*!/*k :.riCtJg^9P iM. . __ j m K L^is M ,^^, %i^. i, *p '^s^.:- 1 / ^P ^^^a^F ^ L-Va L-Va rtfa If \ THEORETICAL MECHANICS STANFORD UNIVERSITY BOOKSTORE Agent of the Publisher Theoretical Mechanics %n (&\tmtntM\j Stxt-tarok BY L. M. HOSKINS Professor of Applied Mathematics in the Leland Stanford Junior University SECOND EDITION ' STANFORD UNIVERSITY, CAL. PUBLISHED BY THE AUTHOR 1903 HALLIDIE ^-^ Copyright, 1900, 1903, By L. M. Hoskins Stanford University Press PREFACE. In a first course in Theoretical Mechanics the primary object to be gained by the student is a thorough grasp of fundamental princi- ples. In most cases it is impossible to go beyond this object in the time available for the course. In the preparation of this text-book, the aim has been to present the fundamental principles in as clear and simple a manner as possible, and to enforce them by a sufficient number of illustrative examples. The examples are for the most part simple applications of the theory presented in the text, many numerical exercises being included. The solution of exercises in- volving numerical data forms a part of the work of the student of which the importance should be emphasized. In the desire to cover as much ground as possible, such work is too apt to be neglected. The mathematical training required for using the book is that usually implied by an elementary knowledge of Differential and Inte- gral Calculus. The most essential portions of Part I (Statics) may, however, be read without such knowledge. As regards arrangement, the aim has been to observe the order of difficulty of the different topics treated, so far as this was possible without doing violence to the logical relations. This is one reason for placing Statics first, instead of treating it in its strictly logical position as a special case of Kinetics. Part II may, however, be read before Part I, if any teacher should prefer this order. As to scope, the primary object has been to meet the needs of students of engineering in American universities and technical col- leges. More has, however, been included than it will usually be possible or desirable to cover in an elementary course. The precise limits of the course must be determined by the judgment of the teacher, taking due account of the relative importance of the various parts of the subject and of the limitations of time. It is hoped that the arrangement of the book is such as to facilitate whatever abridg- ment the teacher may find desirable. In case it is necessary to cut the course down to the narrowest possible limits, the following suggestions as to the most essential parts to retain may be of service. 117270 VI PREFACE. Chapters I-V. These include, besides the introductory general matter, the fundamental principles governing the composition and resolution of forces and the conditions of equilibrium, as applied to a rigid body acted upon by coplanar forces. Chapters XII-XVII ; omitting some of the applications in Chap- ters XIII and XV, and 2 of Chapter XVI. These chapters cover the most essential parts of the theory of the motion of a particle, including a general discussion of the laws of motion and an introduc- tion to the theory of energy. Chapters VI and VII are of practical value to the student of en- gineering, and the course should include at least 1 and 2 of Chapter IX. Chapters XVIII-XX constitute an introduction to the Dynamics of Material Systems, with applications to the rotation of a rigid body. If possible, at least this portion of Part III should be included in the course. In Chapter XIX, 2 (methods of computing moments of inertia of plane areas) is of practical value for its application in Mechanics of Materials, and may be taken independently of any other portion of Part III. If this book is in any degree successful in meeting the needs of students of engineering, it is hoped that it may be of service also to those pursuing the subject for its intrinsic scientific interest or as a preparation for the study of mathematical physics. The opinion is sometimes expressed that the needs of these different classes of stu- dents require essentially different methods of treating the subject. This view, so far as it refers to the fundamental parts of an elementary course, is not shared by the author of this text-book. For all stu- dents, the matter of first importance is the clear understanding of fundamental general principles and the ability to apply them. Stanford University, Cal., October, 1900. PREFACE TO SECOND EDITION. In this edition Chapters I-XXIII are reprinted from the first edition without important change, except the correction of such typographical and other errors as have been detected. Chapter XXIV is new. Stanford University, Cal., August, 1903. CONTENTS. CHAPTER I. Introductory. Preliminary Notions ; Numerical Representation of Quanti- ties ; Vector Quantities. PART I. STATICS. II. Force and Stress. . . . . . .16 Conception of Force ; Classes of Forces ; Numerical Repre- sentation of Forces and of Masses ; Definitions. III. Concurrent Forces. . . . . . .27 Composition and Resolution; Moments; Equilibrium. IV. Composition and Resolution oe Non-concurrent Forces in the Same Plane. . . . . .49 Two Forces; Couples; Any Number of Forces. V. Equilibrium of Coplanar Forces. . . . -65 General Principles ; Applications. VI. Equilibrium of Parts of Bodies and of Systems of Bodies. 92 Equilibrium of Any Part of a Body ; Determination of Inter- nal Forces ; Equilibrium of a System of Bodies ; Stress. VII. Friction. . . . . . . . 106 VIII. Equilibrium of Flexible Cords. . . . .114 IX. Centroids. . . . . . . . .124 Centroid of Parallel Forces; Centroids of Masses, Volumes, Areas and Lines ; Determination of Centroids by Integration. X. Forces in Three Dimensions. . . . .141 Concurrent Forces ; Couples ; Non-concurrent Forces ; Equi- librium. XI. Gravitation. . . . . . . . 155 Attraction Between Particles ; Attractions of Spheres and of Spherical Shells. CONTENTS. PART II. MOTION OF A PARTICLE. CHAPTER PAGE XII. Motion in a Straight Line: Fundamental Principles. . 167 Position, Displacement and Velocity; Velocity-Increment and Acceleration ; Motion and Force. XIII. Motion in a Straight Line: Applications. . . 186 General Method ; Constant Force ; Force Varying with Dis- tance from Fixed Point ; Miscellaneous Problems. XIV. Motion in a Curved Path. . . . . 208 Position, Displacement and Velocity ; Velocity-Increment and Acceleration; Motion and Force ; Simultaneous Motions. XV. Plane Motion of a Particle. . . . .236 Methods of Specifying Motion in a Plane ; Motion Under Any Forces; Resultant Force Constant or Zero; Central Force; Constrained Motion. XVI. Momentum and Impulse. . . . . .277 Rectilinear Motion; Motion in Any Path. XVII. Work and Energy. . . . . . .298 Work in Case of Rectilinear Motion ; Work in Any Motion ; Energy of a Particle ; Energy of a System ; Virtual Work. PART III. MOTION OF SYSTEMS OF PARTICLES AND OF RIGID BODIES. XVIII. Motion of Any System of Particles. . . .321 Motions of Individual Particles and of Center of Mass; An- gular Motion ; Effective Forces ; D'Alembert's Principle. XIX. Moment of Inertia. ...... 331 Rigid Body ; Plane Area. XX. Motion of a Rigid Body: Translation, Rotation About a Fixed Axis. ...... 346 XXI. Any Plane Motion of a Rigid Body. . . . 360 Nature of Plane Motion; Composition and Resolution of Plane Motions ; Dynamics of Plane Motion ; Statics. CONTENTS. XI CHAPTER PAGE XXII. Principle of Impulse and Momentum. . . . 382 Any System of Particles ; Rigid Body Having Motion of Translation or of Rotation ; Resultant Momentum ; Any Plane Motion of Rigid Body. XXIII. Theory of Energy. ...... 405 External and Internal Work ; Energy of Any System of Par- ticles ; Conservation of Energy ; Rigid System ; Principle of Virtual Work. XXIV. Relative Motion. ...... 432 Index. ......... 453 THEORETICAL MECHANICS OF TH-.. CHAPTER I. INTRODUCTORY. i. Preliminary Notions, i. Mechanics Defined. Mechanics is the science which treats of the motions of material bodies. Motions take place in accordance with definite laws. This state- ment means that the motions of bodies depend in an invariable way upon certain definite conditions. The fundamental object of the science of Mechanics is to investigate these conditions and to formulate the laws in accordance with which they determine the motion. In its present form the science of Mechanics rests upon a few fundamental laws of great generality. These laws are generaliza- tions from experience, and in the presentation of the principles of the science they are taken as postulates, propositions not deduc- ible from anything more fundamental. So far as these postulates and the principles deduced from them give a true account of the motions of natural bodies, they constitute a natural science. These fundamental laws, however, involve certain conceptions of matter and of motion which are ideal. This is necessarily the case. The motion of a body can be completely specified only by describ- ing the motion of every ultimate portion of which it is composed ; yet the ultimate structure of matter is wholly unknown. The bodies to which the laws of motion apply are therefore defined in an ideal way. Moreover, the very conception of motion involves the abstract notions of Geometry, with the notion of time added. Because of this ideal character of the laws, the science based upon them is properly called Theoretical Mechanics. 2 THEORETICAL MECHANICS. 2. Material Bodies. A body is any definite portion of matter. No attempt need be made to define matter or to enumerate its properties. A sufficiently definite preliminary notion is supplied by ordinary experience. The characteristics of material bodies which are of importance in a study of their motions are the following : (i) Every body has a definite volume and a definite figure ; (2) every body possesses a definite mass ; (3) bodies exert forces upon one another. 3. Mass. The mass of a body is often briefly defined as its "quantity of matter." These words, however, convey no definite idea of the meaning of mass as a factor in the determination of motion. A satisfactory definition of mass cannot be given in advance of a discussion of the fundamental laws of motion. We conceive of mass as the one invariable characteristic of mat- ter. Every individual portion of matter is regarded as possessing a definite mass whose value is uninfluenced by changes of position or by physical or chemical transformations. The volume and shape of a body may change, the forces it exerts upon other bodies and those which they exert upon it are different under different circumstances ; but its mass is regarded as an absolute constant. 4. Force. A force is an action exerted by one body upon another, tending to change the state of motion of the body acted upon. A force may be conceived as a push or a pull, acting upon a definite portion of a body. Such a push or pull always tends to change the motion of the body ; but this tendency may be counter- acted in whole or in part by the action of other forces. Mechanics is often called the science of motion and force, because of the importance of force in the development of the laws of the science. Force, like mass, is a quantity whose significance cannot be satis- factorily explained except by a full discussion of the fundamental laws of motion. 5. Particle. A body may be conceived to be divided into very small parts, each of which may be called a particle. Ideally, there is no limit to this process of subdivision. For the purposes of mathematical analysis it is often conceived to be carried so far that the linear dimensions of the particles become vanishingly small. INTRODUCTORY. 3 These particles may be conceived in either of two ways, corre- sponding to two different conceptions of the structure of matter. (1) It may be assumed that matter occupies space continuously. By this it is to be understood that every portion of matter whose mass is finite occupies a finite volume. (2) It may be assumed that any definite portion of matter con- sists of particles, each of which possesses finite mass but occupies no finite volume. The hypothesis of continuity does not necessarily imply that there may not be void spaces between the parts of a body. In mathe- matical language its meaning may be stated as follows : Let M be the mass of a body and V its volume ; and let bkM be the mass of a small portion whose volume is A V. Then if A V is taken smaller and smaller so that A Vj V approaches zero, the hypothesis of continuity implies that AM/M also approaches zero ; while the hypothesis of discontinuity implies that AM/M may be finite. In certain discussions it is found convenient to adopt one hypoth- esis, in other cases the other ; so far as results are concerned, it is usually immaterial which is adopted. In the analysis of the motion of a particle, it is regarded as a geometrical point endowed with mass. A body whose linear dimen- sions are small in comparison with the range of its motion is often regarded as a particle. 6. Rigid Body. A rigid body may be defined as a body whose particles do not change their distances from one another. An important part of the science of Theoretical Mechanics deals with bodies which are assumed to satisfy this definition. Actual solid bodies undergo appreciable changes of shape and size ; and if sufficiently small portions could be observed, they would doubtless be found to be in rapid motion, thus departing very far from the condition specified in the definition of rigidity. But dis- regarding the motions of ultimate particles and considering only the motion of a body as a whole, the theory of the motion of an ideal rigid body describes with great accuracy the motion of an actual solid body whose shape remains nearly constant. 7. Position and Motion. In the foregoing definitions and ex- planations it has been assumed that position and motion need no definitions. It is, in fact, doubtful whether any definitions can be given which convey clearer notions than the words themselves. 4 THEORETICAL MECHANICS. By the position of a particle is meant its relation in space to some body taken as a standard of reference. A particle is in motion when its position is changing. A body is in motion when the particles composing it are in motion. What shall be chosen as the body of reference in specifying the position and motion of a particle is a matter of arbitrary choice. Position and motion are thus not absolute but relative. The motion of a particle with reference to one body may be very different from its motion with reference to another. (See Arts. 267, 268.) 8. Kinds of Quantity. The science of Mechanics deals with quantities of four fundamental kinds : time, space, mass, force. In the development of the principles'of the science other quantities are introduced which are derived from these but involve no other elementary conceptions.* 9. Divisions of the Subject. The general subject of Mechanics may be subdivided in various ways. First, the basis of the subdivision may be the nature of the bodies dealt with. On this basis there are the following divisions : (a) Mechanics of a Particle and of systems of particles in general. ( b) Mechanics of Solid Bodies \ including (1) rigid and (2) non- rigid solids. if) Mechanics of Fluids, including (1) liquids and (2) gases. The present work deals mainly with particles and with rigid solids. Second, the basis of subdivision may be the fundamental kinds of quantity involved. On this basis the whole subject, and each of the above divisions, may be divided as follows : {a) Kinematics, treating of motion, without reference to the causes producing or influencing it. Under this there are (1) Pure Kine- matics, treating of motion apart from the idea of mass, and (2) Mass-Kinematics, treating of motion and mass. (b) Dynamics, treating of forces and of their influence upon the motions of bodies. Dynamics, the science of force, is again sub- divided into Statics and Kinetics. * It is pointed out in a later chapter that these four quantities are not always regarded as being all fundamental. In particular, force is often regarded as a quantity which must be defined in terms of mass, space and time. Ordinary experience, however, supplies a tolerably definite notion of force which is wholly independent of any conception of the meaning of the laws of motion. See " The Six Gateways of Knowledge," by Lord Kelvin. INTRODUCTORY. 5 Statics treats of the conditions of equivalence of systems of forces, and especially of the conditions under which forces are balanced so that they do not affect the motions of the bodies acted upon. Kinetics treats of the laws in accordance with which the motions of bodies are influenced by the forces acting upon them and by their masses. Strictly speaking, Kinetics includes Statics ; for from a knowledge of the effects of forces upon the motions of bodies may be derived all principles relating to the equivalence of different systems of forces and to the conditions under which forces are balanced. Statics and Kinetics are, however, often treated as coordinate branches of Dy- namics ; first, because Statics, although a special case, is a case of great importance, and second, because the principles of Statics can be developed to a large extent independently of those of Kinetics. The arrangement of subjects in this book does not strictly follow either of the above classifications, but is designed to meet the needs of beginners by presenting the different topics somewhat in the order of their difficulty. The arrangement adopted is as follows : Introductory chapter, treating of certain principles which have application in various branches of the general subject of Mechanics. Part I. Statics ; the discussion of the subject being limited mainly to systems of forces acting in the same plane upon a particle or a rigid body. Part II. Motion of a Particle ; including Kinematics and Kinetics, but limited mainly to the case of motion in a plane. Part III. Motion of Systems of Particles and of Rigid Bodies ; including Kinematics and Kinetics, but dealing mainly with motion in a plane. The remaining portion of this introductory chapter is devoted to certain principles which, while not strictly included in the subject of Mechanics, are of fundamental importance in the development of the science. 2. The Numerical Representation of Quantities. 10. Comparison of Like Quantities. The magnitudes of any two quantities of the same kind (as two distances, or two intervals of time, or two forces) may be compared by determining their ratio. Such a ratio is an abstract number. 11. Numerical Value of a Quantity. The numerical value of 6 THEORETICAL MECHANICS. a quantity is the ratio of its magnitude to that of a given quantity of the same kind, taken as a standard. This standard quantity is called a unit. Thus, to express a length as a number, some definite length must be chosen as a unit (its numerical value being called one) ; then the numerical value of any given length will be the number expressing its ratio to the unit length. The unit in terms of which quantities of a given kind are ex- pressed is wholly arbitrary. Thus, there are in common use several different units of length, as the foot, the inch, the yard, the mile, the meter ; several units of time, as the second, the minute, the week, the year ; and several units of mass, as the pound, the ton, the ounce, the gram, the kilogram. 12. Numerical Value Depends on Unit. Evidently, the nu- merical value of a quantity gives no idea of its magnitude unless the unit employed is known. Any given quantity may be represented by any number whatever, by properly choosing the unit. Thus, the same distance may be called 10,560 feet, 3,520 yards, or 2 miles. But the ratio of two quantities of the same kind is a definite number, which depends only upon the magnitudes of the quantities, and not upon their numerical values. The ratio of the numerical values of two quantities is the same as the ratio of the quantities themselves if they are expressed in the same unit, but not otherwise. 13. Fundamental and Derived Units. Although, as above stated, the unit in terms of which quantities of any particular kind are expressed numerically may be chosen arbitrarily, it is usually advantageous to choose the units for quantities of different kinds in such a way that certain of them depend upon others. In Physics, the usual practice is to choose arbitrarily the units of time, length and mass, and to make all other units depend upon these. Those units which are chosen arbitrarily are called fundamental, while those which are so defined as to depend upon the fundamental units are called derived units. For example, the unit of area is usually taken as the area of a square whose side has the unit length. It should, however, be clearly understood that this dependence is not necessary, but is merely assumed for convenience. 14. Dimensions of Derived Units. The word dimension is commonly employed to describe the way in which a derived unit depends upon the fundamental units. INTRODUCTORY. 7 Thus, if the unit area is denned as the area of a square whose side has the unit length, the relation may be expressed by the "dimen- sional equation ' ' (unit area) = (unit length) 2 , and the unit area is said to involve two dimensions of length. Sim- ilarly, if the unit volume is denned as the volume of a cube whose edge has the unit length, the unit volume involves three dimensions of length. In the illustrations just given, only one kind of fundamental quantity (length) is employed. There will be occasion in the follow- ing pages to deal with derived units which depend upon two or more fundamental units. Evidently the manner in which a derived unit is made to depend upon the fundamental units is to some extent arbitrary, and by changing the manner of dependence the dimensions of any given kind of quantity may be changed. This subject will be further treated in connection with the various kinds of quantity dealt with in the following pages. 15. Dimensional Equations. The dimensional equation above written is not to be interpreted as an ordinary algebraic equation. It is merely a concise method of expressing the way in which the derived unit (of area) is made to depend upon the fundamental unit (of length). It will be convenient to abbreviate such equations by the use of symbols to denote the various units. The quantities whose units are taken as fundamental in the following pages are usually some or all of the following : length, mass, time, force. In writing dimensional equations the letters L, M, T, F will be used to designate the units of these four kinds of quantity. 3. Vector Quantities. 16. Vector Quantity Defined. A directed straight fine of def- inite length is called a vector. A vector quantity is one possessing direction as well as mag- nitude. Any vector quantity may be represented by a vector. For this purpose the direction of the vector must agree with that of the quan- 8 THEORETICAL MECHANICS. tity to be represented, and its length must represent the magnitude of the quantity. In order that the length of a line may represent the magnitude of any given quantity, a scale must be chosen. That is, a certain length must be taken as equivalent to a unit magnitude of the kind to be represented. The length thus chosen is arbitrary, but when once chosen, the length representing any definite magnitude of the kind in question is determined. A vector may be designated by naming, in proper order, the letters placed at its ends. Thus, if A and B are any two points, AB and BA denote two vectors of equal length but opposite in di- rection. A vector may also be designated by a single symbol, as in the case of ordinary algebraic quantities. Scalar. The processes of ordinary algebra are limited to quan- tities which either do not involve direction or else are restricted to two opposite directions. Such quantities are called scalars. A scalar quantity is completely specified by assigning its magnitude and algebraic sign. 17. Equality of Vectors. Two vectors are said to be equal if they agree not only in magnitude but in direction. Thus, if MN and PQ (Fig. 1) are parallel and of equal length, we may write MN = PQ. If this were an ordinary algebraic equation, it would express only equality of length, and would be equally true if the lines were not parallel ; but as a vector equation it expresses identity of direction as well as equality of length. Thus, referring to Fig. 1, MN=QP is not true as a vector equation. 