the 3' axes; next find the resultant of the forces act- ing in the x axis, and the resultant of those acting in the y axis; finally, get the resultant of these two rectangular resultants; this is the resultant sought. For example, let it be required to determine the resultant of the six forces acting upon the 4 foot board shown in Fig. 15. The computations in outline are scheduled below. The values of the angles which the several forces make with the horizontal were computed from dimensions in the figure; the sum of the x components is + 3.40, and that of the 3' components is— 7.22 pounds. The signs of the sums indicate respectively that the x com- ponent of the resultant R acts toward the right and the y component downward; hence the resultant acts to the right and downward. The angle which R makes with the horizontal is tan~^ (7.22 -T- 3,40 = 2.123) — 64° 47'- The value of the resultant is R = \/34o^ + 7.22^ = 7.98 pounds. 5 lbs. F a cos a sin a Fx Fy 8 4 6 12 7 S 45° 63° 26' 36° 52' 90° 14° 2' I. 0.707 0.447 0.800 0. 0.970 0. 0.707 0.894 0.606 I . 0.242 +8. + 2.83 -2.68 —9.60 0.00 +4.85 + 2.83 + 5-36 — 7.20 — 7.00 — I. 21 +3 -40 — 7.22 § 2. NoNCOPLANAR FORCES. — Before showing how to find the resultant of any number of such forces, we explain (i) how to find the resultant of three rectangular concurrent forces (lines of action at right angles to each other), and (2) how to resolve a force into three nonco- planar rectangular components. (i) If three noncoplanar concurrent forces acting on a rigid body be represented by OA , OB, and OC, then their resultant is represented by the diagonal OD of the par- allelopiped OABC (Fig. 16). (This is called the parallel- opiped law, and OABC is called a parallelopiped fof forces.) For, according to the parallelogram law, OC represents the resultant of two of the forces OA Fig. 16 14 Chap, i and OB, and OD represents the resultant of OC and the third force OC, and (hence, also) the resultant of the three given forces. This law leads to a simple algebraic method for finding the resultant when the three forces are rectangular (at right angles to each other). Thus, let Fi, F2, and F3 (Fig. 17) be the three forces, R their resultant, and di, 62, and 9$ the angles between R and the forces respectively; then R' = Fi^ + F2' + FzS cos 01 = Fi/R, cos 02 = F2/R, cos ^3 = F3/R. For the resultant of Fi and F2 (represented by OC, Fig. 17) equals (Fi^ + Fa^)!, and hence R^ = (Fi^ + F2'') + Fs'' ; also the triangles ODA, ODB, and ODC are right-angled Sit A, B, and C respectively, and hence cos^i = OA/OD = Fi/R, cos ^2 = OB/OD = F2/R, etc. (2) A force can be resolved into three noncoplanar concurrent forces by applying the parallelepiped law inversely. Thus, let OD (Fig. 18) represent the given force F; first, construct any parallel opiped of which OD is a Fig. 17 Fig. 18 Fig. 19 diagonal; then the three edges intersecting at represent forces equivalent to the given force because the resultant of these three forces is, according to the parallelepiped law, represented by OD. Inasmuch as many paral- lelopipeds can be constructed on OD as diagonal, many sets of three forces equivalent to the given force can be found. The practical case is resolution into components along three definite rectangular axes; then there is only one set of components. The com- ponents may be found quite simply by an algebraic method: thus, let F (Fig. 19) be the force to be resolved, a, 13, and 7 the angles between F and the axes, and Fx, Fy, and Fg the x, y, and z components respectively; then, since OX, OY, and OZ are projections of OD on the rectangular axes, Fx = F cos a, Fy = F cos jS, Fz = F cos 7. Sometimes the direction of the force F to be resolved is given by means of two angles, one being the angle between F and one of the desired compo- nents, and the other being the angle which the projection of F on the plane of the other two components makes with one of those two, as for instance a and 4> (Fig. 19). Then F may be resolved best in this way: first, resolve Art. 4 15 it into two components F cos a (along the x axis) and F sin a (in the plane of the y and 2 axes), and then resolve F sin a into components along the y and 2 axes, that is, F sin a sin and F sin a cos c^. ^ny number of noncoplanar concurrent forces can be compounded graphically by means of their force polygon, but this method is not practi- cable generally, because the polygon is not a plane one; however, it could be drawn in " plan and elevation " so as to furnish the resultant sought. The algebraic method is preferable; it is carried out as follows: First, select three rectangular axes of resolution (here called x, y, and 2), with origin at the point of concurrence of the forces to be compounded; next resolve each force into its x, y, and 2 components, and imagine it replaced by them, thus arriving at a set consisting of forces acting in the axes; then find the resultants of the forces in the x, in the y, and in the 2 axis; finally, compound these three resultants, thus finding the resultant sought. For example, let it be required to determine the resultant of the four forces acting on a 4 foot cube (Fig. 20). The forces are concurrent at 0; the 10 and the 15 pound forces act through quarter points of certain edges as shown. The x, y, and 2 components of the 18 and 40 pound forces are ob- viously as scheduled adjoining. Since the 15 pound force is perpendicular to the x axis, its x component equals zero; and since the angle which that force makes with the 2 axis = tan~^ f = 36° 52', its y and 2 components are 15 sin 36° 52' = 9, and 15 cos 36° 52' = 12 pounds respectively as scheduled. The components of the 10 poimd force were de- termined as follows: Since Ya = 5 and YO = 4 feet, the angle which the ID pound force makes with the y axis is tan~^ | = 51° 20'; the y component of the force equals 10 cos 51° 20' = 6.25 as scheduled, and the other rec- tangular component (in the zx plane) equals 10 sin 51° 20' = 7.81 pounds. Fig. 20 F Fz Fv F, 18 40 IS 10 18.00 0.00 0.00 -4.69 0.00 0.00 — 9.00 -6.25 0.00 40.00 — 1 2 . 00 -6.25 + 13-31 -15-25 + 21.75 The angle which this component, acting in Ob, makes with the z axis equals tan~^ f = 36° 52'; hence the x and z components of the 10 pound force equal respectively 7.81 sin 36° 52', or 4,69, and 7.81 cos 36° 52', or 6.25 pounds. The value of the resultant is R = V 13.312 -f 15.252 + 21.752 = 29.7 pounds. l6 Chap, i The signs of the sums of the x, y, and z components show that the result- ant R acts toward the right, downwards and forward. Its angles with the X, y, and z axes are respectively: cos"' (13.31 -r- 29.7) = 63°; cos~^ (15-25 -j- 29.7) = 59°; cos-i (21.75 -^ 29.7) = 43°. 5. Moment of a Force; Couples* § I. The Moment or Torque of a force with respect to a point is the product of the magnitude of the force and the perpendicular distance be- tween its line of action and the point. The perpendicular distance is called the arm of the force with respect to that point, and the point is called an origin or center oj moments. Experience suggests the notion that the moment of a force with respect to a point is a measure of the tendency of the force to rotate the body about a line through the point and perpen- dicular to the plane of the force and the point. Such a notion can be verified quite accurately by means of a simple apparatus represented in Fig. 21. It consists of a board mounted on a horizontal shaft, a heavy body, and the pail which can be suspended horn the board; the shaft rests in ball bearings so that practically no resistance to turning is exerted at the shaft; the board, without the body and the pail, is well balanced so that gravity would not cause it to turn from any position. Now, let the pail containing shot be hung from B, C, D, etc., in succession, the amount of shot being taken so that the heavy body will be supported, OA not being horizontal necessarily. Then in each case the turning effect of the pull at B, C, or D equals the turning effect of the pull at A ; hence the turning effects of the pulls at B, C, D, etc., are equal. And if the moments of these pulls (several weights of pail and shot) about O be computed, then those mo- ments will be found equal too, and therefore moments are measures of turn- ing effects. It follows from the definition of moment that the unit moment is that of a unit force whose arm is a unit length. There are no one-word names for any of these units of moment; the units are called foot-pound, inch- ton, etc., according as the unit length and force are the foot and the pound, the inch and the ton, etc. In a discussion involving the moments of several forces, it is generally convenient to give signs to the moments to indicate the directions (clock- wise or anticlockwise) in which the several forces turn or tend to turn the body to which they are applied about the origin in question. In this book, clockwise rotation is regarded as negative and anti as positive, and rotations are supposed to be viewed from the reader's side of the printed page; * See Art. 8 also. Art. s 17 lOOlbs. 30 A JB... ■-- — ■s. thus the moment of the 200 pound force (Fig. 22) about O is positive and about A negative. Principles of Moments. — If two sets of coplanar forces are equivalent (Art. 2), then the moment-sum* for one set with respect to any point equals that for the other with respect to the same point. This will \ A \ ^ be granted as self-evident by most students; others may be convinced by this: Let Si and ^'2 denote the two equivalent sets, and Sz a third set (coplanar with ''200 lb5. Si and ^2) which could balance Si, and hence also S2. yig. 22 Since Si and Sz would balance, they together would not turn the body on which they act about any line; hence the moment-sums for Si and 53 with respect to any point in the plane are equal in value but opposite in sign. Likewise, the moment-sums for S2. and ^3 with respect to O are equal in value but opposite in sign. Therefore the moment-sums for Si and S2 wath respect to O, being equal to the same thing, are equal. It follows from the foregoing principle that the moment-sum for any set of coplanar forces with respect to any point in their plane equals the moment of their resultant with respect to that point. Also, the moment of a force with respect to a point equals the moment-sum of its components with respect to the same point, the components to be coplanar with the given force and the point. When the moment of a force about a certain point must be computed, and the arm of the force with respect to that point cannot be easily de- termined, then the desired moment can be computed, more easily perhaps, from components of the force, by aid of the preceding principle. Thus, let it be required to compute the moment of the 100 pound force (Fig. 22) about O. The horizontal and vertical components of the force are respectively, 86.6 and 50 pounds; imagining them applied at A makes their arms 3 and 4 feet respectively, hence the desired moment equals — (86.6 X 3) — (50 X 4) = — 459-8 foot-pounds. Sometimes the components can be applied (in imagination) so that one passes through the origin of moments; then its moment equals zero and the desired moment equals the moment of the other component. Thus the horizontal and vertical compo- nents of the 200 pound force equal 89.4 and 178.8 pounds respectively; imagining them applied at C, their arms are o and 3 feet, and so the desired moment equals 178.8 X 3 = 536.4 foot-pounds. f *" Moment-sum " means the algebraic sum of the moments of the forces of a system; torque of a set of forces means the same thing. t Some writers regard the parallelogram law (for forces) as fundamental, and deduce the principle of moments from the law. We give such a deduction of the principle for two forces, — namely, the moment of the resultant of two concurrent forces about any point in their plane equals the algebraic sum of the moments of the forces about the same point (Varignon's theorem). The theorem can be extended easily to any number of coplanar forces, thus proving the principle. Let P and Q (Fig. 24) be two concurrent forces acting — on a body not shown — in i8 Chap, b B r"^ c % V a A V Fig. 23 § 2. Couples (see also Art. 8), — Two equal and parallel forces which are opposite in sense may advantageously be considered together; so con- sidered, they are called a couple. By arm of the couple is meant the per- pendicular distance between the lines of action of the forces; by plane of the couple is meant the plane of those lines of action. The jnoment or torque of the couple about any point in the plane of the couple is the algebraic sum of the moments of the forces about that same point. This sum, or moment, including the sign, is the same for all origins in the plane. For, let F and F (Fig. 23) be the forces of the couple, AB being the arm; then the moments of the couple with respect to the origins a, b, and c are respectively: - F(Aa) + FiBa) = F{AB); + F{Ab) - F(Bb) = F(AB); and + F{Ac) + F(Bc) = F(AB). Since a, b, and c represent all possible origins, the proposition is proved. Since the origin of moments is immaterial, no reference is made to it in speaking of the moment of a couple. Moreover, the value of the moment is computed more simply than as stated in the definition, by multiplying the common magnitude of the forces and the arm of the couple. As just shown, the moment of the couple FF is positive ; that is, the couple would turn the body, if free, counterclockwise about any axis perpendicular to the plane of the couple; obviously, the moment of the couple GG is negative. Sense of a couple refers to the way in which the couple would turn a body; thus we speak of a clockwise sense or a counterclockwise sense. Two coplaner couples whose mojnents (including sign) are equal are equivalent. The following proof is in two parts; namely, (i) when the four forces of the couples are nonparallel, and (2) when they are paral- lel, (i) Let P1P2 (Fig. 25) be one given couple, and Q1Q2 the other, their arms being p and q respectively; then Pp = Qq. We will show that the P couple would balance the reversed Q couple; it will follow that the given couples are equivalent. The area of the parallelogram A BCD = ADp = ACq; therefore, since Pp = Qq, AD/ AC = P/Q. That is, the sides of the parallelogram represent the forces P and Q; and so the diagonal the lines OA and OB respectively, and let 22 be their resultant. Also, let OABC be a parallelogram for the forces, D any origin of moments, and a, 13, and 7 the angles between OD and the forces respectively as marked. Now P sin a, Q sin /3, and R sin 7 are the values of the components of P, Q, and R at right angles to OD, and it is clear from the figure that P sin a -\- Q sin = R sin 7. Multiplying through by OD, we get P • OD sin a + Q • OD sin = R-OD sin 7. But these terms are the moments of P, Q, and R respec- tively; hence the moment of R equals the sum of the moments of P and Q as stated. When D is between P and Q, then a slight modification in the proof is necessary. Art. s 19 AB represents the resultant of Qi reversed and Pi, and the diagonal BA represents the resultant of Q2 reversed and P2. Since the resultants are equal, opposite, and colinear they balance, and so the P couple and the reversed Q couple balance. Hence, etc. (2) When Pi, P2, Qi and Q2 are parallel, and the moments of the two couples are equal, then each couple is equivalent to some third couple, the forces of which intersect Pi, P2, Qi, and Qi, according to (i). Therefore they are equivalent to each other. § 3. A Force and a Couple. — The resultant of a coplaner force and couple is a single force; the resultant is equal to and has the same direction as the force, and its moment about any point on the given force equals the moment of the couple. Proof follows: Let F (Fig. 26) be the given force, and P1P2 the given couple. (If the forces of the given couple are parallel to F, then imagine the couple shifted Fig. 25 Fig. 26 until they are not so parallel.) Now suppose that AB and BC represent the magnitudes and directions of Pi and F respectively; then AC repre- sents the magnitude and direction of the resultant of those two forces. (The line of action of the resultant is R' , parallel to ylC and through the in- tersection of Pi and F.) Let CD equal AB; then AD represents the magni- tude and direction of the resultant of R' and P2, and hence of the three forces Pi, F, and P2. But AD is equal and parallel to BC; hence this final resultant is equal and parallel to F. (The line of action of this final resultant is R, parallel to BC and through the intersection of R' and P2.) Since R is equivalent to F, Pi, and P2, its moment about any point of F equals the sum of the moments of F, Pi, and P2 about that point; but F has no moment about such point, and hence the moment of R equals the sum of the moments of Pi and P2 (the moment of the couple). It follows from the foregoing that a force R can he resolved into a force equal and parallel to R, and a couple whose moment equals that of R about any point on the component force. Thus the moment of the couple component depends on the line of action chosen for the force component. Independent proof of this proposition follows : 20 Chap, i Fig. 27 Let R (Fig. 27) be the force to be resolved, and O a point through which the line of action of the force component is to pass. First we resolve R into two concurrent components, one of which passes through 0; take any point on R (as a) for the point of concurrence and any direction (as ab) for the line of action of the second com- ponent. These components we call Ci and C2 respectively. To determine Ci and C2, we draw AB to represent R, and AC and BC parallel to Ci and Co respectively; then AC = Ci, and CB = C2. Next we resolve Ci at O into two components parallel to C2 and R, which com- ponents we call C3 and C4 respectively. To determine Cs and d, we draw from A a line parallel to C3 and from C a line parallel to d, and so locate D; then AD = C3, and DC = C4. Ob\'iously now C2, C3 and d are equivalent to R, that is, they are components of R; and as required d passes through 0, and C2 and d (equal, parallel, and opposite) constitute a couple. Moreover, according to the principle of moments, the moment of R about any point on d equals that of C2, C3, and d about that point; but the '"'"''^^^f^^M^ moment of d equals zero, hence, etc.* 6. Graphical Composition of Coplanar Nonconcurrent Forces § I. First Method. — When the forces to be compounded are not parallel nor nearly so, then we compoimd any two of the forces, next their resultant and the third force, that resultant and the fourth force, and so on until the resultant of all the forces has been found. For example, consider the forces acting on the retaining wall shown in section in Fig. 28: * (i) Composition of a Force and a Couple and (2) Resolution of a Force into a Force and a Couple can be performed also as follows (student should supply figure): (i) Replace the couple by an equivalent couple whose forces equal the given force, and place the couple so that one of its forces is colinear with and opposite to the given force. These two forces balance; the other force of the new couple remains, and it is the resultant sought. (Study of the steps in the process shows that the resultant force is equal and parallel to the original force, and that the moment of the resultant about a point on the line of action of the original force equals the moment of the couple.) (2) Apply two forces at the given point equal and parallel to the given force and opposite to each other. These two forces along with the given force can be grouped into a force and a couple, and they (the force and couple) are the components sought. (Study of the. steps of the process shows that the component force is equal and parallel to the given force, and the moment of the couple equals that of the given force about the given point.) Fig. 28 Art. 6 21 they consist of its own weight (16,000 pounds per foot of length), the earth pressure on the back (6000 pounds), that on the top of the base (9000 pounds), and that on the bottom of the base. The resultant of the first three forces will now be determined. We draw AB and BC to represent the 6000 and the 16,000 pound forces, and then join A and C; AC represents the magni- tude and direction of the resultant of the two forces, and the line marked R' (parallel to AC and through point i) is the line of action of that resultant. We next draw CD to represent the 9000 pound force, and join A and D; AD represents the magnitude and direction of the resultant of R' and 9000 (and hence also of the three given forces), and the line marked R (through point 2 and parallel to AD) is the line of action of that resultant. It may be noted that the magnitude and the direction of the resultant is found just as for concurrent forces (Art. 4). For nonconcurrent forces it is necessary to draw the lines of action of the intermediate resultants (R', R", etc.), in order to find the line of action of the final resultant, lines which are unnecessary when compounding concurrent forces. When the forces are parallel or nearly so, the foregoing method fails because there is no accessible intersection of the lines of action of two given forces through which to draw the line of action of the first resultant. This difiiculty can be met as follows: Introduce into the given system two equal, opposite, and colinear forces, which will not change the resultant, taking their common line of action somewhat across those of the given forces; then use the first method, compounding first any pair of forces whose intersection is accessible, etc. § 2. Second Method, applicable to any coplanar forces. — We first re- solve each force into two concurrent components, resolving in such a way Fig. 29 that these components, excepting one of the first force and one of the last force, balance or destroy each other; these two remaining components are, in general, concurrent, and so we readily find their resultant, which is also the resultant of the given forces. For example, let Fi, F2, F3, and F4 (Fig. 29) be the forces to be compounded. First we draw a force polygon for the given forces, taking them in any convenient order, as ABCDE; then we take any convenient point O as the common vertex of the tri- angles of resolution. AO and OB represent two components of Fi in mag- nitude and direction, BO and OC two components of F2, etc.; thus this 22 Chap, i resolution gives several pairs of equal and opposite components, OB and BO, OC and CO, OD and DO. The components of Fi are taken to act through point i, those of F^ through 2, those of Fz through 3, etc., ti first point, i, being taken at pleasure on Fi, point 2 where oh intersects F2, point 3 where oc intersects 7^3; etc. Thus the components OB and BO are colinear and they balance; likewise OC and CO, and OD and DO. Only the first and last components AO and OE remain; their resultant is represented hy AE in magnitude and direction, and its line of action is ae (parallel to AE through the intersection of ao and oe). The common vertex of the triangles of resolution (Fig. 29) is the pole of the force polygon; the lines from the pole to the vertexes of the force polygon, 0/1, OB, OC, etc., are rays; the line of action of the several forces, oa, ob, OC, etc., are strings which, considered collectively, is the string or funicular polygon (also called equilibrium polygon, especially when the given forces are balanced or in equilibrium). The rays are sometimes referred to by number, OA being the first, 05 the second, etc.; likewise the strings. In using this second method, the beginner had best reason out the vari- ous steps of the construction somewhat as in the foregoing. After some practice he might use the following aids: (i) The two strings intersecting on the line of action of any force are parallel to the rays drawn to the ends of that side of the force polygon corresponding to that force, thus the strings intersecting on be are ob and oc. (2) The string which joins points in the lines of action of any two forces is parallel to the ray which is drawn to the common point of the two sides of the force polygon corresponding to those forces, or, the string joining points on be and cd is parallel to OC. (3) The bare construction in the second method is simply this: Draw a force and a string polygon for the forces, then draw a line from the begin- ning to the end of the force polygon and a parallel line through the inter- section of the first and last strings; the first line represents the magnitude and direction of the resultant (sense being from the beginning to the end of the force polygon), and the second line is the line of action of the resultant. This second method is not so simple in principle as the first, but in the second there is more opportunity for varying the construction to keep the drawing within convenient limits; thus the pole may be shifted, and the starting point of the string polygon may be taken anywhere on any of the given forces. Though many string polygons may be drawn for a given set of forces, all determine the same line of action of the resultant; that is, the intersections of the first and last strings of all string polygons lie on one straight line, the line of action of the resultant. § 3. When the Force Polygon Closes. — It may seem, at first thought, that the resultant vanishes, or is zero; in general, this conclusion would be wrong, the system actually reducing to a couple. Thus, let Fi, F2, F3, and Fi (Fig. 30) be a force-set whose force polygon ABODE closes; using the first method for compounding, we find that the resultant R" of the Art. 7 23 first three forces is given by ^D in magnitude and in direction, and ad is its l^e of action; hence* i?" is equal, opposite, and parallel to F4, and so t^ , given force-set reduces to a couple (R", Fi). The arm of this couple is tne perpendicular distance between Ft and R", and so the moment of the couple is the product of Fi (or R") and the arm (according to the scale of the space diagram) ; the sense of the couple, clockwise, is apparent from the relative positions and directions of the forces of the couple as seen in the space diagram. In Fig. 31 the composition has been made by the second method; the system reduces to the two components AO (acting in ao) and OE (acting in oe). These components are equal, opposite, and parallel, and so the given force-set reduces to a couple. The arm of the couple is the perpendicular distance between the first and last strings, ao and oe; the moment of the couple is the product of OA or EO (according E ^= .— ^D i"^^ J^A^ Fig. 30 Fig. 31 to the scale of the force diagram) and the arm (according to the scale of the space diagram) ; the sense is apparent from the space diagram. The length of the arm and the magnitude of the forces of the couple depend on the order in which the forces are taken in the force polygon, in the first method; and upon the position chosen for the pole 0, in the second method. But the moment of the couple is independent of all these vari- ations. This fact may be verified by actually compounding a certain force- set (whose force polygon closes) in several ways, making all these different variations and thus arriving at different couples. The couples are all equivalent to the same force-set and so equivalent to each other, and hence their moments are equal (Art. 5). 7. Algebraic Composition of Coplanar Nonconcurrent Forces § I. Parallel Forces.— If the forces be given sign, those in either direc- tion being called positive and those in the other negative, then the alge- braic sum of the forces gives the magnitude and sense of the resultant, the sign of the sum indicating the sense of the resultant. According to the principle of moments (Art. 5), the moment of the resultant about any point equals the algebraic sum of the moments of the forces about that point, and 24 Chap, i 20 lbs. ^30 It > '40 lbs. 1 50 lbs. Fig. 32 this requirement fixes the position or line of action of the resultant. For example, let us find the resultant of the four forces acting on a 10 foot board, as shown in Fig. 32. CalUng upward forces positive, the alge- braic sum is + 20 — 40 — 50 + 30 =—40; hence the resultant equals 40 pounds and acts downward. The algebraic sum of the moments of the forces about the left end of the ^^30 lbs. board, say, is o — 120 — 350 + 270 = — 200 foot-pounds, and hence the moment of the resultant also is — 200 foot-pounds; this fixes the arm of the resultant, 200 H- 40, or 5 feet. Since the resultant acts downwards and its moment about the origin is negative, its line of action must be to the right of the origin, 5 feet. To find the resultant of two parallel forces (a common problem), we may of course use the general method just explained, but the following special results are worth noting. We distinguish two cases: (i) the two forces are alike in sense; (2) they are opposite. In (i) the resultant equals the sum of the forces and agrees with them in sense; in (2) the resultant equals the difference between the two forces and agrees with the larger in sense. In order that the moment of the resultant R may equal the sum of the moments of the forces, P and Q (Fig. 33), then, in case (i), R must lie between the forces, and in case (2) out- side of them and adjacent to the larger force (assumed to be P in the figure). Furthermore, if the distances from R to P and Q be called p and g respec- tively, and that between P and Q be a, then in either case, Rp = Qa and Rg = Pa, or p = Qa/R and g = Pa/R either of which definitely fixes the position of R. Also for either case, Pp = Qg or P/Q = g/p; hence P/Q = BC/AC, that is, the line of action of the resultant of two parallel forces divides any secant intersecting their lines of action into two segments which are inversely proportional to the two forces. If the algebraic sum of a set of parallel forces equals zero, then it may appear to the student that their resultant vanishes or is zero; this does not follow, but the resultant actually is in general a couple. For the re- sultant of all but one of the given forces is a single force equal, opposite, and parallel to the omitted one; but these two are not in general colinear, and so they constitute a couple, the resultant of the system. The couple arrived at depends on which one of the given forces is omitted, but the moment of the couple does not, for that couple is the resultant of the set, hP-H r; -q->l C > <- q -> ^i B < --ip ■) A i P R ^R Ip a Fig. 33 Art. 7 25 A60lb&. 401bvf "20 lbs. SOlbs. Fig. 34 SOlbs. and the moment equals the algebraic sum of the moments of given forces, a definite quantity. For example, let us find the resultant of the five forces acting on a 10 foot board, as shown in Fig. 34. Their algebraic sum is zero, and so their resultant is, presumably, a couple. Compounding all but the 40 pound force, we find that their resultant equals 40 pounds, acts down- ward, 7.5 feet to the right of the left end of the board, and so the resultant is a couple whose moment is + (40 X 2.5) = -|-ioo foot-pounds. Instead of actually determining the forces of the resultant couple as explained, it is usually sufficient to determine the moment of the result- ant couple; this moment equals the algebraic sum of the moments of the given forces about any point. Thus, in the preceding example, after ascer- taining that the resultant is a couple, we compute the moment-sum for the given forces, wdth moment origin at the middle of the board, say, or (20 X 5) - (60 X 3) + (30 X i) - (50 X i) + (40 X 5) = +100 foot-pounds; and then conclude that any couple whose moment equals -fioo foot-pounds may be regarded as the resultant of the system. § 2. NoNPARALLEL FORCES.— As shown in Art. 6, the resultant is in general a single force, given in magnitude and direction by the line joining the be- ginning and end of the force polygon for the forces. It follows, therefore, that the component of that resultant force along any line equals the alge- braic sum of the components of the given forces along that line. From this principle we can get the components of the resultant along any two rec- tangular axes; and from these components the magnitude and direction of the resultant itself can be readily determined by obvious means. Ac- cording to the principle of moments (Art. 5), the moment of the resultant about any point must equal the sum of the moments of the given forces about that point; and this requirement fixes the position or line of action of the resultant. For example, let us find the resultant of the six forces acting on a board, 4 by 4 feet, as shown in Fig. 35. The angles which the forces make with the horizontal and the arms of the forces with respect to the center of the board are recorded in columns 2 and 3 of the schedule on page 26; they could be computed trigono- metrically or could be scaled from a larger drawing. The X and y components of the several forces are recorded in columns 4 and 5 respectively, and the moments of the forces with respect to the center of the board in column 6. The algebraic sums of the x ^6lbs. VIbs. 51b5^ Fig. 35 y components are +3.40 and -7.22 pounds respectively; hence and the R = V'3.40^ -|- 7.22^ = 7.98 pounds The signs of the sums indicate that R acts toward the right and downward; the angle which R makes with the 26 Chap. horizontal is tan~^ (7.22 -J- 3.40), or 64° 47'. The sum of the moments is — 14.14 foot-pounds; and, since the moment of R also equals —14.14, R lies on the right-hand side of the origin of moments (the moment being negative), and its arm is 14.14 -^ 7.98 = 1.77 feet. Thus, R has been completely determined. I 2 3 4 s 6 7 8 9 10 F a a F. Fy M fli 02 Ml il/2 8 I .00 -1-8.00 0.00 - 8.00 I - 8.00 4 45° 0. 71 + 2.83 + 2.83 - 2.84 I - 2.83 6 63° 25' 0.89 -2.68 + 5.36 - S-37 I - 5-36 7 90° 2.00 0.00 — 7.00 -I-14.00 2 -t-14.00 12 36° 52' I .60 — 9.60 — 7.20 — 19.20 2 — 19.20 5 14° 2' 1-45 +4.85 — I. 21 + 7-27 I 2 + 4.85 + 2.42 +3 -40 — 7.22 -14.14 -25.18 -f 1 1 . 06 The moment-sum (—14.14) may be determined also — more simply in this example — by adding the moments of the x and y components of the several forces (Art. 5). This method requires that we take definite lines of action for all the components. Of course any force and its x and y com- ponents must be concurrent (Art. 3). We take the heavy dots (Fig. 35), as the points of concurrence; then the arms of the x components of the forces respectively with respect to the center of the board are as recorded in column 7, and the arms of the y components are as recorded in col- umn 8. The moments of the x components and of the y components are recorded in columns 9 and 10 respectively. The sums of the moments of the X components and of the y components are —25.18 and +11.06 foot- pounds respectively, and hence the moment-sum for the given forces is — 25.18+ 11.06 = — 14.12. If 2/^x and ^Fy for a force-set both equal zero, then any force polygon for the forces would close; hence the resultant of the forces is a couple (Art. 6). Any couple whose moment equals the alge- braic sum of the moments of the given forces about i^bs. any point may be regarded as the resultant. For ex- ample, let us find the resultant of six forces acting on a drawing board 4 by 4 feet, as shown in Fig. 36. In- spection shows that the algebraic sums of the x and the y components equal zero, and so the resultant is a couple. Taking moments about the lower left- hand corner, we get 2i/= —60 X 1.414 — 20 X 1.414 — 18X3— 12X I = —179 foot-pounds, and this is also the value of the moment of the resultant couple. If the forces consist of a number of couples, then 2Fj = 2Fy = o, and the resultant is a couple. But the sum of the moments of these forces 60 lbs. Art. 8 27 Fig. 37 about any point equals the sum of the moments of the couples; hence any couple whose moment equals the sum of the moments of the given couples may be regarded as the resultant. 8. Moment of a Force ; Couples * § I. Moment about a Line. — Art. 5 relates to moments of forces and to couples with special reference to coplanar forces and couples. In some discussions on noncoplanar forces it is convenient to make use of the moment or torque of a force with respect to a line; this is defined as the product of the component of the force perpendicular to the line — the other component being parallel to it — and the distance from the line to the perpendicu- lar component, or to the force (the distances being equal). For example, let F (Fig. 37), acting on a body not shown, be the force, and LL' the line, or axis of moments as it is called. MN is any plane perpendicular to the axis, represented to make the figure plain. OACB is a parallelogram with OC (representing F) as diagonal, and sides perpendicu- lar and parallel to LL'; then OA and OB represent the perpendicular and parallel components {Fi and F2) referred to; and the moment of F about LL' is the product of Fi and PL. The moment of a force with respect to a line is a measure of the tendency of the force to turn the body to which the force is applied about that line. Thus, when the force is parallel to the line the moment is zero, and obvi- ously the force has no tendency to turn the body about the line. Again, when the force is perpendicular to the line the moment of the force about the line equals the product of the force and the perpendicular distance from the line to the force, and it is shown in Art. 5 that this product meas- ures the tendency of the force to turn the body about the line. Finally, when the force F is not parallel nor perpendicular to the axis of moments (Fig. 37), then Fi and F2 together are equivalent to F, and their combined turning effect equals that of F. But F^ has no turning effect; therefore that of Fi and that of F are equal. But it was explained that Fi X LP (the moment of Fi) measures the turning effect of Fi, and therefore that product also measures the turning effect of F. In a discussion involving moments of several forces about a line, it is generally convenient to give signs to the moments to indicate the directions (clockwise or counter) in which the several forces would turn the body about the line if it were free to rotate about that line. Whether a given rotation is clockwise or counter depends on the point of view; in a par- ticular discussion a point of view should be assumed on the line or axis of moments and outside of the body, so that all rotations would be seen look- ing in the same direction. When the axis of moments is also an axis of * See Art. 5 also. 28 Chap, i coordinates, then it is customary to view rotations about that axis from the positive end of the coordinate axis, looking in the negative direction. Principle of Moments. — If two sets of forces are equivalent (Art. 2), then the moment-sum for one set with respect to any line equals the mo- ment-sum for the other set with respect to the same line. This will be granted as self-evident by most students; others may consider this: Let ^1 and 6*2 denote the two equivalent sets of forces, and ^3 a third set which would balance Si and hence also ^2. Since Sx and ^3 would balance, they would not turn the body on which they act about any line; hence the moment-sums for Si and ^'3 with respect to any line are equal in value but opposite in sign. Likewise, the moment-sums for S2 and S3 with respect to that same line are equal in value and opposite in sign. The moment- sums for ^i and S2 being equal to the same thing, are therefore equal. It follows from the preceding that the moment-sum for any set of forces with respect to a given line equals the moment of the resultant of those forces with respect to the same line. Also, the moment of a force about any line equals the moment-sum of its components with respect to the same line. This last principle suggests a second method for computing the moment of a force with respect to a line, more simple than the first method in some cases: Resolve the force into three rectangular components, one of which is parallel to the axis of moments; compute the moment of each of the other two components about the axis, and add the moments alge- braically; this sum equals the moment of the given force. For an example, we compute the moment of a 100 pound force which acts upon a 4 foot cube as shown in Fig. 38, with respect to those edges marked X, F, and Z. The x, y, and z com- ponents of the force are 37.2, 74.2, and 55.7 pounds respectively (see Art. 4); these components must be concurrent with the given force. Taking A as the point of concurrence, the moments are com- puted as follows: -74-2 X 4 + 55-7 X 4 = "74; -37.2 X 4 - 55.7 X ^ = -260; and '37.2 X 4 + 74.2 X 2 = 297 foot-pounds. With point of con- currence taken at B or at any other point in AB, the same result would be obtained for the moment. § 2. Couples (see also Art. 5). — Two couples whose planes are parallel and whose moments, or torques, and senses are the same are equivalent. Proof of this proposition for coplanar couples is given in Art. 5; proof for noncoplanar couples follows. Let Pi and P2 (Fig. 39) be the forces of one couple, Qi and Q2 (not shown) the forces of the other, and p and q the arms of the couples respectively; then by supposition Pp = Qq. According to Art. 5, the (2 couple can be replaced by a couple in its own plane provided that the moment and sense of the new couple equals that of the Q couple. Let Si and ^2 be the forces of that replacing couple, Si and ^2 being chosen 100 >b5 ■37.2 Art. 8 29 Fig. 39 parallel and equal to Pi and P^; then the arm ab of the S couple equals />, and abed is a parallelogram. We now show that the P couple would balance the reversed S couple; it will follow that the P and 6* couples are equiva- lent, and hence also the P and Q couples. The resultant R' of P] and —52 {Si reversed) equals the resultant R" of P2 and —S\ (Si reversed), and R' and R" are parallel and opposite in sense. Moreover, R' lies midway between Pi and 52, and R" lies midway between P2 and Si; therefore each resultant acts through the center of the parallelogram abed, and hence they are colinear. The ' resultants therefore bal- ance, and hence the four forces Pi, P2, —Si, —S2 do also. Therefore, etc. The resultant of any number of couples is a eouple. Proofs of this prop- osition for the case of coplanar couples are given in Arts. 6 and 7. For the case of noncoplanar parallel couples: The given couples can be re- placed by equivalent ones respectively, all in some one plane; the result- ant of these is a couple, and hence the resultant of the given ones is also a couple. For the case of nonparallel couples: Imagine each of the two couples to be replaced by an equivalent couple, and let the four forces of the replacing couples be equal; furthermore, imagine the two new couples so placed (in their respective planes) that a force of one couple will balance a force of the other. See Fig. 40 (perspec- tive), which shows the two replacing couples, there marked P1P2 and P3P4; ex is the angle between the planes of the couples. Since P2 and P4 balance, Pi and P3, constituting a couple, are equivalent to Pi, P2, P3 and P4 and hence to the two original couples. The resultant of any coplanar or parallel couples can be determined very simply; the resultant is any couple parallel to the given couples, its moment being equal to the alge- braic sum of the moments of the given couples. The resultant of nonparallel couples can be determined best from their vectors* by means of this proposi- tion, — The vector of the resultant of any number of couples equals the sum of the vectors of those couples. Proof: Consider first two couples, say the two whose resultant was found in the preceding paragraph. Let ABC (Fig, 41) be an end view of Fig. 40, looking along the line AA'] that is, ABC of * The vector of a given couple is perpendicular to the plane of the couple (exact posi- tion of vector immaterial); its length is equal to the moment of the couple according to some scale understood; and its sense agrees with the sense (rotation) of the couple according to some rule of agreement, as for example the following: Imagine the vector to be a right-handed screw turning with the couple; then the arrowhead on the vector must point in the direction in which the screw advances. Fig. 40 3© Chap, i Fig. 41 is ABC of Fig. 40 in true proportions. Then AM (perpendicular to AB), AN (perpendicular to AC), and AO (perpendicular to BC) are respec- tively the vectors of the two given couples and their resultant, provided that the lengths of the vectors are proportional to the moments of the couples Ffi, Ffz and Ff; let the lengths be in that proportion. Vector AO is the sum of the vectors AM and AN, provided that OMAN is a parallelogram; we now show that it is a parallelogram. Angle MAO = 13; since in the tri- angle MAO and ABC two sides are proportional each to each and the in- cluded angles are equal, the triangles are similar; it follows that OM is perpendicular to AC, or parallel to AN. From similar reasoning, it fol- lows that ON is perpendicular to AB, or parallel to AM. Hence OMAN is a parallelogram. Obviously, if the proposition holds for two couples, it holds for any number. Composition of three couples whose planes are mutually at right angles is an important special case. We take the three planes as coordinate planes, and call the couples whose planes are perpendicular to the x, y, and axes Cx, Cy, and Cg respectively, their vectors Vx, Vy and Vz, and the re- sultant couple C and its vector v. Then v = {v^^ + Vy"^ + ^z^)^; hence Also, if 01, 4>2, and ^3 denote the direction angles of v, then cos ii<- 7' > represented in Fig. 56 under the action ^2 of three loads (its own weight neglected), Fig. 56 ^^'^ supported at A and B; required, the reactions of the two supports. The five forces just mentioned constitute a system in equilibrium; therefore, taking moment origins on Ri and R2 respectively, and assuming that Ri and R2 act upwards, we get SMi = 2000 X 6 + 1000 X 2 — 3000 X 3 + -??2 X 10 = o, and 21/2 = 2000 X16 + 1000 X12 -f 3000 X 7 — i?i X 10 = o. Art. 12 45 The first gives Ri = — $oo pounds, and the second Ri = 6500; the negative sign means that R2 acts downward on the beam and not upward, as as- sumed. As a check on the solution we try whether XF = o; thus, — 2000 — 1000 — 3000 + 6500 — 500 = o. The graphical solution of the foregoing problem is based on the conditions that the force and the string polygon for the forces close; the process of constructing and closing the polygons determines the unknown forces. To illustrate we take the beam shown in Figs. 56 and 57 and determine the reactions. First, the force polygon should be drawn as far as possible, the knowns represented first, thus AB, BC, and CD (Fig. 58) representing the 2000, the 1000, and 3000 pound forces respectively; then the lines of action should be lettered to correspond, ah, he, and cd (Fig. 57). If R2, say, is taken next, it would be lettered DE, and Ri would be EA, since the force 2000 Ibv. 1000 lbs. ^000 lbs 2000lb5-1000lbs. ^OOO'"*- Fig. S7 Fig. 58 Fig. 59 polygon for all must close. It remains now to locate E; this can be done by means of the string polygon. (At this point it may be well for the reader to recall the significance of the strings of a string polygon; see Art. 6.) The polygon may be started at any point on any of the lines of action of the forces of the system; if it be started at i (on ah), then strings oa and oh must be drawn through that point; oc must be drawn from 2 (where oh cuts he), od from 3 (where oe cuts cd), and oe from 4 (where od cuts de) and from 5 (where oa cuts ea) ; hence the closing string oe passes through 4 and 5. Finally, the ray OE, parallel to oe, is drawn, thus determining E; DE represents R2, and EA Ri. Fig. 59 shows another solution; Ri is taken as the fourth force DE\ and R2 as the fifth E'A. § 2. We take this opportunity to mention a class of problems on forces in equilibrium, not parallel necessarily, which cannot be solved by the principles of statics alone, and are therefore called statieally indeterminate problems. A beam resting on more than two supports furnishes a simple illustration; thus, let it be required to determine the reactions of the sup- ports {A, B, and C) on the beam represented in Fig. 60, due to the two 46 Chap, n loads. If not already warned of the difficulty in this problem, some stu- dents would probably write moment equations for the forces in equilibrium (Pi, Pi, Ri, Ri, and Rz), with moment origins at A, B, and C, and then attempt to solve the equations simultaneously for the three unknowns. Such attempt would fail, even though each jP. |Pz equation would be correct, because the I '^ 1 ^ X- three would not be independent — there IR, 'rj 'Rj being only two conditions of equilibrium Fig. 60 ^<^r ^ system of the kind under considera- tion (Art. 10 under iv) — and so the three equations would not determine the three unknowns. Doubters are advised to try to determine Ri, Ro, and R3 in this way in the simple case where the spans and the loads are equal, and the loads are applied at the centers of the spans. How may one determine whether a given problem (a force system in equilibrium with some unknowns required) is statically determinate or indeterminate? A complete answer to the question is beyond the scope of this book; we may remark, however, that statically indeterminate prob- lems commonly arise in connection with structures which have redundant or superfluous parts or supports, by which is meant that some of the parts or supports are not strictly necessary for the equilibrium of the structure. For example, in Fig. 60 one support is superfluous, since the beam on tv/o supports would, if strong enough, support the load. No statically inde- terminate problems are given in this book without notice; but the student may meet a force system in equilibrium containing many unknowns, and he is now reminded that it is futile to write out more equilibrium equations than there are algebraic conditions of equilibrium for the system under consideration (Art. 10), with the expectation that the equations if solved will determine the unknowns. And so it is well to know the number of con- ditions of equilibrium for each class of force systems. 13. Coplanar Nonconcurrent Nonparallel Forces Principles of equilibrium for a force system of this kind are developed in Art. 10 under (iv). Their use will be explained now by applying them to two particular common problems. § I. Typical Problem (iv). — A system of coplanar nonconcurrent non- parallel forces is in equilibrium, and all except two are wholly known; only the line of action of one of these two and a point in that of the other are known, and it is required that these two be determined completely. The algebraic solution of this problem can be effected by means of any one of these sets of equilibrium equations: SF, = ZFy = XM = o; 2i?x = 2M„ = l^Mb = o; 2Ma = ZMt = Slfc = o. Art. 13 47 SO.OOOlbs Fig. 61 For an example, consider the roof truss represented in Fig. 61. It sustains two loads, 3 5, 000 (weight of roof and truss) and 50,000 pounds (wind pressure). The left end of the truss merely rests on a wall, but the right end is fastened to a wall; therefore the reaction of the left-hand wall must be vertical, but that of the other may be inclined. Let it be required to determine these reactions. We call the left reaction A , the right one B, and the inclination of B to the horizontal 6. Then the first set of equilibrium equations gives XMb = +35,000 X 45 + 50,000 X (60 cos 30°) — A X go = o, or A = 46,400 pounds. 2F^ = —Bco5d + 50,000 sin 30° = o, and 2/^y = +B sin 6 — 50,000 cos 30° — 35,000 + 46,400 = o; these solved simultaneously give B = 40,500 pounds and = 51° 54'. For algebraic solutions, it is generally advisable to imagine the second unknown force, whose point of application is known, to be replaced by two (unknown) components. Then the problem is in the form of typical problem (v) (see next page). Thus, in the preceding example the unknowns would be A and, instead of B and 6, B^ and By. After finding Bj, and By, one could easily get B and 6. The graphical solution of this problem is eftected by drawing the force and the string polygons, making both close since the force system is in equilibrium. To illustrate we use the preceding example. We first draw the polygon ABC (Fig. 62) for the known forces, and continue it with a line through C parallel to the left-hand reaction. The end of that line, as yet unknown, is to be marked D; that point once determined, then DA will represent the right-hand reaction. To find D we must construct a string polygon; so we next mark the lines of action of the several forces to agree with the nota- tion in the force polygon, choose a pole 0, and draw the rays OA, OB, and OC. To make use of the known point i of the fourth force (right-hand reac- tion), the string polygon must be begun at that point. The string oa is the one to draw through that point (to ab), and then ob and oc as shown. The string od must pass through points i and 4, and so is determined. Next we draw the ray OD (parallel to od), and thus determine D (the intersection of CD and OD). The following special graphical method is simpler in principle than the pre- ceding method: Let R = the resultant of the wholly known forces, P = the force whose line of action is known, and Q = the force whose point of applica- tion is known. Find R, and then imagine the wholly known forces replaced by R; R, P, and Q would be in equilibrium. Now a balanced three- force 35,000 lb5. Fig. 62 48 Chap, n 50.0001b5.A( Fig. 63 system is concurrent or parallel (Art. 10, § 2); hence if R intersects P, then Q acts through that point of intersection, and if R is parallel to P, then Q is also. If the three forces are concurrent, then determine P and Q from the force triangle for the three forces as explained in Art. 1 1 ; if they are parallel, determine P and Q as explained in Art. 12. To illustrate, we use the data of the foregoing example. First we draw AB and BC (Fig. 63), to represent the two loads; then AC represents the magnitude and direction of their resultant R. The line of action of R is ac, parallel to AC and passing through the intersection of ab and be. (When the wholly known forces are noncon- current it is necessary to construct a string polygon to find a point in the line of action of R, see Art. 6.) We next extend the lines of action of R and P, and join their intersection with the point of application of Q; this line is the line of action of Q. Finally we complete the force triangle AC DA for R, P, and Q; then CD = P and DA = Q. §2. Typical Problem (v). — A system of coplanar nonconcurrent non- parallel forces is in equilibrium, and all the forces except three are wholly known; only the lines of action of these three are known, and their magni- tudes and senses are required.* The algebraic solution of this problem can be effected by means of any one of these three sets of equilibrium equations: SFx = SF„ = 2M = o; 2/^^ = 2Ma = Mlb = o; or l^Ma = ^Ah = ZMc = o. For example, consider the crane represented in Fig. 64. It consists of a post AB, a. boom CD, and a brace EF; the post rests in a depression in the floor below, and against the side of a hole in the floor above. The external forces acting on the crane consist of the load W (8 tons), the weights of the parts named (0.8, 0.9, and i.i tons respectively), and the reactions of the floors. The upper floor exerts a single hori- zontal force on the post; the lower floor exerts two forces on the post, one horizontal and one vertical. Let it be required to de- termine the magnitudes of these reactions. The entire external system of forces just described is in equilibrium. Calling the reactions A, B^, and By respectively, then the first set of equilibrium equations become: 2Ma = — 8 X 20 — 0.9 X II - I.I X 7 + 5x X 18 = o, or B^ = 9.86; XFj, = 9.86 - A = o, or A = 9.86; XFy = By — 8.0 — 0.8 — 0.9 — I.I = o, or By = 10.8 tons. * If the three unknown forces are concurrent or parallel, the problem is indeterminate. '^^''mwi^ Fig. 64 Art. 13 49 0.9tons etons Fig. 65 The general graphical solution is carried out as follows: Let P, Q, and S stand for the three forces whose lines of action only are known. Imagine any two of these, say P and Q, replaced by their resultant R'; one point in that resultant is known, the intersection of P and Q. Then S, R', and the known forces would be in equilibrium, and the given problem has been transformed to typical problem iv. So we first determine S and R', as explained in § i, and then resolve R' into two components parallel to P and Q; these compo- nents are P and Q. To illustrate, we take the preceding example, and we call the two lower reactions P and Q, and the upper one 5 (Fig. 65). The re'sultant R' of P and Q passes through the lower end of the post. We draw the polygon ABCDE for the knowns, and continue it with a line parallel to S, The as yet unknown end of that line is to be marked F; that point once determined, then FA will represent R' , since the polygon for all the forces must close. To find F we must construct a string polygon; so we mark the lines of ac- tion of the several forces to agree with the notation in the force polygon, choose a pole 0, and draw rays OA, OB, OC, OD, and OE. The string polygon must be begun at the lower end of the post, the point of application of FA or R'. The strings to pass through that point are of and oa (Art. 6), and so we draw oa to ab; then ob, oc, od, and oe as shown. Now point i is in of, and point 6 is also; therefore of is deter- mined. The ray OF is drawn next (parallel to of), thus determining i^; then EF and FA represent 5 and R', as already stated. Finally we draw through F a vertical and through A a horizontal; then FG and GA represent the vertical and horizontal reactions (P and Q) of the lower floor. The following special graphical method is simpler in principle than the preced- ing: First we determine the resultant R of the wholly known forces; R and the three partly unknown forces (P, Q, and S) would be in equilibrium. The special condition of equilibrium for four such forces is that the resultant R' of any pair as P and Q balances the other pair; hence R' and the other pair (R and S) are in equilibrium, and so must be con- current or parallel. Next we solve the system R\ R, and S (if concurrent by Art. 11, and if parallel by Art. 12). Finally we complete the force polygon for R, S, P, and Q. For an illus- FiG. 66 50 Chap, n tration we take the preceding example. Let the two lower reactions be called P and Q, and the upper one 5 (Fig. 66). The resultant R of the loads is I0.8 tons acting as shown (construction for R is indicated). The resultant R' acts through point i; and, since R and .S are concurrent at point 2, R' acts through point 2 also. We now draw the force triangle AEFA for R, S, and R' , AE representing R; then EF represents S. Finally we draw lines from A and F parallel to Q and P, thus fixing G\ and then FG represents P, and GA represents Q. 14. Noncoplanar Forces in Equilibrium § I. The pirinciples of equilibrium for noncoplanar forces are set forth in Art. 10 under (v), (vi), and (vii). The three following illustrations deal with concurrent, parallel, and nonconcurrent nonparallel forces respectively. (i) A heavy body W (Fig. 67) weighing 1000 pounds is suspended from a ring over the center of a street 60 feet wide; the ring is supported by three ropes OA,OB, and OC; A and B are points on the face of a building as shown, and C is a point on the face of a building (not shown) on the opposite side of the street, OC being perpendicular to the face of the buildings. Values of the tensions in the ropes are required. There are four forces acting on the ring, — the pull of 1000 pounds, and the pulls of the three ropes which we call L, M, and N respec- tively; this system is concurrent. To deter- mine the unknown forces in it, we use the conditions that the algebraic sums of the com- ponents along three rectangular axes equal zero; as axes we choose a vertical line and two horizontal lines, one parallel and one To get the components of L, M, and N, we need values of certain angles: A'OC' = tan-^ A'C'/OC'= 28° 4'; AOA'= tan-i AA'/OA'= 30° 28'; B'OC^ tan-' B'C'/0C'=s8° 40'; BOB'= tan"' BB'/OB' = 46° 11'. The X, y, and z components, respectively, of L are L cos 30° 28' sin 28° 4'= 0.405 L, L sin 30° 28'= 0.507 L, and L cos 30° 28' cos 28° 4' = 0.760 L; of M they are il/ cos 46° 11' sin 38° 40'= 0.4325 M, M sin 46° ii' = 0.721 M, and M cos 46° 11' cos 38° 40'= 0.5405 M; of N they are o, o, and N\ of the looo-pound pull they are o, 1000, and o. The algebraic sums of the X, y, and z components are — 0.405 L -\- 0.4325 M -\- o -\- o =0 , -{-0.507 L -f 0.721 M -1- o — 1000 = o, — 0.760 L — 0.5405 M + iV -|- o = o. Solving these equations simultaneously, we find that L = 846, M= 792, and N= 1072 pounds. Fig. 67 transverse to the street Art. 14 SI Fig. 68 (ii) A body weighing 1000 pounds is suspended from the ceiling of a room by means of three vertical ropes; the points of attachment at the ceiling lie at the vertices of an equilateral triangle ABC (Fig. 68) whose sides are 10 feet long; W is the projection of the center of gravity of the body upon the ceiling. The tension in each rope is required. We call the tensions in the ropes fastened at A, B, and C, respec- tively, L, M, and N. The four forces acting on the body constitute a parallel system; the conditions of equilibrium for such are that the sums of the moments of the forces about any three coplanar nonparallel axes perpendicular to the forces equal zero. The lines AB, BC, and CA are good lines to choose as axes of moments. With respect to these lines the moment equations are respectively, N X 8.66 — 1000 X 2.10 = o, L X 8.66 — 1000 X 4.15 = o, and M X 8.66 — 1000 X 2.41 =0, 8.66 being the altitude of the triangle. Solu- tion of these equations shows that L = 479, M = 278, and N = 243 pounds. (iii) Fig. 69 shows a velocipede crane. The crane can be run along on a single rail below, tipping being prevented by two overhead rails which guide a horizontal wheel mounted on the top of the crane post. The crane weighs 1.25 tons, and it is balanced so that its center of gravity is in the axis of the post. We will now show how to deter- mine the supporting forces (exerted by the rails) when the crane supports a load of 1.5 tons and the jib is swung out at right angles to the rails to- ward the left : (Fig. 70). There are three support- ing forces or reactions, one on each wheel. Since the lower rail is level, the crane does not tend to roll, and there is no reaction of the rails in their direction. The reaction of the upper rail is directed horizontally and evi- dently as shown; the reaction on each lower wheel has two components as shown. We call these component reactions Ax, Ay, Bx, and 5„, and the 1^ Fig. 70 '■Vi 'v> upper reaction C. The external system of forces acting on the entire crane 52 Chap, ii ■-rV T-Y consist of the reactions named, the weight of the crane, and the load. For noncoplanar nonconcurrent nonparallel systems there are, in general, six con- ditions of equilibrium, but this system has only five because there are no " z forces " (see the figure). The five conditions of equilibrium are 2F, =A, + B,-C = o; (i) 2i^„ =+^„ + ^„- 1.25- 1.5 = 0; (2) 2if^ = ByX4- AyX6 = 0; (3) XMy = ^x X 4 - -^x X 6 = o; (4) ZM, = C X 16 - 1.5 X 6 = o. (5) From (5) it follows that C = 5.625 tons; from (i) and (4), that Bx = 3.375 tons, and Ax = 2.25 ^tons; from (2) and (3), that By = 1.65 tons, and Ay = 1. 10 tons. We now give another solution, making use of the principle that if the forces of a system in equilibrium be represented by vectors, then the projection of the vectors on any plane represents a force system also in equilibrium (see Art. 10 under (vii)). Fig. 71 shows such projections on the x-y, y-z, and z-x planes of Fig. 70. From the y-z pro- jection (side elevation), 2ilf a =ByX 10 — 2.75 X 6 = o, or By = 1.65 tons; and 2ifB = — Ay Xio + 2.75 X 4 = o, or Ay = 1. 10 tons. From the x-y projec- tion (end elevation), ZMa = C X 16 — 1.5 X 6 = o, or C = 5.625 tons. From the z-x projection (plan), 'ZMa = — BxX ^ 10 -\- 5.625 X 6 = o, or Bx = 2.375 tons; and 27lfB = — ^i X 10 + 5.625 X 4 = o, or Ax = 2.25 tons. § 2. A noncoplanar system can gen- erally be solved by means of an equivalent coplanar system. This indirect method is regarded as simpler than the direct one when the forces of the non- coplanar system are nonparallel. The two following examples will illustrate. For one example we use the data of example (i). Instead of ropes OA and OB (Fig. 67), imagine a rope 00' in the plane of those ropes, and also in the same vertical plane with COC. Such a rope fastened to and to the building at 0' would help to support the ring in its place, 8.nd would leave Plan. Fig. 71 [--«■ -7H rB y / ■r^ / V p./ C-e— N 0,/ Il0001b5. A.- Ml T Fig. 72 5-- \ ' / l\/ the tension in OC unchanged. Thus the ring would be acted upon by three forceSj — 1000, iV, and the pull P of the new rope (Fig. 72). A force tri- Art. 14 53 angle, FGHF, for these forces shows that the pull N = 1070 and P = 1460 pounds. We next lay out the ropes OA, OB, and 00' in their true relations, and then we resolve the pull 1460 in the imaginary rope into components along the real ropes. Thus we lay off OQ equal to 1460, and then on the diagonal OQ complete the parallelogram OMQN; and find OM and ON, representing the tensions in the real ropes, 860 and 790 pounds. For another illustration we take a tripod (Fig. 73), shown in plan and eleva- tion. The requirement is to determine the forces acting at the top of each leg of the tripod due to a load of 1000 pounds. On account of this load, each leg is under the action of two forces, one applied at each end of that leg, and so those two forces act along the axis of the leg. We imagine a single leg in the plane of any two, and in the same vertical plane with the third, to replace the two; thus OD to replace OA and OB. Then there would be three forces applied to the pin at 0, namely, the load 1000 pounds, and the supporting forces exerted by OC and OD. So we draw a force triangle for these three forces FGHF; it shows that the push of OC is GH = 565, and that of OD is HF =650 pounds. Next we lay out the other pair of legs and the imaginary one in their true relation 0"A", 0"B", and 0"D", and make 0"P = HF = 650 pounds; then resolve 0"P into two components along the pair 0"A" and 0"B" by means of a parallelogram 0"MPN. Thus we find that 0"M and 0"N represent the pushes of AO and BO, or 340 pounds. CHAPTER III SIMPLE STRUCTURES 15. Simple Frameworks (Truss Type) § I. The frames herein considered consist of straight members, and the axes of all the members lie in one plane; such are called plane frames, and the plane of the axes is called the plane of the frame. In order to make the axes of all members lie in one plane, and the truss symmetrical with respect to that plane, some of the members must be made in parts or with forked ends. For example see Fig. 74, which shows plan and elevation of a joint of a frame at which four members are pinned together, one vertical (double), one diagonal D (single), and two horizontals Hi and Hi (each double). Wooden members are generally bolted together with more or less mortising; steel members are riv- eted together or joined by pins through holes in the members, the axes of pins and holes being perpen- dicular to the plane of the frame. All frames here considered are assumed to be of the pin-connected type; and, furthermore, it is assumed that each member connects only two joints, that is, extends from one joint to another but not also to a third one. In such pin-connected frames, the lines of action of the pin pressures (forces exerted by pins on the members) are in or parallel to the plane of the frame. Thus, the resultant pressure of the pin on the diagonal member D (Fig. 74) is clearly in the plane; the pin exerts on the vertical member two forces which, on account of the symmetrical arrangement, are equal, parallel, and equally distant from the plane, and therefore the resultant of these two forces lies in the plane; and obviously the resultant of the forces exerted by a pin on each horizontal member lies in the plane. Thus all resultant pin pressures will be regarded as lying in the same plane, and we will have only coplanar forces to deal with in the present connection. We assume that the pins are practically frictionless; in that case each pin pressure acts practically normally to the surface of the pin, and so the line of action of each pressure cuts the axis of the corresponding pin. In this and the following articles we assume that the loads are applied to the frame at its joints only, and in such manner that the line of action of each load cuts the axis of the pin at the joint. Then each member, if its own weight 54 Art. is 55 is neglected, is subjected to forces (pin pressures and loads) at its two pin holes only, somewhat as shown in Fig. 75 or Fig. 76, where P' and P" denote pin pressures and L' and L" loads. Let R' denote the resultant of P' and L\ and R" the resultant of P" and L". Since R' and R" balance, each acts along — — — p ?'/ ; NP" Tension Compression R' ^nn^ A ^ ^,^|___ '^ " R' m A B n R" Fig. 75 Fig. 76 the axis of the member, and hence each member is under simple tension or compression. Any two parts of the member, as m and n, exert equal and opposite forces upon each other; A (Figs. 75 and 76) denotes the force exerted on m by n, and B that exerted on n by m. Since A balances R' , and B balances R" , A and B also act along the axis of the member. And obviously, if R' and R" are pushes (the member in compression), then A and B are pushes; and if R' and R" are pulls (member is in tension), then A and B are pulls. And conversely, if A and B are pushes, then the member is in compression; and if pulls, then in tension. By stress in a member is meant either of the two forces which two portions, as m and n, exert upon each other.* We are now ready to explain a method for determining the stresses in the members of a simple truss due to given loads ; we begin with an Example. — Fig. 77 represents a truss supported at each end; the angles equal 60 degrees; it sustains two loads of 2000 pounds each and one of 1000 pounds. First, '°°°1"'^- ^^^^I'"*- it is necessary to ascertam the values of the A 1 \ A reactions A and B. Since all the external /'" \ /"'\ forces acting on the truss (loads and reactions) V \,-. / \, . are in equilibrium, ^Ma ^ B X 40 — 2000 X A/ \ / V/ \ / \b A ' ' 1 i ' I \ 30 — 1000 X 10 — 2000 X 20 = o, or5 = 2750 T ,^°°°|"^^- ^ I pounds; and Svlfs = -^ X 40+ ^^' '^'^ ^^' "^ 1000 X 30 + 2000 X 10 + 2000 X 20 = o, or A = 2250. IIF = 2250 + 2750 — 2000 — 2000 — 1000 = o, which result checks the computed values of A and B. We now direct our attention to the joint A, the small part of truss near A (or " pass a section " about A and consider that part of the truss within the section), and then note all the forces acting on that part (see Fig. 78). There are three such forces, — the reaction 2250 * The term stress is defined variously. Some writers use it to designate the forces which any two different bodies or any two parts of the same body exert upon each other; that is they use it as a general term for an "action and reaction" (Art. 11). Most engineers, how- ever, use the term in a more restricted sense to designate the force which one part of a body exerts upon an adjacent part at the surface of division. 56 Chap, m pounds, and the two forces exerted upon the part under consideration by the remainder of the truss; they are marked Fi and F2, and both are assumed to be pulls.* This part of the truss, as well as every other part, is at rest, and so the three forces are in equilibrium. Determination of the unknown forces Fi and Fi presents typical problem (i) (Art, 11). We choose the algebraic method for solving: 'LFy = F2 sin 60° + 2250 = o, or F2 = —2600; the negative sign indicates that F2 is really a push, that is, the stress is com- pressive. SFi = Fi — 2600 cos 60° = o, or Fi = +1300; the positive sign indicates that the stress is tensile. Passing a section around B, and consider- ing the forces acting on the part of the truss within the section (or " con- sidering forces at joint 5"), we get Fig. 79. The forces are the reaction 2750 pounds and the two forces exerted on the part under consideration by the remainder of the truss; they are marked F3 and F4 and are assumed to b'e pulls. Solution of this three-force system shows that F3 = +1588 (tension), and F4 =— 3177 (compression). Next we might discuss joint C, D, or E and determine two more stresses. Fig. 80 represents joint C and the forces acting upon it so far as known. Stress _ ^ ^ 1 1000 lbs lOOOlbs.l \ \/ K ' ^'=7\ I3p01 b5. Ny 1588^^5. / \ / \ gyjOibsT ZOOollba. 2600 lbs. 1444 lbs- 666 lbs. 31771bs. Fig. 79 Fig. 80 Fig. 81 Fig. 82 in CA was determined to be a tension of 1300 pounds; therefore the part of CA not shown in the figure exerts a pull of 1300 on the part shown as indi- cated. Similarly, the part of CB not shown in the figure exerts a pull of 1588 on the part shown as indicated; F5 and F^ are assumed to be pulls. Solution of this five-force system shows that F5 = +1444 (tension), and Fg = +866 (tension). Taking joint D next, we get Fig. 81, four forces acting on the joint (the load, and the three forces exerted on the joint by the remainder of the truss), DA was found to be under a compression of 2600 pounds, hence the part of DA not shown in the figure acts on the part shown as indicated; CD was found to be under a tension of 1444 pounds, hence the part of DC not shown in the figure acts on the part shown as indicated; F7 is assumed to be a pull. ZFx = o shows that F^ = —2021 (compression); and writing out 2Fj, we find that it equals zero, which is a fair check on the computation. Fig. 82 represents joint E and all the forces acting upon it, as already deter- mined. If 2Fa; = o and SFj, = o for those forces, then the check on the pre- ceding computations is satisfactory. * In simple trusses the kind of stress (tension or compression) in any member is apparent. When the kind is not apparent, we might follow the suggestion in the footnote, page 41. But for uniformity we will always assume the force to be a pull. Then, according to the footnote, the force is actually a pull or a push (and the stress is tensile or compressive), ac- cording as its computed value is positive or negative. 27501bil AxT. IS 57 5000 I lbs. 5000 1 lbs. Directions. — The foregoing method for "analyzing a truss" (determin- ing the stresses in its members) can be formulated into brief directions as follows: (i) Determine the reactions (supporting forces) on the truss if possible. (2) Consider a joint at which there are only two unknown forces, and then determine those two. (3) Repeat (2) again and again until all stresses have been determined. (These directions do not pro\ide for a certain contingency which may arise; see § 2 for a case and directions for meeting it.) We now give illustration of truss analysis by this method but omitting the computations; they should be supplied by the student. The truss shown in Fig. 83 will be used; it is supported at each end, and supports three loads of 5000 pounds as shown. Obviously each reaction equals one-half the total load. On joint A there are three forces (the re- action, and the stresses in AD and AE)\ solving that force system we find that the first stress = 15,000 pounds compression, and the second = 13,000 tension. On joint D there are four forces (the load 5000 ^^' ^ pounds, the stress \n AD =^ 15,000 pounds, and the stresses in DE and DC unknown); solving that system, we find that the stress in DE = 4335 pounds compression, and that in DC = 12,500 compression. On joint E there are four forces (the stress in AE = 13,000 pounds, the stress in DE = 4335 pounds, and the stresses in EC and EG unknown) ; solving the system, we find that the stress in EC = 4335 pounds tension, and that in EG = 8667 tension. § 2. We now explain the contingency or diflSculty mentioned in the fore- going directions and how to meet it; the truss shown in Fig. 84 furnishes an illustration. Following the directions, we determine the reactions Ri and R2, 2800 pounds and 2400. Then we take joint A, and find stresses in AB and -43" to be 3960 (compression) and 2800 (tension) respectively; next we take joint G, and find stresses in GF and GI to be 3400 (compression) and 2400 (tension) respectively. No joint re- mains at which there are only two 600 1 lbs 800, lbs SOOllbb. aooiibs. ->i<- /e'--->l<-— - Fig. 84 unknown stresses, and the difiiculty is already met. Now if in some way we could ascertain the stress in almost any other member, then we could con- tinue to apply the rule. For example, if we knew the stress in HB, HJ, or HI, then consideration of joint H would determine the two unknown stresses there; consideration of joint B would give stresses in BJ and BC; considera- tion of joint C would give stresses in CJ and CD, etc. Now there is a way to -o Chap, in ascertain the stresses in CD, JD, and HI, — hy passing a section through those members, and solving the force system acting upon either portion of the truss. Fig, 85 represents the left-hand portion and all the forces acting ^ upon it; namely, the three loads, the left reaction, , ^7 and the forces which the right-hand part exerts (^i, ^2, ^°°'"' /ss and 53, assumed to be pulls). Solution of this force system presents typical problem (v) (Art. 13). To determine Si, for example, we take moments about the intersection of ^2 and S3 (or joint D), and find ^ Si= 1600 pounds tension. Then having determined z&oolibs. 1200 jibs. s^ we proceed as in the foregoing examples. Fig. 85 In order to determine the stress in any particular member of a truss the following direction may be tried: Imagine the truss separated into two distinct parts (" pass a section " through the truss); pass it in such a way that the member under consideration is one of the members cut by the section, and so that the system of forces acting on one of the two parts is solvable for the desired stress; then solve the system for the desired stress. (The system of forces acting on one part of the truss consists of the loads and reactions on that part, and the forces, or stresses, which the other part exerts upon it. In plane trusses this system is always coplanar; it can be solved if it is concurrent with not more than two unknowns, or if it is non- concurrent with not more than . three unknowns, provided that the three unknowns are not parallel nor concurrent.) Foregoing direction may be applied not only to bridge over the difiiculty sometimes met in connection with directions in § i, but also when it is desired to determine the stress in a particular member quite directly without first computing stresses in several other members. For example, let it be required to determine the stress in BC (Fig. 86), the truss being supported at its ends, span ^£ = 32 feet, rise CG" = 8 feet, and five loads as shown. Obviously Fig. 86 1000 lbs. Fig. 87 each reaction equals 4000 pounds. A section cutting BC, BG, and OF gives a left-hand part of the truss with its external forces as shown in Fig. 87. The force system can be solved for the desired stress; taking moments about the intersection of ^2 and ^3 (joint G), we get - 5i X 8 X cos 26° 34' — 4000 X 16 -f 3000 X 8 = o, or 5i = — 5600, the negative sign indicating that Si is compressive and not tensile, as assumed in the moment equation. Art. 1 6 59 § 3. Warning is here given that not all trusses can be analyzed by the principles of statics alone, as in the preceding; that is to say, there are trusses that are statically indeterminate. Only the so-called complete or per- fect trusses are always statically determinate; beside these there are incom- plete trusses, and trusses with redundant members. A pin-connected triangle (Fig. 88) is the simplest complete truss; it is indeformable and has no superfluous or redundant members. Adding two more members makes a complete truss of two triangles; and each addition of two members as shown extends the truss and leaves it complete. If m = number of members, and j = number of joints, then for a complete truss, m = 2 j — t,. A pin-connected quadrilateral (Fig. 89) is the simplest incomplete truss; it is deformable and requires the addition of one or more members to make it complete. For an incomplete truss, w < 27 — 3. A pin-connected quadrilateral with two diagonal members (Fig. 90) is the simplest truss with a superfluous or redundant member; it is indeformable and would be so with any member removed. For a truss with a redundant member m > 2 j — t,- Figs. 91, 92, and 93 are other examples of the three classes of trusses described. Fig. 88 Fig. 89 Fig. 90 Fig. qi Fig. 92 Fig. 93 In the foregoing it is assumed that the trusses are pin-connected, and that each member can sustain tension or compression as called upon by the loading. For a classification not so restricted as this one, readers are referred to stand- ard works on Structures.* 16. Graphical Analysis of Trusses; Stress Diagrams § I. Graphical methods are especially well adapted for analyzing trusses. As in the algebraic methods of the preceding article, we imagine the truss separated into two parts, and direct our attention to the external forces acting upon either part. Graphical instead of algebraical conditions of equilibrium are then applied to these forces to determine the unknowns. The notation for graphical work described in Art. 2 can be advantageously systemized as follows: Each triangular space in the truss diagram is marked by a lower- case letter, also the space between consecutive lines of action of the loads and reactions (Fig. 94) ; then the two letters on opposite sides of any line serve to * Johnson, Bryan, and Turncaure's Modern Framed Structures. 6o Chap, hi lOOOMbs. 1000 lbs. designate that line, and the same capital letters are used to designate the magnitude of the corresponding force. This scheme of notation is a great help in graphical analyses of trusses. As an illustration we determine the stress in each member of the truss of Fig. 94. Evidently each reaction equals one-half the load, or 2000 pounds. We " pass section " a, and consider the forces acting on the left- hand part of the truss (Fig. 95) ; they are the load 500 pounds, the reaction 2000 pounds, and the stresses cd and da. Since those forces are in equilibrium, their polygon closes; in constructing it, the unknowns will be determined. Beginning with the knowns, yl.B is drawn to represent 2000 pounds, BC to represent 500 pounds ; and then a line from A (or C) parallel to the line of action of one unknown, and a line from C (or A) parallel to the other, are drawn. The last two lines determine D (or D'), and the closed polygon is A BCD A (or A BCD' A) ; hence the forces in the members cd and ad are represented by CD and DA (3000 and 2600 pounds) respectively. It is seen from the force polygon that CD is a push, and DA is a pull; hence the members cd and ad are in com- pression and tension respectively. 500 lbs b|a EOOOlbs D-- Fig. 95 B ->--'A 3000 lbs. Fig. 96 We may next pass section (3, and consider the forces acting on the smaller (and simpler) part of the truss (Fig. 96); they are the load 1000 pounds, the stress 3000 pounds (compressive), and the stresses fe and de. Their force polygon may be drawn thus: DC to represent 3000 pounds (compression), CF to represent 1000 pounds, a line from F parallel to one of the unknowns, and one from D parallel to the other. The last two lines determine E, and the force polygon is DCFED; hence the forces in the members fe and ed are represented by FE and ED (2500 and 866 pounds) respectively. Both members are in compression. We next pass section 7, and consider the forces acting on the smaller part of the truss (Fig. 97) ; they consist of the stress 2600 pounds (tension), the stress 866 pounds (compression), and the stresses eg and ga. Their force polygon may be drawn thus: AD to represent 2600 pounds (tension), DE to represent 866 pounds (compression), a line from E parallel to one of the unknowns, and a line from A parallel to the other. The last two lines determine G, and the force polygon is A DEC A; 866 lbs. ^/a 26001b5.d ^ a ^L .2— a D \? E Fig. 97 AsT. id 6l hence the forces in the members eg and ag are represented by EG and GA (866 and 1732 pounds) respectively. Each member is in tension. On account of the symmetry of the truss and loading, the forces in the remaining mem- bers are now known. In drawing the force polygon for all the external forces on the part of a truss included within a section about a joint, it will be advantageous to repre- sent the forces in the order in which they occur about the joint. A force polygon so drawn will be called a polygon for the joint; and for brevity, if the order taken is clockwise the polygon will be called a clockwise polygon, and if counterclockwise it will be called a counterclockwise polygon. ABC DA (Fig. 95) is a clockwise polygon for joint b of Fig. 94; A BCD' A is a force polygon for the "forces at joint i," but it is not a polygon for the joint, be- cause the forces are not represented in the polygon in the order in which the forces occur about the joint. The student should draw the counterclockwise polygon for the joint, and compare with ABCDA. If the polygons for all the joints of a truss are drawn separately as in the preceding illustration, then the stress in each member will have been repre- sented twice. It is possible to combine the polygons so that it will not be necessary to represent the stress in any mem- ber more than once, thus reducing the number of lines to be drawn. Such a combination of force polygons is called a stress diagram. Fig. 98 is a stress diagram for the truss of Fig. 94 loaded as there shown. Comparing the part of the stress diagram consisting of solid lines with Figs. 95, 96, and 97, it is seen to be a combination of the latter three figures. It will also be observed that the polygons are all clockwise polygons; counterclockwise polygons also could be combined into a stress diagram. Directions for constructing a stress diagram for a truss under given loads: (i) Letter the truss diagram as already explained. (2) Determine the reactions. (In some exceptional cases this stage may or must be omitted; also stage (3). See § 2 for two illustrations.) (3) Construct a force polygon for all the external forces applied to the truss (loads and reactions), representing them in the order in which their points of application occur about the truss, clockwise or counterclockwise. (The part of that polygon representing the loads is called a load line.) (4) On the sides of that polygon construct the polygons for all the joints. They must be clockwise or counterclockwise ones, according as the polygon for the loads and reactions was drawn clockwise or counterclockwise. The first polygon drawn must be for a joint at which but two members are fastened; the joints at the supports are usually such. Next the polygon is drawn for a point at which not more than two stresses are unknown; that is, of all the members fastened at that joint the forces in not more than two are unknown. Then the next joint at which not more than two stresses are unknown is con- 62 Chap, iii sidered, etc., etc. (These directions do not provide for a certain diflSculty which may arise; see § 2 for a case and directions for handling it.) To illustrate the foregoing directions we analyze the truss represented in Fig. 99; it sustains four loads (600, 1000, 1200, and 1800 pounds), and is (000 1800 :s H supported at its ends. Supposing the reactions to have been determined, we draw the force polygon for the loads and reactions ABC DBF A, at the left; it is a clockwise polygon. We may begin by drawing the clockwise polygon for joint I or 2; for the former it is FABGF* Member hg is therefore in com- pression and gf in tension. Next we may draw the clockwise polygon for joint 2, 3, or 4; for the joint 2 it is CDEHC. Member ch is in compression and eh in tension. For joint 3, the polygon is HEFGH, and member gh is in tension. If the work has been correctly and accurately done, the line GH is parallel to gh. § 2. There are exceptional cases not covered by the foregoing directions. In case the reactions cannot be determined in advance, the stress diagram can still be drawn if the truss is statically determinate. Fig. ii 100 represents such a case, the truss being pinned to its supports. The diagram can be constructed by drawing in succession the proper polygons (all clockwise or counter- |r clockwise) for joints i, 2, 3, and 4. Then, if desired, the reactions can be determined by drawing the polygons for joints 5 and 6. Fig. loi represents a case where the reactions can be determined at stage (2) of the analysis, but determina- tion of the reactions is not essential for the construction of the stress diagram. The truss is supported by a shelf A and a tie B. The stress diagram can be constructed Fig. ioi by drawing in succession proper polygons for joints i, 2, 3, 4, and 5. The reaction at B is determined by the polygon for joint 5; that at A by the polygon for joint 6. * The student is urged to make sketches of the bodies (parts of truss) upon which the forces, whose polygons are being drawn, act. A force acting upon the "cut" end of a mem- ber and toward the joint is a pu'ih, and the stress in the member is compressive; if the force acts away from the joint, it is a pull, and the stress is tensile. Fig. 100 Art. i6 ^3 Fig. 1 02 shows a truss the analysis of which is not fully provided for in the directions. Thus, suppose that the reactions have been determined; the polygon for joint i may be drawn first, next that for joint 2, and then that for joint 3. Similarly the polygons for joints i', 2', and 3' can be drawn; but then no joint remains at which there are but two unknown stresses, and so no more polygons can be drawn, as yet. If in any way the number of un- known stresses at a remaining joint could be reduced to two, then the polygon for that joint could be drawn, and the stress diagram could be completed. Thus, if the stress in ij, jm, or mf could be determined, then the polygon for joint 4 could be drawn, and then those for 5, 6, 7, and 8. 