18. Addition of Vectors. Non-parallel vectors cannot be added in the sense in which scalars are added ; but the word addition is used to describe the process of combining two vectors in the follow- ing manner : INTRODUCTORY. Q Let MN and PQ (Fig. 2) represent two vectors ; then by their sum is meant a vector determined as follows : Make AB equal and parallel to MN, and BC equal and par- allel to PQ ; then A C represents the re- quired sum. In fact, there may be writ- ten the equation AB + BC = AC; or MN + PQ = A C. But this must be understood as a vec- tor equation, and not as applying to the lengths MN, PQ and A C If the order of the two vectors is changed their sum remains the same. For if lines equal and parallel to PQ and MN be laid off successively from A , the result will be AB' and B' C, which, with AB and BC, form a parallelogram of which AC is a diagonal. Next, let any three vectors be drawn consecutively, as AB, BC, CD (Fig. 3) ; then AD is called their sum. It is easily seen that the three vectors may be drawn in any order without changing their sum as thus defined. It is also evident B ^]D that AB + BC+ CD = AC + CD = (AB + BC) + CD ; that is, the result of adding the three Fig. 3. vectors is the same as the result of add- ing any one to the sum of the other two. The same construction may be extended to any number of vec- tors. 19. Vector Subtraction. Parallel vectors may be distinguished in direction by signs plus and minus as in Algebra. Thus, if A and B are any two points, the two vectors AB and BA may be desig- nated by prefixing signs plus and minus to the same symbol ; if the symbol p represents the vector AB, /represents the vector BA. The minus sign may thus be used to denote reversal of the di- rection of a vector, so that we may write AB BA. IO THEORETICAL MECHANICS. It may also be used to denote subtraction, in accordance with the following definition : To subtract a vector is to add its negative. With this definition of subtraction, the difference between two vectors may be found as follows : Let it be required to subtract the vector MN from the vector PQ (Fig. 4). Lay off AB = MN and AC = PQ; then PQ~ MN AC AB = AC+BA =BA + AC = BC 20. Vector Equations. From the above definitions of vector addition and subtraction, vector equations may be formed containing any number of terms in each member, and may be transformed in accordance with sim- ple rules. Thus, if A, B, C, D, E, F are any six points, we may write the vector equa- B tion AB + BC+CD = AE + EF+FD; (i) for each member of the m equation is equal to the vector AD. In several particu- lars vector equations may be treated by the same rules as the equations of ordinary Algebra. ( i ) Equal vectors may be added to each member of a vector equa- tion without destroying the equality. Thus, if a vector DM be added to each member of equation (i), each member becomes equal to AM, and therefore the equality still holds. By an extension of this principle, any number of vector equations may be added, mem- ber to member. (2) A term may be transposed from one member to the other by changing its sign. Thus, from equation (1) we may derive the equation INTRODUCTORY. II AB + BC = AE + EF+FD + DC, or AB + BC=AE + EF -\- FD ~ CD, by transposing the term CD and changing its sign. (3) The order of the terms in either member of a vector equa- tion may be changed at pleasure. For, as already pointed out, changing the order in which several vectors are combined does not change their sum. 21. Constant and Variable Vector Quantities. A vector quan- tity may be either constant or variable. A constant vector quantity remains unchanged not only in mag- nitude but in direction. A variable vector quantity changes either in magnitude or in direction, or in both. If a point A is fixed, while another point B moves in any man- ner, the vector AB is variable. If B moves always along the straight line which at a certain instant passes through the two points, the vector varies in magnitude only. If B describes a circle with center at A, the vector varies in direction but not in magnitude. If B describes any other path, the vector varies both in magnitude and in direction. 22. Increment of Variable Vector. If a variable vector has different values at the beginning and end of any interval of time, its increment during that interval is the vector which, added to the initial value, will pro- duce the final value. The increment may be determined, when the initial and final values of the vector are known, by the fol- lowing construction: From any point (Fig. 6), lay oft" OA and OB, representing the initial and final FlG * 6 - values of the variable vector ; then AB represents the required increment. For, there may be written the vector equation OA + AB=OB, showing that AB satisfies the definition of increment just given. The increment is evidently found by subtracting the initial value of the variable vector from the final value. For OBOA=AB. 12 THEORETICAL MECHANICS. In general, if p x denotes the initial value of a vector and p, z its final value, the vector increment is equal to/ 2 p lt the minus sign de- noting vector subtraction (Art. 19). 23. Resolution of a Vector. To resolve a vector is to express it as the sum of any number of vectors. If AB represents the given vector, it can be resolved by constructing a closed polygon of which AB is a side ; thus, AB=AL + LM+ MN + NB y L, M, and N being any three points whatever. Any vectors whose sum is equal to a given vector are called components of that vector. The discussion which follows will be limited mainly to coplanar vectors ; that is, to vectors lying in, or parallel to, the same plane. The foregoing principles are, however, true without this restriction. 24. Resolution into Two Vectors Having Given Directions. A given vector can be expressed as the sum of two vectors parallel to any two lines coplanar with it. For, let AB (Fig. 7) represent the given vector, and MN y PQ the given lines. From A draw a line parallel to MN> and from B a line parallel to PQ. Let C be the point of intersection of these lines ; then A C -\- CB = AB y and AC and CB are the required components. An equally correct construction is to draw A C parallel to PQ, and CB parallel to MN. The result is the same as before, since A C and CB are equal vectors, as are also AC and CB. Evidently only one pair of components can be found which are equivalent to a vector and have assigned directions. Resolved Part of a Vector. If A C and CB (Fig. 8) are at right angles to each other, each is called the resolved part of AB. If AB represents any vector quantity, and 6 is the angle between AB and a given direc- Fig. 8. C r-~ given 25 INTRODUCTORY. 13 ma Fig. E' N 9- tion, the resolved part of AB in that direction is given in magnitude by the product (magnitude AB) X (cos 6). The resolved part of a vector is equal to the algebraic sum of the resolved parts of its components, for any direction of reso- lution. Thus (Fig. 9) AE is the sum of the vectors AB, BC, CD, DE. The resolved parts of these components par- allel to the line MN are A'B', B'C, CD', D'E' ; while the re- solved part of AE is A'E'. Evidently A'E' is the algebraic sum of A'B', B'C, CD', D'E'. 26. Localized Vector. If a vector is associated with a definite point or line in space, it may be called a localized vector. The point at which a vector is localized may be called its posi- tion; and the line in which it is localized its position-line. 27. Moment of Localized Vector. The moment of a localized vector about a point is the product of the magnitude of the vector into the perpendicular distance of its position-line from the given point. The origin of moments is the point about which the moment is taken. The arm of the vector is the perpendicular distance of its position-line from the origin. The plane of the moment is the plane containing the origin of moments and the position-line of the vector. In comparing moments having the same plane (or parallel planes) there are two cases which may be distinguished by signs plus and minus. These cases are illustrated in Fig. 10, in which O is the origin of mo- ments and MN, PQ denote the directions and position-lines of two localized vectors. The relations of MN and PQ to the origin O 14 THEORETICAL MECHANICS. suggest rotations about O in opposite directions. One of these di- rections being taken as plus, the other will be called minus. Which shall be called plus may be decided arbitrarily. For uniformity, it will usually be assumed that the positive direction is counter-clock- wise ; that is, opposite to the direction of rotation of the hands of a watch placed face upward in the plane of the paper. 28. Moment Represented by Area of Triangle. If a triangle be constructed with its vertex at the origin of moments and its base in the position-line of a localized vector, the length of the base being numerically equal to the magnitude of the vector, the area of the triangle is numerically equal to half the moment of the vector. This follows immediately from the definition of moment. 29. Moment of Sum of Two Vectors. If/, q and r represent three localized vectors such that/ + q = r > an d if their position-lines meet in a point, the algebraic sum of the moments of / and q with respect to any origin in their plane is equal to the moment of r with respect to that origin. Let the three vectors p, q and r be represented in magnitude and direction by OA, OB and OC (Fig. 11) respectively, so that Fig. 11. OC is the diagonal of the parallelogram of which OA and OB are sides. Let O be the point of intersection of the three position-lines, and M the origin of moments. Draw MO. Then moment of OA = 2 (area of triangle MOA); " OB = 2 ( " " " MOB); " OC= 2 ( " " " MOC). INTRODUCTORY. 1 5 Draw AA\ BB ', CC ', each perpendicular to MO. The double areas of the three triangles are respectively MO X AA\ MO X BB', MO X CC Now, ; CC = AA' + BB'\ . . . (i) for, drawing AC" parallel to OM, it is seen that A A' = C" C and BB'=CC". Hence area MO A + area MOB = area MOC. . . (2) This proves the proposition for the case in which the moments of the two given vectors have the same sign. The case in which the origin is so taken that the moments have opposite signs is represented in Fig. 1 2. The lettering is the same as in Fig. 1 1 , and the reasoning to be employed is exactly similar. The algebraic sum of the moments is now the numerical difference. Thus, from the geometry of the figure, we have CC = AA' -BB' . . . (Y) instead of equation (1), and therefore area MO A area MOB = area MOC, . (2' ) instead of equation (2). 30. Vector Quantities in Mechanics. Some of the important quantities dealt with in the science of Mechanics are vector quantities. The foregoing brief discussion is given as a useful introduction to the development of the elementary principles of Mechanics. PART I. STATICS. CHAPTER II. FORCE AND STRESS. i. Conception of Force. 31. Motion of a Body if Uninfluenced by Other Bodies. A body remains at rest, or moves in a straight line at a uniform rate, except in so far as its motion is influenced by other bodies. ( New- ton's first law of motion. See Art. 259.) Since no body can be wholly free from the influence of other bodies, it is impossible to verify this law directly. The indirect evidence for its truth is, however, very strong. When a body is observed to depart from a condition of uniform rectilinear motion, it is always found reasonable to attribute this departure to the influence of other bodies. Thus, a stone thrown horizontal^ into the air departs from the condition of uniform motion in a straight line in two ways: its path curves downward, and its rate of hiotioVi changes. The first effect we attribute to the influence of the earth (which we say attracts the body), while the second we attribute partly to the "attraction" of the earth and partly to the influence of the air (which we say resists the motion of the body). 32. Force Defined. When one body influences the motion of another body, the first is said to exert a force upon the second. Force may therefore be defined as follows : A force is an action exerted by one body upon another, tending to change the state of motion of the body acted upon. When an external body exerts a force upon any portion of the human body, the action (if sufficiently intense) is recognized by the senses. In such a case the name pusk or pull 'is given to the force. It is therefore natural to conceive of every force as a push or a pull. This conception will be found useful in deciding what may properly be called forces. FORCE AND STRESS. 17 33. Magnitude and Direction of a Force When the action of forces is recognized by the senses, we form the idea that forces possess different magnitudes. Thus, if two pressures are successively- applied to the same part of the body, one may be recognized as much greater than the other. We thus come to think of a force as possessing a definite magnitude ; although we are not able to compare the magnitudes of forces with precision by means of their immediate effects upon the senses. Experience also gives the idea that a force possesses a definite direction ; but we are not able, from the immediate evidence of the senses, to compare the directions of forces with accuracy. This conception of a force as possessing definite magnitude and direction is verified and made definite when the effects of forces upon the motions of bodies are studied. 34. Force a Vector Quantity. Since a force possesses both a definite magnitude and a definite direction, it is a vector quantity (Art. 16). Like other vector quantities, forces may be represented by directed right lines or vectors. To accomplish this, some length must be chosen to represent the unit force ; then any given force will be represented by a vector whose length is the same multiple of the chosen length that the magnitude of the force is of the unit force. 35. Action and Reaction. If a body A exerts a force upon a second body B, the body B at the same time exerts upon A a force of equal magnitude in the opposite direction. One of the forces of such a pair is often called an action and the other a reaction ; and the principle may be stated in the words * ' to every action there corresponds an equal and contrary reaction." (Newton's third law of motion. See Art. 259.) 36. Stress. A stress consists of two equal and opposite forces constituting the action and reaction between two bodies or portions of matter. Illustrations. Any two bodies attract each other in accordance with the law of gravitation ; the two forces they exert upon each other constitute a stress. Such a stress acts between the earth and the moon, or between any other two heavenly bodies. Two electrified bodies attract (or repel ) each other with equal and opposite forces, constituting a stress. 1 8 THEORETICAL MECHANICS. Two bodies in contact exert upon each other equal and opposite forces at the surface of touching; these forces constitute a stress. 37. Place of Application of a Force. A force applied to a body may act either throughout a certain volume, or upon every part of a certain surface. Thus, the attraction of the earth upon a body ac- cording to the law of gravitation acts upon every part of the mass, and therefore its place of application is the whole space occupied by the body; while the pressure between two bodies which touch each other is distributed over their surface of contact. In both cases the forces are distributed. A concentrated force is a force of finite magnitude applied at a point. Although this definition is ideal, since all forces are in real- ity distributed in their action, either throughout a volume or over an area, the conception of a concentrated force is a useful one. In many cases a relatively large force is applied throughout a small volume or over a small area, and the conception of a concentrated force is approximately realized. By the study of the effects of forces upon the motions of the bodies to which they are applied, the conception is reached that a distributed force is ' ' equivalent to " a concentrated force. It is with the laws of equivalence of systems of concentrated forces that the science of Statics mainly deals; a distributed force being regarded as the limiting case of a system of concentrated forces whose number becomes very great while their individual magnitudes become very small. A force is thus regarded as having a definite point of application and a definite line of action. It is thus a localized vector quantity (Art. 26). 2. Classes of Forces. 38. Conditions Under Which Forces Act. The conditions under which bodies exert forces upon one another are known only from observation and experiment. It will be useful to enumerate a few cases of forces whose laws have been the subject of scientific investigation. 39. Examples of Forces. (1) Gravitation. Any two bodies exert upon each other attractive forces in accordance with the follow- ing law, called Newton' s law of gravitation : FORCE AND STRESS. 1 9 Every particle of matter attracts every other particle with a force which acts in the line joining the two particles, and whose magnitude is proportional directly to the product of their masses and inversely to the square of the distance between them. (2) Electrical forces. If two bodies are charged with electricity, each exerts a force upon the other. The forces are repulsive or attractive, according as the charges are ' ' like " or ' ' unlike ' ' in kind. (3) Magnetic forces. Two magnets exert upon each other forces, both attractive and repulsive ; like poles repelling and unlike attract- ing each other. (4) Electromagnetic forces. Forces are exerted upon each other by a magnet and a wire carrying an electric current; also by two wires carrying electric currents. (5) Molecular and atomic forces. The "atomic theory" of the constitution of matter assumes that a body of definite composition is made up of ultimate particles called molecules, each of which possesses the same physical properties as the whole body; while each molecule is made up of lesser parts called atoms, which do not separately possess the physical properties of the body. The theory assumes that forces (besides the universal force of gravitation) act between the ultimate particles. 40. Action at a Distance. Forces are sometimes classified as "actions at a distance" and "actions by contact." Under the former would be classed gravitational, electrical, and magnetic forces, since they apparently are not exerted by means of any material connection between the bodies concerned. Modern researches have rendered it highly probable that electrical and magnetic forces are in reality actions transmitted through a ' ' medium ' ' which fills all space. Efforts have also been made to show that the force of gravi- tation may be explained in a similar manner. Whether this view be correct or not, the distinction between actions at a distance and actions by contact is practically useful. 41. Passive Resistances. The forces which bodies exert upon one another are often divided into two classes, called respectively active forces and passive resistances. By an active force is meant one which acts independently of the state of motion of the body, and also independently of any other forces which may be applied to it ; while a passive resistance comes into action only to prevent or to resist certain motions of the body. 20 THEORETICAL MECHANICS. Thus, if a body rests upon a horizontal table, it is acted upon by the attraction of the earth, which is an active force, tending to give it motion downward, and having the same magnitude and direction whatever other forces may be applied to the body ; and also by a passive resistance, exerted upward by the table to resist the tendency of the body to move downward. This latter force acts only because the body has a tendency to move downward through the table. If, by means of the hand, there be applied to the body a supporting force of less magnitude than the downward pull of the earth, the resisting force exerted by the table is diminished, its amount being always just sufficient to neutralize the effect of the other forces. If the hand supports the body with an upward force equal to the downward attraction of the earth, the pressure of the table upon it becomes zero. If the upward force applied by the hand becomes still greater, the table exerts no force to resist the tendency of the body to rise. This illustrates the fact that a passive resistance can have any magnitude within certain limits, but that its value cannot go outside these limits. If it were possible to analyze the forces exerted by one body upon another at their surface of contact, and by contiguous portions of the same body upon each other, in such a way as to determine the actions between the ultimate particles and the laws governing these molecular or atomic forces, it is probable that these would be found to have the same essential nature as the forces called active. But for practical purposes the distinction between active forces and passive resistances is often useful. In the following Articles are considered two cases of passive resistances which are of importance in the discussion of the problems of Statics. 42. Pressure Between Bodies in Contact. If two bodies touch each other, each exerts a force upon the other at the surface of contact. These contact forces are passive resistances, and (within certain limits) will take any values necessary to resist certain relative motions of the bodies. But the magnitudes and directions of the forces which the bodies can exert upon each other are subject to certain limitations depending upon the nature of the surfaces and the material composing the bodies. Let A and B (Fig. 1 3) represent the two bodies. If for any reason A has a tendency to slide upon B, this tendency will be FORCE AND STRESS. 21 resisted, and may be wholly neutralized. The rougher the surfaces of contact, the greater the force that can be exerted to resist the sliding. If the surfaces be very smooth, the possible magnitude of the force is very small. Hence we are led to the fol- lowing definition : A perfectly smooth surface is one which can offer no resistance to the sliding of a body upon it. Although this definition cannot be realized in the Fig. 13. case f anv actual body, certain surfaces approach near to the condition of perfect smoothness. If for any reason A has a tendency to move toward B in the direction of the normal to their surface of contact, the latter body is able to exert a resisting force in the direction of the normal. The magnitude which this resisting force can take is limited only by the strength of the material of which B is composed. Such a ' 4 normal resistance ' ' can be exerted by a smooth surface as well as by a rough one. The pressure exerted between rough surfaces will be consid- ered later under the head of friction. Smooth hinge. A hinge joint is often used to connect two bodies, in such a way as to leave them free to move in a certain manner relatively to one another. Such a joint consists of a cylin- drical pin which forms a part of one body, or is rigidly connected with it, and a cy- lindrical hole formed in the other body, into which the pin is inserted. Such a con- nection permits the bodies to turn relatively to each other FlG - I4 ' about the axis of the pin, but (if the pin and hole are of equal diameter) prevents any other rela- tive motion. (See Fig. 14.) In order to permit free motion about the axis, the pin must be slightly smaller than the hole, so that the two bodies are in contact along a straight element common to the two cylindrical surfaces. If these surfaces are smooth, the pressure between them has the direction of their common normal. In the problems of Statics, a smooth hinge is often introduced as a means of connecting bodies. In such a case the pressure between the bodies is to be taken as acting in a line through* the center of the hinge. 22 THEORETICAL MECHANICS. 43. Tension in a Flexible Cord. A flexible cord is one which may be bent. A perfectly flexible cord may be bent without resist- ance. No actual cord possesses perfect flexibility, but the resistance to bending may be very slight. Let AB (Fig. 15) represent a portion of a perfectly flexible cord, and let two equal and opposite forces, Pand P' , be applied to it at A and B. Let C be any point between A and B. The force P tends to cause AC to P^ i\ww\!\ \ \\m _ y iP' move to the left ; this A. C B is prevented by a force Fig. 15. equal and opposite to P exerted at C by CB upon AC. Similarly, P' tends to cause CB to move to the right ; and this is prevented by a force equal and opposite to P' exerted at C by A C upon CB. That is, the two parts of the cord exert upon each other at C equal and opposite forces, constituting a stress (Art. 36). The same is true at any other point between A and B. Such a stress (resisting a tendency of the two portions of the cord to sep- arate in the direction of their length) is called a tension. We have here illustrated a special case of the principle that if a flexible cord is pulled tight between two points, any two adjacent portions exert upon each other forces par- allel to the direction of the cord. This subject will be discussed more fully in a later chapter. It is introduced at this point because in the problems to be discussed in the following chapters a flexible cord is often used as a means of applying a force to a body. Thus, if a cord is attached to a body X (Fig. 16) at a point B y a force of any magnitude applied to the cord at another point A, in the direction BA } will cause the cord to exert an equal force upon the body at B. 3. Numerical Representation of Forces and of Masses Grav- itation System. 44. Relation Between Units of Force and of Mass. Although force and mass are two distinct kinds of quantity, so that it is pos- sible to choose the unit in which each is expressed independently of FORCE AND STRESS. 23 the other, yet it is convenient to choose these units so that one de- pends upon the other. The method of choosing units which will now be described makes the unit mass arbitrary, while the unit of force depends upon that of mass. This is the gravitation system of units. In a later chapter another system will be described. 