1000 I lbs. 1000 Ibi. 1000 lbs I- ig . 1000 lbs. I \ VA\^ 1000 lbs. Fig. 102 The difficulty here pointed out is just like that mentioned under the direc- tions in § I of the preceding article. It may be met by means of the direction in § 2 of that article, which explains how to determine the stress in a par- ticular member quite directly and independently of any stress diagram or polygons for joints. Thus to determine the stress in w/we pass a section as a, and solve the external system of forces (including stresses in the members cut) which acts upon either part of the truss for the desired stress. Then we proceed with the stress diagram as already pointed out. There are other ways of meeting the difficulty presented in this form of truss, but that here explained is quite general and can be applied readily to other forms. We will now explain this matter in detail, using the same truss. Evidently each reaction equals one-half the total load. ABCDEE'D'C'B'A 'FA is a clock- wise polygon for the loads and reactions. The polygon for joint i is FABGF; that for joint 2 is GBCHG; that for joint 3 is FGHIF. The polygons for joints i', 2', and 3' are B'A'FG'B', C'B'G'H'C, and H'G'FI'H' respectively. The forces acting on the part of the truss to the left of section a are the loads at joints i, 2, 5, and 6, the left reaction, and the forces exerted on the left part of the truss by the right (stresses el, Im, and mf). This system may be solved graphically or algebraically; the algebraic method is much the simpler, arms of forces being scaled from the truss drawing. Thus to ascertain the stress mf, we take moments about the intersection of el and Im, and get 1000 X 64 Chap, rn 7.5 + 1000 X 15 + 1000 X 22.5 + 500 X 30 — 4000 X 30 — {mf) X 17.5 = o, or mf = 3425 (tension). Next we represent the stress mf in its proper place in the stress diagram at MF, and then draw the polygon for joint 4; it is MFIJM. Completion presents no difl&culties. 17. Simple Frameworks (Crane Type) The frames here considered, like the trusses of the preceding articles, are plane and symmetrical with respect to the plane of the frame. For example, the crane represented in Fig. 103 consists of a post MN, a boom PQ, and a brace KQ; the boom consists of two pieces between which the post and the brace lie, and the brace is forked at its lower end by means of side pieces and straddles the post. Like the trusses, these frames are assumed to be pin connected, the pins being practically frictionless. Thus each pin pressure lies in the plane of the frame, and the line of action cuts the axis of the pin. Unlike the trusses, these frames may include a member which is pinned to others at more than two points; the loads also on these frames are applied anywhere, not at the joints necessarily. The result of these conditions is that the stress in any member of the frame is generally not a simple tension or compression, the member being bent as well as stretched or shortened. We will not attempt to determine the stresses in the members of these frames but limit the discussions to a determination of the forces which act upon each member, the pin pressures, reactions of supports, etc. In general the pressure of a pin on a member does not act along the axis of that member. Take, for example, the brace (diagonal) (Fig, 103); it is , N Fig. 103 acted upon by three forces, — its own weight W and the pin pressures K and Q. These three forces must be concurrent or parallel (Art. 10, § 2). If they are concurrent, then neither K nor Q is axial or else both are; but obviously both K and Q cannot be axial and then balance W , and so neither acts axiaUy. If they are parallel, then neither K nor Q acts axially. In some consideration of frameworks, the weights of some or all members are negligible in comparison with other forces (loads) which act upon the frame, and so we may have to do with a. member acted upon by only two forces, — pin pressures. On such a member, the pin pressures do act along the Art. 17 65 axis of that member, since the pressures balance each other and so must be colinear (Fig. 103). " Analysis of a crane " means the determination of every force (magnitude and direction) acting on each part or member due to weight of the crane or loads on it or both. The general method of procedure may be briefly summa- rized as follows: (i) Make a sketch of the entire crane, and represent as far as possible all the external forces acting upon it; apply the appropriate con- ditions of equilibrium to the force system, and then determine as many of the unknowns as possible. (2) Make a sketch of a member or of a combina- tion as they are on the crane, and represent as far as possible all the external forces acting on it; then apply the appropriate conditions of eqmUbrium to the force system, and then determine as many of the unknowns as possible. (3) If other forces remain to be determined, then continue as directed in (2), bearing in mind the law of " action and reaction " (Art. 11). We will now give two examples of analysis employing both algebraic and graphic methods. Example (i).— We analyze the crane represented in Fig. 103; the crane is supported at M and N by sockets in the ceihng and floor. MN =18, PQ = 14, MP = NK = 3 feet; it bears a load of 8 tons on the boom at 16 feet from the axis of the post; weights of members neglected. Fig. 104 rA— to ^>r OTO p /i" t\j ^ — .- .,4' /...-M 1 /^ * » / t Py / 7. U tons P 8 tons K V K 7.lltons ■©"z' n!x .0- ^<^'' Nx|Nj N 8 I tons :t; Fig. 104 (at the left) represents the entire crane with all external forces, the senses of the reactions being quite obvious. The solution of this system falls under Art. 13. Silfiv = o gives M = 7.1 1 tons; since 'ZF^ = o, A'^ = 7.1 1 tons; and since SFy = 0, Aj, = 8 tons. We sketch the brace KQ next. Since it is a two-force member, the pin pressures K and Q are axial, equal, and obviously have senses as shown. The common value of K and Q cannot be determined from a consideration of their equihbrium. Next we sketch the boom. Q on the boom and Q on the brace constitute an action and reaction, and so are coHnear, opposite, and equal; the pressure at P is unknown in direction, and in an algebraic solution can be dealt with most easily through its components Px and Pj,, senses guessed at. Solution of this system falls 66 Chap, hi under Art. 13. Since 'LMp= o, Q = 14.05 tons; since 2F;, = o, P^ = 10.67 tons; since XFy = o, Py= -1.14 tons, the negative sign indicating that Py acts downward. Finally, P = V{io.6f+ 1.14^) = 10.73 tons, and the in- clination of P with the horizontal is tan^^ (i-i4 "^ 10.67) = 6° 7'; and now all the forces on each member are determined, those on the post being represented in the figure. Generally, several sketches may be made and considered in several differ- ent orders, each furnishing a complete analysis. For example, we might have taken the entire crane, the boom, and the post; or the brace, the boom, and the entire crane. The student is advised to try these orders and make the analysis. The graphic method of solving the various force systems may be carried out as follows: The system acting on the entire crane consists of four forces, and so the resultant of any pair of the four forces, as Nx and Ny, balances the other pair; therefore that resultant is concurrent with the second pair and acts in the line 1-2 (Fig. 105). So we draw the force triangle ABC A for those three forces (making AB represent 8 tons), and find that BC represents M and CA the resultant of the first pair. Next we resolve CA into components parallel to Nx and Ny, and find that CD and . Ceiling DA represent Ny and Nx respectively. The forces on the boom being three in number (the load, Q, and P), they must be parallel or concurrent, and because two (the load and Q) are concurrent, all must be; thus the line of action of P is determined. So we may draw the force triangle EFGE for the three forces, W^^p^///////yy////////^^^^^ making EF represent 8 tons; thus we find that EG = P and GF = Q. Example (ii). — For another illus- tration, we analyze the hydraulic crane represented in Figure 106. It consists of a hollow post MN (up into which the piston can be projected) a boom PQ, Floor Fig. 106 \rt. 17 67 and a pin-connected frame KPQ. A single roller is mounted on the pin K, and two on the pin P, so that as the piston moves the frame moves with it, all rollers rolling on the post. Thus there are twelve parts: a post, a boom, two struts KP (one on each side of the post), two ties KQ (one on each side), a pin at P, one at Q, one at K, two rollers at P and one at K. We take the load as 10 tons and x= 15 feet, and neglect the weights of the parts. Fig. 107 represents the entire crane, not including the piston, with aU the external forces acting upon it. 2Fx = o shows that M = Nx, and SMat = o shows that M = (10X15)-^^ where h = height of post. L and iV„ cannot be found from this force system; so we try the frame with rollers (Fig. 108) . The external forces acting on it are the load, the piston pressure L, the post pressure Ri against the single roller, and the result- ^'M \ 1^"^- ., Q V) — •! 1 1. 10^ tons h\ Fig. 107 Py =3 lojtons lOitons Fig. 109 ant post pressure Ro against the lower rollers. The solution of this system falls under Art. 13; it shows that L = 10 and Ri = R2= 21.4 tons. Fig. 109 represents the boom alone and the external forces acting upon it, — the load, the piston pressure, the pin pressure Q (acting along the ties because each is a two-force member), and the pin pressure P whose direction is unknown. The solution of this system falls under Art. 13. DeaUng with the unknown components of P (senses guessed at), we find from XMp = o that Q = 25.2 tons; from SF^ = o that Px = 24 tons; and from SF^ = o that Py= 7.2 tons. The pin at K (Fig. no) is subjected to three forces, namely, the pull of the two ties (25.2 tons), the pressure of the roller (21.4), and the force F exerted by the struts (along the axis of the struts, since each is a two- force member). From 2F„ = o we find that F = 7.8 tons. Fig. no also represents the post with all Since 2F„ = o, iV„ = o; SMtv = o gives gives Nx = M. Now we have all the frl2.13ton& Fig. 117 unknowns in the force system presents typical problem (v). 2Fx = o gives Qx = 12.13 tons; SA/q = o gives Py = 0.95; and 2F„ = o gives Qy = 9.85. Ha\ing found the value of Py, we find from ZFy = o for Fig. 115 that Ky = 10.95 tons. To check the analysis, we might supply values of the forces acting on the brace (Fig. 117), and then test whether the force system is balanced, that is, whether S/^i = o, XFy = o, and 2M = o. 18. Cranes. — Continued In this article we show how to analyze three cranes, paying some attention to the forces due to the hoisting rig. Generally, a pulley is an important part of such rig. We assume here that the tensions Ti and T2 (Fig. 118) in the rope or chain on opposite sides of the pulley on which it bears are equal. This assumption impHes perfect flexibility of rope or chain and a frictionless pin supporting the pulley. The pressure P against the pin equals the resultant of those tensions, or 2 T cos ^ a, and it bisects the angle between their lines of action. If the Hnes of action are parallel (a = o), P = 2 T; if they are at right angles {a = 90°), P = 1.414 T, Example (i). — Fig. 119 represents a crane supported in a footstep bearing at the floor and a collar bearing on the wall bracket H. The hoisting rig con- sists of a simple hand winch mounted on the wall at W, a chain, and pulleys as shown. Pulley at G is 12 inches in diameter; the load is one-half ton. The reactions at the supports depend on the hoisting rig, as will be seen from the following: On the entire crane, including the top pulley (Fig. 120), there are acting four forces, namely, the upper reaction U, the lower reactions P, Fig 70 Chap, iir and Py, and the pressure of the chain against the pulley equivalent to two components, one-half ton each, as shown. Taking moments about the lower end, we find H to be 0.0S7 ton; from SF^ = o and SFj, = o, we find that 5i K6=I2' KJ = 5' PKJ =6KJ Fig. 119 Fig. 120 Px = 0.413 and Py = 0.5 ton. All members except the vertical HP are simple tension or compression members. Force polygons for joints G and / show that the stresses are as follows: GK = 0.35 ton (tension); GJ = 1 ton (com- pression); JK = 0.57 ton (compression); JP = i ton (compression). Mem- ber HP is subjected to the reactions of the supports as already computed, and the following forces: a pull of 0.35 ton along KG; a push of 0.57 ton along KJ; and a push of i ton along PJ. Example (ii). — Fig. 121 represents a common type of derrick. It is sup- ported by a footstep at the bottom of post and at the top by two stiff legs ""'^W////////////////////'- Fig. 121 ■A >- ^2T which extend backward to the ground or other base; the spread (angle be- tween their horizontal projections) being 90 degrees so that the derrick can swing about its vertical axis through 270 degrees. Sometimes the derrick is Art. i8 yi supported at the top by a collar bearing held in place by cables extending off to quite remote points on the ground. Obviously the pull on a stiff leg is greatest when the boom is in the same plane with that leg; the pull on a cable is greatest when that cable and the boom are in the same plane and on opposite sides of the post. Let P denote this pull, and a the inclination of the cable to the horizontal or the inclination of the line joining the pivot on the post with the lower end of a stiff leg. Then taking moments of all external forces on the derrick about the footstep bear- ing, we get Ph cos a = Ws, or P = Ws/h cos a (only the weight of the load being taken into account). Calling the horizontal and the vertical reactions at the footstep H and V respectively, we find that H = Ws/h and V = W -{- P sin a = W(i + tan a • s/h). There are seven forces acting on the part shown in Fig. 122, which consists of the crane post, the winch W, the two sheaves S, and a part of the hoisting and topping ropes as shown. The forces are: H, V, and P (already explained) ; Q, the pressure of the boom on the post acting in a direction as yet unknown; ^ W, approximate value of the tension in the hoisting rope; T, which denotes the tension in the topping rope; and 2 T, exerted by the top pulley shackle. Of these seven forces, all except Q and T are already known. To find these we may proceed as follows : Take moments of all the forces about the pin at Q, and thus find T; then take horizontal and vertical components, and thus find the horizontal and vertical components of Q, and finally Q itself. The force system can be solved graphically as follows : First find the line of action of the resultant R of the two forces T and 2 T; then this R and the other five forces constitute a system in equilibrium, which solve for R and Q by methods explained in Art. 13; finally resolve R into its components T and 2 T. Example (iii). — Fig. 123 represents a sheer leg crane. It consists of two front legs AC and BC and a back leg CD, all connected by a horizontal pin at C; the front legs are pin-supported on the ground at A and B, and the back leg ,,^^jf is restrained at the ground by a holding- .^^^/^ h down rail and a long horizontal screw which ^.f^^y // I works in a nut on the lower end D. The ^^^ ^'^'^-^^-.^.^ I ■■ purpose of the screw is to move D, thus .^^-—'^ ^^^^=s=^.. turning the front legs about AB and moving '•=^=-~ ^--^ the load in and out. We will now show how to determine the pressures on the ends of the legs due to their own weights, taking the following data: lengths of front legs 160 feet, distance between their lower ends 50 feet, distance between their upper ends 10 feet, length of back stay 210 feet, weight of each front leg 44 tons, of the back leg 53 tons; we take the crane in its position of greatest overhang, 64 feet. The external forces acting on the crane are the following (see Fig. 124): the three weights, the holding-down force Dy, the push of the screw Dx, the inward pushes At and Bz of the supports at A and B, and the pressures of the _. Chap, ui 72 pins at A and B; each of these pressures is represented by two components in the figure, A^, Ay, and B^, By, respectively. There are six conditions of equilibrium for this system, namely, the sums of the components of the forces along the x, y, and z axes, and the sums of the moments about those axes equal zero. Thus, — ZFy= Ay+By- Dy- SS-44- 44=0 XF, = - A,-\-B^ = o •ZM^= -Ay X 25 + J5^ X 25 + 44 X 15 - 44 X 15 = o 2M„=^xX 25-5xX 25 = S7lf.= I>„X87.6 + 53X 11.8-44X32X 2 = (i) (2) (3) (4) (5) (6) Equation (6) shows that Dj, = 25 tons; (4) shows that Ay = By-, from these results and (2) it follows that Ay and By equal 83 tons. No other unknowns can be determined from the equations; but (3) shows that Az = B^, (5) that A, = Bx, and (i) that A^-\-B^^D^. V^ 44- tons ze^-^^l- 1 75.8:.^ 75.8'--^ ^^, l>^r;%7"^' tons Fig. 125 Fig. 126 To get values of these unknowns we consider the forces acting on the back leg; there are four forces, namely, the weight of the leg (53 tons), the holding- down force Dy (25 tons), the screw pressure Dz, and the pressure of the upper pin at C, represented for convenience by two components which we call Cx and Cy (Fig. 125). This system is in equilibrium and so Slfc = 25 X 151. 6 — D^ X 145-2 + 53 X 75-8 = o, or D^ = 53.8 tons; SF^ = C^ - 53.8 = o, or ^' — 53-8; and I^Fy = Cj, — 25 — 53 = o, or Cy = 78. Returning now to equations (i) and (5), we find that Ax and Bx = 26.9. To get Az and B^ it is necessary to discuss the forces on one of the front legs. There are three forces, — the weight 44 tons, and the pressures at the ends; each of the pres- sures is represented (Fig. 126) by three components, 26.9, 83, and B^ below, and Qx, Qy and Qz above. The system being in equilibrium, we take moments about the vertical line through Q; thus 5^ X 64 — 26.9 X 20 = o, or B^ = 8.41 tons. Inspection shows that Qx = 26.9, Qy = 39, and Qz = 8.41 tons. Art. 1 8 73 lb. 2 tons |<- 75.8'- ■ ->) ; K-J2'>H-32'-i The forces acting on the upper pin (at C) are represented in Fig. 127, by means of their components. We now give another solution of the foregoing example, making use of the principle that if the forces of a system i n equilibrium b e represented by vectors, then the projection of those vectors on Fig. 127 any plane represents a force system also in equilibrium (Art. 10 under (vii)). Projecting the force system represented in Fig. 124 on the three coordinate planes, we get the three systems represented in Fig. 128, — side elevation, end elevation, and plan. From the side elevation, ^Ma = o gives Dy= 2% tons; 2i//> = o shows that Ay = By] and ^Fy shows that Ay-\- By = 166, or ^1, and By = 83 tons. No further numerical result can be obtained from these projected systems. Considering the back leg alone as before, we would find that Z)^ = 53-8 tons; then from the plan A^= Bx obviously, and Ax+Bx= 53.8, or A^ and -Sx = 26.9 tons. Az and Bz would be gotten as before.* * For full information on cranes, see Bottcher's book on that subject, English translation by Tolhausen. Side Elevation 53.8ton5 * — _ [<-?5''>{<25'>l End Elevation CHAPTER IV FRICTION 19. Definitions and General Principles § I. Definitions, Etc. — When one body slides or tends to slide over an- other, then the sliding of the first or its tendency to slide is resisted by the second. Thus,, if A (Fig. 129) is a body which slides or tends to slide toward the right over B, then B is exerting some such force as i? on ^, and the component of R along the surface of contact is the resistance which B offers to the sliding or tendency. Of course A exerts on 5 a force equal and opposite to R; either of these equal forces is called the total reaction between the two bodies. The component of either total reaction along the (plane) surface of contact is called friction, and the component of either along the normal is called normal pressure; they will be denoted by F and N respectively. If the surface of contact of the two bodies is not plane, the force exerted at each elementary part of the surface is the total reaction at that element, and its components in and normal to the element are the friction and the normal pressure at the ele- ment. Friction is called kinetic or static according as sliding does or does not take place. Only static friction is considered here. 'W/////^//M^///M/////'//////' Fig. 129 W////////////. '///.^////////A •'''//////////////y, V///.ii'///////A '5 YW 4 lbs. '■^6' YW Fig. 130 YW The amount of static friction between two bodies depends upon the degree of the tendency to slip. Thus suppose that A (Fig. 130) is a block weighing 10 pounds, upon a horizontal surface B\ that the block is subjected to a hori- zontal pull P, and that the pull must exceed 6 pounds to start the block. Obviously when P = 2 pounds say, then F = 2; when P = 4 pounds, then F = 4; etc., until motion begins. So long as P does not exceed 6 pounds, F equals P; that is, F is passive and changes just as P changes. The inclination of the reaction R also depends on the degree of the tendency to slip. When P = 2 pounds, then the angle NOR = tan-^ f^ = 11° 19'; when P = 4 pounds, 74 Art. 19 75 NOR = tan-' t% =21° 48'; etc., until motion begins. The greatest values of the friction F and the angle NOR obtain when motion impends. The friction corresponding to impending motion is called limiting friction. We wnll denote it by Fm, since it is a maximum value (see Fig. 130). The coefficient of static friction for two surfaces is the ratio of the limiting friction corresponding to any normal pressure between the surfaces and that normal pressure. We will denote it by /i; then M = FJN, or F„, = ixN; also, F > nN. The angle of friction for two surfaces is the angle between the directions of the normal pressure and the total reaction when motion is impending. We will denote it by (see Fig. 130); then tan<^ = Fm/N; hence tan = n. If a block were placed upon an inclined plane, the inclination at which slipping would impend is called the angle of repose for the two rubbing surfaces; it will be denoted by p. The angles of friction and repose for two surfaces are equal; proof follows: Suppose that A \^ (Fig. 131) is on the point of sliding down the incline; two forces act on A , its own weight W and the reaction ^^ R of the plane. Since A is at rest, R and W are colinear, \ffff^ that is, R is vertical; and since motion impends, the <0--^ angle between R and the normal is the angle of fric- ^^^- ^^^ tion (/). It follows, from the geometry of the figure, that and p are equal. The coefficient of static friction for two bodies A and B may be found in several ways: (i) Place ^ on 5 as in Fig. 130, and determine the pull P which will just start A; then fx = P divided by the weight of ^. Or (ii) tilt B, and determine the inclination at which gravity will start A down; then ju equals the tangent of that angle of inclination. In either method several determina- tions must be made to obtain a fair average. Many experiments have been made in these ways, and it has been ascertained that coefficients of static friction depend on the nature of the materials, character of rubbing surfaces and kind of lubricant, if any be used. Early experimenters reported (Coulomb 1871, Rennie 1828, Morin 1834, and others) that the coefficient is independent of the intensity of normal pressure; and although this announcement was clearly subject to the limitation of the range of the experiments performed, yet it was generalized and long accepted as a universal law of friction. But the universality of the law has been questioned; Morin himself pointed out that length of time of contact of the two bodies influences the coefficient; and obviously the coefficient changes when the intensities of pressure get so low that a considerable part of the friction is due to adhesion, or so high as to affect the character of the surfaces in contact. Messiter and Hanson report* prac- * Eng. News, 1895, Vol. 33, page 322. ^ Chap, iv 76 tical constancy of coefficient for yellow pine and spruce. They give the following for planed or sandpapered (i) yellow pine and (2) spruce. (i) n = 0.25 to 0.32; average M = 0.29 for 100 to 1000 lbs. per sq. in. (2) M = 0.18 to 0.53; average M = 0.42 for 100 to i6oolbs.per sq. in. The variation depends on relation of grain of wood to direction of slide. Coefficients of Static Friction (Compiled by Rankine from experiments by Morin and others.) Dry masonry and brickwork o-6 to 0.7 Masonry and brickwork with damp mortar 0.74 Timber on stone about 0.4 Iron on stone ^-3 to 0.7 Timber on timber o-2 to 0.5 Timber on metals o-2 to 0.6 Metals on metals o-i5 to 0.25 Masonry on dry clay o-S^ Masonry on moist clay ^-33 Earth on earth 0-25 to i.o Earth on earth, dry sand, clay, and mixed earth 0.38 to 0.75 Earth on earth, damp clay i-o Earth on earth, wet clay 0.31 Earth on earth, shingle and gravel 0.81 to i.ii § 2. Tractive Force. — Let W = the weight of a body A upon a horizontal surface B (Fig. 132), m = the coefficient of friction for the surfaces in contact, P = cos d -\- lisind cos {d — ) -.6 Fig. 132 4 A V e w Fig. 133 Fig. 134 Fig. 13s The forces acting on A are P, W, and the reaction of the plane whose two components are A'' and (when motion impends) Fm (Fig. 133). Now P cos Q = F„, iV = PF — Psin0, and Fm = m^V; these three equations solved simul- taneously furnish the first stated value of P. The second value can be ob- tained from the first, or by solving the three-force system acting on A as repre- sented in Fig. 134. According to Lami's theorem (Art. 10), P/PF = sin 0/sin (90 -f - 0); hence P = M^ sin 0/cos (6 - (p). Art. 19 77 If the pull P is horizontal (d = o), then P = /iW. If the pull is inclined, but not too much, then the pull P required to start the body may be less than nW. In fact the least value of P obtains when 6 = (f), — "the best angle of traction equals the angle of friction," — and the minimum value of the pull is Wsincj). Proofs follow: (i) Evidently W sincf) -r- cos (0 — 0), the general value of P, changes as 9 changes, and, for a given W and as stated, etc. (ii) Or, let AB (Fig. 135) represent W, BC be parallel to P, and AC he parallel to R; then CA represents R. If be changed, then BC (and P) will change; and evidently P will be least when BC is per- pendicular to CA, that is, when d = (j). And then BC (or P) = W sin 0. § 3. Test for Rest or Motion. — A body is supported so that it can slip and is subjected to given forces; it is required to ascertain whether those forces do cause slipping, and the value of the friction is desired. We assume that the body is at rest, and determine the friction F and the normal pressure N from conditions or equations of equilibrium; then we compare F with fxN. If F is less than /jlN, there is no motion and the computed value of F is correct; if F is greater than fiN, then there is motion and the friction is kinetic, its value being less than jjlN. For example, consider a block of material weighing 100 pounds supported on a horizontal surface, the coeflEicient of friction being ^, and imagine a down push of 200 pounds applied to the block at an angle of 30 degrees with the vertical. N = 100 + 200 cos 30 = 273.2, and for rest, F = 200 sin 30 = 100; ijlN = I X 273.2 = 136.6, and this is the greatest fric- tional resistance which the support can offer so long as N = 273.2. Only 100 pounds are required to prevent motion, and so the body is at rest under the action of friction of that required value. Or, to test for rest or motion, we may make use of the so-called cone of jric- tion for the two bodies in contact, which may be described thus: Let P (Fig. 136 or 137) denote the resultant of all the forces applied to or acting on the body A (whose state is to be investigated) but not including the total reaction of the supporting body B; the point where P cuts the surface of con- tact between A and B, and DOC equal the angle of friction; then the cone generated by revolving OC about OD is the cone of friction. If the line of action of the resultant P does not fall outside the cone (Fig. 136), then there is no slipping; if it does fall outside (Fig. 137), then there is slipping. Proof follows: As already pointed out, the direction of the total reaction i? on a body, which tends to slide over another, depends on the degree of the tendency; the greater the tendency, the greater the inclination of R from the normal; but the in- clination has a limit, that limit being equal to the angle of friction, and it ^o Chap, iv obtains when slipping impends. Therefore when P acts within the cone or along an element of it, then R can incline and completely oppose P (Fig. 136), no matter how large P may be. When P falls outside the cone, R can incline only to an element, and the friction cannot successfully oppose the component of P which tends to move the body (Fig. 137). In the preceding example P is the resultant of the weight of the block 100 pounds, and the applied push 200 pounds. That resultant makes an angle of 10° t,2>' '^^'ith W or the normal. The angle of friction is tan"^ i or 26° 34'; hence P falls inside the cone and, according to the principle of the cone, motion does not ensue. As another application of the cone principle consider Fig. 138, which repre- sents (in plan and elevation) a type of simple hanger. It consists of a fixed vertical rod and a horizontal piece which is forked; there is a hole in each part of the fork so that the piece can be slipped over the rod as shown in the elevation. The hanger, if properly made, will not slip down along the rod on account of its own weight or that of a load unless it be hung quite close to the fork. The mechanics of the device may be explained as follows: Obviously the rod reacts on the hanger at Oi and O2. When slipping impends at these points, the reactions act along OiCi and O2C2 inclined to the normals an amount equal to the angle of friction as shown. The hanger being at rest (by supposition), the third force acting upon it (the load, weight of hanger neglected) must be concurrent with these two reactions; hence to just put the hanger on the point of slipping, the load must be hung from a point in the vertical through C. If the load is hung out beyond C, as at A, the hanger will not slip. For suppose slipping to impend at Oi, then R at Ox would act along OiCi, and R and W would concur at a. To preserve equilibrium, R at O2 must also act through a, which is possible, since O^a is within the cone. Or suppose slipping to impend at O2, then R at O2 would act along O2C2, and R and W would concur at m. To preserve equilibrium, R at Oi must also act through m which is possible. In similar manner, it can be shown that a load hung between the rod and C, as at B, would cause slipping. "SP Fig. 138 20. Friction in Some Mechanical Devices § I. Inclined Plane. — Let a = the inclination of the plane to the horizontal (Fig. 139), p = angle of repose for the plane and a particular body upon it, ) Art. 20 79 as can be shown by means of Lami's theorem (Art. 10) applied to the three forces acting on the body {P, W, and the reaction R of the plane). Thus Pi/W = sin (a + 0)/sin (90 — + 0) ; hence, etc. Pi is a minimum (for given W, a, and ) when 6 = ; then its value is W sin (a + ). For, it is ob- vious that Pi is least when sin (90 — ^ + 6) is greatest, that is, when 4> = d. (ii) When the inclination of the plane is greater than the angle of repose (a > p = ). (iii) When the inclination of the plane is less than the angle of repose (a < p = <^), then the body would not slip down on account of its own weight. The push P required to start the body down is given by P3 = W sin (0 - a)/cos {(j> + d). Ps is a minimum when 6 = —0; then its value is W sin ( — a). When the force P acts along the plane (d = o), then the values of Pi, P2, and P3 are respectively, Fig. 139 W sin (a -\- (/)) COS0 W sin (a — = their common angle of friction, then the force P required to start the wedge inward is given by Pi = W tan (2 -f a). wm/m/m^wm///////?///// Fig. 140 Fig. 141 Fig. 142 Fig. 141 represents the three forces W, Ri, and R2 acting on the block M; also the three forces R2" (= R2), R3, and P acting on the wedge. The angles which Ri, R2, and R3 make with their normal components equal (p, since motion im- pends, by supposition. In Fig. 142, ABCA is a triangle for the forces acting 8o Chap, rv on M, AB representing W; and CBDC is a triangle for the forces acting on the wedge. The given formula for Pi may be derived from these triangles by solving for BD, which represents Pi. From the first triangle (R2 = R2") /W = cos 0/sin (90 - - a - 0), or R2 = R2" = W cos 0/cos (2 + a) ; from the second triangle Pi/{R2 = R%") = sin (2 + a)/cos 0. Therefore Pj = [R^' = R^") sin (2 + a)/cos = IF tan (2 + a). If the wedge angle a is less than 2 0, the wedge will not slip out under any load W even when there is no push P; that is, the wedge is self-locking. The force required to pull the wedge out, that is to lower the load W, must equal W tan (2 — a), when a > (guide at right of M), or W sm (2 — a) -T- cos a, when a < (guide at left of M). In order that the force Q (Fig. 143) may overcome the resistances W, the frictional resistances at the four contacts must be overcome also. If the con- FiG. 143 Fig. 144 Fig. 14s tacts are equally rough and = their common angle of friction, then the force ^1 necessary to start the wedge down is given by 2W Fig. 146 cot (0 + a:) — tan Fig. 144 represents the forces Q, R\ , and R2 acting on the wedge, and the forces acting on M and N . Each of the reactions R makes with its normal component an angle equal to (motion impending). In Fig. 145, ABCA is a triangle for the forces acting on i/, AB representing W . AC DA is a triangle for the forces acting on the wedge. The given formula for Qi can be derived from these triangles by solving them for DA, which represents Qu If the wedge angle 2 a is less than 2 0, then the wedge would not slip out under any pressures W even when there is no push Q; that is, the wedge is self-locking. The force required to pull the wedge out {M and N guided above) is given by '^^ ^^ cot (0 - a) + tan Art. 20 ol § 3. Screw. — Fig. 146 represents a simple jackscrew much used for raising and lowering heavy loads through short distances. In the simpler forms, the screw is turned by means of a lever stuck through a hole in the head H of the screw. There is frictional resistance between the screw and the nut, also between the cap C and the head of the screw, unless the load can turn with the screw. Let P = the (horizontal) force applied to the lever; a = the arm of P with respect to the axis of the screw; W = load on the cap; r = mean radius of the screw, | (ri + ra); a = pitch angle = tan-^ (h ^ 2 7rr), where h = pitch; and = tan-^, where m = coefiEicient of friction. Dis- regarding the friction between the cap and head of the screw, the moment required to raise the load (or move the screw against W) is given by Pia = TFr tan ( -\- a) = o, or cos ((t>-\- a)'E dR = W, and Pia —"EdR sin ( + a) r = o, or r sin ((^ -f a) 2 (ii? = Pia. These two equations combined give Pia = Wr tan (0 + 0;). (2) When the screw tends to descend, dF acts upward as shown at B; and when motion impends, the angle between dR and the vertical is — a. Taking the sum of the vertical components, and the sum of the moments as above, we get equations which yield the required result. To allow for the friction between the cap and the head of the screw, let n = the coefhcient of friction, and R = the effective arm of the friction there with respect to the axis of the screw. (If the surface of contact between the cap 82 Chap, iv and the head were flat and a full circle, R would equal two-thirds the radius of the circle. But the contact is generally a hollow circle, as in Fig. 146, and then R is practically equal to the mean radius.) The friction moment at the cap is ixWR; (i) for raising the load, Pa = Wr tan (0 + a) + fiWR, (2) for lowering the load. Pa = Wr tan ( - a) + fiWR. § 4. Journal in Worn Bearing. — Fig. 147 represents, in section, a. journal in a worn bearing, wear much exaggerated; the contact between the two is along a Hne practically. When the journal is about to turn clockwise and slip, then the bearing exerts a reaction R', making an angle <^ (the angle of fric- tion for the surfaces in contact) with the normal ON; when the journal is about to turn counterclockwise and slip, then the bearing exerts a reaction R" inclined at an angle (}> with ON, but on the other side. If the radius of the journal is r, then the perpendicular from the center to R' and R" equals r sin <^, and the circle of radius r sin with center at the center of the cross section of the journal is tangent to R' and R". This circle is called the friction circle for journal and bearing. For smooth contacts sin (j) nearly equals tan (j> or /z, and hence the radius of the circle practically equals iir. Fig. 147 Fig. 148 We use the friction circle as an aid to fix upon the line of action of the re- action between journal and bearing when motion impends; the line is tangent to the circle. For example, consider the bell crank shown in Fig. 148, the journal being i| inches in diameter and the coefl&cient of friction 0.3; the re- quirement is to determine the least force P, acting as shown, which will over- come Q (that is, start the bell crank to turn clockwise), and the pressure on the bearing then. The radius of the friction circle is f sin tan~^ 0.3 = 0.18 inch. Since there are but three forces acting on the bell crank {P, Q, and R), they are concurrent, that is, R acts through 0; but R is also tangent to the circle as shown, and so its line of action is known. To determine the values of P and R, we draw AB to represent Q by some scale, and lines through A and B parallel to P and R to their intersection C; then BC and CA represent the magnitudes and directions of R and P respectively, (Which one of the two tangent lines to take can be determined by trial. Thus, trying ON, the contact between journal and bearing would be at N, and the tangential or frictional component of the pressure on the journal would Art. 20 83 be as shown, not consistent with the assumed tendency to slipping. Obvi- ously the other tangent is the correct one, and on investigating for the friction component of R when acting at M we find that such component is consistent with the assumed tendency to slip.) The force P which would just permit Q to start the bell crank to turn counter- clockwise could be determined in a similar way. Then R would act along the tangent ON, and P would be represented by C'A. When P has any value between C'A and CA, then slipping does not impend, and the line of action of R cuts the friction circle. When a link L (Fig. 149) of a machine or structure is pinned to other parts or members, and there is slipping or tendency to slipping at the pins, then the pressure exerted by each pin on the link does not necessarily act through the center of the pinhole there. If slipping impends, then the line of action of the pressure is tangent to the friction circle; and if the link is a two-force member (only the two pin pressures acting on it), then the two pressures are colinear and must act along a line which is tangent to both friction circles. Which one of the four tangents to take in a given case depends upon the direc- tion of the tendency to slipping at each pin, and whether the link is under ten- sion or compression. To ascertain the correct tangent, try any one as the line of action of the two pin pressures R, and then investigate the i?'s for their frictional components to ascertain whether the directions of those components are consistent with the directions of slip; only one tangent will satisfy all Fig. 149 the conditions for a given case. For example, suppose that the tendency is for a to increase and /3 to decrease; if the pressures put the link under tension, then the two pressures act along tangent number i at points Ai and A2, and if the pins put the link under compression then the two pressures act along tangent number 2 at points Bi and B^. The deviations of the various tangents (lines of action of the pin pressures) from the axis of the link depend on the diameter of the friction circle and the length of the link. Generally the diameter is so small compared to the length of the link that the deviation is small, and one may safely take the axis of the link as the line of action of the pin pressures so long as the link is at rest and for all states of tendency to slip. § 5. Belt or Coil Friction. — Fig. 150 represents a cyUnder about a part of which a belt or rope is wrapped. If the cylinder is not very smooth, then o . Chap, iv the pulls Pi and P2 may be quite unequal without causing slipping over the cylinder, as may be easily verified by trial. When slipping impends, then the ratio of these pulls depends on the coefficient of friction and on the angle of wrap. If P2 = the larger pull, m = the coefficient of friction, a = the angle of lap expressed in radians, and e = base of the Napierian system of logarithms (2.718), then as proved below. For a given value of Pi, P2 increases very rapidly with a as shown by Fig. 151, which is the polar graph of the foregoing equation, P2 and a being the vari- ables, e= 2.718, M taken as \, and Pi = OA. The following table gives values of the ratio P2/P1 for three values of the coefficient of friction and for twelve values of the angle of lap. Maximum Ratios P2/P1 (Slipping Impending) ** M a a 2 IT 25r 1 1 1 1 1 4 3 2 4 3 1 O.I 1. 17 I 23 1-37 0.7 3.00 4-33 9.00 0.2 1-37 i-Si 1.87 0.8 3SI 5-34 12.34 0.3 1.60 1.87 2.57 0.9 4. II 6.58 16.90 0.4 1.87 2.31 351 I.O 4.81 8.12 23 14 o-S 2. 19 2.8s 4.81 2.0 23- 66. 535- 0.6 2.57 3-51 6.59 30 III. 535- 12,390. Fig. 150 Fig. 151 Fig. 153 Derivation of Formula. — The forces acting upon the part of the belt in con- tact with the cylinder consist of the tensions Pi and P2, the normal pressure, and the friction (Fig. 152). Let p denote the normal pressure per unit length of arc; then the normal pressure on any part whose length is ds (enlarged in Fig. 153) is pds. The friction on that part may be called dF, and the ten- sions P and P + dP. Since the part is at rest, pds = 2 P sin ^ dd = Pdd, or p = P/r; that is, the normal pressure per unit length at any point of the con- tact equals the belt tension there divided by the radius of the cylinder. When slipping impends, dF = fxpds, and since dF = dP, ,„ P. dP ds dP -=- n ~ds, or —- = ju — = fjdd. Art. 20 85 Integration gives [ioge Pj^' = fx |^0j"; hence, loge P2 - log. Pi = ^Jux, or P2 = Pie*^. For an example consider the band-brake shown in Fig. 154. It consists of a rope or other band wrapped part way around a brake wheel W, the two ends of the band being fastened to the brake lever L; the lever is pivoted at Q. Obviously any force as P tightens the band, and if the wheel tends to turn (on account of some turning force, not shown), then P induces friction between wheel and band. We will now show how great a frictional moment (origin in the axis of the wheel) the force P can induce. Let M = the moment, P2 = the larger tension in the brake band (on the side as marked when the wheel tends to rotate as indicated), Pi = the smaller tension, r = radius of the wheel, ai = arm of Pi with respect to Q, (h= arm of P2, and a = arm of P. Consideration of the forces acting on the brake-strap shows that M ={Pi — Pi)r; consideration of forces acting on the lever shows that Pa = Pifli + P2a2. For a given P, M is greatest when slipping impends, and then P2 -V- Pi = e*^. These three equations solved simultaneously show that M = Pa (e'*« - i)r -=- (a^e'"' + ai). For example, let P = 75 pounds, a = 10 feet, n = I, a = 320° (= 5.5 radians), r = 3 feet, ai = 2 feet, and a2 = 9 inches. Then a -5- 2 tt = about 9, and e*"* = 4.1 15 (see table on preceding page); and M" = 75 X 10 (4.1 1 — i) 3 -T- (I X 4-11 + 2) = 765 foot-pounds. Fig. 154 CHAPTER V CENTER OF GRAVITY 21. Center of Gravity of Bodies § I. It is shown in Art. 7 that the resultant of two parallel forces Fi and Fj acting at two points A and B of any body cuts the line A Bin a, point P so that AP/PB = F2/F1 (Fig. 155). This proportion fixes the position of P, and since the proportion is independent of the angle between AB and the forces, P is also, so independent. Therefore ii AB were a rod and Fi and F2 the weights of two bodies suspended from A and B, then the resultant R of Fi and F2 would always pass through the same point even if the tilt of the rod were changed slowly so as to leave the suspending strings parallel. Furthermore, if three parallel forces be applied at definite points A, B, and C of a body (Fig. 155), and if R denotes the resultant of Fi and F2 as before and R' the resultant of R and F3 (and so also the resultant of Fi, Fo, and F3), then CP'/PP' = R/F3. This proportion fixes P' (in CP), and it is independent of the angle between the forces and the plane of ABC. Therefore ii AB and CP be two rods rigidly fastened at P, and Fi, F2, and F3 the weights of bodies suspended from A B, and C, then the resultant of the three forces would always pass through P' if the rods were slowly turned about leaving the strings parallel. And so if any number of parallel forces have definite points of application on a rigid body, the resultant of the forces always passes through some one definite point of the body, or of its extension, when the body is turned about so as not to disturb the parallelism of the forces. This unique point is called the center or centroid of the parallel forces. The forces of gravity on all the constituent particles of a body constitute a parallel force system having definite points of application; therefore all those forces have a centroid. That is, the resultant of the forces of gravity on all the particles of a body (its weight) always passes through some one definite point of the body, or of its extension, no matter how the body is turned about; 86 Art. 21 87 this point is called the center of gravity of the body. The positions of the cen- ters of gravity of many regular bodies are given in Art. 24, and methods for determining those centers of gravity are explained in Art. 23. We now show how to locate the center of gravity of a body (or of a collection of bodies) which consists of simple parts whose weights and centers of gravity are known. Let yl, B, C, etc. (Fig. 156), be the centers of gravity of certain parts of a body (not shown) ; Wi, W2, Ws, etc., the weights of those parts; xi, ji, Zi, the coordinates of A ; X2, }% Z2, the coordi- nates of B, etc. Also let W denote the weight of the whole body, Q its center of gravity, and x, y, z, the coordinates of Q. Since W is the resultant of Wi, W2, W3, etc., the moment of W about any Hne equals the algebraic sum of the moments of Wi, W2, W3, etc., about the same line (Art. 8). moments about the y-axis, we get Fig. 156 Thus, taking Wx = Wixi + W2X2 + W3X3 + •> from which equation x can be determined. Similarly, by taking moments about the x-axis we can get y. To get z, we imagine the body turned until the y-axis is vertical, — the coordinate axes are assumed fixed to the body, — and then take moments about the x-axis; or, what comes to the same thing, we im- agine the forces of gravity (TFi, TF2, TF3, etc.) all turned about their respective points of application until they become parallel to the y-axis, and then take moments with respect to the x-axis. A name for the product of the weight of the body and the ordinate of its center of gravity with respect to a plane will prove convenient; we will call such product the moment of the body with respect to the plane.* Then the equations mentioned can be rendered in the form of a proposition as follows: The moment of a body with respect to any plane equals the algebraic sum of the moments of its parts with respect to that same plane. (i) As an example we determine the coordinates of the center of gravity of a slender wire 43 inches long bent as represented by the heavy line in Fig. 157. If the weight of the wire per unit length is w, say, then the weights of the several straight portions beginning at the left are as listed in the schedule under W. The coordinates of the respective centers of gravity are listed under X, y, and 2; and the moments of the parts with respect to the yz, zx, and xy planes in the last three columns respectively. The coordinates of the center * This moment does not of course have anything to do with turning effect like the ordi- nary moment of a force (with respect to a line or point). To distinguish these moments, the first is sometimes called a statical moment, not very appropriately, however. See also Art. 22 for other statical moments. 88 of gravity of the whole wire are: x= 177.5 ly -^ 43 «' 43 z£) = 3.44 in.; z = 192 2£; -^ 43 «' = 4-47 in- Chap, v 4.13 in.; y = 148 w -^ F X y z Wx Wy Wz 6 8 10 10 4 -2.5 0.0 0.0 10.0 10. 3 6 6 3 — 2 8 8 4 4 8 — 12.5M; 00.0 00.0 50.0 100. 40.0 ow 18 48 60 30 -8 40 w 48 32 00 40 32 43 w 177. Sw 148 w 192 w Fig. 158 (ii) As another example, we determine the center of gravity of a flat sheet of tin consisting of three parts (Fig. 158), namely, a square, a semicircle, and an equilateral triangle. If 5 = the side of the square, and w = the weight of the tin per unit area, then the weights of the parts are as scheduled under W. Obviously, the center of gravity of the square is at the intersection of the diagonals; in Art. 24 it is explained that the center of gravity of a semicircle is 4 r/3 TT distant from the center of the circle where r = radius, and that the center of gravity of a triangle is on the medians of the triangle and \ a distant from any base where a = altitude measured to that base. Hence the coor- dinates of the centers of gravity of the parts are as scheduled under x and y. The moments of the parts with respect to the yz and zx planes are scheduled in the last two columns respectively; hence, for the whole sheet of tin x = 0.571 s^w -^ 1.826 shv = 0.313 s, and y = 0.633 ^"^ "=" 1-826 shv = 0.347 s. Part. W X y Wx Wy Square 1 . 000 s^w •393 • 433 0.505 •50 -.289 0.505 — .212 .500 . 500 s^w .196 -.125 . 500 S^li> Semicircle -•083 Triangle .216 i.826 5''w; .571 s^w . 633 s^w §2. To Determine the Center of Gravity of the Remainder of a Simple Body from which one or more simple parts have been taken, we may use the principles of moments in modified form thus: The moment of the re- Art. 21 89 mainder of a body with respect to any plane equals the moment of the whole minus the moments of the parts taken away. (iii) As an example, we determine the center of gravity of a cylinder of cast iron (specific weight 450 pounds per cubic foot) with a conical recess in one end and a cylindrical hole in the other, shown in section y in Fig. 1 59. The weights of the complete solid cylinder, |/"]<- 4.'.'.>|<— - 5 "->\ of the cone, and of the small cylinder, all as of cast iron, are given under W. The coordinates of the center of gravity of the solid cylinder and of the parts are given under x and y (see Art. 24 for information on cone), and the moments with respect to the yz and zx planes are given in the last two columns. The weight of the actual piece of cast iron is 327.5 — (41 + 26.2) = 260.3 pounds; the moments of the piece equal 1637.5 ~ (205.0 + 78,6) = 1353.9 and 2620 - (61.5 + 314.4) = 2244.1 inch pounds respectively. For the piece of cast iron, therefore, x = 1353.9 -^ 260.3 — 5-2, and y = 2244.1 -r- 260.3 = 8.6 inches. Fig. 159 Part. W * y Wx Wy Cylinder 327-5 41.0 26.2 5 5 3 8 i-S 12 1637-5 205.0 78.6 2620.0 Cone 61. 5 314-4 Hole § 3. Experimental Methods may be resorted to for finding the center of gravity of a body so irregular that the foregoing methods cannot be applied, (i) Method of Suspension: The body is suspended from one point of it, and the direction of the suspending cord is then marked in some way on the body; the operation is repeated for another point of suspension. Since the center of gravity is in the two lines or directions so fixed in the body, it is at their intersection. (2) Method of Balancing: The body is balanced on a straight-edge, and the vertical plane containing the edge is marked on the body; the operation is repeated for two more balancing positions of the body. Since the center of gravity is in the three planes so fixed in the body, it is at their common point. Fig. 160 This method is readily applied to a body in the form of a thin plane plate; for such only two balancings are necessary. (3) Method of Weighing: The weight W of the body is determined, and then it is supported on a knife-edge B (Fig. 160) and on a point support which rests upon a platform scale; the reaction W of the point support is weighed, and the horizontal distance a of the point from the knife-edge is measured; ■ ''///////'//////n///, Chap, v 90 then the horizontal distance from the center of gravity to the knife-edge is W'a/W. In this manner the horizontal distances of the center of gravity from several knife-edge supports can be got and the center of gravity located. • The distance of the center of gravity of a body from the plane through three points of the body can be determined if the body can be supported at the points and if certain weighings can be performed as described. Let A, B, and C (behind B and not shown) be three such points of the body (Fig. 161) ; a = dis- tance of A from the line joining B and C; W= weight of the body; W' = weight recorded by the scale when A, B, and C are at the same level as shown I Fig. 161 Fig. 162 in Fig. 161, and W" = weight recorded by the scale when A is higher than B and C by any amount h (Fig. 162). Then the distance y of the center of gravity from the plane ABC is given by Va2 - h^ w' - W" a. ' h W Proof: From the first position it is plain that W'a = Wx; from the second it follows that W"a cos 6 = W{xcosd - ysinO). Solvin g these simultane- ously we get y = {W - W") (a cote)/W; but cot^ = Va- - h^ -^ h, hence, etc. 22. Centroids of Lines, Surfaces, and Solids §1. Lines, surfaces, and (geometric) solids have no weight, and therefore they have no center of gravity in the strict sense of the term as defined in the preceding article. However, we do speak of the center of gravity of those geo- metric conceptions; and we mean by the term, the center of gravity of the line, surface, or volume materialized, that is, conceived as a homogeneous slender wire, thin plate, or body, respectively. The center of gravity of a line, surface, or solid is sometimes spoken of as the center of gravity of the length (of the line), area (of the surface), and volume (of the solid). The term centroid has been proposed as a substitute for center of gravity when applied to lines, sur- faces, and solids as being more appropriate; the new term is given preference in this book. If a given line, surface, or soHd is imagined as materialized, then we can apply the principle of moments (Art. 21) to it. Thus, \i W = the weight of the whole materialized line, surface, or solid, Wi, Wi, W2, etc., = the weights of all the parts into which we imagine it divided, x = the coordinate of the Art. 23 91 center of gravity of the whole with reference to some convenient reference plane, and Xi, xi, Xi, etc. = the coordinates of the centers of gravity of all the parts respectively, then W-x = PTiXi + Wtx^ + Wzxz + • • • . But the weights W , W\, W2, W3, etc., are proportional to the respective lengths (L, Li, L2, Lz, etc.) or areas {A, Ai, A2, A3, etc.) or volumes (F, Vi, V2, V3, etc.), as the case may be ; and therefore it follows from the preceding equations that for lines, Lx = LiXi + L2X2 + L3X3 + • • • , for surfaces, Ax= ^1X1 + 712X2 + ^3X3+ • • • , and for solids, Vx = FiXi + F2X2 + F3X3 + • • • . The foregoing formulas can be rendered conveniently in a single statement of words or proposition by means of a new term which we now define. The moment of a line, surface, or solid with respect to a plane is the product of the length of the line, area of the surface, or volume of the solid and the coordinate of the centroid of the line, surface, or solid with respect to that plane. (The moment of a plane line or surface with respect to a plane perpendicular to the plane of the line or surface is also called its moment with respect to the line of intersection of the two planes.) The proposition or principle of mo- ments, then, is this: The moment of a line, surface, or solid with respect to any plane equals the algebraic sum of the moments of the parts of that line, surface, or solid into which we imagine the whole divided, with respect to that same plane.* The principle of moments can be used to determine the centroids of all geometrical bodies which can be divided up into parts whose magnitudes and centroids are known. Three examples follow: (i) Let it be required to locate the centroid of the line represented (heavily) in Fig. 163, the curved portion being a circular arc; given that each coordinate of the centroid of the arc is 6.366 inches (Art. 24). Let x denote the x coordinate of the centroid of the line; then taking moments about OY (the length of the line = 35.7 inches), 35.7 x — 10 X o + 10 X 5 + 15.71 X 6.366, or X = 4.20 inches. Obviously, the y coordinate also equals 4.20 inches. ^^^' ^^^ (ii) Let it be required to locate the centroid of the shaded area in Fig. 164, which represents the cross section of a "channel" (a form of steel beam much used in construction). We consider the section as divided into a rectangle, 0.40 by 1 5 inches, and two trapezoids. The distance of the centroid of either trapezoid from its longer base is given by 3 (0.90 + 0.80) -^ 3 (0.90 + 0.40) = 1. 3 1 inches (Art. 24). The second column of the adjoining schedule gives the areas of the parts; the third, the centroidal coordinates with respect to the base * Of course these moments have nothing to do with turning effects like the moment of a force with respect to a line or a point. To distinguish these moments, the former are some- times called statical moments, not very appropriately, however. lO'-'-M 92 Chap, v of the section; and the last, the moments with respect to that base. The dis- tance of the centroid of the entire section from the base is 7.70 -^ 9.8 = 0.79 inch. Part. A y Ay Rectangle 6.0 3-8 0.20 1. 71 1 .20 6.50 Two trapezoids 9.8 • 7-70 Fig. 164 k-^-->k /6" >k-,9'--->l Fig. 165 (iii) Let it be required to locate the centroid of a solid consisting of a cone, a cylinder and a hemisphere as represented in Fig. 165; given that the centroid of the cone is 2 inches from its base, and that of the hemisphere is 3 inches from its base (Art. 24). The volumes of the parts, the x coordinates of their centroids, and the moments with respect to the yz plane are as recorded in the adjoining schedule. The total volume is 4825.5 in^ and the algebraic sum of the moments of the parts is —6433 in*; therefore x = — 6433 -^ 4825.5 = — 1.33 inches, the negative sign indicating that the cen- troid of the whole solid is to the left of 0. Part V X M Cone S36.2 3217.0 1072.3 10 — II S.362 -11.795 Cylinder Hemisphere 4825. s -6,433 Yk" 6'-'-->\<— 6" -H Q § 2. To Determine the Centroid of the Remain- ^ DER OF A Simple Figure from which one or more simple parts have been taken, we may use the prin- ciple of moments in the following modified form. The moment of the remainder of a figure with respect to any plane equals the moment of the whole minus the moments of the parts taken away. For example, let it Fig. 166 be required to determine the centroid of the shaded area in Fig. i66, the part of the square remaining after the triangle and the Art. 23 93 quadrant have been taken away; given that the centroid of the triangle is 2 inches from OY and 4 inches from YC, and that the centroid of the quadrant is 2.54 inches from OX and CX (see Art. 24). The areas, centroidal coordi- nates, and moments appear in the adjoining schedule. The area of the shaded portion is 144 — (36 + 28.27) — 79-73 square inches, and the moments of the shaded part with respect to the y and x axes are 864 — (72+266.9) = 525.1 and 864 — (288+ 71.8) = 494.2 cubic inches respectively. Therefore x = 525.1 -^ 79.73 = 6.59, and y = 494-2 -^ 79-73 = 6.20 inches. Part A X y Ax Ay Souare 144 36 28.27 6 2 9-44 6 8 2-54 864 -72 — 266.9 864 Trianele -288 Quadrant -71.8 5251 494.2 23. Centroids Determined by Integration § I. If it is desired to locate the centroid of a line, a surface, or a solid which cannot be divided into a finite number of simple parts whose lengths, areas, or volumes and centroids are known, and if the line, surface, or solid is "mathe- matically regular," then we imagine the line, surface, or solid divided into an infinitely great number of parts, and apply the principle of moments. To find the sum of the moments of all these elementary parts involves an integra- tion. Thus, let L = the length of a line, x = the x coordinate of its centroid, dLi, dLi, dLz, etc. = the lengths of elementary portions of the line, and Xi, x-2, xz, etc, = the x coordinates of the centroids of those portions respectively, then Lx = dLi • xi + dL2 • X2 + dLs • xz -f = / dL • X, in which dL stands for any of the elementary lengths and x for the x coordinate of the centroid of that dL. Similarly, for areas and volumes; and thus we have these formulas: (i) Lx = jdL . x; (2) Ax= j dA - x; (3) Vx= j dV - x, and corresponding ones for y and i (the y and z codrdinates of the centroid). These formulas can be used to determine x, y, and z if the form of the line, surface, or solid is such that the integrations can be performed. In any par- ticular case, limits of integration must be assigned so that all elementary por- tions are included in the integration (summation).* Six examples illustrating their use follow: * The centroid is a mean point. The ordinate from any plane to the centroid of a line, surface, or solid equals the mean of the ordinates of all the equal elementary portions of the line, surface,, or solid, it being understood that the mean takes into account signs of the ordi- nates. For, let xi, x^, Xi, etc., be the ordinates of the elementary portions and n the number 94 ^^^^- '' (i) Circular arc; radius = r and central angle = 2a (Fig. 167). The radius which bisects the central angle is a line of symmetry, therefore the centroid is on that line; if that line is taken as x axis, then y = o. The length of the arc = 2 ra (a expressed in radians), dL = rdcj), andx = r cos (p; therefore formula (i) becomes 2 rax r dd(t) = 2 r"^ sin a; or x = (r sin a) -i- a. a fJ —a (ii) The preceding problem will now be solved without using polar coordi- nates. Since x^ + / = r\ xdx + ydy = o, or dy =- (x dx)/y. Hence and dL = Vdx^ + ^y2 = gxVi-^ xV/ = dx r/y = dx rj^/r^ - x\ wV Cl'Jv 2 rax = I xdL = 2 r / «y *^ r cos a V r^ — x'^ = 2 r^sina; etc. (iii) The parabolic segment AOBA (Fig. 168); altitude = a and base = b. Evidently the axis of the parabola is a line of symmetry, and therefore it contains the centroid. If that line be taken as the x axis, then y = o. Let x and y be the coordinates of any point P on the parabola ; then the area of the elementary portion shaded is 2 y dx. Since the area of the segment is f ab, and the equation of the parabola is 4 ay^ = b'^x, formula (2) becomes -abx = / 2y dx' X = —p. \ xi dx — - ba?; Jo -Va^o 5 and 2 5 — ^ ^ 'I X = -ba^ -r- -ab = -a. 2 3 of them (infinite); then the mean ordinate is {xi -\- X2 ~\- x^ -{• • • •) -i- n; also, let Q = the length, area, or volimie of the line, surface, or solid, and dQ = the length, area, or volume of the equal elementary portions; then the mean ordinate equals fa + 3:2+ • ndQ •)dQ JdQ-x = X. Art. 23 95 (iv) Circular sector (Fig. 169); radius = r and central angle = 2 a. The radius which bisects the central angle is evidently a line of symmetry, and so the centroid is on that line. If that line is taken as x axis, then y = o. The area of the sector equals r^cc, a. expressed in radians; dA = pd(f)'dp, where p = OP and P is any point in the sector. Therefore formula (2) becomes and r^ax X I pd(j)dp- X = j j pd(j)dp' p cos <^; J— a Jo J— a = ( - /^ sin a j -^ ( r^a j = 2 r sin a 3« (v) Conical or pyramidal solid; altitude = a (Fig. 170). We take the origin of coordinates at the apex, and the x axis perpendicular to the base; OMNO represents the projection of the cone or pyramid on the XY plane. We imagine the sohd divided into plates or laminas parallel to the base; if the area of the base is called ^, say, then the area of the lamina represented is Ax'^/a', and the volume of the lamina is dx • Ax^/a^. And since the volume of the solid is ^ ^a, formula (3) becomes T r" f*" Aa^ -Aax = I {dx' Ax^/a^) x = A/a^ / x^dx = ; 3 Jo Jo 4 Fig. 169 Fig. 171 hence, x = j a, that is, the perpendicular distance from the centroid to the base equals one-fourth the altitude. Evidently, the centroid of every lamina lies on the line joining the apex and the centroid of the base; therefore the centroid of all the laminas (that is, the solid) lies on that line. (vi) Octant of a sphere; radius = r (Fig. 171). Obviously x = y = z; ^ is given by I {dx dy dz)x. Jo Evaluating the integral and substituting for V its value, \ irr^, we find that x= ^r. § 2. Surfaces and Solids of Revolution. — For surfaces, we use formula (2) and select as element the surface described by an elementary part of the 96 Chap, v generating curve MN (Fig, 172). Let the x axis be taken coincident with the axis of revolution; then the area described by a part of the generating curve of length ds is 2 iry ds. The centroid of this area is in the x axis, and its x coor- dinate is the X in the figure; hence if A stands for the area of the surface of revolution, I TT I yds- X, or :*; = —r- j xy ds. Fig. 172 Fig. 173 The limits of integration must be assigned so that each product xy ds will be in- cluded in the integration. For solids, we use formula (3), and take as element that volume generated by an elementary part of the generating plane MPN (Fig. 173) which is included between two lines perpendicular to the axis of revolution. Thus, if the x axis is taken coincident with the axis of revolution, then PQqp generates the elementary volume, or dV = ir {y^} — y-^) dx. Now the centroid of this elementary volume is in the x axis, and its X coordinate is the x in the figure; hence if V denotes the volume of the solid of revolution, then Vx = ir I (yi^ — yi^) dx'X, or x=yj {yi^ — yi^)x dx. The limits of integration are to be assigned so that each product (yi^ — yi^)x dx will be included in the integration.* * Theorems of Pappus and Guldinus. — These relate primarily to the determination of the area and volume of a solid of revolution; they involve the centroid of the generating curve or plane, and are therefore mentioned in this place, (i) The first theorem states that the area of a surface of revolution generated by a plane curve revolved about a line in its plane equals the product of the length of the curve and the circumference of the circle described by the centroid of the curve. Proof: Let MN (Fig. 172) be the generating curve, L = length of the curve, y = the ordinate of the centroid of L from the axis of revolution, and A = area of the surface generated. Then A = \ 2-KydL and yL = \ y dL. Combining these equations we get yl = L 2 ivy, which is the proposition in mathematical form. (2) The second theorem states that the volume of a solid of revolution generated by a plane figure revolved about an axis in the plane equals the product of the area of the figure and the circumference of the circle described by its centroid. Proof: Let MPN (Fig. 173) be the generating plane, a = area of the plane, y = the ordinate of the centroid of a from the axis of revolution, and V = volume of the solid generated. Then V = ( Tr{y2^ — y) dx, and from eq. (2), a'y = j iji — yO dx h (^2 + yi). Combining these equations we get F = o 2 Try, which is the proposition in mathematical form. To illustrate, we determine the area of the surface generated by revolving the circular arc ABC (Fig. 175) about AC, and the volume of the solid generated by revolving the figure Art. 23 97 For an illustration imagine the quadrant XY (Fig. 174) rotated about OX so as to generate a hemisphere. The positions of the centroids of the surface and solid generated could be computed as follows: (i) The area of the hemi- sphere is 2 irr^, X = rsixKl), y = r cos , and ds = r d(t>; hence for the area IT ^ = (2 X -^ 2 TT/-^) I xyds = r J sin ^ cos (f>d, and dx = r cos (t>d(f); hence for the volume X = (tt -4- f^rr^ ) / (js^ — o) ;» (/x = (3 r/2) i cos^ (f) s\n 3' ~ f ^> 3-^^ 2 = f c. Paraboloid of Revolution, formed by revolving a parabola about its axis. Let h = height of the paraboloid, the distance from its apex to the base, then the distance from the centroid of the solid to the base is | /f. CHAPTER VI SUSPENDED CABLES (WIRE, CHAIN, ETC.) 25. Parabolic Cable §1. Symmetrical Case. — When a cable is suspended from two points and it sustains loads uniformly spaced along the horizontal and spaced so closely that the loading is practically continuous, then the curve assumed by the cable is a parabolic arc as will now be shown. The symmetrical case (points of suspension at same level) will be considered first. Let AOB (Fig. 194) be Y ,H — X ' ( D > 'WX X Fig. igs the cable suspended from A and B, w = load per unit (horizontal) length, a = span AB,f = sag, H = tension in cable at lowest point, and T = tension at any other point Q (coordinates x and y). The forces acting on the portion OQ are H,T, and the distributed load wx (Fig. 195); this load acts at mid- length of X. Since the forces are in equilibrium, their moment-sum equals zero for any origin of moments; hence moments about Q give Hy = wx{x/2), , 2H X- = ■ — ■ y, w or y •w (i) This is the standard form of the equation of a parabola; the axis of the parab- ola coincides with the y axis, and the vertex is at O. If we substitute for X and y their values for the point A {x = a/2, and y = f), then we get a2/4 = (2 H/w) f,ox H = wa^/Sf; hence equation (i) may be written x^ a^ 4/ }': or 4/, a2 • (2) A formula for the tension T at any point Q may be arrived at as follows: Let (}> = slope of the curve at Q; then it is plain from Fig. 195 that r sin (^ = wx, and T cos <^ = H. Squaring and adding gives 102 Art. 25 ^°3 At the points of suspension, x = a/ 2, and the value of T at that point is £? ( I + 16 ^;V = * wa (i + -^J- (4) The adjoining table gives values of T/wa for various values of f/a, the sag ratio (denoted by n in the table). n = T/wa = I.O 0-5 0.25 0. 125 0. 1 0.05 0.515 0.559 0.707 I. ir8 1.346 2.550 O.OI 12. 81 The length of cable for any span a and sag /or sag ratio n = f/a. — Let / = length of cable AB and ds = length of an elementary portion; then as in all plane curves, ds^ = dx^ + dy^ = [i + (dy/dx)-] dx^, or ds = [1 + (dy/dxY]^ dx. From the equation of the curve (2), dy/dx = (Sf/a^) x = S nx/a; hence ds = (i + 64nH^/a'^)^ dx. Integrating between proper limits (o and | / for s, and o and | a for x) and then doubling, we get I = a(i(i + i6n^)'+^\ogeUn-\-{i + 16 n^)'])- (5) An approximate formula for I, much more convenient to use than the fore- going one, m^ay be deduced as follows: Expanding the coeflScient of dx above by the binomial theorem, we get (/5 = fi +32^2— 2— 5i2w^-4+ • • • jdx; I ^< i-\--n^-^n^ + )■ (6) and integrating between limits as before we find that 8 „ 3^ 3 5 The following table gives values of l/a by the exact and approximate formulas, equations (5) and (6) respectively, for several sag ratios n = f/a. 1. 01 I .0003 1. 0003 §2. Unsymmetrical Case. — By this is meant a cable suspended from two points not at the same level; see Fig. 196, where ACB represents a cable suspended from A and B. In this case, also, the cable hangs in the arc of a parabola as will be proved presently. Let a = horizontal distance between points of supports (as in §1), b = vertical distance between the points, 6 = angle which AB makes with horizontal ( = tan~^ b/a), x and y = coordi- nates of any point Q of the cable as shown, T = tension at the highest n = 1.0 0.5 0.25 0.125 0. 1 0.05 l/a exact 2.3234 1.4789 1.1478 I . 0402 1.0260 I . 0066 l/a approximate . 1.1417 I. 0401 1.0260 I . 0066 I04 Chap, vi point, V and H' respectively = the two components of T along AY and AB. There are three forces acting on the part AQ, - its load wx, the tension T, and the tension at Q. The moment-sum for these three forces for any origin equals zero; with Q as origin - wx • x/2 + V'x - H' (QP) = o, or V'x - wx^/2 = H' (y cos ^ - x sin 6). (i) This is the equation of a parabola with the axis parallel to the y coordinate axis. To express the equation of the curve in terms of the dimensions a, b, and the vertical sag /i under the middle point of the chord AB: — The forces acting on the entire cable consist of the load wa, the tension at A, and that at B. Their moment-sum with origin at B is wa-a/2 — V'a = o; hence V^ = •wa/2. (2) The forces on the upper half AC consist of the load wa/2, the tension at A, and that at C. Their moment-sum with origin at C is 72 wa'' '^'•£_rf + H'/,cos9 = o; hence ff' = g^— -,- Substituting these values of V and H' in (i) gives ^^ {a — x) = y — re tan d, a^ ,-2 or >• = (4/1 + ^)- -4/1-2 (3) (4) The vertical distance of any point as Q below the chord AB \s y Xxtand; hence if we let y' denote that distance, the foregoing equation can be put into the more convenient form 4/1^ y = a' (a — x). (5) Art. 25 105 The value of the slope at any point of the curve is (differentiating equa- tion 4) dy _ /i 8/1 JC _ =4-- -^ ^-. ax a a a a Let a and /3 = the slope angles at .4 and B respectively (where x = o, and X = a); then tan a = {b-\- 4/i)/a, and tan 6 = (b — 4/i)/a. (6) Let Xo and jo = the coordinates of the vertex of the parabola (where dy/dx = o); then Xo = a{b-\- 4/i)/8/i, and yo = (b + 4/i)-/i6/i. (7) Let H and V respectively = the horizontal and vertical components of T. Then (see Fig. 196) H = H'cosO = wa'^/Sfi, and V = V -\- H' sine = ^ wa {i -\- a tan 0/4/1); and since T^ = H"^ -\- V^, we find that sin 11 wr 2 wi J where «i = sag ratio fi-i- AB = /i -^ asec0. The last expression shows that for given w, a, and Wi, the tension T increases as the angle d is made larger; also that for given w, a, and 6, T increases as Wi is made smaller.* Length of the Parabolic Arc AB (Fig. 196). — Let ai = the length of the chord AB, Wi = sag ratio /i -r- ci, and h = length of the arc AB. Also let ds = length of an elementary portion of the arc; then (/s = [i + {dy/dx)^f dx. From the equation of the curve (4), we can get This last value of dy/dx substituted in the foregoing expression for ds gives ds- ^ ^-wi'-'h'^'ilh'A^n } i + 8wi(i — 2-J2 «i/ 1 — 2 - j + sin > sec 6 dx. Now this equation is in the form ds = {i-\- X)^secd dx, where X = S til (1 — 2 x/a) [2 «i (i — 2 x/a) + sin 6]. * Let MN (Fig. 197) represent the load on the cable ^B,and let MO and NO be parallel to the tangents at A and B (Fig. 196) re- spectively; then OMNO is a force triangle for the three forces act- ing on the cable AB, and OM represents T and NO represents S. It is plain from the figure that OR tan a — OR tan = MN, or OR (tan a - tan /S) = MN. But OR = T cos a = S cos /3, and MN = wa; hence T cos a (tan a — tan 0) = wa = S cos /3 (tan a — tan 18). Substituting the values of tan a. and tan/3 given by equation (6), we find that r cos a = wa^l&Ji. = 5 cos 0. FiG. 197 io6 Chap, vi Unless the sag is relatively large ds and sec d dx are nearly equal at all points along the curve (see Fig. 196); hence (i + X) is nearly equal to i at all points, which means that X is small compared to i. Therefore, we may expand (i _j_ X) by the binomial theorem, and drop all terms except the first few without serious error. Thus we have as a close approximation ds = {i -\- iX-^X^)secddx, and '■ = / (i + |Z- lX^)secddx. Substituting for X and X^ their values, and integrating we finally get /i = fli(i + §cos2 0.«i2- V-«i'). (10) If the approximation made in the derivation of formula (10) is not per- missible in a given case, then one might determine the exact length of the cable AB somewhat as follows when a, b, and/i are given: We first locate the vertex O of the parabola of which the cable is a part from equation (7). The vertex will be found either between A and B, on the cable (Fig. 198), or A'v B'"> Fig. 198 Fig. 199 —Tfi^^ beyond B (Fig. 199). Then we determine the length of the arcs AOA' and BOB' by means of equation (5), §1, and finally the length h of the arcAB from Zi = i AOA' + h BOB' for Fig. 198, or h = ^ AOA' - \ BOB' for Fig. 199. For example take a = 800 feet, h = 300 feet, and /i = 200 feet. Let Xq and jQ = the coordinates of the vertex. From equation (7) (soo + 4 X 200) 800 , (300 + 4X200)2 Xf, = -^^^ = tji^o, and Vo = — 7-7^ = i1°-S- ° 8X200 ^^ 16X200 Hence the cable hangs as shown in Fig. 200. The length AA' = 1348.6 feet according to (5), (a = iioo and n = 378.5 -4- iioo); the length BB' = 530.9 ■^...250'->k- 5S0' — ->< Fig. 200 feet according to (5), (a = 500 and n = 78.5 -f- 500). i X 1348.6 + i X 530.9 = 939.8 feet. Hence AB = Art. 26 107 26. Catenary Cable. §1. Symmetrical Case. — A chain or flexible cable suspended from two points and hanging freely under its own weight or a load uniformly distributed along its length assumes a curve called (common) catenary. Let A and B (Fig. 201) be the points of suspension of such a cable, C its lowest point, Q any other point of the cable, 5 = the length CQ, H = ten- sion at C, T = tension at (), = slope of the curve a.t Q, w = weight of load per unit length of cable, and c = a. length so that cw = H or c = H/w. The forces acting on CQ are H, T, and ws. Since they are in equilibrium, T cos cj) — H, and T sin (f) = ws; hence tan — ws/H = s/c. But tan <^ = dy/dx, therefore Fig. 201 dy/dx = s/c. (i) Now since ds^ = dx"^ + dy^, {ds/dyY = (dx/dyY + i and {ds/dxY = i + {dy/dxY; also ds^ _ c^ _c^ -\- s"^ dyl S' s'^ and ds dx c^-\-s^ (2) Integrating the first one of these equations we get y = (c^ -\- 5^)2 -{- A where ^ is a constant of integration. But y = c where s = o, therefore A = o, and hence >»- = c^ + S-, or s- = y2 _ ^2, (3) or s- = y Integrating the second differential equation we get s X cloge V /■ + 1 = c sinh~^ - the constant of integration being zero (x = o when s = o). From (3) X s = Ic {e'l'^ — e ^1'^) — c sinh - (4) (5) To obtain the cartesian equation of the catenary we combine (3) and (4) or (3) and (5) so as to eliminate 5. Thus squaring (5) and comparing with (3) we get easily (6) X y = \c ie^''^ -\- e ^^'^) = ccosh-. or . = clog,[^±\/(^)'-i] = .cosh-f. (7) The slope angle 4> at any point in terms of the coordinates of the point {x,y,s) is given by ^an<^ = s/c = \ if'" - e-""^') = sinh {x/c). (8) xo8 Chap, vi See equations (i) and (5). And from equations (2) and (3), we get sin0 = s/y and cos0 = c/y. (9) It follows from the equilibrium equation T sin <^ = ws and (9) , that T — wy, (10) that is, the tension at any point Q equals the weight of a length of cable reach- ing from Q to the directrix OX. Hence T increases from C to A. According to the definition of c H = wc. (11) In passing, it may be noted that since T cos ^ = H, the horizontal com- ponent of the tension at any point Q = wc, constant ^ for a given suspended cable. As in the preceding article, let a = span AB (Fig. 202), / = sag, and / = length of cable ACB. Any two of the three dimensions a, I, and / determine the catenary, as will be shown presently. For the point A, X = I a, y = f -{- c, and s = ^ I. Hence substi- tuting in equations (3), (4), and (6) respectively we get Fig. 202 (/ + ,)2 = ,2 + i;2^ or clj=\{lljy-\. (3') \a = c sinh-i (1 //^)^ or 1 ^Ic - sinh-i {\ l/c). (4') and f-\-c = c cosh (^ a/c), or i + (//c) = cosh (| a/c). (6') When I and / are given (3') gives c, and then a may be gotten from (4') or (6'). When a and / are given (6') determines c but the equation cannot be solved directly, — only by trial or by some sunilar method; having thus determined c, I may be gotten from (3') or (4'). When a and / are given, (4') determines c (solution by trial), and then / may be gotten from (3') or (6'). Inasmuch as these trial methods are generally long, computations on some catenary problems may be facilitated by means of diagrams. In Fig. 203 the curves marked A give the relation between J/ a and l/a for values of f/a from o to 0.5 and (corresponding) values of l/a from i to about 1.50. For example, let a = 800 feet and / = 160 feet. Then f/a = 0.20, and the corresponding ordinate (over f/a = 0.20) to curve A reads i.io; hence l/a = I.IO, and / = 800 X i.io = 8S0 feet (length of cable). Most practical catenary problems involve the strength of the wire or cable and the load per unit length of wire. For such problems we have, in ad- dition to (3'), (4') and (6'), T = w(f+c), or T/w=f+c, (11') where T = the greatest tension (at the points of support), which should of Art. 26 109 course not exceed the strength of the wire. Most of these problems can be solved by trial only, unless a diagram is available. For example, given the strength T of a wire, the load per unit length w, and the span a; required the proper length of wire /. Here T/wa = f/a-\- c/a. - (11") This equation and (6') contain only two unknown quantities / and c, and the two equations determine / and c. But they can be solved only by trial. After/ and c have been ascertained, then I may be computed from (3') directly. The curves marked B in Fig. 203 show the relation between f/a and T/wa. I.-30 MS 1.20 O •1- wo I.1>0 1.00 ~ A . 1 J. _i \ — 1 _ - 1 -/ K ■ 1 1 ^ A ■i . .95 1.0 5. 1 \ ■ 9 5 l:^b / ^l.it> "~ ~~ ~ s \ / \ ^ ~~ 1 T / f ^, — " t- 1 \ . r ^ "" 1 \ ^ A I.IO. 1 . w l.-dU / \ V'^'^ - ~~ " L / s i A i( ^ _^ _ _ •!• I ^ /- _ _ V _ _ \ y ^ _ < _ t-85 uf; i \ .Hh i.ih f \ Tr^S 1 ^ ^ ■* "" ~ 1 \ y \ ' \ / iTl " ^ / -X I.IO-^ / N yHO ^. ~ f / \ ^ > 1 4 / N \ .^ , ~ " 1 / / s ^ "-p 1 T" ~f ^ _ _ , "v ^ _ _ — ^ ^ _ _ — — 75 - - - - Ic ^. r ^ — — — — — — — — — — — — — _J — — — — — — y ^ t _ ^ -7n ±. „ '1 1 _ _ _ _ , __ ^ _ ^ u _ _ _ _ 1.15 1.10 O > .05 1.00 .05 .10 .15 .20 .25 .30 .35 .40 .45 Values of -f -^ a Fig. 203 Thus ii f/a = 0.20, as in the preceding illustration, then the corresponding ordinate to curve B (over //a = 0.20) reads 0.85; hence T/wa = 0.85 and T = 0.85 wa*. * Fig. 203 was prepared from plate II of Mr. Thomas' paper mentioned in the footnote at the end of this chapter. (For cases of relatively small sag ratios, see that plate.) The curves marked A might be prepared as follows: (i) Assume different values of ///; (ii) com- pute the corresponding values of c/f by means of (3'); (iii) compute values of l/c IromJ/c = (///) -i- (c//); (iv) compute values of a/c from (4'); (v) compute values of l/a from' l/a = Q/c) -^ {a/c); (vi) compute values of f/a from J/ a = (l/a) -r- (///). Finally plot values of f/a and l/a. (The adjoining schedule gives the computed values for /// = 5 as an illustration.) 1 n HI IV v VI VU vni l/f c/f l/c a/c l/a f/a c/a T/wa 5 2.6250 1.9048 1.6946 1-1237 0.2247 0.5900 0.8147 To plot curves B we would continue the computation, first computing values of c/a from the values of a/c already obtained, and then values of T/ic-a from T/im = f/a + c/a. I'inally values in columns vi and viii of the schedule when plotted furnish the desired graph. no Chap, vi § 2. Unsymmetrical Case (points of suspension not on same level). — The cable uniformly loaded along its length hangs in an arc of a catenary. The vertex C may be on the cable (between the points of suspension A and B) as in Fig. 204, or beyond the lower point of suspension as in Fig. 205. In either r \ \ '^' o -(l-k)a ■ .-..■rHik-Oa ka Fig. 204 Fig. 205 figure, a = the horizontal distance between A and B,b = the vertical distance, 6 = angle between AB and the horizontal, / = sag or vertical distance AC, and I = arc AB. Most problems in this case as in the symmetrical case can be solved only by a trial method; hence diagrams are practically neces- sary in this case also. 3.00 "2.50 I 2.00 1.50 :2 1.00 50 r j;; -t^ r Tr " "IJIT rZ" ~ J 2" t « i i^ 4^ -iT ^ I V 1 ' 1 ^ V^^ r ^ , ^ r J b^a\0 X05 \jg .15 .20 i.25 ' UO ' *35 /40 45/ jptr ^ 1 ^ ^ 2> X ' \ t t ' 7 ^y t * J 4 t ' t W- 5 ^ it J - J -t ^^' ^^ u C ^ ^- - I t ^^' - ^ K ^ \ ^ \ L _ t t yi^ y \ ^ ^ ^ \ > tT ^^t ^% \ t S V V + V t X>^ ^ y ^ \ ^ \ ^ \ \ \ L_ ^^,' %^'^^ \ A ^ ^ \ N- \ \ ^-^ V%^5%'^^ ^ \: ^^ ^ 5 ^ 3^ V"^^^-^,-^ ^^^•^ X ^^ ^^ ^ S ^ S ^%^ ^^\^^Z2^Z,^^ \ 5 "^^ 5 \ ^^ \^^^^^ y^^'l.^'XL !s ^^^s ^^ ^«. ^ ^^;;J^^-%^i-^^'^-^'^ ===========""'^^:-^:-=:p^^^-^^^^^i:^^i^^ ~p:0^'^''^'.!>^O*:iM^^'^.'^1»- .J--' ■V'' i" bi^ S* "''''' '^''^''■' ^''' ^"^ J^ ^ ^O^"" ^.J^ _,^j' ^''_^.' ' |j' ^1 i- .^•a^ - ,,^ '^ - t' -• "] !•' ' '" — \' -- = -s^f=g^-^-§^^^^-r--r ^ -- = " 1.60 .50 1.40 1.30 1.20 > — - 1. 10 .05 .10 .15 .20 .25 .30 Values of f -i- a Fig. 2o6 .35 .40 .45 .50 1.00 In Fig. 2o6 there are two groups of curves relating to this unsymmetrical case; the group occupying the lower right-hand portion consists of graphs showing the relation between //a and Z/o (values at right-hand margin) for ten values of hi a (slope of AB, Figs. 204 and 205). The other group consists Art. 26 III of graphs showing the relation between //a and Tjwa (values at left-hand margin) for the same ten slopes (r = tension at higher point of support and w = weight of cable per unit length). To illustrate, let a = 200 feet, 6 = 40 feet, / = 240 feet, and w = 2 pounds per foot. On the curve for b/a = 0.20 in the lower group, we find the point whose ordinate l/a = 1.20 and note that the abscissa of that point is f/a = 0.385. Hence / = 200 X 0.385 = 77 feet. On the curve (or b/a = 0.20 of the upper group, we find the point whose abscissa is 0.385 and note that its ordinate T/wa = 0.90. Hence T" = 2 X 200 X 0.90 = 360 pounds.* * Fig. 206 was made from certain of the (more extensive) figures in Mr. Robertson's paper mentioned in the footnote at the end of this chapter. The following is an explanation of a method for the construction of such a figure. Let h = arc AC and h = arc BC (Figs 204. and 205); also let yi and y-i = ordinates of A and B respectively, and ka and (t — k)a — the abscissas of A and B. Then for .4, x = ka, y = yi and s = h; for B, x — (k — i)a, y = ^2, and s = h. Hence, substituting in equations (3) and (6) of §r, we get (^M?r and 2!i c = cosh k - ' c and — = cosh (k — i) - ■ c c (3') (6') At the higher point of support A (Fig. 204 or (10) §1, the tension at that point = wyi, and 205), y = yi', hence according to equation tca a \c j \c j This equation, (3') and (6'), constitute the basis of the method. We first assume a value of k, say 0.6, and different values of a/c (say 0.02, 0.04, etc.); then (i) compute values of yjc and y2/c from (6') corresponding to those values of a/c (and k = 0.6); (ii) compute the values of h/c and h/c from (4') to correspond; (iii) compute values of l/c from l/c = (/i/c) + (k/c); (iv) compute values of f/c iromf/c = {yjc) - i. Finally compute- b/a from b/a = [(yi/c) - (^2^)] -^ {a/c) (see Figs. 204 and 205); l/a from l/a = {l/c) ^{a/c)\ f/a ;from f/a = {f/c) -^ {a/c); and T/wa from T/wa = {yi/c) ^ {a/c). (See schedule) 1 a/c u yi/c Ul yi/c IV h/c V h/c VI l/c vn f/c VUl b/a IX l/a X f/a XI T/wa 0.6 + -D 0.4 % S 0.3 ^^1 0. ^' / ^y 0^ y / )^ / r / y /) / f 1.1 1.3 1.5 Values of 1-s-a Fig. 207 J V y -^ / ^. y A ^ y ^ y O.l 0.3 0.5 Values of f-5-a Fig. 20S —r— / / f / \ ( \ \. H r^ V ^ M '^ "■■■ ■*>_■ 1.0 1.2 1.4 Values of T*wa Fig. 209 0.6 0.5 0.4 0.3 0.2 0.1 We now plot values of b/a and l/a as in Fig. 207, and get a curve (marked k = 0.6); plot values of h/a and f/a as in Fig. 208, and get a curve (marked k = 0.6), plot values of b/a and T/wa as in Fig. 209, and get a curve (marked k = 0.6). In a similar manner we take 112 Chap, vi §3. Approximate Solutions of Catenary Problems. — If the cable is suspended from two points at the same level and the sag is small compared to the span so that the slope of the catenary is small at every point, then the load (weight) per unit length of span is nearly constant and equal to the weight of the cable per unit length. Hence the catenary coincides very nearly with a parabola of the given span and sag, and the formulas and re- sults of the preceding article §1 (symmetrical case) may be applied to the case here under consideration without serious error. That the catenary agrees closely with a parabola can be shown otherwise as follows: Expanding the exponentials in the equation of the catenary, (6) §1, we get pX/c — + — + 20^ 3 c^ and X I ^ , x^ oc^ , C 2C^ 3 c"* hence the equation of the catenary may be written 3' = 7 2+^ + X Neglecting the higher powers of the small quantity -, we have as close approximations y = c -\- x'^/2 c, or x"^ = 2 c (y — c). These are equations of a parabola whose axis coincides with the y coordinate axis and vertex c distant above the origin of coordinates. If the supports A and B are not at the same level (Fig. ig6) and the sag / of the cable is small compared to the distance between the points of support, then the slope of the catenary is nearly constant and the load per unit length of horizontal distance is nearly constant {w sec 6, where iv = weight of cable per unit length, and = angle BAX). Hence the catenary coincides very k = 0.7 say, and make computations i, ii, iii, etc., as described; then plot three more curves (Figs. 207, 208 and 209). Then we repeat for still other values of k. From the three sets of curves (Figs. 207, 208, and 209) we pick out sets of corresponding values of l/a, f/a and T/wa for the several values of b/a. Thus for b/a = 0.2, we find the adjoining tabulated values from the curves. b/a ■- = 0.20 k Ua f/a T/wa 0.6 0.7 etc. I IS 1.06 etc. 0.35 0.25 etc. 0.91 1.30 etc. Then we plot these values of //a and l/a and get the curve marked b/a = 0.20 in lower group, Fig. 206; also we plot values oi f/a and T/wa and get the curve marked b/a = 0.20 in the upper group of Fig. 206. In a similar way we tabulate and plot for other values of b/a (0.30, 0.40, etc.) and complete the diagram (Fig. 206). Art. 27 113 nearly with a parabolic arc of the given (oblique) chord AB and sag /i, and the formulas of the preceding article §2 (unsymmetrical case) may be ap- plied to the cable under consideration without serious error, it being under- stood that w of §2 = (weight of cable per unit length) X sec 6. 27. Cables with Concentrated Loads §1. Weight of Cable Negligible. — Let Fig. 210 represent a cable ACB suspended from two given pomts A and B, C being a given point from which a load is suspended. If the cable can be " laid out" in a drawing, the ten- sions m ylC and BC can be determined easily by constructing the force tri- angle PQRF for the load W and the two tensions. PQ = W according to some convenient scale; PR and QR (parallel to ^C and BC respectively) represent the tensions in ^C and BC. Or, if one wishes to avoid graphical methods the two tensions {Ti and T2) may be computed by solving the tri- angle algebraically. Such solution would give Ti = W cos i3/sin (a + |3) and T2 = W cos a/sin (a + /3), where a and /3 are the angles which AC and BC make with the horizontal (Fig. 210). Fig. 210 Fig. 211 When several bodies are suspended from given points on the cable, the cable takes up a definite position, but it is not easy to determine the slopes of the segments of the cable and the tensions. The difficulty lies in the alge- braic computation. For example, consider the case represented in Fig. 211. The given data are shown in the figure; the lengths are drawn to scale, but the inclinations of the segments of the cable may not be correct, being un- known as yet. Let the inclinations be called a, 13, and 7 as shown; and Ti, T2, and 7^3 = the tensions in OA, AB, and BN respectively. At each point of suspension of a load {A or B) there are three forces acting; at A, the load 1000 pounds, Ti, and T^, and at B, the load 2000 pounds, T2, and ^3. Con- sideration of forces at A and of those at B gives respectively Ti cos a^ To cos i3 and Ti sin a - Tz sin /3 = 1000 Ti cos /3 = Ti cos 7 and T2 sin /3 -h 73 sin 7 = 2000. 114 Chap, vi It is plain from the geometry of the figure that 8 cos a + lo cos /3 + 1 2 cos 7 = 20, and 8 sin a + 10 sin j8 — 12 sin 7 = 4. These six equations may be solved simultaneously for the six unknowns (Ti, T2, Ts, a, jS, and 7); the actual solution is not simple. For similar cases with more than two loads, the work of solving the equations increases rapidly with increasing number of loads. Suspended loads can be chosen so as to hold points of suspension {A, B, etc.) in certain definite positions. For instance let it be required to de- termine Wi, W2, etc., to hold a cable in the position shown in Fig. 212. We Fig. 212 may assume any value for one of the weights and then compute the values of the others. Thus taking Wi = 1000 pounds say, then we compute the tension in AB from a force triangle for the three forces acting at A. PQXP is such a triangle, where PQ = 1000 pounds (according to any convenient scale) and PX and QX are parallel to OA and AB respectively; then XQ represents the tension in AB. The next step is to find the value of W2 which corresponds to such tension in AB; so we draw a force triangle for the three forces acting at B one of which is the determined tension in AB. This force triangle is QXRQ, and so RQ represents W2 and XR represents the tension in BC. Finally, we draw the force triangle RXSR for the three forces acting at C, one of which is the determined tension in BC, and thus find that W3 is represented by SR. Obviously any three weights Wi, W2, and TVs in the proportion of PQ, QR, and RS would hold the cable in the specified position. §2. Weight of C.A3le Not Neg- ligible. — It is assumed in the fol- lowing discussion that the cable segments are quite flat so that they are practically parabolic arcs (see preceding article §3). Then the weight of any segment of the cable is practically the same as the weight of a length equal to the chord of the segment. Let ABC (Fig. 213) be a cable supported at A and C, a load being suspended from the cable at its middle point B. Given the span AC = 2 a, the length of the cable =2!, the weight of the cable per unit Art. 27 115 length = w, and the load = W; required the sag (depth of B below AC) and the tension at A. This (apparently) simple problem is determinate but prac- tically unsolvable on account of algebraic difficulties. The equations are easily set up. Thus let ai= the (unknown) length of chord AB,fi = the sag of the cable below the chord as in Fig. 196, S = the tension and ^ = the slope of the cable at B (Fig. 196). Then according to equations (10) and (6) re- spectively of Art. 25, § 2 According to the footnote, on page 105. 5cosi3 = (7£'ai/a)aV8/i- (3) From the three forces acting at B (IF, S, and S), it is plain that 2Ssin^ = W. (4) These four equations determine the unknowns appearing in them, d, /i, S, and /3. Thus by division, the last two give tan jS = 4 Wfi/waia; equating the two values of tan /S and transforming, we get »iZ/.=v/;rM'_,/.. (5) ai wa ai ▼ \ai/ ai This equation and (i) contain only two unknowns, the ratios (a/ai) and (/"i/ai), and the equations determine the ratios. Supposing the ratios determined we may find ai since a is given, and then /i. Exact simultaneous solution of equations (i) and (5) is impossible, but each equation may be graphed and then the coordinates of their intersection would be the desired values of c/ci and/i/ai. The converse of the preceding problem is much simpler. It is this: Given the span AC = 2 a, the chord AB = ax, the sag/i, and the weight of the cable per unit length w; required the load W. Equations (2), (3), and (4) give in succession (3, S, and W. Equation (i) gives the length /.* * For other information on the subjects of this chapter, particularly as related to elec- tric transmission lines, see the following: University of Illinois Bulletin, No. 11 (191 2) by A. Gruell; Transactions American Institute of Electric Engineers, Vol. 30 (191 1), papers by Wm. L. Robertson, Percy H. Thomas, and Harold Pender and H. F. Thompson. These papers contain extensive tables and diagrams, and discuss effects of temperature changes. DYNAMICS DYNAMICS CHAPTER VII RECTILINEAR MOTION 28. Velocity and Acceleration § I. Velocity. — When a point moves so that it traverses or describes equal distances in all equal intervals of time then it is said to move uniformly, and we call the motion uniform. All other motions we call nonuniform. By velocity of a moving point is meant the time-rate at which the point is moving, or describing distance. To express the magnitude of any velocity we must of course compare that velocity to some particular velocity as a standard or unit. Any velocity — that of light, for example — might be taken as standard; but it is more convenient to take the velocity of a point moving uniformly and describing a unit of length in a unit of time for a standard. Thus, we use the foot per second, mile per hour, etc. The word per in these names of velocity- units is quite commonly replaced by the soHdus sign /; thus, foot per second, mile per hour, etc., are abbreviated to ft/sec, mi/hr, etc.* In any uniform motion the velocity may be computed by dividing the dis- tance traversed in any interval of time by the interval. Thus, if i; = the velocity, A5 = the distance traversed, and A/ = the interval of time, then V = As /At. (i) In any nonuniform motion the rate of moving is not constant but changes continuously, as we all realize. Not all, however, have a clear notion of the value of the rate, or velocity, at a particular instant of the motion. To bring this matter up definitely, let us consider the following example: — In a certain launching, the ship moved through the distances given after 5 in the adjoin- ing schedule in the times given after t. / = o 2 4 6 8 10 12 14 16 seconds; 5 = 3.4 9.3 17.3 27.4 39.6 53.4 69.4 88.0 feet. Any displacement divided by the time required for that displacement we regard as the average velocity for that time; thus, in the last 8 seconds the dis- placement is 60.6 feet, and 60.6 -i- 8 = 7.57 ft/sec is the average velocity for that time. (Obviously a constant velocity of this value would produce a dis- placement of 60.6 feet in 8 seconds.) In the adjoining table we have listed * For dimensions of a unit velocity, see Appendix A. 118 Art. 28 119 At (sees.) As (ft.) Va (ft/sec) 8 to 16 = 8 60.6 7-57 8 to 14 = 6 42.0 7.00 8 to 12 = 4 26.0 6.50 8 to 10 = 2 12.2 6.10 the displacements (under A5) for the intervals 8 to 16, 8 to 14, etc. (under A^; and also the average velocities (under Va) for these intervals, respectively. Apparently, the average velocity for the intervals 8 to 9, 8 to 8|, 8 to 8 J, etc., continues to decrease, approaching a definite limit as the interval of time approaches zero. The column of average velocities sug- gests that the definite limit might be about 5.8 feet per second. The exact value of the Umit is the rate at which the ship was moving at the time 8 seconds. Summarizing now: — Let A^ = distance traversed in any interval A/, and Va = average velocity for that interval; then in any kind of rectilinear motion Va = ^s/At. (2) The value of the velocity at a particular instant of the interval is the limiting value of the average velocity for the interval, as the interval is taken smaller and smaller but always including the particular instant. In the calculus notation, this limit is ds/dt; hence ii v = velocity, then V = ds/dt. (3) Here s means the (varying) distance of the moving point from any fixed point in the path. Formula (3) can be used for finding the value of v in any motion in which the relation between s and t is known. Thus, suppose that a point is known to move so that its distance (in feet) from the starting point always equals four times the square of the time (in minutes) from starting, that is, 5 = 4 /^; then V = ds/dt — 8t. This is the general formula for v in this motion; that is, the formula holds for all instants and all positions of the moving point. Thus, at the instant t = 2 minutes, t) = 8X2= 16 feet per minute; at the position 5 = 100 feet, t — V 100 -^ 4 = 5 minutes and z) = 8 X 5 = 40 feet per minute. For another example of the use of formula (3) we consider a "crank and con- necting-rod mechanism " (Fig. 218). OP is the crank, mounted on its shaft at O, PC is the connecting rod, C is the crosshead; (A is a piston and AC the piston rod). When the crank is rotated the crosshead is constrained to move Fig. 218 1 20 Chap, vii in a straight line by the guides G. We will now find a general formula for the velocity of the crosshead (and piston) when the crank rotates uniformly. Let r = the length of crank, / = length of connecting rod, c = r/l, n = number of revolutions of the crank per unit time (assumed constant), co = angle in radians described by crank per unit time (co = 2ivn), s = varying distance of the crosshead from its highest position, 6 = the "crank angle" QOP, and t = time required for the crank to describe the angle 6 {= wt = 2 irnt). Obviously, there is a definite relation between 5 and d (or /), and this relation we need in order to get ds/dl or v. When the crosshead is in its highest posi- tion, its distance from equals / + r; hence for any position, s = {I -\- r) — CQ T OQ, T OQ according as the crank OP is above or below OX. Now CQ = (/2 - r2 sin2 6)'^ = I {i - c^ sin^ 6)^, and 0Q= zLr cos 6; hence 5 = (l -{- r) — I (i — c"^ sin^ d)^ — r cos 6. Differentiating the expression for s with respect to t, we get ds/dt, or v; and remembering that dd/dt = w, we can easily reduce the result to / c sin 2 \ V 2(1 - c2sin2 0)V From this general formula we can get the value of v for any particular case. Thus, let r = 10 inches, / = 30 inches (then c = \), and n = 100 revolutions per minute (w = 2 tt 100 = 628 radians per minute). When the crank is at OPq say, d = 90° and the formula gives v = 6280 inches per minute = 523 feet per minute. The expression ds/dt in equation (3) may be positive or negative; therefore v must be regarded as having the same sign that ds/dt has. Now ds/dt is posi- tive when s increases algebraically, and ds/dt is negative when s decreases algebraically; hence the sign of the velocity of a moving point at any instant is positive or negative according as s is increasing or decreasing then, that is the sign is the same as that of the direction in which the point is moving then. When the mathematical relation between 5 and t is unknown, then equation (3) cannot be used to determine the velocity at a particular instant. But if the displacements of the moving point are known for a number of known intervals beginning or terminating at the instant in question, then a fair approximation to the 4 S ^ ^ desired velocity can be obtained from the values of the ? "? I I average velocity for those intervals as explained in 1 . ! ! i the launching illustration preceding. One may determine the limit of the average velocities approximately by iG- 219 graphical methods. Thus, in Fig. 219 we have plotted the average velocities of the launching example in a manner which is ob- vious and then joined the plotted points by a smooth curve; this curve was extended, as seemed best, to the vertical through point 8. The ordinate 8 A represents approximately the limit sought, that is the velocity at the 8th second. Another graphical method is explained in § 2 of the following article. Art. 28 121 § 2. Acceleration. — A nonuniform motion is said to be accelerated, and the moving point is said to have acceleration. If the velocity changes uni- formly, that is by equal amounts in all equal intervals of time, the motion is uniformly accelerated; if the velocity does not change uniformly, then the motion is nonuniformly accelerated. By acceleration of a moving point is meant the rate at which its velocity is changing. To express the magnitude of any acceleration we must compare that acceleration to some particular acceleration as a standard or unit. Any rate of velocity-change — that of a freely falling body, for example — might be taken as a unit of acceleration but it is more convenient to take the accel- eration of a point whose velocity changes uniformly by one unit (of velocity) in one unit of time. Thus, we have the foot-per-second per second, the mile- per-hour per second, etc. And, abbreviating the word per as before, these would be written ft/sec/sec (also written ft/sec 2), mi/hr/sec, etc.* In a unijormly accelerated motion (u.a.m.), the acceleration may be computed by dividing the velocity-change which takes place in any interval of time by the length of the interval. Thus, if a = acceleration, ^v = the velocity- change and A^ = the interval, then a = ^v/^t. (4) In a nonuniformly accelerated motion the rate of change of the velocity is not constant but it varies continuously from instant to instant. To arrive at a definite notion of the value of the rate or acceleration at a particular instant, let us consider an example. The adjoining schedule gives values of velocity and time taken from a "starting test " of an electric street railway car. t = o V = o I 2.8 5-3 3 7-7 4 9.9 5 6 7 8 9 10 seconds; II. 9 13.7 15.2 16.4 17.3 18.0 miles per hour. Any velocity-change divided by the time required for the change we regard as the average acceleration for that time; thus, during the first six seconds the velocity-change is 13.7 miles per hour, and 13.7 -^ 6 = 2.28 miles per hour per second is the average acceleration for the first six seconds. (Obviously a uniform acceleration of this value would produce in six seconds a velocity- change of 13.7 miles per hour.) At (sees.) Av (mi/hr) a„ (mi/hr/sec) to 6 = 6 13-7 2.28 I to 6 = 5 10.9 2.18 2 to 6 = 4 8.4 2.10 3 to 6 = 3 6.0 2.00 4 to 6 = 2 3-8 1.90 5 to 6 = I 1.8 1.80 * For dimensions of a unit acceleration, see Appendix A. J 22 Chap, vn In the adjoining table we have listed the velocity-changes (under Av) for the intervals o to 6, i to 6, 2 to 6, etc.; and also the average accel- eration (under da) for the same intervals. Obviously the average accelera- tion for the intervals 5I to 6, 5! to 6, etc., continues to decrease, approaching a definite limit as the interval approaches zero. The column of average accelerations suggests that the definite limit might be about 1.7 miles per hour per second. The exact value of the limit is the rate of change of velocity, or the acceleration, when t was 6 seconds. > Summarizing now: — Let Ay = the velocity-change in any interval of time At, and a« = average acceleration for that interval, then in any kind of recti- linear motion Qa = Av/M. (5) The true value of the acceleration at a particular instant of the interval is the limiting value of the average acceleration as the time interval is taken smaller and smaller but always including the particular instant; or in the calculus notation a = dv/dt = dh/dt\ (6) Formulas (6), respectively, can be used for finding the value of a in any rectilinear motion if the relation between v and t or 5 and / are known. Thus suppose that a point is known to move in a straight line so that the velocity (in miles per hour) always equals one-tenth of the square of the time (in seconds) from the start, that is 2; = o.i /-; then a = dv/dt = 0.2 t. This is the general formula for a in this motion ; for instance, at 3 seconds after start- ing a = 0.6 miles per hour per second. For another example of the use of equation (6), we consider the motion of the crosshead of the crank and connecting-rod mechanism described in § i. There we found that / • /, 1 c sin 2 ^ \ V 2(i-c2sin2 0)V Differentiating this with respect to / and remembering that w is constant, we get dv/dt or c cos 2d -\- c^ sin^ d\ a = rw^ ( cos 6 -f- (i-c2sin2 0)^ From this general formula, we can get the value of a for any special case. Thus as in § I, let r = 10 inches, / = 30 inches (then c = I), and n = 100 revolu- tions per minute (c<)=27rioo=628 radians per minute). When the crank is at OPo (Fig. 218), then 6 = go and the formula gives a = — 410 inches per second per second. For meaning of negative sign, see next paragraph. The expression dv/dt, equation (6), may be positive or negative; therefore a must be regarded as having the same sign as has dv/dt. Now dv/dt is positive when the velocity increases algebraically, and dv/dt is negative when the velocity decreases algebraically; therefore the sign of the acceleration of a Art. 28 123 2 3 4- Fig. 220 6Seci. moving point at any instant is positive or negative according as the velocity is increasing or decreasing (algebraically) then. Thus acceleration is posi- tive or negative according as positive or nega- tive velocity is being taken on in the motion. By direction of acceleration is meant the direc- tion of the velocity which is being taken on. When the mathematical relation between v and / (and 5 and /) are unknown, then equation (6) cannot be used to determine the acceleration at a particular instant. But if the values of the velocity are known at a number of known in- stants near the instant in question, then a fair approximation to the accel- eration desired can be obtained from the values of the average acceleration for intervals beginning or terminating at the instant in question, as explained in the car-starting example preceding. Fig. 220 shows a construction for determining the limit of the average acceleration in the example referred to. The ordinate 6 A represents the limit approximately. Another graphic method is explained in the next article under § 2. Note on Rate of a Scalar Quantity. — The foregoing explanations of two particular rates (velocity and acceleration) will now be generalized so that hereafter we will not need to de- rive the expressions or formulas for other rates which will come up for discussion. By a scalar quantity is meant one which has magnitude only, not direction also. An amount of money, the volume of a thing, the population of a city, etc., are scalar quantities. Let X and y denote two scalar quantities which are related to each other so that any change in one produces a change in the other. If all equal changes in x produce equal changes in y, then y is said to vary uniformly with respect to x and y is called a uniform variable. If all equal changes in x produce unequal changes in y, then y is said to vary nonuniformly and y is called a nonuniform variable. If y is a uniform scalar, then the graph which represents the relation between x and y is a straight line obviously, as for example in Fig. 221 where ji and y2 respectively denote values of y corresponding to xi and X2 (values of x). The meaning of "rate of y with respect to a; " or "a;-rate of y " is quite generally understood; it is the change in y per unit change in x. The value of the rate is computed by dividing any change in y by the corresponding change in x. Thus, if Ax and Ay = corresponding changes in x and y {x2 - Xi and y2 - yi), and r = x-mte of y, then r = Ay/Ax. Evidently r is the same for all values of x, that is, the rate of a uniform scalar is constant. Fig. 221 If y is a nonuniform scalar then the graph which represents the relation between x and y is a curved line, as for example in Fig. 221 where vi and y. represent values of y which corre- spond to xi and x^ respectively. Any change in y divided by the corresponding change in x 124 Chap, vii is commonly called the average rate of y with respect to x for the range xi — X\. Thus, if fa = average x-rate of y for the range Ax (= x^ — x\) in x, then Ta = ^yl^x. The average rate is represented by the slope of the chord AB, for tan BAC = ^y/^x. The value of the average x-rate of y depends on the amount of the range A;r. It approaches a definite value as Ax is taken smaller and smaller, x-i approaching .vi for instance. This limit- ing value is taken as the true or instantaneous rate of y at the value y = yi{oix = X\). Thus, if r = x-rate of y at any value of y, then r = lim (Ay /Ax) = dy/dx. The x-rate of y at y = yi is represented by the limit of the slope of the chord ^5 as 5 ap- proaches A, that is, by the slope of the tangent at A. By means of the foregoing formula, we can determine the .x-rate of y provided that we know the precise relation between x and y, that is, the equation y = / (x). In case we do not know this equation but do know values of y corresponding to several values of x, then we can de- termine the x-rate of y at one of the values of x approximately. This approximate value can be obtained from the average rates for ranges of x which begin or terminate at the value of X for which the rate is desired as already explained in some of the preceding examples. § 3. Features of a Motion Determined by Integration. — In the preceding article we showed how to determine the velocity from the dis- tance-time {s-t) law, and the acceleration from the velocity-time {v-t) law. The process, in each case, is one of differentiation. By means of the reverse process, integration, one may determine the s-l from the v-t law, and the v-t from the a-t law. We explain further by means of examples. Suppose that a point moves in a straight line according to the law v = dot + 4. In all cases of rectilinear motion v — ds/dt, or ds = v dt; hence in the present instance, ds = (60 / + 4) dl. Integration gives s = ^ot'^ -\- ^t + C, where C is a constant to be determined from "initial conditions." Let us sup- pose that s is reckoned from the place where the moving point is when t = o, or that 5 = when / = o; then substituting these (simultaneous) values of 5 and t in the equation containing C, we get o = o -|- o + C, or C = o. Hence the s-t law is 5 = 30 f^ + 4 /. We might have integrated "between limits," thus Jds = I (60 / + 4) dt, or 5 = 30 /2 -f 4 /, t/O the lower limits being the simultaneous values of s and / from the given initial conditions. For another example, we will suppose that a point moves in a straight line so that a = cos t, initial conditions being v = 4 when / = o. In all cases of rectilinear motion a = dv/dt, or dv = adt; hence, in this instance, dv = cos tdt. Integration gives t; = sin / + C. Substituting the (initial) simultaneous values of v and / in this equation we find 4 = o + C, or C = 4; hence z) = sin / -f 4 is the law sought. Or, integrating between limits we get COS tdt, or y — 4 = sin t. rdv= r t/4 t/0 Art. 28 125 If the as law for a motion is given, then the v-s law can be found by integrating v dv = a ds, which follows from a = dv/dt = {dv/ds) (ds/dt) = (dv/ds) V. Thus, suppose that in a rectilinear motion a = 25 + 3, initial conditions being ^; = 10 when 5 = 4; then I vdv = I (2 s -{- 7,)ds, or ^v^ = s"^-^ ^s + C. Initial values substituted in the last equation give C = 22, and hence ^ v^ = s^ + 35+ 22. And in any rectilinear motion ads = I vdv = ^ {v^ — v^), where Vi and V2 are values of the velocity when s = Si and 5 = ^2 respectively. The formulas a = dv/dt and v = ds/dt can be used to get "time." These can be written dt = {i/a) dv and dt = (i/v) ds; hence by integration - dv, and 4 — ^1 = / -ds. V, a Js^ V These respectively give the time required for v to change from vi to V2, and for s to change from Si to ^2. Uniformly Accelerated Motion. — Let a = the value of the (constant) accel- eration, and Vo = the velocity at the instant from which time is reckoned, and So = the distance of the moving point from the origin at that instant. (Some- times Vo and So are called initial velocity and initial distance, respectively.) Since a is constant, integration of a = dv/dt gives at once v = at-\- Ci, and from the initial conditions (v = Vo when / = o), Ci = Vo, hence V = at -{- Vq. (i) From V = ds/dt = at -{- Vo we find by integration that s = I at"^ -\- Vot + C2, and the initial conditions {s = So when t = o) make C2 = So', hence 5 = I a/2 + Vot + So. (2) Eliminating t between (i) and (2) we find that 2a(s — So) = v^ — vo"^. (3) If the initial velocity and distance = o, then v = at, 5 = 1 at^, and 2 as = v^. (4) Although uniformly accelerated motions are important practically, the student is advised not to make a special effort to memorize the foregoing formulas (i, 2, 3, and their special forms, 4). But, if he will memorize them, then he should also remember that they are for a special motion, constant acceleration. All students ought to be able to discuss a uniformly accelerated motion nonmathematically — by means of elementary notions somewhat as in the following example: The velocity of a certain train can be reduced by braking from 40 to 20 miles per hour in a distance of 1600 feet. In what dis- 126 Chap, vii tance would braking stop the train from 40 miles per hour, supposing the retardation to be the same at all velocities? Since the velocity changes uni- formly, the average velocity during the reduction from 40 to 20 miles per hour equals one-half of 4c + 20 or 30 miles per hour; and the time required for the reduction of velocity or travel of 1600 feet (= 0.303 miles) is 0.303 -^ 30 = 0.0101 hours, or 36.4 seconds. The time required to stop the train from 40 miles per hour would be twice 36.4 or 72.8 seconds; and, inasmuch as the average velocity during the stoppage would be one-half of (40 -f o) = 20 miles per hour or 29.3 feet per second, the distance travelled in the 72.8 seconds would be 29.3 X 72.8 = 2133 feet. 29. Motion Graphs The features of rectilinear motion, discussed in the preceding article, can be represented nicely by certain curves described in the following: A distance-time (s-t) graph for any motion is a curve drawn "upon" a pair of rectangular reference axes so that the coordinates of any point on the curve represent corresponding, or simulta- neous, values of s and t, where t = the time elapsed from some instant of reckoning (usu- ally taken at the instant of starting), and 5 = the distance of the moving point from some fixed point chosen as origin (usually taken at the place of starting). Fig. 222 is the s-t graph for the launching mentioned in § I of the preceding article. Since the slope of the s-t graph is proportional to ds/dt and v = ds/dt, the slope at any point of the graph represents the velocity at the corresponding instant, according to some scale. The slope scale depends on the scales used for plotting the s-t graph. Thus, in Fig. 222 the scales are i inch of ordinate = 100 feet and i inch of abscissa = 10 seconds, hence, a slope of unity = 100 (feet) ^10 (seconds) = 10 (feet per second). Thus, the velocity at / = 8 seconds, where the slope h BC -^ AC = 0.54, is 5.4 feet per second. Instead of interpreting the slope in this way, that is by a slope scale, we might determine the velocity as follows: draw the tangent line at the point A in question, drop a per- pendicular from any point B in the tangent to the horizontal through A, measure CA and CB according to the proper scales and compute the ratio BC h- AC (as measured); this ratio equals the desired velocity. Thus, in Fig. 222, AC = 5 seconds, CB = 27 feet, and ^ = 27 ^ 5 = 5.4 feet per second.* * Several instruments have been devised recently for drawing a tangent to a plane curve. A very simple one is represented in Fig. 223. It consists of a metal straight-edge A with a por- tion of one side polished to a mirror. OB represents a curve on a piece of paper across which 16 Sees. Fig. 222 Fig. 223 Art. 29 127 The velocity-time (v-t) graph for any rectilinear motion is a curve drawn upon a pair of rectangular reference axes so that the coordinates of any point of the curve represent corresponding, or simultaneous, values of the velocity V and time /. The curve in Fig. 224 is a v-t graph for the car-starting test Fig. 224 I05ec&, mentioned in § 2 of the preceding article. The slope of the v-t graph at any point represents the acceleration at the corresponding instant. To actually determine the acceleration from the graph, the slope must be interpreted by proper scale or be computed in a manner analogous to that explained in the foregoing under distance-time graph. Thus, at the fifth second, the accelera- tion is represented by the slope of the tangent at A; since AC = 2.5 seconds and CB = 4.8 miles per hour, the acceleration = 4.8 -^ 2.5 = 1.92 miles per hour per second. The "area under the curve " (between the curve, the time axis, and any two ordinates) represents the displacement for the interval of time represented by the distance between the ordinates. Proof: Let m = velocity scale-number and n = time scale-number, that is unit ordinate (inch) = m units of velocity (feet per second) ; unit abscissa (inch) = n units of time (seconds). Thus, let X and y be the lengths (inches) of the coordinates of any point P of a v-t curve (Fig. 225) ; then the corresponding values of v and / are my and nx. the straight-edge is laid at random but so that a portion of the curve is reflected from the mirror. The image CD and the curve CO are not smoothly continuous; there is a cusp at C. But if the instrument be turned about C until the cusp disappears, the curve merging smoothly into its image, then the straight-edge A is normal to the curve OB at C. Having located the normal at C, it is easy to draw the tangent. The principle of this instrument is the basis of Wagener's derivator (see Gramberg's Tcchnische Messungcn) by means of which the slope of a curve at any point can be read directly, without drawing the tangent or normal. An auto- graphic form of (mirror) derivator has been devised by A. Elmendorf (see Sci. Am. Suppl. for Feb. 12, 1916). Guillery's "aphegraphe" is another instrument for drawing a tanf^ent to a plane curve. A metal strip or batten must first be fitted to the cuive before the instrument proper can be applied. For full description of the aphegraphe, see Mem. Soc. Ing. Civ. de France, Bull, for .■\pril, 1911, where M. Guillery also explains how he applied his instrument to determine the acceleration-time curves for several mechanisms, and in particular the a-t curve for the "tup" of an impact testing machine during a blow. 128 Chap, vn Further let h and k = the times corresponding to Xi and 3C2 and to Si and 52, the values of s (space) ; and A = area. Then /»^j , rkv dt I r'2 , = 1 ydx = I = — / vdt = A S2 — S1 mn (see preceding article) ; and A {mn) = s^ — Si. Hence {mn) is the scale-number for interpretating the area. Thus, in Fig. 224, one-inch ordinate = 20 miles per hour = 29.3 feet per second, and one-inch abscissa = Fig. 225 ^ seconds; hence one square inch = 29.3 (feet per second) X 5 (seconds) = 146.5 feet. The area may be interpreted more directly by multiplying the average ordinate measured by the scale of ordinates (hence equal to the average velocity for the time interval) by the length of the interval. Thus, in Fig. 224 the average ordinate represents 10.9 miles per hour =16 feet per second, and the time interval is 10 seconds, hence the displacement is 160 feet. The acceleration-time {a-t) graph for any rectilinear motion is a curve drawn upon a pair of rectangular reference axes so that the coordinates of any point of the curve represent corresponding, or simultaneous, values of the ac- celeration a and the time /. The "area under the curve" represents the velocity-change for the time interval represented by the distance between the ordinates. For the area under the curve is given by a dt, and Vi — V\= I adt h Jh (see preceding article). To determine the numerical value of the velocity- change, the area must be interpreted by scale or be computed in a manner analogous to that explained in the foregoing under velocity-time graph. The velocity-distance {v-s) graph for a rectilinear motion is a curve drawn upon a pair of rectangular axes so that the coordinates of any point of the curve represent corresponding, or simultaneous, values of the velocity v and distance s. Fig. 226 is the v-s graph for an air-brake test on a pas- senger train.* The subnormal at any point of the graph represents the acceleration at the corre- sponding instant. For, any subnormal as BC is given by ^C tan BAC = vdv/ds, and from the preceding article a = dv/dt = {dv/ds) {ds/dt) = vdv/ds; hence BC = a. To actually determine the value of a from a subnormal we must use the proper scale, depending on the scales used for plotting the v-s graph. For Fig. 226 one-inch ordinate = 50 miles per hour, and one-inch abscissa = 1000 feet = 0.19 mile; hence the subnormal scale is one inch = 50' -r- 0.19 = 13,150 miles per hour per hour = 3.65 miles per hour per second. The subnormal BC * "Air-brake Tests — Westinghouse." Page 297. 200 400 60O 800 1000 Fig. 226 Art. 29 129 = 0.72 inch; hence the (negative) acceleration at A (when the train had made 600 feet from the place where braking began) was 0.72 X 3.65 = 2.63 miles per hour per second. The acceleration-distance (as) graph for a rectilinear motion is a curve drawn upon a pair of rectangular axes so that the coordinates of any point on the graph represent corresponding, or simultaneous, values of a and s. "Area under the curve " (between the curve, the 5 axis, and ordinates ai and a2) represents one-half the change in the velocity-square corresponding to the change o^ — ai or ^2 — ^i. For, the area is given by ads = I vdv = ^ {v^ — v^). motion is The. reciprocal acceleration-velocity {- -v\ graph for a rectilinear a curve drawn upon rectangular axes so that the coordinates of any point on the curve represent corresponding, or simultaneous, values of i/a and v. "Area under the curve" (between the curve, the v axis, and ordinates i/ai and 1/C2) represents the time required for the acceleration to change from ai to 02, or velocity from Vi to V2- For, the area is given by ^dv= I dt= k- k. I), d Jh The reciprocal velocity-distance {--s\ graph for a rectilinear motion IS a curve drawn upon rectangular axes so that the coordinates of any point on the curve represent corresponding, or simul- taneous, values of i/v and 5. "Area under the curve" (between the curve, the s axis, and ordinates i/i'i and i/v^) represents the time required for the velocity to change from V\ to V2. For, the area is given by I sec. dt = t^ — tu I sec. Example. — A mechanism is to be de- signed for producing a rectiUnear motion whose acceleration-time graph is shown in Fig. 227. There are three distinct laws of acceleration. In the first and last quarter seconds the acceleration is constant and equals 16 feet per second per second; in the second quarter the ac- celeration decreases uniformly from 16 to — 48; and in the third it increases uniformly from —48 to 16. Preliminary to the design it is necessary to find the distance-time law; this we proceed I'A ^/8 "z % % Figs. 227, 228, 229 F I sec. i^o Chap, vii to do, but first we get the velocity-time and distance-time graphs approxi- mately. During the first quarter of a second the velocity changes uniformly, and the change is i6 X j = 4 feet per second; and if the initial velocity is zero, then OA (Fig. 228) is the velocity- time graph for the first quarter second. Since the velocity changes uniformly in the first quarter second, the average velocity equals | (o -f 4) = 2 feet per second, and the displacement during the quarter =2X5 = ^ foot. If the initial distance is zero then and A (Fig. 229) are points on the distance-time graph. In a similar way interme- diate points could be computed. In the second quarter the acceleration varies uniformly. The average acceleration for the interval from | to t^ second is 8 feet per second per second; hence the velocity-change for that interval is 8 X yV = 2 foot per second, and the velocity at / = /^ is 4 -|- ^ = 4.5 feet per second, and B (Fig. 228) is a point on the velocity-time graph. In a similar way, C, D, and intermediate points could be determined. The portion AD is curved, and the average velocity for any interval cannot be ascertained so simply. But estimating the average ordinate for the third eighth of a second to be 4.4, then the displacement for that interval is 4.4 X | = 0.55 feet, and C (Fig. 229) is another point on the distance- time graph. In a similar way we might determine other points approximately. Determination of the graphs for the third and fourth quarter seconds by this method presents no difficulties, so we pass on to a second (mathematical) determination of the graphs. In the first quarter, dv/dt = 16, or dv = 16 dt; hence v -^ i6t+ C. But in accordance with initial conditions assumed, v = o when / = o; hence C = o, and z; = 16 / is the equation of the velocity-time graph for the first quarter. From that equation we find for t = I, v = 4 as before. Since V = ds/dt, ds = vdt = 16 tdt, and s = Sf + C. In accordance with initial conditions assumed, s = o when / = o; hence C = o, and 5 = 8 ^ is the equa- tion of the distance-time graph for the first quarter. From that equation we find for / = I, 5 = I as before; at / = i 5 = | foot; etc. In the second quarter, c = 80 - 256 /, equation of AD (Fig. 227); hence dv = (80 - 256 /) (// or t' = 80 / - 128 /2 -f C. We found that v = 4 when t = i;therefore4 = 80 X i - 128 X tV + C, orC = - 8, andz; =&ot- 128^-8 is the equation of the velocity-time graph for the second quarter. Continu- ing, ds/dt = 80 / - 128 /2 - 8, or 5 = 40 /^ - 42I /^ - S / -h C; but S — 2 when / = I, hence C = f , and 5 = 40 /^ - 42! /^ - 8 / + f is the equation of the distance-time graph for the second quarter. The equations of the graphs for the remaining quarters could be obtained in a similar way. Care must be taken in determining the constants of inte- gration; use no value of / (and corresponding value of v or s) which does not fall within the period to which the equation under consideration pertains. The remaining equations are — rtRT. 30 131 4ft.persec.persec. 1 16 sees. 24 For the third quarter For the fourth quarter a = — 176 + 256 1 c = 16, 2, = — 176 / + 128 /2 + 56 z; = 16 / — 16, 5= -88/2 + 421^+56/- 10 5=8/2-16/+ 8. Graphs for Uniformly Accelerated Motion. — Fig. 230 shows the acceleration- time graph for a rectilinear motion; in the first six seconds a = 4 feet per second per second, in the next ten seconds a || o and in the last 8 seconds a = — 3 feet per second per second (the negative sign meaning retardation). Fig. 231 shows the corresponding velocity-time graph, it being assumed that there is no initial velocity. Fig. 232 shows the corresponding dis- tance-time graph, initial distance being taken as zero. Fig. 233 shows the as and v-s graphs; AB-CD-EF is the former, and OGHJ is the latter. 6 sees. O ■3 6 sees. I6secs. 24 15sec5. 5 \( 72 M- Figs. 230, 231, 232 |D IE IF I 1 ■ — 240' ->k-96'->l Fig. 233 30. Simple Harmonic Motion and a Similar One § I. Simple Harmonic Motion (S.H.M.). — If a point moves uniformly along the circumference of a circle then the motion of the projection of that point on any diameter is called a simple harmonic motion. Obviously the projection {Q) moves to and fro in its path, and travels the length of the diameter twice while the point (P) in the circumference, goes once around. By amplitude of the s.h.m. is meant one-half the length of the path of Q, equal to the radius of the circle, '^y frequency oi the s.h.m. is meant the number of complete (to and fro) oscillations of the moving point Q per unit time, equal to the number of excursions of P around the circumference per unit time. By period of the s.h.m. is meant the time required for one complete to and fro oscillation of the moving point Q, equal to the time required for one excursion of P around the circle. By displacement of the moving point Q is meant its distance from the center of the path; it is regarded as positive or negative according as Q is on the positive or negative side of the center. Let us now consider a simple harmonic motion to ascertain approximately its nature. Suppose that the circle (Fig. 235) to be the path of P, and the vertical diameter, say, the path of Q. The y-t (space-time) and the y-d graphs for the motion of Q can be constructed very easily. We mark any number, say sixteen, equidistant positions of P, and number them consecu- tively and also the positions of Q to correspond (Fig. 234). Then on an exten- sion of the horizontal diameter we lay off any convenient length oT to represent 360° or the period, and divide oT into sixteen equal parts numbering the points of division as shown. Finally we project points o, i, 2, etc., of the circle upon 132 Chap, vu the verticals through the corresponding points o, i, 2, etc., of oT. These projections are on the y-t or y-d graph. The slope of the graph at any point represents the velocity of Q at the corresponding instant; hence the _2/' "^ i ^ \6 A : i '.\w 2 4 6 8\ ; i;0 ; 1 2 uo" e '\<0 12 Fig. 234 velocity of <3 is greatest at the middle of its path, and equals zero at the ends of the path. The arcs o-i, 1-2, 2-3, etc., are equal, and are therefore described by the moving point P in equal intervals of time. The lengths o-i, 1-2, 2-3, etc., in the diameter are described by Q in the same equal intervals of time; hence the average velocities of Q for those intervals are pro- portional to those distances. For comparison, the distances were laid off from O; OA = o-i, OB = 1-2, OC = 2-3, and OD = 3-4. Carefully com- paring these distances, we see t^at the numerical difference between succes- sive average velocities increases; hence the acceleration increases in value as Q moves from o to 4. In fact the acceleration is zero when Q is at the middle, and greatest when at either end of its path (proved below). We now examine s.h.m. more carefully, using the following notation: — r = amplitude (radius of the circle), n = frequency, o) = 2 TTW (abbreviation), co being angle in radians swept out per unit time by OP, X, y, or s = displacement, t = time after some convenient origin as described later, V = velocity of the s.h.m. at the time /, a = acceleration of the s.h.m. at the time t. When time t is reckoned from the instant when Q was at middle of its path and moving in positive direction. — Suppose the circle (Fig. 235) to be the path of P, which moves in sense indicated by arrow, and let us consider the motion of the projection of P on the verti- cal diameter, from now on called V instead of Q. Let 6 = angle XOP; then, since t is time elapsed since P was a.t X, 6 = 2 irnt = co/ and dd/dt = co. It is plain from the figure that y = r sin 0; and since v = dy/dt = r (dd/dt) cos d, Fig. 235 V = rw cos 9 = ru cos co/. (i) These are (general) formulas for v in terms of 6 and / respectively. Art. 30 133 Since cos 9 = sin (d -\- ^ir), v = m sin (5 + § tt). This formula for v sug- gests an easy method for drawing a v-d graph, showing how the velocity varies with d, and hence with /. First we draw an auxiliary circle with radius equal to 2 Trn according to any convenient scale; divide the circumference into any convenient number of equal parts, as sixteen; and number the points of division as in Fig. 236, that is 90° ahead of the numbers in Fig. 234. On an Fig. 236 extension of the horizontal diameter we lay off oT to represent 360°, and sub- divide this into the same number of equal parts (sixteen), numbering as shown; then 01, 02, etc., represent 6 = 22^°, 6 = 45°, etc. Finally we project points o, I, 2, etc., of the circle toward the right to meet corresponding vertical lines through points o, i, 2, etc., of oT. These points of meeting are on the v-9 graph, for the coordinates of any point on the curve are corresponding, or simultaneous, values of d and rw sin (^ + | tt) , or v. Inspection of the v-d graph verifies what was said about the acceleration. It shows clearly that the velocity of the moving point V (Fig. 235) changes more rapidly when V is near the ends of its path than when near the center; hence the acceleration of V is greater near the ends than near the center. Since the v-d graph is also a v-t graph, the slopes of the graph represent, to proper scale, values of the (varying) acceleration. The curve is steepest when 6 = 90° and 270° (when V is at the ends of its path), and horizontal when = o and 180° (when V is at the center of its path); hence again the acceleration is greatest at the ends of the path, and zero at the center. When the moving point V is approaching the center of its path — from either side — then V is getting up speed, and hence the acceleration of V is directed toward the center; when V is receding from the center, then V is slowing down, and hence the acceleration is directed toward the center. Therefore the acceleration is always directed toward the center. A general formula for acceleration in a s.h.m. will now be derived. We take the motion of V (Fig. 235) for that purpose, and let a — the acceleration at any time /. Now = dv/dt, v = rco cos0, and dd/dl — co; hence a= — rorsind = — ror sin cct. (2) These are (general) formulas for a in terms of d and t respectively. Since sin = — sin {6 + tt), a = rw" sin (d -\- ir). This last formula sug- gests an easy method for drawing the a-9 graph, showing how a varies with 134 Chap, vii 6, and hence also with t. First we draw an auxiUary circle (Fig. 237) with radius equal to rw- according to any convenient scale; divide the circumfer- ence into any convenient number of equal parts, say sixteen; and number Fig. 237 them as in the figure, that is 180° ahead of the numbers in Fig. 234. On an extension of the horizontal diameter we lay off OT to represent 360°, and sub- divide OT into sixteen equal parts numbering as shown; then 01, 02, etc., represent d = 22^°, 6 = 45°, etc. Finally, we project points o, i, 2, etc., of the circle horizontally to meet the corresponding vertical lines through points o, I, 2, etc., of the Hne OT. These points of meeting are on the a-d graph, for the coordinates of any point on the curve are corresponding, or simulta- neous, values of 9 and rw^ sin (6 -\-ir), or a. In Fig. 238 the foregoing described distance, velocity, and acceleration graphs are superimposed; the solid curve is the y-d graph, the dashed curve the v-d graph, and dot-dash curve is the a-d graph. Fig. 238 Fig. 239 Time dated from the instant when Q was at the positive end of its path. — We might continue to regard the s.h.m. as taking place in the vertical diameter of Fig. 235, reckoning time from the instant when P was at Y. It will be more convenient to consider the motion of the projection of P on the horizontal diameter; then we measure 6 and / as before. It is easy to show that x=rcosd = rcoso:t; v= —r(j}sm6= — rcosinw/; a= — rw^ cos = — rco^ cos w/. Time dated from the instant when Q was at some intermediate point. — Let / be reckoned from the instant when P (Fig. 239) was at some point as Pq, and let 6 = PoOP and e = XOPo- This latter angle is called angle of lead; but angle of lag when Po is below OX. Now, XOP = + e = co/ + e. In the s.h.m. executed by V, J = r sin (^ + e) ; v = rco cos (0 -f- e); a = — r(j? sin (0 + e). In the s.h.m. executed by ZT, x = r cos (0 -F e) ; z; = — rw sin (5 + e) ; a= — to? cos (0 + e). Art. 30 135 Formulas jor Velocity and A cceleration in Terms oj Displacement. — These do not depend on the way in which time is reckoned. Referring to the fore- going formulas we see that V = 0} Vr^ — s"^ = ior Vi — {s/rY, a co'^s. where 5 stands for displa cement x or y. The graph oi v = 0: Vr^ — s~ is the velocity-displacement graph for any s.h.m.; it is an ellipse. Fig. 241 shows that graph for the motion of the pro- jection of P on the horizontal diameter of the circle. When P is where indi- cated say, the velocity of H is represented by the ordinate HV. The graph of a = — co-5 is the acceleration-displacement graph ; it is a straight line. The diagonal line in the figure is the a-s graph for the motion of H. The acceler- ation of H is represented by the ordinate HA. if Hi H oIh :h jx \ ; fc; — ^ %i. Fig. 241 Fig. 242 Mechanism for Producing a Simple Harmonic Motion. — The mechanism represented in Fig. 242 consists of a crank C and a slotted slider S. The slider is constrained by fixed guides G so that it can be moved to and fro only (verti- cally in this figure). The crank-pin P projects through the slot of the slider; hence if the crank be turned, the crank-pin presses against and moves the slider. If the crank be turned uniformly then every point of the slider exe- cutes a simple harmonic motion. § 2. Crank and Connecting-rod Mechanism (Fig. 243). — When the crank is rotating uniformly, the motion of the crosshead (and piston) resem- FiG. 243 bles a simple harmonic one quite closely, as will be shown presently. Exact formulas for the position, velocity, and acceleration of the crosshead for any position of the crank were derived in Art. 28. The formulas are not simple. 136 Chap, vii The following approximate formulas (i, 2, and 3) are simpler and quite accu- rate, as will be shown. As in Art. 28, let r = length of crank, I = length of connecting rod, c = r/l, n = number of revolutions of crank per unit time (assumed constant), co = angle in radians described by crank per unit time (co = 2 ttw), s = the varying distance of the crosshead from its position most remote from the crank, d = the crank angle PdOP, and t = time required for the crank to describe the angle 6 {Q = oit = 2 -wnt). It follows from the geometry of the figure, as explained in Art. 28, that s= {l-\.r) -l{i- c"^ sin2 6)^ - r cos 6. Now (i — c^ sin^ 0)2 = I — I c^ sin^ 6 — \d^ sin^ 6 — etc. (binomial expansion). And since c is generally less than \, the third and succeeding terms in the series are very small and negligible; hence we have as a good approximation Ui - c2 sin2 0)^ = / (i - i c2 sin^ 0) = / (i - 1 c^ + 1 c^ cos 2 d), and 5 = r (i — COS0) + 4 cr (i — cos 2 0). (i) Now if we differentiate this with respect to t, we get ds/dt or v (velocity of the crosshead), and remembering that dd/dt = co, we finally get V = r(ji (sin + ^ c sin 26). (2) Differentiating again and remembering that w is constant we get dv/dt or a = ru)^ (cos 6 -{- c cos 2 6). (3) Because of our way of measuring s, the positive direction is from the cylinder toward the crank. Positive velocity v means that the crosshead is moving toward the crank, and positive acceleration a means that velocity toward the crank is being added to the velocity. In order to furnish a comparison between the foregoing approximate formu- las and the exact ones of Art. 28, we give in the adjoining table the values of a for the case c = 32 for a few values of the crank angle d (Fig. 243). e a, exact a, approx. 0° + 1.286 rw2 + 1.286 r«2 30 60 90 120 + IO15 +0.357 — 0.298 -0.643 + 1.009 +0.357 -0.286 -0.643 ISO 180 -0.717 -0.714 -0.723 -0.714 To find the acceleration of the crosshead when the crank is at the " head-end dead-center " (crank at OPo) we put 6 = o, and find from either the exact or approximate formula that a = rco^ (i -\- c). Art. 30 137 Length of Rod Length of Crank 0—0- O 1 m w V Rod —o- equals -o- J Cranho- » — o— „ -0-4 i> -o — t> -o— »? -0-5 " ~° — ;; -o— V -o- Q „ -o »- J. //.//. -5 ^ ■yi VUVlll — o — 0-0 -0-0 -0-0 7 8 Fig. 244 To find the acceleration at the "crank-end dead-center" we put 6 = 180°, and find from either the exact or approximate formula that a = — rw- (i — c). To show that the motion of the crosshead C is approximately simple har- monic we show that its motion resembles the motion of Q (Fig. 243) which is a simple harmonic one. In Fig. 244 we have marked nine corresponding posi- tions of Q and C. Thus points o to 8 are the positions of Q when the crank angles are 0°, 225°, 45°, etc., and points O, I, II, III, etc., are the corre- sponding positions of C. In the lower part of Fig. 244 the paths of Q and C (with the points i, 2, 3 and I, II, III, marked upon them) have been brought together for comparison. It is seen that the actual distances described by Q and C in any interval of time are nearly the same, and so the motion of C is nearly the same as that of Q. The three intermediate lines in the figure are paths of C with points corre- sponding to I, 2, 3, etc., for three other lengths of connecting rod. And we see that the longer the rod the more nearly is the motion of the crosshead simply harmonic. To arrive at a more complete comparison of the motions of C and Q, we will derive the formulas for the position, velocity, and acceleration of Q correspond- ing to equations (i), (2), and (3). The variable distance of Q from Po (Fig. 243) we will call z, then s = r (i — cos 6). (4) Differentiating with respect to /, we get for velocity of Q V = rcc sin 6, (5) and differentiating again we get for accel- eration of Q a = ro}"^ cos 9. (6) Now compare (i) and (4), (2) and (5), and (3) and (6) and note that the formulas for the motion of C contain an "extra " term. Each of these terms depends on c (= r/l), or on the "obliquity " of the connecting rod (maximum inclination of the rod to the line of stroke OC). The smaller c (the longer the rod in comparison with the crank), the smaller are the extra terms, and so the longer the rod the more nearly is the motion of the crosshead a simple harmonic one. Fig. 245 138 Chap, vii Fig. 245 presents a comparison of the motion of the crosshead C and the motion of Q. The solid lines refer to the first motion and the dashed lines to the second. Vc is the velocity-distance {v-s) graph and Ac is the acceleration- distance (as) graph for the motion of C. Vq is the velocity-distance graph and Aq is the acceleration-distance graph for the motion of Q. The graphs for C were drawn for a connecting rod three cranks long (c = i -^ 3). For longer rods the graphs for C would come much nearer the graphs for Q. 31. Motion and Force The preceding discussion of motion deals, for the most part, with displace- ment, velocity, and acceleration; it does not refer at all to the forces acting upon the moving bodies. In this article we explain in what manner any rec- tilinear motion of a rigid body depends upon the forces acting upon it. § I. First View and Form of the Fundamental Principle. — In Art. 2 it is explained that the units of force most used by engineers are the so-called gravitation iinits, equal to the earth-pulls on certain things called standards of weight. These units have slightly different values at different places; thus we have the London pound-force, the New York pound-force, etc. Some writers define the pound-force as any force equal to the earth's attraction on the standard pound weight at London or at sea level in latitude 45°, thus making the unit force invariable or an "absolute " one. Besides these units there are others; see § 2 of this article. In Art. 2, we explained also that the word weight is used in at least two senses in common parlance (see footnote, page 4). But we will continue to use it in a single sense, to connote the earth-pull on a body, and we employ a separate word (mass, see § 2 of this article) to connote the amount of substance or stuff in a body. Our two weighing devices, beam-scale and spring-scale, differ in a certain feature which is worth noting here. A beam-scale measures the weight (earth-pull) of a body in terms of the local unit of force, say the pound force for the place where the weighing is done; a spring-scale measures the weight of a body in terms of an invariable unit, say the particular pound force for which the scale was graduated. A beam-scale will not detect the change in the weight of a body with change of place because the magnitude of the unit (pull on the poise) changes just as the weight of the body changes. A spring-scale if sufficiently accurate will detect change in weight with change of place. First-hand knowledge of the relation between motion and the forces acting on the moving body must rest on observation or experiment. Let us consider a simple case of motion, that of a falling body. The motion takes place under the action of the weight of the body and the resistance of the surrounding air. But if the falling body is quite dense, the air resistance is negligible compared to the weight until the velocity becomes quite large. Observations have shown that such a body falls with a constant acceleration of about 32 feet per second per second at moderate velocities, and we infer that any force equal to the Art. 31 139 weight of the body would, if acting alone on that body, produce an acceleration of the value stated. We are now led to inquire what is the effect on a body of an applied force of some other magnitude, say a force equal to double its weight or one-half its weight? If we could intensify or dilute the earth-pull upon a body by a (gravity) lens or screen, then we could make a body fall under a force differing from its own weight and ascertain the answer to our question by observing the fall. Unfortunately for our purpose, we cannot so concentrate or dilute the force of gravity but we can dilute it indirectly by means of an "Atwood ma- chine," designed for that purpose. The essential parts of that machine are a light pulley P mounted on a smooth horizontal axle (Fig. 246), some blocks of metal which can be suspended as shown by a light flexible cord, and a timing device for getting the acceleration of A and B when the system is allowed to move. Neglecting the small influence of the pulley, axle, and cord, we regard A and B as the body moved and the difference in their weights (Wb — Wa) as the driving force. Experiments with this ma- chine show that A and B move with constant acceleration, and when runs are made with various driving forces — all metal pieces being used each time — then the acceleraUons in the ^^^ ., differeni runs are directly proportional to the driving forces. In this machine the driving force can be made very small but it cannot be made larger than the weight of all the metal pieces. It would seem that the force- acceleration relation stated holds even for driving forces larger than the weight of the body moved; and we assume that when any forces are applied successively to the same body so as to make it move in g, straight line, then the accelerations are proportional to the forces respectively. Or, if F and F' = the magnitudes of two forces applied to any body in succession, and a and a' = the accelerations respectively, then F/F' = a/a'. If IF = the weight of the body, g = the acceleration due to gravity (IF), F and a as above, then the foregoing principle gives also F/W = a/g, or as it is more commonly written, F = (W/g) a. Generally, a moving body is under the influence of more than one force. When the body moves in a straight line, the resultant of all the forces acting upon it is a single force acting in the direction of the acceleration (proved in Art. 35). Therefore the resultant has no component at right angles to the line of motion; or, the algebraic sum of the components of all the forces acting on the body along any line at right angles to the path equals zero. Thus, if the path is taken as an X axis and two lines at right angles to each other and to the path as y and z axes, then SF„=o, 2F, = 0, and ^F, = R, where SPx, "^Fy, and llF^ stand for the algebraic sums of the x, y, and 2 com- 140 Chap, vii ponents of all the forces acting on the body, and R denotes their resultant. Furthermore, as proved in Art. 35, W (i) J? ^^ K = — a. Any unit of force may be used for R and W in equation (i), and any unit for g and a. When a gravitational unit of force is used — such are most conven- ient in engineering calculations — then, strictly, the numerical value of g used should correspond to the "locality " of the unit-force used. That is, when one is about to make a calculation by means of equation (i), implying the New York pound-force say, then he should use for g its value for New York. As already stated, the variation in g is negligible in most engineering calculations, and we generally use 32.2 feet per second per second or even 32 for simplicity. Non-gravitational units, the dyne for example, may be used in equation (i). But when such units are preferred, then equation (2) is to be preferred in place of equation (i). Examples. — When a body moves in a straight line and if all the forces act- ing on it are known so that R can be computed, then the acceleration can be determined easily by means of equation (i). If the acceleration is known then we can determine R easily, and from R we can find out something about the forces acting on the body. I. A (Fig. 247) represents a body being dragged along a rough horizontal surface 5 by a pull P acting as shown. Suppose that the body weighs 100 pounds, P = 40 pounds, and the friction resistance = 10 pounds. We will find the acceleration of A and the normal component of the force exerted between A and B. The forces acting on A are represented in Fig. 248, N de- noting the normal component of the reaction oi B on A, friction being the other component. Resolving at right angles to the path, we get N + 40 sin 20° = 100, or iV = 86.3 pounds. Resolving along the path, we get i? = 40 cos 20° — 10 = 27.6. Equation (i) gives 27.6 = (100 -r- 32.2) a, or a — 8.9 feet per second per second. A _- too I /b3. Fig. 247 1^ /o Fig. 248 40, lbs. Fig. 249 Fig. 250 2. A (Fig. 249) represents a body being dragged up the rough inclined plane 5 by a pull P equal to 50 pounds; A weighs 60 pounds and the coefficient of friction for A and B is I. We determine the acceleration. Three forces act on A, namely the weight, the pull, and the reaction of B. The last force is represented by two components (TV and F) in Fig. 250. Resolving at right angles to the path, we get N = 60 cos 30° = 52 ; hence F = ^2 -^ 4= 1^ pounds. Resolving along the path, we get 7? = 50 — 13 — 60 sin 30° = 7 pounds; hence 7 = (60 -^ 32.2) a, or a = 3.75 feet per second per second. Art. 31 141 3. A certain passenger elevator gets up speed at the rate of 4 feet per second per second, and can be stopped at the rate of 8 feet per second per second. We discuss the pressure on the shoes of a standing passenger weighing 160 pounds, during an ascent. The forces acting on the man are his own weight and the pressure P of the floor on his shoes (upward). During acceleration the resultant of these forces is upward, hence P is larger than 160 pounds and R = P — 160. Equation (i) becomes P — 160 = (160 -f- 32) X 4 = 20, or P — 180 pounds. During the next period, constant speed, a = o and P — 160. During retardation the acceleration is downward and hence R gilso. There- fore R = 160 — P = (160 -h 32) X 8 = 40, or P = 120 pounds. 4. We determine the reaction of the car (Fig. 251) on A during the period of getting up speed at the rate of 2 feet per second per second; A weighs 1000 pounds. We suppose the floor of the car so rough that A does not slip. There are two forces acting on A (Fig. 252), its own weight and the pressure P of the floor. This latter force must be inclined as shown to furnish a component on A in the direction of the acceleration. Resolving at right angles to the path, we get P cos 6 = 1000; resolving along the path, we get R = P sind = (1000 -^ 32.2) X 2. Solving these two simultaneously we find that P = 1002 pounds and ^ = 3° Zi' • (The horizontal component of P is friction. To prevent slipping the floor must be rough enough to furnish such a resistance.) ■lOOOIbi ^3-1 Fig. 251 ^6'>i. Fig. 252 Fig. 253 150 Ibi, ^ /so lbs. / N\ ^p^^ Fig. 254 Fig. 2SS 5. A box (Fig. 253) containing a body A slides down a rough inclined plane B whose inclination is 40°. The box weighs 300 pounds, A weighs 150 pounds, the coefficient of kinetic friction "between " box and plane is J, and A is per- fectly smooth. We determine the acceleration of A and box, and the pressures between them. Fig. 254 represents all external forces acting on A and box, the reaction of the plane B being represented by two components, F and N. Re- solving at right angles to the path, we get N = 150 cos 40° + 300 cos 40° = 345, hence F = 345 -f- 5 = 69. Resolving along the path, we get R = 300 sin 40° + 150 sin 40° — 69 = 220 ^ (450 -^ 32.2) a, or a = 15.75 feet per second per second. Fig. 255 represents all the forces acting on A , where P and Q are the pressures exerted by the front and bottom of the box respectively. Resolving along the direction of the motion we get 7? = 150 sin 40° — P = (150-7-32.2)15.75, or P = 23.1. Resolving along the normal we get () = 150 cos 40° =115 pounds. 6. A body slides down a plane under the influence of gravity and the re- action of the plane only; required the acceleration. Let W = weight of the body; fi = coefficient of friction; a = inclination of the plane; N — normal 142 Chap, vii pressure; and F = friction. Then resolving forces normally to the path we get iV = M^ cos a; therefore F = ijlN = /AV cos a. Resolving along the path we get R = W sina — F = W (sin a — /jl cos a) = {W -^ g) a, or a = g (sin a — (J. cos a) . If the plane is perfectly smooth n = o, and a — g sin a. § 2. Second View and Form of the Fundamental Principle. — Physi- cists avoid the (common) double meaning of the word weight by employing the word mass to connote amount of material, substance, or stuff, in a body, and weight to connote the earth-pull on the body. Such usage is followed in this book. Material is measured in different ways; for example, Hquids gen- erally by gallon, earthwork by cubic yard, cloth by square yard, brick by thou- sand, iron by ton, etc. But mass means amount of substance as measured by a beam-scale. Our standards of mass (commonly and legally called " stand- ards of weight ") are the pound and the kilogram. These are certain pieces of metal preserved in London and Paris respectively. The mass of a body, measured as just explained, does not change with change of locality, and this is in accordance with our conception of material, substance, or stuff. The force-acceleration relation, F = (W/g) a, can be put into an alternative form which is preferable from some points of view. Thus suppose that two bodies whose weights at the same place are Wi and W2 are subjected to equal forces F; let g = the acceleration due to gravity at the place and ai and a^ = the accelerations produced by the two forces F. Then F = (Wi/g) Ci = Wi/g) 02, or ai/a2 = W2/W1. That is, the accelerations of the two bodies are inversely as their weights at the same place; and since the masses of two bodies are proportional to the weights (at the same place), the accelerations of the two bodies are inversely proportional to their masses. This relation and that between the accelerations produced in a body by two different forces acting singly can be expressed in one statement as follows: — Whenever a force acts upon a body so as to make it move in a straight line, then the acceleration produced is proportional to the force directly and to the mass of the body inversely, or a ^F -^ m. This proportion- ality can be put into the form of an equation, F = Kma, where i^T is a proportionality factor whose value depends on the units used for expressing magnitudes of F, m, and a. This is the alternative form mentioned. We may fix the value of K in two ways: — (i) choose units of F, m, and a at pleasure, and deduce the value of X; or (2) choose a value of K and units for any two of the quantities F, m, and a, and then deduce the proper unit for the third quantity. On plan (i) we choose, for example, the pound-force, the pound-mass, and the foot per second per second as units for F, m, and a, and Art. 31 143 then determine K by reference to any motion in which F, m, and a are known. The motion of a falling body is such a one. Thus when a body "weighing " say 10 pounds falls, then F = 10 pounds, m = 10 pounds, and a = about 32.2 feet per second per second, and we have 10 = X" X 10 X 32-2, or K = i -i- 32.2. On plan (2) we take K equal to unity for simplicity, and then (i) choose units of m and a at pleasure, and deduce the proper unit of F; or (ii) choose units of F and a at pleasure, and deduce the proper unit of m. (i) Physicists take the gram as unit mass, and the centimeter per second per second as unit of acceleration; then the corresponding unit of force {K = i) is such a force as would give to the gram an acceleration of one centimeter per second per second. They call this force the dyne, (ii) If we take the pound as unit of force, the foot per second per second as unit of acceleration, then the corre- sponding unit of mass (K = i) is such a mass which will sustain an acceleration of one foot per second per second under the action of a force of one pound. This unit of mass has no generally accepted name, but it is sometimes called "engineers' unit of mass," also "slug " and "gee-pound." A set of units for which A' = i is called a systematic set of units, also a kinetic set. We will always use systematic units and thus always have F = ma, or when several forces make a body move in a straight line, R = ma. (2) where R denotes the resultant of those forces. For a falling body R = W and a = g; thus when systematic units are used W = mg, or m = W/g. (3) Therefore R = (W/g) a as in § i. To arrive at a notion of the magnitude of the unfamiliar units dyne (force) and slug (mass), let us consider the well-known force-mass-acceleration rela- tion in the case of a falling body. A body whose mass is one gram, falling at Paris, falls under the action of a force (earth-pull) of one Paris gram, and has an acceleration of 981 centimeters per second per second. Hence a force of 0.001019 (= I -H 981) Paris grams would give to a body whose mass is one gram an acceleration of one centimeter per second per second. Therefore that force is the dyne, that is I dyne = 0.001019 Paris grams (force). A body whose mass is one pound, falling at London, falls under the action of a force (earth-pull) of one London pound, and has an acceleration of 32.2 feet per second per second. Hence a force of one London pound would give to a body whose mass is 32.2 pounds an acceleration of one foot per second per second. Therefore, that mass is the slug, that is . I slug = 32.2 pounds (mass). CHAPTER VIII CURVILINEAR MOTION 32. Velocity and Acceleration § I. Velocity. — In common parlance, velocity of a moving point at a certain instant means the rate at which the point is describing distance then. So understood, velocity has magnitude and sign only, and is therefore a scalar quantity. In the preceding chapter (on rectilinear motion) we used the word in this sense; in the present chapter we use the word in a broader sense — so that it is a vector quantity whose magnitude is the rate at which the moving point is describing distance at the instant in question and whose direction is the same as that of the motion then. If 5 = the (varying) distance of the moving point from some fixed origin in the path, the distance being measured along the path, then the magnitude of the velocity at any instant equals the value of ds/dt for that instant. Or if V = magnitude of velocity, V = ds/dt. (i) If the point is moving uniformly, then the rate at which distance is described is constant, and is given by As/^t, where As is the distance described in any interval At. The direction of the motion at any instant (and the direction of the velocity then) is along the tangent to the path at the position of the moving point at that instant. To illustrate, imagine a lo-foot wheel mounted on a horizontal axis which points north and south, and suppose that the wheel is rotating at 180 revolutions per minute clockwise when viewed from the south. When a certain point on the rim is in its highest position then the velocity of the point has a magnitude of 2 tt 5 X 180 = 5655 feet per minute, and the direction of the velocity is horizontal from west to east. The magnitude part of a velocity is called speed by some writers; we follow this usage. Thus in the preceding illustration the speed is 5655 feet per minute; while the wheel turns, the speed of the point is constant but the velocity changes in direction. § 2. Acceleration. — The acceleration of a moving point at any instant is the rate at which its velocity — not speed — is changing then. If V de- notes the (varying) velocity of a moving point and v the (varying) speed, then the definition states that the acceleration is dV/dt and not dv/dt. Inasmuch as most readers are unfamiliar with the rate of a vector quantity — the rate chapters in most books on differential calculus deal with rates of scalar quanti- ties only — we explain in considerable detail just what is meant by the rate ot 144 Art. 32 H5 change of a velocity, but first we explain for subsequent use a motion graph called Hodo graph. — This is a curve which shows how the velocity of a moving point varies. It is constructed by laying off vectors from a point to represent successive velocities, and then the free ends of the vectors are joined by a smooth curve. The curve is the hodograph for the motion. Thus, suppose that A BCD (Fig. 256 ) is the path of a moving point P, and that the vectors at A, B,C, and D represent the velocities of P when at A, B, C, and D respec- tively. If 0'A\ O'B', O'C, and O'D' (Fig. 257) are drawn (from any point O') Fig. 256 a bed Fig. 258 to represent the velocities respectively, then the curve A'B'C'D' is the hodo- graph for the motion of P from ^ to D. The increment or change in the velocity while P moves from /I to D say is represented by the vector A'D' (in magnitude and direction). The change in the speed = length O'D' — length O'A'. (The hodograph should not be confused with the speed- time curve. The latter is represented in Fig. 25S where ab, he, and cd represent the times required for P to move from A to B, B to C, and C to D respectively, and the ordinates over a, b, c, and d represent the speeds 2XA,B, C, and D.) Fig. 259 Fig. 260 1.8 2,0 2.2 2.4Secs. Fig. 261 We are now ready to explain the meaning of rate of change of velocity; we base our explanation on a simple case of curvilinear motion. Suppose that a point P starts at Q (Fig. 259) and describes the circle shown in such a way that the distance traversed (in feet) equals double the cube of the time after start- ing (in seconds), or 5 = 2 t^. Required the acceleration say, when t = 2.4 seconds, or 5 = 2 X 2.4^ = 27.65 feet. The curve in Fig. 260 is the hodograph of the motion for the interval from / = 1.6 to i = 2.6, containing the instant 146 Chap, viii in question. It was constructed from the adjoining schedule, computed from s = 2t^,d = s/20 (radians) = 2.865 ^ (degrees), and v = ds/dt = 6 P. t (sec.) s (ft.) e (deg.) V (ft/sec) 1.6 8.192 23-5 15-36 1.8 I I . 664 33-4 19.44 2.0 16.000 45-8 24.00 2.2 21 . 296 61.0 29.04 2.4 27.648 79.2 34-56 2.6 35-152 100. 7 40.56 Vectors O'A', O'B', O'C , etc., represent the velocities of F when / = 1.6, 1.8, 2 .0, etc. , as marked. Vectors A 'E' ,B'E', C'E' , and D'E' represent the velocity- increments for the intervals 1.6 to 2.4, 1.8 to 2.4, 2.0 to 2.4, and 2.2 to 2.4 seconds respectively. The magnitudes of these increments were scaled from the original hodograph drawing (the scale of which was one inch = 5 feet per second) and are recorded in the adjoining schedule under A V. At (sec.) AV (ft/sec) AV/At (ft/sec/sec) Av (ft/sec) Av/At (ft /sec /sec) 1.6 to 2.4 = 0.8 1.8 to 2.4 = 0.6 2.0 to 2.4 = 0.4 2.2 to 2.4 = 0.2 28.75 25-15 19-55 11-45 35-9 41.9 48-9 57-2 19.20 15.12 10.56 5-52 24.0 25.2 26.4 27.6 Now the magnitude of the average acceleration for the interval 1.6 to 2.4 seconds is 28.75 "^ ^-^ = 35-9 feet per second per second, and the direction of that average acceleration is A'E'. The magnitudes of the average accelera- tions for the intervals 1.8 to 2.4, 2.0 to 2.4 and 2.2 to 2.4 also are given under AF/A/; the directions of those average accelerations are respectively B'E', C'E', and D'E'. Now the acceleration when / = 2.4 seconds is the limit of these average accelerations, as At is taken smaller and smaller but always terminating at / = 2.4. The magnitude of this limit, which is the magnitude of the accelera- tion sought, is the limit of the magnitudes of the average accelerations. This limit can be found approximately by plotting as in Fig. 261, where ordinates equal to the computed average accelerations were erected at the proper points on the time-base, thus determining the solid curve abed. Any other ordinate represents an average acceleration for an interval terminating at / = 2.4 seconds, thus the ordinate at 2.1 represents the average acceleration for the interval 2.1 to 2.4. The curve abed may be extended a short distance without error, and therefore the ordinate over / = 2.4 is approximately the limit of the values 35.9, 41. Q, etc., and it represents closely the magnitude of the accelera- tion at ^ = 2.4 seconds. This ordinate scales 66.5 feet per second per second Art. 32 147 on the original drawing already mentioned. The direction of this acceleration is the limit of the directions of the average acceleration, and obviously this limit is the tangent to the hodograph at £'. On the original drawing the angle between this tangent and the horizontal is 24 degrees. For emphasis by contrast we will determine the way in which the speed changes during the motion under consideration. Speed-increments are listed under H-o in the schedule; average rates of change of speed for the respective time-intervals are listed under A17 A/. The luniting value of these averages, as A^ is taken smaller and smaller but always termmating at / = 2.4, is about 28 feet per second per second, and this is the rate at which the speed changes (dv/dt) at / = 2.4 seconds. We now generalize the foregoing. Let A B (Fig. 262) be the path of a moving point P, and let O'A' and O'B' be the velocities of P when at .4 and B respec- tively. Then vector A'B' is the velocity-in- crement for the interval A/ while P moves from A to B; (chord A'B') -^ A/ is the mag- nitude of the average acceleration for the in- terval, and the direction A'B' is the direction of the average acceleration. The magnitude of the (instantaneous) acceleration of P when ^^^- ^^^ passing A is the limit of (chord .1 'B') 4- A/, as B is taken closer and closer to A] and the direction of the acceleration is the limit of the direction of A'B' as B approaches A,or B' approaches A'. Now hm (chord A'B') -^ A^ = lim (arc A'B') -i- A/ = ds' /dt where ds' is the elementary portion of the hodo- graph at A', and s' is the distance of P' (the point in the hodograph corre- sponding to P) from any fixed origin on the hodograph; and the limiting direction of the chord A'B' is the tangent at A'. Finally, the acceleration of P is a vector quantity equal to ds' /dt and parallel to the tangent to the hodograph at the point P' corresponding to P. It should be noted that the acceleration of P is not directed along the tan- gent to the path but always toward the concave side of the path. It may be noted also that since the velocity of P' equals ds'/dt and is directed along the tangent to the hodograph at P', the acceleration of P is the same as the velocity of its rorresponding point P', it being understood that s' (distance on the hodograph) must be measured by the scale of the hodograph diagram. As an example of the use of our final result, that the acceleration of P is given by the velocity of its corresponding point in the hodograph, we determine the acceleration of a point which describes a circle at a constant speed. Let P (Fig. 263) be the point, r = radius of the circle, and v = the speed of P. The hodograph is a circle whose radius equals v\ A' corresponds to A and P' to P; and hence A 'O'P' equals 6. We measure the distance 5 (traversed by P) from A, and s' (traversed by P') from A'. Then s'/v = s/r, or s' = sv/r. Now 148 Chap, vin the velocity of P' equals ds' /dt = (ds/dt) (v/r) = v'^/r, and the velocity of P' is directed along the tangent at P' (parallel to the radius OP); hence the acceleration of P is directed from P to O and its magnitude is v^/r. The method for determining acceleration used in the preceding example is diflScult to apply in most motions. Why then was the method developed at length? To make plain the meaning of acceleration in curvilinear motion and particularly to show students, in an elementary way, that acceleration in curvilinear motion does not equal dv/dt and is not directed along the tan- gent to the path in general. Thus in the preceding example it was found that the magnitude of the acceleration is v^/r, whereas dv/dt = o since v is constant; also it was found that the acceleration is directed along the normal to the path. In the motion discussed at length (where 5 = 2 /^), it was found that the magnitude of the acceleration when / = 2.4 seconds is about 66.5 feet per second per second; but, smce v = ds/dt = 6 f, dv/dt = 12 t = 28.8 for t = 2.4.