45. Unit Mass. The unit mass is taken as the quantity of mat- ter in a certain piece of platinum arbitrarily chosen and established as the standard by act of the British Parliament. This unit mass is called one pound. 46. Weight. The weight of a body is the force with which the earth attracts it, in accordance with the law of gravitation. From this law (Art. 39) it follows that the weights of bodies in the same locality at the surface of the earth are proportional to their masses. But if the bodies are in different positions on the earth, or at differ- ent elevations above the surface, their weights may not be propor- tional to their masses. In problems relating to bodies near the earth, the weight of a body has often to be included among the forces applied to it. By the law of gravitation every particle of the earth attracts every par- ticle of the body. These several forces may for many purposes be regarded as equivalent to a single force applied at a certain point called the center of gravity of the body. The truth of this state- ment is for the present assumed without proof. (See Chapter IX.) 47. Unit Force. The gravitation unit of force is a force equal to the weight of a unit mass at the earth's surface. A pound force is a force equal to the weight of a pound mass at the earth's surface. The pound force as thus defined has not the same value for all positions on the surface of the earth, since the weight of any given body varies if it be taken to different localities. The variations are, however, comparatively small, and for many purposes unimportant. The pound force may be made wholly definite by specifying a certain place at which its value is to be determined. It is unfortunate that the word pound is used in two senses, ap- plying both to a unit force and to a unit mass. The usage is, how- ever, so common that it is important for the student to become fa- miliar with it. The double meaning arises from the fact that the 24 THEORETICAL MECHANICS. most convenient as well as the most accurate method of comparing the masses of bodies is by determining the ratio of their weights. 48. Comparison of Forces. The magnitude of a force in terms of the gravitation unit, or pound force, may be found by determin- ing how many pounds mass it will support against the attraction of the earth. This is the method employed, directly or indirectly, in many machines for testing the strength of materials. 49. Measurement of Masses. The mass of a body may be determined by comparing its weight with that of a body or bodies of known mass. This is the method adopted in ' ' weighing ' ' a body on an ordinary balance consisting of a lever or a system of levers. The weight of a body as thus determined is a correct indication of its mass, provided the masses of the standard bodies or ' ' weights ' ' are accurately known; for if the same experiment be conducted in two different localities, the weights of the standard bodies will change in the same ratio as the weights of the bodies balanced against them. If the weight of a body is determined by a spring balance, the result may be different if the weighing is done in different localities; for in this case the process of weighing consists, not in comparing the weight of one body with that of another, but in determining di- rectly the pull exerted by the earth * upon the body by ascertaining how much this pull will stretch a spring. If the weight of the body changes, therefore, as it may if its location is changed, this change will be indicated by a corresponding variation in the amount of stretching of the spring. 50. Metric System. In the gravitation system of estimating forces and masses, the unit mass is wholly arbitrary. Besides the units above described, which owe their establishment to British law and custom, there is another important set, based upon the French gram or the kilogram. A gram is very nearly equal to the mass of a cubic centimeter of pure water at the temperature of maximum density. This relation is not exact, however, and the gram, like the pound, must be regarded as an arbitrary unit, whose value is to be * What is actually measured is the ' ' apparent ' ' pull of the earth upon the body, which differs from the true attractive force for reasons which cannot be here discussed, the chief of them being the diurnal rotation of the earth. See Art. 311. FORCE AND STRESS. 25 determined by reference to a certain standard body. For ordinary purposes the kilogram, equal to 1,000 grams, is often more con- venient than the gram. The gravitation unit of force, in the French or metric system, may be taken as the weight of a body of one kilogram mass. For brevity this unit may be called a force of one kilogram. To make the unit definite, the position must be specified, since the weight of the same body is different in different places. The following is the relation between the pound and the kilo- gram : 1 kilogram = 2.2046 pounds. 1 pound = 0.45359 kilogram. The gravitation system of units of force and mass, although the most convenient in many practical applications, is not the best for the purposes of pure science. The discussion of other systems must, however, be deferred. 4. Definitions. 51. Concurrent and Non-concurrent Forces. Forces acting upon a body are concurrent when they have the same point of appli- cation. When applied at different points they are non- concurrent. 52. Coplanar Forces are those whose lines of action lie in the same plane. The following pages treat mainly of coplanar systems. 53. A Couple is a system consisting of two forces, equal in mag- nitude, opposite in direction, and having different lines of action. The perpendicular distance between the two lines of action is called the arm of the couple. 54. Equivalent Systems of Forces. Two systems of forces are equivalent if either may be substituted for the other without change of effect. From this definition it follows that two systems, each of which is equivalent to a third system, are equivalent to each other. 55. Resultant. A single force that is equivalent to a given sys- tem of forces is called the resultant of that system. It will be shown subsequently that a system of forces may not be equivalent to any 26 THEORETICAL MECHANICS. single force. When such is the case, the simplest system equivalent to the given system may be called its resultant. Any forces having a given force for their resultant are called components of that force. 56. Composition and Resolution of Forces. Having given any system of forces, the process of finding an equivalent system is called the composition of forces if the system determined is simpler than the given system ; if the reverse is the case, the process is called the resolution of forces. The process of finding the resultant of any given forces is the most important case of composition ; while the process of finding two or more forces, which together are equivalent to a single given force, is the most common case of resolution. 57. Equilibrium. A system of forces applied to a body is in equilibrium if the state of motion of the body is not changed by their action. The term equilibrium is also applied to the body upon which the forces act ; a body is said to be in equilibrium if its state of motion is unchanged by the action of all applied forces. CHAPTER III. CONCURRENT FORCES. i. Composition and Resolution of Concurrent Forces. 58. Forces Having the Same Line of Action. The resultant of two concurrent forces acting in the same direction is a force in that direction equal to their sum. The resultant of two concurrent forces acting in opposite directions is a force whose magnitude is equal to the difference between the magnitudes of the given forces, and whose direction is that of the greater. No proof need be given of the truth of these statements. They must be regarded as following immediately from our conception of force and from our experience regarding the effects of forces. They are special cases of the general principle of the parallelogram of forces (Art. 59). These principles may be generalized in the following manner : ( 1 ) The resultant of any number ot concurrent forces having the same direction is a force in that direction whose magnitude is equal to the sum of the magnitudes of the given forces. (2) The resultant of any concurrent forces having the same line of action is found by combining those having one direction into a single force and those having the opposite direction into another force, and then combining the'se partial resultants according to the rule already stated. If signs plus and minus are used to distinguish the two opposite directions along the common line of action of the forces, the foregoing principles may be included in the following general rule: The resultant of any concurrent forces having the same line of action is a force equal to their algebraic sum. 59. Resultant of Any Two Concurrent Forces. If any two concurrent forces be represented in magnitude and direction by lines drawn from the same point, their resultant is represented in magnitude and direction by the diagonal (drawn from that point) of the parallel- ogram of which the two lines are adjacent sides. 28 THEORETICAL MECHANICS. Thus let OA and OB (Fig. 17) represent the two given forces, and OC the diagonal of the parallelogram of which OA and OB are sides ; then OC represents the resultant. This principle is known as the par- allelogram of forces. Many proofs of the proposition have been offered, but none of these has been generally accept- ed as satisfactory. Perhaps the most satisfactory view is that which regards this principle as one of the fundamental laws of force, depending for its verifica- tion upon experience. The matter is further considered in Chapter XIV. 60. Triangle of Forces. Since the opposite sides of a parallel- ogram are equal, the magnitude and direction of the resultant of two forces may be found by constructing a triangle instead of a parallelogram. Thus, any two forces being given, let them be represented in magnitude and direction by lines OA and A C laid off consecutively (Fig. 17) ; then OC represents the magnitude and direction of the resultant. For, if OB be drawn equal and parallel to AC, OC is a diagonal of the parallelogram of which OA and OB are adjacent sides. The lines OB, BC are not needed for the determination of OC. Evidently, the principle of the par- allelogram of forces is equivalent to the . following : If A, B, C are three points so chosen Fig. 18. that AB, BC represent, in magnitude and direction, two concurrent forces, their resultant is represented in magnitude and direction by A C. This principle is known as the triangle of forces. The point of application of the resultant is the same as that of the components. 61. Computation of Resultant of Two Concurrent Forces. Let P and Q represent the magnitudes of two concurrent forces, and 6 the angle between their lines of action. [The angle is to be measured between the positive directions of the lines of action. Thus, if XX' and YY' (Fig. 19) are the lines of action, the directions CONCURRENT FORCES. 2 9 being as indicated by the arrows, is the angle X'OY'.] Laying offAB, BC to represent Pand Q, A C will represent the resultant R. By elementary Trigonometry, the magnitude of the resultant is given by the formula R* = P* + Q>+ 2 pQ cos e. To find the direction of the resultant, let angle CAB = a, angle ACB = $\ then sin a sin fi sin 6 or sin a Q P Q P sin 6 ; sin 8 sin 0. R R Examples. 1. A particle is acted upon by forces of 10 lbs. and 20 lbs. whose lines of action include an angle of 25. Determine the magnitude and direc- tion of a single force which would produce the same effect. Ans. A force of 29.4 lbs. making an angle of 16 42' with the force of 10 lbs. 2. Two men pull a body horizon- tally by means of ropes. One exerts a force of 28 lbs. directly north, the other a force of 42 lbs. directed N. 42 E. What single force would be equivalent to the two ? 3. Two persons lifting a body ex- ert forces of 44 lbs. and 60 lbs. in di- rections inclined 28 to the vertical on opposite sides, force would produce the same effect ? Ans. A force of 92. 1 lbs. at angle 32 40' with force of 44 lbs. Fig. 19. What single 62. Resultant of Any Number of Concurrent Forces. The resultant of any number of concurrent forces may be found by first finding the resultant of two of them, then combining this resultant with a third force, and so on. Thus, if AB and BC (Fig. 20) represent two of the forces in magnitude and direction, their resultant is represented by A C Draw CD to represent a third force; the resultant of AC and CD is AD, which is therefore the resultant of the three forces represented by 30 THEORETICAL MECHANICS. AB> BC, and CD. If DE represents a fourth force, the resultant of AD and DE is AE, which is therefore the resultant of AB, BC, CD and DE. This process may be ex- Z?/^^^ tended to include any number of / ^^"^^^^ forces. Evidently the lines A C and Jr '"/ A are not nee ded m tne con " / ^ +?"* / struction. / *"" ^ __ -p The above process may be de- a ^y-- ~/~^r^^ scribed as follows : "^^^^^^^ If any number of concurrent forces D are represented in magnitude and di- Fig. 20. rection by lines forming sides of a continuously described polygon, the line drawn from the starting point to close the polygon represents the resultant in magnitude and direction. This principle is called the polygon of forces. The sides of the polygon cannot, of course, represent the lines of action of the forces, since these all intersect in the point at which the forces are applied. 63. Resultant as Vector Sum. The process of finding the result- ant of several forces by constructing the polygon is evidently the same as the process of vector addition (Art. 18). That is, The resultant of any number of concurrent forces is a force equal to their vector sum, its point of application being the same as that of the given forces. 64. Vector Diagrams and Space Diagrams. In combining forces by geometrical construction, they must be represented in magnitude and direction, and also in line of action. In most cases it is desir- able, in order to prevent confusion, to draw two separate diagrams, one showing the lines of action of the forces, the other representing them in magnitude and direction only. These two classes of dia- grams may be called space diagrams and vector diagrams respec- tively. 65. Computation of Resultant by Force Polygon. The re- sultant of any coplanar concurrent forces may be computed by graphical construction, by drawing the force polygon to scale and measuring the line representing the resultant. Or, the length and direction of the closing side of the force polygon may be computed CONCURRENT FORCES. 3 1 trigonometrically from the force polygon. A more convenient trig- onometric method will be given later. 66. Resolution of a Force Into Any Number of Compo- nents. A force may be resolved into any number of components by means of the polygon of forces. If the magnitude and direction of the given force be represented by a line, this line may be made one side of a polygon, and the other sides will repre- sent, in magnitude and direction, forces of which the given force is the result- ant. Thus, let the given force be rep- resented by the vector AB (Fig. 21). Choose any number of points (as C, D> E), and draw lines forming the poly- gon AC DEB. Then AQ CD, DE } EB represent, in magnitude and direc- tion, four forces whose resultant is AB. Fig. 21. A force may obviously be replaced by other forces in an infinite number of ways. Some special cases of the general problem of resolution will now be, considered. 67. Resolution of a Force Into Two Components. This is a special case of the general problem of resolution into any number of components. If a force be represented in magnitude and direc- tion by a vector AB, and if C be any point whatever, the vectors A C, CB represent two forces which together are equivalent to the given force. Since an infinite number of triangles may be drawn, each having any given line as one side, it follows that a force may be replaced by two forces in an infinite number of ways. To make the problem determinate, certain conditions must be specified which the components are to satisfy. The following cases furnish determinate problems. Each should be solved both by geo- metrical construction and by algebraic computation. (1) Let a force be resolved into two components whose lines of action are given. Solution, (a) Geometrical. Let AB (Fig. 22) represent the given force in magnitude and direction, and let the two components have lines of action parallel to XX and YY. From A draw a line parallel to XX, and from B a line parallel to YY, C being the point of intersection of these two lines; then AC and CB represent 32 THEORETICAL MECHANICS. the required components in magnitude and direction, tion may evidently be varied by drawing from A a YY and from B a line parallel to XX, giving A C and C'B as the two com- ponents. The result is the same as before, since AC and C'B are equal vectors, as are also A C and CB. () Algebraic. From the given data there are known in the triangle ACB the side AB and all the angles. Hence, if AB = R, AC=Q, CB=P, ACB = i8o 0, ABC=a y BAC=frwe have (as in Art. 6i) The construc- line parallel to q _ p sin a sin /3 A' sin 6 from which P and Q may be computed. (2) Let the two components be given in magnitude only. (3) Let the magnitude of one and the direction of the other be given. (4) Let one component be wholly given. Examples. 1. A force of 50 lbs. is equivalent to two forces whose directions are inclined to that of the given force at angles of 1 2 and 86 respec- tively; determine their magnitudes. Ans. 50.6 lbs. and 10.5 lbs. 2. A force of 83 lbs. is to be replaced by two components, one of which is a force of 36 lbs. at right angles to the given force. De- termine the magnitude and direction of the other component. 3. Resolve a force of 200 lbs. into two components, of 130 lbs. and 98 lbs. respectively. Determine the directions of the two com- ponents. An s. Angle between resultant and greater component = 24 34'. 4. A force of 1 50 lbs. is to be resolved into two components, one acting at an angle of 20 with the given force, the other having a magnitude of 80 lbs. Determine completely the two components. 5. A force of 150 lbs. is to be resolved into two components, one acting at an angle of 20 with the given force. What is the least possible magnitude of the other? Ans. 51.31 lbs. CONCURRENT FORCES. 33 68. Resolved Part of a Force Definition .If a force is equiv- alent to two components at right angles to each other, each is called a resolved part of the given force. The resolved part of a force represented by AB (Fig. 23) in the direction of a line MN is represented in magnitude and direction by A'B', the orthographic projection of AB upon MN Proposition. The resolved part, in a given direction, of the resultant of any concurrent forces, is equal to the algebraic sum of the resolved parts of the components in that di- ft' j\r rection. Fig. 23. Since the resultant is the vector sum of the components, this proposition is a special case of that stated in Art. 25. Algebraic expression for the resolved part. From the definition it follows that the resolved part of a force of magnitude P in a direc- tion making an angle with that of the force is equal to P cos 6. If 6 lies between 90 and 270 its cosine is negative. Hence if 6 is measured from the positive direction of the line along which the resolution is made, the product P cos 6 gives the resolved part with proper sign, whatever the value of the angle. Let X and Y be the resolved parts of a force P in two directions at right angles to each other, being the angle between P and JC, and 7r/2 6 the angle between P and Y. Then X = Pcos 6; Y = P cos (tt/ 2 6) = Psin 6; P* = X' 1 + F 2 ; a X " a Y cos v = ; sin v = . P P From these equations we may compute the resolved parts when the force and the directions of resolution are given; or we may compute the magnitude and direction of the force when its resolved parts in two directions at right angles to each other are known. 34 theoretical mechanics. Examples. i. Let 20 lbs. and 40 lbs. be the resolved parts of a force in two directions at right angles to each other. Determine the magnitude and direction of the force. 2. Compute the resolved part of a force of 235 lbs. in a direction inclined 25 to that of the force ; also in a direction inclined 160 to that of the force. 69. Algebraic Computation of Resultant of Any Concurrent Forces. From the principle of the triangle of forces, it 13 possible to compute the magnitude and direction of the resultant of any known concurrent forces. Thus, the resultant of any two may be computed by the solution of a triangle ; this resultant may be com- bined with a third force and their resultant computed in a similar manner; and the process may be continued until the resultant of the whole system is determined. This process would, however, be very laborious, and in most cases the following method is to be preferred. Choose a pair of rectangular axes in the plane of the forces, and let the given forces be P x , making angles a x , ft with the axes ; P 2 , making angles a 2 , ft 2 with the axes ; etc. Replace each force by its resolved parts parallel to the two axes ; P x being replaced by com- ponents P x cos a x > P x cos ft; P t by components P 2 cos a. it P 2 cos f& % \ etc. The given forces are thus replaced by two systems of collinear forces. The resultants of these two systems are respectively P x cos a x -\- P. z cos a , -f- . . . = X, and P x cos ft + P, cos , + . . . == Y. The resultant of the whole system is equal to the resultant of X and Y. Let R denote this resultant and a, b the angles it makes with the x- and ^/-axes respectively; then r* = x* + y * x /, Y cos a = ; cos . R R Examples. 1. Find the resultant of the following concurrent forces: 23 lbs. directed N. 40 E. ; 42 lbs. S. 20 E. ; 86 lbs. due E. ; 56 lbs. N. 33 W. CONCURRENT FORCES. 35 2. A body at a point O is pulled equally by three men in direc- tions OA, OB and 0C } such that AOB = BOC=6o a and AOC = 120 . What single force would produce the same effect ? Ans. If P = pull exerted by one man, the resultant is a force 2P in the direction OB. 3. A body of 145 lbs. mass is acted upon by a horizontal force of 63 lbs. What single force would be equivalent to the horizontal force and the weight of the body ? 2. Moments of Concurrent Forces, 70. Moment of a Force. The moment of a force with respect to a point is the product of the magnitude of the force into the perpen- dicular distance of its line of action from the given point. This is a particular case of the definition of the moment of a localized vector, already given (Art. 27). The sign of the moment of a force will usually be assumed to follow the convention there adopted. Moment about an axis. A line through the origin of moments, perpendicular to the plane containing the origin and the line of action of the force, may be called an axis of moments. The moment of the force with respect to the origin may also be regarded as its moment with respect to this axis. 71. Moment of Resultant of Two Concurrent Forces. The moment of the resultant of two concurrent forces with reference to a point in their plane is equal to the algebraic sum of their separate moments with reference to that point. Since the resultant of two concurrent forces is a force equal to their vector sum applied at their common point of application, this proposition is a particular case of that proved in Art. 29 for localized vectors. It was there assumed that the two vectors were not parallel. If, however, they are parallel and applied at the same point, the truth of the proposition is evident. 72. Moment of Resultant of Any Number of Concurrent Forces. The moment of the resultant of any number of concurrent forces with reference to any point in their plane is equal to the algebraic sum of their separate moments with reference to that point. The proposition having been proved for two forces, let the result- ant of any two of the given forces be combined with a third ; since the proposition is true for these two forces and their resultant, it is true 36 THEORETICAL MECHANICS. for the three forces and their resultant. Similarly, it is true for this resultant and a fourth force, and is therefore true for the four forces and their resultant. The same line of reasoning may be extended to include any number of forces. Examples. i. Compute the moments of the four forces described in Ex. i, Art. 69, the origin of moments being a point 8 ft. directly north from the point of application of the forces. 2. Determine the line of action of the resultant of two forces of 18 lbs. and 12 lbs. whose lines of action include an angle of 6o. Compute the moments of the two forces with respect to an origin on the action-line of the resultant, 4 ft. from the point of application of the forces. Test the truth of the above proposition. 3. In Ex. 2, compute the moments of the two forces and their resultant with respect to an origin on the line of action of the force of 1 2 lbs. , 5 ft. from the point of application. Ans. 77.94 ft. -lbs; o; 77.94 ft. -lbs. 4. Prove that the moments of two concurrent forces are equal in magnitude for any origin on the line of action of the resultant. 3. Equilibrium of Concurrent Forces. 