* Fig. 263 33. Components of Velocity and Acceleration § I. Components of Velocity. — Velocity, like any other vector quantity, can be resolved into components. For our purpose components parallel to axes of coordinates (as x, y, and s) are most useful; such components are called * Note on Rate of Change of a Vector Quantity. — We shall have to deal with the rates of vector quantities other than velocity. Therefore we now generalize our notions on the rate of this vector quantity (velocit}') just arrived at so as to prepare for the rates of these other vector quantities for future use. Let OA, OB, OC, etc. (Fig. 264), represent successive values of any vector p, in magnitude and direction, vector OB represent- ing p at time ti, OB at time to, OC at time /s, etc. The changes in p during the intervals ti to t^, ti to tz, /i to t^, etc., are represented by the vectors AB, AC, AD, etc. The average rate of change in the vector p during any of these intervals may be found by dividing the change by the time; thus for the interval /i to tt the average rate = AB -r- (^2 — /i), and this rate is a vector whose direction is AB. For the interval t\ to tz, the average rate = AC -i- {tz — /i) and the direction of the rate is AC. In general, both the magnitude and the direction of the average rate of a vector depend on the length of the interval for which the average rate is taken or computed. By true or instantaneous rate of change of the vector at the time ti, say, is meant the limit of the average rate AB -^ {t2 — /i) as ^2 is taken closer and closer to ti. The magnitude of this limit = limit of chord AB -h (t-i — ti) = limit of arc AB -r- {t2 — /i) = dS/dt where dS = elementary portion of the arc ; the direction of the hmit is the direction of the tangent to the arc at A. Imagine a point P to move in the curve AD so that the vector OP represents the vector p at each instant. The velocity of P = dS/dt and its direction at any instant is tangent to the curve at the point where P is at the instant; hence the time-rate of p is the same as the velocity of P (the moving end of p). Fig. 264 Art. ss 149 Axial Components. — Let x, y, and s = the (changing) coordinates of a moving point P, and v^, Vy, and v^ = the components of the velocity of P parallel to the x, y, and s coordinate axes respectively; then •i'l = dx/dt, Vy = dy/dt, v^ = dz/dt. (i) (Proof follows.) These formulas state that each axial component of the velocity at any instant equals the rate at which the corresponding coordinate of the moving point is changing then. In the following derivation of the for- mulas we assume for simplicity that the path of the moving point is a plane curve — in the xy plane; proof can be extended readily to include the case of a tortuous, or twisted, path. Let P (Fig. 265) be the moving point, v = the magnitude of the velocity of P, and a = the angle which the tangent at P makes with the x axis. Then v^ = v cos a, and Vy = v sin a. But v = ds/dt, cos a. = dx/ds, and sin a = dy/ds; hence _ ds dx _ dx , _dsdv _dy "'"dlTs ~Tt' ^"^"^ '"'~ dtds ~ dl' Fig. 265 Fig. 266 For an example, we determine the x and y components of the velocity of a point P which moves in the circle of Fig. 266 according to the law 5=2 t'^, s being in feet and / in seconds. (This is the motion discussed at length in the preceding article.) It is plain from the figure that x = 20 cos = 20 cos (5/20) = 20 cos (o.i^^); hence i)x = — 20 sin (o.i/^) 0.3 1- = — 6f sin {p.if). When / = 2 seconds, say, zj^ = — 6 X 4 sin (0.8 radians) = — 24 sin 45.8° = — 17.2 feet per second. The negative sign means that the component of the velocity is directed toward the left. In a similar way it can be shown that Vy= 6/^ cos {o.if). Other Components. — The velocity of a moving point P is directed along the tangent to its path at the point where P is at the instant under consideration; hence, the tangential component of the velocity equals the velocity itself, and the velocity has no normal component (along the normal to the path). For formulas for components of velocity along and perpendicular to the radius- vector of the moving point see Hoskins' "Theoretical Mechanics," Ziwet's, or any other standard work on that subject. § 2. Components of Acceleration. — Acceleration is a vector quantity, and can be resolved into components therefore. The most useful components for I50 Chap, vtri our purposes are: — (i) Those parallel to axes of coordinates {x, y, and z), called axial components; (2) those parallel to the tangent and normal to the path of the moving point at the place where the point is at the instant in question.* Axial Components. — Let a^, ay, and az = the axial components of the acceleration of a moving point P, and as in § i let Vx, Vy, and Vz = the (varying) axial components of the velocity of P, then flx = dVx/dt, Oy = dvy/dl, a^ = dvjdt. (2) (Two proofs follow.) These formulas state that each axial component of the acceleration of P at any instant equals the rate at which the corresponding axial component of the velocity of P is changing then. Since Vx = dx/dt, Vy = dy/dt, and v^ = dz/dt, we have also ax = d^x/df, ay = d'-y/df^, a, = dh/df^. (3) In the following proof it is assumed for simplicity, that the path of the moving point is a plane curve ^ — in the xy plane. The proof can be extended readily to include the case of a tortuous or twisted path. LetP (Fig. 267) be the moving point, and O'P^ (Fig. 268) be parallel and equal to the velocity v\ Fig. 268 Fig. 267 then P' is the point "corresponding " to P, and the direction of the accelera- tion of P is tangent to the hodograph at P' as indicated. Let a = the mag- nitude of the acceleration, and a' = the angle between the acceleration and the X axis. Then ax = a cos a and ay = a sin a'. But a = ds'/dt, where ds' denotes elementary length on the hodograph (see Art. 32); and since the coordinates of P' are Vx and Vy, cos a' = dvx/ds', and sin a' = dvy/ds'. Hence _ ds' dvx _ dvx , _ ^ ^ _ ^ 4. ^'~l[tdl'~dt' ^"""^ '''' ~ dt ds' ~ dt • ' y * For discussion of components along and perpen- dicular to the radius-vector drawn from any fixed origin to the moving point see texts referred to in § i . t The following is an alternative proof: — Let AB (Fig. 269) be a portion of the path of the moving point P, and let O'A' and O'B' represent the velocities of P ^ '^ when at A and B. Then A'B' represents the change ^^^' ^9 jj^ ^Yic velocity while P moves from A to B, and A'M and A'N represent the x and y components of this velocity-change. Let A'Q, tangent to the hodograph at ^', represent the acceleration of P when at A. Then 1 N , \B' 1 / /-A 1 . A 1 i // / \'/ 1 "■ iM 1 ax a cos a A'B' , ,.,\r. ,• A'B'cosB'A'M .. A'M hm ^ lim (cos B'A'M) = hm rr = hm But A'M = O'X - O'X' = increment in the x component of the velocity = Avx] «as =«: lim (Avi/At) = dVx/dt. In a similar way one could prove that a„= dvy/dt. hence Art. ,-i$ 151 For an example we determine the x and y components of the acceleration in the motion of the preceding example (see Fig. 266). In that example it was shown that the general value of the x component of the velocity (true for any instant) is Vx = — 6 /- sin (o.i t^) ; hence dvx/dt (or Qx) = — 12 t sin (o.i i^) — 1.8 /^ cos (o.i l^). And when / = 2.4 seconds, say, Gx — — 29.4 feet per second per second. In a similar way the value of Cy can be found from the general expresssion for Vy. Tangential and Normal Componenls. — They will be denoted by at and a„ respectively; other notation as before, and r = radius of curvature of the path at the point occupied by the moving point at the instant in question. Then at = dv/dt = d'^s/dt", and c„ = v'^/r. (4) (Two proofs follow.) These formulas respectively state that at = the rate at which the speed (magnitude of the velocity) changes, and that a„ is propor- tional to the square of the speed directly and to the radius of curvature inversely. Where the speed is increasing, dv/dt is positive and at has the same direction as the velocity; where the speed is decreasing, dv/dt is negative and at is opposite to the velocity in direction. The normal acceleration a„ is always directed from the mo\ang point toward the center of curvature. (The words tangential and normal refer to the tangent and normal to the path at the point where the moving point is at the instant in question.) Fig. 270 Fig. 271 Let AB (Fig. 270) be the path of a moving point P,v = velocity of P at ^, V -\- ^v = its velocity at B, and Ad = the angle between the normals (and the tangents) to the path at A and B. Also let O'A' and O'B' be equal to and par- allel to V and v + Av respectively; then A' and B' are on the hodograph and angle A'O'B' = Ad. The acceleration of P when at A is parallel to the tan- gent A'Q. Let A'Q represent c; then A'AI and A'N respectively represent the tangential and normal components of a. Hence at = acos(f) = -77 cos <^, dt and a„ = a sin 4> To continue the proof, we need to recall certain formulas from calculus. Let CC (Fig. 271) be any curve, O a convenient "pole," p and p + Ap the radius vectors of C and C, A7 the angle COC, Al the arc CC, yp the angle between OC and the tangent at C. From calculus, siiwp = p dy/dl and cos 1/' = dp/dl. 152 Chap, vin These formulas when applied to the hodograph (Fig. 270) become sm4> — V dd/ds' and cos 4> = dv/ds'. Hence ds' dv dv , ds' dd dB dd ds v^ ^ dt ds dt dt ds dt ds dt r For an example we determine the tangential and normal components of the acceleration in the motion of the two preceding examples. Since 5=2^^, V = 6 t" and dv/dt = 12/ = a«; at / = 2.4 seconds, say, at = 28.8 feet per second per second. Also a„ = v'^/r = 36 ^^,20 = 1.8 /^; at / = 2.4, a„ = 59.7 feet per second per second. f The {resultant) acceleration can be obtained from its axial or tangential and normal components. Thus a = V a/ + Cy^ + flz'- = Vat^ + an\ The angles which a makes with the x, y, and z axes are given respectively by cos~^ (a^c/a), cos~^ {ay/ a), and cos~^ {az/a). The angle which a makes with the normal equals tan"^ {at/ an). From a = {at^ + a„^)5 it appears that a does not equal at {= dv/dt) in general; onJ^ when an = o. And an{= "v^/r) = o only when v = o or r ^ oc, that is, where the moving point reverses direction of motion or where the radius of curvature is infinitely great. * The following is an alternative proof: — Let AB (Fig. 272) be a portion of the path of the moving point P, and O'A' and O'B' represent the velocities of P when at A and B respec- tively. Then the curve A'B' is the hodograph for .IB; the chord A'B' represents the change in the velocity while P moves from A to B; and the tangent A' a represents the accelera- tion a of P when at A. Let v = the magnitude of the velocity a,t A, v -\- Av = that at B, Ad — the angle between the normals (and the tangents) at A and B, and the angle be- tween the acceleration and the velocity at A. Then at = a cos(^ = ,. A'B',. , ^,,„„ ,. A'B' coiEA'B' ,. A'E ,. (ii -}- Ai.) cos Ai* - » hm lim (cos, EAB) = lim = nm — — = lim — = A^ Ai At At ,. AvcosA6 — v(i — cosA9) ,. Av .„ ,. sin-|A9 dv i ,. Ad ^^ dv hm -r = lim -- cos A0 — 2V hm — — — = -j. ti hm — A9 = -y At At At dt 2 At dt Referring to the figure it will be seen that a„ = a sin = ,. A'B'.. .. ^.,^,. ,. A'B'^mEA'B' .. EB' (j^±Av)^nA9 hm-— — lim (smEA'B') = hm- — = hm— -— = hm — = At At At At ,. sinAff , ,. Av . ,„ ,. Ad , dO \ ds -fi V hm — h lim -— sin Ad = v hm— - + o = v-j-= "»- -7; = — • At At At dt r dt r t Some students find it diflficult or impossible to realize that acceleration of a point in curvilinear motion has a component along the normal to the path at the place where the moving point is at the instant in question, notwithstanding detailed calculations (as on pages 145 and 146) for a specific case and mathematical derivation of the general formulas for the normal component of acceleration. Let us consider the matter from the perplexed student's own standpoint. He may ask, " If the moving point has an acceleration along the normal, why does it not acquire velocity along the normal ? " If he will grant that velocity cannot be acquired instantaneously but only with lapse of time, then it is easy to show that velocity is acquired along the normal. Thus to take a concrete case, suppose that a point P is moving in the curve (Fig. 272); con- ART. 33 153 Simple Harmonic Motion (see Art. 30). — The fact that the components of the velocity and acceleration — along any line — of a moving point P equal the velocity and acceleration of the projection of the point on that same line, enables one to get the formulas for velocity and acceleration in a simple harmonic motion very easily. Thus let P, Fig. 273, be a point describ- ing the circle uniformly, and Q its projection on the horizontal diameter; then the motion of () is a simple harmonic one (Art. 30). Let the amplitude of the s.h.m. (radius of the circle) = 2 feet, and the frequency of the s.h.m. (revolutions of P per unit time) = 100 vibrations per minute. Then the velocity ofP=27rX2Xioo = i26o feet per minute =21 feet per second, directed along the tangent at P as shown; and the acceleration of P = 21- -^ 2 = 220 feet per second per second, directed along the radius PO. Now when PO makes an angle d = 30° say, then the velocity of Q is 21 sin 30° = 10.5 feet per second; the acceleration oiQ = 220 cos 30° = iSo feet per second per second, directed toward O whether P is travelling clockwise or counter clock- wise. Evidently the greatest velocity of Q obtains when () is at 0; that value equals 21 feet per second. The greatest acceleration of Q obtains when Q is at either end of its path; that value is 220 feet per second per second. Fig. 273 Fig. 274 General formulas for velocity and acceleration in simple harmonic motion can be as easily derived. Let r = amplitude, n = frequency. Then the velocity of P = 2 irrn and its acceleration = 4 ir-r-n^ H- r = 4 Trhi^r. Hence velocity and acceleration of Q are respectively (see Fig. 273) — 2 irrn sin d and — 4 ir'^n-r cos 6. § 3. Projectile Without Air Resistance. — Let u = the velocity of projection (initial velocity), and a = the angle of projection (angle between direction of projection and the horizontal), x and y = the coordinates of the projectile P (Fig. 274) at any time t after projection, v = the velocity of P, and sider the interval while P moves from A to B. Let O'A' and O'B' represent the velocities at A and B respectively; then A'B' represents the velocity acquired by P in the interval, and this acquired velocity has a component not only along the normals at A and B, but along any other normal to AB. Or, the student may say, "Since the velocity is always directed along the tangent — and hence has no normal component — there cannot be an accel- eration along the normal." But here is a case which may convince him: a ball thrown obliquely into the air. The acceleration of the ball, due to gravity, is at all times vertically down, and this acceleration has for every position of the ball a component along the normal at that position. (Strictly the acceleration is not quite vertical by reason of air resistance, but neglecting this fact is of no consequence here.) 154 Chap, vni a = the acceleration of P at the time /. The only force acting on the pro- jectile during flight is gravity. Hence the acceleration of the projectile is vertically downwards at all times and equal to g (Art. 34), ox a^ = o and dy = — g. Since there is no x acceleration, the x velocity remains constant during the flight, and we find that value from the initial conditions (m, a) to be Vx = ucosa. (i) The y velocity is decreased at all times by the y acceleration — g. In the interval /, that decrease is gt, and since the initial y velocity is u sin a, the v velocity at any time / is given by Vy = u sin a — gt. (2) Since the x velocity remains constant, the x displacement in the interval / is given by X = u cos a ' t. (3) The y velocity varies uniformly with the time; hence the average y velocity for the interval / is | [(w sin a + {u sin a — gt)] = w sin a — | gt. The y dis- placement for the interval equals the product of the average y velocity and the time or y = usina • t — ^ gf^. (4) Foregoing results determine the velocity and position at any time /. They may be arrived at more directly by integrating the given equations dvx , dvu Thus integrating the first equation we find that Vx = C\, where Ci is a constant of integration whose value for reasons already stated is u cos a. Integrating the second equation we find that Vy = — gt -{- C2 where C2 is another constant of integration. From the initial conditions Vy = m sin a when / = o, and on substituting these values of Vy and t in the last equation we find that C2 = M sin a; thus Vy = — gt -\- u sin a as before. Now integrating Vx = dx/dt = u cos a, we get x = u cos a • / -f C3. From initial conditions x = o when / = o; therefore v = o -\- Cz, or C3 = o, and x = u cos a • ^ as before. Inte- grating Vy = dy/dt =— gt -]r usm. a, we get y = — \ gf^ -{- usina- 1 -\- d. From initial conditions y = o when / = o; therefore o = o -{- o -|- C4 or Ci = o, and y = — \ gt^ -\- u sina ' t as before. The trajectory (path of the projectile) is a portion of a parabola as can be shown from the equation of the trajectory. To arrive at the equation we may combine equations (3) and (4) so as to eliminate /. Thus we find that y 2u'^ cos^ a = xu^ sin 2 a — gx^. (5) Range aftd Greatest Height. — At the end X of the range, y = o; hence the time of flight is given by u sin at — ^ gt^ = o, or t = (2 w sin a)/g. The range R equals the value of x in equation (3) when / = the value just found; thus R = (u^ sin 2 a) -7- g. Art. 34 155 R also equals the value of x in equation (5) when y = o. The formula for R shows that the range is greatest — for a given velocity of projection — when a = 45°. That greatest value is u'^/g. At the highest point of the trajectory Vy = o; hence the time of flight to that point is given by m sin a — g/ = o, or / = {u sin a) -^ g. The height H of the trajectory equals the value of y in equation (4) when t = the value just found ; thus H = ^ {usin a)2 -^ g. H also equals the value of y in equation (5) when x = ^ R. 34. Motion of the Center of Gravity of a Body In Art. 3 1 we found that any rectilinear motion of a body depends in a very simple way upon the forces acting on the body. The relation between the motion of the center of gravity of a body (whether rigid or not) which has any sort of motion however complicated is also quite simply related to the forces exerted on the body as we shall see presently. § I. A Particle is a body so small that its dimensions are negligible in comparison with the range of its motion. In any motion of a particle no dis- tinction need be made between the displacements (velocities or accelerations) of different points of the particle, for they are equal, or practically so; and by displacement (velocity or acceleration) of the particle is meant the dis- placement (velocity or acceleration) of any point of the particle. "Laws of Motion.'" — i. When no force is exerted upon a particle then it remains at rest or continues to move uniformly in a straight line. 2. When a single force is exerted upon a particle^ then it is accelerated; the direction of the acceleration is the same as the direction of the force, and its magnitude is propor- tional to the force directly and to the mass of the particle inversely. 3. When one particle exerts a force upon another, then the latter exerts one on the former; and the two forces are equal, colinear, and opposite. These are essentially Newton'' s Laws of Motion. The form of statement here used differs however from that in which he announced them (1687). They are based on observation and experience. Newton was led to them through his study of the motions of heavenly bodies. No other moving bodies have been so accurately and extensively observed, and the agreement of the laws and those motions constitutes the best evidence of the correctness of the laws. Law 3 has already been referred to (page 43, footnote). This law is doubted by some beginners in this subject. The doubt is sometimes expressed in this way: "When a horse pulls on a cart, then, if the cart pulls back on the horse an equal amount (as the law states), why is it that they generally move for- ward? " Close attention to the forces which act on the horse and on the cart should clear up this doubt. There are three forces exerted on the horse, — • his weight (exerted by the earth), the pull of the cart, and the reaction exerted by the roadway on his hoofs. When the horizontal (forward) component of iS6 Chap, viii the reaction on his hoofs exceeds the pull back by the cart then the horse starts forward. There are three forces acting on the cart, — • its weight (exerted by the earth), the pull of the horse, and the reaction of the roadway on the wheels. When the pull exceeds the horizontal (backward) reaction of the roadway then the cart starts forward. Or, the motion of horse and cart together may be explained like this: There are four forces acting upon the pair, — the weight of the horse, that of the cart, the reaction of the roadway on the horse, and that on the cart; the horse and cart start to move when the horizontal component of the reaction of the roadway on the horse exceeds that on the cart. Law 2 is discussed at length in Art. 3 1 for the case of rectilinear motion, but is not referred to there as a "law." It covers curvilinear motion, as well as rectilinear, inasmuch as no reference to kind of motion is made in the law. We cannot give a real illustration of a particle moving under the action of a single force. But miagine a particle projected in some way, and then sub- jected to a single force inclined to the direction of projection; the particle would move in a curved path. (A ball in flight through the air is a near ap- proach to our imagined illustration. This ball is acted upon by two forces, earth-pull and air resistance; at moderate velocities the latter may be neg- ligible in comparison with the former.) The law states that the direction of the acceleration of the particle agrees at each instant with the direction of the force, and that the magnitude of the acceleration is directly proportional to the force and inversely proportional to the mass of the particle (a ) -^ gr = sin , or ' tan = v^/gr* * This formula, or some modification of it, is used to determine the proper elevation of the outer rail on railroad curves, except as noted below. The following is a practical rule deduced from the formula: "The correct superelevation for any curve is equal to the middle ordinate of a chord [of the curve] whose length in feet is 1.6 times the speed of the train in miles per hour." On the Pennsylvania Railroad the rule is modified as follows: " No speed greater than 50 miles per hour should be assumed in determining the superelevation by the above method even though higher speed may be made. No superelevation exceeding 7 inches is permissible and none exceeding 6 inches should be used except at special locations on passenger tracks." The formula was deduced on the basis that resultant flange pressure should = zero. The same formula is arrived at by making ties of the track perpendicular to the resultant pressure between the floor of the car and any object resting upon it, or perpendicular to a plumb Une suspended in the car. - P CHAPTER IX TRANSLATION AND ROTATION 35. Translation A translation is such a motion of a rigid body that each straight Hne of the body remains fixed in direction; there is no turning about of any line of the body. The coupling or side rods of a locomotive (connecting the driving wheels on either side) have a translatory motion when the locomotive is running on a straight track. It should be noticed that our definition does not require rectilinear motion of each point of the moving body. But rectilinear translations are most common, and such translations have been quite fully discussed in Art. 31. The motions of all points of a body in translation are alike. For, let A and B be any two points of the body, and A' and B' be the positions of those points in space at a certain instant and A" and B" their positions at a later instant. By definition of translation the Hnes A'B' and A"B" are parallel; and since the lines are equal in length the figure A'B' B"A" is a parallelogram, and A' A" and B'B" (the displacements of A and B respectively) are equal and parallel. Since the displacements of all points of the moving body for any interval, long or short, are equal and parallel, the velocities of all points at any instant are alike, and hence also the accelerations. By displacement, velocity, and acceleration of a body having a motion of translation is meant the dis- placement, velocity, and acceleration respectively of any one of its points. The general principle of Art. 34, relating to the motion of the mass-center of a body moving in any way, when applied to a translation, takes this form: the algebraic sum of the components — along any line — of the external forces acting on the body equals the product of the mass of the body and the com- ponent of the acceleration of the body along that line. This gives three in- dependent "equations of motion," namely, 2/^x = Ma., ^Fy= May, 2F. = Ma„ where x, y, and z denote three noncoplanar lines of resolution. The resultant of all the external forces acting on a body having a motion of trans- lation is a single force; its line of action passes through the mass-center, the force is directed like the acceleration of the body, and its magnitude equals the product of the mass of the body and the acceleration.* Assuming that the resultant is a single force, most students will accede to the second statement in the foregoing * The student is reminded that the resultant of a system of forces is a force, a couple, or a pair of noncoplanar forces (see Chapter I). 163 164 Chap, ix proposition, on the basis of their experience; for, they will say, if the resultant did not pass through the mass-center, the body would turn and not have a translatory notion. But it can be demonstrated as follows: Let Fig. 281 represent the body and points i, 2, 3, etc., its constituent particles; the external forces acting on the body are not shown. Suppose that the accelera- tion is directed, say, toward the right, and let a = the magnitude of that acceleration, and m\, W2, m^, etc. = the masses of the particles respectively. Then the resultants of all the forces acting on the several particles equal re- spectively mia, in^a, m^a, etc., all directed like the acceleration, as represented in the figure. Now this system of imaginary forces (resultants) is equivalent to all the real forces, external and internal, acting on the system of particles; and the resultant of the imaginary system and that of the real system are identical in magnitude, line of action, and sense. But the internal forces occur in pairs of equal, colinear, and opposite forces (Art. 34), and so constitute a balanced system and contribute nothing to the resultant of the real system. Hence, the resultant of the external system and that of the imaginary system are identical. We proceed now to ascertain the resultant from the latter system. A Fig. 23i Fig. 282 The imaginary system (I) consists of parallel forces proportional to the masses of the particles, and the lines of action of the forces pass through the particles respectively. The system of earth-pulls (gravity, G) likewise con- sists of parallel forces proportional to the masses of the particles, and the lines of action of these pulls pass through the particles respectively. Hence systems T and G are very similar; and if we imagine the body turned so that the line AB (parallel to a) in Fig. 281 is vertical (Fig. 282) then systems / and G are still more alike. The difference is in the magnitudes of corresponding forces; the forces of / are respectively proportional to the forces of G. It follows that the line of action of the resultants of systems / and G coincide (in the body) ; but the resultant of system G passes through the mass-center of the body; and hence the resultant of system / (and the resultant of the external system) also passes through the mass-center. From Fig. 281 it is obvious that the resultant of the external system is a single force directed like the acceleration, and equals Wifl -\- m2(i + • • • = o^m = Ala. The algebraic sum of the moments, or torque, of all the external forces about any line through the mass-center equals zero, for the resultant of those forces has no Art. 35 165 ZOOOIbs. moment about such line. This principle gives three independent m.oment equations: r. = o, r, = o, r. = o, (i) where T^, Ty, and Tz denote the moment-sums for three noncoplanar lines through the mass-center. Or we may take moments about any three lines and equate the torques of the external forces about the lines to the moments of the resultant {Ma) about the same lines respectively. Examples. — i. A (Fig. 283) is a rectangular prism weighing 2000 pounds. The car is being started at 4 feet per second per second. Required the pressure of the car on the bottom of the prism. There are only two forces acting on the prism, — its own weight and the required pressure P. See the figure where P is shown resolved into two components (Pi and P2) at the base of the prism. The (unknown) distance from the point of ap- plication of P to the center of the base is de- noted by X. HFy = Pi — 2000 = May — o, or Pi = 2000; HF^ = P2 = (2000/32.2) 4 = 248. Hence P = V(2ooo2 -|- 248^) = 2015 pounds, and the inclination of P to the vertical = tan~^ (248/2000) = 8° 25'. To determine x we take the torque, of the forces acting on the prism, about the horizontal line through the mass-center and perpendicular to the direction of motion and equate to zero. Thus 248 X 2.5 — 2000 x = o, or X = 0.31 feet = 3.72 inches. (P2 = 24S pounds is friction, and the floor and prism must be rough enough to develop such a value, to prevent the slipping, here assumed not to occur. Thus the coefficient of friction must be not less than 248 -^ 2000 = 0.124 or about one-eighth. If the coefiicient were less than one-eighth, the friction developed under the prism, say 200 pounds, could not give the prism an acceleration of 4 feet per second per second, only 3.22. Hence the prism would eventually be left behind. The prism is not "thrown off by the force of inertia " in such a case, as some would describe the phe- nomenon, but the car slips out from under the prism.) 2. C and C (Fig. 284) are two parallel cranks, their shafts being connected mechanically so that they rotate together with equal speeds and in the same direction. P is a bar pinned to the cranks. We discuss the forces acting on B when the mechanism is in motion. There are three such forces ; the weight of B and the pressures of the pins on B. We will neglect the weight, or assume that the plane of the cranks is horizontal so that the bar lies upon the cranks and the supporting forces balance the weight. If the bar is uniform then it seems reasonable to assume that the pin pressures Q are parallel; if so they must be equal since the algebraic sum of their moments about the mass-center of B equals zero. -/^ ^ ^^^ /^ Fig. 284 1 66 Chap, ix Moreover, the resultant of the pin pressures = <3 + Q = Ma, where M = mass of the rod and a = its acceleration, and the pressures act in the direction of a. The acceleration of the bar is the same as that of the center of either pin P. If the cranks be made to turn uniformly, then the acceleration is in the direc- tion PO and it equals V"/r (Art. 32), where v = velocity of P and r — PO; hence 2 Q = Mv^/r = (W/g) {fir), or(^ = \ Wv^gr. 3. Imagine a locomotive raised up off its track, and that steam is "turned on" so that the drivers are made to rotate at constant speed. If the connect- ing rod on one side be detached — the drivers being driven from the other side — then the side rod on the first side would be under the action of pin pressures just like those discussed in the preceding example. Each pressure equals | Mv^/r, directed along its crank radius and toward the crank shaft. (The weight of the rod induces pressures equal to | PF upwards.) When the locomotive is running on its track, then there is superimposed upon the motion of the side rod just discussed the forward (or backward) motion of the locomotive as a whole. The velocity of the side rod equals the vector sum of v and the velocity of the locomotive ; and the acceleration of the rod equals the vector sum of the acceleration V'/r and that of the locomotive. Now when the velocity of the locomotive is constant its acceleration is zero, and the acceleration of the side rod is still V"/r and parallel to the cranks and directed as explained in example 2, Hence, even when the locomotive is running on a track, the pin pressures on the (lone) side rod are as when the locomotive is "jacked up " and running. Let V = speed of locomotive, R = radius of driving wheels; then v = Vr/R, and the pin pressures = ^ (W/g) r V^/R^ (weight of rod neglected). For example, let W =275 pounds, r — i foot, R = 2.75 feet, and F = 60 miles per hour = 88 feet per second; then the pin pressures = ^ (275/32.2) X i X (88 -^ 2.75)^ = 4425 pounds. Locomotive Side Rod. — We give here another solution of the side rod prob- lem (see preceding examples). In Fig. 285 each pin pressure on the rod is represented by two components, hori- zontal and vertical. The vertical com- ponents are equal since the sum of the moments of all the forces acting on the rod (pressures and weight) about the center of gravity (at mid-length of the rod) equals zero; hence both vertical com- ponents are denoted by the same letter The horizontal components are A^i and A2. Let a = the total, or abso- lute, acceleration of any and every point of the rod when the cranks make any angle 6 with the downward vertical, and cz^ and ay = the horizontal and verti- cal components of a. Then ■X": — X2 = Max, and Fig. 285 Y. 2Y-W = May, or Y = ^W + May). Presently we show how to find ax and ay for any position of the cranks. Then Art. 35 167 from the above we can determine A'l - X2 and F. The values of Xx and Xg depend upon the load or pull on the locomotivCj and how it is distributed among the driving wheels. But Y does not depend on the pull, only on W and Qy. We now discuss the motion of one of the crank pins with the view of obtain- ing formulas for a^ and Oy. Let V = the velocity of the locomotive, A = its acceleration, R = radius of the driving wheels, and r = length of the cranks {CP, Fig. 286). It will be convenient to refer the motion of the crank-pin P to the coordinate axes shown; OY is the position occupied by the crank when P was in its lowest position. Let 5 be the distance of C from OY, and X and y the coordinates of P. Then and s = Rd, X = s — r sind, y = R — r cos 6 Fig. 286 Now a^ = d'^x/dl'^ and Qy = d-y/dt"^, and for use below V = ds/dt = R dd/dt, or dd/dt = V/R. Thus dx ds ^ dd ,, V - = --rcose^-=V-rcose.-^V^i - ^ cos dj ; dVf r ^ , r.r . ^ dd '^^^V-r'''V-^^R''''^'dt -■<^--u R^ ^cosdj-\-—rsme; dV r . r dd r V^ Thus it is seen that a^ and Oy depend on the velocity and acceleration of the locomotive. The largest values of c^ and a„ obtain at high speed, and then the A terms (in the expressions for a^ and ay) are small and negligible compared to the V terms. So when we neglect these terms or when the acceleration of the locomotive is zero, then a^ = (V/Py r sin 6, and ay = (V/Py r cos 6. When the rod is in its lowest position, 6^0,0^ = 0, a-y = {V/RYr, Xi = X2, and F = I W -^ ^ (W/g) (F/7?)V; the forces F- act upward on the rod. When 6 = 90°, a^ = (V/R)h, a^ = o; the resultant of the two forces X acts toward the right and equals (W/g) (V/R)h, and F = | W. When the rod is in its highest position d = 180°, a^ = o, ay = - {V/R)h; Xi = X2, and F = | W -^ (W/g) {V/R)h; for high speeds F acts down on the rod. When 6 = 270°, flj = —(V/R)h, Qy = o; the resultant of Xi and Xi acts toward the left and equals {W/g) {V/R)h, and F = | W. s r 1 68 Chap, ix 36. Moment of Inertia and Radius of Gyration §1. General Principles, Etc. — Perhaps every student has observed that the effort required to start a body to rotating about a fixed axis seems to depend not only on the mass of the body but also on the remoteness of the material of the body from the axis of rotation. Fig. 287 represents a simple apparatus by means of which one can roughly "sense " -^ this fact. It consists of a vertical shaft 5 to which a grooved pulley P and cross arm A are fastened rigidly, and a heavy body B which can be clamped on the cross arm. The pull or turning effort may be applied by means of a cord wrapped about the pulley. It is shown Fig. 287 .^ ^^^ following article that this "rotational inertia " of a body is proportional to the "moment of inertia " of the body about the axis, and this article is devoted to a discussion of moment of inertia, as a preparation for the following article.* The moment of inertia of a body with respect to a line is the sum of the prod- ucts obtained by multiplying the mass of each particle of the body by the square of its distance from the line. Or, if / = moment of inertia, mi, mi, m^, etc. = the masses of the particles, and n, r^, r^, etc., their distances respectively from the line or axis, then / = miTx^ + nhr2^ + • • • = Swr^; or if the body is continuous, then /= jdM-r^, (i) where dM denotes the mass of any elementary portion and r its distance from the line about which moment of inertia is taken. The elementary portion must be chosen so that each point of it is equally distant from the line, else there is doubt as to what distance to take for r. It is plain from the foregoing formulas that a unit of moment of inertia de- pends upon the units of mass and distance used. There is no single-word name for any unit of moment of inertia. Each unit is described by stating the units of mass and distance involved in it, and in accordance with the "make-up " * Euler (1707-83) introduced the term "moment of inertia," and he explained its appro- priateness (in his "Theoria Motus Corporum SoHdorum," p. 167) somewhat as follows: The choice of the name, moment of inertia (Ger. tragheits-moment), is based on analogies in the equations of motion for translations and rotations. In a translation the acceleration is proportional directly to the "accelerating force" and inversely to the mass, or "inertia," of the moving body; and in a rotation the angular acceleration is proportional directly to the moment of the accelerating force and inversely to a quantity, Xmr~, depending on the mass or inertia. This quantity, to complete a similarity, we may call "moment of inertia." Then we have for translations and rotations respectively, linear acceleration = (force) /(inertia or mass); and angular acceleration = (moment of force)/(moment of inertia). Art. 36 169 of the unit. Thus, when the pound and the foot are used as units of mass and length respectively, then the unit of moment of inertia is called a pound-foot square; when the slug (about 32.2 pounds) and the foot are used, then the unit moment of inertia is called slug-foot square.* The moment of inertia of any right prism — cross section of any form — with respect to any line parallel to the axis of the prism can be computed in a special way, preferred by some. Thus if we take as elementary portion a filament of the prism parallel to the axis, then dM = {adA) 8 where a = the altitude of the prism, dA — the cross section of the filament, and 8 = density; and p 1= a8 I dAr\ (2) This integral (extending over the area of the cross section) is called the moment of inertia of the cross section about the line specified (see appendix B). Since a moment of inertia is one dimension in mass and two in length, it can be expressed as the product of a mass and a length squared; it is sometimes convenient to so express it. The radius of gyration of a body with respect to a line is such a length whose square multiplied by the mass of the body equals the momxent of inertia of the body with respect to that line. That is, if k and / denote the radius of gyration and moment of inertia of the body with respect to any axis and M = its mass, then ¥M = I or k= y/JjM. (3) The radius of gyration may be viewed as follows : If we imagine all the material of a body concentrated into a point so located that the moment of inertia of the material point about the line in question equals the moment of inertia of the body about that line, then the distance between the line and the point equals the radius of gyration of the body about that line. The material point is sometimes called the center of gyration of the body for the particular line. To furnish still another view of radius of gyration we call attention to the fact that the square of the radius of gyration of a homogeneous body with respect to any line is the mean of the squares of the distances of all the equal elementary parts of the body from that line. For let r^, r^, etc., be the dis- tances from the elements, dM, to the axis, and let n denote their number (in- finite). Then the mean of the squares is W + rs^ -f •••)/« = {n^dM + r^^dM + - • • )/ndM = I/M = k\ Obviously the radius of gyration of a body with respect to a line is intermediate between the distances from the line to the nearest and most remote particle of the body. This fact will assist in estimating the radius of gyration of a body. Examples. — i. Required to show that the moment of inertia of a slender rod about a line through the center and inclined at an angle with the rod is -x^MP sin^ a, where M = mass, I = length, and a = angle between the line and the axis of the rod. Let a — the cross section of the rod, 5 = the density, and * For dimensions of a unit of moment of inertia, see Appendix A. lyo Chap, ix X = the distance of any elementary portion from the middle of the rod AB (Fig. 288). Then dM = 5 {a dx), and the distance of the element from CD = X sin a. Hence I = I 8adx' x^ sm^ a = 8a sm^ a — = ■ ' J-U ISA-u 3 4 and this reduces to yV MP sin^ a, since 8al = M. Fig. 288 Fig. 289 2. Required to show that the moment of inertia of a right parallelopiped about a central axis parallel to an edge equals yV M (a- + b-) where M = mass of the parallelopiped and a and b = the lengths of the edges which are perpen- dicular to that axis. See Fig. 289 where the s axis is the one to which this moment of inertia corresponds. We take for elementary portion a volume dxdydz; its mass = 8 (dxdydz), and the square of its distance from the z axis = x^ +3'^. Hence I I (x2 + v2) dx dy dz=— {a?b + a¥) = etc. •a/2 J -b/2 Jo 12 3, Required to show that the moment of inertia of a right circular cylinder with respect to its axis is | Mr"^, where M = its mass and r = the radius of its base. We use the special method for prisms (see equation 2) and choose polar coordinates (see Fig. 290) ; then dA = pdddp and dM = 8 {ap dd dp) ; hence I=a8 fdAp' = a8 r P^^ dp , '" ''^'^' "" 'de = etc. Fig. 290 Fig. 291 4. Required to show that the moment of inertia of a sphere about a diameter is I Mr^ where M = its mass and r = its radius. We might begin with equation (i) but we will use a special method, making use of the result found in exam- ple 3. We conceive the sphere made of laminas perpendicular to the diameter in question; determine the moment of inertia of each lamina; and then add the moments of inertia of the laminas. Let XX' (Fig. 291) be the diameter in Art. 36 171 question, and PQ a section of one of the laminas; then the mass of the lamina is b (iry'^dx). According to example 3 the moment of inertia of this lamina (cylinder) about its axis (XX') is ^ 5 (iry^ dx)y^. Hence the moment of inertia of the sphere is J A 5 (tt/ dx) = ^ ttS J_ (r2 - ^2)2 dx = j% dw r^ = etc. § 2. Parallel Axis Theorem; Reduction or Transformation Formu- las. — There is a simple relation between the moments of inertia (and the radii of gyration) of a body with respect to parallel lines one of which passes through the mass-center of the body. By means of this relation we can simplify many calculations of moment of inertia, and avoid integrations (see examples following); it may be stated as follows: The moment of inertia of a body with respect to any line equals its moment of inertia with respect to a parallel line passing through the mass-center plus the product of the mass of the body and the square of the distance between the lines. Or, if / = the first moment of inertia, 7 = the second (for the line through the mass-center), M = mass, and d = the distance between the parallel lines, I = l+Md\ (4) Proof. — Let (Fig. 292) be the mass-center, and P any other point of the body (not shown), LL the line about which the mo- ment of inertia is /, and OZ a parallel line (through the mass-center) about which the moment of inertia is /. Distance between these parallel lines is d. For convenience we take x and y axes through O, the former in the plane of the two parallel lines and the latter perpendicular to that plane. Let x, y, and z = the coordinates of P. The square of the dis- tance of P from the z axis equals x'^ + y'^, hence 7= I dM (x"^ -\- y^) . The square of the distance of P from the line LL equals (d — x)^ + y-, hence / = J[{d - xy+y']dM = Jix^ + y')dM + d'JdM -2dJxdM. Now the first of the last three integrals = /, and the second one = Md"^. If now we show that the third = o, then formula (4) is proved. The third integral is proportional to the moment of the body with respect to the yz plane; but this plane contains the mass-center, and hence the moment equals zero (Arts. 21 and 23). Thus, if Tf = weight of the body, Fig. 292 JxdM= CxdW/g = (i/g)Wx. If we divide both sides of equation (4) by M, we get I/M = I/M -f d^, or k^ = k^-\- d^ (5) 172 Chap. IX that is, the square of the radius of gyration of a body with respect to any line equals the square of Us radius of gyration with respect to a parallel line passing through the mass-center plus the square of the distance between the two lines. According to (5) k is always greater than d; that is, the radius of gyration of a body with respect to a line is always greater than the distance from the line to the center of gravity of the body. But, if the dimensions of the cross sections of the body perpendicular to the line in question are small com- pared to d, then k/d is small compared to i , and k equals d approximately (see example 2). In such a case the moment of inertia is approximately equal to Md\ Examples. — i. Required the moment of inertia of a prism of cast iron (weighing 450 pounds per cubic foot) 6 inches X 9 inches X 3 feet with respect to one of the long edges. The block weighs 507 pounds. According to example 2, § I, the moment of inertia of the block with respect to the line through the mass-center parallel to the long edge is 507 (6^ + 9-) -^- 12 = 4940 pounds- inches^. The square of the distance from a long edge to the mass-center = 29.25 inches^; hence the moment of inertia desired = 4940 -f 507 X 29.25 = 19,760 pound-inches^ = 4.27 slug-feet-. 2. Required the radius of gyration of a round steel rod i inch in diameter with respect to a line 1 2 inches from the axis of the rod. According to example 3, § I, the square of the radius of gyration of the rod with respect to its axis is 5 0.5^ = 0.125 inches^. According to equation (5) the radius of gyration desired = V (0.125 + 144) = 12.01, nearly the same as the distance from the line of reference to the mass-center of the rod. 3. It is required to show that the moment of inertia of a right circular cone with respect to a line through the apex and parallel to the base = j\ M {r^ + 4 c^) where Af — mass of the cone, r = radius of its base, and a = its altitude. We conceive the cone as made of laminas parallel to the base, find the moment of inertia of each lamina with respect to the specified line, and then add all the moments. For con- venience we take the axis of the cone as the ^'-coordinate axis, and the line for which the moment of inertia is required as the X axis (Fig. 293). The moment of inertia of the lamina Fig. 293 indicated about a diameter is j dM • x"^ where dM = the mass of the lamina and x = its radius. Hence its moment of inertia about the x axis = J dMx"^ + dMy"^ (see equation 3), and the moment of inertia of the en- tire cone = / {\dMx^ + dMy^), the limits being assigned so as to include all laminas. We choose to integrate with respect to y, and so must express dM and X in terms of y. From similar triangles in the figure x/y = r/a, or X = ry/a; obviously dM = dirx^dy = 8t (r^y^/a^) dy where 5 = density. Hence C Trr^hfdy , C'^Trr^v^dv irr^Sa , irr^a^ I = I ^^ -f / V-^ = 1 — - = etc. Jo 4 a* Jo a 20 s l«— r -> Y *-x^/ a \ 1 \ /P / ^ / 1 X Aet. 36 173 Composite Body. — By this term is meant a body which one naturally con- ceives as consisting of finite parts, for example, a flywheel which consists of a hub, several spokes, and a rim. The moment of inertia of such a body with respect to any line can be computed by adding the moments of inertia of all the component parts with respect to that same line. The radius of gyration of a composite body does not equal the sum of the radii of gyration of the component parts. It can be determined from equation (3), where / = moment of inertia of the whole body and M = its mass. § 3. Radius of Gyration of Some Homogeneous Bodies. — Let k = radius of gyration, a subscript with k referring to the axis with respect to which k is taken; thus kx means radius of gyration with respect to the x axis. Also M = mass and 8 = density. Straight Slender Rod. — Let / = its length, a = angle between the rod and the axis. Then about an axis through the mass-center k^ = tV l^ sin^ a; about an axis through one end of the rod k^ = ^ P sin^ a. Slender Rod Bent into a Circular Arc (Fig. 294). — Let r = radius of the arc, then kx- ^ I ''" [i — (sin a cos a)/a\, and ky"^ = | r^ [i + (sin a cos a) /a]. The divisor a must be expressed in radians (i degree = 0.0175 radians). j^2 = ^2 (^^j^g 2 axis is through O and perpendicular to the plane of the arc). 0-H^ P'iG. 295 Fig. 296 Fig. 297 Right Paralleloplped (Fig. 295).— The axis OX contains the mass-center, and is parallel to the edge c; kx^ — tV (^^ + ^")- Right Circular Cylinder (Fig. 296). — Both axes OX and OY contain the mass-center, r = radius and a = altitude; then kx'^hr^ V = tV (3 '-' + «')• Hollow Right Circular Cylinder (Fig. 297). — Let R = outer radius, r = inner radius, and a = altitude; then ^^^.^ iY kx' = h(R' + r'); ky'^ = HR' + r^~ + U')- I Right Rectangular Pyramid (Fig. 298). — The x axis ^ contains the mass-center and is parallel to the edge a; ^__^_—^,- M = I ahhb. Fig. 298 iy4 Chap, ix Right Circular Cone (Fig. 299). — The x axis contains the mass-center, and is parallel to the base; M = ^ irr'^ad. Frustum of a Cone. — Let R = radius of larger base, r = radius of smaller base, and a = altitude. For the axis of the frustum sphere. — Let r = radius. For a diameter k^=ir^; I^^Sirr^b. ■z Fig. 299 Fig. 300 Hollow Sphere. — Let R = outer and r = inner radius. For a diameter yfe2= |(^5_^)_^ (/23_^). I ^ ^^^Tr{R'-r>)8. Ellipsoid. — Let 2 a, 2 b, and 2 c = length of axes. For the axis whose length = 2 c, k^ = l (a2 _|_ j2) . / = ^^ Trahcb (a2 + b""). Paraboloid Generated by Revolving a Parabola about its Axis. — Let r = radius of base and h = its height. For the axis of revolution Ring (Fig. 300). — The x axis contains the mass-center and is parallel to the plane of the ring; the y axis is the axis of the ring. K''-hR'+l r-; h = Tf'Rr'd (R' + I r'). ky' = 7^2 _|_ I ^.2. 7^ ^ 2 ^2J^r2^ (7^2 _^ I ^2), § 4. Experimental Determination of Moment of Inertia. — When the body is so irregular in shape that the moment of inertia desired cannot be computed easily, then an experimental method may be simpler. There are several experimental methods available. By Gravity Pendulum. — This method is available if the body can be suspended and oscillated, like a pendulum, about an axis coinciding with or parallel to the line with respect to which the moment of inertia is desired. Let T = the time of one complete (to and fro) oscillation, c = distance from the mass-center to the axis of suspension, W = weight of the pendulum, g = acceleration due to gravity, k = radius of gyration, and I = moment of inertia about the axis of suspension; then T — VcW k= — Vcg and / = 5- • (i) 2 TT 4 TT'^ Art. 36 ^75 Above formulas are based on t he fo rmula for the time of oscillation or period of a pendulum T = 2tv Vk^cg (see Art. 39). If the axis of suspen- sion does not coincide with the Hne about which the moment of inertia is desired, then it remains to ''transfer" / to that Hne (see § 2). The desired moment of inertia can be determined without any time obser- vation as follows: From the same axis about which the suspended body oscillates suspend a " mathematical pendulum," a very small bob with cord suspension (see Art. 39) ; adjust the length of the cord so that the periods (times of oscillation) of bob and body become equal; then k = Vd, and / = Wd/g, . (2) where / = the distance from the center of the bob to the axis of suspension and k, W, c, I have the same meaning as above. The foregoing result is based on the fact that k^/c (for the pendulum) equals the length / of the mathematical pendulum (see Art. 39). By Torsion Potdulum. — The torsion pendulum here referred to would consist of an elastic wire suspended in a vertical position, the lower end being fashioned or terminated in a disk so that objects, whose moments of inertia are to be determined, may be suspended on the wire and made to oscillate about its axis. Let / = the (observed) period (time of one oscilla- tion) of the bare pendulum, h = the (observed) period of the pendulum when it is loaded with a body A which is so regular i.\ shape (as a cube or cyUnder) that its moment of inertia about the axis of oscillation can be computed easily, and h = the (observed) period of the pendulum when it is loaded with the body B whose moment of inertia is desired; further let h = the (com- puted) moment of inertia of A and I2 = the m.oment B about the axis of suspension. B should be suspended so that the axis of suspension coincides with or is parallel to the line (of B) about which the moment of inertia is desired. Then l2=h{k-t)^{h-t). (3) This result is based on the fact that the square of the period of a torsion pendulum is proportional to the moment of inertia of the pendulum with respect to the axis of oscillation. Thus, if / = the moment of inertia of the bare pendulum, and C the proportionality factor, then f = CI, k^ = C (/ + /i), and /o- = C (/ + A). These three equations may be combined so as to eliminate C and I and thus give equation (3). If B cannot be suspended so as to make the axis of oscillation and the line (of B) about which the moment of inertia of B is desired coincident, then it remains to reduce, or transform, h to that line (see § 20 of this article). iy6 Chap, ix 37. Rotation § I. A rotation is such a motion of a rigid body that one line of the body or of the extension of the body remains fixed. The fixed line is the axis of the rotation. The motion of the flywheel of a stationary engine is one of rotatiorv^ and the axis of rotation is the axis of the shaft on which the wheel is mounted; the motion of an ordinary clock pendulum is one of rotation, and the axis of rotation is the horizontal line through the point of support and perpendicular to the axis of the pendulum. Obviously all points of a rotating body, except those on the axis if any, describe circles whose centers are in the axis and whose planes are perpendicular to the axis. The plane in which the mass-center of the body moves will be called the plane of the rotation, and the intersection of the axis of rotation and the plane of rotation will be called center of rotation. All points of the body on any line parallel to the axis move alike; hence the motion of the projection of the line on the plane of the motion represents that of all the points, and the motion of the body itself is represented by the motion of its projection. By angular displacement of a rotating body during any time interval is meant the angle described during that interval by any line of the body perpen- dicular to the axis of rotation. Obviously all such lines de- scribe equal angles in the same interval, and we select a line which cuts the axis. Let the irregular outline (Fig. 301) rep- resent a rotating body, the plane of rotation being that of the paper, and the center of rotation. Let P be any point and 6 the angle XOP, OX being any fixed line of reference. Fig. 301 ^g customarily, 6 is regarded as positive or negative according as OX when turned about toward OP moves counter clockwise or clockwise. If di and 6i denote initial and final values of 6 corresponding to any rotation, then the angular displacement = Oo — 6\. The angular velocity of a rotating body is the time-rate at which its angular displacement occurs; or, otherwise stated, it is the time-rate at which any line of the body perpendicular to the axis describes angle. The time-rate at which OP describes angle, or the time-rate (of change) of 6 is dd/dt (see Art. 28, Note). Hence, if co denotes angular velocity, CO = de/dt. (i) Any angular displacement divided by the duration of that displacement gives the average angular velocity for that duration or interval of time. If the body is rotating uniformly (describing equal angles in all equal intervals of time), then the average velocity is also the actual velocity. The formulas for angular velocity imply as wm/* the angular velocity of a body rotating uniformly and making a unit angular displacement in each unit time. There are several such units; thus, one revolution per minute, one degree per hour, one radian per second, etc. The last is the one usually used * For dimensions of units of angular velocity and acceleration, see Appendix A, Art. 37 177 herein. An angular velocity must be regarded as having sign, the same as that of dd/dt. Since dd/dt is positive or negative according as d increases or de- creases algebraically, the angular velocity of a rotating body at any instant is positive or negative according as it is turning in the counter clockwise or clockwise direction at that time. The angular acceleration of a rotating body is the time-rate (of change) of its angular velocity. If, as in the preceding, co denotes the angular velocity, then the general expression for the time-rate of the angular velocity is do:/dt; hence if a denotes the angular acceleration, a = dio/di = d^d/dtK (2) The change in angular velocity which takes place during any interval of time divided by the length of the interval gives the average angular acceleration for that interval. If the velocity changes uniformly, then this average accelera- tion is also the actual acceleration. The foregoing formulas imply as unit * the angular acceleration of a body whose angular velocity is changing uniformly and so that unit angular velocity- change occurs in each unit time. One revolution per second per second, one radian per second per second, etc., are such units. An angular acceleration must be regarded as having sign, — the same as that of dw/dt. Since dw/dt is positive or negative according as co increases or decreases algebraically, an angular acceleration is positive or negative according as the angular velocity is increasing or decreasing (algebraically). There are simple relations between the linear velocity v and linear acceleration a of any point P of a rotating body and the angular velocity and acceleration of the body. Let r — the distance of P from the axis of rotation, s = distance travelled by P in any time from some fixed point in the path of P, and 6 — the angle described by the radius to P in that same time. Then s =^ rd ii dhe expressed in radians; ds/dt — r dd/dt, or V — roi. Differentiating again, we find that dv/dt = r do^/dt, or at = ra; also «„ (= v'^/r) = rco^. Here at and a„ mean the tangential and normal components of the acceleration of P (Art. 34). § 2. Equation of Motion. — We have already called attention to the fact (Art. 36, footnote) that in the case of rotation the angular acceleration is proportional to the algebraic sum of the moments of all the external forces acting on the body directly and to the moment of inertia of the body inversely, both moments being about the axis of rotation. Or, if To and Iq be used to denote these moments, and a = the angular acceleration, then a is proportional to {Tq -^ Iq) ; and, if systematic units (Art. 4) be used then To = ha = Mko'a, (3) * For dimensions of units of angular velocity and acceleration, see Appendix A. 178 Chap, ix where M = mass of the body and ko = its radius of gyration about the axis of rotation. If W/g be written for M (Art. 4, § 2), then any unit of force (in- cluding W), any unit of length, and any unit of time may be used in (3). The foregoing is called the equation of motion for a rotation; it may be de- rived from a consideration of the torque, about the axis of rotation, of all the forces acting on each particle of the body. Let P' (Fig. 302) represent a particle of the rotating body not shown, m' = its mass, and a' = its acceleration. Then the resultant of all the forces acting on P' = m'a' , and the tangential, normal, and axial components of this force are m'o/, m'an , and o respectively. Similarly the tangential, normal, and axial components of the resultant of all the forces acting on the Fig. 302 second particle P" are m"at", m"an", and o. All the radial or normal components are directed toward the axis of rotation, and all the tangential components clockwise or counter clockwise. Now the torque of all the forces acting on P' equals the torque of m'at and m'an ; this torque = m'at'r' . Similarly the torque of all the forces acting on P" = m"ai"r" . Hence the torque of the forces acting on all the particles equals m'at'r' + m"at"r" + • • • = m'r'ar' + m"r"ar" + • • • = cvSwr^ = ah. Now the system of forces acting on all the particles consists of internal and external forces. The internal forces jointly have no torque since they consist of pairs of coUnear, equal, and opposite forces. Hence, the torque of the ex- ternal forces equals Ua. Examples. — i . Fig. 303 represents a circular disk of cast iron 4 inches thick and 3 feet in diameter. It is supported on a fixed horizontal shaft 3 inches in diameter. A cord is wrapped around the disk, and then a pull P = ^...j'_.^ 100 pounds is applied to the cord as shown. What is the angular acceleration of the disk? The external forces acting on the disk and cord are the weight of the disk and cord P, and the reaction of the shaft. Only one of these, P, has a moment about the axis of rota- tion. We are assuming that the disk is homogeneous so that the center of gravity is in the axis of rotation, and that the shaft is ' '^°"^ frictionless. Tq of equation (3) is therefore 100 X 1.5 = 150 foot-pounds. Now the square of the radius of gyration of the disk about the axis of rotation is I (1.52 + 0.125^) = 1. 133 feet^ (Art. 36). And since the weight of the disk is .T053 pounds, its moment of inertia about the axis of rotation is (1053 -^ 32.2) 1.J33 = 37-0 slug-feet^. Hence the angular acceleration of the disk is 150 -^ 37.0 = 4.0 radians per second per second. 2. Suppose that a turning force P in the preceding example is supplied not "by hand " but by means of a body suspended from the cord, and suppose that the body weighs 100 pounds. Obviously the system (disk and suspended body) moves with acceleration ; hence the two forces acting on the body (gravity and the pull P of the cord) are not equal or balanced but have a resultant Art. 37 179 downward (direction of the acceleration of the body). That resultant is 100 — P, and it equals the product of the mass and acceleration of the body, or 100 — P = (100 -V- 32.2) X a where a = the acceleration. The torque on the disk is P X i-S, and 1.5 P = /a = 37.0 a. But a = the tangential acceleration of any point on the rim of the disk = 1.5 X a, or a = 1.5 a. These three equations 100 - P = (100/32.2) a, 1.5 P = 37.0 a, and a =1.5 a, solved simultaneously give a = 3.41 radians per second per second, less than in example i as was to be expected, because the pull P in this example is less than 100 pounds. The value of P as obtained from the foregoing equations is 84.1 pounds. 3. In Fig. 304 we take weight of A = 64 pounds, of P = 96 pounds, and of pulley C = 144 pounds; assume coefficient of friction under B = I ior sliding, axle friction zero; take diameter of pulley = 2 feet 6 inches, and the radius of gyration of the pulley about the axis of rotation = 10.6 inches. We show how to determine the acceleration of the system. Let a = acceleration of A and B, and a = (angular) acceleration of the pulley. Obviously a = 1.25 a. Let us now consider the forces acting on each body A, B, and C. On A there are two, — gravity (64 pounds) and the pull of the cord Pi (see Fig. 305). On Fig. 304 96 /bs. B ¥ F Fig. 306 Fig. 307 B there are three, — gravity (96 pounds), the pull of the cord P2, and the re- action of the supporting surface D (see Fig. 306 where this latter force is represented by two components N and F). On the pulley there are three forces, — gravity (144 pounds), the reaction Q of the axle, and the pressure of the cord. Since the mass of. the cord is negligible, the tension at any point of the cord from A to the pulley is Pi, and at any point from B to the pulley it is P2. Hence the pressure of the cord against the pulley equals the resultant of Pi and P2 (Fig. 307), and that pressure is equivalent to Pi and P2. Therefore the equation of motion becomes (Pi - P2) 1.25 = (i44 ^ 32) (10.6 -^ i2ya = 4.5 X 0.778 X a = 3.5 a. Since the acceleration of B is toward the right, the resultant force on it acts in that direction and equals P2 - P = P2 - ^ N = P2 - ^, 96 = Po - 19.2; and hence P2 - 19-2 = (96 "^ 32) and 64 - Pi = 2 a, l8o Chap, ix together with a = 1.25 a, we j&nd that a = 6.19 feet per second per second, and a = 4.95 radians per second per second. The equations also show that Pi = 51.62 pounds, and P2 = 37.77 pounds. 38. Axle Reactions § I. Simple Cases. — Rotating bodies are commonly supported by shafts upon or with which the bodies rotate. In such a case, axle reaction means the force which the shaft exerts upon the rotating body. To determine such a force we make use of the principle of the motion of the mass-center. The principle states (Art. 34) that the algebraic sum of the components — along any line — of all the external forces acting on a body, moving in any way, equals the product of the mass of the body and the component of the accelera- tion of the mass-center along that line. In general, the principle furnishes three independent equations, one for each of three rectangular lines of resolu- tion. If the mass-center of the (rotating) body does not lie in the axis of rota- tion then there are three lines of resolution which are generally more convenient to use than any others, and these we now describe. Let the circle (Fig. 30S) be the path of the mass-center of a rotating body (not shown), O be the center of rotation (intersection of the axis of rotation and plane of the path of the mass-center), and C be the mass-center. Then the three convenient lines are the axis of rotation, the line OC, and a line perpendicular to the first two. The directions of these lines are called re- FiG. 30S spectively axial, radial or normal (OC being a radius and normal of the circle), and tangential (the third line being parallel to the tan- gent at C). Now let I1F<,I!F„, and ^Fa = the algebraic sums of the tangen- tial, normal, and axial components of all the external forces acting on any rotating body; at and a„ = the tangential and normal components of the ac- celeration of its mass-center — the axial component of the acceleration equals zero; and M = the mass. Then 2F, = Alat, :SF„ = Man, 2F, = o. (i) Systematic units (Art. 31) must be used in the foregoing. If W/g be substituted for M (Art. 31) then any unit may be used for force (including weight), any unit for length, and any unit for time. Let r = radius of the circle described by the mass-center, v = velocity of the mass-center, a = angular acceleration, and co = angular velocity of the rotating body at the instant under consideration; then (see Art. 37, §1) at = ra, and a„ — v'^/r = rco^, and we may use these in equations (i). If the mass-center of the rotating body is in the axis of rotation, then the Art. 38 181 Fig. 309 acceleration of the mass-center is always zero, and the algebraic sum of the components of the external forces along any line equals zero. Examples. — i. AB (Fig. 309) is a bar of wrought iron 1.5 inches (perpen- dicular to paper) X 4 inches X 6 feet, suspended from a horizontal axis at A. Suppose that the bar is made to rotate and is then left to it- r, self rotating under the influence of gravity, the axle reaction, and the initial velocity given to it. Suppose further that the initial velocity was such that when the bar gets into the position shown, the angular velocity is 60 revolutions per minute. Required the axle reaction in the position shown. The only forces acting on the body are its weight W = 120 pounds, and the axle reaction represented by two components Ri and R^. We neglect the axle friction; then the lines of actions of Ri and R-i cut the axis of rotation, and the equation of motion (Art. 37) becomes IF (2 sin 35°) = la. Now I = (IF/32.2) ^2 = (120/32.2) 7.01; hence a = 5.26 radians per second per second, and Cj = 2 X 5.26 = 10.52 feet per second per second. The angular velocity, 60 revolutions per minute, equals 6.28 radians per second; hence a„ = 2 X 6.28^ = 78.8 feet per second per second. Finally, equations (i) become 120 sin 35° — Ri= (120/32.2) 10.52 = 39.2, and R2 — 120 cos 35° = (120/32.2) 78.8 = 294. From the first Ri = 29.7, and from the second R2 = 392 pounds. 2. AB (Fig. 310) is a simple brake for retarding the motion of the drum C and suspended body IF. Let IF = 2000 pounds, weight of the drum = 1800 pounds, radius of gyration of drum about axis of rotation =2.5 feet, coeflBcient of Fig. 310 friction " between " brake and drum = 0.5. Suppose that IF is descending and the brake pull P is 1000 pounds. Required the axle reaction on the drum. Fig. 311 shows all the forces acting on the drum, — its own weight (1800 poimds), the brake pressure represented by two components N (normal pressure) and F (friction), the pull T of the rope, and the axle reaction represented by two components Ri and R2. From a consideration of the forces acting on the brake it is plain that N = (1000 X 6.5) -^ 1.5 = 4333 pounds; and hence F = 0.5 X 4333 = 2167 pounds. Now in order to get T we write out the equations of motion of the drum and the suspended body. Since F is greater than the l82 Chap, ix weight of the body the velocities of drum and body are being decreased; hence T is greater than W but less than F. If a = the acceleration of the drum and a = the acceleration of the suspended body, then the equations of motion are 2167 y. 2) ~ T^ ^ i ^ (1800/32.2) 2.52a, and T — 2000 = (2000/32.2) a. These equations and a = T)^^^ solved simultaneously, give T = 2103 pounds. Since the acceleration of the mass-center of the drum equals zero, 2167 — Ri^ o, or Ri = 2167, and R2 — 4333 — 1800 — 2103 = o, or R2 = 8236 pounds. Therefore the axle reaction = V{2i6'j'^ + 8237') = 8500 pounds inclined up- wards and to the left at an angle of 14I degrees with the vertical. § 2. Non-simple Cases. — Axle reactions cannot be determined by means of equations (i) in some cases; moment equations must be resorted to. The moment equation To = ha (Art. 37) is available for all cases but additional ones may be needed. It will be recalled that To is the torque of all the external forces about the axis of rotation; we will presently deduce expressions for the torque about other lines, — first for a body of any shape and then for bodies of certain common, symmetrical shapes. Body of any Shape. — Let the axis of rotation be taken as the z axis of an x-y-z coordinate frame, the x axis containing the mass-center on the positive k>x4 I < S Q . X I / Mrcc Z t| ' symmetry Z 'Mru)' Fig. 312 Fig. 313 Fig. 314 Fig. 315 side of the axis. The following six equations apply to this general case: (i) 2F, = - MFco2, (2) llFy = M?a, T,= -a fzx dM + aj2 Cyz dM, (4) Ty= - a I yzdM -or \ zx dM, (5) (3) 2F. = 0, T,= ha = Mk'^a. (6) Equations (i), (2), and (3) follow from Art. 38 (page 180); (6) is equation (3) of Art. 37, the notation being changed to agree more appropriately with Fig. 312; (4) and (5) will be deduced immediately. The method used for arriving at (4) and (5) is like that used to get (6). Art. 38 183 The moment of the resultant of all the forces acting on each particle is com- puted; then such moments for all the particles are summed; and since the internal forces contribute nothing to it, this sum is the value of the torque of the external forces about the axis of moments being used. Let P (Fig. 312) represent any particle of the rotating body, r the radius of the circle described by P, e the varying angle PAB, (x, y, z) the coordinates of P, m its mass, and a its acceleration. The resultant of all the forces acting on the particle P equals ma. Components of this resultant along and perpendicular to PA equal mrur and mra. The first component acts in the (radial) direction PA , and the second in the direction of the tangential component of a. The moments of the resultant ma about the x and y axes respectively equal the sums of the moments of its components; these sums are — mra cos 6s + mror sin 6z — — mazx + mcJ'yz, and — mra sin dz — mroi^ cos 6z = — mayz — moihx. If now we add all such moments about the x axis, and then those about the y axis, for all the particles comprising the body, we arrive at — aLmzx-{- orllmyz, and —allmyz — (j?'Lmzx\ and these reduce, for a continuous body, to the right-hand members of (4) and (5). (i) The body is homogeneous, has a plane of symmetry, and rotates about an axis perpendicular to that plane. In this case, (4) and (5) become Tx = o and Ty = o. (4') (5') For: — The xy plane (Fig. 312) now coincides with the plane of symmetry; hence, for any particle whose coordinates are (.v, y, 2) there is a corresponding one whose coordinates are (x, y, —z). And because of the (assumed) homo- geneity of the body the masses of the particles may be taken equal. It fol- lows that Swsx = o and Hmyz = o for the two particles. Therefore, these summations extended to all the pairs of particles comprising the body equal zero, and hence Tx = Ty = o SiS stated. Let Fig. 313 represent the plane of the symmetry section of the body, C the mass-center, the center of rotation, Q a point on OC extended so that OQ = k-/r. In general, the resultant of the external forces is a single force acting in the xy plane, and through Q. The components of the resultant along and perpendicular to OC are indicated in the figure; the first always acts from Q toward and the second in the direction of the tangential accel- eration of C. Proof of this statement is left for the student to supply. The fact that the described components satisfy equations (i), (2), (3), (4'), (5'), and (6) may be regarded as sufficient proof. When the angular velocity is constant (a = 6), the resultant of the external forces is a force directed along the radius CO and in that direction; its value is Mroi^. When the mass-center is in the axis of rotation (r = o), the re- sultant is a couple; its plane is perpendicular to the axis of rotation and its moment is /««. When both a and the r = o, the resultant is nil. 184 Chap- IX (ii) The body is homogeneous, has a line of symmetry, and rotates about an axis parallel to that line. In this case, (4) and (5) become, as in (i), Tx = o and T, = o (4") (5") For: — Let Pi and P2 (Fig. 314) be any symmetrical pair of particles (so that the line P1P2 is perpendicular to the axis of symmetry and is bisected by that axis). Also let the coordinates of Pi and P2 be (xiyiZi) and (x2}'2Z2) respectively. Evidently these coordinates are related as follows: h (xi + X2) = r, yi= - y2, and Zi = Z2. On account of homogeneity, the masses, nii and m^, of the Pi and P2 may be taken equal; hence for the two particles Hmyz = niiyiZi + m2y2Z2 = wiyiZi — miyiZi = o, and Smzx = miZiXi + ^W2Z2a-2 = miZiXi + WiZi(2 r — Xi) = 2 miZi?. It follows that, if the summations be extended to all the pairs of particles constituting the body, then fyz dM = o, and jzx dM = rjz dM = rMz (see Art. 23), where s denotes the 2 coordinate of the mass-center; but the mass-center is in the XOY plane, and hence i = o. Thus finally, we see from (4), (5), and the foregoing results that T^, and T^ = o as stated. All the remarks under case (i) about the resultant of the external forces hold in this case. (iii) The body is homogeneous, has a plane of symmetry, and rotates about an axis in that plane. In this case, (4) and (5) become r, = - afzx dM, and Ty = - '^""^zx dM. (4'") (5'") Yox: — The xz plane coincides with the plane of symmetry in this case; hence for any particle whose coordinates are (x, y, z) there is a corresponding one whose coordinates are {x, - y, z). It follows thatjjs dM^o, and, from (4) and (5), that T^ and Ty have the values stated above. In general, the resultant of the external forces is a single force as in cases (i) and (ii); but in this case, the resultant acts not in the xy plane but in a parallel plane, the z coordinate of which is jzxdM ^ Mr. See Fig. 315; Q' is the point where the Une of action of the resultant pierces the plane of symmetry. Proof of the foregoing statements is left for the student to supply. The fact that the described resultant satisfies (i), (2), (3), (4'"), (s'"), and (6) may be regarded as sufficient proof.* Centrifugal Action and Dynamic Balance. — It is common experience that a rotating body unless well "balanced" exerts forces upon its shaft — and thus upon the bearings supporting the shaft — which are due in part to the *For some remarks on special cases, see page i88b. Art. 39 185 velocity of rotation. Such parts or components of the forces are said also to be due to the "centrifugal action " of the rotating body; and the components are called " centrifugal " forces or pulls. A rotating machine part so shaped or loaded that it exerts no resultant centrifugal pull on its shaft or bearhigs is said to be in "running " or "dynamic balance." The common method of designing for running balance consists in arranging or proportioning the various (simple) parts of the rotating body, a motor crank shaft for example, so that the centrifugal pulls of all the various parts will neutralize.* Even after careful design and manufacture of a crank shaft, say, the running balance may be imperfect. Then mechanical methodsf are resorted to for completing the task. The conditions for dynamic balance can be stated with reference to equa- tions (i) to (6):— (a) The mass-center of the body must be in the axis of rotation {r= o); this insures "standing" or "static balance." (b) The "products of inertia" (see Art. 57) J yzdM &nd jzxdM must equal zero; this with (a) insures dynamic balance. For:— When these conditions are fulfilled and the angular velocity is constant, the right-hand members of equations (i) to (6) equal zero, that is they are equilibrium equations; hence, the bearing pressures are independent of the motion. | The axis of rotation of a body which is dynamically balanced for that axis is sometimes called a " free axis " of the body; for, if the body could be rotated about that axis and then left to itself entirely free from all external forces, even gravity, it would continue to rotate about that axis. 39. Pendulums § I. Gravity Pendulum. — By this term is meant the common pendulum, that is a body suspended on a horizontal axis so that it can be made to oscillate freely under the influence of gravity. A real pendulum is sometimes called a compound or physical pendulum to distinguish it from an imaginary one con- sisting of a mass-point or particle suspended by a massless cord; this latter is called a simple or mathematical pendulum. Let T = the period or time of one complete or double (to and fro) oscillation, k = the radius of gyration of the pendulum with respect to the axis of suspension, c = distance from the center of gravity of the pendulum to that axis, and 2 jS = the angle swept out by the pendulum in one single oscillation. Then, as will be shown presently, the period is given closely by T =2T Vk'/cg, (i) * For a good treatment of dynamic balancing, see the book by Dunkerly or Sharpe on Balancing of Engines. t For a description of an ingenious balancing machine, Akimofif's, see American Machinist for May 18, 1Q16. X For a theory of balancing of engines based on these conditions extended see Lorenz, Tecknische Mcchanik, Band I. 1 86 Chap, ix provided that j8 is small.* Since jS does not appear in this formula the period of any pendulum is independent of /3; that is all small oscillations of a pendulum have equal periods or, as we say, they are isochronous. When g is expressed in feet per second per second then k and c should be expressed in feet; T will be in seconds. For the derivation of equation (i) let OG (Fig. 316a) be a pendulum in any swinging position, O the center of suspension, G the center of gravity; let W = the weight of the pendulum, c = OG, and 6 the (varying) angle which OG makes with the vertical, regarded as positive when the pendulum is on the right side of the vertical, as shown. There are three forces acting on the pendulum, — gravity, the supporting force at the knife edge, and the pressure of the sur- rounding air. The moment of the first force about the axis of suspension is Wc sin 6; the moments of the other two forces we take as negligible. Hence the resultant torque on the pendulum in any position = Wc sin 6 practically. The angular acceleration = 'ds or Ftds where Fi means working or tangential component of F; and the work done by F in the displacement from A to B = ( Ft ds, limits of integration to be assigned so as to include all elementary works Ft ds in the motion from A to B. It is worth noting that if the force F acts normally to the path at all points, then Ft = o always, and the formula gives zero for the work done by F, as it should. 189 'A 190 Chap, x The unit work is the work done by a force whose working component equals unit force acting through unit distance.* The unit of work depends upon the units used for force and distance; thus we have the foot-pound, centimeter- dyne, etc. The second unit is also called erg; and 10^ ergs = i joule. The horse-power-hour and the watt-hour are larger units of work. They are the amounts of work done in one hour at the rates of one horse-power and one watt, respectively (see Art. 41); thus, One horse-power-hour = 1,980,000 foot-pounds, and One watt-hour = 3600 joules. When the works done by several forces are under discussion, it may be convenient to give signs to their works according to this commonly used rule: When the working component acts in the direction of motion, the work of the force is regarded as positive; when opposite to the direction of motion then the work is regarded as negative. The formula J F cos cf) ds, with the lower and upper limits of integration to correspond to the initial and final positions A and B, respectively, observes this rule of signs for work, if s is measured posi- tive in the direction of motion from some fixed origin to P, and 4> is measured from the " positive tangent" around to the line of action of the force as shown in the figure. Forces which do positive work are sometimes called efforts; those which do negative work are called resistances. "f Work Diagram. — If values of Ft and 5 be plotted on two rectangular axes (Fig. 319) for all positions of the point of application of F, then the curve joining the consecutive plotted points might be called a " work- ing force-space" (Ft-s) curve. The portion of the diagram "under the curve" (between the curve, the 5-axis, and any two ordinates) is called the work diagram for the force F for ^i»a->i " " '"I * the displacement corresponding to the bounding ordinates. „ The area of a work diagram represents the work done by the force during the displacement corresponding to the bounding ordinates. Proof: Let m = the force scale-number, and n = the space scale- number; that is, unit ordinate (inch) = m units of Ft (pounds) and unit abscissa (inch) = n units of 5 (feet). Also let A = area; then /»ij nhp^s I C^ r^ J work A= I ydx= I = — / Ftds= Jx, Ja m n mnja mn Hence, A {mn) = work; that is, A = work according to the scale number mn to be used for interpreting the area. By average working component of F is meant a value of Ft which multiplied by the distance 52 — Si, or b — a, gives the work done by F. Obviously, that average working force is represented by the average ordinate to the curve of * For dimensions of unit work, see Appendix A. t The (negative) work done by a resistance on a body is often referred to as (positive) work done by the body against the resistance. Art. 40 191 the work diagram. When that curve is straight, that is, when Ft varies uni- formly with respect to s, then the average working component equals the mean of the initial and final values. Fig. 320 is a fac-simile of a record made by the traction dynamometer (a spring balance essentially) in a certain train test. Abscissas represent distances travelled by the train, and ordinates represent " draw-bar pulls" (the pulls between the tender and first car of the train). Thus, the figure is a work diagram. To determine the area of such a diagram as this we first draw in an average curve "by eye," and then ascertain the area under this curve in any convenient way. \//^vMv***^^'^^VV'^^^ •3 tons 6 ins. = / Mile. Fig. 320 V V Fig. 322 § 2. Some Special Cases. — (i) The work done by a force which is con- stant in magnitude and direction equals the product of the force and the pro- jection of the displacement of its point of application upon the line of action of the force. For, let F = the force, APB (Fig. 321) the path of its point of application, ^ = the (variable) angle between ¥ and the direction of the motion of the point of application P. Then the work done by F is I F cos (j)ds = F j ds cos , where ds is an elementary portion of the path. Now ds cos is the projection of the element ds upon F, or upon any line parallel to F, and Cds cos <^ is the sum of the projections of all the elements of APB upon the line. But the sum of the projections = the projection of APB = the projection of the chord AB. (ii) The work done by gravity upon a body in any motion equals the product of its weight and the vertical distance described by the center of gravity; the work is positive or negative according as the center of gravity has descended or ascended. Let Wi, w^, etc., denote the weights of the particles of the body; yii y2 , etc., their distances above some datum plane — below which the body does not descend — at the beginning of motion; and ji" , jo", etc., their dis- tances above that plane at the end of the motion (see Fig. 322) where a' a" is the path of the first particle, b'b" that of the second, etc.). Also let W denote the weight of the body, and y' and y" the initial and final heights of its center of gravity above the plane. Then the works done by gravity on the particles , respectively, are Wi {yi — y/'), K'2 iyi — y^"), etc., and the sum of these works can be written {wiyi + W2y2 + . . : ) -{wiyi" + W2y-i' +...). 192 AP. X The first term of this sum = Wy', and the second = Wy" (see Art. 21); hence the sum of these works done on all the particles equals Wy' - Wy" = Wiy'- y"). (iii) The algebraic sum of the works done by any number of forces having a common point of application during any displacement of that point equals the work done by their resultant during that displacement. For, let F', F", F'", etc., = the forces, R = their resultant, and F/, Ft", F"', etc., and Rt = the components of the forces and of the resultant, respectively, along the tangent to the path of the point of application. Now Ft + F" + F"' + • • • = Rt (Art. 4). Hence Ft' ds + Ft" ds + F/" ds -\- - • - = Rtds, and fFt'ds^ fFt"ds-\- CFt'"ds-\- • • • = fRtds; that is, the sum of the works done by the forces equals the work done by their resultant. (iv) The work done by a pair of equal, colinear, and opposite forces in any displacement of their points of application equals X2 r^i Fdr or - I F dr according as the forces tend to separate or draw the points of application together; F = the common magnitude of the two forces — not constant necessarily — , r = the distance between their points of appHcation, and n and r2 = initial and final values of r. Let A and B (Fig. 323) be the points of application of the two forces — acting on a body ^ V not shown — at any intermediate stage of the y^- ^"_;:::^:^^^;^ displacement, and suppose that the path of A is ■|p^^-^ /By" ^i^^2 and that of B is BA. Let x',y' be the /y' y. coordinates of A, and x", y" those of B. (For simplicity in figure we have taken the paths of A Fig ^2^ V J ^ and B as coplanar. The following proof could be extended to cover the case of any paths. The paths are not necessarily due to the forces F alone; but since we are concerned with the work done by these two forces only, no mention is made of any other forces concerned with the motion.) According to the preceding paragraph the work done by either force F in any displacement equals the sum of the works done by the X and y components of F in that same displacement. Hence in an elementary displacement ds the work of F on ^ = (-F cos dx' - F sin 6 dy'), and the work oi F on B = {F cos 6 dx" -\- FsinO dy"). The work done by both forces F in the elementary displacement is F[cosd (dx" - dx') + sin 6 {dy" - dy')]. It will readily be seen from the figure that (x" - x')- + (y"- y')^ = r'; and, by differentiation, we find that (x"- x') {dx"- dx') + {y"- y') {dy" - dy') = rdr. Art. 41 195 Dividing by r and transforming we find that cos 6 {dx" - dx') + sin d {dy" - dy') = dr. Hence, the work done in the elementary displacement is F dr, and the work done in the displacement from AiBi to A1B2 equals the integral of F dr between the limits as stated. Obviously, changing the senses of the forces F (in the figure) so that they tend to make A and B approach, changes the sign of the total work done by the forces. 41. Energy When the state or condition of a body is such that it can do work against forces applied to it, the body is said to possess energy. For example, a stretched spring can do work against forces applied to it if they are such that it may contract, and a body in motion can do work against an applied force which tends to stop it; the spring and the body, therefore, possess energy. The amount of energy possessed by a body at any instant is the amount of work which it can do against applied forces while its state or condition changes from that of the instant to an assumed standard state or condition. The meaning of the standard condition is explained in subsequent articles. The unit of energy must, in accordance with the above, be the same as the unit of work. Thus we have the foot-pound, foot-ton, centimeter-dyne (or erg), the joule, horse-power-hour, watt-hour, etc. (see preceding article).* § I. Mechanical Energy. — Energ>^ is classified into kinds depending on the state or condition of the body, in virtue of which it has energy. Kinetic Energy of a body is energy which the body has by virtue of its velocity. The amount of kinetic energy possessed by a particle at any instant is the work which it can do while the velocity changes from its value at that instant to some other value taken as a standard. It is customary to take zero velocity as the standard one; this being understood, then the amount of kinetic energy possessed by a particle is the work which the particle can do in '' giving up all its velocity." The kinetic energy of a single particle whose mass and velocity are m and v, respectively, equals | mv'^. Proof: Let Fi, Fi, F3, etc., be the forces which act on the particle P (Fig. 324) and eventually stop it; and let AB he the path, A the beginning (where velocity = Vi) and B the end where velocity = o. Then we are to prove that the work done by the particle on the neighboring particles or bodies (which exert the forces Fi, F2, F3, etc.) equals | mvi^, during the motion. Now, the work done by the forces Fi, F2, F3, etc., on the particle is / Fi cos (f>ids ^ I Fa cos (f>2 ds +'' ' = / (Fi cos <^i -f- F2 cos 02 + • • • ) ds, * For dimensions of a unit of energy, see Appendix A. 194 Chap, x where 4>i, 02, 03, etc., are the angles between Fi. F2, F3, etc., and the direction of motion (Art. 40). Since the particle P exerts forces on its neighbors, equal and opposite to Fi, F2, etc., the work done by the particle on its neighbors is / (Fi cos 01 + F2 cos 02 + • • • ) ds. But Fi cos 01 + F2 cos 02 + • • • = Jnat = m dv/dt, where at is the tangential component of the acceleration of the particle; hence the work done by P is — / m (dv/dt) ds ^ — j m {ds/dt) dv = — j mv dv = ^ mv^. The kinetic energy of a body (a collection or system or particles) is the sum of the kinetic energies of the constituent particles of the body. We will now evaluate this sum for certain common cases, — namely, (i) translation, (ii) rotation, and (iii) combined translation and rotation. (i) In translatory motion all particles of the moving body have at each instant equal velocities; hence, the simi of the kinetic energies of the particles is \ ntiv- + I WoZ'- + • • • = 2 "^^ (2^0, where Wi, nh, etc., = the masses of the particles and v = their common velocity at the instant imder consideration. Or, if M = the mass of the body and E = energy, then E = ^Mv^ = i(W/g)v\ (i) If 32.2 is written for g, then v should be expressed in feet per second. E will be in foot-pounds, foot-tons, etc., according as W is expressed in pounds, or tons, etc. (ii) In a rotation about a fixed axis the velocity of any particle of the body equals the product of the angular velocity of the body, expressed in radians per unit time, and the distance from the particle to the axis of rotation (Art. 37). Hence, the sum of the kinetic energies of the particles of the body is I mi (ri co)2 -f i 1712 (r^f^y + • • • = | oj^ Zmr-, where co = the angular velocity of the body at the instant under consideration, and ri, ^2, etc.,= the distances of the particles respectively from the axis of rotation. But I,mr^ = the moment of inertia of the body about the axis of rotation; hence, the kinetic energy is given by £ = i /co2 = i Mk^o:' = h (W/g) kW', (2) where I = the moment of inertia described, and k = the radius of gyration of the body about the axis of rotation. If 32.2 is written for g, then k should be expressed in feet and co in radians per second (w — 2 -wn where n = revolu- tions per second). Then E will be in foot-pounds, foot-tons, etc., according as W is expressed in pounds, tons, etc. (iii) A body which has a combined plane translation and rotation (Art. 50), Uke a wheel rolling, has kinetic energy given by £ = i M^^ + Hco^ = I {W/g) v' + i (W/g) ¥oi\ (3) Art. 41 195 where M = mass of the body, v = velocity of the center of gravity, / = the moment of inertia of the body with respect to an axis through the center of gravity perpendicular to the plane of the motion, k = radius of gyration with respect to the same axis, and w = the angular velocity of the motion. Proof of this formula is given in Art. 51. The portions ^ Mv^ and | /co^ of the kinetic energy are sometimes called the translational and rotational components, respectively. As an example of the use of the preceding formula we find the kinetic energy of a cylindrical disk, 6 feet in diameter and 400 pounds in weight, which is rolling so that the center has a velocity of 4 feet per second. M = 400 -i- 32.2 = 12.4 slugs; the square of the radius of gyration of the disk is | 3^ = 4.5 feet^ (see Art. 36); and cxi, the rate at which the wheel is turning, is 4 -^3 radians per second. Hence £ = - 12.4 X 4" + - 12.4 X 4.5(4 -^ 3)'" = 148.8 foot-pounds. Potential Energy. — A body may possess energy which is not due to velocity. Thus two mutually attracting bodies can do work against forces applied to either or both if allowed to move so that they approach each other; and, as stated, a compressed or stretched spring can do work against applied forces if permitted to resmne its natural length. The "change of condition or state" in the first case is a change in configuration, a change in the positions of the bodies relative to each other; and, in the second case, if we conceive of the spring as consisting of discrete particles, the change is also one in configuration (of the particles). Energy of a system of particles dependent on configura- tion of the system is called energy of configuration, and potential energy more commonly. The amount of potential energy possessed by a system in any configuration is the work which it can do in passing from that configuration to any other taken as a standard, it being understood that no other change of condition takes place. The standard configuration may be chosen at pleasure, but it is convenient to so select it that in all other configurations considered the poten- tial energy is positive. A most common case of potential energy is that of the earth and an elevated body. In this case, standard configuration means one in which the body and earth are as near together as possible. Practically, it is necessary to regard the earth as fixed and the energy as resident in the elevated body. The amount of potential energy of an elevated body is just equal to the work which gravity would do upon the body during the descent into the standard or lowest position, and this work is given by Wh (see preceding article), where W = the weight of the body and h = the distance through wliich the center of gravity of the body can descend. § 2. Other Forms of Energy. — Kinetic energy and potential energy are often called mechanical energy. It is the opinion of some that all energy Iq6 Chap, x is mechanical, and some think that it is all kinetic. Whether either of these views be correct, it is practically necessary to recognize other forms. A mere enumeration of these with brief remarks is sufficient for the present purpose, since we shall deal mostly with energy known to be mechanical. Thermal Energy. — A hot body may do work under favorable conditions; thus, if such a one is placed in a boiler containing water, the water will be heated and a part may be converted into steam which may drive a steam engine, that is, do work. By giving up its heat the hot body has done work, and, hence, by definition, it possessed energy in its heated state. Not only is this fact well known, but also the fact that a given quantity of heat represents a definite amount of energy. Thus, one British thermal unit (B.T.U.), which is the amount of heat required to raise the temperature of one pound of water one Fahrenheit degree, = 778 foot-pounds. And one (small) calorie, which is the amount of heat required to raise the temperature of one gram of water one Centigrade degree, = 4.187 X 10^ ergs (at 15 degrees). Based on the molecular hy^Dothesis the common theory is that heat is due to the vibratory motion of molecules, that is, thermal energy is kinetic. Chemical Energy. — Many substances combine chemically, and their com- bination gives evidence that they possessed energy. Thus, coal and oxygen combine and produce heat which, as we have seen, is a form of energy. We rightly say, therefore, that the coal and oxygen before combination possessed energy. Based on this molecular hyj^othesis the theory of chemical energy in cases where heat is generated in the chemical combination is that internal (molecular) forces of the substances do work during the combination, and, hence (see Art. 43), increase the kinetic energy of the molecules. According to this explanation the energy before combination is potential; and after, kinetic. Electrical Energy. — If a charged storage battery be connected with a motor, work may be done by the latter. As the work is done, the electrical condition of the battery changes, and we therefore ascribe the energy to the batter}'. The energy is called electrical because it is due to a change of electrical condi- tion. The nature of electrical energy is even less understood than that of thermal energy, and no commonly accepted explanation of it has yet been made^ 42. Power § I . In common parlance the word power has many meanings (see diction- ary). Thus we hear of the power of a giant, power of example, power of the press, etc. And of things mechanical, we hear such expressions as a powerful derrick, a powerful cannon, a powerful pump, etc. On reflection we note that the adjective in these three expressions probably does not refer to the same feature of the derrick, cannon, and pump. A derrick is probably called power- ful because it can lift a very heavy body, or exert a very great (lifting) force. A cannon is generally called powerful because it can project a heavy shot Art. 42 197 with great velocity, and we shall see presently such performance depenas on the energy which the gun can impart to the shot. A pump is probably called powerful because it can elevate or transport a large quantity of liquid in a short space of time, or perform much work per unit time. Use of the word power in the sense of force was very common in engineering literature at one time. Such usage is comparatively rare now, but not obso- lete. Thus we read of the " tractive power of a locomotive" to denote pull in the bulletins of the American Locomotive Company. (But Goss in his Loco- motive Performance, and Henderson in his Locomotive Operation, seem to prefer tractive force ; and in Locomotive Tests and Exhibits, of the Pennsylvania Rail- road System at the Louisiana Purchase Exposition, we find " tractive effort " to denote that pull.) The other two uses of the word power to denote (i) work or energy, and (ii) rate at which work is done or energy is transmitted or transformed are c[uite common. Thus in the same text-book we find: (i) " the actual power utilized is one-half the energy available," and (ii) "the power of the plant is about 470 horse-power" (258,500 foot-pounds per second, see below). And in another book there appear: (i) " the power of the rotating shaft could be converted into electrical energy," and (ii) '"the power is here measured in kilowatts " (one kilowatt equals 10'" ergs per second, see below). It seems probable that this double usage of the word power in engineering literature will persist. In common with most authors, even those quoted above, we will define power in a single sense, namely, — as the rate at which work is done. Units of Power * like units of work may be classed into gravitational, which vary slightly with locality, and absolute. Thus, the foot-pound per minute and the kilogram-meter per second are units of the first class ; also the (practical) EngUsh and American horse-power = 550 foot-pounds per second = 33,000 foot-pounds per minute, Continental horse-power ==^75 kilogram-meters per second = 4500 kilogram-meters per minute. The dyne-centimeter (or erg) per second is a unit of the second class; also the watt which is 10'^ ergs per second, and the (practical) kilowatt — 1000 watts = io^° ergs per second. The Bureau of Standards has recently decided to adopt the English and American horse-power as the exact equivalent of 746 watts, thus making this horse-power an absolute unit. " Thus defined it is the rate of work ex- pressed by 550 foot-pounds per second at 50° latitude and sea level, approxi- mately the location of London, where the original experiments were made by James Watt to determine its value. The ' continental horsepower ' is similarly most conveniently defined as 736 watts, equivalent to 75 kilogram-meters per second at latitude 52° 30', or Berlin." f * For dimensions of a unit of power see Appendix A. t Circular of the Bureau of Standards, No. 34. 198 c^^- '^ § 2. Measurement of Power. — There is only one instrument in common use which measures power directly, the wattmeter. It measures electric power and reads in watts, hence the name wattmeter. Power other than elec- trical is generally measured indirectly by measuring the amount of work done or energy transmitted in a certain length of time; this work or energy divided by the time gives the average power for the period. And to measure the work or energy generally requires the measurement of a force; this force multi- pUed by the distance through which it acts (as explained later) gives the work or energy. Thus most appliances for ascertaining power measure force first of all, and so are properly called dynamometers (force-measurers). Dynamom- eters are of two kinds, — absorption and transmission. Those of the first kind absorb or waste the energy which they measure, and those of the second kind transmit the energy or nearly all of it. A great many dynamometers have been devised. Only one of each kind is here described.* Prony Brake. — k simple form is shown in Fig. 325. AA are two bearing blocks which bear against the face of the pulley on the shaft of the motor or other machine whose power is to be measured; BC is the beam, one end of which is supported on a post D which rests on the platform of a weighing scale; BB are nuts by means of which the pressures between the pulley and the bear- ing blocks may be changed and consequently the frictional drag also when the pulley is turning. The drag on the brake tends to depress the end C when the pulley is rotating as indicated. ' w/!nu//f» ' - Fig. 325 Fig. 326 Let 5 = the reading of the scale when the pulley is rotating at the desired speed, the brake then absorbing the energy which is to be measured; n = the revolutions of pulley per unit time; a = the horizontal distance from the support of C to the center of the shaft; and X = a correction explained below. Then the power equals P = (5 - X) 2 -Kan. (i) If S and X are expressed in pounds, a in feet, and n in revolutions per minute, then P = 0.000190 (5 — X) an horse-power. (2) The meaning of X will appear from the following derivation of formula (i). Let F = the total frictional drag on the pulley while the energy to be measured * For full descriptions of many others see Flathers' Dynamometers or Carpenter and Deder- ichs' Experimental Engineering. Art. 42 199 is being absorbed and d = diameter of pulley. The work done on the pulley by this friction per revolution is F ird, and per unit time the work is Fir dn. Now let W = weight of the brake, and w = weight of D; then it is plain from Fig. 326 that \Fd + Wb = (S - w) a, or Fd = 2[S - {w + Wb/a)] a; and hence P = [S - (w -\- Wb/a)] 2 wan. This last equation is like (i) except that X replaces w + Wb/a. Now obvi- ously Wb/a is the pressure on the scale due to W; hence X is that portion of S due to W and w. X can be determined directly as follows: Loosen the screws BB and insert a small roller between the top of pulley and the upper block A, but without shifting C; then read the scale. That reading = X, for the pressure on the scale then = w + Wb/a. H.-j^f^^-y --V, t....^x....J a k-b-H V-'b -H Fig. 327 Fig. 328 Tatham Dynamometer. — This consists of four pulleys, A, B, C and D (Fig. 327), two levers E and F, a weighing beam G, and a belt HIJK. Pulleys A and B are mounted on the frame of the dynamometer; pulleys C and D are idlers and are mounted on the levers which, in turn, are supported on knife edges resting on the frame and by knife-edge Hnks L and M suspended from the weighing beam, all as shown; the weighing beam is supported from the frame at N. The dimensions are such that the straight portions of the belt are vertical, and H and K are vertically below the knife-edge supports of the levers. The shafts of A and B extend backwards to connect with machines between which the energy to be measured is transmitted. In all cases, the connections to machines should be made so that the tension in I is greater than that in /(/"tight" and / "slack"); and, if possible, the machine whose power is to be determined should be connected to or be on the shaft of A . When the dynamometer is in operation, then L and M pull on the weighing beam; and, if the beam be balanced by the poise, then the scale-reading gives the difference in tensions of / and /, or P2 - P\ (see Fig. 328). Let 5 = the 200 Chap, x scale-reading, n = revolutions per unit time of pulley A, D ^ diameter of the pulley plus thickness of belt, and P = power; then P = STrDn. For Sir D is the work done by the belt on A in one turn of A ; and, hence, the work done per unit time is Sir Dn. Fig. 328 shows the forces acting on the various parts, and makes plain how the poise measures P2 — Pi- Thus, from the right-hand lever Qi = Pic/b; from the left-hand lever Q2 = P2c/b; and from the weighing beam Wx = (Q2 -Qi)a= (P2 - Pi) ac/b. Hence, P2 - Pi = {Wb/ac) x. Now, Wb/ac is a constant, and so it is possible to graduate the scale beam (mark values of X on it), so that the readings will give the corresponding values of P2 — Pi. (No mention has been made of the weights of the parts. These are counter- balanced by a balancing weight on the scale beam as in an ordinary platform scale.) § 3. Indicator; Locomotive Power. — To determine the work done in the cylinder of a steam or gas engine per stroke or per unit time, use is made of an instrument called an indicator. The indicator makes a diagram or " card" from which the intensity of the pressure on either side of the piston at any point of a stroke can be read. Fig. 329 represents, in principle, the original form of indicator as used by James Watt (1736-1819). A is a cylinder; 5 is a piston working against a coil spring C whose upper end is fixed; D is a, pencil which presses against the card or paper E; F is a frame, movable right and left in suitable sHdes, for holding the paper or card. When the piston is moved the pencil simply makes a vertical line on the card; when the frame is moved the pencil makes a horizontal line. To take a diagram the cylinder of the instrument is connected with one end of the cylinder of the engine to be indicated, and the frame is connected to the cross-head of the engine with suitable reducing device so that the frame gets a motion just like that of the piston but greatly reduced. When the instrument is connected up, as just described, then the pencil describes a curve, something like GHIJG, the upper portion GHI being drawn during the forward stroke and the lower portion IJG during the return. The ordinates to the curve from the line of zero pressure K represent pressure per unit area in the cylinder, the scale of ordinates depending on the stiffness of the spring of course. The horizontal width of the diagram represents the stroke of the piston. Fig. 330 is a facsimile of an indicator card; the solid curve pertains to one end of the cylinder, and the dotted curve to the other end; AB is the line of zero pressure. The area AC DEB A represents the work done on one side of the piston (per unit area) during the forward stroke, and the area BEFCAB represents the work done on it during the return stroke. But the first work is Art. 42 2°^ positive, and the second negative; hence the work done on that side of the piston during both strokes is represented by the area enclosed by the curve CDEFC. Similarly, the area of the dotted curve represents the work done upon the other side of the piston (per unit area) during a to-and-fro stroke. The mean heights of these areas represent pressures per unit area which are called mean effective ^iq. 330 pressures, one for the head-end and one for the crank-end of the cylinder. Let pi = mean effective pressure for the head- end, p2 = that for the crank-end, A = area of cross-section of the cylinder, A' = area of cross-section of the piston rod, and I = length of stroke. Then the work done by the steam in the head-end during two consecutive strokes = piAl; that done by the steam in the crank-end = p2 {A — A') I, and the total work done is the sum of these expressions. The average of the mean effective pressures {pi and P2) for the two ends of the cylinder is sometimes called the mean effective pressure (for the cylinder). Let p = this mean effective pressure (per unit area) ; a = the average of the areas of the two sides of the piston, or what amounts to the same thing, the area of the cross-section of the cylinder minus one-half the area of the cross- section of the piston rod; n = the number of strokes of the piston per unit time; and / = length of stroke, as before. Then, as will be shown presently, the work done on the piston per double stroke is 2 pal closely; and, hence, the work done per unit time, or the power, is F = plan. (i) If the customary units are used, namely, p in pounds per square inch, / in feet, a in square inches, and n in strokes per minute, then P, above, is in foot-pounds per minute; and a^^^ P = — horse-powers. 33,000 To justify 2 pal: —As already explained, the work done in the cylinder per double stroke equals p^^i _[_ p^ (^1 - A') I. This can be written as follows: [{Pi + P2)A-P2A']1, or ^ X l(Pi -^ P2) (a - j-^^A'Y Now pi = p2 nearly, and therefore p2 ^ (pi + P2) = ^ nearly. Hence the work done per double stroke equals approximately 2 X I {pi-h P2) {A -h ^') I or 2 pal. For a single-expansion, two-cylinder locomotive, P = 2 plan. Let 5 = the " piston speed," the actual distance which a piston describes in its cylinder per unit time; then s = In and P = 2 pas. (2) 202 Chap, x With customary units for p, a, and 5 (pounds per square inch, square inches, and feet per minute respectively) the foregoing formula gives P in foot-pounds per minute. Since the piston speed and the velocity of the locomotive are related, it is possible to express the indicated power of a locomotive in terms of its velocity. Thus let v = the velocity of the locomotive, and D= diam- eter of the driving wheels; then one turn of the drivers means a displacement of the locomotive equal to irD and a displacement of the piston relative to its cylinder equal to 2 /. Hence v/s = xD/2 /, or 5 = (2 l/irD) v. Substitut- ing for s in the preceding formula for P, we find that 4 pal I (3) 100 80 •£60 -o « ;q-4o where c? = diameter of the cylinder. (Strictly d = the diameter of a circle whose area equals the area of the cross-section of the cyUnder minus one-half the area of the cross-section of the piston rod.) With pounds per square inch for p, inches for d, I, and D, and feet per minute for v, the foregoing formula gives P in foot-pounds per minute. Both formulas for P show that the power of a locomotive is zero at starting, and would increase exactly with the velocity if the mean effective pressure were the same at all speeds. The mean effective pressure depends upon the boiler pressure obviously, and on the cut-olT and piston speed.* The American Locomotive Company has adopted the line A BCD (Fig. 331), as expressing the variation of mean effective pressure with change of piston speed, for the manner of running (cut-ofif, etc.) which engine men usually employ. Thus, for all speeds up to 250 feet per minute, the mean effective pressure is taken at 85 per cent of the boiler pressure; at 500 feet per minute, it is taken at about 65 per cent, etc. Let po = boiler pressure and K = ratio of mean effective to boiler pressure, which may be called speed coefficient for convenience; so that p = Kpo. Then the formula for indicated power of the locomotive can be written p _ 2 A^o^^. (4) Thus, for a given boiler pressure the power varies as Ks. The line OEFGH (Fig. 331) is a graph of the preceding equation, the maximum value of P being called 100 per cent. It appears, then, that for the American Locomotive Company speed coefficients, the power increases uniformly up to a piston speed of 250 feet per minute, then less rapidly up to a maximum value at about 700 feet per minute, then remains nearly constant up to about 1000 feet per minute, and then diminishes. * See Fig. 42 in Goss' High Pressures in Locomotive Service, which shows clearly how the mean effective pressure varied in a test made by him. ?0 X 'f ~G* ■~lr A^S -.B / \X < /e \ — ^ N9 1 >* >s. / ^^ "v^^ / ^s ^D / ^ I 600 Feet per 1?00 Minute. Art. 43 203 43. Principles of Work and Energy § I. Principle of Work and Kinetic Energy. — In any displacement of a single particle the forces acting upon it, if any, do more or less work; and, in general, the velocity of the particle is changed, and, hence, the kinetic energy also. There is a simple relation between the total work done upon the particle by all the forces acting upon it in the displacement and the change in the kinetic energy as we will now show. Let P (Fig. 332) be the particle; m = its mass; OAB be its path (not a plane curve necessarily); Vi = its velocity at A, and v^ = its velocity at B; R = the resultant of all the forces acting on P; and Rt = the component of R along the tangent to the path at P. Then the work done by all the forces during an elementary dis- placement ds is Rt ds. But Rt = mat = ^ dv/dt, where a = tan- gential component of the acceleration of P. Hence the work done on P in the displacement ds is m (dv/dt) ds = m (ds/dt) dv = mv dv; and the work done in the total displacement AB is mvdv = ^ mvi^ — \ mv^. Now ^ mvz^ is the kinetic energy of the particle at B, and | mvi^ is its kinetic energy at A ; hence | mvz^ — § nivi^ is the increment in the kinetic energy of P. Thus we have the simple relation, — in any displacement of a particle, the work done by all the forces acting upon it equals the increment in the kinetic energy of the particle. If the total work done is positive then the increment in the kinetic energy is positive also, and there is a real gain and increase in velocity; if the total work done upon the particle is negative, then the increment in the kinetic energy is negative and there is a loss and decrease in velocity. Let Pi, P2, P3, etc., be the particles of any body (not rigid necessarily). In any displacement of the body, work done by forces acting upon Pi — increment in kinetic energy of Pi, (< a <( i( li " p __ <' '< << " 'I p II a (( a (I a jj c( a u a ic n -f 3 — -13, etc. = etc. Adding we get total work done on all particles = sum of increments in their kinetic energies = increment in kinetic energy of the system. That is, in any displacement of any body the total work done upon it by all the external and internal forces acting upon it equals the increment in the kinetic energy of the body. In a displacement of a rigid body the total work done by the internal forces equals zero. Proof: — Consider any internal force exerted, say, on Pi by P2; Pi exerts an equal, opposite, and colinear force on Po. Since the body is rigid the distance between the points of application {Pi and P2) of these two forces 204 Chap, x does not change, and hence (Art. 40) the total work done by these two forces equals zero. But all the internal forces occur in such pairs; hence, the total work done by all the internal forces equals zero, as stated. Thus we have the principle, — in any displacement of a rigid body the total work done upon the body by the external forces acting upon it equals the increment in the kinetic energy of the body. From these principles it follows that the rate at which work is done upon a body equals the rate at which it gains kinetic energy. But the rate at which work is done is power; so we may state that the combined power of all the forces doing work on a body at any instant equals the rate at which it is gaining kinetic energy then. The foregoing principles written out mathematically would take the form: work done = increment in kinetic energy. Since work is of the form force X distance or space, we may state that the " space-efifect " of force is kinetic energy. (The " time-effect" of force is momentum, see Art. 45.) The fore- going principles are especially well adapted for ascertaining the change in velocity — velocity-square, rather — when it is possible to compute the total work done on the body under consideration for the space in which the change takes place. By their means we may ascertain also something about the forces or displacement which accompany any given change in the kinetic energy of a body. We illustrate by means of some Examples. — i. .4 (Fig. 2>2>z) is a body weighing 400 pounds. It is dragged along a rough horizontal plane B by a force P, inclined as shown; P = 80 pounds. The coefficient of friction is about i/io. What is t)ie ^ velocity acquired from rest in the first 10 feet? In the first ^'-- 20 feet? The normal pressure between A and B = 400 — 80 •"f/iii/inii' B Fig. 333 sin 20° = 372.6 pounds; hence, the friction = 37.3 pounds. Now' we know all the forces acting on A . Gravity (400 pounds) does no work on A ; the work done by P during a displacement of ID feet = (80 cos 20°) X 10 = 752 foot-pounds; the reaction of B on A does work = — 37.3 X 10 = — 373 foot-pounds. Hence, the total work done on yl = y52 — 373 = 379 foot-pounds; and this is also the amount of the gain in the kinetic energy of A during 10 feet of displacement. Let Vi = the velocity (in feet per second) at the end of the first 10 feet; then the kinetic energy of A at the end of the first 10 feet = | (400/32.2) ^r = 6.21 Vi^ foot- pounds. Hence 6.21 Vi^ = 379, or Vi = 7.81 feet per second. Let V2 = the velocity of A at the end of the first 20 feet; then the energy of A there = 6.21 V2^. Since the work done on A during the first 20 feet = 758 foot-pounds, 6.21 V2^ = 758, or V2 = ii.o feet per second. Such a problem can be solved also by first finding the acceleration. Thus, since the resultant force acting on ^ = 80 cos 20° - 37.3 = 37.9 pounds, the acceleration = 37.9 ^ (400/32.2) = 3.05 feet per second per second. The time for describing the first 10 feet = the velocity acquired -^ the acceleration = ^1/3-05 = 0-328 vi. The distance = the average velocity X the time; that Art. 43 205 Fig. 334 is, 10 = I vi X 0.328111, or Vi = 7.81 feet per second as before. Obviously, the first method is more direct than the second. 2. A piece of timber 12" X 12" X 16' is suspended by means of two parallel ropes as shown in position A'B' (Fig. 334). The ropes are 10 feet long and the timber weighs Soo pounds. It is raised into the position AB, two feet above A'B', and then allowed to swing. What are its kinetic energy and velocity when it reaches its lowest position? The forces acting on the timber during its descent are gravity, the pulls of the ropes, and air pressure. We neglect the last. At each instant the pulls are normal to the direction of the displacement of their respective points of application; there- fore the pulls do no work. The work done by gravity during the descent = 800 X 2 = 1600 foot-pounds. Since this is the total work done on the timber, the kinetic energy of the timber inUts lowest position = 1600 foot-pounds. Now the timber has a motion of translation — no turning — , and therefore at each instant all points of the timber have identical velocities (Art. 35). Hence, ii v = the velocity in the lowest position, then f (800/32.2) v~ = 1600, or D = 11.35 feet per second. 3. A certain flywheel and its shaft weigh 400 pounds; the radius of gyration of both with respect to the axis of rotation = 10 inches. The wheel is set to rotating at 100 revolutions per minute, and is then left to itself, coming to rest under the influence of axle friction and air resistance after making 84 turns. Required, the average torque of the resistances. The moment of inertia of the wheel and shaft, about the axis, = (400/32.2) (10/12)^ = 8.64 slug-feet^. The angular velocity, 100 revolutions per minute, = 2 tt 100/60 = 10.47 radians per second. Hence, the kinetic energy of this wheel and shaft, when released, = | 8.64 X 10.47' ~ 474 foot-pounds. Besides the forces mentioned above, gravity and the normal pressure of the bearings act on the wheel and shaft, but these do no work during the stoppage. Let M = average torque of the resistances in foot-pounds; then the work done by them during the stoppage is — M 2^84. = —528 M foot-pounds. This equals the gain in the kinetic energy of the wheel; that is, $28 M = 474, or M = 0.90 foot-pounds. 4. A (Fig, 335) is a sheave supported on a smooth horizontal shaft. A is 3 feet in diameter, and its radius of gyration with respect to the axis of rotation = 9 inches. The weights of A, B, and C are 100, 200, and 300 pounds, respectively. The system is released and allowed to move under the influence of gravity and the resistances brought into action. Required the velocity of the suspended weights when they have moved through 10 feet. The system moves under the action of the following external forces, — gravity, axle reaction, air resistance, and the internal reactions between sheave and rope and the fibers of the rope. If the rope is quite flexible then the forces A C 5 Fig. 335 2o6 Chap, x in the rope do little work; this will be neglected. If the rope does not slip on the sheave, then no work is done by the reaction between rope and sheave. Thus, little or no work is done by the internal forces. The work done by air resistance is small unless the speeds of the moving bodies get high; it will be neglected. The work done by the frictional component of the axle reaction per turn isil/2 tt, where M is the frictional moment which we will assume has been found to be 10 inch-pounds. In the displacement under consideration, ID feet for B and C, the wheel makes 10/3 tt turns. Hence, the total work done by friction = (10 X 2x) (10/3 x) = 66.7 inch-pounds = 5.6 foot-pounds. Gravity does no work on A; or\. B and C its work = 300 X 10 - 200 X 10 = 1000 foot-pounds. We neglect its work on the rope as small. Hence, the total work done on the system = 1000 - 5.6 = 994.4 foot-pounds. Now let V = the required velocity in feet per second; then the angular velocity of the wheel =.v-^ 1.5 = 0.6677; radians per second. The kinetic energy of the system equals I 300 „ . I 200 „ I . N2 -^5 — ■v~-\-- V -\- -I {o.ob'jv)^, 232.2 232.2 2 where 7 = moment of inertia of the sheave. Now / = (100/32.2) X (9/12)^ = 1.75 slug-feet 2. Hence, the kinetic energy of the system = 8.16 z)^ foot- pounds. Thus the work-energy equation is 994.4 = 8.16 v-; hence v ^ 11 feet per second. 5. A certain pair of car wheels and their axle weigh 2000 pounds. Their diameter is 33 inches and the radius of gyration of wheels and axle is 9 inches. They are rolled along a level track until their speed is 60 revolutions per min- ute, and are then left under the influence of the rolling resistance of the track, coming to rest after rolling a distance of 1000 feet. (Data not from an actual experiment.) Required, the average rolling resistance. When released, the angular velocity of the wheels = one revolution per second = 6.28 radians per second, and the linear velocity of their centers = tt 33/12 = 8.64 feet per second. Hence, the kinetic energy = I 2000 _ „ , o , I 2000 X 8.64- + (9/12)- X 6.282 = 30J0 foot-pounds. 2 32.2 2 32.2 This is also the value of the work done by the rolling resistance, air resistance neglected. Hence, the rolling resistance is equivalent to a constant pull-back of 3010/1000 = 3 pounds. § 2. Moving Trains. — We will now apply the principles of work and energy to some train problems. First, we briefly consider the forces directly concerned with the motion of a train consisting of engine, tender, and cars. For convenience we regard the train as consisting of two parts, namely, the locomotive (engine and tender) and the cars; notation as in Art. 42, § 3. Locomotive. — For simplicity we regard the locomotive as being driven by an imaginary (forward) force F equivalent to the steam pressures. To be Art. 43 207 equivalent the work done by F per unit time (or power of F) must equal the indicated power of the locomotive, or Fv must equal pd^{l/D) v; hence, F = p dH/D. This force F we will call the cylinder efort of the locomotive. The resist- ances to motion experienced by a locomotive running alone on a straight and level track may be put into three groups: — (i) Those which arise through its action as a machine, consisting of friction in the working mechanism (valves and gear, cross-head, piston, crank pins, and journals of driving-wheels) ; (ii) those which arise through its action as a vehicle, like the resistances exper- ienced by the cars (see below) ; (iii) the air resistance. For convenience we may regard all the resistances in each group lumped, as it were, into a single resistance acting backward on the locomotive. We call them machine resist- ance, vehicle resistance, and frontal resistance, respectively; and we desig- nate them by Rm, Rv, and Rf. The sum of these three is called locomotive resistance, and will be denoted by Ri. Thus, we regard a moving locomotive as under the action of the following forces (see Fig. 336) : gravity, the support- T ^ \ \ ^J-^^r Fig. 336 till ing forces of the track (having no components along the rails), the draw-bar pull T, the locomotive resistance {Rm + -^w + Rf), and the cylinder effort F. The actual external forces are shown in Fig. 337. c y^ \ t t 11 n n Fig. 337 t tilt If the velocity of the locomotive is constant and the track is straight and level, then for any run of length L the work-energy equation is [{p dH/D) -T - Ri]L = o; hence T = {p dH/D) - Ri. 2o8 Chap, x If the velocity is changing, then the power equation is where v = velocity of locomotive and a = its acceleration. Hence T = ip dH/D) - Ri- Ma. If the locomotive is running on a grade then the grade resistance Rg must be included in an obvious way. According to the American Locomotive Company {Bulletin, No. looi), the resistances in pounds are as follows: Rf = 0.24 V~, where V is velocity in miles per hour; R„ = 22.2 X weight on drivers, in tons; and R^, = the same as for cars (see further on). The Cars. — The cars are urged forward by only one force, the pull of the tender on the first car; this is called draw-bar pull. The cars are retarded by several forces, namely: The rolling resistance of the rails upon the treads of the car wheels; the journal friction at the axles of the wheels; the air resist- ance; and miscellaneous forces, due to oscillation and concussion. The " laws" of these separate resistances are known only in a very general way. Because of lack of knowledge of these separate items of resistance, and, for convenience, it is customary to " lump" them into a single equivalent resist- ance, called train resistance. Thus we may imagine trains to be without actual track, journal, air, etc., resistance, but subjected to this equivalent force, conceived as a single pull backward on the train. A train of cars, then, may be regarded as moving under the action of four forces, namely, the draw-bar pull, the train resistance, gravity, and a supporting force exerted by the track, having no components along the rails. Many experiments have been made to determine train resistance, special " dynamometer cars" (equipped with instruments for measuring and record- ing speed of train, draw-bar pull, steam pressure, wind velocity and direction, etc.) being used for that purpose now-a-days. The methods for determining train resistance are very simple in principle. One method is this: — the loco- motive drags the cars along a straight, level track at a constant speed; the draw-bar pull and the speed are measured. Then the (total) train resistance for that speed equals the draw-bar pull. But level stretches of track are not always convenient of access, and constant speeds are not easily maintained. For an experiment on a grade let H = the ascent or descent of the center of gravity of the train during the experiment, L = the length of the run, W = weight of cars, T = average draw-bar pull, R = average train resistance. Then the grade resistance is Rg = ±WH/L, according as the train i ascending or descending the grade, and the work-energy equation is {T-Rt- Rg) L = E, IS Art. 43 209 where E is the gain in kinetic energy of the cars during the run, to be regarded as negative if there is a loss of kinetic energy. Hence Rt = T-Rg- E/L. This gives average train resistance for the speeds of the run, or, perhaps, the train resistance for the average speed of the run. Another method is based on the power equation (the rate at which work is done on the cars equals the rate at which they gain kinetic energy); this is dt\2 Mv^ ) = Mva, where M = mass of cars, v = velocity, and a = acceleration. Hence Rt = T - Rg- Ma. If the train is being retarded then a should be regarded as negative. There are many practical difi&culties in carrying out experiments as suggested; dis- cussion of these is not appropriate here.* Obviously, train resistance depends upon many conditions, as state of track and rolling-stock, weather and wind, and velocity of train. It is practically impossible to express the influence of all these conditions in a formula for train resistance. For a long time a favorite formula was the so-called Engineering News formula, r = 2 + j F, where r = train resistance in pounds per ton (weight of cars), and V = velocity of train in miles per hour. Recent experiments have shown very clearly that train resistance (per ton) depends very much on the loading of the cars, being much less for heavily loaded cars than for empties, and not so much on velocity as formerly belie\'ed. The American Locomotive Company in Bulletin No. looi states that " The best data available shows that the resistance varies from about 2.5 to 3 pounds for 72-ton cars to 6 to 8 pounds for 20-ton cars" (see Fig. 338); and "for speeds from 5 to 10 up to 30 to 35 miles per hour the resistance is practically constant." Schmidt, in the bulletin already mentioned, gives formulas for train resistance (per ton) for trains consisting of cars of different average weights; also the following as an approximation _ V -\- 39.6 — 0.031 w 4.08 -{- 0.152 w where r = train resistance in pounds per ton, V = velocity in miles per hour and w = average weight of cars in tons. * See Schmidt's Freight Train Resistance, University of Illinois Bulletin No. 43. 20 40 60 Tons per Car. Fig. 338 2IO Chap. X Examples. — i. A certain locomotive (engine and tender) weighs 178.5 tons, 106 tons on the drivers. There are two cyhnders, 23 inches (diameter) X 32 inches (stroke); the drivers are 63 inches in diameter; and boiler pressure is 200 pounds per square inch. Required the maximum draw-bar pull which this locomotive can exert on a level track at 20 miles per hour. The cylinder effort is KpodH/D = {K 200 X 232 X 32) -^ 63 = if 53,800 pounds. Now the piston speed s = 2 vl/ir D (see preceding article) = (2 X 20 X 32) -^ (it X 63) = 6.465 miles per hour = 569 feet per minute. The speed coefficient (see Fig. 331) is about 0.60; hence the cylinder effort is 0.60 X 53,800 = 32,300 pounds. The frontal resistance = 0.24 X 20' = 96 pounds; the machine resistance = 22.2 X 106 = 2350 pounds; the vehicle resistance is about 4 pounds per ton or 4 X (178.5 — 106) = 290. Hence, the total locomotive resistance is about 2740 pounds, and the maximum draw-bar pull = 32,300 — 2740 = 29,560 pounds about. 2. A freight train consists of 30 cars, average weight with load being 60 tons. What is the " resistance" of this train at 20 miles per hour? According to P. R. R. (Fig. 338), the resistance is about 3.5 pounds per ton or 6300 pounds total; according to C. B. & Q., it is about 2.5 pounds per ton or 4500 pounds total. According to Schmidt's formula, it is about 4.4 pounds per ton or 7920 pounds total. 3. The locomotive (example i) pulls the train (example 2) along a straight track. Required to show graphically how the cylinder effort and the various resistances vary with the velocity, assuming laws of resistances, etc., as in the preceding examples. As in example i the cylinder effort F= iiT 53,800; the piston speed 5 = 2 vI/ttD = 28.45 ^ ^^^t per minute where v is velocity of locomotive in miles per hour. Thus we have V = 5 10 15 20 25 30 35 40 s = 142 284 426 569 710 853 995 1138 K = o.&s •85 .825 •714 .604 •493 •417 •353 .306 F = 22.9 22.9 22.2 19.2 16.2 13.2 II. 2 9-5 8.2 the value of K having been taken from the curve A BCD (Fig. 331). We next compute the values of the cylinder effort from F = K 26.9 (tons) for the velocities v. Plotting the results we get curve i (Fig. 339). As in example i we take machine resistance as 2350 pounds, and vehicle resistance (for locomotive) as 290 pounds at all velocities. The frontal resistance (given by Rf = 0.24 1)2) varies. At d = o, i?/ = o; and at z) = 35 miles per hour, R^ = 385 about pounds. Inasmuch as R/ is small compared with the other elements of locomotive resistance, we will take a mean value of R/, say 200 pounds, and regard it as correct for all velocities. Then the locomotive resistance Ri = 2350 -|- 290 -|- 200 = 2840 pounds. Plotting this we get the straight line number 2. Taking train resistance at 3I pounds per ton, as in example 2, gives Rt = 6300 pounds, assumed to be independent of fiRT. 44 211 velocity for the range from 5 to 35 miles per hour. This plotted gives line number 3. The ordinate between lines i and 3 at any velocity represents the net or resultant driving or accelerating force on the train at that velocity. Thus, at 20 miles per hour that ordinate scales about 23,500 pounds. Lbs. 50 000 40 000 30 000 20 000 roooo ^ / B ^ •«^ ^"^/-v ^^ C D ^ 6300 lbs ■3 f iRi= 284-0 lbs\ 10 20 E5 30 35 Miles per Hour Fig. 339 4. Referring to the train of the preceding example: — Required to show how its acceleration varies with the velocity. Under the preceding example it was explained that any ordinate between lines i and 3 represents net driving force on the train. Hence the acceleration at any velocity = such ordinate (to scale) divided by the mass of the train. Thus at 20 miles per hour, say, the ordinate represents about 23,500 pounds, and the acceleration = 23,500 4- (3^957^00/32.2) = 0.1914 feet per second per second. In this way the accelerations at other velocities might be computed, and then the curve of accelerations determined. This curve would resemble curve i; indeed the accelerations are proportional to the ordinates from line 3 to line i. 44. Efficiency; Hoists § I. Efficiency of Machines. — Among the machines and appliances used in the industries there are some whose function is the conversion or transmission of energy. For example, — an electric dynamo which converts mechanical into electrical energy; a steam engine which converts energy of steam into (kinetic) energy of its flywheel; a line-shaft which transmits energy from one place in a shop to one or more other places; etc. In this article, "machine " means the kind of machine or appliance just described. The amount of energy supplied to a machine in any interval of time, for conversion or transmission, is called the input for that time; the amount of energy con- verted into the desired form or transmitted to the desired place is called the output. Experience has shown that output is always less than input; that 212 Chap.x is a machine does not convert or transmit the entire input. The difiference between output and input, for the same interval of time of course, is called lost energy or loss simply. By efficiency, in this connection, is meant the ratio of output to input; that is if e = efficiency, then e — output -i- input. Most machines are designed for a definite rate of working or for a certain load called its full load. Then we speak of the efficiency of a machine at full load, half-load, quarter over-load, etc., these efficiencies being different generally. The two following tables are given to furnish some notion of the efiiciencies of the more common machines.* Full-load Efficiency of Efficiency of Some Machine Elements* — Per cent P'^'" cent Hydraulic turbines 60-85 Common bearing, singly 96-98 impulse wheels 75-85 Common bearmg, long Imes of shaftmg. . . 95 Roller bearings 9^ Steam boilers 50-75 Ballbearings .. .. 99 engines 5-20 Spur gear cast teeth, mcludmg beanngs. . 93 turbines . . 5-20 Spear gear cut teeth, including bearings. . 96 Bevel gear cast teeth, including bearings. 92 Gas and oil engines 16-30 Bevel gear cut teeth, including bearings. . 95 Belting 96-98 Electric dynamos 80-92 Pin-connected chains, as used on bicycles. 95-97 motors 75-90 High-grade transmission chains 97-99 transformers. . . . 50-95 * From Kimball and Barr's Elements of Machine Design. The efficiency of a combination of machines. A, B, C, etc., A transmitting to B, B toC, etc., is the product of the efficiencies of the individual machines. For, let 61, €2, ez, etc. = the efficiencies of the separate machines A,B,C, etc., and e = the efficiency of the group. Then if £ = the input for A, the output of ^ = exE = the input for B; the output oi B = CieiE = the input for C; the output of C = eze^eiE = the input for D; etc. Hence, the output of the last machine -^ the input of the first = {eie^ez . . . )E -^ E = exe^ez . . . , or e = e\' ei' ez . . . . For example, if a dynamo is run by a steam engine, then the efficiency of the combination or set = the product of their separate efficiencies, say 0.20 X 0.90 = 0.18 or 18 per cent. § 2. Hoisting Appliances, Etc. — There are certain rather simple ap- pHances by means of which a given force can overcome a relatively large resistance; as, for example, the lever, the wedge, the screw, the pulley, etc. Such an appliance is generally operated by means of a single force, which we * For detailed information see IMead's Water Power Engineering, from which most values in the first table were taken; Gebhardt's Steam Power Plant Engineering; and Franklin and Esty's Elements of Electrical Engineering. Art. 44 ^•'■3 call driving force and denote by F; it is called effort also. The force which the appliance is desired to overcome we call resisting force and denote by R, or when the force is a weight, by t^; it is called resistance also. In many appliances (hereafter called "common ") all equal displacements of the point of application of the driving force result in equal displacements of the point of application of the resisting force; and generally these displacements re- spectively take place along the lines of action of the driving and resisting forces. These displacements, or their components along the lines of action of the forces respectively if the displacements are inclined to the forces, we will denote by a and b respectively. In a common appliance, the work done by the driving force and (by the appUance) against the resisting force are respectively Fa and Rb; hence the efficiency is given by e^ Rb^ Fa. (i) Let Fo = the effort which would be required to overcome the resistance R if the machine were frictionless; then Foa = Rb. Substituting in (i) we find that efficiency is given also by e = Fo-^F. (i') Let Ro = the resistance which F could overcome if the machine were friction- less; then Fa = Rob. Substituting in (i) we find that efficiency is given also by e = R-^Re. (i ) Most of the appliances now under discussion can be operated backward as well as directly. For example, the lever, the wedge, the screw, etc., can be used to lower a heavy body as well as to raise it. Some of these appliances, which can be run either way, will run backward without direct assistance when loaded; that is the load will overcome the internal friction. Such appliances are said to overhaul. Some will not run backward unassisted; that is the load cannot overcome the internal friction. Such appUances are said to self-lock. An appliance will overhaul or self-lock according as its (direct) * efficiency is greater or less than one-half, if the works done in overcoming friction in a forward and in an equal backward motion are equal (usual case). Proof: — As before, let F = the effort, R = the (useful) re- sistance, a and b = corresponding displacements of F and R, and w = the work done against friction, all in forward motion of the appliance. Then Fa = Rb -\- w. Now if the efficiency (forward motion) is greater than one- half, then more than one-half of the work Fa is expended usefully (against R) ; that is Rb is greater than w, and hence R could overcome the friction in backward motion. If the efficiency (forward motion) is less than one-half. * When a machine is run backwards it is said to have reversed efficiency, by (considerable) extension of the definition of efficiency. In such case the load (on the hoist, for example) is the effort, and the applied force is regarded as the useful resistance. In case the machine selflocks so that the applied force (P say) must assist the load, then by considerable stretch of imagination — P is regarded as the useful resistance; the computed (reversed) efficiency is negative. 214 Chap, x then less one-half the work Fa is done against R; that is Rh is less than w, and hence R could not overcome friction unassisted in backward motion. By mechanical advantage of an appliance is meant the ratio of the resisting to the driving force when the appliance is operating steadily, at constant speed. Thus, see equation (i), mechanical advantage is given by R/F = e a/b (2) Obviously the value of the ratio a/h does not depend on the loss or wasted work; that is, it is independent of the efficiency. (The ratio depends solely on the geometrical proportions of the appliance.)* Hence we may assume e = I and write a/b = R'/F' where R'/F' means the mechanical advantage of the appUance if it were without friction. Finally, R/F = e R'/F' (2') or, (mechanical advantage) = (efficiency c) X (mechanical advantage at e =1). In some appliances or mechanisms the driving and (useful) resisting forces {F and R) are applied at a wheel (pulley, gear, etc.), and it is more convenient in the discussion to deal with the torques, or moments of the forces about the shaft axes respectively, than with the forces. Let Ti and T2 denote those torques, of the driving and resisting forces respectively, and a and /5 corre- sponding angular displacements, in radians, of the wheels to which Ti and T2 are applied. The works done by the force F and against R during the dis- placements a and /3 are Tia and T^^. Hence, the efficiency is e = T-S/Txa, and Ts/ri = ea/^. (3) Reasoning as in the preceding paragraph we conclude that a/jS = T2/T1 where T2 /Ti means the ratio of the resisting torque to the driv- ing torque if the appliance were frictionless, e = i. Hence T,/T, = e T^'/T,'. (3') We may call the ratio of the resisting and driving torques, the mechanical advantage of torque; then the foregoing result may be stated as follows: (mechanical advantage of torque) = (efficiency e) X (me- chanical advantage of torque at e = i). Examples. — i. The pitch of the screw-jack (Fig. 340) is h, the mean radius of the screw thread is r, the length of the lever is /. What is the efficiency of the jack when it is overcoming (raising) IF? It is shown in Art. 20 that the force required at the end of the lever to start the screw is P ==^ W (r/l) tan (^ + a), where * When the displacements a and b are not inclined to the forces F and R respectively, then the ratio a/b is sometimes called the velocity ratio of the appliance, for the velocities of the points of apphcation of F and R are as a to b. Thus we have for such cases mechanical advantage = efficiency X velocity ratio. High mechanical advantage requires high velocity ratio, b small compared to a; thus the adage "what is gained in force is lost in velocity." Art. 44 215 Fig. 341 (f) is the angle of static friction and a is the pitch angle, tan~^ {h/2 irr). Hence the force required to raise the load W uniformly is given by the same expression if is taken to denote the angle of kinetic friction (see Art. 45). Let it be so understood in what follows. If the screw were frictionless then the force required would be Po = W (r/l) tan a; and hence the efficiency is e = tan a -^ tan ((^ + a). 2. Fig. 341 represents a double purchase crab, for hoisting. Hand cranks can be applied on the ends of the shaft B or C The hoisting rope winds on the drum. The crank is 18 inches long; the drum is 10 and the rope f inches in diameter; gears A' and B are 20 and B' and C are 4 inches in diameter. Required the mechanical advantage of the appliance when a crank is used on C. Evidently, angular displacements of gears A ' and B' respectively are as i to 5 ; also the angular displacements of gears B and C. Hence, for one turn of A', C makes 5 X 5 == 25 turns. For one turn of A', b = 10.75 tt, ^.nd a (the corresponding displacement of the point of application of the effort) = 36X X 25 = 900 TT. Therefore, the velocity ratio of the appliance is 83.8. If the efficiency is So per cent say, the mechanical advantage is 0.80 X 83.8 = 67. See equation (2). This result could be obtained from equation (2') as follows: Let P' = effort at the crank handle, T' = tooth pressure between gears ^-I' and B', T = tooth pressure between gears B and C, 6' = obliquity * of T', d = * Fig. 342 represents parts of two gears Oi and O2. Imagine the tops of the teeth cut ofi and the remaining hollows filled in so that the toothed wheels become friction wheels, transmitting power by friction (developed by press- ing the wheels together). Evidently the diameters of these (imaginar\') friction wheels could be chosen so that the ratio i of the angular velocities of the wheels would be equal to / that of the gears. The intersections of the faces of such wheels, of the particular diameters mentioned, and a plane perpendicular to the shafts are the "pitch circles" of the gears. The common point of the two pitch circles is the Q "pitch point." To insure constant ratio of angular ve- locities of two gears in mesh, the form of the teeth must be such that the normal to the surfaces of two teeth where they touch passes through the pitch point. (For proof, see any standard book on Mechanism.) The angle be- tween this normal and the common tangent of the pitch circles is called " obliquity " of the normal. In some gears this obliquity remains constant, as the gears turn; in others, it varies. In the figure, is the pitch point, OT the common tangent to the pitch circles, AO the normal for the teeth touching at A , and BO the normal for the teeth touching at B; the obliquity of these normals are denoted by 0' and d". If the teeth were frictionless, the pressure between two teeth (in contact) would act along Fig. 342 2l6 Chap, x obliquity of T, and R' = resistance, — all on the supposition that the appliance is frictionless, or e = i. Then, considering torques on the three shafts it is plain that P' X i8 = r X 2 cos ^, r X lo cos ^ = r' X 2 cos 6', r X locos^' = R' X 5.375; and from these it follows that R'/P', the mechanical advantage at e = i, is 83.8. Hence the (true) mechanical advantage is, as before, 83.8 X 0.80 = 67. 3. Fig. 343 shows the operating mechanism for a lift-bridge, but there is /■Tower posf of fixed span Truss of liff span ffack Equalizing sinaff '^0 gearing at opposite end of span Socket for., iL-A capstan ^ ^ ^TZc^f^FTi^ IThfi s^ T-.' • IT r^ z. Truss of lift span '\ |<- Liff span ''■■Tower post of fixed span Fig. 343 the normal, at the contact pohit, and hence through the pitch point. Let TV denote the resultant of all the pressures on the frictionless teeth of either gear; obviously iV passes through 0. Imagine iV resolved at O into components along and perf)endicular to OT, and let N t denote the first component. Then the (frictionless) torques exerted by the driving and driven gears on each other are respectively iVt I dx and A"t \ d-i, (i) where d\ and dt are the diameters of the pitch circles of the gears respectively. On account of the tooth friction, each tooth pressure is inclined to the normal at the contact, by an amount equal to the angle of friction <^ (see Art. 45). At two teeth which are approaching, the pair touching at A, the directions of the frictional forces are such that., the obliquity (to OT) of the tooth pressure R' is Q' + 4>. At two teeth which are receding, the pair touching at B, the directions of the frictions are such that the obliquity (to OT) of the tooth pressure R" is B" — <^. In either case, the Hne of action of the tooth pressure cuts the line joining the centers of the gears between the pitch point O and the center O2 of the driven gear. It follows that the line of action of the resultant of all the tooth pres- sures on either gear cuts the line O1O2 in a point C between and O2. Let .V denote the distance from C to 0, R the mentioned resultant, and Rt the compo- nent of R parallel to OT. Then the torque exerted by the driving and the driven gears on each other are respectively Rtildi-x) and Rtihdi+x). If e' is the efficiency of the gears alone, shafts frictionless, then did-2 (i — e')) approximately. (2) 2 di -\- d2 For, let Ti be a torque applied to drive the driving gear, and Tj a resisting torque applied to the driven gear; then e' = Tsrfi ^ T,d2, Ti = Rt ih di + x), T-i = Rt {h d. - x). These three equations yield (2). Art. 44 217 a duplicate mechanism on the other end of the Hft span not shown. All this mechanism rests on and moves with the lift span. There are fixed tower posts adjacent to the lift span as shown, and on these posts there are fixed vertical racks which engage spur pinions GG of the operating mechanism. The pinions are driven by the motors or hand capstans, and thus the Hft span is raised or lowered. The intermediate drive consists of: a motor pinion A number of teeth, 18, diameter, 5 . 14 inches cross shaft gear B 126 36.00 cross shaft bevel pinions CC 20 10 .00 counter shaft bevel gears DD 60 30.00 counter shaft spur pinions EE 15 8.36 operating shaft spur gear FF 52 29 .00 operating shaft pinion GG 15 11 -94 bevel pinion on hand operator A' 16 12.72 bevel gear on hand operator B' 24 19.08 The lift span is accurately counterweighted so that no work is done against gravity either in raising or lowering; but vertical pressure must be developed between the racks and the pinions G to overcome the (internal) friction in the counter weight mechanism. It is estimated that a vertical pressure of 5000 pounds is required at each pinion for lifting the bridge. How great a driving torque, at each motor, is required to develop this pressure at each pinion G? We will first neglect losses in mechanism. Let 2 T/ be the driving torque at the motor and T2' the resisting torque at the pinion G. Then the torque to the cross shaft = 2 T/ -j%^-, torque to the counter shaft = Ti ~\%^ |§, and torque to the operating shaft = Ti' VV' In tI = 36.9 T/. But 36.9 r/ = T2', or T2 jTi = 36.9. We take the efficiencies of the gears to be as follows: A and 5, 96 per cent; C and D, 92 per cent; E and F, 96 per cent. Then the efficiency of the transmission from the motor to the pinion G is 0.96 X 0.92 X 0.96 = 0.85, or 85 per cent. Hence the ratio of the actual driving and resisting torques is T^lT^ = 0.85 X 36.9 = 31.4. The lever arm of the vertical pressure of the rack against the pinion G is 5.97 + a; (see footnote on page 216). Since di (diameter of the pitch circle of the rack) is infinite, equation (2) as written is not usable; it can be written which, since d\l di = o in the present problem, becomes % _ 1 ~ 2 di (i — e') = 2 X 11.94 (i — 0.98, say) = 0.12 inches. Therefore, the actual resisting torque on the pinion is 5000 X 6.09 = 30,450 inch-pounds = 2537 foot-pounds. Hence Ti = 2537 -^ 31.4 = 80 foot-pounds, and the required motor torque is 160 foot-pounds. 2l8 Chap, x -^ / * i :^ ^ \ 1 J ^ V ^ ^ 1 M U...R-.J u ' T| Fig. 344 Pulley. — Fig. 344 represents a simple pulley with part of a rope or chain upon it. Let S = tension in leading or off side of the rope, and T = tension in the following or on side. Then evidently S is greater than r, for 5 overcomes not only T but also the friction at the pin and the "rigidity" of the rope. The resistance due to pin friction = the product of the coefiScient of axle friction (see Art. 45) and the pressure on the pin; this pressure = S -\- T. Hence, if / = coefficient and r = radius of the pin, the work done against friction per revolution of the pulley = 2 tt r/ (5 + T). The work done in bending or unbending (inelastic) rope over the pulley is proportional to the amount of rope so bent per revolution (that is 2 ttR), and it seems to be proportional also to the tension, to the area of the cross section of the rope and inversely proportional to the radius R. Thus the work of bending = C2TrRT d^/R = CnrTd'-, where d = diameter of the rope and C is an experimental coefficient depending on the kind of rope and perhaps other elements. Likewise the work of unbending (at off side) = CiirSd^. Now, if we equate the work done by the effort 5 to the work done against rigidity, the resistance T, and the axle friction, and then simplify the resulting equation, we get / . j2n ^/ , ,r , ^J2\ ^V--^r-^r)^^V^-^r^^r} This equation can be written in the following approximately correct form, — 5 = (i + 2/^ + 2c|)r = irr, where K is an abbreviation for i + 2 fr/R + 2 C d^/R. According to experi- ments by Eytelwein, C equals about 0.23 when d and R are expressed in inches. The American Bridge Company made some experiments to determine C and K for such pulleys and rope as are in common use in tackle for construction work, and found that C depends not only on kind of rope, as expected, but also on the size of rope. The following table is taken from their report.* Dimensions and coefficients Hemp Wire Diameter of rooe, d li I§ if 2 3I 4A 5l 6 i I li if 0.23 0.20 0.19 0.17 1.20 I. 21 1.23 1.24 3 Center pin to center rope, R 7l Diameter of oin. 2 /* 2h Coefficient C 0.9 Coefficient K 1. 16 These values of C and K are higher than those usually employed. C = 0.08 to 0.22 is advised for hemp rope,t and " K = 1.06 to 1.07 may be considered maximum practical values since generally K = 1.02 to 1.04."^ * Trans. Am. Soc. C. E., 1903, Vol. 51, p. 161; also Eng. Rec, 1903, Vol. 48, p. 307. t Hiitte, Taschenhuch (Twentieth Edition), Vol. i, p. 247. X Bottcher-Tolhausen, Cranes, p. 15. Art. 44 219 Tackle.— When a fixed pulley (Fig. 345), for example, is used for lift- ing, P = KW or W/P = i/K; for lowering, W = KP, or P/IV = i/K. When a movable pulley (Fig. 346) is used for lifting, IF = P + T = P + P/K, or W/P = (i + K)/K; for lowering, W = P -\- S = P -\- KP, or W/P = i + A'. In a similar way we can determine the mechanical advantage of any com- bination of pulleys in terms of K. For example, consider the tackle repre- sented in Fig. 347. There are two separate pulleys in each block A and '' /(//////////{ ^i -'//(///////////^ i, Fig. 345 Fig. 346 B. The pulleys in a block are generally alike in size but are here represented unlike for clearness. Let P = the applied pull and W = load. Pi, P2, P?. and P4 = the tensions as indicated in Fig. 348. When the tackle is used for lifting. Pi = P/K, P2 = Pi/K = P/K\ P3 = PoJK = P/X^ and P4 = P3/K = P/K\ And since PF = Pi -^ P2 + P3 + P4, we have also P P P P P or W/P = (A'3 -f K^ + A' -f i)/A4. When the tackle (Fig. 347) is used to lower the load, Pi = A"P, P2 = /CPi = K^P, Pz = KP^ - A^P, Pi = KPi = A'4P. -/////////////////////. -//////////////////■ w//'/jum> ■w////(////M. Fig. 347 And since Uw ! jw W Fig. 348 TF = Pi -f P2 + P3 + P4, we have TF = AP + A^P + A''P + A^P = P (A -j- A^ -f K^ -\- K^) or W/P = A (i -F A + A2 + A^). 220 Chap, x Special {Chain) Hoists. — Fig. 349 represents a Weston differential hoist. The upper block contains two pulleys differing sUghtly in diameter; they are fastened together. The lower block contains only one pulley. The pulley grooves have pockets into which the links of the chain fit; thus slipping of the chain is prevented. The chain is endless and is reeved as shown. If there were no lost work, then the tension in each portion of the chain to block B would equal one-half the load (Fig. 350), and the pulls on the block A would be as indicated in the figure. Now if R and r = the distances from the center of the pin in block A to the axis of the chain as indicated then moments about the axis of the pin give PoR + h ^Vr = \ WR, or W = Pq 2 RliR - r)\ the ratio, HV^o = 2R/{R — r) may be made very large by making R — r small. The mechanical advantage is IF W 2 R e = P Po R-r' where P = the actual force required to raise W and e = efficiency. These hoists are made of various capacities up to W = 3 tons; their efficiencies are relatively low, from about 25 to 40 per cent according to the manufactiu-ers' lists. In the so-called Duplex and Triplex hoists the upper blocks are screw- geared and spur-geared respectively. At full load the efficiency of these hoists varies from about 30 to 40 and from 70 to 80 per cent. Example. — We will now show how to apply some of the preceding prin- ciples and formulas in a computation relating to the operating machinery of the vertical lift bridge represented in Figs. 351 and 352. The Hft span when down in place rests on two piers. When up it is balanced by two counter- weights as shown. Each counterweight is suspended by means of two pairs of one-inch cable; each pair of cables extends upwards from the counterweight, over a sheave and downward to a point of attachment on the lift span. At each corner of the lift span there is a spirally grooved drum carrying two one- half-inch cables. Each cable has one end attached to its drum; the other end of the up-haul cable is attached to a point vertically above at the top of the tower, and the other end of the down-haul cable is similarly attached at the base of the tower. As the drum is revolved, one cable is wound upon it and the other is paid out. The tw^o drums at either end of the span are mounted upon a single cross-shaft A, which carries a bevel gear B. The gears BB mesh with bevel pinions DD mounted on the longitudinal shaft C which also carries a bevel gear E. E meshes with a bevel pinion F on a vertical shaft which carries a capstan head. This capstan head takes a horizontal lever by means of which a man operates the mechanism. To lift the bridge he rotates the capstan headed shaft in the proper direction and drives the drums; they wind the up-haul cable upon themselves and pay out the down-haul cable as already' described. This winding up necessitates upward motion of the bridge. Art. 44 221 The length of the lever (radius of circle in which the man walks as he oper- ates) is 6 feet. The pinions F and D are alike; each is 6.86 inches in diameter and has 21 teeth. The bevel wheels E and B are also alike; each is 16.87 inches in diameter and has 53 teeth. The drums are 18 inches, the sheaves 54 inches, and the sheave shafts are 3^ inches in diameter. The lift span weighs 68,000 pounds and each counterweight weighs one-half that amount. Thus the span would be perfectly balanced, if the mechanism were frictionless and the cables without stiffness and weight, and no effort would be required to operate the bridge. Drurriia 1-^ Ouf/m_e_ ofUft^ ^PPJl^ [A Drum%\ Drum Drum Diagram of Operating Machinery. Tower Brat Sheave^ Counter- weight- ■5 4-" Sheave Court terweiqht 7777777rr777777mTmmr, Side Elevation. Fig. 351 Cross Section. Fig. 352 In the following computation the weight of cables is neglected. Then the tension in each counterweight cable on the counterweight side of the sheaves is one-fourth of 34,000 pounds or 8500 pounds. When the span is being lifted, the tension on the other or following side of the sheave is less than 8500 pounds. Call that tension T\ then 8500 = KT, or r = 8500 -^ K 222 • Chap, x (see under " pulley" above). We will take K == 1.06; then T = 8020 pounds, and hence the hft on the span due to counterweights = 8 X 8020 = 64,160 pounds. This leaves 68,000 — 64,160 = 3840 pounds to be furnished by the four up-haul cables, or 960 pounds apiece. Let a and b — respectively any corresponding displacements of the effort at the hand lever and the resistance 960 pounds at each drum; then b 3.43 21 18 Hence, if the mechanism were frictionless the effort Pq required to produce 960 pounds tension in one rope = 960 H- 50 = 19.2 pounds; and the effort P required to produce that tension by means of the actual mechanism = 19.2 -^ e, where e = the efi&ciency of that part of the mechanism which transmits from P to the resistance 960 pounds. The efficiency of each pair of gears and necessary bearings we take as 0.95; the efficiency of a drum about i -h 1.03 = 0.97; hence e = 0.95 X 0.95 X 0.97 = 0.875. Therefore P = 19.2 -^ 0.875 == 22 pounds, and the effort (at the lever) required to develop a tension of 960 pounds at the four driuns = 4 X 22 = 88 pounds. The computation can also be made as follows: — We regard the total force Q exerted at the hand lever and the force of gravity on the counterweights as two efforts which overcome the (useful) resistance (gravity on the lift span) and the wasteful resistances in the entire mechanism. For any rise b of the lift-span the counterweights descend an equal distance and the hand-lever effort works through a distance 50 b; and since the efficiency of a sheave = I -^ 1.06 = 0.944, we have QX SobX 0.875 -f 2 X 34,000 XbX 0.944 -= 68,000 X 6, or (J = 87 pounds. 45. Kinetic Friction § I. Kinetic Friction, or Friction of Motion, is the friction between two bodies when sliding actually occurs. The coefficient of kinetic friction for two bodies is the ratio of the kinetic friction to the corresponding normal pressure between them. The angle of kinetic friction is the angle between the normal pressure and the total pressure (resultant of the normal pressure and the kinetic friction). One of the so-called laws of friction states that the kinetic coefficient is less than the static coefficient (Art. 19), and implies that there is a sudden or abrupt change in the values of the coefficients. Experiments by Jenkin and Ewing* on the kinetic coefficients at speeds as low as 0.0002 foot per second (about f foot per hour) lead them to conclude that "it is highly probable that the kinetic coefficient gradually increases when the velocity becomes extremely small, so as to pass without discontinuity into the static coefficient." Experiments by Kimballf also indicate that there is no abrupt * Phil. Trans. Roy. Soc, 1877, Vol. 167, Part 2. t Am. Jour. Set., 1877, Vol. 13, p. 353. Art. 45 223 change from static to kinetic coefficient. Moreover, they show that the kinetic coefficient may be greater than the static. Galton and Westinghouse experiments* indicate that the coefficient for dry surfaces probably decreases progressively from the value of the static coefficient as the velocity increases. See the following table of Coefficients of Friction at Various Speeds Cast-iron Brake Shoes on Steel-tired Wheels Velocity Coefficients Number of tests Miles per hour Feet per second Max mum Minimum Mean o-f- 0+ 0.330 10 14-5 281 0. 161 .242 54 20 29 240 ■133 .192 69 30 44 196 .098 .164 94 40 59 194 .088 .140 70 50 73 153 .050 .116 55 60 88 123 .058 .074 12 The foregoing coefficients are based on observations taken very soon after application of the brakes. The wide variation from minimimi to maximum value at any velocity was due in part to the different intensities of pressure employed at that velocity; in general the coefficient decreases with increase in intensity until seizing occurs. Continual rubbing of dry surfaces abraids them and decreases the coefficient of friction. This effect is clearly shown in the following table, also taken from the Galton and Westinghouse experiments. Coefficient of Friction as Affected by Time of Rubbing Cast-iron Brake Shoes on Steel-tired Wheels Miles per hour Time after applying brakes 0+ 5 seconds 10 seconds 15 seconds 20 seconds 20 27 37 47 60 0.182 .171 •152 .132 .072 0.152 .130 .096 .080 .063 0133 .119 ,083 .070 .058 O.I16 .081 .069 0.099 .072 The discrepancies between the two foregoing tables are due in part to the fact that the values for time o + in the second table are based on compara- tively few experiments. The following table gives coefficients for several different materials; it shows also influence of intensity of pressure, velocity and water lubrication. Proc. Inst. Mcch. Engrs., 1S79, p. 170. 2 24 Chap, x Coefficients of Friction * Various Brake-shoe Materials on Steel-tired Wheels Materials Cast iron. Cast iron. Oak Oak Poplar. . . . Poplar . . . . Cast iron. Cast iron. Oak Oak Poplar. . . . Poplar. . . . Pressure, pounds per square inch lO 40 10 40 10 40 20 80 40 120 40 120 Velocity, miles per hour 43 36 60 43 32 30 037 073 041 070 15 0.37 •30 55 40 72 53 28 26 032 055 038 053 Lubrication none none none none none none water water water water water water * From Experiments by Ernest Wilson, Engr. News, 1909, Vol. 62, p. 736. Coefficients of Kinetic Friction (Rough Averages) Compiled by Rankine from Experiments by Morin and others Wood on wood, dry . . soapy Metals on oak, dry . . . wet . . . soapy . Metals on elm, dry. . . Hemp on oak, dry. . . . wet. . . . 25-0.50 2 5- -6 24- . 26 2 2 - .25 53 33 Leather on oak Leather on metals, dry. . . wet . . greasy oily . . . Metals on metals, dry. . . . wet . . . . ). 27-0.38 ■56 •36 ■23 ■15 .15-0.2 •3? The following table gives coefficients for eight ship launchings and hence for similar cases. The lubricant used seems to have been mainly tallow. Coefficients from Launching Data.* Load, tons per square foot Coefficient of friction Load, tons per square foot Coefficient of friction 0.50 0.79 1.03 1.29 0.0532 .0487 .0400 .0407 1.62 2.05 3.56 4 50 0.0370 .0324 .0257 .0217 * From Peabody's Naval Architecture. § 2. Pivot and Journal Friction. — Pivots. — Let W = load, ju = coeffi- cient of friction, (i) In the case of a flat pivot (Fig. 353) the average pressure per unit area of contact is W/tR^. On any element of area dA the normal pressure = (W/irR^) dA (supposing that the total pressure is uniformly dis- tributed), and the frictional resistance on the element = iJ.(W/irR^)dA. The Art. 45 225 moment of this resistance about the axis of the shaft = iJ.(W/irR^)dA-p. take dA = pdd-dp; then the total resisting moment = We r Tm Jo Jo w -iddp'^dp pW-R. 3 Thus the actual resistance may be regarded as a single force = pW with an arm — ^R; and, for example, the work done against friction per revolution or the power lost may be computed simply on that basis. Thus the work done per revolution = | irpWR, and the power lost = | irpWRn where n = number of revolutions per unit time. (ii) In a similar way we might determine the resisting (frictional) moment in a collar bearing pivot (Fig. 354). We would find the moment to be 1pW{R^-r^)-^ {R^-r^). Hence we may regard the resistance as a single force = pW with an arm f {R^ - r') ^ (i?2 _ r'~). Fig. 353 Fig. 354 Fig. 355 Fig. 356 (iii) In the conical pivot (Fig. 355), the total normal pressure, and hence the friction too, is increased by wedge action. Let p = the intensity of normal pressure at any point of the contact, regarded as constant. Then the normal pressure on an elementary area dA = pdA . Since the friction has no vertical component, the vertical component of the normal pressures on all the elemen- tary areas = W; that is, W pdA • sina = W = pA sin a, or p = P A sin a But A sin a = the horizontal projection of the actual surface of contact. Hence the intensity of the normal pressure is independent of a, the pivot angle. For Fig. 355, p = W/tR-; hence the normal pressure on the elemen- tary area dA is {W/irR-)dA and the frictional resistance = p(W/TrR^)dA. The moment of this resistance about the axis of the shaft = p{W/TR^)dA-p, and the entire resisting moment = the integral of this expression. For sim- plicity in integration, imagine dA to be of such shape that its horizontal projection equals pdd-dp (see Fig. 355). Then sin a-dA = pdd-dp, and the resisting moment = ^^pWddp^dp^ pW irR^ sin a sin a 3 Jr»2 7r pL Jo '-R. 226 Chap, x Hence we may regard the resistance as a single force p.W/sm a with an arm ^ R (iv) In a similar way we may compute the resisting (frictional) moment in the case of a frustrated conical pivot (Fig. 356). We would find that the resisting moment = IjW 2 R^ — r^ sin a 3 i?" — r^ Hence we may regard the friction as a single force = nW/sin a with an arm I (R^ - f^)/{R^ - r^). Journal Friction. — We do not attempt to compute the normal pressure and frictional resistance at each point of a journal bearing and then the resisting moment as in the case of pivots. So-called coefficients of journal friction have been determined from direct experiments on journal friction. This coefficient is the ratio of the frictional resistance to the pressure between journal and the bearing. Thus in a certain experiment there were 20 babbitt bearings sus- taining a 2yV-inch shaft; the load per bearing was 2,000 pounds, and it was found that 1 1 24 watts were required to run the shaft at 350 revolutions per minute. All the power was used to overcome the journal friction. Since 11 24 watts = 49,600 foot-pounds per minute and 350 revolutions per minute corresponds to a (shaft) surface velocity of 223 feet per minute, the total frictional resistance = 49,600 -T- 223 = 222 pounds or I I.I pounds per bearing. Hence the coefficient of journal friction in this particular instance was ii.i ^ 2000 = 0.0055. The pressure between a journal and its bearing is not uniformly distributed over the surface of contact. By nominal intensity of pressure ("pressure" for brevity) is meant the whole pressure divided by the product of the length and diameter of the bearing. Thus in the experiment just mentioned, the length of each bearing was 9!^ inches; hence the nominal intensity was 2000 (2tV X 9§i) = 90 pounds per square inch. It has been found from numerous experiments that coefiicients of journal friction depend on (i) the method of lubrication, (ii) the lubricant, (iii) its temperature, (iv) the velocity of rubbing, and (v) intensity of pressure on the bearing. (i) Tower* and Goodmanf report the following relative coefficients as showing effect of the method of applying the lubricant: Method Bath Saturated pad Ordinary pad . Siphon Tower I .00 6.48 7.06 Goodman I .00 1.32 2.21 4.20 * Proc. Inst. Mech. Engrs., 1883. t Mechanics Applied to Engineering, 1896. Art. 45 227 (ii) The following table (according to Tower) indicates how the coefficient depends on the lubricant. Numbers are relative. sperm oil Rape oil Mineral oil Lard oil Mineral grease 1. 00 I .06 1.29 1-35 2.17 The journal was steel; gun metal brass embracing somewhat less than one- half the circimiference of the journal; speed 300 revolutions per minute; nom- inal loads from 100 to 310 pounds per square inch; oil temperature 90° F.; bath lubrication. Tower states also: "the numbers represent the relative thickness or body of the various oils, and also in their order, though perhaps not in their numerical proportions, their relative weight-carrying power. Thus sperm oil, wliich has the highest lubricating power, has the least weight-carry- ing power; and though the best oil for light loads would be inferior to the thicker oils if heavy pressures or high temperatures were to be encountered, (iii) The coefficient decreases with increased temperature (see Figs. 357 and 359). But if the temperature gets so high as to lower the viscosity greatly, then the lubricant gets squeezed out and the coefficient increases, (iv) In general the coefficient increases with increase of speed (see Figs. 358 and 360 and accom- panying explanations). But at the lower speeds the coefficient may decrease with increase of velocity (see Fig. 358). (v) The coefficient decreases with increase of intensity of pressure (see Figs. 358 and 361). But the intensity may become so great that the lubricant gets squeezed out, and then the co- efficient increases and seizing occurs. 50 75 100 125 150 Degrees Fahrenheit. All ValoclHes, 846 fr.permin. Fig. 357 \ in-^ ■ \ |4.^JS 5;^ l\ ^ Z Li-- ■ \y <^ -^ bO lA-Z is^ -^ 356 ) 100 200 300 400 500 600 700 80O Feet per Minute. All Temperatures, 77° F. Fig. 3s8 Figs. 357 and 358 are after Stribeck.* Gas-motor oil and ring oiler were used in the experiments. Figs. 359, 360, and 361 are after Lasche.f They show respectively how the coefficient depends on temperature, velocity, and * Z.. dV . d. I., 1902, Vol. 46, p. 1341. \ Z.dV. d. /., 1902, Vol. 46, p. 1881. 227a Chap, x pressure. The lubrication was forced; journal and bearing combinations as follows: Number Journal . Bearing. I steel white metal II nickel steel white metal III nickel steel IV nickel steel bronze V wrought iron white metal The heavy line in each figure represents the average law for the five combina- tions, and the other two curves relate to the two combinations departing most from the average result. 