73. General Condition of Equilibrium. From the definitions of equilibrium (Art. 57) and of resultant (Art. 55) it follows imme- diately that if a system of forces is in equilibrium the resultant is zero ; and conversely, if the resultant is zero the system is in equilibrium. This is the general condition of equilibrium. From it may be derived various special conditions, B adapted to the discussion of different classes of problems. 74. Geometrical Condition of Equi- librium. Force polygon. The figure formed by drawing in succession vec- tors representing any number of forces in magnitude and direction is called a force polygon for those forces. Thus, Fig. 25 shows a force polygon for five forces represented by the vectors AB, BC, CD, DE, EF, CONCURRENT FORCES. 37 In general the initial and final points (as A and F, Fig. 25) do not coincide. In case they do coincide the polygon is said to close. The order in which the vectors representing the forces are drawn does not affect the relative positions of the initial and final points. Condition of equilibrium. If A and F are the initial and final points of a force polygon drawn for any given forces, AF represents their resultant in magnitude and direction. Hence the resultant will be zero if F coincides with A> but not otherwise. Therefore, If any number of concurrent forces form a system in equi- librium y their force polygon is closed. And conversely, if the force polygon closes, the system is in equilibrium. 75. Moment-Condition of Equilibrium. Proposition. If any number of concurrent forces are in equilibrium, the algebraic sum of their moments with respect to any point in their plane is zero. It has been shown that the moment of the resultant, with respect to any origin, is equal to the algebraic sum of the moments of the components with respect to that origin. If the system is in equi- librium the resultant is zero, therefore its moment is zero whatever the origin; which proves the proposition. 76. Equations of Equilibrium. From the principles stated in the last two Articles, any number of equations may be written which must be satisfied by a system of concurrent forces in equilibrium. These equations are of two kinds: (a) The sum of the resolved parts of the given forces in any direction must be zero. For, since the force polygon must close for equilibrium, the alge- braic sum of the projections of its sides upon any line must equal zero. (b) The sum of the moments must be zero for any origin. Although each of these two general conditions leads to an infinite number of equations, only two of these can be independent. This will be seen by considering what any one equation implies. (1) If the sum of the resolved parts in any direction is zero, the resultant force, if one exists, must be perpendicular to that direction. (2) Jf the sum of the moments is zero with respect to any point, the resultant force, if one exists, must act in a line passing through that point. It is now evident that there will be equilibrium if either of the following conditions is satisfied: 38 THEORETICAL MECHANICS. I. The sum of the resolved parts of the forces is zero in each of two directions. II. The sum of the moments is zero with respect to each of two points not collinear with the point of application of the forces. III. The sum of the moments is zero for one origin (not coincid- ing with the point of application of the forces), and the sum of the resolved parts is zero for any one direction not perpendicular to the line joining the origin of moments with the point of application. 77. Algebraic Deduction of Conditions of Equilibrium. With the notation of Art. 69, we have X = P x cos a x -\- P 2 cos a , -f . . . , K==P l cos/8 I + />cos0, + . . . , R = XX 1 + Y\ The general condition of equilibrium, R = o, requires that X = o and Y == o. For unless X 2 and Y 1 are separately equal to zero, their sum cannot equal zero unless one of them is negative; but this would make X or Y imaginary. It follows that the condition ft == o is equivalent to the two conditions P x cos a x -j- P 2 cos a % + . . . = o ; . ( 1 ) P x cos j3 x + P t cos /3, + ... =0. . (2) Since the axes along which the forces are resolved may be any pair of rectangular axes in the plane of the forces, it follows that for equilibrium the sum of the resolved parts of the forces in any direc- tion must be zero. This agrees with Art. 76. It might be shown algebraically that if two equations such as (1) and (2) are satisfied, every equation obtained by resolving the forces in any direction or by taking their moments about any origin must also be satisfied. 78. Solution of Problems in Equilibrium. In solving prob- lems in equilibrium algebraically, it is necessary to write as many independent equations as there are unknown quantities to be deter- mined. From the above principles of equilibrium for concurrent forces, two independent equations can always be written. If the CONCURRENT FORCES. 39 number of unknown quantities is greater than two, the principles of equilibrium alone are not sufficient for their determination. In such a case the problem, if determinate, involves something more than the principles of equilibrium. From the preceding discussion it is evident that the two inde- pendent statical equations can always be written with certainty. If the solution of the problem presents difficulty it lies usually in the algebraic solution of the equations. In writing the two independent equations of equilibrium in either of the three allowable ways (Art. 76), simplification is usually pos- sible by taking advantage of the following obvious principles : (a) If the resolution is made in a direction perpendicular to a force, that force does not enter the resulting equation. (b) If moments are taken about an origin lying on the line of action of a force, that force does not enter the resulting equation. In many cases the ''geometrical condition of equilibrium," that the force polygon must close, leads to a short solution without the use of the equations of equilibrium. It is thus seen that any problem may be analyzed in two ways, geometrically and algebraically. In solving the examples that fol- low, the student is advised to use both methods. The geometrical method frequently has the advantage of giving a rapid solution, while it also gives a clear view of the relations of the forces. On the other hand, the algebraic method has the advantage of affording an exhaustive treatment applicable to any determinate problem. 79. Applications. The methods of applying the foregoing prin- ciples will be illustrated by the solution of the following problems. I. A body of 40 lbs. mass is suspended by a cord from a ring which is supported by two cords making angles of 30 and 70 with the vertical. Determine the tensions in the cords.* Two systems of forces in equilibrium are presented here, (a) two forces acting upon the suspended body (its weight of 40 lbs. directed downward and an upward force exerted by the string), and (b) three forces acting upon the ring (exerted by the three cords). The conditions of equilibrium are to be applied to each system sep- arately. * In all cases in which cords are introduced they are understood to be perfectly flexible. (Art. 43.) 4 THEORETICAL MECHANICS. (a) The upward force exerted by the cord CD upon the body D (Fig. 26) must equal the downward force of 40 lbs. due to gravity. (b) The cord CD must pull down- ward upon the ring C with a force equal to its upward pull upon D. Hence, of the three forces acting upon the ring, one is known completely and the direc- tions of the other two are known. Geometrical solution. The vectors representing the three forces applied to the ring must form a closed triangle. Draw LM(Fig. 26) vertically downward to represent the tension in CD, MN parallel to CB, LN parallel to AC', then MN and NL must represent the forces exerted upon the ring by the cords BC and AC respectively. From the given data, LM= 40 lbs., angle LMN= 30, angle MLN '- 70, angle LNM 8o. Hence, by trigonometry, Fig. 26. LN sin 3o c or LN LM sin 30 MN sin 7o c a 5 sin 8o 0.9848 LM sin 8o c X 40 lbs. = 20. 3 lbs. ; MN = LM sin 70 Q-9397 sin 8o ~~ 0.9848 X 40 lbs. = 38. 2 lbs. Algebraic solution. Let P and Q denote the tensions in AC and CB, and let the independent equations of equilibrium be formed by resolving forces horizontally and vertically. Then (taking up- ward and toward the right as positive directions) P cos 20 -f- Q cos 6o = o ; P cos 70 -j- Q cos 30 40 lbs. == o. Or 0.9397 P-f 0.5000 2 = o; o. 3420 P -f- o. 8660 Q = 40 lbs. Solving, P= 20.3 lbs., Q= 38.2 lbs. CONCURRENT FORCES. 41 II. A body of 60 lbs. mass rests against a smooth plane surface inclined 25 to the horizontal. It is supported partly by the plane and partly by a cord inclined 20 to the plane. What are the magni- tudes of the supporting forces ? ( Fig. The forces acting upon the body are ^^/^^^ three, all known in direction (the weight ^-^^^ acting vertically downward, the pull of ^^^^^5 the cord along its length, and the pres- ^T_ i sure of the smooth plane normally to its . Fig. 27. surface), and one known in magnitude. Geometrical solution. The triangle of forces can be drawn and the unknown forces determined as in the preceding problem. Algebraic solution. In writing the equations of equilibrium, it will be advantageous to resolve forces parallel and perpendicular to the plane. This gives an equation not involving the normal pressure. Let P = tension in cord, and Q = pressure exerted by plane. Resolving parallel to the plane, P cos 20 60 cos 6 5 = 0. Resolving perpendicularly to the plane, P cos 70 -{- Q 60 cos 25 = o. Solving, P = 27. o lbs. , Q = 45. 1 lbs. Examples. 1. A body of 20 lbs. mass resting upon a horizontal surface is acted upon by gravity and by a horizontal force of 5 lbs. The sup- porting body exerts upon it such a force as to hold it at rest. Determine the magnitude and direction of this force. Ans. 20.6 lbs., inclined 14 2 to the normal. 2. A body of 120 lbs. mass, suspended by a flexible string, is pulled horizontally with a force of 45 lbs. Determine the direction of the suspending cord and the tension sus- tained by it. (Fig. 28.) 3. A body of 70 lbs. mass is suspended by a flexible cord from a ring which is supported by two cords making angles of 25 and 6o respectively with the vertical. Determine the > tensions in the strings. Fig. 28. 4. A body of P lbs. mass is suspended by 42 THEORETICAL MECHANICS. a cord from a ring which is supported by two cords making angles a and fi with the vertical. Determine the tensions in the cords. Ans. Psm a /sin (a -\- P) an d P sin /3/sin (a -f /3). 5. A body of 70 lbs. mass is suspended as in Ex. 3, and is also pulled horizontally by means of an at- tached cord with a force of 10 lbs. De- termine the tension in each cord. (Fig. 29.) [This presents two sets of concurrent forces, one set acting on the suspended body, the other acting on the ring.] 6. If the horizontal force in Ex. 5 be gradually increased, one of the cords will finally cease to act. For what value of the horizontal force will this occur ? In this limiting case, what tensions are sustained by the cords ? Ans. Horizontal force = 121.24 lbs. Tensions =140 lbs. and o. 7. In Fig. 30 the cords a, b, c and d are inclined to the vertical at angles of 70, 30, 20 and o respectively, while f y g and h are horizontal. The mass of the body P is 60 lbs. , and the horizontal force exerted by the cord h is 20 lbs. Determine the tension in each of the cords. [Apply conditions of equilibrium separately to the forces concurrent at Z, at M, at N and at P, beginning with P.] Ans. Tension in a = 49.7 lbs. ; in c, 63.8 lbs.; in g, 20 lbs. 80. Equilibrium of a Particle on a Smooth Surface. If a particle rests against a smooth surface of any form, the pressure ex- erted upon the particle by the surface is a "passive resistance" (Art. 41) whose direction is known, being that of the normal to the surface. If the surface is a plane, the direction of the pressure is the same whatever the position of the particle ; but for a curved surface the direction of the pressure exerted by the surface depends upon the position. To solve a problem relating to the equilibrium of a particle on a smooth surface, the general equations of equilibrium may be written in the usual manner ; but among the forces to be included in the CONCURRENT FORCES. 43 o system is the unknown pressure due to the surface. The direction of this reaction is known if the position of the particle is known ; otherwise it must be expressed in terms of the variable coordinates of the surface. The consideration of this problem will be confined to the case in which the particle and the ap- plied forces are restricted to a plane curve lying in the given surface. Let the particle be restricted to the plane curve shown in Fig. 31, and let the equation of this curve re- ferred to rectangular axes OX, OY be known. If N denotes the mag- nitude of the unknown normal pressure acting on the particle, and cf> the angle between the ^r-axis and the direction of N> we have ^-component of JV iVcos ; j-component of N = N sin (/>. But since N has the direction of the normal to the curve, dy_ dyjdx Fig. 31. COS (j) sin cj> ds Vi 4. (dy/dxf dx 1 ds 1 7 1 -f- (dy/dx) 1 The value of dyjdx in terms of x and y can be found from the equa- tion of the curve. In many cases the direction-angles of N can be expressed more conveniently in some other manner. The method of solving problems of this kind will now be illus- trated. I. A smooth wire bent into the form of a circle and placed with a diameter vertical carries a bead of W lbs. mass, to which is applied a horizontal force of P lbs. In what position can the bead be in equilibrium ? Algebraic solution. Let the radius drawn to the bead make with the vertical an angle 0, and let N denote the pressure of the wire upon the bead, its direction being radial. (Fig. 32.) 44 THEORETICAL MECHANICS. Resolving forces horizontally, P N sin = o. Resolving vertically, WNcos0 = o. Solving these equations for the unknown quantities N and 0, we have tan0 =P/W; Fig. 32. N = P/sin 6 = V P* + W\ Geometrical solution. The par- ticle is in equilibrium under the action of three forces, P, W and N. Two of these (P and W") are known ; hence the force triangle can be drawn completely, thus determining the mag- nitude and direction of N. From the triangle it is evident that N acts toward the center ; that its magni- tude is VP 2 + W 2 ; and that tan = P/W. II. A particle whose mass is Wlbs. rests upon a smooth hori- zontal plane ; to it are attached a string AB passing over a smooth peg at B and sustaining a body of mass P lbs. , and a string A C passing over a smooth peg at C and sustaining a body of mass Q lbs. Required the direction of AC for equi- librium. Algebraic solution. The particle is acted upon by four forces : its weight, W lbs. ; a force of P lbs. due to the tension in the string AB ; a force of Q lbs. due to the tension in the string A C ; and the normal pressure N due to the plane. Let denote the angle between A C and the horizontal. Re- solving forces horizontally and vertically, P + Q cos = o ; N W + Q sin = o. CONCURRENT FORCES. 45 Solving for N and 0, cos 6 = P/Q ; N= WVQ* P\ The geometrical solution by the force triangle is obvious. III. To determine the position of equilibrium of a body resting on the surface of a smooth elliptic cylinder and acted upon by a force of known magnitude directed along the tangent to the ellipse. Let the mass of the body be W lbs. , and let the tangential force be due to the tension in a string which passes over a smooth peg and sustains a body of known mass P lbs., as in Fig. 34. The body is acted upon by three forces : its weight W lbs. acting vertically downward ; the pull of the cord, its magnitude being P lbs. and its direction that of the tangent to the ellipse ; and the normal pres- sure exerted by the smooth surface, its magnitude N being unknown. If (f> denotes the angle between the normal and the axis of x, measured as shown in the figure, the equations obtained by resolving along the tangent and normal are Fig. 34. From (1), From (2), W cos P=o; N W sin (j> = o. cos = PI W. N= Wmt=y W* P x (0 (2) (3) The coordinates of the point for which cos = - ds The equation of the ellipse is V 1 -f {dyfdxf y = I. (4) 4 6 THEORETICAL MECHANICS. Differentiating, dy dx 2 ' ay which substituted in the value of cos gives & 2 x P cos = - = . . (5) Va'y' 1 + b\x l W The coordinates of the position of equilibrium must satisfy equations (4) and (5). Solving, ?. / P l a l . ^ P' l a l -I- (#" /") ,; = /~ (H^ 2 -P 2 ) 2 ^ >/> 2 tf 2 + (W* P*)6 2 ' Examples. 1. A body of Wlbs. mass, resting on a smooth plane surface inclined at angle 6 to the horizontal, is supported by the pressure of the plane and another force acting at angle a with the plane. De- termine the supporting forces. Ans. Normal pressure = W cos (6 -\- a) /cos a. 2. In what direction must a force of P lbs. be applied to a body whose mass is W lbs. to hold it at rest on a smooth plane inclined at angle a to the horizontal ? Ans. If 6 = angle between Pand the plane, cos 6 = ( IV/P) sin a. 3. A certain body may be supported on a smooth inclined plane by a force P acting at an angle a with the horizontal, or by a force Q acting at an angle /3 with the horizontal. Required the mass W of the body and the inclination 6 of the plane to the horizontal. Ans.X*ne = Pc0%a - Qc0sli . W= PQ * W ~ a) . Q sin j3 P sin a P cos a (2 cos /3 4. A certain body may be supported on a smooth inclined plane by a horizontal force of 10 lbs., or by a force of 8 lbs. acting along the plane. Required the mass of the body and the inclination of the plane. Ans. 13^ lbs.; 36 52'. 5. A body whose mass is 10 lbs. is supported on a smooth in- clined plane by a force of 2 lbs. acting along the plane and a hori- zontal force of 5 lbs. Determine the inclination of the plane. 6. A smooth wire, bent into the form of a parabola with axis ver- tical and vertex downward, carries a bead of mass W lbs. which is CONCURRENT FORCES. 47 pulled horizontally by a force of P lbs. Determine the position of equilibrium. Ans. If the equation of the parabola is x 2 = qmy, the coordi- nates of the position of equilibrium are x = (2PI W)m, y = (pyw^m. 7. A heavy bead is placed upon a smooth circular wire in a vertical plane, and is pulled by a string which passes over a smooth pulley at the highest point of the circle and carries a body of known weight. Find the position of equilibrium. 8. Two particles whose masses are P lbs. and Q lbs. are con- nected by a flexible string of length / which passes over a smooth circular cylinder of diameter d with axis horizontal. Required the position of equilibrium. (Fig. 35.) [Apply the conditions of equilibrium to each particle separately, remembering that the tension in the string is uniform throughout its length, so that it exerts equal forces on the two particles. Both particles, or only one of them, may be in contact with the cylinder in the posi- tion of equilibrium ; these two cases must be treated separately.] 9. In the case described in Ex. 8, let the masses of the particles be 2 lbs. and 3 lbs. , the length of the string 1 ft. , and the diameter of the cylinder 18 ins. Determine the position of equilibrium, the tension in the string, and the reac- tions exerted on the particles by the cylinder. 10. A string attached to a fixed point A and passing over a smooth peg B, carries at the free end a body of known weight P, and a body of weight W is suspended at a point C by means of another string. The length A C and the positions A and B being known, it is required to determine the position of equilibrium. (Fig. 36.) [The solution involves an equation of the third degree.] 1 1. In Ex. 10, let the weight W be suspended from a smooth ring sliding on the string A CB. Determine the position of equilibrium. 12. Given three concurrent forces of magnitudes P y Q, R, the angles between P and Q, Q and R, R and P, respectively, being r t p y q. Prove that the square of the resultant is equal to P' 1 + Q*+ R 2 + 2PQ cos r + 2QR cos/ + 2RPcos q. Fig. 35. WO Fig. 36. THEORETICAL MECHANICS. 13. AB and AC (Fig. 37) are smooth rods lying in a vertical plane. Heavy rings of known masses P lbs. and Q lbs. slide upon the rods, being connected by a flexible string MN. Required the inclination of MN to the horizontal in the position of equilibrium. 14. Two particles whose masses are P lbs. and Q lbs. respectively rest against smooth inclined planes whose inclinations to the horizontal are a and /3. The particles are connected by a flexible cord which passes over a smooth peg placed vertically above If/ is Fig. 37. the intersection of the planes, the length of the cord and h the ver- tical distance of the peg above the in- tersection of the planes, prove that the angles 6 and made by the cord with the planes in the position of equilibrium are determined by the equations p sin a n sin /3 cos a cos j3 __/ cos cos (j) sin 6 sin h 15. Determine the magnitude and direction of the least force that will sustain a particle of mass W kilograms on a smooth plane inclined at an angle a with the horizon. Com- pute results if W = 45 and a = 36 . 16. A particle on a smooth plane inclined at angle a to the horizon is acted on by a force directed toward a fixed point and varying inversely as the square of the distance from that point. De- termine the position of equilibrium. 17. Two smooth rings of 14 lbs. and 18 lbs., connected by a flex- ible string 2 ft. long, slide on a smooth vertical circular wire of radius 5 ft. Determine the position of equilibrium. Arts. Angle between string and horizontal = i 28'. CHAPTER IV. COMPOSITION AND RESOLUTION OF NON-CONCURRENT FORCES IN THE SAME PLANE. i. Two Non-Concurrent Forces. 81. Rigid Body. In the discussions which follow, the systems of forces are in most cases regarded as applied to a rigid body. A rigid body may be defined as one whose particles do not change their positions relative to one another under any applied forces. No known body satisfies this condition strictly, even for forces of small magnitude, but in the solution of problems of practical importance most solid bodies may, with slight error, be regarded as rigid. After a body has assumed a form of equilibrium under applied forces it may, in applying the principles of Statics, be treated as a rigid body without error. 82. Change of Point of Application. The effect of a force upon the motion of a rigid body will be the same, at whatever point in its line of action it is applied, if the particle upon which it acts is rigidly connected with the body. This proposition, which is amply justified by experience, is funda- mental to the development of the principles of Statics. In applying the principle, we are at liberty to assume a point of application outside the actual body, the latter being ideally extended to any desired limits. 83. Collinear Forces. Two forces having the same line of action, applied to the same rigid body at any points of that line, may be combined as if they were concurrent ; since by Art. 82 both may be treated as if applied at any one point in their common line of action. As a particular case, two forces applied to a rigid body balance each other if they are equal in magnitude, opposite in direction, and collinear. Hence 'such a pair of forces may be introduced into a system without changing its effect. 84. Resultant of Two Non-Parallel Forces. If two coplanar forces are not parallel, their lines of action must intersect, and the 4 50 THEORETICAL MECHANICS. point of intersection may be treated as their common point of appli- cation. They may therefore be treated as concurrent forces, and their resultant may be determined as in Art. 59. Hence the following proposition may be stated: The resultant of two non-parallel forces acting in the same plane on a rigid body is a force equal to their vector sum, and its line of action passes through the point of intersection of the lines of action of the given forces. Its point of application may be any point of this line. Thus, if two forces are applied to a body at points M and N (Fig. 39), their lines of action being MS and NT, each may be assumed to act at R, the point of inter- section of these lines. The resultant must therefore be a force which may be taken as acting at R ; that is, its line of action must pass through R. When the magnitude and direction of the resultant have been found by con- structing the vector triangle, the line of action becomes known, and any point of this line may be regarded as the point of application. By an extension of the above reasoning, any number of forces whose lines of action meet in a point may be treated as if that point were 7> their common point of application; in y'l \ other words, as if they were concur- / , rent. This is true even though the / j point of intersection of the lines of 'aJl. / \C action falls outside the limits of the A ^5 V body. Thus, if the forces P lt P ti P 3 /p 1 p\ are applied to the bar AC (Fig. 40) * O at points A, B and C, their direc- Fig. 40. tions being such that their lines of action intersect at a point D, their effect is the same as if all were applied at D } the latter point being regarded as rigidly connected with the bar. 85. Resolution of a Force Into Two Non-Parallel Compo- nents. Let M (Fig. 41) be the point of application of a force, and MS its line of action. Any point in this line, as R, may be COMPOSITION OF NON-CONCURRENT FORCES. 