0.015 ^ g 0.010 it a. 0.005 \ V ^- '^^, ^*>._ *«^ 0.015 0.010 0.005 y ^''' „»-" "H 0.03 50 75 100 150 175 Temperature, Degrees F. All Pre55ur3s, 92. Slbs.persq.in. All Velocities, I97ftpermin. Fig. 359 ) 1000 2000 3000 4000 5000 Velocity, ft.permin. All Pressures, 9Z.5lbs.persq. in. All Temperatures, 112° F. Fig. 360 50 100 150 Pressure, Ibs.per sq.in . All Temperatures, II Z" F. All Velocities, 197 ft.permin. Fig. 361 Roller and Ball Bearings generally develop less resistance to turning than ordinary bearings. For descriptions of roller and ball bearings and information on their coefficients of resistance, students are referred to works on machine design; see also Kent's Pocket Book, or the American Civil Engineer's Pocket Book for coefl&cients of resistance. But to furnish some notion of relative values the two following tables are given. In the experiments from which the first was compiled* the speed was 560 revolutions per minute, and the loads were from 113 to 456 pounds. In the ball bearingsf the balls were | inch in diameter and they ran in grooves, cross sections of which were arcs of circles whose radius was equal to two-thirds the diameter of the balls; the circle of the centers of the balls was 4 inches in diameter. Coefficients of Journal Friction Different Bearings Ball Bearings Diameter of journal . Flexible rollers Solid rollers Babbitt bearing 2if 0.018 .014 .032 .025 0.022 .021 .027 0.043 .082 .096 .107 Load in pounds Revolutions per minute 6S 38s 780 840 2,420 4.500 10,800 0.0033 .0017 • 0015 0.0035 .0018 .0015 0.0037 .0019 .0015 .0011 * Benjamin, Machinery, N. Y., 1905, Vol. 12, p. 62. t Stribeck, Z. d V. d. I., igoi, Vol. 45, p. 121. Art. 45 22711^ Tests to determine the coefficients of friction for ball, flexible roller, and babbitt bearings for line shafts have been made at the University of Wiscon- sin. The diameter of the shaft was 2^^ inches, the speed 150 to 450 revolu- tions per minute, the load 700 to 2250 pounds per bearing, (33 to 100 pounds per square inch for the babbitt bearings); the extreme (natural) variation of the temperature of the lubricants was from 65 to no degrees Fahrenheit. For absolute values of the coefficients for the various conditions named, see report of the tests.* The relative value of the coefficients for four condi- tions are given in the following table : RELATIVE VALUES OF COEFFICIENTS OF JOURNAL FRICTION, AT LOAD OF ABOUT 1200 POUNDS PER BEARING. Peripheral Speed Lubricant Temperature. BaH bearings. F]exible roller. Babbitt 100 ft/min. 77 deg. 100 deg. I 2-5 3-6 300 ft/min. 77 deg. I 2.7 4 5 100 deg. I 3 4 * Thomas, Maurer, and Kelso, Jour. Am. Soc. Mech. Engrs. for March, 1914. CHAPTER XI MOMENTUM AND IMPULSE* 46. Linear Momentum and Impulse § I. (Linear) Momentum. — By momentum of a moving particle is meant the product of its mass and velocity. We regard momentum as having direc- tion, namely, that of velocity; thus, momentum is a vector quantity. By momentum of a collection of particles is meant the vector-sum of the momen- tums of the particles. For example, let m' and m" = the masses of two particles (Fig. 362), v' and v" = the velocities of the par- ticles at a certain instant, and suppose that AB = m'v' and BC = m"v" according to some convenient scale; then' AC represents the ^ momentum of the two particles. Since the component of the vector AC along any line equals the algebraic sum of the components of the vectors AB and BC along that line, it follows that the component of the momentum of a pair of particles along any line equals the algebraic sum of the components of their momentums along that line. Obviously, this proposition can be extended to a collection of any number of particles. A simple expression for this component can be arrived at as follows: Let m', m" , m' ", etc. = the masses of the particles; v' , v", etc. = their velocities; and v'^, -a" x, etc. = the components of these velocities along any Une X. Then the component of the momentum of the collection along this line = ^'^'^ _|_ yn"v"x + . . . Now if x' , x", etc. = the x coordinates of the moving particles, and x = the x coordinate of the mass-center, all at the same instant, then m'x' + m"x" + . . . = xHm (Art. 34); and differentiating with respect to /, we get m'dx'/dt -\- ni"dx"/dt -|- . . . = {dx/dt)llm, or m'v'^ + m"v"x +....= VjJZm = Mvx, where M = 2m = the mass of the collection. That is, the x component of the momentum of the collection of particles equals the product of the mass of the collection and the x component of the velocity of the mass-center. Hence, the component momentum is just' the same as though all the material of the body were concentrated at the mass-center. In the case of a body having a motion of translation, all the particles have at any instant velocities which are equal in magnitude and the same in direc- tion (Art. 35). Hence the momentums of the particles are parallel, and their * This chapter is not prerequisite to Chapter XII. 228 Art. 46 229 vector sum = m'v + m"v + . , . = vTtm = Mv, where v = their common velocity and M = the mass of the body. The definition of momentum impUes that the unit of momentum equals the momentum of a body of unit mass moving with unit velocity. The magnitude of the unit, therefore, depends on the units of mass and velocity used. No single word has been generally accepted for any unit of momentum. The dimensional formula for momentum is F'T' (see appendix A), that is, a unit momentmn is one dimension in force and one in time. Hence, any unit of momentum may be and commonly is called by names of the units of force and time used. Thus the unit of momentum in the C.G.S. system is called the dyne-second; in the ''engineers' system," the pound (force) -second, etc. In Art. 34 it is explained that the acceleration of the mass-center of any collection of particles does not depend at all on the forces which the particles exert upon each other but on the external forces; also that the algebraic sum of the components of the external forces along any line equals the product of the mass of the system and the component of the acceleration of the mass- center along that line, that is, n + i^"x+ . . . = Mi., (i) where F'x, P"x, etc., are the components of the external forces along a line x, and ax is the x component of the acceleration of the mass-center. Now ax equals the rate at which the x component of the velocity of the mass-center changes, that is, ax — dvx/dt, where Vx is the x component of the velocity of the mass-center; hence, Max = M dvx/dt = d{Mvx)/dt; and finally F'x + F"x + . . . = d{,Mvx)/dt (2) But Mvx is the x component of the momentum of the system, and d{Mv^/dt is the rate at which that component changes; hence the algebraic sum of the components of the external forces along any line x equals the rate at which the x component of the linear momentum changes. The principle just arrived at (equation 2) was derived from the law of motion of the mass-center (equation i), and it is essentially an alternative form of the law. But practically the former seems to apply more simply in certain cases as the following examples show. Fig. 363 represents a jet of water impinging against a flat plate. Required the pressure of the jet upon the plate. Let W = the weight of water impinging per unit time, v = the velocity of the water in the jet, and a = the angle between the jet and the plate as indicated. We suppose that the water does not rebound from the plate with any considerable velocity; then the momen- tum of the water after striking has no component normal to the plate. The momentum of an amount of water equal to W before striking is {W/g)v, and the component of that momentum along the normal to the plate = (W/g)v sin a; hence the change in the (normal) component momentum is iW/g)v sin a. This change takes place in unit time; therefore, it is the rate at which 230 Chap, xi momentum along the normal is changed, and also the value of the normal pressure of the plate against the jet. The jet exerts an equal (normal) pressure against the plate. If the plate is rough, then the water also exerts a frictional force on the plate. -w/m//w^/w7i^;/w,. Fig. 363 For another example, we will determine the pressure on a bend in a pipe by water flowing through it at constant velocity. Let W = the weight of the water flowing past any section of the pipe per unit time; v = velocity of the water, assumed to be the same at all points of inlet and outlet cross sections of the bend; and a = the angle of the bend (Fig. 364). Also let A^ = the time required for the body of water AB to move into the position A'B'. The momentum of the body of water at the beginning of the interval = that of AA' -\- that of A'B; its momentum at the end of the interval = that oi A'B -\- that of BB' . Hence the change in the momentum of the body of water in the time Ai = momentum of BB' — momentum of AA'. These momentums respectively are in the direction BB' and AA'; each equals {WM/g)v. Hence the change of momentum under consideration is represented by the vector MN where OM and ON represent the two mo- mentums just mentioned. But MN = 2{0M) sm\a; hence the change = 2{WM/g)v sin ^ a, and the rate at which the change occurs = 2{W/g)v sin ^ a. The direction of this rate is MN; it bisects the angle a. This rate of change of momentum is maintained by the forces acting on the body of water in A 'B. Those forces consist of gravity G, the pressures Pi and P2 (of the water) on the front and rear faces of the body, and the pressure P of the bend upon it. Their resultant R = 2{W/g)v sin | a, and R bisects a. If R, G, Pi and P2 are known then P can be determined. For it is such a force which compounded with G, Pi and Pi gives R. The pressure of the water on the bend = — P. For another example, we take the jet propeller of a ship. This consists essentially of a pump which takes in water from the sea and ejects it from nozzles toward the rear (to propel the ship forward). Let W = weight of water so ejected per unit time, v = velocity of the ship, and V — velocity of the ejected water relative to the ship. The absolute velocity of the jet (rela- tive to the sea) = V — v. Hence the amount of momentum produced by the pimiping plant (pump, pipes, etc.) per unit time = (W/g) (V — v). The direction of this is horizontal and backward; hence the plant exerts a force on the body of water within the passages at any instant equal to (W/g) {V — v); the water exerts an equal force forward on the passages. Art. 46 231 If the algebraic sum of the components — along any line — of the external forces acting on a body equals zero, then the rate of change of the component momentum (along that Une) equals zero; hence, if the sum remains zero for any interval of time, the component momentum remains constant. This is known as the principle of conservation of linear momentum. It follows that if there are no external forces acting on the body, its linear momentum remains constant. The grand illustration of this principle is furnished by the solar system. Even the nearest stars exert no appreciable attractions on the solar system, and so the members of the system move under the action of their mutual attractions only. Accordingly, the component of the momentum of the system along any line does not change; the linear momentum is, therefore, constant in amount and direction. It follows that the mass-center of the system moves uniformly, and in a straight line. § 2. (Linear) Impulse. — If the magnitude and direction of a force are constant for any interval of time, then the product of the magnitude of the force and the interval is called the impulse of the force for that interval. If the magnitude varies, then the impulse for any interval equals the sum of the impulses for all the elementary periods of time which make up the interval ; that is impulse = lim [F'At + F"At + . . . . ] = j Fdt, where F = the varying force. If the direction of the force varies, we regard the impulse for any elementary portion of time as a vector quantity having the direction of the force, and then in principle we add (vectorially) the ele- mentary impulses for all the portions of time which make up the interval. That is to say, we integrate Fdt vectorially, arriving at a definite vector quantity. Units of impulse depend on the units of force and time used.* There is no current single- word name for any unit of impulse. Each unit is named by the names of the units of force and time involved in it. Thus, in the C.G.S. system the unit of ; impulse is the dyne-second; in the "engineers' system" the unit of impulse is the pound (force) -second. Fig* "365 It is evident (Fig. 365) that the elementary inpulse F dt is the resultant of the impulses of the x and y components of F (or .v, y, and 2 components, if preferred). Hence the x, y, and z components respec- tively of the impulse of F equal the impulses of the components of the force F. If we integrate equation (2) over any interval ti - ti, say, we get f'^F'^dt + rV".(f^ -f- . . . = MvJ' - M^J = A (M^x), (3) where vj and vj' = the x velocities of the center of gravity of the system at times ti and fe respectively. Equation (3) can be put into the following * See appendix A. 232 Chap, xi principle of {linear) impulse and momentum. The algebraic sum of the com- ponents — along any line — of the impulses of the external forces acting on any system of particles equals the increment in the component of the mo- mentum of the system along that same line, the sum and the increment re- ferring to any interval of time. The principle of impulse and momentum answers such questions as, — how much velocity in a given time? or how much time to produce a certain ve- locity? For example, it is required to ascertain how much time is required to give a velocity of 40 feet per second to a certain body by sliding it along a horizontal rail by means of a constant push of 20 pounds, the body weighing 100 pounds and the frictional resistance of the rail being 8 pounds. The ex- ternal forces acting on the body are gravity, the push, and the reaction of the rail, the horizontal and vertical components of which are friction and the " normal pressure." Only the impulses of the push and friction have com- ponents along the line of motion; hence 20 / — 8 / = (100/32.2) 40, where / = the required time. Therefore / = 10.3 seconds. Solution of such a problem by earlier methods of this book would be as follows: Let a = the acceleration; then a = (20 — 8) -^ (100/32.2) = 3.86 feet per second per second. Hence / = 40 -^ 3.86 — 10.3 seconds. 47. Impact or Collision § I. Blow. — Momentum of a blow, energy of a blow, and especially force of a blow are terms generally used more or less vaguely. But when one of the two colliding bodies is fixed, then the first two terms are taken to mean the momentum and the kinetic energy respectively of the moving body just before the impact, perfectly definite quantities. If the motion is one of translation, these are Mv = {W/g)v and \ Mv^ = | {W/g)v' respectively. If in a numerical case we write g = 32.2 (feet per second per second), v should be expressed in feet per second; W may be expressed in any force unit. If the pound is used, then the momentum is in pound-seconds and the energy in foot-pounds. Force of a blow means the pressure which two colliding bodies exert upon each other. The pressure changes during the collision. Analysis of this variation is beyond the scope of this book. We will deal only with average values of the force of a blow. In the first place, it should be noted that there are two average values of the force of a given blow, — a space-average and a time-average. We explain the distinction by means of an example, but we choose the simpler case of a varying horizontal pull dragging a body along a smooth horizontal surface instead of a blow. Let us suppose first that the pull varies uniformly with respect to time, from a zero value to 40 pounds in 20 seconds (see Fig. 366). Then the time-average is represented by the average ordinate to the line which shows how the force varies with respect to Art. 47 233 the time; hence it is 20 pounds. We wish to find now how the force varies with respect to distance. Let P = the value of the pull at any time t after starting; then the law of force is P = 2 ^. Also let M = mass of the body; a and v respectively = the acceleration and velocity at any time t, and s = the displacement up to that time. Then a = -r7 = ir7i, V = Yji", and 5 = — r-^ t\ The total displacement (^i) in the 20 seconds = (1/3 M) 8000. It follows from the last equation that t = (3 Ms)^\ hence P = 2 (3 Ms)^. This equation determines the graph shown in Fig. 367, from which it is ap- parent that the space-average force is more than 20 pounds, or the time-aver- 40 lbs — ^^ ■30 /" •20 / ■\oy 5 10 15 SOsec Fig. 366 •-1000 -e-3M Fig. 367 age. The space-average equals the area under the curve divided by the base of the area. The area is / '' P ds^ I "2 isMs)^ ds = 1.5 (3M)^5i*; hence the space-average = 1.5 (3 AT)^ ^i^ = 30 pounds. It will be observed that the space-average is that constant force whose work equals the work done by the (real) varying force (see Art. 40). Like- wise the time-average is that constant force whose impulse equals the impulse of the (real) varying force. Hence the space-average equals the quotient of the work done by the force (equal to the kinetic energy produced by the force) and the distance through which the force acted; and the time-average equals the quotient of the impulse of the force (equal to the momentum produced by the force) and the duration of the impulse. Crushers or crusher gages are small cylinders of copper or lead — one inch diameter and one inch high are common proportions — used in a way de- scribed presently to determine the energy and force of a blow. Fig. 3 68 shows several energy-compression curves for i\- by i|-inch lead cylinders.* Curve A-B IS a so-called static curve, and was obtained by crushing a lead cylinder in an ordinary testing machine at 0.05 inches per minute. The amount of compressing force and the amount of the compression were observed at fre- quent stages during the test, and from these observations the amount of work * American Machinist, Vol. i^, Part i, p. 436 (1910). 234 Chap, xi done on the cylinder up to each stage was computed. Amounts of compression and corresponding amounts of work were plotted to determine the curve. Curve C is a static curve but for a higher speed. D is a so-called dynamic curve. It was obtained from drop or impact tests in which each crusher was subjected to a blow from a "hammer" dropped upon it. The hammer c o 1- O- e o U A t ". AU t -c _ — ^^ X\'< .1 ^- 1— ^ ^ •D ^^ ,»— — \J ^ .^ ' -— ^ ■^ .0 / ,^ ^ ^^^ -^ .8 / /• .^ ^ J / ^^ '^ ^ 4 ^ ^>' .6 l/t 7^ .4 T .2 • 10000 20000 30000 Energ-y, inch-pounds. Fig. 368 40000 50000 weighed 1330 pounds, and the maximum drop used was 38 inches. For each test the amount of compression c was observed and the amount of work 1330 {h + c) was computed, where //- = height of drop to the cylinder; and this compression and work were plotted for one point on the curve. Curve E-F is a dynamic curve obtained from tests in which the hammers were lighter and the drops higher than for D. Crushers are used as follows to determine the energy of a blow, as of a steam hammer for example. The crusher is placed on the anvil and subjected to the blow. Then the amount of the compression of the crusher is measured, and the corresponding energy is read off from the appropriate compression-energy curve (previously determined from tests on crushers like the one used). The space-average force of such a blow equals the quotient of the energy of the blow and the compression unless cjh is not a small fraction ; in that case the space- average = W Qi -\- c) -^ c. A very skillful use of the crusher was made by Lieut. B. W. Dunn to deter- mine not only the average but the actual value of the force of a blow at any instant during the impact.* He devised apparatus which made a photo- graphic record (space-time graph. Art. 29) of the motion of a hammer during the impact. From that graph he deduced the velocity-time, and from this the retardation-time graph; then the force of the blow F at any instant from /r = p[7 -|- {W/g) a, where W = weight of hammer and a = retardation at the instant. The order of measurements involved in this apparatus is indicated by these circumstances of one test: amount of compression of crusher about * Journal of the Franklin Institute, Vol. 144, p. 321 (1897). Art. 47 235 ^ inch and time of impact about xmiTT second; the weight of hammer was ^T, pounds and the drop 15 inches. For the copper crushers used the maxi- mum pressure occurred just before the end of the compression, and its value was slightly less than twice the space-average. § 2. Motion after Collision. — In this section we discuss the changes of motion of one or both colliding bodies due to the collision in certain compara- tively simple cases. In most cases of collision the pressures which the colliding bodies exert on each other are enormous compared with other forces acting on the bodies. For example, the space-average pressure between two billiard balls colliding with velocity of 8 feet per second is about 1300 pounds. There- fore in discussing changes of motion of the bodies during collision we may neglect the other (ordinary) forces acting on the bodies, gravity for example; that is we regard the two bodies jointly as under the action of no external forces. Hence, according to the principle of conservation (Art. 46), the momentum of the two bodies jointly is not changed by the impact. If the centers of gravity of two bodies about to collide are moving along the same straight line, then the collision or impact is called direct; if otherwise, oblique. If the pressures which two colliding bodies exert upon each other during impact are directed along the line joining their centers of gravity, then the impact is called central; if otherwise, eccentric. These are the kinds of impact called simple, above. Direct Central Impact. — We assume that the bodies have motions of trans- lation before impact. Since the impact is supposed to be central, the pressure (of impact) on each body acts through the center of gravity of that body and does not turn it. Hence the motion of each body after collision is one of translation. Let A and B be the two bodies, Ml and M2 = their masses, Ui and ih. = their velocities just before impact, and Vi and V2 = their velocities just after impact respectively. We regard these velocities as having sign; velocity in one direction (along the line of motion) being positive, and that in the other being negative. Then the momentum of the two bodies before impact = MiUi -f- M2M2, and after impact it = MiVi -\- M2V2. Since the momentums before and after impact are equal, we have MiVi + M2V2 = MiUx + 7I/2M2. (i) The foregoing expressions are correct whether A and B are moving in the same or opposite directions before or after the impact. Thus, if both are moving toward the right before impact, at 8 and 10 feet per second say, their momen- tum is 8 ^1/1 -|- 10 7I/2; but if A is moving toward the right and B toward the left, their momentum is 8 Mi — 10 M2. It has been learned experimentally that when two spheres A and B collide directly and centrally the velocity of separation is always less than and opposite' to the velocity of approach, and the ratio of these two velocities seems to 236 Chap, xi depend only on the material of the two spheres. The ratio of the velocity of separation to that of approach (signs disregarded) is called coefficient of restitu- tion; it is generally denoted by e. The following are approximate values of e for a few materials, glass II, ivory |, steel and cork |, wood about \, clay and putty o. Now the velocity of approach equals Ui — ih (or W2 — Ui), — the first with reference to A (regarded as fixed) and the second with reference to B (re- garded as fixed) — , and the velocity of separation is Vi — Vo (or v^ — Vi). Since these velocities are opposite in direction, we have — {vi — v^/{ui — U2) = e, or — {vi — V2) = e (th — U2). (2) Equations (i) and (2) solved simultaneously for the final velocities Vi and V2 give v. = u^-(i+e)j^^j^^{u,-u.);v2 = i^-{i + e)j^-^^^^ (3) If one of the colHding bodies is fixed, say B, then 1^ = o, and M2 is the mass of B and its supports, infinitely great. Thus we have Vy = — eui. Oblique Central Impact. — We assume as before that the bodies A and B have a motion of translation before impact; then the pressure on each during the impact acts through the center of gravity and produces no turning. Let Ui and 1/2 = the velocities of A and B before impact; Vi and F2 their velocities after impact; Ui and M2 = the components of Ui and U2 along the fine of impact pressure (joining the centers of gravity of A and B when in con- tact); V]_ and V2 = the components of Vi and F2 ^' along that line; and Wi and W2 = the components Pjg . of Z7i and U2 at right angles to that line. See Fig. 369 which represents one of several possible ways of oblique collision. Since the impact pressure on either body has no component transversely to the line of pressure XX, the component of the momentum of either body at right angles to XX is not changed. Hence the transverse component of the velocity of either body is not changed by the impact. The longitudinal components are changed as in direct impact, and vi and V2 are given by equations (3). The final velocities Fi and F2, therefore, are determined, Vi by its components Vi and Wi, and V2 by its components % and W2. Loss of Energy in Impact. — Let L = the loss of kinetic energy; then L = (i M,U^' + i M2U2') - (^ MiFi^ + \ M2V2'). Now f/i^ = wi2 -f wi2, C/2' = ^2^ + ■W2\ Vi" = Vi" + Wi", and Fg^ = V2^ + W2^; hence Z = i MM' - V) + h M2W - ^2^). Substituting for Vi and V2 their values from equation (3) and simplifying we get . . „, M1M2 , s. Art. 48 237 For perfectly elastic bodies (e = i), L = o. For other bodies (i — e-) is not zero but a positive quantity; and since (th — ito) is not zero, L is always a finite positive quantity. That is, in every collision of bodies not perfectly elastic there is loss of kinetic energy. If the bodies are without elasticity (e = o), the loss = | [{M,M2)/(Mi + M.)] (ih - u^y. The foregoing is essentially Newton's analysis of impact. Several more recent analyses have been made independent of any coefificient of restitution but taking into account the vibrations set up in the colliding bodies. On account of the difficulties of the problem they include only impact of spheres and cylinders end on. Explanation of these analyses fall beyond the scope of this book.* 48. Angular Momentum and Impulse § I. Angular Momentum. — The linear momentum of a moving particle is a vector quantity, as explained in Art. 46 ; the magnitude of the momentum is mv (where m = mass of the particle and v = its velocity), and the direction is that of the velocity. We go farther now and assign position to the mo- mentum and to the momentum-vector. The position, or position-line, of the momentum of a moving particle is the line through the particle in the direction of the velocity. Thus the Hnear momentum of a particle is a " locaUzed " vector quantity, — like a concentrated force, which has magnitude, direc- tion and a definite position, or line of action as it is more commonly called. We apply the term moment of momentum to a product which is analogous to the product which we call moment of a force about a line. Thus the moment of momentum of a moving particle about a line (or angular momentimi as it is also called) is the product of the component of the momentum perpendicular to the Une -- the other component being parallel to it — and the distance from the line to the per- pendicular component. (Compare definition of moment of a force about a line, Art. 8.) For example, let (Fig. 370) be the position of the moving particle at a given instant, OC the direc- ( tion of its velocity, and OABC a parallelogram Fig. 370 whose sides are parallel and perpendicular to the line LL\ an axis of moments. (QQ is a plane perpendicular to LL' represented to make the figure more plain.) Then according to some scale OC represents the momentum mv, and OA and OB represent components of mv perpendicular ajid parallel to LL' respectively. The angular momen- tum of the particle about LL' is OA X PL. It follows from the definition of * See Love's Theory of Elasticity, Vol. 2; Nature, Vol. 88, p. 531 (1912) for an instructive paper by Prof. Hopkinson, on "The Pressure of a Blow"; also Journal of the Franklin In- slitutc, Vol. 172, p. 22 (1911) for an account of some determinations of the time of impact of metal spheres. 238 Chap, xi the term, that the angular momentum of a particle about a line parallel to its momentum is zero; and about a line perpendicular to its momentum the angular momentum is the product of the momentum and the distance from the line to the particle. There is another method for computing the angular momentum of a mov- ing particle about a line which is more simple generally than that described in the definition of angular momentum. It is as follows: we resolve the momentum into three rectangular components, one of which is parallel to the axis ot moments — then the other two are perpendicular to the axis — , and add the moments of the two perpendicular components about the line; the sum equals the angular momentum of the particle. Proof: Imagine the momentum OC (Fig. 370) resolved first into two rectangular components OA and OB as before, and then OA into any two rectangular components per- pendicular to LV. These last two are not shown in the figure but their relations to OA and the axis LL' are shown in projection on the plane QQ in Fig. 371. The moment of the component O'M about LL' is O'AI X L'm ' B U^ X Fig. 371 Fig. 372 = O'M X O'L' sin fi = O'M sin fx X O'L'. The moment of the component O'N is O'N X L'n = O'N X O'L' sin 7 = O'N sin 7 X O'L'. Hence the sum of the moments = (O'M sin fi + O'N sin 7) O'L' = O'A' sin a X O'L' = O'A' X O'L' sin a = O'A' X L'P' which is the angular momentum of the particle as defined. *■ By angular momentum of any collection of particles (body) about a line is meant the algebraic sum of the angular momentums of the particles about that line. In the case of a rigid body rotating about a fixed axis, the angular momentum of the body about the axis of rotation can be computed quite easily. Thus let Wi, fth, etc., = the masses of the particles of the body; r\, 1-2, etc., = the distances of the particles respectively from the axis of rotation; and CO = the angular velocity of the body. Then the linear velocities of the particles are respectively rico, r2co, etc. (Art. 37), and their linear momentums are Wiriw, m^roco, etc. These momentums are perpendicular to the axis of moments; hence the angular momentums are ntiViuri, nhrooir-i, etc. And since these are of the same sign, the angular momentum of the body is Wi^i^co + miV^cji + . . . = coSwr- = co/, where / = the moment of inertia of the body about the axis of rotation (Art. 36). A general formula for the angular momentum of a body about a line can be Art. 48 239 arrived at as follows: Let P (Fig. 372) be one of the particles of the body, OX the line about which to compute the angular momentum, and PD = the velocity of P. Let OXYZ be a set of fixed coordinate axes; x, j, and z = the (varying) coordinates of P; m = mass of P; d = velocity of P; v^c, Vy, and v^ = the axial components of v (represented by PA , PB, and PC respectively). Then to some scale, PD represents the momentum mv of the particle, and PA, PB, and PC represent the axial components of the momentum; these equal mv^, mvy, and mv^ respectively. Hence the angular momentum of P about OZ 'J/) Y. A'^y -^-x- R> is mVyX — mV:,y, and the angular momentum of the entire body is 2 {mVyX — mVxy). We will now ascertain how the angular momentum of a body about any line depends on the forces concerned in the motion. Let P, Fig. 373, be one of the particles of a body, OX a fixed line about which the angular momentum in taken, R = the resultant of all the forces acting on this particle, v = its velocity, and a = its acceleration. Further, let the coordinates of P at any particular instant under consideration be .v, y, and s referred to axes one of which is the line OX; R^, Ry, and R^ = the axial components of R; v^, Vy, and Vz = the axial components of v; a^, Oy, and Oz = the axial components of a; and Tz = the torque of all the forces acting on P about the z axis. Then T^ = RyX - R^y (Art. 8); and since Rx = ma^ and Ry = may (Art. 34), Tg = niGyX — maxy. Now imagine one equation like the last written down for each particle of the body. The sum of the left-hand members equals the sum of the right-hand members of course. To the first sum the internal forces (exerted by the particles upon each other) contribute nothing because these internal forces occur in pairs, the forces of each being colinear, ecj^ual, and opposite, and so the moments of such two forces cancel. Therefore, the first sum is also the torque of the external forces about the 2 axis. Thus, we have Fig. 373 SPj = 2 {mayX — maxy), (i) where SP^ = the torque of all the external forces, acting on the body, about the 2 axis. The second sum, 2 {mayX — maxy), equals the rate at which the angular momentum of the body about the z axis is changing. We prove this by differentiating the expression for angular momentum about the 2 axis, S {mVyX — mVxy), with respect to the time; thus -J- 2 {mVyX — mVxy) = 2 at m (dvy , dx\ (dvx , dyW 240 Chap, xi Now dvy/dt = ay, dx/dt = Vx, dvx/dt = d, and dy/dt = Vy-, and substitution of these equivalents of the four derivatives in the long equation gives -7- 2 {mVyX — mvxy)= 2 {mayX — maxj), at which was to be proved. Thus finally we have the important principle that the torque oj the external forces, acting on any body, about any line equals the rate at which the angular momentum of the body about that line is changing, or • 2r, = dh/dt, (2) where the line in question is called z, and hz = the angular momentum of the body about that line. For an example we will apply the foregoing principle to determine the torque of the water flowing through the water motor (Barker Mill) repre- sented in Fig. 374. Essentially, the motor consists of a horizontal cylinder AB, mounted on a vertical pivot C, and an inlet D connected by a water-tight sleeve joint to a feed pipe E. On opposite sides of the cylinder and near its ends there are orifices or nozzles through which the water escapes horizontally. The water turns the motor in the opposite direction. Let W = the weight of water escaping per unit time, v = the velocity of escape relative to the orifices, and CO = the angular velocity of the motor. The amount of water which escapes in a short interval of time A/ is W M; and, since the absolute velocity of escape = t> — rco (Art. 53), the angular momentum of this water about the axis of rotation is {WM/g) (v — rw) r. Hence the rate at which the motor gives angular momentum to the water is (W/g) {v - rco)r, and this equals the torque of the motor on the water; also the torque of the water on the motor. ' If the torque — about any line — of the external forces acting on a body equals zero, then the rate of change of the angular momentum of the body about that line equals zero; hence, if the torque remains zero for any interval of time, then the angular momentum remains constant. This is known as the principle of the conservation of angular momentum. It can be well illustrated by means of the apparatus on which the man (Fig. 375) is standing. It consists of a metal plate A supported on balls in suitable circular races in A and B so that A can be rotated about the line C with very little friction resistance; B is fixed. Imagine that a man has mounted the plate A and holds a balancing pole as shown, all being at rest; then the angular momentum of the man-plate- pole system about CC equals zero. Now suppose that the man exerts himself in any way, to move the pole about for example, but touches nothing except A and the pole. The only external forces acting on the system are gravity, reactions of the balls on A , and the air pressure. The first has no torque about C; the other two very little and are negligible here. Hence there is no external Art. 48 241 torque about C, and the angular momentum of the system about C equals zero always. This is strikingly illustrated if the man, without moving his feet on the plate, trys to rotate the pole (over his head as shown) about C. In doing so, he and A begin to rotate in the opposite direction. If / and /' = the moments of inertia of man (and .1) and the pole respectively about C, and co and w' = their angular velocities at any instant, then the principle requires that the angular momentums /w and Ico' shall be equal (and opposite). Or, imagine the man-plate-pole system is given an angular velocity by external means (the man holding the rod as shown, say), and then left to itself. If now the man should change the pole into a vertical position before him, he would reduce the moment of inertia of the system (about C) very materially; and since the angular momentum must remain constant, the angular velocity of the system would increase accordingly. The grand illustration of the principle of conservation of angular momentum is furnished by the solar system. The system moves under the influence of no external forces; hence the angular momentum of the system about any line remains constant. The angular momentum about a certain line through the mass-center of the system is greater than that about any other such line. The Une is known as the invariable axis of the system — a plane perpendicular to it as the invariable plane — and ''is the nearest approach to an absolutely fixed direction yet known." Center of Percussion. — Fig. 376 represents a body OC suspended like a pendulum; C is the center of suspension, and C is the center of gravity or mass- center of the body. Let R = the reaction of the axle supporting the pendulum. J-4 = E !lD I k - r >i<- r — H Fig. 374 B rO Ry C Fig. 376 and P = the time average force of a blow applied as shown. In general, R would not be vertical during the blow; so let R^ and Ry = the horizontal and vertical components of the time-average of R during the blow. The value of R^ depends not only on the force of the blow P but also on the arm of the blow with respect to the axis of suspension. It will be shown presently that 242 Chap, xi if the arm has a certain value, then Rx equals zero. The point Q in OC (ex- tended) and in the line of action of a blow applied as just explained so that there is no component axle reaction parallel to the blow, is called the center of percussion of the body for the particular axis of suspension. {Q is the point that was called center of oscillation in Art. 39.) The distance of the center of percussion from the axis of suspension equals q = k'^/c = c -\-k /c, where k = the radius of gyration of the pendulum about the axis of suspension, c = the distance from the center of gravity to that axis, and k = the radius of gyration about a Une through the mass-center and parallel to the axis of suspension. To develop the expression for q given above let M = the mass of the body, p = the arm of P about the axis of suspension, co = the angular velocity of the body produced by the blow, and At = the duration of the blow. By the end of the blow the velocity of C will be ceo, and practically horizontal; hence, according to Art. 46, P - Rx = Mco:/At. The only force which has a torque about during the blow is P; hence Pp = M¥ui/M. These two equations solved simultaneously for R^ give R^ = P (1 — cp/P); therefore, if ^ = k^/c, Rx = o which was to be shown. Every American boy has batted a baseball a few times in such a way that the bat "stung " his hands; and he soon learned that such stinging is a result of impact near his hands or quite near the big end of the bat; in fact, quite remote from the center of percussion of the bat (with reference to the particular axis of rotation about which the bat was being swung at the instant of impact). Such a blow also results in rapid vibrations of the material of the bat which cause the sting. Large pendulums are used in certain impact testing machines for striking a blow. To avoid the impulsive reaction at the suspension and vibrations in the pendulum, they are always so arranged that the line of action of the blow passes through the center of percussion of the pendulum. § 2. Angular Impulse. — If the line of action of a force is fixed in posi- tion then the angular impulse of that force for any interval about any line is the moment of the impulse of the force for the interval about that Hne. The moment of an impulse is computed just like moment of a force (Art. 8) or angular momentum; that is, we resolve the impulse into two components, one parallel and one perpendicular to the line and then we take the product of the perpendicular component and the distance from it to the line. If the line of action of the force changes then the angular impulse of the force about any line for any interval is the algebraic sum of the angular impulses for all the elementary portions of time which comprise the interval. Thus let F = the force, i" — /' = the interval, 6 = the angle between the line of action F Art. 48 243 and the line, and p = the perpendicular distance between the two lines. Then the angular impulse is Xt" nt" F dt-?,md- p= \ F sin d - p dt. Since F sinO • p — the torque of the force about the line in question, the angular impulse of the force may also be regarded as the time-integral of the torque of the force. Hence, if T — the torque of the force about the line at any instant then the angular momentum for the interval equals Tdt. £ h' Now let us integrate equation (2) over any interval /" — /' say; then r ^T,dt, or 2 r T,dt, = ///' - hj = A/?„ (3) in which JiJ and lij' denote the angular momentums of the body about the z axis at the times t' and t" respectively. Equation (3) can be put into the fol- lowing principle of angular impulse and momentum: The sum of the angular impulses of all the external forces acting on a body about any line equals the increment in the angular momentum of the body about that line. 49. Gyrostat § I. General Description. — The words gyroscope and gyrostat are generally used synonymously but sometimes a distinction is made, as follows: A gyrostat consists of a wheel and axle, both being symmetrical to the axis of the axle, and mounted so that they may be rotated about that axis; a gyro- scope consists of a gyrostat mounted in a frame which can be rotated. Fig. 377 represents a common form of gyroscope; the gyrostat (wheel W and axle A A') is supported by a ring R which can be rotated about the axis BB'; the axle BB' is supported by the /^^^-■^ forked pillar F which can be rotated about the axis CC. l\^5^/\^^. . Thus the wheel can be rotated about its center into t^^^^&A^^ any desired position. The gyroscope seems to have ^ IWV>^-/^)"^b' been designed for illustrating principles of composition ^^^y^JJJ of rotations (Art. 54). In 1852 Foucault (French phy- x.^-^^ sicist) made an interesting application of the instru- — i-"^-s^i=P" ment; by its means he practically made visible the k:;--_1_--^ rotation of the earth. More recently the gyroscope has p^^ been made use of in several connections, — to steer a torpedo, to serve as a substitute, unaffected by the iron of the ship, for the ordinary (magnetic) mariner's compass, to stabilize a mono-rail car, and to steady a ship in a rough sea; it has been proposed also to stabiHze flying machines by means of a gyroscope. When its wheel is spinning, a gyroscope possesses properties which seem 244 Chap, xi peculiar to students as yet uninformed in the matter, inasmuch as it does not always respond as expected to efforts made to change its motion or position. For example, if a gyroscope like that represented in Fig. 377, well made and practically frictionless at all bearings and pivots, be grasped by the pillar and then moved about in any way, the axle of the wheel remains fixed in direction in spite of any attempt to alter it. The (gimbal) method of support makes it impossible to exert any resultant torque on the gyrostat (by way of the pillar) about any line through the center; and hence, as will be proved later, the direction of the axle cannot be thus changed. It is this property of per- manence of direction of the spin-axis of a gimbal-supported gyrostat which is made use of in the self-steering torpedo. For another example, consider the effect of a torque applied directly to the gyrostat. A vertical force, say, applied at A would turn the gyrostat when not spinning about the axis B. But when spinning, that force U would rotate the spin-axis about the axis C, the direction of rotation depending upon the direction of spin. When the gyrostat is spinning in the direction indicated by the arrow co, then such force U would rotate the spin-axis about C in the direction indicated by the arrow fx. Again, a horizontal force applied at A, say, would turn the gyrostat when not spinning about the axis C. But w^hen spinning, such force L would rotate the spin-axis about BB'; and in the direc- tion indicated by the arrow X if the spin is as indicated. This behavior of a spinning gyrostat under the action of torque is exhibited more strikingly by a gyroscope represented plainly in Fig. 378. The wheel may be spun on the axle A; the gyrostat and its frame may be rotated about the axis BB'; and all may be rotated about the axis CC. W is the weight which can be clamped on the stem A' to balance or unbalance the frame with respect to the axis BB'. Now imagine W clamped so that the frame (with W and the gyrostat) is unbalanced. Then if the gyrostat is set spinning and the frame be released in the position shown, say, the frame will not rotate about BB' but about CC. The direction of this rotation depends on the direction of spin and on the di- rection of the torque of gravity about BB'. If, for example, W is clamped quite near BB' so that the torque of gravity is clockwise as seen from B and the spin is as indicated, then A rotates toward B. This rotation persists except in so far as it is interfered with by friction at the pivots, and air resistance. We might recite still other peculiar performances of a gyrostat but the fore- going suffice for our purpose. Professor Perry's book on "Spinning Tops" would be found interesting in this connection. Fig. 378 Art. 49 245 Any such rotation of the axis of a spinning gyrostat is called a precessional motion or precession of the axis or of the gyrostat; the axis and the gyrostat are said to precess. We will call precession normal or obiique according as the axis precesses about a line perpendicular or inclined to the axis. It may not be clear from the foregoing examples of precession how to predict the direction of precession that would result by applying a given torque to a gyro- stat with a given spin. The following is a simple rule for predicting; it is based on the dynamics of the whole matter as will be seen later: " When forces act upon a spinning body tending to cause rotation about any other axis than the spinning axis, the spinning axis sets itself in better agreement with the new (other) axis of rotation; perfect agreement would mean perfect parallelism, the direction of rotation being the same." (From "Spinning Tops".) Or, what amounts to the same thing, the precession is such as to turn the spin- vector* toward the couple or torque-vector. The following is an incomplete proof of the foregoing rule. Further ex- planation is given in the next section and in Art. 56. Fig. 379 represents a gyrostat pivoted at O so that it can be rotated freely about that point; we sup- pose the center of gravity of the gyrostat to be at O. Imagine that the gyrostat is -TIX at rest, not spinning, in the position shown, y^i^ ~^^^ and that a downward force is applied to ^ the axle on the left-hand side of and downward. The torque makes the gyrostat rotate about the axis OB, that is the torque produces angular momentum about that axis. The amount of an- gular momentum produced is proportional to the torque and to the duration of its action (see Art. 48). This angular momentum may be represented by a vector on OB, the length of the vector representing the amount of the angular momentum and the arrow-head pointing so as to agree with the direction of rotation, according to the usual convention, that is, forward in this case. Now imagine that the axis of the gyrostat is at rest in the position shown but the wheel spinning, say, counter-clockwise when viewed from the right. The an- gular momentum of the spinning gyrostat about its axisf would be represented by a vector on OA pointing in the direction OA; let 01 be that vector. * A spin-vector is a vector on the axis of spin, its arrow-head pointing to the place from which the spin appears counter-clockwise; or — what amounts to the same thing — the arrow- head points in the direction along which the axis would advance if it were a right-hand screw turning in a fixed nut. The length of the vector — immaterial in the present connection — represents the angular velocity of spin to some convenient scale. Likewise the couple-vector (see Art. 8) is a vector perpendicular to the plane of the couple pointing to the place from which the rotation, which the couple tends to produce, would appear counter-clockwise; or — what amounts to the same thing — the arrow-head points in the direction along which the vector would advance if it were a right-handed screw turned by the couple in a fixed nut. t This angular momentum is greater than that for any other line, and hence may bf regarded as the total or resultant angular momentum of the gyrostat (see Art. 55). 246 Chap, xi Now suppose that the torque already described comes into action, and let OJ represent the angular momentum which it would produce in a short interval of time. This angular momentum added to the original angular momentum gives OR as the resultant angular momentum of the gyrostat at the end of the interval. It seems, therefore, that the spin-axis would coincide with OR at the end of the interval; indeed, that axis does approach OR, that is the spin- axis turns toward the torque-axis as stated in the rule which v/e undertook to prove. The approach just mentioned is not a direct one; the gyrostat yields slightly to the torque just as though there were no spin; that is the wheel rises (in this instance) slightly. This is only the first (small) swing of a rapid oscillation of the spin-axis — nutation as it is called — which accompanies the (more prominent) precession of the spin-axis toward the torque-axis. The (unavoid- able) friction at the pivot O rapidly damps this oscillation so that the oscilla- tion generally escapes notice. The mentioned rise of the spin-axis may be explained as follows: In the approach of that axis toward OR the gyrostat rotates about OC, due to which it acquires angular momentum about OC, clockwise when viewed from above; but since there is no torque about OC, the gyrostat can acquire no (resultant) angular momentum about that line (see Art. 48 on conservation) ; hence the spin-axis rises so that at each instant the component along OC of the angular momentum due to spin just equals the angular momentum due to the rotation about OC. There is another item of gyrostat behavior worth noting here. Suppose that the gyrostat shown in Fig. 378 to be precessing as already explained. If the precession be hurried, say by means of a horizontal push applied at .4', the center of gravity of the frame (with gyrostat and weight) rises; if the precession be retarded, the center of gravity descends. This behavior is in accordance with the rule for predicting precession. In the first case we have a torque about CC; the torque vector is in the direction OC; the spin- vector is in the direction OA'; and in accordance with the rule OA' turns toward OC , that is the center of gravity rises. In the second case we have a torque about CC but the torque- vector is OC; and the spin- vector OA' turns toward that vector, that is the center of gravity descends. Thus we may state as another rule: Hurry a precession, the gyrostat rises or opposes the torque which causes the precession; retard a precession, the gyrostat falls, or yields to the torque which causes the precession. Self -steering Torpedo. — The gyroscope of such a torpedo is linked to appro- priate valves of a compressed air engine in such a way that any turning of the spin- axis toward either side of the torpedo causes the engine to turn the (vertical) rudder of the torpedo in the opposite direction. Prior to projection of a torpedo, the gimbals are locked so as to hold the spin-axis of the gyrostat parallel (or inchned at any desired angle) to the axis of the torpedo. During the discharge of the torpedo, the gyrostat is automatically set spinning and the gimbals are unlocked. During the flight, the spin-axis continues to point Art. 49 247 in its original direction ; any deviation of the torpedo from its intended course changes the incHnation of the spin-axis relative to the torpedo; simultaneously the gyroscope actuates the rudder as explained, and the torpedo is deflected back toward its proper direction. Like a common pendulum swinging to its lowest position, the torpedo swings beyond a mean direction, and is then swung back again by the rudder. And this oscillation is kept up during the flight so that the actual path of the torpedo is a zigzag, about two feet wide. A gyro- stat (wheel and axle) weighing 2 pounds and rotating at 2500 revolutions per minute has been made to serve the purpose just described. Gyro-compass. — For our purpose we may regard a gyro-compass as con- sisting essentially of a gyrostat (wheel and axle), the axle supported in a ring or case, and the ring suspended from above. See A, Fig. 380. Such a com- pass, when the gyrostat is spinning, sets its spin-axis into the plane of the meridian at the place where the compass happens to be. Imagine such a compass to be set up at the equator with its spin-axis pointing east and west, and suppose that the direction of spin is counter-clockwise when viewed from the west. The rotating earth carries the gyrostat eastward; the spin-axis would remain parallel to its original position if the gyrostat were supported in frictionless gimbals, and would in time be positioned as shown at B. Now consider the gyrostat as shown at B, supported not in gimbals but suspended from above as in the gyro- compass. The supporting force (above) and the force of gravity would have a torque counter-clockwise as viewed from the north; thus the torque vector would point toward the reader. The spin- vector points to the right; hence the torque would turn the end of the spin-axis marked n from the west toward the north. Of course the action is not precisely as outUned above, that is the spin-axis does not remain parallel to its original position for a time and then yield to the influence of the torque mentioned. The action is really continuous; the slightest rotation of the compass with the earth from the position A induces the gravity torque, and the spin-axis begins to turn toward the meridian as described. Though the restraint of the support (fine wire in the Sperry and mercury float in the Anschutz compass) is very small, the gravity torque is so small that the turning of the spin-axis into the meridian is very slow. Like a magnetic compass the gyro-compass swings beyond the meridian from a deflected position and oscillates for a time. In the Anschutz type the period of a free oscillation is about i hour and 20 minutes. Special damping ar- rangements reduce the oscillations to zero (from a deflected position of 40 degrees) in about one and one-half hours. The spin is maintained electrically, at about 20,000 revolutions per minute. Mono-rail Car. — A car on a single rail can be rendered stable even if the 248 Chap, xi center of gravity of the car is above the rail by means of a suitable gyroscope apparatus. Fig. 381 represents the germ of one type of such apparatus. A A' is the spin-axis, Z, is a lever rigidly fastened to the axle BB' by means of which the gyrostat can be made to precess about BB'. Imagine the car to be standing or travelling in an upright position, the gyrostat spinning, and a man standing on the car so that he may grasp and operate the lever. Now suppose that the car is tilted, as by a wind against either side. The car exerts tilting forces on the gyrostat axle at B and B', the torque-vector of which is parallel A' • =c= 4 r>B' cM t^.c Fig. 381 ^s^^^^ to the rail; hence (see the stated rule) the spin-axis begins to set itself parallel to the rail, that is it precesses about BB'. The axle BB' exerts (righting) reactions on the car but if the man will hurry the precession, the (heavy, rapidly spinning) gyrostat will rise against the tilting forces and carry the car back with it toward the vertical position. It is conceivable that a skillful operator could put the car back into its vertical position in one swing, but in general he would swing the car beyond the vertical, then back again and after a few oscillations, into its vertical position. Gyro-stabilizers as now built automatically perform the function of the man of the preceding explanation, and they include two gyrostats, spinning in opposite directions, to enable the car to run on a ciu-ve. The gyrostat wheels of a certain Brennan mono-rail car (40 feet long and weighing 22 tons) are 3I feet in diameter; each weighs f tons, and spins at 3000 revolutions per minute (in a vacuum to avoid air friction). Such a car has taken curves of 105 feet radius at a speed of 7 miles per hour without appreciable disturbance of the level of the car floor. The spin is maintained by electric means; in fact each gyro-wheel is made the armature of a motor and this is driven by a generator on the car. Schlick Gyro-stabilizer for Reducing the Rolling of a Ship. — This is repre- sented in Fig. 382. The gyrostat is mounted in a rigid frame F which is sup- ported in bearings B and B' fixed on the ship. Thus the wheel can be spun about A A' and the axle A A' can precess about BB'. P is a brake pulley by means of which this precession can be controlled. Explanation of the steady- ing action of this device is beyond the scope of this article. Such a stabilizer has been tried out in a ship no feet long, 12 feet wide and of 58 tons displace- ment. The gyro- wheel weighed iioo pounds, was i meter in diameter, and Art. 49 249 was spun at 1600 revolutions per minute. In still water the ship would settle down from a heel of 20 degrees to one of | degree in about 20 single oscillations; the period was about 4I seconds. The stabilizer produced the same extinction in less than three oscillations of 6 seconds period. (See London Engineering, Vol. 83, p. 448 (1907)). §2. Rate or Normal Precession; Determination of Forces. — In the preceding section, we discussed the effect of a torque on a spinning gyro- stat in a qualitative way; we will now discuss the matter quantitatively. Let / = the moment of inertia of the gyrostat about the axis of spin and CO = the angular velocity of spin; then /co — the angular momentum of the gyrostat about that axis (Art. 48). If T = the applied torque, the angular momentum produced by it in the element of time dt is T dt, and the angular approach of the spin-axis toward the torque-axis in that time is lOR (Fig. 379) = tan-i (r dt/Iw) = T dt/Iw. The rate at which this angle is described, that is the angular velocity of precession — generally denoted by O — is n = {IOR)/dt = r//co. If the torque is applied so that its vector is always perpendicular to the axis of spin OA, then there is no torque about OA and hence co is constant; if also the magnitude of the torque is constant, then it follows from the preceding formula that O is constant. That is, in the case assumed, the velocities of spin and precession are constant. The case is quite analogous to that of a moving particle subjected to a constant force whose line of action is always perpendicular to the direction of motion and in a given plane. Such a force does not change the magnitude of the velocity but continually changes the direction of it; indeed, the particle describes a circle with constant speed (Art. 34). Let P (Fig. 383) be the particle, m = its mass, v = its velocity, F = the force, PQ be the path and r = the radius of the circle. The linear momentum = mv; the angle. POQ - i.-ja^'^ > ^^I through which the vector mv is turned in any time / is ,|__7_t^v$r:r^R vt/r. Since r = mv^/F (see Art. 34), the angle = tF/mv. Hence the rate at which F turns the linear momentum vector is F/mv, a result strictly analogous with T/Ioj, the rate at which the torque T turns the angular momentum ^^^ ^ vector /o). The result can be arrived at, independently of Art. 34, in a way to bring out the analogy still more. We may regard F constant in direction for an element of time dt. During that time it produces an amount of momentum, in its own direction PO, equal to F dt. Let PJ represent this momentum and PI the initial momentum mv. At the end of the interval the (resultant) momentum is represented by PR. Hence the change in the direction of the momentum is I PR = {F dt) -=- (mv), and the rate at which the change occurs is the change divided by dt, that is F/mv. The Forces Acting on a Gyrostat Precessing Normally at Constant Speed. — We will now determine certain conditions which the forces in such a case 250 Chap, xi always fulfill. Incidentally, we give an alternative derivation of the formula 12 = T/Iw. We take the gyrostat represented by two projections in Fig. 384. ^A^ is the axis of spin, the perpendicular to the paper at is the axis of pre- cession, and Q is the mass-center of the gyrostat. The assumed directions of spin and precession are indicated by the curved arrows oj and 12 respectively. Fig. 384 For the investigation we shall use two sets of coordinate axes, one fixed and one moving. The fixed set is OX, OY , and OZ, the latter not shown; OZ is taken coincident with the precession-axis, and OX and OY in the plane in .which the spin-axis moves. The moving set consists of NA, NB, and NC; NA is the spin-axis (as already stated), NB is the common perpendicular to the axes of spin and precession, and NC is perpendicular to iV.4 and NB. Let /' = the moment of inertia of the gyrostat about the axis NC, e = the distance (ON) between the axes of spin and precession, = the (varying) angle which the spin-axis makes with OX, P be any particle of the gyrostat, m = its mass, r — its distance from the axis of spin, 6 = the (changing) angle BNP, a, b, and c = the co()rdinates of P with respect to the moving axes, and X, y, and 2 = its coordinates with respect to the fixed axes. It follows from the trigonometric relations in the figure that X = a cos 4) — (b -{■ e) sin = a cos 4> — r cos sin — e sin 0, y = c sin + (6 -f e) cos = a sin 4- '' cos <^ cos 6 -\- e cos <^, and z = c = f sin ^. Dififerentiating these expressions with respect to time (and noting that a, r, and e are constants, and that dd/dt = w and dip/df = 12), we get the following values of the x, y, and z components of the velocity of P: Vx = {cw — al2) sin ^ — (6 -f e) 12 cos 0, Vy = (al2 — ceo) cos — (6 + e) 12 sin 0, and Vn = b(j3. The angular momentums of P about the axes OX, OY, and OZ respectively are (see Art. 48) in{vzy — VyZ), m(vxZ — Vzx), and m(vyX — Vxy). Art. 49 ^5^ If we substitute in these expressions for V:,, Vy, and v^, their values as just de- duced, then sum up for all the particles of the gyrostat, we arrive at the follow- ing simple expressions for the angular momentums of the gyrostat about the X, y, and z axes respectively :* hx = loi cos 0, hy = lu sin <^, and hz = /'O. Differentiating these expressions for h^,, hy, and h, with respect to time (and remembering that co and 9, are assumed to be constant), we find that the rates at which the angular momentums change are dhx/dt = — /a;12 sin (/>, dhy/dt = Io£l cos 0, and dhjdt = o. Now consider the instant, or position of the gyrostat, when the spin-axis NA is parallel to the x axis. Then = o, and the rates respectively equal o, /wS], and o; these respectively are also the torques {T^, Ty, and T^ of the ex- ternal forces about the %, y, and z axes, when = o (Art. 48) ; that is Tx = o, Ty = Ioi% and Tz = o. By means of these results and the following paragraph, it is shown in § 3 that T, = 0, Tp = o, and T = /cofi, (i) where Ts, Tp, and T denote the torques of all the external forces about the axis of spin, the axis of precession, and their common perpendicular. For further information, we will now resort to the principle of the motion of the mass-center (Art. 34, page 159). The mass-center describes a circle at constant speed; hence the acceleration of that point is always directed from the mass- center to the center of the circle, and its value is rli^ where r denotes radius of the circle. Now let M = mass of the gyrostat, Rr, Rp, and R3 = the sums of the components of all the external forces along the radius, the precession axis, and the perpendicular to these two lines; then according to the principle named above Rr = MrO^, Rp = o, and i?3 = o. (2) The six equations in (i) and (2) are the certain conditions referred to at the bottom of page 249; they are applied in the following Examples. — (i) Fig. 385 represents a side and end view of the armature of the motor of an electric locomotive. The armature shaft is parallel to the ties of the track. We will discuss the forces acting on the armature when the locomotive is rounding a curve. Inasmuch as we are not now concerned with the driving of the locomotive by this motor we will assume that the armature is spinning but under no load, the locomotive being driven around the curve by another locomotive. And for simplicity, we assume that there is no eleva- tion of the outer rail, so that the precession of the armature is normal; that is, * In reducing the summations, the student should note that •Sm(b'- + c'-) = I, ^mia' + h"-) = I', and Sm&c = I,mca = T^mab = ^mb = '^mc = o. 252 Chap, xi we take the angle between the axis of spin and the (vertical) precession-axis to be 90 degrees. We take the weight of the armature = 8000 pounds, its radius of gyration =15 inches, its speed =750 revolutions per minute, 1 1 -- h — / ^^000/b^-^ Q,=0/P, = Q,= Fig. 385 W P Fig. 386 distance between centers of bearings = 4 feet, the radius of the curve = 2000 feet, and the speed of the car = 30 miles per hour (= 44 feet per second). Then / = (8000/32.2) (15/12)2 = 388 slug-feet,2 w = 750 X 2 7r/6o = 78.54 radians per second, and fi = 44/2000 = 0.022 radians per second. The forces acting on the armature are gravity and the reactions P and Q of the bearings on the armature shaft. We neglect axle friction and imagine each reaction resolved into three components, vertical, parallel to the armature shaft, and parallel to the rails. We distinguish these components by the sub- scripts 1,2, and 3, respectively (see the figure). If the center of the curve is on the right, then evidently the armature presses outward against the bearing P and hence Qo = o. Since the sum of the component forces along the rails = o, P3 and Q3 must be equal and opposite, or else equal zero. Since the torque about the axis of precession must = o, P3 and Q3 = o. According to equation (2), P2 = (8000/32.2) (44-/2000) = 240 pounds. The torque of all the forces acting on the gyrostat about the common per- pendicular to the spin and precession axes equals IwQ = 388 X 78.54 X 0.022 = 670 foot-pounds. If the direction of spin is the same as the direction of rotation of the car wheels, then the torque is clockwise seen from the rear; hence Pi(2ooo + 2) + Qi(2ooo — 2) — 8000 X 2000 = 670. We have also Pi -\- Qi = 8000; hence, solving these two equations simul- taneously, we find Pi = 8000 , 670 + -^ = 4168, and Qi = 8000 670 3832 pounds. Art. 49 253 If the armature were not spinning (co = o), or the car were running on a straight track (^ = o) then /c■ 2 ft/sec/sec ^ 30.4 fi-Aec/sec ^""'"^A Q w///m////////////////mMm^' ^ Fig. 393 directed like a'\ the second component we describe by means of two com- ponents, as in a rotation about a fixed axis (see Art. 37), one of which (the normal component) is directed along QA and the other (the tangential com- ponent) is at right angles to QA. The normal component equals rur {r = AQ) and is always directed from <3 to ^, toward the base point or center of the rotational component; the tangential component equals m, and obviously its sense depends on the sense of the angular acceleration. For a numerical example let us consider the motion of the bar AB (Fig. 391) the ends of which slide along the hues OA and OB. Let the length of the bar --= 6 feet, and the velocity and acceleration of ^4 = 6 feet per second and 2 teet Art. 50 259 per second per second respectively (both toward the right) when = 30 degrees. Required the velocity and acceleration of P, 4 feet from A. It is plain from the figure that 6 cos^ = x; hence, — 6 sin dd/dt = dx/dt, or — 6 sin d'c^ = v, (i) where co = the angular velocity of the bar and v = velocity of A at any instant. Differentiating the last equation with respect to time we get - 6 (oj cos d'dd/dt + sin d'doi/dt) = dv/dt, or 6 (a;2 cos 6 -\- aimd) = a, (2) where a = the angular acceleration of the bar and a = the acceleration of A Sit any instant. Now when 6 = 30", (i) gives co = — 2 radians per second, and (2) gives a = — 7.6 radians per second per second. The negative signs mean that CO and a are counter-clockwise, clockwise having been taken as positive for 6. Finally, the velocity components of P are t> = 6, and 4 X w = — 8 feet per second as shown in Fig. 392; the acceleration components of P are a = 2, 4X0;=— 30.4, and 4 X co^ = 16 feet per second per second as shown in Fig. 393- § 3. Any uniplanar displacement of a body can be accomplished by means of a single rotation. Thus consider the displacement of ABC from the position AiBiCito A2B2C2 (Fig.394). The points can be brought from ^1 to ^2 by means Fig. 394 Fig. 395 B^ B,r- ^T — -vB I — ^ — Fig. 396 of a rotation of AB about any point on the perpendicular bisector aO (of .4 1^2); and B can be brought from Bi to B2 by means of a single rotation oi AB about any point on the perpendicular bisector bO (of B1B2). If the intersection of the bisectors is taken for the center of rotation of both A and B, then the amounts of the rotations (angles A1OA2 and B1OB2) are equal; hence, the line AB (and body ABC) can be displaced from one position to any other (uni- planar displacement) by means of a single rotation as stated. In case the two bisectors coincide (Fig. 395), then the angles Bi and B2 are equal and hence the lines AiBi and .42^2 extended intersect on the bisector ab extended; this extension is the center of rotation C which would disi)lace AB from AiBi to A2B2. In case the bisectors are parallel (Fig. 396) the center of 26o Chap. xi7 rotation is " at infinity," and the displacement is a translation; thus a uniplanar translation may be regarded as a rotation about a center at infinity. The actual continuous motion of AB from one position AiBi to an- other A2B2 (in which A and B describe smooth curves) can be closely duplicated by a succession of rotations of AB from AiBi (Fig. 389) into successive inter- mediate positions A'B', A"B", etc., until .42^2 is reached. Each small rota- tion is made about a definite center 0', 0", etc. (not shown). The closer these intermediate positions are taken (and the more numerous and closer the centers of rotation 0', 0" , etc.) the more nearly do the successive rotations reproduce the actual continuous motion. "In the limit," the actual motion is repro- duced by the rotations, the centers of rotation forming a continuous line. Thus we may regard any uniplanar motion of a body as consisting of a con- tinuous rotation about a center which, in general, is continuously moving. The position of the center about which the moving body is rotating at any instant is called the instantaneous center of the motion for the particular instant or position (of the body) under consideration, and the line through that center and perpendicular to the plane of the motion is called the instantaneous axis of the motion for that instant. In general, the instantaneous center moves about in the body and in space. Its path in the body is called body centrode; its path in space the space cen- trode. Thus, in the case of a wheel rolling on a plane, the instantaneous center at any instant is the point of contact between the wheel and plane; the successive instantaneous centers on the wheel trace or mark out the circum- ference and this line is the body centrode; the successive instantaneous centers in space trace or mark out the track and this line is the space centrode. It can be shown that any plane motion may be regarded as a rolling of the body centrode on the space centrode. Now in a rotation about a fixed axis the velocities of all points of the body are proportional to the distances of the points from the axis of rotation, and the velocities are respectively normal to the perpendiculars from the points to the axis (Art. 37); the velocity of any particular point is given by v = rw, where v = the velocity of the point, r = the distance of the point from the axis, and co = the angular velocity of the body. So too, in the case of a uni- planar motion, the velocities of all points of the body at any particular instant are proportional to the distances of the points from the instantaneous axis (corresponding to that instant); the velocities are respectively normal to the perpendiculars from the points to the instantaneous axis; and the velocity v of any particular point is given by z^ = rw, where r = the distance from the point to the axis and co = the angular velocity of the body. By means of the foregoing velocity relations, we can locate the instantaneous center for any given position of the moving body if the directions of the veloci- ties of two of its points are given; and then if the value of one velocity is given we can compute the angular velocity of the body and the velocity of any other point. For an example we will consider the connecting rod of an engine {BC, Art. 51 261 Fig. 397), in the position shown, the speed being 100 revolutions per minute. Since the velocity of the point B of the rod is along the tangent to the crank- pin circle at B, the instantaneous center of the connecting rod is on the normal to the tangent at B, that is on AB or. its extension; and since the velocity of the point C of the rod is along AC, the instantaneous center is on the normal C' \ i to AC. Hence the instantaneous center is at the -pic intersection O. Now velocity oi B — 2 t Y. AB (to scale) X 100 = 2000 feet per minute; hence, the angular velocity of the rod = 2000 -^ OB (to scale) = 185 radians per minute. The velocity of C = OC (to scale) X 185 =1110 feet per minute. 51. Kinetics of Plane Motion § I. General Principles. — From the principle of the motion of the mass-center (Art. 34) we may write at once ^F, = Ma,, HFy = May, and SF, = o; (i) where SF^, 'ZFy, and HF^ = the algebraic sums of the components of the ex- ternal forces acting on the body along three rectangular lines, the third one being at right angles to the plane of the motion, a, and ay respectively = the X and y components of the acceleration of the mass-center, and M = the mass of the body. In addition to the above, we have another simple relation (established later), f = la = Mk'-a (2) where T denotes the torque of all the external forces about the line through the mass-center and perpendicular to the plane of the motion, J = the moment of inertia of the body about the line just mentioned, ~k = the radius of gyration of the body about that hne, and a = the angular acceleration of the moving body. Systematic units (Art. 31) must be used in equations (i) and (2). But we may substitute W/g for M (where W = the weight of the body and g = the acceleration due to gravity) and then use any convenient units for force (and weight), length, and time. To derive equation (2), let Fig. 398 represent the moving body, C be the mass-center, a = the acceleration of C, 00 and a = the angular velocity and acceleration respectively of the body. Further, let Pi, P2, etc., be particles of the body; mi, m^, etc., = their masses; Vi, r^, etc., = their distances from the line through C and perpendicular to the plane of the motion; and Ri, R2, etc., = the resultants respectively of all the forces acting on Pi, P^, etc. We will regard the motion as consisting of a translation Uke the motion of C and a rotation about the " base axis " through C. Then the acceleration of Pi can be regarded as consisting of three components, a, ria, and rico- as indicated; likewise the acceleration of P2 can be regarded as consisting of three com- 262 Chap, xn ponents, a, r^a, riw-; etc. Therefore, the resultant Rx consists of three com- ponents wia, wiria, and Wirio;- directed like the corresponding accelerations; similarly, the resultant Ri consists of three components n^fl, nhT^a, and m^r-ior directed like the corresponding accelerations; etc. Now the torque of all the forces acting on Pi = the torque of the (three) components of Ri, similarly, the torque of all the forces acting on P2 = the torque of the (three) com- ponents of R2; etc. Hence, the torque of all the forces acting on all the r.a Fig. 3g8 Fig. 399 particles (external and internal forces acting on the body) = the torque of the components (as nia, mra, and mm^) of all the resultants Ri, R2, etc. Since the internal forces occur in pairs of equal, opposite, and colinear forces, they contribute nothing collectively to the first torque just mentioned. It is plain from the figure that the normal components mirioj-, nhVoj^-, etc., have no torque about the (base) axis. Since the resultant of the components WiO, fn^, etc., passes through the mass-center (Art. 35), they have no torque about the axis. The torque of the remaining set of components is Wi^iari + miTiari + = al (see Art. 36). Hence, we have T = /a, or equation (2). Three Special Cases. — (i) When the velocity of the mass-center is con- stant in amount and direction {a = o), the torque of the external forces about any line perpendicular to the plane of the motion equals la. iii) When the angular velocity is constant (a = o), the torque of the external forces about a line through the mass-center and perpendicular to the plane of the motion equals zero. {Hi) When a and a = o, the torque of the external forces about any line perpendicular to the plane of the motion equals zero. Examples. — i. Required the value of P for starting a wheel (Fig. 399a) or stopping it (Fig. 399b). The figures show the wheel rolling toward the right. In the two figures respectively, the angular accelerations are clock- wise and counter-clockwise; hence the friction F on the wheels act as shown, and F = \D = Mk~a. And since the accelerations a of the mass-center are toward the right and left respectively, P - F = Ma iox each figure. Also a = ra. From] these equations, it follows that P = M (i + k'^/r'^) a. Thus the "effective inertia " of a rolling wheel is i + k'^/r'^ times its inertia when skidding, for in the latter case P — Ma. Art. 51 263 2. It is required to discuss the rolling of a homogeneous cylinder on an inclined plane. Let the weight of the cylinder = 200 pounds, the diameter of its bases = 3 feet, and the inclination of the plane = 25 degrees. Further, we assume that the cylinder and plane do not distort each other, so that there is only line-contact between them and no "rolling resistance" (Art. 52); also that the surfaces in contact are sufficiently rough to prevent slipping so that the roUing is perfect. There are only two external forces acting on the rolling Wj ^ Fig. 400 Fig. 401 cylinder, its own weight and the reaction of the plane, but the latter is repre- sented by two components, N and F, in Fig. 400. Since the mass-center moves in a line parallel to the incline, ax = a, and ay = o; hence equations (i) become 200 sin 25° — F = (200 -^ 32.2) a, N — 200 cos 25° = o, and = 0. The second equation shows that iV = 181 pounds. The first equation con- tains two unknowns (F and a) and does not furnish the value of either of them; so we resort to equation (2). Since ^ = | 1.5- = 1.125 (see Art. 36), equa- tion (2) becomes F X 1.5 = (200 -^ 32.2) 1. 125 X oc. Now we have two equations but three unknowns, and so we need an ad- ditional equation; this is given by the (simple) relation between a and a. Since there is no slipping, the displacement s of the mass-center in any interval of time and the angular displacement 6 of the cyUnder for that interval are related thus: .j = 1.5 (6 in radians and s in feet); hence d^s/df^ = i-5 d^O/df^, or a = 1.5 a. Substituting 1.5 a for a in the first equation and then solving simultaneously V\^ith the fourth, we find that a = 6.05 radians per second per second (o = 9.07 feet per second per second) and F = 28. 2 pounds. 3. It is required to discuss the forces acting on a rolling wheel whose center of gravity is not in the axis of the wheel, the speed of rolling being maintained uniform by a suitable horizontal force P (Fig. 401). Let W = weight of the wheel, r — radius, and c = the distance from its center A to the center of gravity C; further let 6 = the angle between AC and the horizontal in the position of the wheel under consideration. There are three forces acting on the wheel, P, W, and the reaction of the roadway (represented for convenience by two components A^ and F). Equations (i) become P- F = (Pf /g) ~a, and N - W = {W/g) ay. 264 ^^p- ^^ Since the angular velocity is constant, a = o, and equation (2) becomes F (r + c sin 6) — Nc cos d — Pc sin d = o. These equations contain five unknowns {P, F, N, Ux, and ay), and so we need other equations. Obviously the relations between a^, ay, and d furnish the additional equations. To determine these let us regard the rolling as con- sisting of a translation with A as base point and a rotation about A . Then since A moves uniformly, the acceleration of the translational component = o; and there being no angular acceleration, the acceleration of the rotational component of the motion of C is wholly radial (along CA) and equals cor. Hence a equals cw^ and is directed from C to .4 ; and ' a^ = ceo- cos 0, and ay = — coi"^ sinQ. Substituting these values of a^ and ay in the first two equations, and solv- ing them simultaneously with the third we find that For CO we may write 2 ttw, where n = the number of turns of the wheel per unit time. It follows from the foregoing results that P and F are always opposite; that P and F act as shown whenever the center of gravity C is on the left of the vertical through the center A id between — 90° and + 90°) ; that P and F act opposite to the directions indicated in the figure when C is on the right of the vertical through A ; that N always acts upward unless cor sin 6 is greater than g; that the greatest value of N obtains when C is vertically be- low A {d -=- 90°) and then N = W (i + coi'^/g). This excess TFccoVg over W in the value of N is called " hammer blow " in locomotive parlance, but the hammer blow of a locomotive driving wheel depends also upon the side rods attached to the wheel (see Art. 35). Independence of Translation and Rotation. — Referring to equations (i), page 261, it will be noted that they contain no term depending on the rota- tion of the body about the mass-center; therefore, they show that the motion of the mass-center is entirely independent of the rotation about that point. And as already pointed out (Art. 34), the acceleration of the mass-center is the same as though the entire body were concentrated at the mass-center and all the external forces were applied at that point parallel to their actual lines of action. Equation (2) contains no term depending on the motion of the mass-center; therefore, the rotation of the body about the mass-center is independent of any motion of the mass-center itself. And on comparing equations (2) with the equation of motion for rotations about fixed axes (Art. 37), it becomes plain that the external forces produce rotation about a free (moving) axis through the mass-center as though that axis were fixed. Thus we have complete independence of translation (of mass-center) and rotation (about mass-center). Art. 51 265 To illustrate we will apply the principle of independence to explain center of percussion; Art. 48 includes an explanation based on other principles. Let AB (Fig. 402) be a prismatic bar lying on a horizontal surface, and C its center of gravity. Now imagine the bar to be struck a blow in the line F. The only other forces acting on the bar are gravity and the ra supporting force of the surface; these produce no appreciable \ effect on the motion during the blow. The motion produced, [ therefore, consists of a translation as though the blow F acted -k- through the mass-center, and a rotation about the mass-center __^ as though the mass-center were fixed. Any point beyond C gets a velocity toward the right due to the translation, and a velocity toward the left due to the rotation. For some par- Fig. 402 ticular point these two velocities are equal and opposite, and hence if the bar were pivoted there, the pivot would feel no pressure from the bar during the blow. For such a point, G is the center of percussion. Let us now find where this pivot point is. For that purpose let M = mass of the bar, k = its radius of gyration about the line through C perpendicular to the supporting surface,/ = the arm of the blow F about the mass-center, R be the pivot point, r = its distance from C, a = the average acceleration of the mass-center, a = the average angular acceleration of the body during the blow, and M = the duration of the blow. The velocities of R due to the translation and rotation respectively equal a At and raAt. Now a = F/M and a = Ff/Mk^l therefore, for the pivot point we have (f/M) At = r{Ff/Mt) At, or fr = t. r2 That is, r = yfe //. For a given pivot the distance of the center of percussion from the center of gravity is/ = ^ /r, which agrees with the result reached in Art. 48. Kinetic Energy of a Body with Plane Motion. — Let M = the mass of the body, W = its weight, 7 = its moment of inertia about a line through the mass-center perpendicular to the plane of the motion, k = its radius of gyra- tion about the same Une, v = the velocity of the mass-center, and co = the angular velocity of the body. Then the kinetic energy of the body equals im' + hr<^'=h (w/gw + \ {wig)k i^\ (I) The latter is the more convenient form generally for use in a numerical case. If g is taken as 32.2 (feet per second per second), then the foot and second should be adhered to as units of length and time; co should be expressed in radians per unit time. If W be ex-pressed in pounds, tons, etc., then the result will be in foot-pounds, foot-tons, etc. 266 Chap, xn Fig. 403 The first term of (i) equals the kinetic energy which the body would have if its motion were one of translation with velocity equal to v; and the second term equals the kinetic energy which it would have if its motion were one of rota- tion about a fixed axis through the mass-center and perpendicular to the plane of the motion. Hence the kinetic energy of a body with any plane motion may be regarded as consisting of two parts; they are called trans- lational and rotational. The following is a derivation of the preceding formula after the view that a plane motion is a combined translation and a rotation (Art. 50, § 2). Let Fig. 403 represent the moving body, its mass-center, and P any other point of the body. Also let r = the distance of P from the fine through perpendicular to the plane of the motion, and v = the velocity of P. Then v is the resultant of v and rw as indicated. The angle QPS = 90 — (iS — 0), where /3 and 6 are the angles which v and OP respectively make with the X axis. Therefore ^)2 — ^2 _|_ ^2^2 _|_ 2 ^/-co sin (^ — 6), and the kinetic energy of the entire body (2 ^ mv~) equals I v'^^m + I w-Swr- + 2 I'co (sin /SSwr cos 6 — cos ^ Hmr sin 9). Now r cos 6 and r sin 0, respectively, equal the x and y coordinate of P. Hence Swr COS0 = 2wx = xHm (see page 158), x denoting the x coordinate of the mass-center; and since x = o, '^mr cos 9 = o. Similarly, Swr sin 9 ^ o. Hence the foregoing expression for the kinetic energy reduces to The follo\Adng is a derivation based on the view that any plane motion con- sists of a succession of instantaneous rotations (Art. 50, § 3). Let / ^ the moment of inertia of the body about that Hne which is the (instantaneous) axis of rotation at the instant in question, d = the distance from that axis to the mass-center, p = the distance of any point P of the body from the axis, z^ = velocity of P (as before), and co = angular velocity of the body. Then v = pw, and the kinetic energy of the body is SI mv^ = \ w-2wp2 = i /aj2 (2) This is a much simpler ex-pression than (i) but not so convenient to use gener- ally, because / refers to an axis not fixed in the body. It remains to reduce (2) to (i). According to the parallel axis theorem (Art. 36, § 2), / = / + Md-\ hence |/a;2 1 /co2 + i M{d • a))2 = 1 /co2 + i .1/^' Art. 51 267 For an example we will compute the kinetic energy of a solid cylinder rolling on a plane surface. Let W = weight of cylinder, D = its diameter, and n = number of turns of the cylinder per unit time. Then M = W/g, V = tDh, 7=1 iW/g)D~ (see Art. 36), and co =- 2 im. Hence the kinetic energy of the cylinder equals i (W/g)TrWhl' + i (W/g)TrWhi\ Thus it appears that two-thirds of the energy is translational and one-third is rotational. § 2. Dynamics of a Simple Moving Vehicle. ^ Let W = weight of the body of the vehicle and its load, if any; w = the weight of each wheel (in- cluding one-half of the axle if the wheels are rigidly mounted on their axles) ; k = radius of gyration of wheel (with one-half of axle in case mentioned); r = radius of wheel; n = number of wheels; and v = velocity of the vehicle. The kinetic energy of each wheel is \ (w/gy + h (Wg)k' (v/ry = i (w/g) (i + k^r^y. Hence the kinetic energy of the entire vehicle is 1 VW nw 47"^ g -^)} Comparing this expression with that for the kinetic energy of a body with a motion of translation, we see that the motion of the entire vehicle may be regarded as one of translation provided that the weight of the vehicle is taken equal to W + nw (i -\- k^/r-). For modern freight cars r = 16.5 inches and k = 9.5 inches (about); hence k-/r^ = 0.35. Therefore the "effective in- ertia " of the wheels when rolling is about one-third greater than when at rest or skidding. Height of Draw Bar. — Fig. 404 represents a vehicle, as a railroad car, being dragged on a level track by a pull P. The other external forces acting on the W / N •77777?7;77T^'777777777777777m777777777777/l^. Fig. 404 H ^ ^- ■ra^ -tr ■h Fig. 406 "-5^5^ \ Q car are gravity iyV -f- nw) and the reactions of the rails on the wheels (each represented by its horizontal and vertical components). In Fig. 405 there are represented all the external forces acting on one wheel, in Fig. 406 those acting on the car body. The pressures between axles and bearings are repre- sented by their horizontal and vertical components; axle friction is disregarded. Let a = the acceleration of the car; then the angular acceleration of the 268 Chap, xii wheels = a/r. Consideration of the forces on the wheel, equation (2), page 261, shows that Fr = -h?