51 regarded as the point of application, and the force may be replaced by- two components acting at R. The magnitudes and directions of the two components must be such that their vector sum is equal to the given force. 86. Resultant of Two Parallel Forces. In determining the resultant of two parallel forces, two cases must be considered, according as the forces act in the same direction or in opposite directions. (1) Forces having the same direc- tion. Let P and Q denote the magnitudes of two forces having the same direction, A and B (Fig. 42) being any points in their lines of action. Let two forces, each of magnitude F, be applied Fig. 41. tR m Fig. 42. to the body, acting in opposite directions along the line AB; call these forces F' and F" . The system of four forces (P, Q, F\ F") is equivalent to the given system of two forces (P, Q) t since F' and F" balance each other. Let the force F' be com- bined with P (giving a resultant R) and the force F" with Q (giving a resultant R"). The line of action of R' passes through A, and its direction is that of the diagonal of a parallelogram with sides drawn from A, proportional and parallel to P and F. The line of action of R" passes through B, and is found by construct- ing a parallelogram with sides drawn from B, proportional and parallel to Q and F. The forces R and R' may be regarded as applied at D, the intersection of their lines of action ; R is therefore 52 THEORETICAL MECHANICS. equivalent to two forces equal and parallel to P and F' applied at D, and R" is equivalent to two forces equal and parallel to Q and F" applied at D; these four forces applied at D being thus equiv- alent to the given forces P and Q. But since the forces F' and F'' balance each other, the system is equivalent to two collinear forces of magnitudes P and Q, or to a single force P -f Q, acting at D parallel to the given forces. This single force is therefore the resultant of the two given forces. It remains to determine the position of the line of action of this resultant relative to the lines of action of the given forces. Let C be the point in which this line intersects AB; then by similar tri- angles, F/P = AC/ CD, and F/Q = CB/CD. Combining these two equations, AC/CB -= Q/P; that is, the line of action of the resultant divides AB into segments inversely proportional to Pa.nd Q. (2) Forces having opposite directions. The case in which P and Q have opposite directions is represented in Fig. 43 ; the reasoning r; p /b D R Fig. 43. employed in the preceding case applies equally to this, the two fig- ures being lettered in a corresponding manner. The magnitude of the resultant is found to be the difference (or algebraic sum) of P and Q. Also, the point D and therefore the line CD will fall outside the space included between the lines of action of P and Q and on the side of the greater of these forces. In this case the point C divides AB externally into segments inversely proportional to P and Q. COMPOSITION OF NON-CONCURRENT FORCES. 53 The examples which follow are designed to illustrate the method of finding the resultant of two parallel forces, or of finding two par- allel forces that are equivalent to a single force. The reader who is already acquainted with the principles of equilibrium (to be developed later) will notice that several of the examples can be solved by means of these principles. It is, however, desirable that they should be analyzed by the principles of composition and resolution of forces already developed; the resultant of two forces being regarded as a single force which may replace them without affecting the state of the body as regards rest or motion. Thus, in example 3 of the following list, the bar AB may be acted upon by any number of forces not specified; these, together with the two upward forces exerted by the strings, maintain the bar in its condition of rest. Now suppose the action of these two forces ceases; what single force may be applied with the same result ? Examples. 1. Parallel forces of 40 lbs. and 65 lbs. acting in the same direc- tion are applied at the ends of a bar 1 2 ft. long. Find the magnitude, direction and line of action of a single force which would produce the same effect. 2. Assuming the two forces in the preceding example to act in opposite directions, determine their resultant completely. 3. A heavy bar AB (Fig. 44), 8 ft. long, is supported by vertical cords attached at A and B, which pass over smooth pegs at C and D Q C D and sustain bodies P and Q of masses 100 lbs. and 60 lbs. respectively. If the two supporting cords are replaced by a single cord, at what point must it be attached, and what force must it exert? D? QO 4. A heavy bar AB, 12 ft. long, Fig. 44. is found to balance on a smooth sup- port at C, 7.5 ft. from one end, the upward pressure exerted by the support being 48 lbs. If the same bar rests on two smooth supports at the ends, what will be the supporting forces ? 5. A heavy bar 18 ft. long is supported at the ends by equal vertical forces of 40 lbs. If the supports are moved to points 4 ft. from one end and 6 ft. from the other respectively, what are the sup- porting forces ? 54 THEORETICAL MECHANICS. 6. The bar AB (Fig. 45) carries a load P of 20 lbs. hanging by a string from the end A, and is supported by a string attached at C, 4 ft. from A, the tension of the supporting string being found by a spring balance to be 40 lbs. If the load P is removed, at what point must the string be attached in order to support the bar ? What tension will B it sustain ? I Ans. 8 ft. from A; 20 lbs. 7. It is found that a bar AB, 12 r 1 p ft. long, may be supported by a force ' of 18 lbs. applied at the middle point. IG * 45 * If it rests in a horizontal position against smooth supports at A and C, 4 ft. apart, the support at A being above and that at C below the bar, what will be the supporting forces ? Ans. 27 lbs. upward at C, 9 lbs. downward at A. 87. Two Equal and Opposite Forces. If two forces P and Q are equal in magnitude, opposite in direction, and have different lines of action, the construction above given (Art. 86) for finding their resultant fails, for the reason that the two lines AD and BD (Fig. 43) become parallel. If it is attempted to apply the general rule for determining the resultant of two opposite forces, it is found that the magnitude of the resultant P Q is zero ; while the equation for determining its line of action becomes (see Fig. 43) AC/CB= Q/P= 1, which can be true only if AC and CB are infinite. Mathematically interpreted, these results mean that the resultant is a force of magni- tude zero, acting in a line infinitely distant from the lines of action of the given forces. Evidently P and Q in this case form a couple (Art. 53). A fur- ther discussion of couples will be given later. 88. Moment of Resultant of Two Forces. Proposition. The algebraic sum of the moments of any two coplanar forces with respect to a point in their plane is equal to the moment of their resultant with respect to that point. In the proof of this proposition two cases must be considered, according as the two given forces are or are not parallel. {a) Non-parallel forces. For two non-parallel forces, the prop- COMPOSITION OF NON-CONCURRENT FORCES. 55 >R Q O osition just stated is a special case of that proved in Art. 29 for any- localized vectors. For the resultant of two non-parallel forces is a force equal to their vector sum, acting in a line through the intersection of their lines of action.* (fr) Parallel forces. Let P and Q represent the magnitudes of any two parallel forces and R that of their re- sultant. Take any point O (Fig. 46) as origin of moments, and draw through O a line perpendicular to the forces, intersecting their lines of action at A, B and C respectively. Let OA = /, OB = q } OC = r. From Art. 86 we have P _BC _ rq Q AC p-r Reducing, Pp + Qq = (P + Q)r = Rr. Fig. 46. If the forces P and Q have opposite directions, the demonstra- tion needs modification. (See Fig. 47.) The equation is P Q BC AC P Pp-Qq=(P-Q)r=Rr. In both cases, the reasoning is easily adapted to the case in which the origin of moments falls between the lines of action of the two forces P and Q. In all cases, the re- sult is expressed by the proposition y(R Py Qj^ that the moment of the resultant of two parallel forces is equal to the algebraic sum of their moments. The proof of the proposition for Fig. 47. parallel forces may be put in another form, as follows : Referring to Art. 86, let any point (Figs. 42 and 43) be taken as origin of moments. Having proved that the moment of the resultant of two non-parallel * Any two non-parallel forces may, in fact, as already shown, be treated as if concurrent, so that this case reduces to that treated in Art. 71. 56 THEORETICAL MECHANICS. forces is equal to the algebraic sum of their separate moments, we may write the following equations: mom. of R ' = mom. of P -j- mom. of F' ; mom. of R" = mom. of Q -\- mom. of F" ; mom. of R = mom of R' -j- mom. of R" = mom. of P -f- mom. of Q -f- mom. ofi**' -f- mom. of ^ = mom. of P -f- mom. of (2 (since the moments of /^' and i 7 " are equal in magnitude but oppo- site in sign). 89. Moment of a Couple. The proposition just proved is mean- ingless when applied to the case of two equal and opposite forces, since two such forces are not equivalent to a single resultant force. The moment of a couple will therefore be made a matter of defini- tion, as follows : The moment of a couple is the algebraic sum of the moments of its two forces. With this definition it may readily be shown that the moment of a couple has the same value for every origin in its plane, and is equal to the product of the magnitude of either force into the perpen- dicular distance between the lines of action. 2. Couples. 90. Equivalent Couples in the Same Plane. Any two couples in the same plane are equivalent if their moments are equal in mag- nitude and sign. To prove this, consider first the resultant of two couples whose moments are equal in magnitude but opposite in sign. Let P de- note the magnitude of the forces of one couple and p the length of the arm or perpendicular distance between the lines of action ; and let Q, q denote the like quantities for the second couple, these quan- tities being so related that Pp = Of Let the lines of action and directions of the four forces be as shown COMPOSITION OF NON-CONCURRENT FORCES. 57 in Fig. 48, so that the moments have opposite signs. It may be shown that their resultant is zero. Notice first that the lengths of the sides of the parallelogram Fig. 48. ABCD formed by the four lines of action are proportional to P and Q. For area ABCD = (AB) q = (AD) -p ; and since Pp = Qq y AB/AD = Q/P. The resultant of the force P acting along AD and the force Q acting along AB must therefore act along the diagonal AC of the parallelogram ABCD; and the resultant of the force P acting along CB and the force Q acting along CD must also act along the diag- onal CA. But these resultants are equal, opposite, and collinear; hence their resultant is zero. The resultant of the two couples is therefore zero. Next, starting with the given couple (P, p) } let two pairs of equal and opposite forces of magnitude Q be assumed to act in the lines AB, CD (Fig. 49), so taken that Qq = Pp. These four forces form two couples ; one of these counterbalances the given couple as shown above, hence the other must be equivalent to the given couple. That is, the given couple (P, p) is equivalent to the couple (Q, q) whose moment is equal to its own in magnitude and sign. Since the lines AB and DC may be any two parallel lines inter- secting the lines of action of the given couple, it follows that any two 58 THEORETICAL MECHANICS. coplanar couples whose forces are not parallel are equivalent if their moments are algebraically equal. But two couples whose forces are parallel and moments equal are Fig. 49. k2P *b yftP py-. equivalent, because each may be proved equivalent to a third couple ; so that the proposition holds for any coplanar couples. 91. Couples in Parallel Planes. Couples in parallel planes are equivalent if their moments are equal in magnitude and sign. Let AB (Fig. 50) be a line perpendicular to the forces of a couple, intersecting their lines of action in A and B. Let CD be any line equal and parallel to AB, not &P necessarily in the plane of the couple. Introduce at C two equal and opposite forces, each equal and parallel to P, the force of the given couple ; also two similar forces at D. The force at A and the upward force at D may be replaced by an upward force 2 P acting Y^> \fp at the middle point of AD. The force Fig. 50. at B and the downward force at C may be replaced by a downward force iP acting at the middle point of CB. These two forces 2P balance each other, leaving two forces, an upward force at C and a down- ward force at D, each equal to P. These form a couple equivalent to the given couple. Since the couple acting at 7 and D is equivalent to any couple in the same plane having an equal moment; and since CD might be 2P 4P D COMPOSITION OF NON-CONCURRENT FORCES. 59 taken in any plane parallel to that of the given couple, it follows that any two couples in parallel planes are equivalent if their mo- ments are equal. 92. Equivalence of Couples General Result. The results of the last two Articles may be stated in one general proposition, as follows : Any two couples in the same plane or in parallel planes are equivalent if their moments are equal in magnitude and sign. 93. Resultant of Coplanar Couples. The resultant of any num- ber of coplanar couples is a couple whose moment is equal to the algebraic sum of the moments of the given couples. Choose any two parallel lines in the plane of the couples, as MN, M'N' (Fig. 51), and let each of the given couples be replaced by an equivalent couple with forces acting in these lines. Let the common arm of these substituted couples be /, and let their forces be P x , P. 2 , P 6 , etc., their values being either pos- itive or negative according to the direc- tions of the forces, and being such that P x p, P 2 p, . . . are algebraically equal to the moments of the given couples. The resultant of the forces acting in MN is a force equal to their algebraic sum P x + P while the resultant of the forces acting in M'N' has the same magnitude but the opposite direction. Hence the whole system is equivalent to a couple of moment (/> + />+/>,+ . . . )p, which is equal to Pit + P*P + P*f+ , that is, to the algebraic sum of the moments of the given couples. The proposition obviously holds for couples lying in any parallel planes. 94. Resultant of Force and Couple. A force and a couple acting in the same plane or in parallel planes are equivalent to a sin- gle force. Let P be the magnitude of the given force and G the moment of 60 THEORETICAL MECHANICS. the given couple. Replace the couple by an equivalent couple in a plane parallel to its own, containing the given force P. The magni- tude, direction and line of action of one of the forces of this equiva- lent couple may be chosen arbitrarily; let them be so chosen that this force counterbalances the given force P. The other force of the couple must be equal in magnitude and direction to P, and its line of action must be such that the moment of the couple is algebraically equal to G; its line of action is therefore at a distance G/P from that of the given force. This last force is the resultant of the given force and couple. Examples. i. A force of 50 lbs. acting in any assumed line, and a couple coplanar with it whose arm is 29 ft. and whose forces have the magnitude 10 lbs., have what resultant? A g 2. A bar AB, resting horizontally upon a smooth support at B, is held in p! equilibrium by two opposite horizontal forces applied at the end A in lines 6 * 1G ' 52, ins. apart. The supporting force at B is 5 lbs. and the horizontal forces are each 50 lbs. If the horizontal forces cease to act and the support is removed, what single force will support the bar in equilibrium ? (Fig- 52.) 3. Any Number of Coplanar Forces. 95. Resultant of Any Number of Forces. It may now be shown that the resultant of any number of coplanar forces is a single force equal to their vector sum, unless this sum is zero; in which case the resultant is a couple. Of three or more forces, there must be two which do not form a couple. Any such pair may be replaced by their resultant, which is a single force equal to their vector sum (Arts. 84, 86), thus reduc- ing the number of forces by one. This process may be repeated until the system has been reduced to an equivalent system of two forces. If these two are not equal and opposite, they are equivalent to a single force equal to their vector sum and therefore to the vector sum of the given forces. If they are equal and opposite, they form a couple which is the resultant of the given system; the vector sum of all the forces being zero. COMPOSITION OF NON-CONCURRENT FORCES. 6 1 96. Moment of Resultant of Any Number of Forces. In a similar manner the principle of moments, already proved for two forces, may be extended to any number of coplanar forces. That is, it may be shown that the sum of the moments of any number of co- planar forces about any origin in their plane is equal to the moment of their resultant about that origin. Combining two of the forces which are not equal and opposite, the sum of their moments is (by Art. 88) equal to the moment of their resultant. Combining this resultant with another force not equal and opposite to it, the sum of their moments is equal to the moment of their resultant. This process may be continued until the number of forces is reduced to two. If these two are not equal and opposite, the sum of their moments (equal to the sum of the mo- ments of the given forces) is equal to the moment of the single force which is their resultant and the resultant of the given system. If the two forces are equal and opposite, they form a couple which is the resultant of the given system, and whose moment is equal to the sum of the moments of the given forces. Hence the proposition is proved. 97. Computation of Resultant. From the foregoing principles, the resultant of any system of coplanar forces may be computed, whether this resultant be a single force or a couple. (1) If the resultant is a single force, its magnitude and direction may be found as if the forces were concurrent (Art. 65 or 69) ; the position of its line of action must be such that its moment about any assumed point is equal to the sum of the moments of the given forces about that point. Thus, if P denotes the magnitude of the resultant force, and G the sum of the moments of the given forces, the line of action of the resultant must be at a distance G/P from the origin of moments. This condition, together with the requirement that the sign of the moment of the resultant is the same as that of the sum of the moments of the given forces, completely determines the line of action of the resultant. (2) If the vector sum of the given forces is zero, the resultant of the corresponding concurrent system will be zero. The resultant of the given system is then a couple, whose moment may be found by computing the sum of the moments of the given forces about any point. The value of this moment will be the same, whatever point be chosen as origin. 62 THEORETICAL MECHANICS. Let the following examples be solved by the direct application of these principles. Examples. [Let the magnitude, direction and point of application of a force be specified by the following notation : x, y denote the rectangular coordinates of its point of application ; a the angle its direction makes with the positive direction of the^r-axis; Pits magnitude.] i. Find the resultant of the following forces: P x = 20 lbs., X x = 2 ft., y x =55 6-ft., a x = o; P 2 = s o" x 2 = 3 " 7 2 = 7" a 2 =i8o; P 3 = So " *,= 5" 7 3 = 7" * = 90- 2. Find the resultant of a system of parallel forces whose magni- tudes and directions are 10 lbs., 24 lbs., 15 lbs., 3 lbs., 48 lbs.; the successive distances between their lines of action being 2 ft. , 3 ft. , 8 ft, 7 ft. 3. Find the resultant of the following forces: P x = 40 lbs. , x x = 4 ft., /i = 6 ft., a x = 0; P 2 =2 7 " *i = 2 " 72 = 14 - a 2 = 180; ^3=13 " x* = 6 M 73 = 8 i( a 3 = 180; P<=i6 " x i = 10 " 74 = " a, = 90; P 5 =20 " *5 = 6 (< 75 = 2 " 5= 90; P 6 =36 " x 6 = 12 4< 7e = 3 " a 6 = 270. 98. Reduction to Force and Couple. The general results above found regarding the resultant of any coplanar forces may be deduced by a somewhat different line of reason- ing, as follows: Let P u P 2 , . . . , represent the magnitudes of any coplanar forces applied to a rigid body. Choose any point O in the plane of the forces (Fig. 53), and let the perpendicular distances from to the lines of action of the sev- J / eral forces be p lt p. iy etc. Consider first some one force as P x , Suppose Fig. 53. two equal and opposite forces, parallel to P x and equal to it in magnitude, to be applied to the body at 0. One of these forms, with the given force, a couple of moment P x p x . Hence the given force P x is equiv- alent to a couple of moment P x p x and a force equal and parallel to P x applied at O. If this process is repeated for each of the given COMPOSITION OF NON-CONCURRENT FORCES. 63 forces, it is seen that the given system is equivalent to the following two systems: (1) A system of concurrent forces, applied at O, equal and par- allel to P x , P t , .... (2) A system of couples whose moments are P x p x , P 2 p 2 , . . . . The resultant of this system of concurrent forces may be found as in Art. 69, while the couples may be combined as in Art. 93. Hence the following proposition: Any system of coplanar forces is equivalent to a single force equal to their vector sum, applied at any chosen point; together with a couple whose moment is the algebraic sum of the moments of the given forces with respect to that point. 99. Computation of the Force and Couple. Let the point O at which the concurrent forces are assumed to act be taken as origin of coordinates, and let the angles made with the x- and jj/-axes by any force P be denoted by a, /3, with proper suffix. Let the force and couple to which the system is reduced be speci- fied by the following notation: R = magnitude of force; a = angle between R and ^r-axis; b = angle between R and j/-axis; X y Y = axial components of R ; G = moment of couple. Then X = R cos a P x cos a x + P* cos a 2 + Y = R cos b = P x cos & + P 2 cos & + . . . (0 R = VX 2 +..y; (2) cos a = XI R ; cos b = Y/R (3) Also, by Art. 98, ff/VA + /^%+ .'>. ' <4) The values of R, a and b completely determine the required force ; and the value of G determines the moment of the couple. It is seen that the magnitude and direction of the force R are the same wherever the point O be taken; while in general the value of G depends upon the position of the point at which the concurrent forces are assumed to act. 64 THEORETICAL MECHANICS. Let the following examples be solved by this process. Examples. 1. Find a force and couple equivalent to the following forces: P x = 70 lbs., x x = 4 ft., y x = 8. ft., a l = 6o; P 2 = 4 o " x, = 6 ' y t = 6 " ,= 45 ; P. A =25 " x 3 = o " y 3 10 " a 3 =i2o\ 2. Reduce each of the systems of forces given in examples 1 and 3, Art. 97, to a couple and a force applied at the origin of coordinates. 100. Resultant Force or Resultant Couple. The force and couple to which any system may be reduced by the above process may in general be combined into a simpler resultant. Since in par- ticular cases one or both the quantities R and G determined by the above process may reduce to zero, the following four cases must be considered : (1) Suppose neither R nor G is equal to zero. In this case the force and couple may be combined into a single force (Art. 94) hav- ing the same magnitude and direction as R; its line of action being distant G/R from O. (2) Suppose G = o. In this case the whole system reduces to the single force R, its line of action passing through the assumed point O. [This case always results if the given system is equivalent to a single resultant force, and if the point happens to be chosen upon its line of action. ] (3) Let R = o. In this case the resultant of the given system is a couple whose moment is G. (4) Let R = o, G = o. In this case the given system is equiv- alent to no force. In other words the given forces exactly balance each other and the system is in equilibrium.* Examples. In each of the examples of Art. 99, determine completely the resultant force or resultant couple. * In the last three Articles we have reached by means of algebraic analysis the same results deduced previously by geometrical reasoning. The student will find it profitable to become familiar with both methods of treatment. CHAPTER V. EQUILIBRIUM OF COPLANAR FORCES. i. General Principles of Equilibrium. 101. Meaning of Equilibrium. The word equilibrium is used with reference both to forces and to bodies. (See Art. 57.) Any number of forces form a system in equilibrium if their com- bined action produces no effect upon the motion of the body to which they are applied. A rigid body is in equilibrium if all external forces acting upon it form a system in equilibrium. 102. General Condition of Equilibrium. It follows from the definition of equilibrium that the necessary and sufficient condition of equilibrium for any system of forces is that their resultant is zero. This general condition implies subordinate conditions which are now to be considered. These subordinate conditions may be deduced in two ways : First, by the direct application of the principles regarding the re- sultant deduced in Arts. 95 and 96 ; or second, by applying the re- sults of the algebraic discussion of Arts. 98-100. It will be useful to consider both methods. 103. Equations of Equilibrium. It has been shown (Art. 95) that the resultant is either a force or a couple. If a force, it is equal to the vector sum of the given forces, and the resolved part of the resultant in any chosen direction must equal the algebraic sum of the resolved parts of the given forces in that direction. If the re- sultant is a couple, the vector sum of the forces is zero, and the algebraic sum of their resolved parts in any direction is zero. In either case, the sum of the moments of the given forces about any origin is equal to the moment of the resultant (Art. 96). Hence, if the resultant is zero, the following conditions must be satisfied: (1) The sum of the resolved parts of the given forces in any direction is zero. (2) The sum of the moments of the given forces about any point is zero. 66 THEORETICAL MECHANICS. From these two principles, an infinite number of equations may be written ; since the forces may be resolved in any direction, and any point may be taken as origin of moments. It may be shown, however, that only three of these equations can be independent. 104. Three Independent Equations. In order to determine how many of the possible equations of equilibrium can be indepen- dent, consider how much is implied by any one of them. (a) If the sum of the resolved parts of the forces is zero for any given direction of resolution, the resultant force, if one exists, must be perpendicular to that direction. There may, however, be a re- sultant couple. (d) If the sum of the moments is zero for any chosen origin, there can be no resultant couple, since the moment of a couple is not zero for any origin ; if there is a resultant force, its line of action must pass through the origin. These principles lead immediately to the following propositions: (1) There will be equilibrium if the sum of the resolved parts is zero for each of two directions of resolution, and the sum of the mo- ments is zero for one origin. For, by principle (a), there can be no resultant force, since such a force would need to be perpendicular to both directions of resolu- tion; and by principle (J?) there can be no resultant couple, since the moment of a couple cannot be zero. (2) There will be equilibrium if the sum of the moments is zero for each of the two origins, and the sum of the resolved parts is zero in any one direction, not perpendicular to the line joining the two origins. For, by principle (b), there can be no resultant couple, and the resultant force (if one exists) must act in a line through the two origins of moments ; while by principle (a), the resultant force (if there is one) must be perpendicular to the direction of resolution. (3) There will be equilibrium if the sum of the moments is zero for each of three origins not in the same straight fine. For, by principle (&), the resultant force (if one exists) must act in a line containing the three origins; and there can be no resultant couple, since the moment of a couple cannot be zero. Since, therefore, three equations can be written, which, if satisfied, insure that the resultant is zero, it follows that all possible equations obtained in accordance with the principles of Art. 103 must be true EQUILIBRIUM OF NON-CONCURRENT FORCES. 67 if those three are true. Hence three, and only three, of the equa- tions of equilibrium can be independent. 105. Conditions of Equilibrium Deduced Algebraically. It was shown in Art. 98 that any system of coplanar forces can be reduced to a force and a couple; the force being applied at any chosen point and being equal to the vector sum of the given forces, while the mo- ment of the couple is equal to the algebraic sum of the moments of the given forces about the chosen point. Using the notation of Art. 99, the equations for computing the force and couple are X = R cos a = P x cos a l -f- P y cos a , -f . . . ; Y = R sin /,+ .... The analysis of the four possible cases, given in Art. 100, shows that there will be equilibrium if R and G are both zero, but not otherwise. Now the condition R = o requires that X = o and Y = o, un- less either X' 2 or Y 2 is negative, that is, unless X or Y is imaginary. Butif/^,/^, . . . are real forces, X and Y must be real. For equilibrium, therefore, it is necessary that the three equa- tions P x cos a x -f" P% cos a 2 "T" > (0 /\ cos /+/>, cos ,+ . . . =0, . . (2) 4 (l 2 - b 2 ), that is if a 2 > 4/* 3b 2 . Numerical case. Suppose AB = / = 8 ft. ; AC = a = 10 ft. ; AD b = 6 ft. Then P = R Wb e= w 1 2 v 7 2 - Wa b 2 2VI 2 - b 2 Va 2 - -b 2 2|/28 M/= 0.567 W\ 21/7 ^='0.945 W\ y-p a ^ [i -1/7 1 = 0.244 rr. This value of Q being positive, equilibrium is possible, be foreseen from the condition above deduced ; for This might 00 hence 4 / 2 -3^= 148; If, however, we take b = 7.5 ft., the other data remaining unchanged, the value of Q is negative, and the solution fails. Geometrical analysis. The resultant of P and W acts through the intersection of their lines of action (F, Fig. 57) ; hence for equi- librium the resultant of Q and R must act through F. Since Q must act upward, the resultant of Q and R acts in some line between A C and the vertical; hence this line cannot pass through Sunless .Flies 76 THEORETICAL MECHANICS. above A C. In the limiting case in which A C passes through F, we have, from the geometrical relations, a 2 = 4 / 2 3&\ Hence if a* > \P 3# 2 , equilibrium is impossible. This agrees with the result reached above. Examples. i. A bar of mass W lbs. and length / rests with one end upon a smooth horizontal floor and the other against a smooth vertical wall, being held in equilibrium by a horizontal string attached at a point distant b from the upper end. The center of gravity is at a distance a from the upper end, and the inclination to the horizontal is 0. Determine completely all forces acting upon the bar. Ans. Tension in string = IV (/ a)/ b tan 0. 2. In Ex. i, let W = 20 lbs., = 40, a == 0.6/, b = 0.3/; de- termine all forces completely. Ans. Reaction at lower end = 20 lbs. ; at upper end, 31.79 lbs. ; tension = 31.79 lbs. 3. A bar of mass 1 5 lbs. , whose center of gravity is at its middle point, rests with its ends upon two smooth planes inclined to the horizontal at angles of 36 and 45 respectively. Determine the in- clination of the bar to the horizontal when in equilibrium; also the pressures exerted upon it by the supporting planes. Ans. io 39'; 8.93 lbs. ; 10.74 lbs. 4. A bar of known mass and length being supported as in the preceding example, let its center of gravity be at any point of its length, and let a and /3 be the angles made by the planes with the horizontal. Determine the position of equilibrium and the support- ing pressures. 5. A person weighing 160 lbs., standing upon a scale platform, pulls upon a suspended rope in a direction inclined io to the ver- tical. The scale beam shows his apparent weight to be 140 lbs. With what force does he pull the rope ? Ans. 20. 3 lbs. 6. A bar AB, 16 ft. long, is supported in a horizontal position by a smooth hinge at A and a smooth plane at B, the inclination of the plane to the horizontal being 20. Weights of 20 lbs., 18 lbs. and 28 lbs. are suspended from the body by cords attached at points whose distances from A are 6 ft., 10 ft. and 15 ft. Determine the supporting forces exerted by the plane and hinge. (The pressure exerted by the hinge may have any direction, but always acts through the center of the hinge. See Art. 42.) Ans. Pressure at B = 47.89 lbs. ; pressure at A = 26.63 lbs. EQUILIBRIUM OF NON-CONCURRENT FORCES. 77 7. A bar whose mass, center of gravity and length are known, is supported by strings attached to the ends. In order that the bar may rest in a horizontal position, what condition must be satisfied by the directions of the strings ? Solve geometrically. 8. A bar AB whose center of gravity is at a distance from the end A equal to one-third the length, is supported as in the preced- ing example. If the string at A is inclined 30 to the vertical, what must be the inclination of the string at B ? Determine the tensions in the strings, the mass of the bar being W lbs. Ans. 49 7'; 0.77 W '; 0.509 W. 9. A bar AB, whose center of gravity is at its middle point, and whose mass is 1 2 lbs. , is supported in a horizontal position by strings attached to the ends, and sustains loads of 16 lbs. and 20 lbs. at A and B respectively If the string at A is inclined 45 to the hori- zontal, what is the direction of the string at B ? What tensions are sustained by the strings ? Ans. 49 47' from horizontal; 31.12 lbs.; 34.05 lbs. 10. A bar is supported at a given angle with the horizontal by strings attached at the ends. Show geometrically the relations that must be satisfied by the directions of the strings. Deduce also an algebraic expression for the relation between the angles made by the strings and the bar with the horizontal. Ans. Let a and b = segments into which the length of the bar is divided by the center of gravity ; 0, a, /3 = angles made with hor- izontal by the bar and strings. Then a tan a b tan /3 = {a -J- b) tan d. 11. A bar of mass 20 lbs., whose center of gravity is at its middle point, is supported as in the preceding example, its inclination to the horizontal being 30. The cord attached at the lower end being in- clined 6o to the horizontal, determine the inclination of the other cord, and the tensions in the two cords. Ans. 6 = 70 54'; tensions = 13.23 lbs. and 8.66 lbs. 12. A bar 6 ins. long, whose mass is 2 lbs. and whose center of gravity is 4 ins. from one end, rests inside a smooth hemispherical bowl of radius 4 ins. What is its inclination to the horizontal when in equilibrium, and what are the pressures upon its ends ? no. Equilibrium of Connected Bodies. If two rigid bodies in equilibrium are connected in any way, as by a smooth hinge or by simple contact, the conditions of equilibrium may be applied to each separately. The forces which the bodies exert upon each other will enter the equations as unknown quantities ; and are related to each other in accordance with Newton's third law (Art. 35). That is, A and B being any two bodies, the force which A exerts upon B is 78 THEORETICAL MECHANICS. equal, opposite and collinear with the force which B exerts upon A. This will be illustrated by the following problems. I. Two heavy bars AC and BC (Fig. 58), connected by a smooth hinge at C, are attached to a fixed body by smooth hinges at A and B. Determine the pres- \qXc- A/ sures exerted upon the bars by the ^X^N. yfyB hinges A and B, and the pressure of ^xNv>^^ each upon the other at C ^Sr Solution. In the triangle A B C, p IG _ 8 denote the three angles by A, B, C, and the corresponding opposite sides by a, b> c; and let c be horizontal. Let the center of gravity of each bar be at its middle point. Let W = weight of AC; W" = weight of <7; P' resultant pressure upon A C at A ; H' == horizontal component of P' ; V == vertical component of P'; P" bs= resultant pressure upon BC at B; H" = horizontal component of P" ; V" = vertical component of P" \ P= pressure exerted upon AC by BCat C; ( P= pressure exerted upon BC by AC at C;) H = horizontal component of P; V = vertical component of P. (The positive direction is toward the right for horizontal forces and upward for vertical forces. ) Three independent equations of equilibrium for A C may be ob- tained as follows: Resolving horizontally, ff> -f tf=o. . . . . (1) Resolving vertically, V + V W = o. . . . (2) Taking moments about A, Vb cos A 4- Hb sin A y 2 W'b cos A = o. . (3) Three independent equations of equilibrium for BC may be ob- tained in a similar manner : EQUILIBRIUM OF NON-CONCURRENT FORCES. 79 H" H = o (4) V" V- W" = 0. . . . (5) Va cos B Ha sin 2>> -f % W"a cos = o. . (6) These six equations suffice to determine the six quantities H, V> H\ V, H", V" . The solution may proceed as follows: From (3) and (6), H and V can be found ; the remaining un- known quantities can then be determined at once from the other four equations. The results are as follows: H = V = H' = IV' + W" 2 (tan A -h tan B) W tan B W" tan A m 2 (tan A + tan B) W + W" 2 (tan A -f- tan B)\ ( W + W"~) tan A , r = 2 2 (tan ^ tan i?) H" = V" w w 2 (tan A -f tan i>' ) IT* ( ^ ; + W") tan ^ 2 2 (tan A + tan i? ) II. A heavy bar, AB (Fig. 59), is supported at A by a smooth hinge and leans against a smooth circular cylinder which rests in the angle between a horizontal floor and a vertical wall. Determine completely all forces acting upon the bar and upon the cylinder. Solution. Let the known data be the weight of the bar, W lbs. ; the weight of the cylinder, W lbs.; the distance of the center of gravity of the bar from A, a ft. ; the radius of the cylin- 8o THEORETICAL MECHANICS. der, r ft. ; the distance AC, b ft. The angle BA C is fixed by the known data ; in fact, if BA C = 6, tan (0/ 2 ) = r/(b r). The unknown forces are the pressure on the bar at A, unknown in magnitude and direction ; the pressure upon the bar at its point of contact with the cylinder (and the equal and opposite pressure upon the cylinder), known in direction but not in magnitude ; and the pressures upon the cylinder at its points of contact with the floor and wall, known in direction but not in magnitude. Let Q denote the magnitude of the force exerted by the cylinder upon the bar at E ; P the magnitude of the force exerted upon the bar by the hinge at A ; H and V the horizontal and vertical compo- nents of P. Three independent equations of equilibrium for the bar may be written as follows : Taking moments about A, remembering that AE = AF = b r, we have Q(b r) Wa cos = o. . . . (i) Resolving horizontally, H Q sin B = a . . . . (2) Resolving vertically, V + Q cos 6 W = o. . . . (3) These three equations suffice for the determination of Q, H and V Thus, from (1), Q _ Wa cos 6 b r From (2), H = -Q sin 6 = Wa sin b~ 0COS - r From (3), V-- = W gcos e = w\i a cos * >01 r _ Turning now to the cylinder, let i denote the horizontal pres- sure exerted by the wall and S the vertical pressure exerted by the floor. Since the lines of action of the four forces acting upon the cyl- inder intersect in a point, only two independent equations can be written. These may be the following : Resolving horizontally, Q sin R = o (4) Resolving vertically, Q cos 6 -f 5 W = o. c . . (5) EQUILIBRIUM OF NON-CONCURRENT FORCES. 81 From (4), From (5), R = Q sin = Wa sin 6 cos S = W + Q cos = W + b r Wa cos 2 6 Geometrical solution. The lines of action of the three forces acting upon the bar (P, Q and W) must meet in a point ; this point is found by prolonging the known lines of action of W and Q. Since W is known, the force-triangle for the three forces can be drawn, and is shown in Fig. 60. The forces P and Q are' thus determined. The forces acting upon the cylinder are four in number : a force of magnitude Q> opposite to the force Q acting upon the bar; a force W completely known; forces R and 5, whose lines of action are known. The four being in equilibrium, their force-polygon must close, and can be completely drawn from the data now known t is the quadrilateral whose sides are marked W\ 0i - and R in Fig. 60. In any numerical case, the figure may be drawn to scale and the magnitudes of the required forces found by measurement from the figure. Or, the angles may all be determined from the given data, and the lengths of the sides representing the magnitudes of the required forces may be computed trigonometrically. ! Examples. ra- Fig. 60. 1. In Fig. 59, let A C ='20 ins., dius of cylinder = 8 ins., AD = 7 ins., W= 12 lbs., W = 20 lbs. Deter- mine all the forces acting upon the bar and cylinder (a) algebraically and () geometrically. 2. Find completely the forces acting upon the bar and cylinder in Fig. 61, the bar being hinged at A, and the sur- faces of contact all being smooth. Let 82 THEORETICAL MECHANICS. the weight of the bar be 10 lbs. ; the weight of the cylinder 15 lbs. ; the radius of the cylinder 12 ins.; CA (vertical) 24 ins.; distance of center of gravity of bar from A, 8 ins. Ans. Pressure between bar and cylinder = 1.924 lbs. 3. Three uniform bars AB, BC, CD, of equal length and mass, are connected by smooth hinges at B and C AB is hinged to a fixed support at A, and CD to a fixed support at D. AB and CD are horizontal and BC is vertical, the system being held in equilib- rium by a vertical string attached to A B at a point two-thirds the distance from A to B. Determine all forces acting upon each bar. 4. Two uniform bars, equal in length and mass, connected by a smooth hinge, rest upon a smooth cylinder whose axis is horizontal. Determine the position of equilibrium, and the magnitude and direc- tion of every force acting upon each bar. in. Simple Machines. It is often desired to exert upon some body a force of such magnitude or direction that means are not available for applying it directly. In such a case it may be possible to accomplish the desired object by indirect means. Any body or system of connected bodies by which such an object is accomplished is called a machine* The bodies constituting a machine are usually either actually or nearly in equilibrium. The relations subsisting among the ap- plied forces may therefore be determined from the principles of Statics. A machine is made to accomplish its object by applying to it forces tending to give it a certain definite motion. A force which aids this motion is called an effort, and a force which opposes it a resist- ance. A force which merely guides the motion without any tendency to accelerate or to retard it, is a constraint. The equal and opposite reaction to the force which the machine is designed to exert is a resistance ; and since the overcoming of this resistance is the primary object of the machine, it is called a useful resistance. The useful resistance may be called the load. To determine the relation between the effort and the load is the funda- mental statical problem presented by a machine. Besides the useful resistance there are other forces which act in opposition to the effort and diminish the effectiveness of the machine * Machines considered as devices for the transmission of energy are treated in Chapters XVII and' XXIII. EQUILIBRIUM OF NON-CONCURRENT FORCES. 83 in overcoming the useful resistance. These may be called wasteful resistances. The most important wasteful resistances are the fric- tional forces exerted by the bodies by whose agency the motion of the machine is guided in the desired manner. In the present dis- cussion friction will be disregarded. The method of estimating its effect will be considered in Chapter VII. A machine is said to yield a mechanical advantage if the load is greater than the effort. If the load is less than the effort there is a mechanical disadvantage. The method of applying the principles of equilibrium to machines is illustrated by the following simple examples. Other examples are given in later chapters. Lever. A lever is a rigid bar, movable about a fixed axis. The effort and load may be applied at any points of the bar and in any directions, but are usually taken as acting in a plane perpendicular to the fixed axis. The fixed axis or support about which the lever turns is called the fulcrum. The portions of the bar between the fulcrum and the points of application of the effort and the load are called the arms. The lever may be either straight or bent; and the forces applied to it may or may not be parallel. The application of the principles of equilibrium to the lever is so simple, when friction is neglected, that the solution of the problem ^ will be omitted. Fixed Pulley. A pulley con- sists of a wheel which can rotate freely about an axis, together with a flexible cord wrapped about some part of the circumference of the wheel. The axis about which the wheel revolves may or may not be stationary; so that pulleys may be classed as fixed and movable. Fig. 62 represents a fixed pul- ley, the effort and load being ap- plied to the same cord, passing around a portion of the circumfer- ence of the wheel. The relation between the effort (P) and the load ( W) may be found by applying the principles of equilibrium to the Fig. 62. 8 4 THEORETICAL MECHANICS. system of bodies * consisting of the wheel and the adjacent portion of the cord. At A there is a tension in the cord equal to P, and at B a tension equal to W. The only other force acting upon the system considered is that exerted by the axle about which the wheel revolves; and if there is no friction between the wheel and the axle, this force must act in aline through the center of the wheel. Taking moments about this center, we have (calling a the radius of the wheel ) Pa = Wa ; or P = W. Let R be the pressure exerted by the axle upon the wheel; then R must be equal and opposite to the resultant of P and IV. Its line of action therefore bisects the angle between the two straight portions of the cord; or, calling this angle 0, we have R = 2 W cos (0/2). Evidently there is no mechanical advantage in this case; the fixed pulley merely serves as a means of changing the direction of appli- cation of a force. Movable Pulley. A simple movable pulley is represented in Fig. 63. The relation between the effort (P) and the load ( IV) is determined by considering the system :on- sisting of the wheel and the adjacent por- tions of the two cords, limited by the points A, B and C. Neglecting friction and the weight of the pulley, the principles of equi- librium lead to the following results : Taking moments about the center of the pulley, it is seen that the tensions in the cord at A and B are equal and each equal to P. Resolving forces vertically, it is seen that the tension in the cord at C (which is equal to W) must be equal to the sum of the tensions at A and B, or 2P; that is, P= WI2. * Although we have thus far applied the principles of equilibrium only to rigid bodies, it is allowable to treat as rigid any body or connected system of bodies, every part of which is in equilibrium. See Chapter VI. EQUILIBRIUM OF NON-CONCURRENT FORCES. 85 Sy 'Stems of Pulleys. By combining fixed and movable pulleys in various ways, the mechanical advantage may be made very great, being limited only by the prejudicial resistances due to friction and the lack of perfect flexibility of the cords. Fig. 64 shows a system very commonly employed in rais- ing heavy weights. Two sets of wheels are employed (shown at A and ), each set consisting of several pulleys mounted on a common axis but revolving independently of one another. One set (A ) is attached to a fixed support, while to the other set, which is movable, the load is applied.* Neglecting friction and the rigidity of the cord, the relation between P and W is easily obtained. The tension has the same value in all parts of the cord, as may be seen by apply- ing the principle of moments to each wheel separately, taking origin at the center. If n is the number of "plies," or portions of rope supporting the lower set of pulleys, we have (neglecting the weight of the pulleys and of the cord, and assuming that the straight por- tions of the cord are all parallel) nP=W- P=W\n. Examples. 1. The Dar AB (Fig. 65) is free to rotate about a smooth pin at C. The effort P and R the resistance W act at angles d and with ** C e Fig. 65. * In the system of pulleys as actually constructed, the wheels of each set are usually of equal diameter and mounted side by side upon the same axle. The different portions of the cord are therefore not all coplanar. Each of the separate bodies (each pulley and the block carrying the pulleys) is how- ever acted upon by forces which are practically coplanar. 86 THEORETICAL MECHANICS. the bar. Determine the relation between P and W, and compute the magnitude and direction of the force exerted upon the bar by the pin. 2. In Ex. i, let AC =3 ft-, CB = 2 ft., 6 = 6o, $ = 3 o, W = 85 lbs. Determine P and the magnitude and direction of the pressure exerted upon the pin. Ans. P 32.7 lbs. 3. If P and W (Fig. 62) act in direc- tions inclined 90 to each other, what is the pressure on the axle ? 4. In case of a movable pulley, let the effort act at an angle a with the hori- zontal. What must be the direction of the cord between the pulley and the fixed point of attachment of the cord ? What is the relation between Pand W ? Ans. P= W/2sma. 5. Determine the relation between P and W ( Fig. 63) taking account of the weight of the pulley. 6. Determine the relation between P and W in case of a system such as shown in Fig. 64, the number of mov- able pulleys being 3. 7. In Fig. 64, what must be the strength of the cord if a weight of 5,000 lbs. is to be raised ? 8. In Fig. 66, let the weight W be 1,000 lbs. Determine the tension in each cord and the value of P. 9. Solve Ex. 8, taking into account the weights of the pulleys, 5 lbs. each. 112. Balance. Since the weights of bodies in the same locality are proportional to their masses, the mass of any body may be deter- mined by comparing its weight with that of a body whose mass is known. The essential parts of a balance for comparing the weights of bodies are shown in Fig. 67. The beam L can turn freely in a ver- tical plane about a fixed point C. At A and B are suspended the " scale-pans ' ' M and N, upon which are placed the bodies whose weights are to be compared. The center of gravity of the whole apparatus * is at some point D. When the scale-pans are vacant, * In determining this center oi gravity, the scale-pans are to be regarded as particles located at their points of suspension A and B. Fig. 66. EQUILIBRIUM OF NON-CONCURRENT FORCES. 