- , or F = — -a. g r g r^ We have also (according to equations i, page 261) e-. = ^a, or e = f (. + !>■ Consideration of the forces acting on the car body shows that P — nQ = (W/g)a, or la. When applied high up on the car, P tends to raise the rear end, decreasing the rear vertical axle pressures and increasing the forward vertical axle pres- sures. When applied low, P produces the opposite effect. Obviously, when applied in some certain line, P has no such effect on the vertical axle pressures. We will now locate that line; let h = its height above the plane of the axes of the axles, and H = the height of the center of gravity of the car body and its load above that plane. When the car is at rest {P and Q = o), the (vertical) pressures of the axles on the car body take on certain values. If, when P (and nQ) act on the car body, their resultant acts through the center of gravity, then those forces do not tend to rotate the car body and do not affect vertical pressures of or on the axles already mentioned. Thus, to provide against extra loading or unloading of axles by P (draw-bar effect), the moments of P and nQ about the transverse horizontal line through the center of gravity of the car body (and load) should balance. That is, we should have P{H - h) = nQH,' or H h = + (nw/W) (i + k^/r^) 52, Rolling Resistance § I. Rollers. — In the present connection a roller is taken to differ from a wheel (of a vehicle) in that the latter sustains its load indirectly through its axle, while the former has no axle but takes its load directly. When a roller (or wheel) is made to roll, it experiences more or less resistance from the track (or roadway) upon which it rolls. Obviously the amount of this resist- ance depends in large part on the nature of the surfaces in contact and on the amount of the pressure between them. In the case of an inelastic roadway {A, Fig. 407) the roller leaves a rut, and there is a continual expenditure of energy in thus (permanently) deforming the track as well as against friction due to actual rubbing between roller and track. In the case of an elastic roadway (B, Fig. 407) also, there is rubbing between the roller and the deforming and recovering portions of the track and consequently friction Art. 52 269 loss.* In any case there is expenditure of energy against the (internal) friction in portions of the roller and track which are deforming or recovering.! Let R = the resultant reaction of the track on the roller. Obviously the point of ap- plication of R is on the surface (or arc) of contact between wheel and roadway; and it will be shown presently that this point is in front of the vertical diameter of the roller, the roadway supposed to be horizontal. The distance from this point to the diameter is called the coefficient of rolling re- sistance; we will denote it by c, and express numerical values of the coefficient in inches. Obviously the coefficient of rolling resistance depends on the nature of the wheel and roadway, and is greater for yielding surfaces than for rigid ones. It would seem that the coefficient depends on the load but in certain cases at least the coefficient is not influenced much by it. The coefficient is claimed to be independent of the radius of the roller; also that it varies as the square root of the radius. The precise way in which the coefficient varies with the conditions named has not been established. Below we give some of the meager ex-perimental data relating to the matter. Coulomb seems to have made the first ex-periments to determine coefficients of rolUng resistance. The following are his results for Lignum Vit.e Rollers on Oak "Pieces" Load. Diameter = 2.18 ins. Diameter = 6.55 ins. 220.5 lbs. 1102.7 2205. c = 0.0174 ins. .0205 .0196 c = 0.0196 ins. .0197 .0196 For the circumstances of these ex-periments, it appears that the coefficient does not vary much with the load or the diameter of the roller. For elm rollers 6.55 and 13. 11 inches in diameter on oak pieces. Coulomb found c = 0.0327 inches-l The following also are quoted from Morin's Mechanics: Oak Rollers (diameter = 7.87 inches) on Poplar Strips Width of strips. Load. Coefficient c. 0.97 ins. 3-94 409 lbs. 400 0.00637 ins. .00287 * See Phil. Trans. Roy. Soc, Vol. 166, Part i (1876), for experiments on rollers rolling on a rubber roadway, by Prof. Osborne Reynolds. t See Jour, and Trans. Soc. of Engrs. (London), Vol. 3, p. 180 (1912) for an analysis of this element of rollinp; resistance, by Prof. Herbert Chatley. X From English translation by Joseph Bennett of Morin's Mechanics, p. 339. 270 Chap, xii In these experiments, increasing the length of bearing from 0.97 to 2.94 (about triple) more than halved the coefficient. Thus it appears that the coefficient depends on the loading per unit length of contact between roller and roadway. But the coefficient probably does not decrease indefinitely with increase of length of contact. For some conditions the coefficient seems to vary as the square root of the radius of the roller, that is c = is another coefficient and r = radius of the roller. Dupuit gives the following average values: Wood on wood = 0.0069 Iron on moist wood .0063 Iron on iron .0044 Wheel on macadam .19 For the conditions of his experiments,* Prof. C. L. Crandall takes the co- efficient of rolling resistance as proportional to the square root of the radius, that is c = V r. Roller plates used were i| inches thick; rollers i, 2, 3 and 4 inches in diameter, all i| inches long except the first whose length was i inch. Plates and rollers were used as they came from the plane and lathe; were not polished or filed. Loads varied from 350 to 2500 pounds per linear inch in contact. The coefficient did not seem to vary much with load; with materials it varied as follows: Cast iron = 0.0063 Wrought iron .0120 Steel -0073 These values refer to cast-iron plates; for wrought-iron plates they should be increased about 13 per cent, and for steel plates they should be decreased by that amount. Fig. 408 represents in principle the device used by Coulomb to determine the coefficient of rolling resistance. W = weight of roller, Wi and W2 = weights of suspended bodies as shown. By adjusting the difference between Wi and W2 the roller was made to roll quite uniformly. When rolling at constant speed, the reaction R of the track on the roller is vertical, and R = W -\- Wi + W^- Also there is no resultant torque on the roller; hence the moment of R must be counter-clockwise (in this illustration), and so the point of application of R is in front of the vertical diameter of the roller (as stated). It follows that (IF2 — W\) / = Re = (W + Wi -hW2)c; or c = / (W2 - Wi)/iW + Wi-\- W2), from which c can be computed easily. * Trans. Am. Soc. C.E., Vol. 32, p. 99 (1894). Art. 52 27 1 Fig. 409 represents in principle the device used by Crandall. There were two rollers under load (and a third one to preserve stability only), and three plates as shown. The lower plate was supported on the weighing table of a testing machine; load was applied on the upper plate; and then the middle plate was subjected to a force P sufficient to start the plate. Thus the middle plate was subjected to the reactions of the two main rollers, inclined as shown. I Plate I n w Plate \-^^ I ' ' — 1-^> "'^'//////////////A' Plate — I ■'^'//?^/i(J/////W///////7//tp////////////m/^///////' Fig. 409 Fig. 410 Let R = these reactions (nearly equal), and Q = their inclination to the vertical. Then, evidently, P = 2 i? sin ^ = 2 Rclr^ and Rco&d = W or R = W nearly; hence P = 2 Wc/r and c = i Pr/W. Rollers are generally used for moving a heavy load as shown in Fig. 410. Let r = radius of rollers, c = their coefficient of rolHng resistance (assumed same for top and bottom contacts), Ri, R2, etc., = the reactions of the rollers, 6 = their 'inclinations to the vertical, W = load, and P = the pull required to mo\'e the load. Then since 6 is small, (i?i + -^2 + • . . ) = W (nearly) ; and since sin d = c/r, P = (Ri -{- R2 -\- . . . ) c/r. Hence P = Wc/r. § 2. Rolling Wheel. — The general nature of rolling resistance in the case of a wheel is hke that against a roller. A rolHng wheel of a vehicle ex- periences axle friction as well as rolling resistance,* and few experiments have been made to determine them separately. For cast-iron wheels 20 inches in diameter on cast-iron rails Weisbach and Rittinger, respectively, found for the coefficient of rolling resistance c = 0.0183 and 0.0193 inches.f For an iron railroad wheel 39.4 inches in diameter, Pambour gives c = 0.0196 to 0.0216 inches. (i) Wheel without Axle Friction. — We assume the velocity to be constant. Of course a force must be applied to the wheel to maintain the velocity; we assume it to be applied to the axle of the wheel as shown in Fig. 411, and, for simplicity, that the axle is frictionless. Let D = diameter of wheel, P = driving force, W = weight of wheel and load upon it, R = reaction of roadway, and Rh and 7?„ = the horizontal and vertical components of R (see Fig. 411b and c). Rh is the "rolling resistance." * See Baker's Roads and Pavements for full information on total resistance to traction oi vehicles (due to roUinR resistance and axle friction). t Coxe's translation of Weisbach's Mechanics. 272 Chap, xii Since there is no angular acceleration, the (resultant) torque on the wheel equals zero (see Art. 51) and R acts through the center of the wheel. It Fig. 411 follows that the line of action of R cuts the rim of the wheel in front of the vertical diameter of the wheel as shown, and of course within the arc of con- tact of wheel and roadway. The distance of this point A on the rim to the vertical diameter is the coefficient of rolling resistance, already denoted by c Because the torque of the three forces (P, W, and R) is zero, and the vertical distance between A and is ^ D nearly, P • | Z) = Wc; and since Rh = P, R^ = P= W 2 c/D. (2) The work required to overcome the rolling resistance per turn of the wheel is equal to the work done by the driving force P per turn. But this latter is plainly PirD, or W 2 TTC. (3) (ii) Wheel with Axle Friction. — Fig. 412 represents the wheel, of a horse- drawn vehicle say, moving at a constant speed toward the right. In addition to the foregoing notation, let d = diameter of axle, W = weight of wheel, Q = resultant pressure of the axle on the wheel, Qh and Qv the horizontal and ver- tical components of Q, and / the coefficient of axle or journal friction (see Art. 45). R and Q act somewhat as shown but their inclinations to the ver- tical are much exaggerated. Fig. 412b shows R resolved at A into its hori- zontal and vertical components, and Q resolved at its point of application into its normal and frictional components N and F; F — fQ. Fig. 412c shows Q replaced by Qh and Qv at the center of the wheel, and a couple C; the moment oi C is F ^ d. Since the speed is constant, Rh = Qh and Rh i D^ R,c + F\d. {Continued on page 27 g.) Art. 53 273 53. Relative Motion § I. Motion Relative to a Point. — We can specify position of a point only by means of a set of reference axes or some other equivalent base de- scribed or implied in the specification. Thus when we say that Chicago is io| degrees west and 3 degrees north of Washington — the cities regarded as points — we are really specifying the position of the former city with refer- ence to the meridian and the parallel through Washington. But we say briefly that the specification is relative to Washington. So too when we say that a moving ship A is 40 miles east and 50 miles north of another ship -B at a certain instant, we are specifying position of A by means of the meridian and the parallel through B at the instant in question; but we say that the specifi- cation is relative to B, the coordinate axes being understood. Being small compared to the distances mentioned (40 and 50 miles), the ships were regarded as mere points. If, however, the ships were at close quarters, then to describe the position of A relative to B we would specify the position of at least two points in A (bow and stern for example) relative to axes fixed in B, as indicated in Fig. 415, say. Even if B were turning about, we would still use those axes to specify subsequent positions of A relative to B. For the present we will deal with position (and ' ^^^ motion) of points (or bodies regarded as mere points) relative to another base point — not body — and it should be understood that the coordinate axes, though moving with the base point, remain fixed in direction. Let the points o, i, 2, 3, etc. (Fig. 416), on the lines aa and bb be the positions 5 b < 5 ^ 3 / ly ^ ^ 4 a'^ / oV »». J 5 a w ^ z 3 4 1 4 E Fig. 416 Fig. 417 (relative to a lighthouse), say, of two ships A and B, at 12, i, 2, . . . o'clock of a certain day; then the lines are the paths (relative to the lighthouse) of the ships. From these paths we have made the following tabulation of coordinates of the positions of A relative to B. These coordinates if plotted on rectangular axes, representing a parallel and a meridian, determine the path of A relative to B (Fig. 417). Time (hours) East (degrees) North (degrees) I 2 3 4 S 10 15 21 30 37 40 20 19 16 10 2 -6 274 Chap, xn Taking points o; i, 2, etc., on the line cc (Fig. 416) as the positions (relative to the lighthouse) of a third ship C at the hours mentioned, we have the follow- ing tabulation of the coordinates of the positions of A relative to C from which the path of A relative to C (Fig. 418) was constructed. Thus it is clear that in general the path of a moving point depends on the point of reference or base point. Time (hours) East (degrees) North (degrees) I 2 •J 4 5 2 -3 4 12 18 20 12 20 23 22 IS For another illustration, imagine a table (Fig. 419), several balls (for bear- ings) on the table, a drawing board resting on the balls, another set of balls on the board, and a sheet of plate glass resting on these latter balls. Imagine Glass Board ■ Stiff Wire fohold Pencil "a" A, B, a, and b are Lead Pencils. Fig. 419 also a pencil fastened to the glass so that its point A presses against the board, and another pencil fastened to the board so that its point B presses against the glass. When either board or glass or both are rolled about on the balls, the pencils (if suitable and properly mounted) trace lines on the board and glass. If the edges of the board and glass (or coordinates axes through A and B) are kept in fixed directions, then A traces its path relative to B, and B traces its path relative to ^. By velocity of a point relative to another point is meant the rate at which the first point is traversing its path relative to the second point at the instant in question. We regard this velocity as having direction, that of the tangent to the (relative) path at the point corresponding to the instant in question. By acceleration of a point relative to another point is meant the rate at which the velocity of the first point relative to the second is changing at the instant in question. Motion of Two Points Relative to Each Other. — The velocities and acceler- ations of two points relative to each other are equal and opposite. This proposition may be illustrated by comparing the path of ship A, of the fore- going illustration, relative to B (Fig. 417) with the path of B relative to A (Fig. 420), constructed from the following tabulation of the positions of B relative to A at the hours 12, i, 2, etc. Thus for the hour from 2 to 3 o'clock, Time (hours) . . . East (degrees) , . North (degrees) I 2 3 4 — 10 -15 — 21 -30 -37 — 20 -19 -16 — 10 — 2 5 -40 6 for example, the displacement of A relative to B is represented by vector 2-3 of Fig. 417, and the displacement of B relative to A is represented by vector Art. 53 275 ■-E Fig. 420 2-3 of Fig. 420. Apparently these vectors are equal and parallel (also opposite) ; and it seems that such displacement vectors would be equal, parallel, and opposite for any interval of time. If this be true, then it follows that the rates at which these displacements occur (the relative velocities) are equal and opposite at each instant; and if the velocities are always equal and op- posite then their rates of change (the relative acceler- ations) are also equal and opposite at each instant. To prove that displacements such as mentioned in the preceding illustration are equal and opposite, we will use the glass-board illustration. Suppose that the pencils A and B are attached at the middle points of the glass and board respectively, and that at a certain instant glass and board are in the positions shown at (i) in Fig. 421, and at a later instant in positions shown at (2); the table is not shown. Ai and Bi and A2 and A) B2 are the corresponding positions of the pencil points. During this displacement, A will have traced some such line as A'A2 and B the line B'B2. A^A' is equal and parallel to B1B2; hence A1B1B2A' is a parallelogram, and A'B2 and AiBi are equal and parallel. BiB' is equal and parallel to ^1^12; hence BiA^A^B' is a par- allelogram, and B'A2 and BiAi are equal and parallel. It follows that A'B2B'A2 is a parallelogram, and so A'A2 and B'B2 are equal and parallel. That is, the displacement of A relative to B (chord A'Ai) is equal and parallel to the displacement of B relative to A (chord B'Bi). Obviously the senses of the displacements are opposite. Motions of Two Points Relative to a Third Point. — For convenience we re- gard the third point as fixed, and call velocities and accelerations relative to that point as absolute. To illustrate this case we will modify the glass- board apparatus as follows: Imagine another pencil a rigidly fastened to the glass plate so it presses against the table as shown directly under A, and B extended downward so that its lower end b presses on the table. Then when the glass and board are moved about without turning, a and b draw the paths of A and B relative to any (third) point as C on the table; and as already stated, A and B draw their paths relative to each other. In this case two problems arise: (a) Given the velocity (or acceleration) of a point relative to a second point, and the absolute velocity (or acceleration) of the second; required the absolute velocity (or acceleration) of the first point, (b) Given the absolute velocities (or accelerations) of two points; required the velocity (or acceleration) of either of the two points relative to the other. (a) To do this problem we merely need to add (vectorially) , or compound, the velocity (or acceleration) of the first point relative to the second and the abso- 421 276 Chap, xii lute velocity (or acceleration) of the second; the sum is the desired quantity. To justify this solution we first show that the (vector) sum of the displace- ment of the first point relative to the second and the absolute displacement of the second point equals the absolute displacement of the first, all dis- placements being taken for any interval of time. It will follow that the rela- tive and absolute velocities (and accelerations) are related as above stated. Referring to our glass-board-table device, let A , B, and C be the three points respectively. Let (i) and (2), Fig. 421, be the positions of glass and board at the beginning and end of any interval, as before. Then A'Ao is the dis- placement of A relative to B as explained; ^1-62 is the absolute displacement of B; and A1A2 is the absolute displacement of ^. As already shown, the quadrilaterals in the figure are parallelograms; hence the vector sum oi A'A2 and B1B2 equals A1A2. (b) Let A and B be the first two points and C the third, and the velocity (or acceleration) of A relative to B the desired quantity. According to (a), the absolute velocity (or acceler- ation) of A = the vector sum of the velocity (or acceleration) of A relative to B and the absolute velocity (or acceleration) of B. Therefore the (desired) velocity (or acceleration) is such a velocity (or acceleration) which when added vectorially to the absolute velocity (or acceleration) of 5 = the absolute velocity (or acceleration) of A. For example let Va and Vb (Fig. 422) be the absolute velocities (or accelerations) of A and B; then if OM and ON be drawn to represent Va and % respectively, NM will represent the velocity (or acceleration) of A relative to B. The problem can be solved also on the basis of the principle that if we add equal velocities (or accelerations) to the absolute velocities (or accelerations) of the two points we do not change the velocities (or accelerations) of either of the points relative to the other. Thus, taking the preceding example, we will add to Va and Vb a velocity equal and opposite to Vb (Fig. 423); then the new Vb = o and the new Va = NM. Since now B is at rest relative to C, the new velocity of A relative to C is also the velocity of A relative to B. § 2. Motion or a Point Relative to a Body. — As explained in § i, we specify the positions of a moving point relative to another moving point by means of reference axes of fixed directions through the second point, but its positions relative to a moving body by means of reference axes fixed in the body. See illustrations of the ships. Then the path of a point relative to a body is the fine through the successive positions of the point relative to the body. Thus, to illustrate, consider again the glass-board-table appara- tus (Fig. 419). When both the glass and board are rolled about in any way, /■/ Vb 4 ^^^. *-jb. / ho 1 -Vb N^ / •C ^J>- / ^*A Fig. ■ 423 Art. 53 277 the pencil A traces a line on the board, and that line is the path of A relative to the board. By velocity of a point relative to a moving body is meant the rate at which the point traverses its path relative to the body at the instant in question. By acceleration of a point relative to a moving body is meant the rate at which the velocity of the point relative to the body is changing at the instant in question. When a point P is moving relative to a moving body B then the absolute velocity of P equals the vector sum of its relative velocity and the absolute velocity of that point of B with which P coincides at the instant in question. For simplicity of proof we take the pencil A of the glass-board-table apparatus as the moving point P and the board as the moving body B. Since P and B have plane motion, the proof is not general. Let Bdi (Fig. 424) be the position of B at Fig. 424 a particular time /i, and Bdi the position of 5 at a later time ^2; also Pi and P2 respectively, the positions of P at those times. Let M be the point of B with which P coincides at time ti. At time /i, M is at Mi (under Pi) ; and at time /2, M is at M2. Then for the interval t^ — ti the absolute displacement of P is PiP2; the relative displacement of P is M2P2; and the absolute displace- ment of M is M1M2. Obviously P1P2 = M'2P2 + MiM^ (vectorially). Since this relation holds for any interval, the rates at which these displacements occur (velocities) are related in the same way; that is, the absolute velocity of P = its relative velocity + the velocity of M. When a point P is moving relative to a moving body B then the absolute accelera- tion of P equals the vector sum of three accelerations, namely — the relative accelera- tion of P, the absolute acceleration of that point of B with which P coincides at the instant in question, and a so-called complimentary acceleration. The compli- mentary acceleration equals twice the product of the relative velocity of P and the angular velocity of B at the instant in question; its direction is the same 278 Chap, xii as that of the Hnear velocity of p where Pp is a vector representing the relative velocity of P due to the angular velocity of B. For simphcity again we restrict the proof to plane motions. Let Pipi (Fig. 424) = the relative velocity of P at the time h, and If iWi ^ the absolute velocity of M at that instant. The vector simi of these two velocities equals the absolute velocity of P at the time h. Making OA' and OB' to represent these velocities respectively, we get the diagonal OC to represent the absolute velocity of P at the time h. Let N be the point of the board with which P coincides at the time k; N is under P2 then. The velocity of N (at time k) equals the vector simi of the velocity of i/2 and the velocity of TV "about " M2. Now the velocity of A' about I/2 equals the product of M-2.N and the angular velocity of the board (at time h), or Ar X C02, where Ar = MoN and C02 = the angular velocity. The direction of this velocity Ar • C02 is perpen- dicular to M2N as indicated (assuming that C02 is counter-clockwise). OB and bB" are equal and parallel to Mim^ and Ar'C02 respectively; hence OB" is the velocity of N at time h. Now let Po_p-i (= OA") be the relative velocity of P at time h- Then the diagonal OC" of the parallelogram on OA" and OB" is the absolute velocity of P at time to. Therefore C'C" is the incre- ment in the absolute velocity of P for the interval h — h- It follows readily from the geometry of the figure that C'C" = A'A" + B'B", (i) vectorial addition being understood here and in the following. Now let Mia = P\pi and the angle between these vectors equal the angular displacement Ad of the board during the interval /2 — k- Then the increment in the relative velocity of P for that interval equals the difference between the vectors Mia and Pipi- Oa is equal and parallel to Mia; hence aA" is that difference. Therefore A'A" = A'a+ Avr = 2Vr sin i A9 + Avr, where Vr means relative velocity of P at time h. Since Ob is equal and parallel to Mimi (velocity of M at time /o), B'b is the increment in the velocity of M during the interval h — h', and since bB" = Ar'Ui, B'B" = A%n + Ar • C02, where Vm means velocity of M. Substituting the foregoing values of A'A" and B'B" in equation (i), we get C'C" = Azv + Avr^ +2Vr sin | A^ -j- Ar - on. (2) Now let At = h — h, and ^ approach ti, then we get ,. C'C" ,. At.. , ,. Av^n , ,. Ad ... Ar hm— 7— = lim -— - + hm— - + Vr lim -— + lim — (02. At At At At At The left-hand member is the absolute acceleration of P; the first term of the right-hand member is the relative acceleration of P; the second term is the acceleration of M. Lim {A6/ At) = wi, the angular velocity of the board at Art. 53 279 time/i; hence the third term = ZJrCOi. Lim (Ar/A/)co2 = lim (Ar/At) X limw2 = Vr^i. Hence the third and fourth terms are equal in magnitude, and if their directions — they are vectors — are parallel, then their smn = 2 VrWi. The direction of the third term is the limiting direction of A' a, perpendicular to OA' or Vr obviously. The direction of the fourth term is the limiting direction of bB" or Nc. Now Nc is always (as /2 approaches h) perpendicular to M2P2; and since 1/2^2 is the relative displacement of P, the Umiting direction of M2P2 is Pipi (or Vr). Hence the limiting direction of Nc is perpendicular to Vr. Thus the sum of the last two terms = 2 Vrcoi, and it has the direction mentioned, perpendicular to tv- And this sum is the so-called complimentary acceleration; it is called also acceleration of Coriolis after him who first discovered the rela- tion between the accelerations under discussion. {Continued from page 272!) From the last equation on that page and Rv = Qv ^- W, it follows that Rh = Qh={Q. + W')^ + F^- (4) In most cases of vehicles, Qv is nearly equal to Q, and W is neghgible com- pared to Qv ; therefore we may write as a close approximation The work required to overcome rolling resistance and axle friction is fur- nished by Qh. Per turn of the wheel, that work is QtjvD = ((?„ + W) 2 TC + Qhd = TT (2 c + /J) (?. (5) CHAPTER XIII THREE DIMENSIONAL (SOLID) MOTION 54. Body with a Fixed Point; Kinematics § I. Spherical Motion means motion of a rigid body with only one point of the body fixed. Each point of the body, excepting the fixed one, moves on the surface of a sphere, whence the name spherical motion. Any spherical displacement of a body can be accomplished by means of a rotation about some line of the body passing through the fixed point, and fixed in space. Proof: — Evidently, we mav describe any position of the body by describing the positions of two of its points, not in line with the fixed point. Let A and B denote two such points, equally distant from the fLxed point 0; then during any motion of the body, A and B move on the surface of the same sphere. Let OAiBi be one position of the body, and OA2B2 another. Then we are to prove that the points A and B could be brought from AiBi to .42^2 by means of a single rotation about some fixed line through 0. Let the lines .4 1^1 (Fig. 425) and A2B2 be arcs of great circles of the sphere mentioned; these arcs are equal since A and B are points of a rigid body. The fines A1A2 and B1B2 are arcs of great circles; M _ and N bisect these arcs; MR and NR are great Fig a2^ circles perpendicular to ^1-42 and B1B2 respectively. In general two such great circles do not coincide but intersect at two points, R and S. The diameter ROS is the axis, rotation about which would produce the given displacement, proven presently. Let AiR, A2R, BiR, and B2R be arcs of great circles. Since A1A2R and B1B2R are isosceles triangles, AiR = A2R and BiR = BoR; and, as already stated, AiBi = .42^2. Hence the trian- gles RAiBi and RA2B2 are equal, and the angle AiRBi = A2RB2. Finally, A1RA2 = A1RB2 — A2RB2 = A1RB2 - AiRBi = B1RB2. Hence a rotation of the great circles A^R and BiR about RS of an amoimt equal to the angle A1RA2 would displace A from ^1 to ^2 and B from Bi to B2. Imagine any actual continuous spherical motion of a body, in which the two points A and B of the body are displaced from ^1 to yl2 and Bi to B2 re- spectively. Let A', A", etc., be several intermediate positions of A, and let B', B", etc., be corresponding intermediate positions of B. As already shown, the displacements of AB from AiB^ to A'B', from A'B' to A"B", from A"B" 280 Art. 54 281 to A"'B"', etc., might be accomplished by single rotations about definite fixed lines R'OS', R'VS", R"'OS"', etc. If a large number of intermediate positions A'B', A"B", etc., be assumed, and if the successive rotations be accomplished in times equal to the times required for the actual displace- ments in the continuous motion, then the succession of rotations would closely resemble the actual continuous motion. The more numerous the interme- diate positions, and the more numerous the succession of single rotations, the more closely would the succession resemble the actual motion. "In the limit," the succession would reproduce the actual motion; hence we may regard any spherical motion of a body as consisting of a continuous rotation about a line through the fixed point, the fine continually shifting about in the body and in space. The Kne about which the body is rotating at any instant is the instantaneous axis (of rotation) at that instant. At any particular instant of a spherical motion, the body is rotating about the instantaneous axis at a definite rate; this rate is called the angular velocity of the body at that instant. We will, generally, denote magnitude of angular velocity by co. In a rotation about a fixed axis, the (linear) velocity of any point of the body equals the product of the angular velocity and the perpen- dicular distance (or radius) from the point to the axis; and the direction of the Unear velocity is perpendicular to the plane of the radius and the axis. So too in a spherical motion, the linear velocity of any point of the body at any instant equals the product of the angular velocity at that instant and the radius (perpendicular from the point to the instantaneous axis for that in- stant); the direction of that velocity is perpendicular to the plane of the radius and the axis. Any angular velocity oj may be represented by means of a vector laid off on the corresponding instantaneous axis; the length of the vector is made equal to co according to some convenient scale, and the sense of the vector indicates the direction of the rotation according to some convention. We will always associate direction with angular velocity in the way just described; that is, we regard angular velocity as a vector quantity. In a spherical mo- tion, angular velocity changes in direction continuously; it may or may not change in amount too. In any case, the rate at which the (vector) angular velocity is changing at any instant is called the angular acceleration at that instant. (See page 148 for note on rate of change of a vector quantity.) This rate or acceleration has a definite amount and direction at each instant, and hence is a vector quantity too. We will use a to denote the magnitude of an angular acceleration. § 2. Composition and Resolution op Angular Velocities. — Imagine a body P to be rotating about a line li fixed in a body A ; and that A is rotating about a line h>, intersecting h and fixed in a body B (Fig. 426). For conve- nience we call the motion of P relative to B its absolute motion, and we regard this absolute motion as a resultant motion consisting of the (component) rotations about h and h- We will show presently that the absolute motion 282 Chap, xiii of P is spherical, and that the angular velocity of that motion equals the vector sum of the angular velocity of P relative to A and that of A relative toB. That the absolute motion is spherical will be conceded as almost self-evident; for the point (of P) does not move at all, being a point of B, and O appears to be the only point of P which is fixed. But more on this matter later. Let Oa and Oh (Fig. 427) be the two lines h and h at the instant in question, and let the angular velocities (of rotation) about those lines be coi and C02 re- spectively, and in the directions indicated. Let C, not shown, be any point of the body P in the plane of the lines h and h (or paper). If C is taken above Fig. 426 Oa, then the rotation coi alone brings C up out of the paper; if below, then cui depresses C. If C is above Oh, then C02 alone brings C up out of the paper; if below, 002 depresses C. Hence if C is in either acute angle between h and h, the two rotations give C displacements in opposite directions. Let Co be such a point C, and so chosen too that the two displacements of Co in an ele- ment of time dt would be equal. If ri and r^ = the distances of Co from h and h respectively, these displacements = ricoi dt and r^2 dt. Hence, /iwi = ^2^2, or OCo sin a-on = OCo sin /3 • wo; and sin a- coi = sin j8 • ooo, or sin a/ sin (3 = 0)2/0)1. (i) Let D be any other point on OCo; then its displacements due to oji and 0)2 in the time dt are respectively (OD sin a) wi dt and {OD sin /3)o)2 dt. But these are equal, since sin a • oji = sin/3 • 0)2; hence all points on OCo have zero velocity at the instant in question. Evidently, there are no other points in the body P whose velocity is zero at the instant; hence the state of motion of P is a rotation about OCo, a Hne fixed by equation (i). Let 0) = the angular velocity of the rotation of P (about OCo); Q (Fig. 427) be any point of P in the plane of the paper; q, gi, and g2 = distances of Q from OCo, h, and k respectively. Then the displacements of Q due to o)i and 0)2 are respectively gio:i dt and ^2^2 dt. These displacements for Q as chosen are in the same direction; hence the total or resultant displacement = (910)1 -f- 92W2) dt, and the linear velocity of Q (displacement per unit time) = ^lOJi + 5'2Wp,. Now the angular velocity of the body P equals the linear velocity of Art. 54 283 its point Q divided by the distance of Q from the (instantaneous) axis of rota- tion OCo, or CO = (^lOJi + q-iOii) -^ q. It follows from the trigonometry of the figure that gi = q cos a + (OR) sin a, and q-z = q cos /3 — (OR) sin /3. Substituting these values of qi and q^ in the expression for co and noting equa- tion (i), we arrive at CO — wi cos a + C02 cos /3. (2) Equations (i) and (2) respectively enable us to determine the axis (OCo) of the resultant of two angular velocities coi and coo and the amount of the result- ant angular velocity co. By means of equations (i) and (2) we can show that co is the vector-sum of coi and C02. Let OM and ON (Fig. 428), on h and k (Figs. 426 and 427), repre- sent 0)1 and C02, and let OMNR be a parallelogram. Then coi sin MOR = C02 sin NOR. Comparing this with equation (i), we see that MOR = a and NOR = /3; hence the parallelogram construction gives a diagonal which coincides with J^-z Fig. 428 Fig. 429 the instantaneous axis of the absolute motion. It will be readily seen from the parallelogram that OR = coi cos a -\- C02 cos (3 ; hence the length of the diagonal gives the magnitude of the angular velocity (see equation 2). Reverting now to the proposition that the absolute motion of P is spher- ical, we note that the axis of instantaneous rotation OCo (Fig. 427) is always in the plane of /i and h and hence is not fixed. Therefore there is only one fixed point of P, the intersection of /i and h. Obviously, the foregoing analysis could be extended to a case of simulta- neous rotations about three or more concurrent axes /i, I2, h, etc. Hence, in any case, the resultant motion is spherical, and the resultant angular velocity is given by the vector-sum of the component angular velocities. Conversely, the angular velocity of any spherical motion can be resolved into any number 284 Chap, xiii of concurrent components, and the vector-sum of the components is equal to the given velocity. § 3. Velocity of Any Point of the Moving Body. — Let P (Fig. 429) be any point of a moving body (not shown), fixed at 0; x, y, and z the (chang- ing) coordinates of P with reference to fixed axes OX, OY and OZ; co^, co^, and o3z = the components of the angular velocity of the body with respect to those axes; v = the linear velocity of P; and Vx, Vy, and v^ = the components of v along those axes. Then as will be proved presently I'x = ZOOy — ycOz, Vy = X(j3z — SWx, Vz = JCOx " XOiy. (3) If the body were rotating about the x axis only, then P would be describing a circle about X, and the velocity of P would be XP X co^. This velocity has no X component, and it is plain from the figure that the y and z components of that velocity respectively are — zwx and jco^. These component velocities of P due to angular velocity cox are scheduled below; also the component velocities due to angular velocities cOy and w^. It is plain from the schedule that the total component velocities due to the three angular velocities are as given by equations (3). Rotation about OX produces OY produces OZ produces 55. Body with a Fixed Point; Kinetics § I. Angular Momentum. — We will now explain what is meant by an- gular momentum of a body about or with respect to its fixed point. (For meaning of angular momentum about a line see Art. 48.) By angular mo- mentum of a particle about a point is meant the moment about that point of its momentum. Thus if m and v are the mass and velocity of a particle and p is the perpendicular distance from the point to the line drawn through the particle and in the direction of its motion at the instant in question, then the angular momentum, about the point, of the particle at the instant is mvp. (See Fig. 430.) This angular momentum is represented by a vector which we call r, through the point 0, perpendicular to the plane of O and the mo- mentum vector mv; the length of r is made equal (according to some conven- ient scale) to mvp, and the arrow on the vector is fixed in accordance with the usual rule. By angular momentum of a body about its fixed point is meant the resultant of the angular momentums about that point of all its particles. Thus the angular momentum, about the fixed point of the body, is represented by a definite vector R, namely, the resultant of the vectors which represent the angular momentums, about the point, of the particles of the body. If the vector representing the angular momentum of a body about the fixed point be resolved into a (rectangular) component along any line through the point, such component vector will represent the angular momentum of Vx = 0, Vy = — ZOJx, and Vz = ycox- Vx = ZiOy, Vy = 0, and Vz = — XClOy. Vx = - yo:z, Vy = XWz, and Vz = 0. Art. 55 285 the body about the line. We prove this proposition for a particle, and then extend the proof to a collection of particles, — that is a body. Let P (Fig. 430) be one of the particles of the body, not shown, the fixed point, OX the line, r the vector representing the angular momen- tum of P about 0, r^ the component of r along OX, p the perpendicular from O to the vector mv, or \^^^^ PQ, and a, /3, and 7 the direction angles of v. Then ^\ Vi = mvp cos XON = mvp (cos y - y — cos 13- z) ^ p ^ = m (vcosy ' y — vcos^ ' z) = mvzy — mvyZ. ^^ y_..-- " But this last expression is the value of the angu- y\q,. 430 lar momentum of P about the line OX (Art. 48). Al^o Sr^ = 2w {v^y — Vyz) or R^ = h^, where R^ denotes the component of R along OX and /?x the angular momentum of the body about OX; thus the proposition has been proved. Obviously the greatest value of Rx is R, and hence Jh is greatest when OX coincides with R. That is to say, with respect to lines through the fixed point of a body, the angular momentum is greatest for the line coinciding with the vector which represents the angular momentum of the body about the point; this greatest angular momentum, about a line, equals the angular momentum about the point. For brevity, we will call these (equal) angular momentums the or resultant angular momentum of the body; we will denote it by h. Finally h^, hy, and hz will be our symbols, not only for the angular momentums of the body about x, y, and z axes through the fixed point but for the components of h along those lines. Referring to the foregoing or Art. 48, it will be seen that hx = ^m{vzy-VyZ), hy = '^m{vxZ - v^x), h=^m{vyX - v^y). (i) These expressions for the component angular momentums can be transformed into the following (involving angular and not linear velocities) : hx = + Ix^x — JzWy — Jy(Jiz, hy = — J zWx + ly/dO. Also let R' = the component of R along the fixed line OU; then R' = R cos <>. The rate at which R' changes is dR'/dt = - i? sin (/) • d/dl = - 7? sin <^ • «. Now when R coincides with OU, = o, and the rate of change of R' = o. When R is per- pendicular to OU, = ^TT, and the rate of change of /?' = - i?w. If w is positive (rotation counter-clockwise), then - Ru is negative; and this means that R' is decreasing (obvious from the figure). If w is negative (rotation clockwise), then — Rw is positive; and this means that R' is increasing (obvious from the figure). 288 Chap, xm § 3. Kinetic Energy. — As in § 3 of Art. 54, let co^, o)y, and co^ = the axial components of the angular velocity of the moving body at any particular instant; x, y, and z = the coordinates of some particle P of the body then; and Vx, Vy, and Vz = the axial components of the velocity of P. If m = the mass of the particle, then its kinetic energy at the instant in question is Now if we substitute for Vx, Vy and Vz their values from equations (3) of Art. 54, we arrive at a new expression for the kinetic energy of P; and if we sum up such expressions for all the particles of the body, we find that the kinetic energy of the body is x^x J Z^xWy, + 2 I^J^v' + \ ^^? ~ J^yOiz — JyOiz(Jix where Ix, ly, and I^ are the moments of inertia of the moving body about the X, y, and 2 axes respectively, and Jx, Jy, and J ^ are the products of inertia of the body with respect to the pairs of coordinate planes intersecting in the X, y, and z axes respectively (Art. 57) all at the instant in question. That is Jx = 2wyz, Jy — Zmzx, J^ = Swxy. The products of inertia may be zero ; then the kinetic energy equals 56. Gyrostat § I. Steady Oblique Precession. — Let OC (Fig. 434) be the spin-axis of a gyrostat, OZ a fixed axis about which OC is rotating or precessing at a steady rate, d = the constant angle ZOC, w = the angular velocity of spin, and fi = the angular velocity of precession. Let OX and OY be fixed axes, perpendicular to each other and to OZ; OA an axis on the plane of ZOC and perpendicular to OC; and OB perpendicular to OA and OC. OA and OB are moving axes, rotating with OC about OZ. The motion of the gyrostat consists of the com- ponent rotations co and ^ about OC and OZ re- spectively. The resultant of those components is a rotation about the diagonal of the parallelogram on the vectors Oc and Oz representing w and Q, (Art. 54), and the angular velocity of that resultant rotation is represented by that diagonal. Hence, the components of the angular velocity of the gyrostat along OA, OB, and OC are respectively — 12 sin Q, o, and 00 -f 12 cos 6 = n, n being an abbreviation for co + fi cos d. It may be well to note the distinction between the velocity of spin co and n. The spin velocity is the angular velocity of the gyrostat relative to the moving Fig. 434 Art. $6 289 frame OABC; it is the product of 2 tt and the number of times per unit time which a point on the gyrostat pierces the plane ZOC. The angular velocity n is the component of the absolute angular velocity of the gyrostat along the fixed line with which OC happens to coincide at the instant in question. Since OA, OB, and OC are principal axes of the gyrostat (Art. 57), the angular mo- mentums of the gyrostat about these axes are respectively — ^ 12 sin 6, o, and C (co + 12 cos 6) = Cn, where A and C denote the moments of inertia of the gyrostat about OA and OC respectively. Since the entire frame of axes OABC is rotating about OZ with angular velocity 12, the components of that velocity along OA, OB, and OC are re- spectively, — 12 sin 6, o, and 12 cos 9. Substituting now in equation (3) of Art. 55, we get as the required values of the torques of the external forces about the axes OA, OB, and OC respectively Ti = o, Tz = Cn 12 sin - /1 12- sin d cos 6, and T3 = o. Therefore for steady spin and precession, there must be no torque about any line in the plane of the spin and precession axes {Ti and Tz = o) but a torque equal to Cn 12 sin - ^ 12- sin 6 cos d about a line perpendicular to those axes and through the fixed point. Let us now consider whether a gyrostat may precess steadily under the in- fluence of gravity and the pivot reaction only. Let W = the weight of the gyrostat, and h = the distance of its center of gravity from the pivot; then the torques of gravity about OA, OB, and OC are respectively o, Wh sin 6, and o. We assume that the pivot is so well made that the torques of the reaction about the lines mentioned equal zero practically. Hence the gyrostat is not subjected to any torque about OA and OC but to a torque Wh sin d about OB. If now the quantities W, h, d, etc., be given such values that Wh sin 6 = Cnn sind - A 12= sin d cos 9, then all the conditions for steady precession will be satisfied. Evidently such values can be assigned, in general. Indeed if we solve the preceding equation for 12, we get Cn± V(C^n^- 4 AWh cos d) 12 = 2 A cos 6 from which it is plain that in general there are two possible velocities of pre- cession for a given gyrostat, spin 00, and inclination 6. But if C~n" = 4 AWh cos 6, then there is only one value of 12; and if CV' < 4 AWh cos d, then 12 is imaginary, and the gyrostat will not precess under the conditions imposed. 2QO Chap, xin A gyrostat whose center of gravity is at the pivot will precess steadily for certain conditions of impressed spin, precession, and obliquity. For, suppose that spin, precession, and obliquity are so arranged that Cn = A^cosd; then Ts = o, that is no torque is required to maintain the precession. Hence the gyrostat, with center of gravity at the pivot, would continue to precess. It may be instructive to consider the follov/ing alternative analysis. Let hi, hi, and h be the angular momentums about OA, OB, and OC respectively; and hx, hy, and h those about OX, OY, and OZ. Then hx = {h cos 6 + hs sin 9) cos xf/, hy = {hi cos d + hs sin 6) sin ^, and h = —hi sin- d -\- h cos 6. Now hi, h, and hz are constant (see preceding page for their values) ; the only variable in the right-hand members is xl/; and since the x, y, and z axes are fixed in direction, the rates at which h^, hy, and hz change are respectively — (hi cos + ^3 sin d) Q, sin xj/, (hi cos 6 + hs sin d) 12 cos \}/, and o. Now consider the instant when OC lies in the plane ZOX; then i/' = o, and the rates are o, (hi cos 6 -\- h sin 6) 2, and o. Hence the torques about OX and OZ are zero, and that about OY (the perpendicular to the axes of spin and precession) is (/?i cos ^ + /?3 sin 0)12= - A^^sind cosd -{- CnQsind. This result agrees with that arrived at by the first method. Gyrostat in a Case. — The foregoing analysis must be modified for a spin- ning gyrostat in a frame or case which does not spin but merely precesses with the axis of spin. Let y4 and C be moments of inertia of the spinning part as before, and A' and C the corresponding moments of inertia of the case. Then the angular momentums of the case about the axes OA, OB, and OC (Fig. 434) are respectively -^'12 sine, o, and C^cosO. These may be added to the earlier expressions for corresponding momentums of the spinning part to arrive at values of the angular momentums of the entire gyroscope. Then substituting in equation (3) of Art. 55 as before, we find that the necessary torques about OA, OB, and OC for steady precession are respectively Ti = o, Ti= (Cw + C 12 cos 0)12 sine- (^ + ^') ^'sin^cose, and Tz = o, where w = w -f 12 cos 6 as before. § 2. Unsteady Oblique Precession. — Imagine a gyrostat to have been started spinning in some way, and then released and left to itself on a frictionless pivot under the action of the pivot reaction and gravity. The subsequent motion will now be investigated. Let Wo = the angular velocity of spin, and 0o = the angle between the axis Ar; ^6 291 of spin and the vertical at release. Let Fig. 434 represent the gyrostat at some instant after its release; w = the velocity of spin (velocity of the gyro- stat relative to the plane ZOC) ; Q = the velocity of turning of the plane ZOC (which we will continue to call velocity of precession); and d = the angle ZOC. We do not assume w, fi, and d to be constants. At the instant of release, when there is not yet any motion of the axes OABC, the total angular velocity of the gyrostat is coo. At a later instant the angular velocity of the gyrostat is the resultant of its velocity of spin 00 (relative to the frame OABC) and the angular velocity of the frame. Now this latter velocity has the following components along OA, OB, and OC respectively, Osin^, d* and 12 cos 0; hence the (resultant) angular velocity of the gyrostat has the following com- ponents along OA , OB, and OC respectively, — 12 sin 6, 6, and cu + 12 cos 9 = n, where n is an abbreviation for 00 + cos 6 as in §1. And the angular mo- mentums about those same lines are — Ailsind, Bd, and Cn. According to equation (3), Art. 55, the rate at which the angular momentum about OC is changing is Cn-^A9. sin d - 6 - Bd^smd; but this rate equals zero since there is no torque about OC. And because A = B, Cn = o; hence n is constant, and therefore always equals its initial value, that is, n — o. This does not mean that the spin velocity, co, is constant. Since there is no torque about the (fixed) axis OZ, the angular momentum about that line remains constant; thus that angular momentum at any instant equals its initial value, or A 12 sin 6' X sin + Ccoo cos 6 = Cojq cos 60. This equation shows that 12 = Ccoo (cos do - cos d)~ A sin2 d, (i) from which one may compute the velocity of the plane ZOC, or the velocity of precession. Investigation of the (nutational) motion of the spin-axis in the (azimuthal) plane ZOC can be made simplest by means of the princii)le of work and kinetic energy (Art. 43). From the instant of release of the gyrostat to any subse- quent instant, gravity does an amount of work on the gyrostat equal to the product of the weight and the vertical descent of the center of gravity. If, * According to this ifluxional) notation, a symbol with a dot over it means the time rate of the quantity represented by the symbol; thus d means dd/dl, s means ds/dl, etc. 292 Chap, xril as in § I, PF = the weight, and h = the distance from the pivot to the center of gravity, then the work done by gravity is Wh {cos da - cos e). The initial kinetic energy of the gyrostat is \ Coo,<- m i Y Fig. 7 6" ±. Fig. 8 that the radius of gyration of a single angle about the line XX is 1.85 inches, and that the area of one section is 9 inches 2. Hence the moment of inertia of the pair about XX = 2 (9 X 1.852), and the radius of gyration of the pair is V ' 2 (9 X 1.85-) 2X9 = 1.85 inches. Three Rectangular Axes Theorem. —The moment of inertia of an area with respect to any polar axis (perpendicular to the area) equals the sum of the moments of inertia of the area with respect to any two rectangular axes which intersect the polar axes and he in the area. If the rectangular axes and the polar axis be regarded as x, y, and z coordinates axes respectively, then the theorem can be written /, = /, + /,. (5) To prove this theorem let x and y = the coordinates of the element dA (Fig. 8). Then the distance of dA from the z axis is {x^ + /)^ and hence /, = p/l(x2 + /) =fdA ' .v'' +JdA . / = /^ + /^. If equation (5) be divided by ^, we get at once h^ = K^ + ky\ (6) where K, ky, and k^ denote the radiuses of gyration with respect to the x, y, and 2 axes respectively. 314 Appendix b Graphical Determination of the Moment of Inertia of a Plane Area. — If the area is quite irregular in shape so that it cannot be divided into simple parts whose moments of inertia are known, then the method now to be ex- plained may be resorted to for finding the moment of inertia of the irregular area about any Hne in its plane. This method is merely graphical integra- tion. Let the area at the left in Fig. 9 be the irregular area and XX the line 1 1 k- lY 2" 1 \ 1 A io ■'- \ X Fig. 10 about which the moment of inertia of the area is required, width of the area, parallel to XX, at any point of the figure; distance of the point or width from XX. Then Let w and y the the / (w dy) y^ = j (wy^) dy = \ w' dy, where w' is merely an abbreviation for wy-. Now suppose we multiply several widths w by the square of the corresponding distances y, lay off the products wy^ to any convenient scale from a perpendicular to XX as shown, and then draw a smooth curve through the ends of the ordinates or distances wy^ or w'. The area between this smooth curve and the perpendicular equals / w' dy, and hence it represents /. Evidently, the modified area, as we may call it, must be interpreted according to some scale as we will explain in connection with w y wy- w y wy^ 2.00 1-54 1-25 2.40 1 .96 0.25 0. 122 1.30 I 5° 2.92 1. 91 0.50 0.477 0.93 I -75 2.8s 1.83 0.75 1.03 2.00 1. 71 1 .00 I. 71 A numerical example. — Instead of an irregular figure, we take a regular one so that we can compute its moment of inertia by an exact method also, and thus check the graphical method. We will compute the moment of inertia of a circular quadrant (Fig. 10) of 2-inch radius about one of the straight sides. We have taken nine widths w, see adjoining table. At the right-hand of the figure we have laid off the products wy^, or w', to the scale i inch = 5 inches^. Appendix b 315 The new area is 0.25 inches^. Since the scale of the quadrant is i inch = 2.5 inches, the scale of the new area is i square inch = 5 X 2.5 = 12.5 inches*. Hence, the construction gives 0.25 X 12.5 = 3.12 inches'* as the moment of inertia desired. The exact formula i^^Trr*) gives 3.142 inches*. § 2. Formulas for Moment of Inertia and Radius of Gyration for SOME Special Cases. — In the following, / and k are symbols for moment of inertia and radius of gyration respectively. Only a few formulas for k are stated; in any case k can be computed from V//area. Rectangle. — Let b = base and h = altitude. About a line through the center parallel to b, / = tV b¥- About a line through the center parallel to h, I = tV hb\ About the base b, I = I b¥. About the side h, I = \ hb\ About a diagonal, 1 = 1 bW/Qf- + h"^)- About a line through the center perpendicular to the rectangle, / = yV (^^^ + ^^^)- Square. — Make b = h m foregoing. The moment of inertia for all axes in the plane of the square and passing through the center is ^^ h^, where h is the length of one side of the square. Hollow Rectangle. — Let B and b = outer and inner breadths, and H and h = outer and inner heights. About an axis parallel to B and b and passing through the center, I = tV (BH^ - bh^). Triangle. — Let b = base and h = altitude. About the base, / = ^\ bh^. About a line through the centroid, and parallel to the base, / = 5^ bh^. About a line through the vertex and parallel to the base, 1 = 1 bh^. Regular Polygon. — Let A = area, R = radius of circumscribed circle, r = radius of inscribed circle, and s = length of a side. About any axis through the center and in the plane of the polygon, / = 0^4 A (6 R^ — s"') = 5^5 y4(i2 r^ + 5^). About a line perpendicular to the plane of the polygon passing through the center, / = double the preceding /. Trapezoid. — Let B = long base, b = short base, and h = altitude. About the long base, / = yV (^ + 3 W^^- About the short base, / = tV (3 ^ + ^)^^- About a line through the centroid and parallel to the bases. Circle. — Let d = diameter and r = radius. About a diameter, I = ^^Tj-d* = lirr*; k = \d = \r. About a line through the center and per- pendicular to the circle, / = ^^i^d'^ = ^tt/-*; ^ = V| J = Vf r. Semicircle. — Let d = diameter and r = radius. About the bounding diameter or about the line of symmetry, / = jis 7r• Fig. 2 324 i3~(5)- Compute the moment of the 40 lb. force (Fig. i) about point 2, making use of the principle of moments. 14. A certain chimney is 150 ft. high and weighs 137,500 lbs. Suppose that it is subjected to a horizontal wind pressure of 54,000 lbs., uniformly distributed along its height. Determine where the hne of action of the resultant of the weight and pressure cuts the ground. 15. Fig. 3 represents the cross section of a masonry dam. It weighs 150 lbs/ft' and the water pressure against it is 112,500 lbs. per foot length of dam. The resultant pressure acts at right angles to the face of the dam and 20 ft. above its base. The center of gravity of the cross section is 11.46 ft. from the face of the dam and 24 ft. above the base. Find where the resultant of the weight and the pressure cuts the base. i6-(5). Imagine a clockwise couple of 2 ft-lbs. to act on the square board of Fig. i. Then compound the couple and the 40 lb. force. i7-(5). Fig. 4 represents a 3 ft. pulley on the end of a shaft; the pulley is subjected to a pull of 100 lbs. applied tangentially as shown. Resolve the force into a force acting through the center of the pulley and a couple. i8-(6). Compound the four forces (wind pressures) represented in Fig. 5. (Be prepared to give the incUnation of the resultant and the point where the line of action cuts the floor.) i9-(6). Fig. 6 represents one-half of an arch and certain loads applied to it. Pi = 4000, P2 = 5000, P3 = 6000, and P4 = 10,000 /S,000 53,000lb5. SZOOOIbs. I8,500,^^_ I8,500i^^_ .^Sl 37,000/bs. < gQ' - •>1 (<--7-->|<-5^>K 76- ■>f J/ 8 •*! Fig. 5 Fig. 7 lbs.; their inclinations are 0°, 3°, 8°, and 12° respectively; the coordinates of points I, 2, 3, and 4 are (1.6, o.i), (4.9, 0.7), (8.4, 2.1), and (12.8, 4.8), all in feet. Com- pound the four load by the second method. (Specify the line of action of the result- ant by means of the angle between it and the x axis and the intercept on that axis.) 20-(7). Determine the resultant of the locomotive wheel-loads (Fig. 7). 2i-(7). Determine the resultant of the loads described in Prob. 19 algebraically. 22-(8). Compute the moments of each of the forces represented in Fig. 2 about the X, y, and z axes. 325 23-(8). Determine the resultant of the three couples acting on the 4 ft. cube repre- sented in Fig. 8. (Specify the plane of the resultant by means of the angles which a normal to the plane makes with the edges of the cube.^) 24-(9). Determine the resultant of all except the 300 lb. forces (Fig. 8). 25-(io). State what you can about the resultant in the following cases: (a) A system of coplanar concurrent forces for which SF„ = 0; for which SMa = 0; for which the force polygon closes. (b) A system of noncoplanar concurrent forces for which SF^ = o. (c) A system of coplanar parallel forces for which SF^, = o; for which zMa = o; for which the force polygon closes. (d) A system of coplanar nonconcurrent nonparallel forces for which ZF^ =0; for which zFx = 2F„ = 0; for which SM„ = sMb = o. Fig. 8 Fig. 9 Fig. 10 26-(ii). A and B (Fig. 9) are two smooth cylinders supported by two planes as shown. A weighs 200 lbs. and B 100 lbs.; the diameter of ^ is 6 ft. and of B 10 ft.; a = 30°. Determine the pressures on the planes and that between the cylinders. 27-(ii). Fig. 10 represents two wedges; a = 70° and = 40°. A push P of 1000 lbs. can sustain what load Q if all rubbing surfaces are smooth? 28-(ii). The chains AB and AC (Fig. 11) are 5 ft. long. When BC = 8 ft. and the suspended load W = 2 tons, what is the tension on each chain? If the safe pull for each chain is 3 tons, how large may the spread BC be? Fig. II 29-(ii). Two bars AB and CD (Fig. 12) are connected by a pin at A and to a floor by pins B and C. BC = 8 ft., AB = AC = $ ft., and AD = 8 ft. A weight of 100 lbs. is suspended from D. Determine the pin pressures at yl, B, and C. 30-(ii). A carrier is arranged as shown in Fig. 13. The bar AB connecting the axles of the wheels is 24 ins. long. The bars AC and CB are each 30 ins. long There 326 is a load of 1200 lbs. at C. g///////////////////////////////////////////// C E A" D r ^ ^A B4 Fig. 14 be stalled in the position shown, ^ = 60 degrees, and the steam pressure 150 lbs /in^. Determine the push on the connect- ing rod BC and the pressure against the cross-head guide D. 33-(ii). The beU-crank ^BC (Fig. 16) Determine the compression in AB and the tension in ^C and BC. 3i-(ii). AB (Fig. 14) is a bar suspended from a ceiling by means of vertical ropes AC and BD. The middle points E and F are connected by another rope. AB = AC = BD = 8 f t. A vertical force P is applied at the middle G, deflects the ropes as shown by the dotted hues, and raises the bar. How large must P be to support the bar (weighing 1000 lbs.) 6 ins. above its original position? 32-(ii). The cylinder of the steam engine (Fig. 15) is 10 ins. in diameter, the crank AB \s $ ins. long, and the connecting rod 5C is 15 ins. long. Assume the engine to -\ v~~M& ^1 '^^ \ A / Fig. is Fig. 16 is pinned to a wall at ^ ; a cyUnder G is suspended by means of a cord from D as shown; BD = 4 ins. The cylinder weighs 80 lbs. and is smooth. Determine all the forces which act upon the bell-crank. 34-(ii). Fig. 17 represents a riveting machine operated by compressed air. It consists of a rigid frame F on which the air cyUnder C is mounted; P is the piston; AB is the piston rod pinned to the piston at A so that the rod can be rotated somewhat about A inside of the (hollow) piston; the toggle link BD is pinned to the frame at D; the toggle Unk BE is pinned to the plunger Q (movable iji a vertical guide on the frame) at E; HE are the rivet dies between which the rivet is squeezed. AB = ig ins.; BD = 13 ins.; BE = 10 ins.; the diameter of the cylinder is 10 ins. Assume the air pressure to be 100 lbs/in^ and then determine the pressure at the pins D and E, the pressure against the guide, and the pressure on the rivet. (To "lay out " this mechanism begin at D, then fix A, then B, and then E.) Solve the problem when A is advanced 2 ins. from the position shown. -Air 2000\lbs. 1 500\lbi. iyuiD5.pern : ^ V D 4' Fig. 18 Fig. 17 3S-(i2). The beam AB (Fig. 18) is supported at C and D, and it sustains three loads as shown. The beam weighs 50 lbs. per lineal feet. Determine each supporting force, or reaction. 327 36-(i2). Fig. 19 represents a shutter dam; AB is the shutter, and CD and CE are braces. The shutter and the braces are pinned together at C; the shutter rests against an inclined stop at A; brace CD is pinned to a bed plate at D; brace CE rests against a bed socket at E. The shutter is 4 ft. wide and its length AB = 12 ft. The water pressure is 16,000 lbs., and its "center" is at F, 4 ft. from A. Determine the reactions at D and E due to the water pressure. 37~(i3)- Fig. 20 represents a truss supported by a shelf 5 on a wall and a horizontal tie A ; AB = g ft. and BC = 12 ft. Determine the reactions at A and B due to the loads. 38-(i3). AB (Fig. 21) is a beam supported by a rod CD and a pin a.t A; AB = git. AC = 3 ft, AD = 8 ft., 400 lbs. and the lOGO/is. U-2'6 Fig. 19 and ^E = 5 ft. The beam weighs Determine load, = W 1000 lbs. the pull at C and the pressure at A. 39-(i3). The crane represented in Fig. 22 is sup- ported by two floors as shown. £ is a hole in the upper floor and F is a cylindrical socket in the lower floor. The crane weighs 5 tons and its center of gravity is 2 ft. to the left of the axis of the post. Determine the pressures on the floors when the load W is 5000 lbs. 40-(i3). A and B (Fig. 23) are two horizontal pegs in a wall; they are 3 and 6 ft. above the floor respectively, and the horizontal distance between them is 4 ft. A smooth straight bar CD, 15 ft. long and weighing 200 lbs., is placed under A and over B with its lower end on the floor, but is not sprung into that position. Determine all the pressures on the bar, due to its own weight. 41. AB (Fig. 24) is a bar 12 ft. long fastened to the floor at ^ by a pin and it rests at C on a smooth cylinder 4 ft. in diameter. The center of the cylinder is 6 ft. to the right of A and is connected by a horizontal cord to the bar at D. A weight of 100 lbs. is hung on the free end of the bar. What is the pressure between the bar and the cylinder; between the cyUnder and the floor; what is the tension in the cord; j.; 2o'- >i ^^^ what is the pressure exerted by the pin on the bar A ? Consider the cyUnder and the bar as weightless. 42. AB (Fig. 25) is a bar 20 ins. long, and weighs 10 lbs. It rests on a peg C and against a ?vj .jf._. Fig. 22 328 smooth wall at ^ , as shown. What vertical force applied at B will preserve the equilibrium of the bar? 43. If the weight of the bar in Prob. 42 is 12 lbs. and a load weighing 4 lbs. is suspended at B, at what angle must the bar be placed to insure equilibrium? /77777777777777777777P777777? Fig. 23 Fig. 24. Fig. 25 44-(i4). Figs. 26 and 27 are two outline views of a steam shovel; the former repre- sents a dumping and the latter a digging position. A is the "^-frame," B the boom, and D the dipper. The pin P (axis perpendicular to the paper) is seated on the upper half of a "fifth wheel " which permits swinging of the boom about the vertical axis FQ. Two engines are mounted on the boom, — the main engine which operates the hoisting drum, and the thrusting engine which operates the pinion meshing with a rack on the bottom of the dipper handle. •i/f|<-3'6"*l*--#'6"-*' Fig. 26 Fig. 27 Many of the parts of a shovel are most severely stressed when the dipper is en- countering an unyielding obstruction in the bank. We indicate how some of these stresses may be determined and then ask the student to determine others. The actual resistance of the bank against the dipper cannot be determined with certainty because the line of action of the resistance is generally unknown. It doubt- less depends largely on the direction in which the cutting edge of the dipper tends to move in the bank, determined mainly by the pull of the hoisting rope and the thrust on the dipper handle. Some designers assume that the hne of action of the resist- ance for the digging position shown in Fig. 27 is about along the bottom of the dipper. Making this assumption and analyzing the system of forces acting on the dipper and 329 its handle (resistance of the bank, hoisting pull, weight of dipper and handle, and thrust on the handle) we find that the resistance is about 20,000 lbs. We might proceed now and determine the pressures developed at various points in the structure and mechanism on account of this bank resistance. For instance, analysis of all the external forces acting on the boom, dipper and handle, main and thrusting engines (resistance of bank, pull of front guys G', pin pressure at base of boom, and weight of parts under consideration) shows that the pull of the guys is about 22,000 lbs. The student should now determine the stress in each leg of the /I -frame and that in each back guy G". (These guys are fastened to the car at points 9 ft. apart. 22 1 ft. from the base of the ^ -frame and on the same level with that base.) 45-(i4). Suppose that the shovel is digging as shown in Fig. 27, but with the boom at right angles to the track as shown in Fig. 26. The pull of the front guys is 22,000 s. as in the preceding problem. Determine the stress in each leg of the /1-frame, and the stress in each back guy. \ Fig. 28 Fig. 29 46-(i5). The truss represented in Fig. 28 is supported at A and D; CE = 12 ft., Pi = 1000 lbs. and P2 = 2000 lbs. Deter- mine the amount and kind of stress in each member. 47~(i5)- The truss represented in Fig. 29 is supported at F and D; BF = CE = 12 ft.. Pi = P2 = 2000 lbs., and P3 = P4 = 1000 lbs. Determine the amount and kind of stress in each member. 48-(i5). The truss represented in Fig 30 is supported at A and E; each load P = 1000 lbs Fig. 30 Determine the amount and kind of stress in each member. Fig. 31 49-(i5). The truss represented in Fig. 31 is supported at each end; span = 80 ft. 330 Fig. 32 I, 2, 3, 6 and the points 3, 4, 5, vertices of parallelograms. Draw a stress dia- gram for the truss loaded as shown, and make a record of the stresses in the members. 52-(i6). Solve Prob. 47 graphically. 53-(i7). Fig. ;^^ represents a crane consisting of three members, a boom AC, a. brace AD, and a post BF. The crane is supported at E and F by two floors. The load W= 5 tons. Determine all forces acting on each member. 54-(i7). The crane represented in Fig. 34 consists of a post AB, a. boom CD, and braces DE and FG. The crane is supported by sockets at A and B as shown. The boom passes freely through a smooth slot in the post at H so that and rise = 20 ft. ; consecutive points on AG are equidistant; DI is perpendicular to ^G; H bisects AI, and / bisects GI; each load = 1000 lbs. Determine the amount and kind of stress in IK, and be prepared to describe how to determine the stress in every other member. 5o-(i6). Solve Prob. 46 graphically. Si-(i6). The truss represented in Fig. 32 is supported at each end. The points 7 are at the - II' A — -;??9Z -^ B D -*r- ^ w Floor ^1 I _]^ \j Floor J;_ Fig. 33 any reaction existing there will be xC vertical. The counterweight at O is I ton, the load W is \ ton, and the latter is 21 ft. from the axis of the post. Determine all the forces which act upon each member. 55-(i7). Fig. 35 represents a certain type of hydrauhc crane. It consists of a post AB, an hydraulic cylinder C mounted on the post, a large sleeve 5" m B Fig. 34 which can be slipped along the post, two rol- lers D and E mounted on the sleeve, a boom EF, and a tie rod FG. When water (under pressure) is admitted to the cylinder, the pistons are pushed upward; the upper one bears against the sleeve, and rolls the entire part DEFG up along the post. Let the load W= 2 tons and suppose that it is 10 ft. out from the axis of the post; then determine all the forces which act upon each pin {D, E, and G). Fig. 35 331 56-(i7). Solve Prob. 54 but take into account the weights of the members of the crane as follows: post AB = 0.7 ton, boom CD = 0.5 ton, brace DE = 0.3 ton, and brace FG = 0.6 ton. Middle of boom is 5 ft. 6 ins. from axis of post. 57-(i7). Solve Prob. 53 but take into account the weights of the members which are as follows: post BF= 0.5 ton, brace AD = 0.2 ton, and boom AC = 0.7 ton. The boom is 18 ft. long; its center of gravity is 2 ft. 6 ins. from B. 58-(i8). Fig. 36 represents a crane supported by a foot-step bearing at B and a collar-bearing at C. B can furnish horizontal and vertical support, and C can furnish - ^^^''' 6^ >-To Hoist Fig. 36 Fig. 37 horizontal support only. The pulleys E and F are i ft. in diameter; the noisting cable enters the post at F, descends through the post, over pulley G, and to the hoist as shown. The counter-weight H is 2 tons and the load 4 tons. Determine all the forces which act upon each member. S9-(i8). The crane represented in Fig. 37 consists of a post AB, a boom CD, and a tie rod DE. The pulley at D and the winding drum at G are i ft. in diameter. The load W is I ton. DE = 12 ft. Determine all the forces which act on each member. 6o-(i8). Imagine the winding drum (Prob. 59) to be mounted in bearings at H (supported by the brace CD) instead of at G. Then solve. B c A Cord t B mmmmmm cc wmmm/// B C Fig. 38 Fig. 39 Fig. 40 Fig. 41 6i-(i9). A (Fig. 38) weighs 100 lbs. and B 200 lbs. A, B, and C are very rough. Make separate sketches of A and B and represent all the forces which act on each body when P = 20 lbs. (not large enough to produce any shpping). 62-(i9). A (Fig. 39) weighs 100 lbs. and B 200 lbs. For A and B, ft = I; for B and C, fi = f. How large must P be to cause shpping? 63~(i9)- -4 (Fig. 40) weighs 100 lbs.; the surfaces in contact are very rough; P = 50 lbs., and a = 20°. Determine the friction F and the normal pressure N. 64-(i9). A (Fig. 40) weighs 100 lbs.; a = 20°, and /x = 0.6. How large must P be to start A ? How large is F when slipping impends? 65-(i9). A (Fig. 40) weighs 100 lbs., a = 40°, /j = 0.6, P = 200 lbs. Does P move A ? 66-(2o). Same as Prob. 63 but refer to Fig. 41. Same as Prob. 64 but refer to Fig. 41. Same as Prob. 65 but refer to Fig. 41, Fig. 42 represents a double-wedge device for raising and lowering a heavy load W* The device consists of wedges A and B and bearing blocks C and D; W = 200,000 lbs. The coefficient of friction is 0.5. How large are the required pushes P to raise the load? How large are the required pulls to lower the load? (First consider C and determine jX.-^w,M^^^,„.Z,v?w.jj^,^^^v^^y the forces acting upon it.) 7o-(2o). Fig. 43 represents, somewhat conventionalized, ■ ^^ an adjusting device used in making the closure (insertion of the last few members) of a large cantilever bridge (Beaver River) .f The mechanical elements are a double wedge IF, a screw 5", and a lever L. The accessories are a head piece ZT, two struts A, and two wedge-blocks B\ they are pin-connected as shown. C and C are two portions of the bridge member to be connected; they are under compres- sion P and pin-bear against the wedge blocks B. The nut, which bears against the head piece, can be turned by means of the lever, and the screw and wedge raised or lowered. Raising the wedge separates the wedge blocks and parts C and C Determine the necessary moment (of force) on the lever for raising the wedge against pressures P = 1,235,000 lbs., assuming that the struts A are vertical and the following data: mean diameter of screw = 4^ ins.; pitch of screw = I in.; bevel of wedge (each side) Fig. 43 = I in 10; mean radius of nut where it bears on the head piece=9 ins.; coefficient of friction for all rubbing surfaces = J. (Consider first a wedge-block, and determine all the forces which act upon it.) 7i-(2o). Fig. 44 represents a screw toggle used in the erection of a steel arch (Niagara Falls and Clifton Bridge). t It consists of four multiple links pinned to- gether as shown, a right-and-left screw 5 with nuts N, and a lever L. The toggle is supported by the anchor rod R and brace B. The "pulling end " of the toggle was connected to the arch under construction, supplying the supporting force P. Assume mean diameter of screw = 2 ins., pitch = 5 in., coefficient of friction = 0.3 ; also that now the diagonal MM = 16 ft., 4 ins., and the diagonal NN = 4 ft., 4 ins. De- FlG. 44 termine the couple on the lever which will shorten NN; which will lengthen NN. 72-(2o). Solve Prob. 34, taking into account the friction at all rubbing surfaces (pins, piston, and guide). Pins A and E are 2 ins. and pins B and D are 3 ins. in * Engifieering News, July '15, 1911. f Engineering Record, June 10, 1911. t Skinner's Details of Bridge Construction. 333 diameter. The coefficient of friction is i. (Solve graphically and make drawing of riveter full size or larger.) 73-(2o). Fig. 45 represents a band-brake. The diameter of the wheel is i ft., 8 ins., the angle of lap = 255°, P = 60 lbs., and the coefficient of friction is i; the wheel is turning clockwise. Compute the fractional moment and the pull on the pins A and B. Solve for the case when the wheel is turning in the other direction. 74-(2i). Fig. 46 represents a crank-arm for a shaft, by plan and elevation — dotted Unes to be disregarded. Locate the center of gravity of the arm. 75-(2i). Solve Prob. 74 but change width at thin end as shown by dotted lines. (See Obelisk, Art. 24.) 76-(2i). Fig. 47 represents a connecting rod for a steam engine by plan and elevation the distances of the center of gravity from the center of each hole. 1 4'-- - Fig. 45 The rod is i J ins. thick except as noted. Determine Jl. i_ [<.-4"-H< ■7"-->k-J"-H a- 7: el CM ~CM. ■MCM CM — T • "tn i I t"'. / "1 /3' -^,//'K2'H cvi J: -■X CM Fig. 46 Fig. 47 77-(2i). Fig. 48 represents a thin plate into which holes were punched at A and B, and the pieces glued on at C and D respectively. Area of hole A = 4 ins.^; that of 5 = 2 ins^. Locate the center of gravity of the modified plate. Fig. 48 K" 4" ->i !<■• 4"->i »o ■-inj 1 '_¥ ^ Fig. 50 78-(2i). Fig. 49 represents a five-sided prism made from a rectangular prism by bevelling one edge as shown. There is a cylindrical hole 2 ins. in diameter in the piece, its axis being parallel to the line AB, 4 ins. from the face ABODE, and i^ m. below the face ABG. Find the center of gravity of the modified prism. 79-(2 2). Fig. 50 is the cross section of a steel beam "built- up " of two angles 5X4XI ins. and a plate 8 X f in. The centroid of each angle is 1.57 ins. from the back of the shorter leg. Determine the position of the centroid of the entire section. 8o-(2 2). Fig. 51 represents the cross section of a machine part. Determine the position of the centroid of the crosssection. 8i-(23). Prove that the distance of the centroid of a triangle from its base equals one-third the altitude. 82-(23). Prove that the distance from the centroid to the base of a paraboloid of rr volution formed by revolving a pa- rabola about its axis equals one-third the altitude a (see Fig. 168, page 94). 334 83-(24). Prove construction (i), page 98, for locating the centroid of a trapezoid; also construction (2), page 98, for locating the centroid of any quadrilateral. 84-(2 5). A cord is supported at two points on the same level 30 ft. apart, and its lowest point is 8 ft. below the level of the supports. If the load is 20 lbs. per hori- zontal ft., what are the tensions at the supports and at the lowest point? 85-(26). A cable is to be suspended between two points at the same level 200 ft. apart; the sag is to be 80 feet. Determine the length of the cable. 86-(27). A rope 100 ft. long is suspended from two points A and B at the same level 80 ft. apart. A body weighing 1000 lbs. is suspended from a point Cx ft. dis- tant from A. Determine the tension in AC when x = 20, 30, 40, 50, 60, 70, and 80 ft. Make a graph showing how the tension varies with x. 87-(28). Fig. 52 is a chronographic record of the launching of the U.S.S. Cali- fornia {Transactions of Naval Architects and Marine Engineers, Vol. 12). Determine FINISH 74 75 Seconds. / Revolution of Chronograph CyfMer iv 40-ft. Travel of Ship (Time between Notches = g6econd } Fig. 52 the velocity of the ship at the twentieth second in the following three ways: first, from the average velocities for at least four intervals after the instant ; second, from the average velocities for at least four intervals before the instant; third, from the average velocities for the half-seconds immediately before and after the instant. 88-(28). The following velocities (feet per second) were computed from the chron- 335 ographic record (Fig. 52) by taking the mean of the average velocities for the half- seconds immediately preceding and following the instants or times listed below. t= 15 16 17 18 19 20 21 22 23 24 v= 2.50 3.00 3.55 4.20 4.80 5.45 6.10 6.75 7.45 8.15 Compute the acceleration for / = 16 sees. 89-(28). Reduce a sprint of 100 yds. in 10 sees, to miles per hour. Compare the retardation of a train at 4 mi/hr/sec with the retardation of gravity on a ball thrown vertically upward. 90-(28) . A point P moves in a straight line so that s = 2fi — ^f, where 5 (in feet) equals the distance of P from a fixed origin in the path at any time / (in minutes). Determine the velocity and acceleration when / = i min.; when t = 2 mins. Inter- pret the negative signs. 9i-(28). A certain point P of a mechanism is made to move in a straight Une by means of a crank in such a way that 5 = 3 cos 2 (9, where 5 = the (varying) distance in feet of P from a fixed origin in the path of P, and e = the (varying) angle which the crank makes with a fixed line of reference. The crank rotates uniformly at 100 rev/min. Determine position, velocity, and acceleration of P when e = 60°. Interpret signs of the results. 92-(28). In a certain "gunnery experiment" the shot was fired through screens placed 150 ft. apart. The times (in seconds) of piercing were observed with the following results: screen I 2 3 4 5 6 7 time . 0666 o- 1343 0.2031 0. 2729 0-3439 0.4161 Determine the velocity at the fourth screen. 93-(28). A point P moves in a straight line so that = 4—2/, where a is in feet per minute per minute and t in minutes. When t = o, v = o and s = o. De- termine general formulas for v and s. What are v and 5 when / = 4 ? when / = 5 ? 94-(28). A certain electric train can get up full speed of 24 mi/hr in a distance of 150 ft., and can stop from full speed in a distance of 100 ft. What is the shortest time in minutes in which the train can make a run between two stations 650 ft. apart, the train starting from one station and coming to full stop at the other ? (Assume that the starting and stopping are accomplished uniformly with respect to time.) 95-(28). A certain train can be retarded at a rate of 4 mi/hr/sec by braking. Determine the times (in seconds) and the distances (in feet) in which the train can be stopped from 10, 20, 30, and 40 mi/hr. (Assume that the retardation is the same at all speeds.) 96-(29). Draw the distance-time and velocity-time graphs for the interval from 15 to 24 sees, of the launching mentioned in Prob. 87, and determine the velocity and acceleration at the twentieth second from the graphs. 97-(29). Fig. 53 shows the acceleration-time graph for a certain rectilinear motion. When t = o,v and s = o. Construct the v-t and s-l graphs. 98-(29). Make a sketch of the velocity-time graph for the train-run described in Prob. 94, calling the lengths of the three periods /i, /2, and h respectively. Then use the principle that " area under the curve " represents distance travelled to find values of h, t^, and (3, and finally the time for the entire run. 336 99~(3o)- The period of a certain simple harmonic motion is 8 sees., and the ampli- tude is 6 ins. What is the maximum velocity ? the maximum acceleration ? For the motion from one extreme point in the path to the center, what is the average velocity? the average acceleration? ioo-(3o). Four particles, Qi, Qi, Q3, and Q^, are describing simple harmonic motions in AB (Fig. 54); the period of each motion is 8 sees. At a certain instant the four 5 sees. B Y 3 X B >/////////>////////r'f^i'^w/MW////w, Fig. 53 Fig. 54 Fig. 55 particles are at points i, 2, 3, and 4 respectively; Qy and ^3 are moving toward the right and Q^ and Q^^ are moving toward the left. Write out the expressions for the X coordinates of the moving points / sees, after the instant mentioned. {AB =.12 ins., and is divided into sixths by the points.) ioi-(3i). A (Fig. 55) weighs 200 lbs., B weighs 100 lbs.; the coefficient of friction " under " .