87 the point D will be vertically below C in the position of equilibrium. In this position let CA and CB make equal angles with the horizon- tal, and let the lengths A C and CB be each equal to a. If bodies of equal mass (and weight) are placed in the two scale-pans, the position of equilibrium will be unchanged. Let a body of mass P be placed at M, and a body of mass Q at N, P being greater than Q, and let the beam assume a position of equilibrium, 6 being the angle between CD and the vertical in this position. If initially the lines CA and CB make with the horizontal the angle a, their inclinations to the horizontal in the new position Fig. 67. of equilibrium are a + and a 6 respectively. Let CD = h, and let G = weight of balance. Taking moments about C, Pa cos (a Solving for 0, 0) tan 6 Qa cos (a 6) Gh sin 6 = o. (P Q)a cos a (P+ Q)a sin a + Gh In order that a small difference between P and Q may be detected with certainty, a small value of P Q should cause a relatively great value of 6. This sensitiveness depends upon the values of a, h and a. For given values of a and a, the sensitiveness increases as h de- creases. For given values of a and h, the sensitiveness increases as a decreases. 88 THEORETICAL MECHANICS. A very high degree of sensitiveness can be obtained only at the expense of stability. That is, if a very small value of P Q pro- duces a large value of 0, the beam cannot readily be brought into a condition of equilibrium, but will oscillate between positions far from the position of rest. In the practical use of the balance, the body whose mass is to be determined is placed in one scale- pan, and standard "weights" (bodies of known mass) are placed in the other in sufficient quantity so that the position of equilibrium is the same as when both scale- pans are vacant. In order that the indications of the balance may be correct when used in this manner, the horizontal projections of AC and BC (Fig. 67) must be equal. These horizontal projections may be called the arms of the balance. If the arms are unequal, this fact may be detected by interchanging the weights in the pans. Thus, let a body whose true weight is x lbs. be balanced against standard weights in a balance whose arms are a and b. Let the apparent weight of x be P lbs. when it acts with the arm a, and Q lbs. when it acts with the arm b. Then xa Pb\ xb = Qa. From these equations, x=yPQ; a/b=Vp/Q. Examples. 1. In Fig. 67, let the distances AC and CB be each 15 ins., CD 6 ins., the angles ACD and BCD each 8o, and the weight of the balance 4 lbs. What is the position of equilibrium if weights of 10 lbs. and 10 lbs. -J- }i oz. are placed in the two scale-pans? Ans. 6 = 5'+. 2. A body appears to weigh 18 oz. when placed in one scale-pan and 18^ oz. when placed in the other. What is its true weight, and what is the ratio between the lengths of the arms ? Ans. 18.248 oz.; 0.986. Miscellaneous Examples. 1. A bar AB rests with the end A upon a smooth horizontal plane, and leans against a smooth cylindrical peg at a given distance above the plane. It is held from slipping by a horizontal string A C attached at a fixed point C. Determine all forces acting upon the bar. EQUILIBRIUM OF NON-CONCURRENT FORCES. 89 2. In Ex. i, let the height of the peg above the plane be 18 ins., the distance of the center of gravity of the bar from A 12 ins., the inclination of the bar to the plane 25, its weight 35 lbs. Determine all forces acting upon the bar. Ans. Vertical pressure at A = 26.90 lbs.; pressure of peg = 8.94 lbs.; tension = 3.78 lbs. 3. Let the data be as in Ex. 1, except that the horizontal string A C passes over a smooth pulley at C and sustains a body of known weight. Determine the position of equilibrium and all forces acting upon the bar. [The solution leads to a cubic equation.] 4. Solve Ex. 3 with the following numerical data : Weight of bar, 10 kilogr. ; weight of suspended body, 2 kilogr. ; height of peg above plane, 50 cm. ; distance of center of gravity of bar from A, 50 cm. Ans. Angle of bar with horizontal = 77 56' or 28 30'. 5. A uniform bar AB of known mass and length is supported in a horizontal position by a smooth hinge at A and a cord BC inclined to the bar at a known angle and fixed at C. Two bodies of known mass are suspended from the bar at points midway between the center of gravity and the ends A and B respectively. Determine all forces acting on the bar. 6. Solve Ex. 5 with the following numerical data : Weight of bar, 18 lbs.; suspended weights, 3 kilogr. and 5 kilogr.; inclination of string to bar, 37 (measured from prolongation of AB). Ans. Tension = 14.26 kilogr.; hinge reaction = 13.67 kilogr. inclined 146 20' to AB. 7. A bar AB of weight W is supported by a smooth hinge at A and a string attached at B. The string passes over a smooth peg at C and supports a body of weight P. ACis horizontal and equal to AB, and the center of gravity of the bar is at its middle point. De- termine the position of equilibrium, and all forces acting on the bar. CAB P Ans. cos = 2 2W *.V 2W' 1 P l 8. Solve Ex. 7, taking W = 28 lbs., P= 6 kilogr. Ans. CAB = 21 54 or 241 20'. Hinge reaction = 15.25 lbs. or 36.53 lbs. 9. In Ex. 7, let AC and AB be unequal, A C being still horizontal. Deduce a cubic equation for determining cos 0, where 6 is the angle between the bar and the horizontal. 10. Solve Ex. 9, taking AC= 2 {AB), W= P = 12 kilogr. Ans. 6 = 14 55', or 2 1 5 42'. 11. Modify Ex. 7 by taking AC = AB, but not horizontal. 12. A smooth cylinder of radius a is placed with its axis hori- zontal and parallel to a smooth vertical wall at distance h. A uni- form bar of weight W and length 2/ rests against the cylinder and 90 THEORETICAL MECHANICS. with its lower end against the wall. Determine the position of equi- librium. Ans. If = angle between bar and vertical, / sin 3 -f- a cos = k. 13. In Ex. 12, let the radius of the cylinder be very small compared with h and /. Solve with the following data : W = 15 kilogr. , h = 6 ins. , / = 12 ins. Determine the position of equilibrium and all forces acting on the bar. 14. A uniform smooth bar of length 2/ and weight W rests in a hemispherical bowl of radius a. Assuming the length to be so great that the upper end projects beyond the edge of the bowl, determine the position of equilibrium and all forces acting on the bar. Ans. The angle between the bar and the horizontal is given by the equation cos = (/ V P + 320 2 )/ 8a. 15. Solve Ex. 14 with the following numerical data: W = 12 kilogr. , / = o. 45 met. , a = o. 40 met. Ans. = 30 28'; pressure at lower end = 7.05 kilogr. 16. A heavy bar is supported by a string attached at any two points and passing over a smooth peg. Determine the position of equilibrium and the tension in the string. Ans. The peg divides the string into segments whose lengths are directly proportional to the distances of the center of gravity of the bar from the points of attachment of the string. 17. In the preceding example let the weight of the bar be 12 lbs., the distances of the center of gravity from the points of attach- ment of the strings 2 ft. and 3 ft., the length of the string 7.5 ft. Determine the tension in the string in the position of equilibrium. Ans. 7.89 lbs. 18. What horizontal force is necessary to pull a carriage wheel over a smooth obstacle, the radius of the wheel being a, its weight W, and the height of the obstacle h ? 19. A heavy uniform beam, movable in a vertical plane about a smooth hinge at one end, is sustained by a cord attached to the other end. The angle between the bar and the vertical being fixed, deter- mine what direction of the cord will cause the least pressure on the hinge. Determine the corresponding values of the unknown forces acting on the bar. Ans. Let a = angle of bar with vertical, = angle of string with bar, then tan = }4 tan a. 20. A heavy beam rests with one end against a smooth inclined plane ; to the other end is attached a cord which passes over a smooth pulley and sustains a given weight. Determine the position of equi- librium. 21. The roof-truss shown in skeleton in Fig. 68 is supported by a smooth horizontal surface at A and a smooth hinge at B. It is re- EQUILIBRIUM OF NON-CONCURRENT FORCES. 91 quired to sustain the wind pressure on a portion of the roof of width 1 2 ft. and length A C or CB. Assume the pressure to be normal to the roof and uni- formly equal to 20 lbs. per sq. ft. Compute the sup- porting forces at A and B when the wind is from j\_ the right ; also when it is from the left. 22. Take data as in Ex. 17, but assume the peg rough, so that the tensions in the two portions of the string may be unequal. If the bar is in equi- librium when the peg divides the string into segments 1.9 ft. and 5.6 ft. in length, what are the tensions in the two portions of the cord ? Ans. 10.46 lbs. and 4. 18 lbs. 23. The lines of action of three forces intersect in points A, By and C. Prove that, if their resultant is a couple, they are propor- tional to the vectors AB, BC, CA. 24. A uniform bar of weight W rests with its ends against two smooth planes whose inclinations to the horizontal are a and /3, that of the bar being 6. A weight P is suspended from a point distant one-third the length of the bar from one end. Required the value of P for equilibrium. CHAPTER VI. EQUILIBRIUM OF PARTS OF BODIES AND OF SYSTEMS OF BODIES. i. Equilibrium of Any Part of a Body. 113. External and Internal Forces Acting on a Body. The conditions of equilibrium have thus far been applied only to a par- ticle or to a rigid body regarded as a whole. In case two or more bodies were considered, each was regarded as presenting a separate problem in equilibrium. For certain purposes, however, it is desir- able to confine the attention to a portion of a rigid body; and for other purposes it is found convenient to treat two or more bodies as together forming a system. As a preliminary to the discussion of problems from these points of view, it is useful to recur to the funda- mental conception of force. By the definition (Art. 32), a force is an action exerted by one body upon another. Again, by a fundamental law (Art. 35), forces always act in pairs ; so that when any one body exerts a force upon another, the second body exerts an equal and opposite force upon the first. Thus every force concerns two bodies ; although in the pre- ceding chapters the attention has in every case been directed espe- cially to a single body, that upon which the force acts, and the body exerting the force has not always been specified. The word body, in the definition of force, must be understood to mean any portion of matter. The two portions may be separate from each other (as in most of the cases hitherto discussed), or they may be parts of a single body. This leads to the following classifi- cation of the forces applied to a body: An external force is one exerted upon the body in question by some other body. An internal force is one exerted upon one portion of the body by another portion of the same body. 114. Conditions of Equilibrium for a Rigid Body as a Whole. The discussions of the composition, resolution and equilibrium of forces, as thus far given, have related to external forces applied to the same rigid body, internal forces not appearing in the equations or conditions of equilibrium. EQUILIBRIUM OF PARTS OF BODIES. 93 Thus, although it is generally true that adjacent portions of a rigid body exert forces upon each other, these forces do not come into consideration when the problem relates solely to the equilibrium of the body as a whole. It may, however, be desired to gain some in- formation regarding these internal forces. It will then be necessary to direct the attention to a portion of the body, and to consider the relations among all the forces which act upon this portion. 115. Conditions of Equilibrium for a Portion of a Body. Since the body to which the conditions of equilibrium apply may be any portion of matter whose particles are rigidly connected, it may be any connected portion of a rigid body. But in writing the equa- tions of equilibrium for a portion of a body, certain forces must be included which are omitted if the equilibrium of the whole body is considered. These are the forces exerted upon the portion in ques- tion by other portions of the same body. Such forces are internal when the whole body is considered, but external when one part is considered. To illustrate, let AB (Fig. 69) represent a bar acted upon by two forces of equal magnitude applied at the ends in opposite directions parallel to the length of the bar. These forces are exerted upon AB by some other bodies not specified. If the whole bar be considered, the external forces are the two forces named. But suppose the body under consideration is AC, a portion of AB ; the external forces act- ing upon this body are ( 1 ) a force ap- plied at A (already mentioned) and A (2) a force applied at C (exerted upon AC by OB). This latter force is in- Fig. 69. ternal to the bar AB but external to AC In applying the conditions of equilibrium to AB, the only forces to enter the equations are those exerted at A and B as already mentioned ; but if the equations of equilibrium for A C are to be written, the forces to be included are the force at A and the force exerted upon AC by CB at C 2. Determination of Internal Forces. 116. General Method of Determining Internal Forces. Little is in general known regarding the forces which contiguous portions of a body exert upon each other. But when the external forces are 94 THEORETICAL MECHANICS. known, the internal forces can be partly determined by applying the conditions of equilibrium to portions of the body. Thus, to take the simplest and most important case, suppose it is known that not only the whole body, but every portion of it, is in equilibrium.* For such cases the following proposition may be stated: If a body, every part of which is in equilibrium, be conceived as made up of two parts, X and Y, then the internal forces exerted by X upon Y, together with the external forces acting on Y, form a system in equilibrium. Thus, by applying the conditions of equilibrium to the body Y, the resultant of all the forces exerted by X upon Y may be determined ; it being equal and opposite to the resultant of the external forces applied to Y. The individual forces exerted by the particles of X upon those of Y cannot, however, be determined, these forces being in general very numerous and acting in various unknown directions. Fig. 70. % \l Examples. 1. A beam 8 ft. long, weighing 40 lbs. per linear foot, rests hori- zontally upon supports at the ends. Dividing the beam by a trans- verse plane 3 ft. from one end, determine the resultant of the forces exerted by each portion upon the other. The forces acting upon the left por- tion of the beam (Fig. 71) are (1) the supporting force of 160 lbs.; (2) the weight of three linear feet of the beam, equivalent to a force of 1 20 lbs. applied 1.5 ft. from the left end ; and (3) the forces exerted by the other portion of the beam. The resultant of the first two is an upward force of 40 lbs. acting in a line 4. 5 ft. to the left of the left support. Hence the forces exerted by Fig. 71. * It is to be carefully noted that the equilibrium of a body does not neces- sarily imply the equilibrium of every portion of it. Thus, if the external forces applied to the whole body are balanced, it may still have a motion of rotation, in which case the forces acting upon a portion of the body will in general be unbalanced. This will be apparent when the rotation of a rigid body has been studied. In very many cases, especially such as arise in en- gineering practice, the bodies are not only in equilibrium, but at rest. In such cases all parts of the body are also in equilibrium. EQUILIBRIUM OF PARTS OF BODIES. 95 the other portion of the beam have for their resultant a downward force of 40 lbs. acting in a line 4.5 ft. to the left of the left support. 2. A beam 12 ft. long, of uniform density and cross-section, weighing 20 lbs. per linear foot, rests horizontally upon supports at the ends. Conceiving the beam divided by a transverse plane through the middle of the length, determine the resultant of the forces exerted by each portion upon the other. 3. In Ex. 2, let the dividing plane be 3 ft. from one end. Deter- mine the resultant of the forces exerted by each portion upon the other. 4. Let the same beam rest upon a single support at the middle. Determine the resultant of the forces exerted upon each other by the two portions described in Ex. 2 ; also by the two portions described in Ex. 3. 5. Let the same beam rest upon supports, one of which is 2 ft. from one end and the other 4 ft. from the other end. Answer the questions asked in examples 2 and 3. 6. A uniform beam 17 ft. long, weighing 120 lbs., rests horizon- tally upon supports at the ends and sustains two loads : a load of 40 lbs. 3 ft. from the left end, and a load of 80 lbs. 6 ft. from the right end. Taking a transverse section 4 ft. from the left end, compute the resultant of the forces exerted by one of the two portions of the beam upon the other. Ans. 52. 94 lbs. ; 3.3 ft. from left end and 20.3 ft. from right end. 7. In Ex. 6, let the transverse section be taken 4 ft. from the right end. Answer the same question. Fig. 72. 117. Jointed Frame. Engineering practice has to deal with structures made up of straight bars connected at the ends by hinge joints. (See Art. 42.) A structure thus made is called a. jointed frame or truss. The determination of the internal forces in such a frame is an important problem in engineering. Fig. 72 represents in skeleton a simple form of roof truss, resting upon supports at the ends. Such a truss is acted upon by external forces consisting of the weight of its own members, the weight of 96 THEORETICAL MECHANICS. the roof proper, sometimes the weight of snow and the pressure of wind, and the supporting forces at the ends. Each of the bars composing the truss is subjected to certain forces tending to break it; this tendency is resisted by the internal forces called into action be- tween the contiguous portions of the bar. If these internal forces exceed certain values, the material is injured and the bar may be broken; it is therefore important to be able to determine their mag- nitudes before designing the members. To solve this problem the general method of Art. 116 is em- ployed. But in order to simplify the problem and make it com- 'pletely determinate, certain assumptions are made which are only approximately true. (a) It is assumed that the hinges are without friction. The effect of this may be seen by referring to Fig. 14 (p. 21), which repre- sents a portion of one bar. The connection with other bars is made by means of a pin A which passes through holes in the ends of the bars. The pin being assumed smooth, the pressure of the pin upon the bar is normal to the cylindrical surfaces of both pin and bar, and there- fore acts through the axes of these cylinders. (b) It is assumed that all external forces acting upon the truss are applied to the pins. If this is the case, the only forces acting on any bar are those exerted by the two pins at its ends. These two forces being in equilibrium, must be equal and opposite and must have the same line of action, which is therefore the line joining the centers of the hinges. If a transverse section be taken dividing the bar into two parts, the principle of Art. 1 16 shows that the resultant force exerted by each part upon the other must have the same line of action as the forces exerted upon the ends of the bar by the pins. 118. Determination of Internal Forces in Jointed Frame. The method of determining internal forces in a jointed frame will now be explained by reference to a numerical problem. Fig. 73 shows the dimensions of a roof truss supporting three vertical loads at upper joints, and supported upon horizontal surfaces at the ends. It is required to determine the internal forces in each of the bars. No loads will be considered except those shown. The first step in the solution is (a) The determination of the supporting forces. Assuming these to be vertical, each must be equal to half the total load supported, i.e., to 375 lbs. EQUILIBRIUM OF PARTS OF BODIES. 97 (b) Determination of internal forces. Let a section be taken, as MN y dividing the truss into two parts, X and Y, and let the conditions of equilibrium be applied to either part, as X. This body X is acted upon by an upward force of 375 lbs. at the support ; a downward load of 250 lbs.; and the forces exerted by Y. In ac- cordance with the assumptions made in Art. 117, these forces exerted by Y are three, acting along the lines of the three members cut by the section MN. The magnitudes of these forces are, however, un- known. If three independent equations of equilibrium be written for the body X, the only unknown quantities which will enter them are these three unknown force-magnitudes, which may therefore be determined by solving the equations. Let P x , P 2 and P z denote the magnitudes of the three unknown forces acting upon X (Fig. 74), and let them be assumed to act Fig. 74. 6 f (X) D) r^* A /^ / / M S 01 1 500 #JE7 Fig. 75- away from the body X. (Their actual directions will then be known from the signs of their numerical values.) Let the three equations be written as follows : 7 9 8 THEORETICAL MECHANICS. Taking origin of moments at A, P x and P 3 are eliminated. The arm of P 2 is 5^/2 ft, and the equation is p i X sV 2 -250X 10 = 0. . . . (1) Next taking origin of moments at B, P 2 and P. A are eliminated, and the arm of P l is 2^/5 ft. Hence P 1 X 2^/5 375 X 10 = o. . . . (2) Again, taking moments about C, P x and P 2 are eliminated, and the arm of P % is 10 ft Hence P., X 10 + 250 X 10-375X 20 = 0. . . (3) Solving these equations, P x = - 838. 5 lbs ; P 2 = + 353.6 lbs ; P, = + 500.0 lbs. The signs of these results show that P 2 and P 3 act away from the body Jf, while P x acts toward X. The forces exerted by X upon Y are equal in magnitude to P x , P 2 and P. s , but opposite in direction. Since the independent equations of equilibrium may be written in many different ways, the details of the solution may be varied indef- initely. It is often advantageous to use resolution equations instead of moment equations.* The following examples may be solved by application of the fore- going principles. Examples. 1. Determine the internal forces in all the bars of the truss shown in Fig. 73. 2. Let the truss shown in Fig. 76 sustain loads of 400 lbs. and 800 lbs. due to wind pres- sure, acting nor- mally to the roof and applied at the joints as shown. Suppose the truss * In the above discussion the aim has been merely to exhibit clearly the genera.' principle s upon which the determination of internal forces in framed structures is based. For a full treatment the reader must consult treatises devoted especially to the subject. EQUILIBRIUM OF A SYSTEM OF BODIES. 99 supported at one end by a smooth horizontal surface and at the other end by a hinge. Compute the internal forces in all the bars of the truss, neglecting all loads except those specified. [First determine the supporting forces at the ends by considering the equilibrium of the truss as a whole. Then apply the above method to the computation of the internal forces. ] 3. Equilibrium of a System of Bodies. 119. External and Internal Forces Acting on a System of Bodies. For certain purposes it is found convenient to group a number of bodies or particles together in applying the principles of Statics, such a group being called a system. When a system of bodies is considered, a force acting upon any body of the system is called internal or external according as the body exerting the force is or is not a member of the system. For example, every terrestrial body is attracted by the earth in accordance with the law of gravitation. This attraction is external if the body is considered by itself. If, however, the body and the earth are regarded as forming a system, the force mentioned is in- ternal. Thus, whether the force is external or internal depends upon what has been arbitrarily chosen as the system to be discussed. 