-1 is \, that under 5 is }; P = 300 lbs. Determine the acceleration of -4 and B, and the tension in the rope connecting them. B •»r/j/>///^//j////'/////, Fig. 56 1 I B I ''mTTTTTTTTT^TTTTTTTmr. Fig. 57 B • 'WW/Wr ' ^M////U- X'- Fig. 58 i02-(3i). Suppose that the supporting surface in the preceding problem is not horizontal but inclined at 30 degrees to the horizontal. Then solve. io3-(3i). A (Fig. 56) weighs 50 lbs. and B weighs 100 lbs.; the pull P gives A and B an acceleration of 2 ft/sec/sec. Determine the magnitude and direction (referred to the horizontal) of the pressure between A and B. io4-(3i). Two bodies are connected somewhat as two cars, and are placed on a plane inclined at 30 degrees to the horizontal. The lower one weighs 600 lbs. and is smooth, that is, there is no resistance to its sliding on the plane. The upper one weighs 1000 lbs., and the coefficient friction under it is y"^. With what acceleration will the bodies slide down when released ? WiU there be tension or pressure at the con- nection ? What is its value ? io5-(3i). The weights of .1, B, and C (Fig. 57) are 50, 100, and 200 lbs. respec- tively. Contacts between .4, B, and C are very rough; between C and D very smooth; P = 100 lbs. Determine the forces which the bodies exert upon each other. Sketch each body separately, showing the forces acting on it. io6-(3i). A (Fig. 58) weighs 100 lbs., and B weighs 200 lbs. The coefficient of kinetic friction under -B is |; the coefficient of static friction under A is ■^\. When P = 75 lbs., will A slip? How great is the friction under .4? How large a force P would- just make .4 slip? io7-(3i). A (Fig. 56) weighs 50 lbs., and B weighs 100 lbs. C is perfectly smooth; the coefl&cient of static friction "between" A and .B is ^; the angle between the 337 top of B and the horizontal is 25 degrees. How great may P be without making A slip on 5 ? io8-(3i). A (Fig. 59) weighs 100 lbs., and B weighs 50 lbs. The coefficient of friction under A is \. Neglect the inertia of the pulley and the friction at its axle, and find the acceleration of A and B, and the tension in the cord. (The re- sultant of the three forces acting on A is T— 20, where T = tension; and the resultant of the two forces acting on 5 is 50 — T. Now write the equations of motion, R = {W/g)a, for A and B, and solve them simultaneously for a and T.) io9-(3i). Show that the acceleration of the suspended bodies and the tension in the cord of the Atwood machine (Fig. 60) are respectively (2 W1W2) (iK-Il) and T Wi + IF2 Wi^Wi when the inertia of the pulley and the axle friction are negligible. iio-(3i). Fig. 61 represents a simple engine, without connecting rod. Stroke = 18 ins:, speed = 150 r.p.m. Piston and rod weigh 120 lbs. When x = 3 ins., steam A '^w//mhr^[ i ' B Fig. 59 Fig. 60 Fig. 61 M pressure = 2000 lbs. Determine the pressure of crank pin P on the piston rod. When the piston is advanced 6 ins. beyond the position shown (x = — 3 ins.), the steam pressure is still 2000 lbs. Determine the pressure of the pin on the piston rod for this position. iii-(3i). Suppose that Fig. 61 represents an air compressor, steam being changed to air and the crank turning clockwise. Determine the pressures of the crank pin for the two positions mentioned in the preceding problem (x =+ 3 ins. and x =— 3 ins.). ii2-(3i). Fig. 62 represents, in prmciple, a certain "throw " test- ing machine for subjecting a metal specimen to rapid changes of direct stress (tension and compression). S is the specimen, firmly screwed into two bosses M and N. W is a weight firmly fastened to the lower boss. The parts named can be oscillated in the vertical guides G by means of an ordinary crank-connecting rod mechanism (OP-PC). When the machine is not running, the specimen is sub- jected to a tension equal to the weight of N and W. When the machine is running, the stress on the specimen changes continuously.* Let OP = I in., PC = g ins., weight of N and W = 25 lbs., and speed of crank = 2000 rev/min. Determine the stress on the spec- imen at each end of a stroke or oscillation, and at the middle of the stroke. ii3-(3i)- Take data except speed as in preceding problem. Determine the speed * For detailed description see Phil. Trans. Roy. Soc, Ser. A, Vol. 199 (1902). g-In^g 1 Fig. 62 338 which would make the stress on the specimen equal to zero at the upper end of the stroke. What would the stress be at the lower end at that speed ? ii4-(32). A point P starts at A (Fig. 63), and moves in the circle as indicated traversing distance s so that s = 2 fi, where t is time after starting in seconds and 5 is in feet; radius OA = 20 ft. Draw the hodograph for the first 3 sees. Then de- termine the average accelerations for the intervals i to 3, 1.5 to 3, 2 to 3, 2.5 to 3. Next determine the magnitude and direction of the acceleration when t = 3 from these average accelerations. Y Y f / \ A M y )-^ Fig. 63 Fig. 64 ii5-(33). Determine Vx, Vy, v, Ox, Oy, at, an, and a for the motion described in Prob. 114 and when / = 3 sees. ii6-(33). The point Q (Fig. 64) on the rim of a wheel rolling in a straight line describes a curve known as cycloid. Let v' = velocity of the center of the wheel C, a' = the acceleration of C, and R = radius of the wheel. Find formulas for the x and y components of the velocity and acceleration of Q when in the position shown. (Let 5 = the abscissa of C, and x and y = the coordinates of Q. Then x = s — R sin d, and y = R {i — cos 6) ; also 5 = Rd.) f - Fig. 66 ii7-(33). A point Q describes a simple harmonic motion; the frequency = 100 (to and fro) oscillations per minute and the amplitude = 3 ft. Determine the average accelerations of Q for the following distances traversed : first 6 ins. from one end of its path; second 6 ins.; third 6 ins.; and first 18 ins. ii8-(33). A 16-inch gun can give a projectile weighing 2400 lbs., a muzzle velocity of 1465 mi/hr, and at an elevation of 15 degrees can throw it approximately 9 mi. Compute the range for the velocity and elevation stated neglecting air resistance, and compare with the actual range. ii9-(34). A cylinder C (Fig. 65) is suspended by a cord and rests against a smooth inclined plane P as shown. The cylinder weighs 20 lbs.; its diameter is one foot. The plane is rotated at 30 rev/min about the vertical axis AB. Determine the tension in the cord and the pressure against the plane. i2o-(34). CD (Fig. 66) is a vertical axis about which E can be rotated. A is a 339 5' S" B V i^- D T . ■3' * Fig. 67 — T body resting on E, and B is suspended by means of a cord fastened to A as shown. A weighs 10 lbs. and B weighs 20 lbs. Suppose that E makes 30 rev/min; then compute the pressure at the stop S. The centers of A and B are 5 and 3 ft. from CD respectively. (Neglect friction under A, a,t B and the pulley axle.) i2i-(34). Suppose that ,1 and B in Prob. 120 are rough, the coefi&cients of static friction being j for each. What rate of rotation would lift B? i22-(34). T (Fig. 67)' is a horizontal whirling table. A and B are spheres connected by an elastic cord, the tension in which is 30 lbs. when the table is at rest. A weighs ir lbs. and B weighs 40 lbs. What are the pressures of the stops S' and S" against the spheres when the table is rotated about CD at 20 rev/min ? i23-(35). Suppose that the floor of the car and A (Fig. 283, Art. 35) are very rough so that A will not slip on the car; then ascertain how great an acceleration of the car would result in tipping of A . i24-(35). Suppose that the coefBcient of friction in Prob. 123 is i. If the ap- plied push on the car is gradually increased, thus increasing the acceleration grad- ually, will .1 sUp or tip eventually ? Fig. 68 Fig. 69 2" W///////Ay/^^//////////////A Fig. 70 -^zft^MI^ i2S-(35). The Scotch cross-head (Fig. 61) described in Prob. no presses against the stuffing box and on the cylinder by reason of the weight of the cross-head and the pressure of the crank-pin on it. Suppose that the center of gravity of the cross- head is 1 5 ins. from the center of the slot, the center of the piston is 24 ins. from the same point, and the center of the stuffing box is 13 ins. from 0. Determine the pressures mentioned when the circumstances are as in Prob. no (steam pressure = 2000 lbs., etc.). i26-(36). Show that the moment of inertia of the slender wire AB (Fig. 68) about the x-axis is § Mr\\ — (sin a. cos a)/a\, where M = mass of the wire. i27-(36). Show that the moment of inertia of a right circular cone about its axis is I'ij Mr'', where M = the mass of the cone and r = the radius of its base. i28-(36). Show that the moment of inertia of the ring or torus (Fig. 69) about the 2-axis is M (R^ + 5 r^), where M = the mass of the ring. i29-(36). The length of a homogeneous right elliptic prism is /, and the semi- axes of its cross section are a and b. Prove that the radius of gyration of the prism with respect to a line through its center of gravity parallel to the axis b is i30-(36). Fig. 70 is a section of a cast-iron flywheel; there are six spokes. The cross section of each spoke is elliptical, the axes of the ellipse being 2 inches and 5^ ins. long. Compute the moments of inertia of rim, spokes, and hub with respect to the axis of the wheel; also the radius of gyration of the wheel about that axis. 340 i3i-(37)- In order to produce a tension of loo lbs. in the cord of Ex. 2, Art. 37, how heavy must the suspended body be ? i32-(37)- A, B, and C (Fig. 71) weigh 100 lbs., 30 lbs., and 34.4 lbs., respectively. The diameter of C = 2 ft. 3 ins., and the radius of gyration of C about the axis of rotation = i f t. ; = 30 deg. Friction under .1 , when the system is moving, = ID lbs. Determine the acceleration of A, B, and C, and the tensions, the system having started without initial velocity. (Neglect axle friction.) i33~(37)- A, B, and C (Fig. 72) weigh 50 lbs., 100 lbs., and 150 lbs. respectively. C is a solid disk of cast iron 16 ins. in diameter. Determine the acceleration of A, B, C, and also the pulls of the cord on A and B. (Neglect axle friction.) Fig. 71 Fig. 72 Fig. 74 i34-(37). AB (Fig. 73) is a brake for regulating the descent of the suspended body C. C weighs 1000 lbs., the drum 2000 lbs., the diameter of the drum =12 ft., that of the brake wheel = 14 ft., a = 4 ft., 6 = 6 ins., and the radius of gyration of the entire rotating system about the axis of rotation = 4 ft. When P = 100 lbs. and the coefficient of brake friction is |, what is the acceleration of C? (Neglect axle friction.) 135^(37) • The wheel A (Fig. 74) is a solid cylinder weighing 1000 lbs. and its diameter is 8 ft. It is desired to arrange a brake BC as shown, by means of which the speed of the wheel may be reduced from 100 rev/min to zero in 10 sees. The coefficient of friction at D = \; the available pull P is 100 lbs. Determine the ratio a/h. (Neglect axle friction.) i36-(38). Determine the magnitude and direction of the axle reactions in Probs. 131 and 132; in Probs. 133 and 134. 137^(38) • Iri Fig. 74, a = 6 ft. and & = 6 ins.; the wheel weighs 400 lbs. The coefficient of brake friction = |. When the wheel is turning clockwise, a push Pof 120 lbs. is appUed. Determine the axle reaction. i38-(38). A (Fig. 75) is a rigid piece which can be rotated about the vertical axis BC. D is a vertical bar pinned to A at E, and rests against A at F; the bar is 14 ins. long and weighs 20 lbs. The speed of '^ rotation is 100 rev/min. Determine the pressures on D. i3g-(38). In Ex. i of Art. 38, § 2, take P as applied at F and solve; then as appUed at G and solve. i4o-(39) . Compute the length to the nearest hundredth inch of the simple seconds pendulum for your locality. i4i-(4o). The body C (Fig. 76) weighs 50 lbs. It is dragged up the plane by P (=40 lbs.) and Q {= 20 lbs.). The frictional resistance is 5 lbs.; a = 30°. Com- » ■ ■ > , 341 pute the work done on C by each force acting on it while C is moved from A to B, a distance of i s ft. i42-(4o). ABC (Fig. 77) is a smooth rail in the form of a vertical semicircle of 4 ft. radius. D is a. body, weighing 50 lbs., which can be made to slide along the rail. P is a force of 150 lbs. always inclined 30 deg. to the horizontal; Q is a force of 40 lbs. always directed along the tangent. Compute the work done on D by all the forces acting on it while D is moved from .1 to B. Fig. 76 Fig. 77 i43-(4o). Solve the preceding problem on the supposition that P is always di- rected toward B. i44-(4o). In order to retard the motion of a launching ship, ropes were fastened to it and to points on the shore, so that the ship broke many of the ropes as it pro- gressed. In order to estimate the retarding effect of each rope broken, tension tests were made on samples of the rope (7-in. manilla). Fig. 78 shows the average ten- sion-stretch curve for these tests. The average strength of the samples was about V a in 0) «0 ^ ^^ 2.0 ^^ ^ ^ ^^ y^ • y y 1.0 / / / / / 10000 Pu\l, 20000 Pounds. 30000 Fig. 78 32,500 lbs. It was assumed that the efficiency of the knots used would be about 80 per cent, and therefore that the ropes would fail at about 26,000 lbs. On the basis of this assumption and the curve, it was estimated that each rope (20 ft. long) would do 60,000 ft-lbs. of work on the ship before breaking. Can you check this estimate? (Data taken from Trans. Soc. Nav. Archls. and Mar. Engrs., 1903, p. 295-) i45-(4i). Show that the rotational part of the kinetic energy of a rolling sphere is two-sevenths of its total kinetic energy. 342 i46-(4i). A certain freight car with its load weighs 60 tons. Each pair of wheel* with its axle weighs 1800 lbs., and the radius of gyration of a pair and axle with respect to the axis of the axle is 0.81 ft.; the diameter of the wheels is ^:^ ins. De- termine the ratio of the rotational part of the kinetic energy of the moving car (and load) to the trans- lational part. i47-(42). In the American Machinist for Dec. 2, 1909, there appears a communication in which an alleged " favdt in brake dynamometers " is pointed out and explained. The writer states that on several occasions he got ridiculous results with a Prony brake. The "enigma" became clear to him when he en- countered a " paradox " in his experimental work, described by him as follows: In Fig. 79 S represents a shaft, mounted in two bearings BB', carrying two levers, arms AA', each exactly 50 inches long from center of shaft to the fulcrums MM', respectively, and firmly keyed to the shaft. At K is represented a coun- tecweight which balanced the two lever arms and brought the center of gravity about the center line of the shaft S; T represents a platform scale and W represents a weight, which weighed 100 pounds when placed on the scales T. When W of loo-pounds weight was hung on the fulcrum N', the scales just balanced at no pounds. At first the paradox almost paralyzed the brain, but on closer examination the mystery was easily solved, as follows: Considering A and A' to be firmly keyed to the shaft S, then the two arms and shaft S become practically one solid mass. Therefore, when an}^ weight IF is placed on the fulcrum M' of the lever A', the whole mass will tend to rotate about a line passing through the points of support M and B, with a moment of W times the lever arm X'. The shaft 5 at the point B' will be fetched forcibly up against the top of the box or bearing cap of bearing B', which will resist the rotation of the mass about MB, with a balancing moment equal to WX', or a reaction on the bearing cap equal to WX' WX ~T^ . or -y- • yTy^W 'IOOlbs. Fig. 79 Now, it is evident that the resultant of these two forces is a downward vertical force C at the point C equal to IF + {WX)/Y which load is distributed between the points of support B and M inversely proportional to their respective distances from the point C Hence the load on the scales T will be represented by ZC Z ( WX\ F-I-Z'^V + Z V Y j Hence a weight of 100 pounds on the fulcrum M will produce a load on the scales T equal to V + Z I , iooZ\ instead of 100 pounds as generally believed. The above condition obtains, more or less, in the vast majorities of dynamometers, and is sometimes so exaggerated as to make the re- sults positively ridiculous. In the case of a motor test let IF represent the tangential pull on the armature, an equal upward pull on the opposite side of the shaft might tend to balance the error, or it might tend to make matters worse, depending upon the position of the other 343 points in the diagram, but wherever W may fall the results will be most erroneous. For instance, suppose that IF happens to fall on the line MB at W, then it is evident that the weight exerted on the scales T will be equal to ^^^^,msteadof — as generally accepted. Show that the writer is mistaken in his assertion that ( loo H :^ — ) does not = loo, and that WZ V + Z does not = NW v}//////7iJ/J// ///////////J///////////////m777777^, V + Z and, hence, that his explanation of the " enigma " does not explain. i4o-(42). Fig. 80 represents Durand's dynamometer. A, B,C, and D are sprocket wheels of equal diameter; A and B are mounted on a beam XYT which is carried by the well-known Emery steel-plate support or knife-edge at E. The knife-edge rests on the standard R. Sprocket wheels C and D are mounted on R. The bars ^^ are fastened rig- idly to the beam, and engage loosely with a pin on R, thus limiting rotation of the beam. The sprocket chain passes over A, under D, over B, under C, and up to A. The shafts for C and D are extended forward and back; and on these extensions puUeys may be mounted, or universal joint couplings may be attached, for the receipt and delivery of power. (For detailed descrip- Fig. 80 tion see American Machinist for June 20, 1907.) OEO'T is horizontal; PQ and IH are vertical; MN and KL are inclined at an angle of 27 deg. with the vertical; OR = O'E = 12 ins.; and ET = 24 ins. Suppose that an electric motor on the shaft of C turns counter-clockwise at 100 rev/min, and transmits to a machine on the shaft of D, and that a weight of 40 lbs. at T keeps the beam XY balanced. What is the power of the motor? i49-(42). Assume that the law of mean effective pressure and piston speed is represented by the dotted line in Fig. 331 of the text, so that p = ^o[o.9S - (7-^-^11,000)], where p = mean effective pressure, po = boiler pressure, and s = piston speed in feet per minute. Then derive a formula for indicated locomotive power. Find piston speed at which power is maximum. Also graph your formula in the figure, calling the maximum power 100 per cent. i5<>-(42). Let D = diameter of the driving wheels of a locomotive in inches; / = stroke in inches ; d = diameter of the cylinder in inches ; po = boiler pressure in pounds per square inch; and V = velocity of the locomotive in miles per hour. Assume that the mean effective pressure varies as described in the preceding problem. Derive a formula for the indicated power of the locomotive in horse powers for any velocity V. iSi-(43)- A certain body weighs 400 lbs., and is dragged along a rough hori- zontal plane by a force of 80 lbs. The force is inclined 20 deg. upward from the horizontal; the coefficient of friction between the body and plane is about j\. 344 At a certain point in the motion, the velocity of ^ is 5 ft/sec. What is the velocity of A 10 ft. beyond the point? i52-(43). For the purposes of comparing the " running quahties " of certain freight car trucks, they were tested substantially as follows: Each one was made to roU down a steep incline to give it " initial velocity," and then it passed onto a moderate upgrade; the velocity was measured at two points on the upgrade; then the loss of kinetic energy was computed. These losses furnished a comparison. The up- grade was 0.38 per cent, and the points at which velocities were measured were 257.2 ft. apart. One of these trucks (four-wheeled) weighed 18,150 lbs.; each pair of wheels and axle 1800 lbs. The diameter of wheels was ^;i ins.; the radius of gyration of a pair and axle was 0.81 ft. In one test the velocities at the two points were 14.95 ^.nd 11.05 ft/sec. Determine the average ''truck resistance," a single imaginary force equivalent to actual resistances, not including gravity, on the truck. (Experiments by Prof. L. E. Endsley for American Steel Foundries.) i53-(43). The suspended body C (Fig. 81) weighs 10 lbs. The coefficient of friction under the brake is ^; n = 4^ ins., ^2 = 6 ins., a = 2 ft., and b = 1 it. C is allowed to descend 6 ft., thus turning the wheel, and then the brake is put on, with P = 20 lbs. How much farther will C descend? (Neglect axle friction.) i54-(43). A, B, and C respectively (Fig. 82) weigh 100, 30, and 64.5 lbs. The diameter of C = 30 ins., and its radius of gyration about the axis of rota- tion = I ft.; = 30 deg. The friction under A = 10 lbs. Determine the velocity of the system when A has moved through 10 ft. from rest. i55-(43). Copy Fig. 339 (pertaining to Exs. 2 and 3, Art. 43, § 2) using scale I in. = 10,000 lbs. and 5 mi/hr. (a) Make a graph in your copy which will show how the accelerating force is apportioned between the locomotive and the cars. What is there in your finished figure which represents draw-bar pull? (b) Modify your figure for the case of the train when on an upgrade of 0.5 per cent. i56-(43). Make graphs showing how the total train resistance in pounds varies with the velocity in miles per hour according to Schmidt's formula and the Engi- neering News formula for the train described in Ex. 2, Art. 43, § 2. i57-(43). Make a new figure (as for problem 155) assuming that the train re- sistance varies according to Schmidt's formula. First assume level track; then modify the diagram for the case of an upgrade of 0.5 per cent. i58-(43). Referring to the preceding problem with train on upgrade: (a) Make a graph showing how the acceleration changes with velocity, (h) Find the time re- quired for the velocity to change from 10 to 20 mi/hr. (See § 3, Art. 28.) i59-(43). Make a graph showing how the velocity of the train of the preceding problem (on the upgrade) changes with the time (in seconds) during the run men- tioned. i6o-(43). Make a graph showing how the distance covered by the train of the preceding problem (on an upgrade) changes with the time. 345 i6i-(44)- Fig- 83 represents in outline a certain small vertical-lift bridge. The lifting span is counterweighted as shown. At the center of the span there is a cross- shaft having on each end a drum long enough to provide for two up-haul and two down-haul cables. From each drum the cables are led to deflecting sheaves at each Sheave Floor Spur C5eCT/-><'ag™< ■ Pinion Bevel Gears-^^^f, ^Drum Drum- Cross %5hafr Section at Drums. Sheave Sheave ^;;;^?p^ ] heave ,/'■ \ 1 — -' 1 \ \ 1 \ \ • '^ > ^ Pre 55urt ' , ?■■""■ • — ^ "^ 1 000 soo 600 c 400 t 200 •a IZ & 4 O Seconds before StriKing. 0.5 1.0 1.5 Seconds after Striking. 2.0 Fig. 88 Compute the time-average and the space-average force which stopped the loco- motive, neglecting the effect of the so-called train resistance. Estimate the train resistance from the retardation of the locomotive just before the impact, and then recompute the averages just mentioned. Measure the area under the pressure curve and interpret it. Does the shape of the curve suggest any improvement in the buffer? i75-(48). The power of an operating hydraulic turbine equals the product of the angular velocity of the turbine and the rate at which angular momentum (about the axis of rotation) of the flowing water is changed in its passage through the tur- bine. Prove. i76-(48). A certain homogeneous prism is 2 X 6 X 36 ins. in dimensions. It is mounted so that it can oscillate like a common pendulum about either of two axes of suspension. Both axes contain the center of one small face of the prism; one axis is parallel to the 2-in. edges, and the other is parallel to the 6-in. edges. Locate the center of percussion for each of these axes. i77-(4g). Describe the gyrostatic reaction which a screw-propelled ship sustains when pitching (in a rough sea). i78-(49). In the General Electric Review, Vol. IX, pages 117 and 118, there appears the following: "The spin of a precessing body increases the centrifugal force about 351 the axis of precession. Take the case of a wheel spinning about a horizontal axis supported at one end which is precessing about a vertical axis through the point of support. The total centrifugal force is which equals the ordinary centrifugal force WV^/gR plus the additional centrifugal force due to spin (gyroscopic centrifugal force) {WV^k'^p-)/{gR2r'^). W = weight of the gyroscope, k — its radius of gyration, R = the radius of the circle of precession, r = the radius of the spinning wheel, V = the hnear velocity of the precession, v = the peripheral velocity of the wheel, and p = the ratio v/V.'" Presumably, R means the radius of the circle described by the mass-center of the wheel. Ascertain in your own way whether any force, appropriately called centrifugal force, has the value above stated in the case in question. i79-(49). On page 144 of the journal mentioned in the preceding problem there appears this statement. "The total vertical force on the outside rail [car wheels running around a curve] due to gyroscopic action will therefore be (3 WV^k"^) -i- {2gRrx)." W = the weight of a pair of wheels and axle (presumably), k = radius of gyration of the pair and axle (about their axis) , r = the radius of the wheels, R = radius of the curve, x = gage of the track, and V = the velocity of the car. Can you prove the statement? i8o-(5o). A wheel 6 ft. in diameter rolls on a straight track. At a certain instant the velocity and acceleration of its center are 10 ft/sec. and 4 ft /sec/sec. Deter- mine the acceleration of the lowest point of the wheel at the instant in question. i8i-(5i). When a slender body, such as a pole, chimney, etc., is tipped over from an upright position, the motion is one of rotation about the point of contact of the body and the surface which supports the body until slip occurs at the contact or the lower end Ufts from the surface. Assume that the slender body is hinged to the supporting surface so that it cannot slip or Hft, and then determine the vertical and horizontal components {V and H) of the supporting force for various positions of the tipping body. Draw curves showing how V and H vary with the angular displace- ment of the pole from the vertical. How could you ascertain whether slip or hft would occur first? i82-(5i). Referring to the preceding problem, assume that the pole is supported on the ground, and that sUp cannot occur during tipping. The lower end of the pole will lift when a certain degree of tip is reached; afterwards the pole moves under the influence of gravity only. Until the pole strikes ground, it rotates with the angular velocity which it had at the instant when the contact was broken, and the center of gravity moves in a parabolic path due to its initial velocity (when the contact was broken) and action of gravity. Determine the distance from the (original) point of support of the pole to where it first strikes the ground. i83-(52). In Fig. 410, the load W = 18,000 lbs.; the diameter of the roUers = 15 ins.; the coefficient of rolling resistance "under" the rollers = 0.020, that "over" the rollers = 0.025. How large a force P is required to move the load? Determine the two forces which act upon a roller supposing that the load is distributed equally among the rollers. i84-(52). Referring to Prob. 162: The rollers used were 3 ins. in diameter; about 2000 were used. They were of steel 2 ft. long and rolled between steel plates above 352 and below. Assume that yoiir computed result in Prob. 162 is the value of the pull actually exerted on the building when moving on a level stretch. Then compute the average coefficient of rolling resistance. i85-(53). Two men A and B are walking at a speed of 4 mi/hr along east and west and north and south paths respectively. Compute the velocity of A relative to B when A is walking northward and B eastward; when A is walking northward and B westward. i86-(s3). The disk (Fig. 89) is 4 ft. in diameter and is rotating uniformly about O at one rev/sec. A point P is moving uniformly along the diameter AB from A toward B at a speed of 4 ft/sec. Determine the absolute velocity of P when midway between A and 0; when midway between and B. i87-(53). Suppose that P (see preceding problem) is moving from C toward A; the angle — 150°, and when P reaches A its speed is 6 ft /sec (along CA). What is the absolute velocity of P then? Fig. 89 i88-(53). A certain square is 6 X 6 ft., and its corners are lettered A, B, C, and D in succession around the perimeter. The square is rotating uniformly about a Une through .4 perpendicular to its plane at one rev /sec; a point P is moving uniformly along CD and in that direction at 6 ft/sec. Determine the absolute velocity and acceleration of P when it reaches the mid position between C and D. i89-(54). The sphere (Fig. 90) is suspended from the end of a vertical shaft OZ by means of the rod OC extending into and rigidly fastened to the sphere. The shaft and the rod are connected by a Hooke's (flexible) joint. When the shaft is rotated it exerts a torque on the rod which in turn makes the sphere roll around on the cone. Assume that the sphere is 2 ft. in diameter, R = 4 ft., / = 8 ft., and that the shaft makes 1 50 rev/min. Determine the angular velocity of the sphere, and the X, y, and z components of that velocity. i9o-(55). Referring to the preceding problem, suppose that the sphere is cast iron (weighing 450 lbs/ft^). Then compute the angular momentimi of the sphere and determine the rate at which the angular momentum is changing. i9i-(55). Suppose that there is no "rolling resistance" (Art. 52) between sphere and cone. Then determine the following: normal pressure and friction between cone and sphere; the torque which the shaft must exert on the rod; and the x, y, and z components of the supporting force at 0. i92-(56). Fig. 91 represents in principle the Griffin Mill for grinding cement. The cross piece of the (upright) frame supports the upper (vertical) shaft S by means of a thrust ball bearing. The large pulley P is rigidly fastened to the shaft. The 353 pulley hub HH is extended downward and is restrained laterally by the guides GG, thus virtually forming an extension of the shaft. The "roll" is rigidly fastened to the "roll shaft" and both are suspended on a cylindrical seat on the inside of the hub of the pulley as shown. Thus the roll and its shaft can oscillate like a common pen- dulum about a perpendicular to the paper at 0. K is a cross head rigidly fastened to the roll shaft but slipping in vertical guides on the hub when the roU and its shaft oscUlate like a common pendulum. The "die" is a hard metal ring between which and the roll the grinding of the cement takes place as explained presently. When the mill is idle, the roU shaft hangs in a vertical position; if the pulley be rotated the guides in the hub exert a torque on the cross head, and the roll shaft is made to rotate in the vertical position with the pulley. When it is desired to start the mill for grinding, the roll is first puUed outward "with an iron hook," and then the power is turned on at the pulley. The roll shaft rotates with the pulley; promptly, the roll begins and continues to roll on the die (ring), a great pressure being developed between roll and die. Material to be ground is fed into the mill so that some is caught between the roll and the die and then pul- verized. Suitable paddles on the lower side of the roU continually toss the material which collects in the recess of the base; eventually it is caught be- tween roU and die. It will be noted that the roll and its shaft constitute a large gyrostat. We now propose the problem of determining the pressure between the roll and the ring when the mill is operating. The makers (Bradley Pulverizer Co.) state it to be about 15,000 lbs. for their giant size when run at a pulley speed of 165 to 1 70 rev/min. The follow- ing data, approximated in some cases, was taken from drawings furnished by the makers of the mill. The die is 40 ins. in diameter (inside), 8 ins. high; from the plane of its top to the point of suspension is 5 ft. 4I ins. The roU weighs 880 lbs.; its larger diameter is 24 ins. The roll shaft weighs 600 lbs. ; its length over all is 6 ft. 9^ ins.; its point of suspension is 6 ins. from the upper end; its diameter varies from ' 5 1 ins. at the cross head to 6| ins. at the roll but the ends in the cross head and roll are tapered. For simpUcity, make the following approximations: roll-shaft uniform diameter is 5I ins., smaller diameter of roll = 22 ins., and its thickness is 8 ins. As a further close approximation for locating center of gravity and determining required moments of inertia, assume that the roll is a cylinder 23 ins. in diameter and 8 ins. thick (with 5I ins. hole for the roll shaft). i93-(57). A certain right cone with a circular base is homogeneous; the diameter of its base is 4 ft.; the altitude is 6 ft.; and half the apex angle is 20°. Determine the radius of gyration of the cone with respect to an element of its curved surface. > ///w////w/ //////////^/////Y/w/w/////////^' Fig. 91 PROBLEMS; SET B The number in parentheses following a problem mimber refers lo the article {in the body of this book) which pertains lo that problem. In many of the problems, the data are not fully given. The missing ones are indicated in the statement or figure; they are to be supplied. 2oi-(3). Fig. loi represents a wood frame to which forces are appUed as indi- cated; the forces are applied by means of ropes tied to nails in the frame. By in- *- 4 > ■ 1 ' ' 1 ■ 1 ' 1 ___, — L— r-- — ,--j---, — • 1 1 L__^--L_-. ' 1 1 1 1 1 1 1 1 Fig. ioi spection of the figure, estimate where you might drive the nail and how you would pull the rope to get the same effect as the 50 and 70 lb. forces give. Determine com- pletely * the resultant of the 70 and 80 lb. forces by means of the parallelogram law. 202-(3). Determine completely the resultant of the 40 and 50 lb. forces by means of the triangle law. (Remember that the arrow-heads on the sides of the triangle are not confluent, do not point the same way around.) 203-(3). Determine completely the resultant of the 60 and 70 lb. forces alge- braically. 204-(3). Determine completely the resultant of the 40 and 70 lb. forces. 205-(3). Resolve fully the 50 lb. force into two components parallel to the 60 and 80 lb. forces. 2o6-(3). Resolve fully the 60 lb. force into two components, one perpendicular to that force and one horizontal. 207-(3). Resolve fully the 60 lb. force into two components, one perpendicular to that force and one vertical. 2o8-(3). Resolve fully the 60 lb. force into two components, one horizontal and one vertical. 209-(3). Resolve fully the 60 lb. force into two components, one applied at the upper right-hand corner and one at the lower left-hand corner of the square. * In many of these problems it is required that one or more forces be determined wholly or in part. Complete determination requires, not only the magnitude of the force, but also its direction and at least one point in its line of action; statement also of magnitude, line of action, and sense is complete. In order to make the line of action of a required force seem more real, the student should decide upon an appropriate point of application; for example a nail or hook in the body to which one could ti^. a cord or rope for applying the iorce. 354 355 2io-(3). Resolve fully the 60 lb. force into three components, applied along the two sides and bottom of the frame. 2ii-(3). Resolve fully the 60 lb. force into two components, one of 120 lbs. applied at point I and one horizontal. 2i2-(3). Is it possible to apply three forces along the sides of a triangular board so that they will balance? Is it possible to apply four unequal forces along the sides of a square board so that they will balance? 2i3-(4)- Compound completely the 30, 40, 60, and 90 lb. forces (Fig. loi) graphi- cally. (Do not draw the force polygon in the space diagram; use standard notation.) 2i4-(4). Compound completely the 40, 50, 70, and 100 lb. forces algebraically. (Specify the direction of the resultant by means of the acute angle between its line of action and the horizontal.) 2i5-(4). Resolve fully the 100 lb. force (Fig. 102) into x, y, and z components. 2i6-(4). Compound completely the forces applied to the cube (Fig. 102). YU--- 4 - ->i I jo Fig. 102 2i7-(5)- Compute directly (force times arm) the moment of the 50 lb. force (Fig. loi) about point i. 2i8-(5). Compute indirectly (by means of the principle of moments) the moment of the 60 lb. force (Fig. loi) about point 2. 2i9-(5)- Without actually locating the line of action of the resultant of the horizontal and vertical forces in Fig. loi, ascertain where the line of action cuts a side of the frame. (Use principle of moments.) 2 20-(5). Fig. 103 is a cross section of a "rolling dam." AB is the sheath rigidly fastened to the cylinder which can be rolled upward on two inclined racks CD, one at either end of the dam. The figure shows the dam resting on the bed at A and against the rack at B. In that position the horizontal and vertical com- ponents, Pi and Pi respectively, of the water pressure are 180 and 30 tons. The weight W is 70 tons. Without compu- ting the resultant of these three forces, \^vv\\w\\vw Fig. 103 Fig. 104 ascertain how far from D its line of action intersects the Hne CD. 22i-(s). Resolve each of the forces shown in Fig. 104 into a force applied at the center of the pulley and a couple. 222-(5). The forces (Fig. 105) arc applied at right angles to the cranks. Re- solve Pi into a force at A and a couple, and P2 into a force at B and a couple. 356 223-(6). Fig. io6 is a half-section of a building. The four forces are wind pres- sures, perpendicular to and applied at mid-points of the portions oi, T2, etc. Deter- R= 100 lbs. ^ 1 n B.r" 1 .-f na 1 ±- 7JA' \y A 1 > W" < ^ y^ ^i- 100 lbs. \/(6,9) -X 1 ^ p 4 i5 Fig. 106 Fig. 105 mine completely the resultant of the four forces graphically. (Note where the line of action cuts 'il and 45. The figures in parentheses are coordinates of the points I, 2, and 3 with respect to O.) 2 24-(6). Determine completely the resultant of the three forces described in Prob. 220. 2 2 5-(7). Determine completely the resultant of the six forces applied to a plank as shown in Fig. 107. 226-(7). What force applied at the middle of the plank (Fig. 107) and couple are equivalent to the three forces acting upward? 227-(7). Resolve fully the 20 lb. force (Fig. 107) into •p J two components applied at A and B\ applied at B and C. Determine completely the resultant of the four forces described in A B > C <- ^ < 4- •■> 1 1 'Y' 1 1 -+- 1 _)_ 1 — 1 I -+- _1_ 1 -- - 1 1. 1 _J_ 1 . _J_ 1 __L_ - .4'....-— H 228-(7). Fig. 108. 2 29-(8). Compute the moments of each force (Fig. 102) about the coordinate axes. 230-(8). Fig. 109 represents a cube with several forces applied to it. Specify as com- pletely as possible the resultant of the two couples. 23i-(8). Resolve the couple represented in Fig. 105 into two components, one whose plane is perpendicular to AB, and one whose plane is parallel to AB. 232-(9). Determine completely the resultant of the forces excluding the two Fig, Fig. 109 100 lb. forces applied to the cube shown in Fig. 109. (Specify the coordinates of the point where the line of action of the resultant pierces the XY plane.) 357 no 233-(q)- Fig. no represents a three throw crank shaft, consecutive cranks being 120° apart. When the shaft is made to rotate rapidly, as in a balancing machine, the cranks are subjected to equal air pressures. Assume that these are perpendicular to the .cranks respectively and have (equal) 4 in. arms; then reduce these three forces to a force applied at B and two couples. 234-(io). (a) Two couples are in equiUbrium; what can you say about them? (h) Can a force and a couple be in equilibrium? (c) Three parallel forces are in equilib- rium; what can you say about the middle force? {d) Is it possible for four forces to be in equilibrium if three are parallel? (c) Four forces are in equilibrium and two are known to constitute a couple; what can you say about the other two? (/) What can you say about the resultant of a set of coplanar nonconcurrent forces whose force polygon closes? surfaces are smooth; a, /3, and 7 are respectively , and deg.;(2 = 1000 lbs. Determine the magnitude of P required for equilibrium. 236-(ii). The sphere (Fig. 112) weighs 200 lbs.; a, /8, and 7 are respectively , , and deg.; all contact sur- faces are smooth. Determine the pressures on the wedge M and the tension in the rope. 237-(ii). A homogeneous beam 20ft. long weighs 50 lbs. At the upper end it rests held at the lower end by a rope ft. Ill Find the angle between the beam and the against a smooth vertical wall and is long which is also attached to the wall wall, and determine the pull in the rope. 238-(ii). Two right circular cylinders are supported in a box 18 ins. wide as shown in Fig. 113. A weighs smooth. Find all of the forces acting on each cylin- der and properly represent on separate sketches. 239-(ii). A cylindrical cask is rolled slowly along a level floor by means of a push applied at the top. If d is the diameter, W the weight of the cask and is and B weighs lbs. All surfaces are 17777777777777777777777777^^ Fig. 113 Fig. 114 the inclination of the force with the horizontal, determine the least push required to move the cask over a cleat h units high. 24o-(ii). AB (Fig. 114) is a rigid beam; two hooks are pinned to it at A and B as shown; CD and CE are rods pinned to the hooks and to each other; the hooks engage a heavy body W. AB = 14 ft., CD = CE = 8 ft., W = 4 tons. Deter- mine the tension in each rod and all forces acting on one hook. (Neglect weight of parts.) 358 t' 24i-(ii). In Fig. IIS ^5 =4 ft., AC = CD = 8 ft.; P = 200 lbs.; and the cylinder E weighs 100 lbs. A, B, and E are pin joints. The surfaces at D and F are smooth. Determine the forces acting on the cyl- inder and those acting on the bell-crank ACD. 242-(ii). Fig. 116 shows Ridley's apparatus for withdrawing the mold from a freshly cast concrete pile. The device also compresses the concrete into the space vacated by the mold. Pin joints are used at A, B, C, and D. Assume P = 10,000 lbs.; AB = 30 ins.; BC = 18 ins.; and CC = BD = 24 ins. Find the force at C and the push on the pile. c^ ; ^C Earth ' : .■'.■■ >■ iiiLil ^l^/r>-v^ riG. 115 Fig. 116 -•>k-3'->l Fig. 117 243-(ii). The links in Fig. 117 are pinned at A,B, and C. A and C are rigid supports. Assume P = 100 lbs. Find the reactions (on the links) at A and C. 244-(i2). A horizontal beam 20 ft. long is sup- ported by a knife-edge ft. from the left end and is pinned at ft. from the right end. It sustains loads of 5000 lbs. at the left end, 1000 lbs. at 10 ft. from the left end, 2000 lbs. at the right end, and a uniformly distributed load of 1000 lbs. per ft. between the left end and left support. Determine the reactions of the pin and knife-edge. 245-(i2). The diagrammatic sketch in Fig. 118 represents a lever system for a scales. Assume W = 250 lbs.; a = 12 ins.; 5 = 8 ins.; c = g = ins.; (i = /z = 10 Ins.; e = i\ ins.; / = 15 ins. Determine P. r<--a->t< b---;H { B e—c-fd ■ -*- I I i -*kc->k- .d— >i<— d--->k-c->i . ! I k--e>W — f ^g>l<—-~ h -»l Fig. 118 - ->k— 5'->S Fig. 119 246-(i3). The frame shown in Fig. 119 is pinned at B and rests against a smooth ^ . , . wall at ^. P = 1000 lbs. and Q = 250 lbs. Find the •^1 p reactions at A and B. 247-(i3). The beam in Fig. 120 weighs 50 lbs. per ft. and is 14 ft. long. S is a smooth support; .4 is a ^___^ ^ JF = loolbs. perft.; P = '<"4-** ,^^^ iKo . rn _ K-TT - K^ ft • AF = ■< h Determine Fig. 120 "<#^ pin joint. Q = lbs. 1000 lbs.; C£> = £F = 6 ft.; AF = i ft. the reactions at A and B. 248-(i3). A light bar 9 ft. long is fastened to the floor at its lower end by a hori- zontal pin perpendicular to the bar. The angle between the bar and the floor is 359 100 lbs. deg.; its upper end rests against a smooth vertical wall. Loads of 300 and 200 lbs. are hung at distances of 4 and 7 ft. respectively from the lower end. Calculate the reactions at the ends of the bar. 24g-(i3). The uniform bar AB (Fig. 121) is pinned to the floor at A and sup- ports a load of 100 lbs. at 5. The diameter of the smooth cylinder supporting the bar is ft. Determine the forces on the bar, in the rope AC, and on the cyl- inder. 250-(i3). The truss shown in Fig. 122 is pinned at .1 and supported by a smooth roller at B. Q = 5000 lbs.; W = 20,000 lbs.; the rolling friction (along the track) = 500 lbs. Compute the reactions at A and B when the car is stationary in the middle of the truss. (Mass. Civil Service examination.) 25i-(i3). The travelling wall crane (Fig. 123) rolls on three rails A, B, and C. Fig. 123 The weight of the crane is i ton; the load P is 3 tons; the center of gravity of the crane is 7 ft. from the vertical through A. Neglect flange pressures, and determine the reactions on the rails. 252-(i3). In a wall there are two hori- zontal pegs, A and B; A is 3 ft. above the floor, B is 6 ft., and the horizontal dis- tance between them is ft. A straight bar 15 ft. long weighing 200 lbs. is placed with its lower end on the floor, touching the underside of A and the upper side of B. The bar is not sprung into place, and all surfaces are smooth. A weight of 100 lbs. is hung from the upper end of the bar. Determine the pressure on the floor, on A , and on B. 253-(i3). A frame for building and raising a concrete wall is sketched in Fig. 124. The truss B is supy^orted on a trunnion D and by the telescoping piston C of a pneumatic jack F. The dotted lines show the wall and jack in an ex- treme position. D is 10 ft. vertically and 12 ft. horizontally from E; it is 6 inches from the upper surface IH of the truss; C is 3 ft. from that upper surface; the Fig. 124 360 -b ^ f HCK a 0^ e ■H projections of C and D on IH are 13 ft. apart. The truss weighs 2 tons; its center of gravity is i ft. below IH and 8 ft. to the right of D. The load (wall shown) weighs tons; its center of gravity is ins. above IH and ft. to right of D. When the truss is inclined to the horizontal deg. as shown, how large are the pressures at C and D? (Solve graphically.) 2 54-(i3). Fig. 125 illustrates an 'A apparatus for testing the strength of jl3J-^ i ,^fe? *^^ small concrete beams. The force produced by pouring shot into the bucket L is multiplied by the lever system and imposed (in equal amounts) on the specimen B at PP, points equally distant from the center of the beam. ^ is a knife-edge penetrating the beam slightly and furnishes a support for the upper two levers; the Determine the ratio of P to L. 126 carries a pulley at B weighing 150 lbs. a "W 77777777777777m7777m77, 77777) Fig. 125 other bearings on the beam are rollers. 255-(i4). The vertical shaft in Fig and one at E weighing 80 lbs. The radius of pulley B is 20 ins., of pulley E 15 ins. The center of gravity i of S is 2 ins. from the axis of the shaft; f of £ is J in. Fi = 200 lbs.; Fi = 50 lbs.; and W = 200 lbs. Find the reactions on the bearings at A and when the shaft is in the position shown. 256-(i5). The truss represented in Fig. 127 is supported by pins at .-1 and B. Each load P = 2000 lbs. Determine the amount and kind of stress in each member. 257-(i5). The truss represented in Fig. 128 is supported by a pin at E and by a .'^- y P \ -/2.'--> <--/2'-4 Determine Each load Fig. 127 horizontal tie at .4. Each load P\ = 1000 lbs.; each loadP2 = 500 lbs the amount and kind of stress in each member. 258-(is). The truss represented in Fig. 129 is supported as shown Pi = 500 lbs.; each load Pi = 100 lbs. Determine the amount and kind of stress in AB, AG, and BC. 259-(i5). The structure represented in Fig. 130 is a steel head frame for hoisting ore from a mine. The frame is pinned at A and is anchored at B so that either an upward or a down- ward reaction can occur at that point. The load is 10 tons. Determine the amount and kind of stress in each member of the frame. 26o-(i5). The truss represented in Fig. 131 rests on a smooth support at A and Fig. 129 361 is held by a pin at E. Each load Pi = 500 lbs.; each load Pi = 1000 lbs.; each load Pz = 2000 lbs. Determine the amount and kind of stress in each member. 26i-(i5). The truss represented in Fig. 132 is held by a pin at joint B and by a Fig. 130 Hf 40'- ^ Fig. 131 horizontal tie at joint A. Each load = 1000 lbs. Solve for the stress in each member. 262-(i6). The truss represented in Fig. 133 is supported at each end. Each load Px = 1000 lbs., and P2 = 2000 lbs. Determine the amount and kind of stress in each member graphically. 263-(i6). The truss represented in Fig. 134 is held by a pin at the right end and ->i<-- --^-- —A<— -->! A.k--^ — Fig. 133 Fig. 134 rests upon a smooth support at the left end. Each load Pi = 1000 lbs.; Pi = 2000 lbs. Solve graphically for the stress in each member. 264-(i7). The crane represented in Fig. 135 is pinned at B and is held by the guy rope CA. The load W is 20 tons. PG = 3 ft. Determine the tension in the guy rope and all forces acting on each member of the crane. S' \<-/0'^.'yP--/z' >K-<9->i Fig. 13s T w Fig. 136 26s-(i7). The crane represented in Fig. 136 rests in a socket at A and bears against the smooth side of the hole in the floor at D. There are pins at B, C, and P. 362 B The load W is 4000 lbs., the counterweight E weighs 5000 lbs. Determine all the forces which act upon each member of the crane. 266-(i7). The crane represented in Fig. 137 rests in a socket at A and passes through an opening in the floor at B. The post AB passes freely through a smooth slot in the boom at C, so that any reaction existing there will be horizontal. The load PF is 3 tons and is 6 ft. out from C. Determine all the forces which act upon each member of the crane. 267-(i7). The crane represented in Fig. 138 rests in a socket at A and is horizontally restrained by the ceiling at B. The load PF is 4 tons. The weights of the members are: post AB = 0.7 ton, brace DE = 0.5 ton, boom CF = 0.6 ton. The center of gravity of the boom is 9 ft. to the right of C. For the other members the position of the center of gravity is midway between the ends. Determine all the forces which act upon each member. Fig. 137 WW77' Fig. 138 •'jmiiii. ->K--6-^ ^•W/WWrr. Fig. 139 268-(i7). The crane pinned at B and is sup- ported horizontally at A. The load IF is i ton. De- termine the amount and kind of stress in each of the members C£, DE, and BE, and determine all forces acting upon the post and boom. 26g-(i8). The crane represented in Fig. 139 is 140 shown in Fig. 140 rests in a socket at A and passes through a hole in the floor at B, the sides of the hole affording horizontal support. r ■- -f^^ ^ Fig. 141 The diameter of each pulley is 2 ft to pulley on the post at B. on each member. Fig. 142 The rope is vertical from F to D and is fastened The load W is 8 tons. Determine all forces acting 363 to; Fig. 143 27o-(i8). In Fig. 141 the members are pinned to the wall at A and at i5, and to each other at C. The diameter of the pulley is 2 ft.; the load W is 1000 lbs. Deter- mine the forces which act upon each member of the structure. 27i-(i8). The crane shown in Fig. 142 is pinned at A and is horizontally re- strained by the floor at B. The diameter of the pulley at Z) is 2 ft., of the drum on the floor, i ft. The load W is 2 tons. Determine all the forces which act upon each- member. 272-(i9). A, B, and C (Fig. 143) "respectively weigh 100, 200, and 300 lbs. The surfaces in contact are so rough that a force P of lbs. causes no slip. Represent in a separate sketch of each body all the forces acting on it when P has the stated value. 273-(i9). A body is on a horizontal surface and is subjected to a force which does not make it move; the body weighs 200 lbs.; the force is incUned deg. with the horizontal and equals 200 lbs. Determine the normal pressure and friction on the body when the force is a pull; when a push. 2 74-(i9). Suppose that the coefficient of friction for A and B (Fig. 143) is ; for B and C ; and for C and D . What is the least value of P which would cause a slip? 275-(i9). The coefficient of friction between A and B (Fig. 132, Art. 19) is and e =30°. Determine the least pull P and push P which would cause slip. 276-(i9). Compare the values of the pull P required to cause slip (Fig. 132, Art. 19) when = o, e = 30, and 9 = 50°. 277-(i9)- A straight bar rests in a vertical plane with one end on a rough hori- zontal floor and the other against a smooth vertical wall. The coefficient of fric- tion for floor and bar is . At what minimum angle between bar and floor would the bar rest? 278-(i9). Suppose that the bar of the preceding problem weighs 100 lbs., and is set at an angle of deg. Determine the necessary downward force applied at the upper end to cause slip of the bar. 279-(i9). The ladder AB (Fig. 144) is 40 ft. long and weighs 100 lbs. The co- efficient of friction at A (between the upper rung and the pole) is ^; at 5 j. Com- pute the force P required to overcome gravity and friction in the position shown. -ffod 'to Fig. Fig. 144 riG. 145 28o-(i9). Fig. 145 represents the cross section of a dam, a sluice gate G, and a log sluice or trough AB (shown in section at S). Water is shown passing over the gate and down the sluice permitting the passage of logs. The sluice is made ad- justable to the water level. The front wheels at A rest against vertical rails, and 364 the wheels at B on indined rails. The end A is raised or lowered by means of a vertical rod operated from above by a suitable winze. The log sluice weighs 10 tons, AB = 80 ft. Determine the pull at the rod required to overcome the weight of the sluice and the friction at A and B in the position shown. (The diameter of the wheels A and B is small compared to AB; so regard the sluice as slipping on the two ends (like the ladder in the preceding problem) and take the equivalent coeffi- cient of friction as tV-) 28i-(2o). A (Fig. 146) weighs 100 lbs. 30' when /3 = o and P = lbs., A does not move, (a) Compute the fric- tion on A and determine its direction. (b) Solve when /? = 40°. 282-(2o). A (Fig. 147) weighs 100 lbs. ; the coefficient of friction under A is J. What force P is required to pull .1 down? 283-(2o). The coefficients of friction between A and B and B and C (Fig. 148) are \. W weighs 10 tons. How great must P be to start the wedge B? 284-(2o). A (Fig. 149) weighs 10,000 lbs.; the coefficient of friction for all con- tacts is ^. What value of P is required for starting A up the plane? Figs. 146, 147 St BL Fig. 149 Fig. 150 28s-(2o). In Fig. 150 W = 7000 lbs.; the coefficient of friction for all contacts is \. Determine the values of horizontal forces applied to wedges B required for raising W . 286-(2o). The required forces mentioned in the preceding problem are supplied by means of the right-and-left screw, Fig. 1 50. The mean diameter of screw threads is ins., and the number of threads per inch is Determine the torque required for raising W . Fig. 151 287-(2i), Each parallel opiped {A, B, C, Fig. 151) is homogeneous; their weights are respectively 8, 42, and 16 lbs. Determine the distances of their center of gravity from the coordinate planes. 288-(2i). Fig. 152 represents a bent wire. Determine the coordinates of its center of gravity. 365 289-(2i). Fig. 153 represents a cube with two grooves cut along the medians of two sides as shown. The grooves are 2 ins. wide and i in. deep. Determine the coordinates of the center of gravity of the grooved cube. t<- 4">t i< 4">t Ok- 8"- -MX Fig. 154 _x 29a-(22). Determine the coordinates of the centroid of the shaded area in Fig. 154. (See Art. 24 for centroid of semicircular area.) 29i-(22). The angles Fig. 155 are 6X4X5. Locate the centroid of the entire shaded area. 292-(2 2). Determine the centroid of the shaded area in Fig. 156. The area of the hole is 8 sq. ins. and the coordinates of its centroid are x = 2> and y = ^ ins. 293-(23). Imagine an ellipsoid separated into halves by one of its planes of symmetry. Determine the position of the cen- troid of one of these halves. 294-(24). Prove the statements on page 100 about the cen- troid of a right circular cylinder. 29S~(25)- A certain wire weighs lbs/ft, and can sustain a pull of lbs. with safety. It is to be suspended between two points on the same level and 1000 ft. apart. Assume that the suspended wire will be parabolic, and compute the shortest piece of wire that may be used. 296-(26). Solve the preceding problem but on the assumption that the sus- pended wire takes the catenary form. 297-(27). A rope is to be suspended at its two ends from two points AB on the same level and 40 ft. apart. Heavy weights are to be hung from knots on the rope so that the rope will assume the form of half an octagon. What weights will hold the rope to the desired form? 298-(28). The following data were secured from the launching of the battleship Conneclicut {Engineering News for Dec. 22, 1904): Time /, 5 10 15 20 25 30 35 40 45 50 sec. Distances, 18 60 132 225 325 420 520 610 680 720 ft. Determine the velocity when t = sec; when / = sec. 299-(28). A point moves in a straight line in accordance with the law s = ^ — 40 t, where s is distance in feet from a given point in the path and t i? time in seconds. By calculus find the velocity when t = sec. What is the average velocity for the second preceding the instant named? for the second following the instant? 366 3oo-(28). Express a velocity of 60 mi/hr in feet per second. A certain electric car gets up such a velocity in sees. Express that average starting acceleration in foot and second unit; compare with "gravity." 30i-(28). In a certain run the velocity of an electric car changed in the following manner: Timet, o 10 15 20 25 30 35 40 50 114 125 sec. Velocity^), o 15 21 24.5 26.5 28.5 29.5 31 32.5 21 o mi/hr. The car coasted during the interval 50-114 and was braked during the interval 114- 125. (Sheldon and Hausman, Electric Traction and Transmission Engineering, p. 63.) Find the acceleration when t — sec; when / = sec. 302-(28). In a certain rectilinear motion v = /- — 10 /, where v is velocity in feet per minute and / is time in minutes. Determine by calculus the acceleration when t = min. What is the average acceleration for the minute preceding the instant named? for the minute following the instant? 303- (28). A point moves in a straight line in accordance with the law 5 = 2 sin (0.05 / + ). where 5 is in inches, / in seconds, and the angle in radians. Deter- mine the velocity and acceleration when t = ; when / = . Interpret the signs of your results. 304-(28), A body has an "initial velocity" (when / = o) of 15 ft /sec and an acceleration expressed by a = 90 / — 24 /-, the units being foot and second, (a) How far does the body move in the interval / = to / = ? (b) What are the velocities at the instants named? (c) What is the average velocity for the interval? 305-(28). In a certain series of tests on emergency stops with an automobile the following data were obtained: Speed of auto, 6 10 15 20 25 30 mi/hr. Stopping distance, 1.67 6.00 9.50 36.8 42.0 47.5 ft. {Engineering News for Sept. 7 and Oct. 10, 191 2.) Compare the average retardations. 3o6-(29). Draw the distance-time curve for the data in Prob. 298 and deter- mine the velocity when / = sec; when /= sec. What can you state con- cerning the acceleration at / = 30 sec? 307-(29). Draw the velocity-time graph for the data in Prob. 301. Determine the acceleration when / = sec. What is distance covered from / = to / = sec? 3o8-(29). A train can get up a speed of 60 mi/hr in 5 min., and stop in 0.5 mi. About midway between two stations 10 mi. apart a bad piece of track one mile long necessitates reduction of speed to 10 mi/hr. Assuming that acceleration and re- tardation can be applied uniformly with respect to time, determine the time between stations. (Sketch the velocity-time graph before calculating.) 309-(29). In the preceding problem how much time was lost on account of de- fective track? 3io-(29). In testing automatic safety cushions placed at the bottom of elevator shafts in the Wool worth Building, N. Y., the following data were obtained: Distance from top of air cushion, o 20 40 60 80 100 120 130 135 137 ft. Pressure on bottom of car, 4 4 7, 10 12 9 9 9 8 o Ib/in^, Downward velocity of elevator, 168 168 157 140 116 92 67 52 32 oft/sec. 367 Plot the distance-velocity curve, and find the acceleration when the elevator had fallen ft. in the cushion. {Engineering Record for Sept. 5, 1914-) 31 1-(29). From the data in the preceding problem find the time consumed in fall- ing ft. in the air cushion. (Hint. Plot the reciprocals of the velocities against the corresponding distances.) 3ii-(3i). The following quotation from Engineering News, Dec. 3, 1914, relates to tall building elevators: "I and some of my associates have been subjected to retarda- tions as high as 72 ft /sec/sec without much discomfort. The stress in the ankles is quite noticeable. In air cushion practice it is customary to allow for retardations of five or even six times that of gravity; i.e., retardations up to nearly 200 ft /sec /sec, and it is considered that even this high retardation will not be injurious to hfe. There are several instances on record where it has been sustained without serious injury." Calculate in the case of a man standing upright in an elevator undergoing a retarda- tion of six times gravity, the pressure on the soles of his feet, and the stress at his neck. (Take the weight of his head as 7 per cent of his total weight.) 3i2-(3o). The period of a certain simple harmonic motion is ' sec, and its amplitude is ins. Compute the maximum velocity, the velocity at a quarter point, and the average velocity for one-half the length of path. Ditto for acceleration. 3i2-(3i). Take the weight of the elevator in Prob. 310 as 7500 lbs., and the area of its bottom as 30 ft'. Determine the amount and direction of the resultant of all forces acting on the elevator when it was ft. below the top of the air cushion. What forces make up the resultant? Are the records consistent? 3i3-(3i). Imagine the surface C (Fig. 58) tilted so that its inclination to the horizontal is 20 deg.; that the coefficient of static friction between A and B is 0.5; that the coefficient of kinetic friction between B and the incline is 0.2. A and B weigh 100 and lbs respectively, and P = lbs. Find the acceleration of the system, and represent the forces acting on each body in separate sketches. 3i4-(3i). In the preceding problem determine the force P which would make A slip on B. 3i5-(3i). Bodies .4 and B, Fig. 157, weigh 40 and 60 lbs. respectively. The co- efficient of friction under A is 0.2; that under 5 is 0.25. When P = lbs., the forces on the spring connecting the bodies have what value? ^VWV ww\ 'vtttttttttttttttttttttttttttttpttttmtp Fig. 158 Fig. 159 3i6-(3i). The box shown in Fig. 158 weighs 120 lbs. and the body A 80 lbs.; the system is moving to the right on a floor from an initial impulse. The coefficient of friction between A and the box is zero and that between the box and floor is Find the forces on the (like) springs. 3i7-(3i). A body weighing 24 lbs. is projected up a 30 deg. incline at a velocity of 20 ft/sec. The coefficient of friction between body and incUne is . Find the position of the body after 5 sec; after 20 sec 3i8-(3i). A (Fig. 159) weighs 160 lbs. and B lbs. The coefficient of kinetic friction between A and the incUne is , and between B and the incline 0.25. De- 368 (Consider the rope very flexible and neglect termine the acceleration of the system, tho mass of rope and pulley,) 3i9-(3i)- A, B, and C (Fig. i6o) weigh 60, 30, and 10 lbs. respectively, and the coefl&cient of friction between A and D is . Neglect the stiff- ness of the rope, its mass, and that of the sheave. Find the ten- sion in the rope between B and C. 32o-(3i). The apparatus in Fig. 218 (page 119) is being used to compress air. The piston is 10 ins. in diameter and weighs 40 lbs.; the lengths of crank and connecting rod are 3 and 10 ins. respectively; the crank rotates at 100 r.p.m.; the pressure on the top of piston is 50 lbs/in-. Find the stress in the top of the piston rod when 6 = deg. 3 2 1-(3 2) . A particle is moving at a constant speed of 2 ins/sec along an ellipse whose axes are and ins. long. Determine the amount and direction of the acceler- ation of the particle at the instant it is passing one end of the er axis. 322-(33). P (Fig. 266) moves in the circle (diameter = ft.) with constant speed of one revolution in sec. When the angle POX is deg., what are Vx, i'„, Cr, and «„? FlG. 160 "x, ^-y, "J, U- 323-(33). A projectile is discharged from a gun with a velocity of 644 ft/sec, at an elevation of 30 deg. Neglecting the resistance of the air, compute (o) the extreme range of the projectile measured on a horizontal plane through the muzzle of the gun, and Q}) the maximum elevation of the projectile above this plane. (U. S. Civil Service examination.) 324-(34). A smooth sphere weighing 2 lbs. (Fig. 161) is hi a box rigidly fastened to an arm which can be rotated about a horizontal axis. By means of suitable forces applied to the arm the system is made to get up speed uniformly so that in every second the speed is increased by 2 rev /sec. At the instant when the speed reaches rev /sec, the arm is inclined as shown. What are the values of all the forces acting on the sphere then? Represent them in a separate sketch of the sphere. 32 5-(34). CD (Fig. 162) is a rough incUned plane rotating about the vertical axis AB at a constant speed of 200 rev/min. The body E weighs icx) lbs. and its center of gravity is 3 ft. from AB. Determine the friction and normal pressure on E. 326-(34). AB (Fig. 163) is a board lying upon a table. C is a vertical peg in the table top projecting upward through a suitable hole in the board. The board weighs 20 lbs. The table top (and board) are spun about C at 400 rev/min. Determine the stress at the smallest section of the board. /or- -1 A zk ^ <:- — \ ---.>U- A Fig. 162 Fig. 163 A table with a circular top has a smooth border along the edge and 327-(34) is mounted so that it can be rotated about the vertical through its center. Suppose p 369 that a straight rod lies on the top. When the table is rotated the rod rolls to and its ends bear against the border. Take the diameter of the top — 9 ft., length of rod = 6 ft., weight of rod = 10 lbs., and speed of rotation = 200 rev/min. Compute the pressures at the ends of the rod. 328-(34). Fig. 164 represents a block A upon which is mounted a drum B. A light rope is wound around the drum so that a horizontal pull P can be applied as shown. The weight of A is 960, and that of B is 640 lbs. The coefficient of friction under A is .. When P = lbs., the acceleration of A has ' '/h;m;w//wwh///w- what value? I'^i^- 164 329-(34). A Ipcomotive is running along a level track. The horizontal component of the reactions of the rails on the drivers are forward or back according to certain circumstances; on the other wheels it is opposite to the direction of motion. Call the first force P; the second Q. What is the relation between P and Q (i) when the loco- motive speed is constant? (//) when it is being increased? (iii) when it is being lessened? What does a locomotive do to a bridge over which it is moving in the three cases? 33o-(34)- What is the nature of the action of a travelling crane (Fig. 165) on its i:^^;^A track (a) when it is starting to "travel" (run down the yard)? (b) When it is starting to "traverse" (crab A runs over the bridge B)? (c) When hoisting of the load begins? 33i-(34). Take weights of parts of crane (preceding problem) as follows: bridge A, ^^7777T7777777777m777777777777m7777777/77!7^^77777777h- tOUS; Crab B, tOnS ; load C, ^^°* ^^5 tons. Take acceleration of travel 0.7 ft/sec/ sec, and acceleration of traverse 0.4 ft/sec/sec; then calculate the wheel reactions on track so far as possible for commencements of travel and traverse. (Neglect swing of load. What is effect of this error?) 332-(34). The following is an extract from a description of the Sheepshead Bay Motor Racetrack (Engineering News for Aug. 19, 1915): "There are two parallel stra?ghtaway stretches connected by two turns of 180 deg. each. Each turn con- sists of a circular arc of about 135 deg. connected by 'spirals' to the straightaway stretches; the radius to the inner edge of the circular track is 850 ft. The outer edges of the circular turns are given a maximum super elevation of 25 ft. 6 in., computed for a speed of 96 mi/hr by the common railway formula C = dv^/gR. The width d was taken in 14 ft. strips commencing at the inside of the track, and the super ele- vation computed for speeds of 40, 52I, 65, yyf , and 96 mi/hr. This gives a cross section theoretically of a parabolic curve." Prove the last statement, and show how the formula gives 25 ft. 6 in. 333-(35). Imagine the surface C (Fig. 56) tilted so that the top of B is horizontal; assume the angle of tilt to be deg., the surface C to be frictionless; that A and B respectively weigh and lbs. When the body B is released (P = o), ^ and B slide down the plane together. Then the pressure between A and B has what value and direction? 334~(3S)- Fig- 166 represents a bar which rests upon a horizontal surface in the plane of the paper. The bar weighs 160 lbs. and the coefficient of friction between the bar and table is 0.5. The two forces shown acting on the bar are horizontal. De- 370 ^P= 100 lbs. =500 Ibs^ I Fig. i66 B termine the magnitude, direction and point of application of a third horizontal force, which, when acting with the two forces shown, will give the bar a motion of trans- lation with an acceleration to the right along AB of 6o ft/sec/sec. 33S~(35)- B (Fig. 167) is a straight post which rests upon the front edge of the car A. With what acceleration must .1 be made to move along a level track in order that B will remain in the position shown? 336-(35). Assume that the car of the preceding problem moves up a 30 deg. incline with a uniform acceleration of ft/sec/sec. At what angle with the vertical must the post be inclined in order that, as before, it may maintain its position? 337-(35)- A homogeneous cylinder weighing 100 lbs. is drawn up an inclined plane by a force of 200 lbs. acting parallel to the plane. The cylinder is 4 ft. long and 2 ft. in diameter; it rests on end with its length normal to the plane. The coefficient of friction between cyHnder and plane is 0.2; the angle of inclination of the plane is deg. to the horizontal. Determine the limits between which the point of application of the 200 lb. force must lie in order that the cylinder may not tip over. 338-(36). A solid piece of cast iron consists of a right circular cylinder 4 ft. in diameter and 10 ft. long, and a right circular cone 4 ft. in diameter and ft. long, placed end to end. Determine the moment of inertia and radius of gyration of the body with respect to the common axis of cone and cylinder. 339^(36). The moment of inertia of a sphere with respect to a diameter is given by i Mr-. What is the moment of inertia of a cast-iron sphere, ins. in diameter, with respect to a tangent line? 340-(36). Determine the moment of inertia and radius of gyration of the cast-iron pulley represented in Fig. 168 with respect to its own axis. 34i-(37). "The 16-inch gun has a range of about 42.4 miles, and it travels this dis- tance in about^^5 seconds. If the gun were pointed exactly ndrth or south the lateral de- viation of the projectile, due to the earth's rotation, would amount to 525 feet." (5a- cntific American for May 22, 191 5.) Show how to calculate the stated deviation. 342-(37). The disk of a certain steam tur- FiG. 168 bine is ft. in diameter, and it is run at rev/min. What is the rim speed in miles per hour? Compare the acceleration of a point on the rim with gravity. A small bolt is screwed radially into the disk at the rim; compare the tension in the shank of the bolt just under the head with the weight of the head . 343-(37). -4 and B (Fig. 159) weigh 160, and lbs. respectively. The weight of the pulley is lbs., its diameter is 6 ft., and its radius of gyration 2 ft. The coefficient of friction under .-1 is ; under B it is \. Determine the acceleration of A and B and the tension in both parts of the rope. 344-(37). Fig. 169 represents a drum with a brake attachment. A rope is wound 371 around the drum and a load of 2000 lbs. hung as shown. The force P on the brake arm is 1000 lbs. The coefficient of friction between brake and drum is ; the diameter of the drum is 6 ft., its weight is 1800 lbs., and its _ ,^, >k>^^'>|| radius of gyration is 2.5 ft. The load is released and i__ i ^ =J 1 l-rr/^) mc. w 169 allowed to fall until the drum has attained a speed of 100 rev/min. Then the brake is applied and the system is brought to rest. How much time is required for the braking? 345-(37)- The drum shown in Fig. 170 rests upon a floor and against a low vertical wall as shown. A rope is wound around the axle and passes off horizontally over the wall. The diameter of the drum is 4 ft.; the diameter of the axle is 3 ft.; the radius of gyration of the drum and axle is 20 ins., and their combined weight is 180 lbs. The coefficient of friction between drum and floor is and between drum and wall 0.2. The pull P = 200 lbs. Deter- mine the angular acceleration of the drum, and all forces which act upon it. 346-(37). Fig. 171 represents a brake device for regulating the speed at which a load is lowered. It consists of a ring solidly fixed in a horizontal plane, through the center of which passes a vertical shaft. This shaft carries a cross- arm upon which slide two blocks -4.-1. The rope which carries the load passes over a pulley and is wound around the shaft as shown. When the load starts to descend, the shaft rotates and the blocks AA, shding out to the ends of the cross-arm, bear against the inner side of the ring. The weight of each of the blocks is 64 lbs., the distance from the axis of the shaft to the center of gravity of each block when bearing against the ring is 20 ins., the diameter of the shaft is 6 ins., the coefficient of friction between block and ring is , and the load W is 1600 lbs. Neglecting the weight of the vertical shaft and the pulley, find the maximum velocity which the load will attain in descending. 347-(38)- Referring to Prob. 343: enumerate the forces acting on the pulley. Tell how the reaction of the pulley shaft is related to the other forces acting on the pulley. 348-(38). Referring to Prob. 344: determine the value of the axle reaction on the drum during the braking period. 349-(38). The frame (Fig. 162) can be rotated about the vertical shaft AB. The shaft is 12 ft. long; AD ^ % ft., CD = 10 ft., and BC = 2 ft. The weights of these members are respectively 500, 200, and 400 lbs. 72 is 2 X 4 ft., and perpendicular to paper, i ft.; it weighs 300 lbs., and is placed at mid-length of CD. The entire system is rotated at 1800 rev/min. Determine all forces on each member. 35o-(3o)- The length of a simple seconds pendulum at a certain place is 3.56 ft. Find the length of a pendulum which at the same place swings from one side to the other in 5 sees. (U. S. Civil Service examination.) 3Si-(4o). The body C (Fig. 76) is moved up the plane by a horizontal force P (= lbs.) and Q (= lbs.). The frictional resistance is 10 lbs.; the incUnation of the plane is deg., and the body weighs 40 lbs. Compute the work done on C by each force during a displacement from A to B, 20 ft. What is the algebraic sum of these works ("total work")? 372 352-(4o). C (Fig. 172) is a bead on a circular wire ABD; it is subjected to four forces F, P, Q, and 5; F = 10 lbs. and is always horizontal; P = 40 lbs. and is always directed toward D; Q varies in magnitude and direction so that the simultaneous values of a, /3, and Q are as follows: 60° 15 30 45 75 90 /3 = Q = 5 is a tangential pull and its value (in pounds) is 40 s~, where 5 is the arc AC in feet. Compute the amount of work done by each force for the displacement of C from A to B. 3S3-(4o). ABC is a. right triangle; AC is the hypothenuse, AB = 8 and BC = 6 ft. A certain small body is made to move along AB by several forces, one of which is always directed toward C and equals 10 lbs. How much work does this force do while the body is made to move from A to B? 354-(4i). When the pulley described in Prob. 340 is rotating at 200 rev/min., what is its kinetic energy (in foot-pounds)? 35S-(4i). When a solid (circular) cylinder is rolling on a straight roadway, what portion of its total kinetic energy is "translational"? 356-(42). Two tests were run on a certain steam engine. In the first test the fly-wheel spokes were exposed to the air; in the second test they were enclosed so as to reduce the air resistance as much as possible. The first test gave an indicated power of 12.30 Continental horse-power; the second of 7.88 Continental horse-power. Assuming that energy is worth one cent per kilowatt hour, find the gross (money) saving due to enclosing the fly-wheel. 357-(42). What power is required to move a block weighing 1200 lbs. up a 30 deg. incline at a uniform speed of 100 ft /sec if the coefficient of friction between block and plane is 0.2? 358-(42). S and S' (Fig. 173) are two portions of a shaft. Arms A and A' are rigidly attached to the adjacent ends of the shaft as shown. The ends of the arms are furnished with hooks which are connected by two like coil springs as shown. Thus it is pos- sible to transmit energy from one portion of the shaft to the other; indeed the device illus- trates, in principle, a "transmission dynamom- eter." Let length of each arm = ins., natural length of each spring = 8 ins., stiffness of each spring = 40 lbs/in. (40 lbs. pull required for each inch of stretch). When the shaft is rotating at 200 rev/min, the angle between the arms is power of transmission? 359-(42). Put your solution of the preceding problem into general terms, using the following notation: a = length of each arm in inches, b = natural length of each spring in inches, p = stiffness of spring in pounds per inch, n = speed in revolu- tions per minute, d = angle between arms in degrees. 36o-(43). A flywheel of a 4 h-p riveting machine fluctuates between 60 and 90 r.p.m. Every two seconds an operation occurs which requires | of all the energy sup- deg. Fig. 173 What is the horse- 373 plied for two seconds. Find the moment of inertia of the wheel. (U. S. Civil Service examination.) 36i-(43). A particle, under the action of gravity alone, moves from rest from the highest point on the outer surface of a smooth sphere whose diameter is lo ft. Neglecting friction and the small force necessary to start the particle, find: (a) at what point the particle leaves the sphere, {b) what the velocity is at the instant of leaving. (U. S. Civil Service examination.) 362-(43). A lo-inch rifle has a barrel 45 ft. long and shoots a shell weighing 800 lbs.; the cross section area of the bore is 80 in-. The powder pressure varies from 50,000 lbs/in- at the instant of detonation to 5000 lbs/in- when the projectile leaves the muzzle. Assume the variation of pressure to be uniform with respect to the dis- placement of the projectile, and neglect the effect of friction and recoil; then deter- mine the maximum power developed during a discharge, and the muzzle velocity of the projectile. 363-(43). A car coasts down a 2 per cent grade, starting from a point 1000 ft. from the bottom. As soon as the level track is reached the brakes are set, locking the wheels. The total weight of the car is 3200 lbs. There are two pairs of wheels, each pair, with the axle, weighing 320 lbs., and having a radius of gyration of i ft.; the diameter of the wheels is 3 ft. The total rolling resistance is 10 lbs. and the coeffi- cient of friction between wheels and track is 0.2. Determine how far along the level track the car will go after the brakes are set. 364-(43). A flying airplane is subjected to two external forces — gravity and air pressure. It is convenient for purposes of analysis to regard the latter in the three parts which act on the wings, the propeller, and the remainder of the machine. In "simple flight " (horizontal or inclined straight path, in still air or directly with or against the wind), the air reaction on the wings is dealt with in two components — perpendicular and parallel to the line of flight; the first is called "lift" L, and the second "drag" (formerly "drift") D. The reaction of the air on the propeller can be regarded as a single forward force or "thrust" T, and a couple C opposing the rotation of the propeller. The third force or "body resistance" R is directed nearly along the Une of flight. (See Fig. 174.) Fig. 174 The lift and drag depend upon velocity v of the airplane relative to the air, wing area A, density of air 8, "angle of incidence" a (angle between chord of wing and direction of wind, type of wing section, and "aspect ratio" (ratio of length to breadth of wing); and in biplanes, on the "gap" (distance between the planes) and to a lesser extent on other details. Thus L = KdAv'^ and D = kbAv^, 374 where K and k are experimental (lift and drag) coefficients. They are "absolute" or abstract coefficients, permitting the use of any systematic units in the formulas; for example, pound for force, square foot for area, foot per second for velocity, slug per cubic foot for density. (The formulas are also written without the density factor, the coefficients being made to suit certain convenient units like the pound, square foot, and mile per hour.) The body resistance varies approximately as the velocity square, size and shape of body, and other structural details. In any case, the formula can be written R = Cv^, where C is a coefficient which can be estimated by the expert designer. In horizontal flight the weight of the airplane is balanced by the lift and the vertical component of the thrust. But this component is neglected in ordinary calculations. Thus we have KhAv" = W or v'^=W^ K5A. In uniform horizontal flight the total resistance D + R and the thrust T are equal, and the work done against the total resistance per unit time and the thrust power are equal. If the airplane is moving in still air, this thrust or useful power is P = Tv = {D + R)v. The following data refer to a particular monoplane: weight = looo lbs.; area of wings = 155 ft^; C = 0.02; and for angle of incidence a = —2 o 2 4 6 8 10 12 14 16 18 deg. K = 0.000 0.070 0.150 0.235 0.305 0.380 0.435 0-475 0-495 o-5oo 0.480 k = 0.017 0.014 0.014 0.017 0.024 0.033 0.043 0.052 0.063 0.078 O.IOO (i) Plot values of K, k, and K/k on an « base, using scales of i in. = 4 deg., and i in. = o.i. (ii) Compute values of v for the stated values of a. (iii) Compute values of D and R for the computed values of v. Plot values of D, R, and D -\- R on the same v base, using scales i in. = 20 ft/sec, and i in. = 100 lbs. (iv) Compute values of P for the computed values of v. Reduce these values of P to horse-power, and then plot horse-power required on the velocity base used in (iii). (The horse-power available plotted on a velocity base gives a curve convex upward, intersecting the horse-power required curve in two points. The velocities corresponding to these two points are the minimum and maximum velocities of hori- zontal ffight for the airplane.) 365-(43). When the motor of a flying airplane is cut out, the plane soon takes on a gUde at an angle and speed depending mainly on the set of the elevator. Then the propeller is driven by the thrust which is now opposed to the motion of the airplane; but this thrust is small compared to the drag. Show that where d = the angle of the glide. Compute values of 9 and v for the values of a given in the preceding problem, and then plot cot 6 on the v base. 366-(43). A locomotive weighing 200 tons pulls a train of 30 cars, each weighing 50 tons. At 10 mi/hr the locomotive resistance is 10 lbs/ton, and the train resistance is 4 lbs/ton. What net horse-power is necessary to maintain that speed on a i per cent grade? 367-(43). Find the total work done at the drawbar of a locomotive in starting a 200-ton train against a one per cent grade from rest to 30 mi/hr in 300 ft. when the 375 frictional resistances are lo lbs/ton (assumed the same for all speeds). (Mass. Civil Service examination.) 368-(43). The engine of a certain locomotive weighs ii8 tons, io6 on the drivers; the loaded tender weighs 60 tons. The cylinders are 24 (diameter) by 30 ins. (length), and the drivers are 60 ins. in diameter; the boiler pressure is 200 lbs/in-. What is the highest velocity at which this locomotive can haul a train weighing looo tons on a level track? (Assume that the train resistance follows the Engineering News formula, and that the mean effective follows the law p = 200(0.95 — (7 5 -i- 11,000)]; see Prob. 149.) 369-(44). Fig. 175 shows plan of a capstan and hauling tackle, and an elevation -^-A IS'- c-S" Arm i Fig. 17s of the barrel of the capstan. The sweep (arm) to which the horse is hitched at ^ is 11 ft. long; there is one sheave in each block B and C. Assume that the horse can exert a prolonged, steady pull of lbs., and make an estimate (supported by calculation) of the pull which can be exerted at the load. (Neglect friction on the barrel.) 37o-(44). Fig. 176 represents in plan and elevation a coal shipping station for loading barges at a Mississippi River " landing." The coal is brought in railway cars up on to the trestle to the supply track, and then dumped into a hopper from which it is discharged as required into a transfer car. This car runs on a track, extending toward the river and up the cradle to a dumping platform, whence it discharges its load into the barge below. The loading plant is operated by means of three cables, two to handle the transfer car, and one to handle the cradle. The winding drums for the cables are mounted in the "operating house," and are driven by gearing from an electric 35 horse-power motor. The hauling cable A extends from the drum A vertically downwards to a sheave, thence horizontally to another sheave in the middle of the track, thence along the center of the track down the incline and through the cradle, passing under a 10- inch pulley (not clearly indicated) and a 30-inch pulley, and then up to a 36-inch pulley at the track level; the end of the cable is attached to the transfer car. The back-haul cable B extends vertically down from drum 5 to a sheave, thence horizon- tally, and by turns over two sheaves, to a point in the upper end of the inclined track; thence down the center of the track and over the deck of the cradle to the car. The cradle-cable C leads from its drum downwards to a sheave, thence along one side of the incline to the upper end of the cradle; it crosses under by means of two sheaves and then extends along the opposite side of the incline to a point of attachment on the trestle which carries the overhead track. Normally, the cradle will be held in position by a ratchet brake on the first pair wheels. Diameter of all sheaves not given above or indicated in the figure is 30 ins., diameter of the cables is H in. The cradle weighs about 22 tons, the transfer car empty about 8 tons, and loaded about 20 tons. 376 o t--. o 377 L -^ B ±3 \ — ^ C B 3 Fig. 177 Compute: (a) the torque required at the drum A for pulling the loaded transfer car up the cradle from a "standing start"; (6) the torque required at the drum B to pull the empty car up the 5 per cent grade from a standing start; (c) the torque re- quired at the drum C for pulling the cradle up its track, transfer car not on the cradle. 37i-(45). A (Fig. 177) is a long board supported in a horizontal position; 5 is a Week provided with end guides so that it can be sHd along the board without turning; lid C is a heavy piece that can be slid across the board. Suppose that B — and with it C — is moved to the left with a velocity of 4 ft /sec, and C is moved across the board, in the direction indicated, with a velocity of 0.2 ft /sec. (0) What is the direction of the friction exerted by .1 on C? {b) What is the amount of that force if the weight of C — all supported hy A — is 10 lbs., and the coefficient of ki- netic friction between A and B is 0.4? (c) If the coefficient between B and C is 0.4, what is the amount of friction between them? (j:' i m n-*-. '•p::iaiA '>^( ti^-.