120. Conditions of Equilibrium for a System of Bodies. It will now be shown that if every body in a system is in equilibrium, the external forces acting on the system as a whole satisfy the same conditions as if the system were a rigid body in equilibrium. The equations of equilibrium may be written for each body of the system separately. Let a resolution equation be written for each body, the direction of resolution being the same for all ; and let a moment equation be written for each body, the same origin of mo- ments being taken for all. Adding all the resolution equations, the resulting equation shows that The algebraic sum of the resolved parts of all forces acting upon the bodies is equal to zero, whatever the direction of resolution. Similarly, the addition of the moment equations shows that The algebraic sum of the moments of all the forces acting upon the bodies is equal to zero, whatever the origin of moments. Now, these equations include both external and internal forces. But since the forces exerted by any two of the bodies upon each IOO THEORETICAL MECHANICS. other are equal and opposite and have the same line of action, the sum of the resolved parts of two such forces in any direction is zero, and the sum of their moments about any origin is zero. Hence the sum of the resolved parts of all internal forces in any direction is zero, and the sum of the moments of all internal forces about any origin is zero. The following proposition may therefore be stated : If every body of a system is in equilibrium, the external forces applied to the system satisfy the following conditions : (a) The sum of their resolved parts in any direction is zero. (J?) The sum of their moments is zero for any origin* That is, the external forces satisfy the same conditions as if the system were a rigid body. The converse of this proposition is, however, not generally true. It cannot be stated that when the external forces for the system satisfy the above conditions, the forces acting on each separate body also satisfy the conditions of equilibrium. Although the general principle just deduced is often useful in the solution of problems, a complete solution usually requires in addition the employment of some at least of the equations of equilibrium for the bodies taken separately. 121. Application. Two homogeneous smooth cylinders rest, with axes horizontal, against each other and against two inclined planes making given angles with the horizontal. Required to determine the position of equilibrium. Regarding the two cylinders as a system, the external forces are the weights of the cylinders (taken as acting vertically through their centers), and the pressures exerted by the supporting planes (acting normally to the surfaces). The pressures exerted by the cylinders upon each other are internal to the system. Two solutions will be given ; one partly geometrical, the other algebraic. (See Fig. 77.) Let r' and r" denote the radii of the cylinders ; W and W" their weights ; 6 the angle between the horizontal and the plane containing the axes of the cylinders ; a and /3 the angles made by the planes with the horizontal ; R ' and R " the normal pressures exerted by the planes. (1) Geometrical solution. Applying the conditions of equi- librium to the external forces acting upon the system, notice that the resultant of W and W" is a vertical force of magnitude W -f- W" > EQUILIBRIUM OF A SYSTEM OF BODIES. IOI and is in equilibrium with R' and R", Since their directions are known, their magnitudes may be determined by means of a triangle. Fig. 77. Draw LN (Fig. 78) equal and parallel to W -f- W'\ and make NH and HL parallel respectively to R " and R ' ; then NH repre- sents R" in magnitude and direction, and HL represents R '. Also, the angle HLN = a, and HNL = j3. Hence R' R" sin /3 sin a or R' = (W -f W 7 ") FT' + W" sin (a -f- /3) sin /3 sin (a + ) sin (a -(- /3) Next, to determine 0, notice that the re- sultant of W and R' acts in a line through A, and the resultant of W" and 7?" acts in a line through i>. Since these two result- ants balance each other, each must act along the line AB. Now if, in Fig. 78, the point M be so taken that LM = W and MN = W'\ the resultant of R ' and W is rep- resented in magnitude and direction by HM, while MH represents 102 THEORETICAL MECHANICS. the magnitude and direction of the resultant of W" and R" . Hence, for equilibrium, the line AB must be parallel to HM. In the triangle LMH we have angle at L = a ; angle at H 90 (a + 0) ; angle at J/ = 90 -f ; W R' hence or sin (90 a 0) sin (90 -J" 0) W R' cos (a -f- 0) cos Solving for 0, cos (a + 0) _ W _ FT' sin (a + g) cos0 " tf' "~ JF' + IF* ' sin /3 ' r W 7 " cotan a W cotan /3 or, finally, tan a . (2) Algebraic solution. Applying the algebraic conditions of equilibrium to the four forces W, W" , R' and R" } three inde- pendent equations may be written as follows : Resolving horizontally, R' sin a R" sin ft = o. . . . (1) Resolving vertically, R' cos a + R" cos ( W + W") = o. . (2) Taking moments about .5, and noticing that the arm of R' is (r' -f r")cos(a + 0), ^' (r* + r") cosO R' (r' + r^ cos (a + 0) = o. (3) These three equations contain three unknown quantities, R', R" and 0. From equations (1) and (2), R' and R" maybe found ; and then can be determined from (3). The results will agree with those found above. If it is desired to determine the pressure between the two cylin- ders, the conditions of equilibrium must be applied to one of the cylinders alone. 4. Stress. 122. External and Internal Stresses. A stress has already been defined as consisting of two equal and opposite forces exerted by two portions of matter upon each other (Art. 36). The two forces of a stress always constitute an ' ' action ' ' and its ' ' reaction. ' ' STRESS. IO3 When any body or system of bodies is under consideration, the stresses with which the system is concerned may be either external or internal. An external stress is one exerted between two portions of matter, one of which belongs to the body or system while the other does not. An internal stress is one exerted between two por- tions of matter both of which belong to the body or system. This classification expresses no distinction as to the nature of the stresses. It is merely a distinction made for convenience in dealing with the problems of Mechanics. 123. Kinds of Internal Stress in a Body. Considering any two adjacent portions of a body, separated by a plane surface, the resultant stress between these two portions may have any direction. Whatever this direction may be, let each of the forces of the stress be resolved into two components, one normal to the surface of separa- tion and the other parallel to it. The normal components of the forces constitute a normal stress, and the other components a tan- gential stress. The normal stress may be either tensile or compressive. The former resists a tendency of the two portions of the body to separate ; the latter resists a tendency of the two portions to move toward each other. The tangential stress is called a shearing stress. It resists a tend- ency of the two portions of the body to slide past each other along the surface separating them. To illustrate the two kinds of normal stress, reference may be made to the problem discussed in Art. 118. The roof truss being regarded as consisting of two parts, X and K, it was found that Y exerts upon X three forces whose magnitudes and directions are shown in Fig. 7 5 ; while X exerts upon Y forces equal and opposite to these. It is thus seen that normal stresses act in every trans- verse section of each bar ; the stress in DC being compressive, while those in BC and BE are tensile stresses. The nature of the normal stress in any bar of a jointed frame may thus readily be found by de- termining the direction of the internal force exerted by one part of the structure upon the other. Examples. 1. Determine the kind of stress in each of the bars of the truss shown in Fig. 73, due to the loads described in Ex. 1, Art. 118. 104 THEORETICAL MECHANICS. 2. In Ex. 2, Art. 118, determine the kind of stress in each member of the truss. 124. Kinds of Stress Between Bodies. If two bodies are in contact, the forces which they exert upon each other at the surface of contact constitute stresses which may be classified in the same manner as the internal stresses in a body. The names tension, com- pression and shear are, however, confined to the internal stresses. The normal stress between two bodies in contact is usually pres- sure ; that is, the forces have such directions as to resist a tendency of the bodies to move toward each other. If the bodies for any reason have a tendency to separate, they are in general able to exert only very slight forces to resist this tendency. The tangential stress between two bodies at their surface of con- tact can have any magnitude up to a certain limit. Thus, if a body rests upon a horizontal surface, and if a horizontal force is applied to it, the supporting body exerts a force in the opposite direction. If the applied force does not become too great, the opposing force will always just equal it and the body will remain in equilibrium. A tan- gential stress thus acts between the two bodies. The forces compos- ing such a stress are of the kind to which the name friction is ap- plied. The subject of friction is treated in Chapter VII. Besides the stresses acting between bodies at their surfaces of con- tact, there are other stresses which act between bodies not in contact and without any apparent material connection. These are the so- called * actions at a distance ' ' already mentioned (Art. 40). Such stresses are frequently called attractions and repulsions, and are anal- ogous to tensile and compressive stresses respectively. 125. Strain. The external forces applied to a body usually tend to cause the different parts of the body to move relatively to each other. This tendency is opposed by the internal forces called into action. No natural body, however, remains wholly rigid under ap- plied forces, but the parts move, at least slightly, relatively to one another. Strain is the name given to the deformation which a body under- goes under the action of applied forces. In this work we shall not usually be concerned with strain, the bodies considered being regarded as rigid. Any ordinary solid body will assume a form of equilibrium under the applied forces (if they STRESS. IO5 are not such as to cause rupture) and may then be regarded as a strictly rigid body. Examples. 1. A straight bar of uniform cross-section and density, 14 ft. long and weighing 1 20 lbs. , rests in a horizontal position upon smooth supports at the ends. Conceive the bar to be divided by a plane perpendicular to the length, 4 ft. from one end. Determine (a) the resultant tangential stress and (b) the resultant normal stress between the two portions at the surface of separation. 2. With the data of Ex. 1, take a transverse section through the middle of the bar, and determine the resultant tangential stress and resultant normal stress between the two portions. 3. A body weighing 50 lbs. is at rest upon a rough horizontal surface while acted upon by a force of 10 lbs. directed at an angle of 30 upward from the horizontal. Describe completely the resultant stress acting between the given body and the supporting body. 4. A uniform bar AB, 12 ft. long, weighing 18 lbs., is supported at A by a smooth hinge and at B by a string inclined 30 to the vertical. Conceiving the bar divided in the middle by a transverse plane, determine (a) the resultant normal stress, (b) the resultant tangential stress, and (V) the resultant stress, acting between the two portions of the body. CHAPTER VII. FRICTION. 126. Smooth and Rough Surfaces. The conception of a per- fectly smooth surface has frequently been employed in the foregoing discussions. Such a surface was defined in Art. 42. The surfaces of actual bodies are, however, always more or less rough. If any two bodies are in contact, each exerts upon the other a force, the direction of which depends upon various conditions. Let the surface of one of the bodies be a plane, and consider the force P which this body exerts upon the other. Whatever be the direction of this force, let it be resolved into two components N and T, the former perpendicular to the plane surface and the latter parallel to it. (Fig. 79 shows P % N and T in magnitude and direction. The lines of action of P and N are also shown ; but T will act along a line lying in the surface of contact. ) If the plane surface be very rough, the magnitude of T may be considerable in comparison with N; but if the surface be made smoother, the magni- tude of the tangential force which can be exerted becomes smaller. We are thus led, as in Art. Fig. 79. 42, to define a perfectly smooth surface as one which can exert no force parallel to itself upon any body placed in contact with it. In other words, the resultant pressure exerted by a smooth surface upon any body in contact with it must have the direction of the normal to the surface. Although the surface of a body can never be made to satisfy this definition of perfect smoothness, there are cases which approach near to it. The surface of a sheet of ice is often nearly smooth, and the difficulty of walking on such a sheet is due to the fact that only a small tangential force can be exerted by the ice upon a body in con- tact with it. The forces exerted by two bodies upon each other at their surface of contact are of the kind called passive resistances (Art. 41). They are called into action to resist a tendency of the two bodies, due to any cause, to move relatively to each other. Thus, suppose a body FRICTION. 107 to rest upon a horizontal surface and to be acted upon by the attrac- tion of the earth ; i. c, by a downward force equal to the weight of the body. The pressure exerted upon it by the table is called into action to resist this downward force and is exactly equal and opposite to it. Let a horizontal force be now applied to the body. If this force be not too great the body will remain at rest, and in order to hold it at rest the supporting body exerts upon it, in addition to the vertical force already mentioned, a force equal and opposite to the horizontal applied force. If the magnitude of the horizontal force be gradually changed from zero to any value (up to a certain limit), or if its direction be changed, the resisting force changes in a cor- responding manner so as always to be equal and opposite to it. The resultant force exerted by the supporting body is made up of the upward force resisting the weight of the supported body, and the horizontal force just described. If additional forces be applied to the body, equal and opposite forces will be exerted by the supporting body ; but the magnitude and direction of the force that can be resisted are always subject to limitation, depending upon the material composing the bodies and the nature of their surfaces. 127. Friction. The name friction is given to the tangential component of the force exerted by one body upon another at their surface of contact. The frictional forces exerted by the two bodies upon each other are equal and opposite and constitute a stress. Experiments show that friction acts according to certain laws. In stating these laws, the term "applied forces" will be used to desig- nate all forces acting upon the body in question, except the "passive resistance" exerted by the other body at their surface of contact. 128. Laws of Friction. The following may be stated as the laws or principles to be employed in discussing the force of friction exerted upon a body by another in contact with it. (1) If the body is in equilibrium the friction is equal and oppo- site to the tangential component of the resultant of the applied forces. (2) If the tangential component of the resultant of the applied forces becomes greater than a certain limiting value, the friction can- not become great enough to balance it. The value of the friction when sliding is about to take place is called the limiting friction. (3) The magnitude of the limiting friction bears a constant ratio to that of the normal pressure between the two bodies. (4) The ratio of the limiting friction to the normal pressure is 108 THEORETICAL MECHANICS. independent of the area of contact of the two bodies, if the touching surfaces are uniform in character. (5) If the body is sliding along the surface of contact the friction is independent of the velocity and proportional to the normal pres- sure. In this case, the ratio of the friction to the normal pressure is found to be less than when sliding is about to occur. Of these five laws, (1) and (2) are exact. The first is, indeed, a direct result of the principles of equilibrium. Laws (3), (4) and (5) are only approximate, but are nearly enough true for most practical applications. 129. Coefficient of Friction. The ratio of the magnitude of the limiting friction to that of the normal pressure between the bodies is called the coefficient of friction. That is, if jjl is the coefficient of friction, fi = FIN, in which F denotes the limiting friction and N the normal pressure. Law (3) states that //, is constant for two given bodies, whatever the magnitude of N. If the pressure between the two bodies be- comes nearly great enough to crush the material of which either is composed, it is found that /jl is considerably greater than for small pressures. Within certain limits of normal pressure, however, its value is nearly constant, and it will be regarded as constant in the applications which follow. It is to be noted that the value of /x depends upon the nature of both the bodies concerned. 130. Angle of Friction. The total pressure exerted by either body upon the other is the resultant of the normal pressure and the friction. If this resultant be determined by the triangle of forces it is seen that its direction makes with the normal an angle whose tan- gent is the ratio of the friction to the normal pressure. This angle has its greatest value when the friction is as great as possible. The angle of friction is the angle between the resultant pressure and the normal when sliding is about to take place. If this angle is denoted by , it follows from the definition that tan (j> = F/N= /*. That is, the tangent of the angle of friction is equal to the coefficient of friction. FRICTION. IO9 131. Experimental Determination of Coefficient of Friction. The value of the coefficient of friction may often be determined very simply. Thus, let a body be placed upon a horizontal plane surface and let a force be applied horizontally just great enough so that the body is on the point of sliding. This force is equal and opposite to the limiting friction. Its magnitude may be determined by a spring balance. The normal pressure is in this case equal to the weight of the body, which may also be determined by the spring balance. The ratio of these two forces is the coefficient of friction. Another method of determining the coefficient of friction is as follows : Let a body be placed upon an inclined plane, and let no forces act upon it except its weight and the pressure of the supporting plane. This supporting pressure is exactly equal and opposite to the weight so long as the body is in equilibrium ; the angle it makes with the normal is therefore equal to the inclination of the plane to the horizontal. Let this inclination be gradually increased until the body is on the point of sliding. The resultant pressure exerted by the plane now makes with the normal an angle equal to the angle of friction. Hence, if the inclination of the plane in this limiting posi- tion be measured, the coefficient of friction can be found from the relation fjL = tan . Many experiments have been made to determine the value of the coefficient of friction for different materials both with and without lubricants. The range of the results is indicated by the following values * of /x : Wood on wood, dry . 0.25 to 0.5 " " " soaped . 0.2 Metals on oak, dry . 2.5 to 0.6 m " wet . 0.24 to 0.26 " " " soaped 0.2 Metals on metals, dry 0.15 to 0.2 << i< a wet 0.3 Smooth surfaces occasionally 1 ubricated 0.07 to 0.08 " V thoroughly t * 0.03 to 0.036 Ex AMPLES. 1. A body of lV\bs. mass is at rest upon a plane making angle with the horizontal. A cord attached to this body runs parallel to the plane, passes over a smooth pulley and sustains a weight of P See Encyclopaedia Britannica, Vol. XV, p. 765. IIO THEORETICAL MECHANICS. lbs. Determine the magnitude and direction of the friction, the normal pressure, and the total pressure, exerted by the plane upon the body. 2. In Ex. i, let W = 50 lbs., P = 40 lbs., = 32 , and sup- pose the body is just about to slide up the plane. Determine the coefficient of friction. Ans. ^ = 0.318 ; <= 17 40'. 3. In Ex. 1, if the coefficient of friction is 0.4 and is 30 , what must be the value of P (in terms of W) in order that the body may just slide up the plane? What value of P will just allow sliding down the plane? Ans. 0.846 W '; 0.154 IV. 4. A body of 30 lbs. mass, resting upon a plane inclined 45 to the horizontal, is pulled horizontally by a force P. If the coefficient of friction is o. 2, between what limits may the value of P vary and still permit the body to remain at rest? Ans. 20 lbs. and 45 lbs. 5. A straight bar of length /, whose center of gravity is distant a from the lower end, rests upon a horizontal surface and leans against a vertical wall. If the coefficient of friction between the floor and bar is /x, and the wall is perfectly smooth, what angle with the verti- cal may the bar make without sliding? Ans. tan -1 (fil/a). 6. If the conditions are as in Ex. 5, except that the vertical wall is rough, the coefficient of friction between the wall and the bar being /x', what is the position of incipient sliding ? 7. What are the limiting positions of equilibrium for a heavy bead on a circular wire ring whose plane is vertical ? ( Express the result in terms of the coefficient of friction.) 8. What are the limiting positions of equilibrium of a uniform bar placed wholly within a spherical bowl ? ( Express in terms of the angle of friction.) Ans. Let 2a = angle subtended at center by bar, 6 = its angle with the horizontal in the limiting position ; then tan 6 = % [tan (a -j- cj>) tan (a <)]. 9. A bar A B whose mass is 6 lbs. and whose center of gravity is 4 ft. from the end A is supported in a horizontal position by a string attached at A and a peg 7 ft. from A. The coefficient of fric- tion between the bar and the peg is 0.3. If the bar is about to slide in the direction AB, determine the direction of the string, the ten- sion it sustains, and the supporting pressure exerted by the peg. Ans. Tension = 2.77 lbs.; resistance of peg = 3.58 lbs.; angle between string and bar = 68 12'. 10. In the preceding case, between what limits must the direction of the string lie in order that there may be equilibrium ? Determine the corresponding limiting values of the supporting forces. n. Solve the preceding problem in the following general case: Let W = weight of bar, a = distance of its center of gravity from A, b = distance of peg from A, fi = coefficient of friction. FRICTION. Ill 132. Friction in Machines. In order that a simple machine may produce its desired effect, its motion must be constrained, that is, guided in a definite manner. This is accomplished by means of bodies which touch the machine parts at certain points and exert constraining forces upon them. Besides the forces which produce the desired constraint, there are usually frictional forces which resist any tendency of the bodies to slide over each other at their surfaces of contact. In general friction is a wasteful resistance ; but, as will be seen, it may in certain cases act with the effort and thus aid the useful object of the machine. The effect of friction on the operation of a machine may be estimated by means of the laws of friction above stated. The method of applying them is illustrated in the following simple case. 133. Fixed Pulley with Friction. If a pulley is mounted freely upon a cylindrical axle, the frictional force exerted by the axle upon the wheel always acts in such a way as to oppose what- ever rotation the wheel has or tends to have by reason of the action of other forces. Referring to Fig. 80, let a de- note the radius of the pulley; r the radius of the axle ; P the effort ; W the load ; R the resultant pressure exerted by the axle on the pulley ; N the normal component of R ; F the friction, or tangential component of R ; fi the coefficient of friction ; cf) the angle of friction. The cylindrical hole into which the axle fits is slightly larger than the axle, but they may be regarded as practically equal in size. If the wheel is on the point of moving with the effort, the equation of moments, taking origin at center of wheel, is Pa Wa Fr = o, Fig. 80. or P= W+Fr/a, (1) which shows a mechanical disadvantage. 112 THEORETICAL MECHANICS. If the wheel is on the point of moving against the effort, the direction of F is reversed, and the equation is P = WFr/a, .... (2) showing a mechanical advantage. The value of F in case of incipient motion may be determined as follows: In this limiting case, F = N tan , the total resistance R acting at an angle (j> with the normal to the surfaces at the point of contact. But this total resistance must act in a line through the intersection of the lines of action of P and W. Let be the value of the angle CAB (Fig. 80); then sin sin r a cosec a . '. sin =1 - sin sin a. (3) a Since R, P and W are in equilibrium, we have R P W sin 2a sin (a -f- 0) sin (a 0) D P sin 2a W s\n 2) / (a r sin <) or P/lV=(a r sin )/(a -\- r sin $), according as there is incipient motion with P or against P. Show that the first of these equations is included as a special case in the above general solution. 3. With data as in Ex. 1 , except that P and W are parallel, de- termine the relation between P and W iox both cases of incipient motion. Determine also the value of /^and of R in each case. Ans. Pj W '= 0.952 or 1.050. 8 CHAPTER VIII. EQUILIBRIUM OF FLEXIBLE CORDS. 134. Natural Bodies Not Perfectly Rigid. A perfectly rigid body could not be deformed in any degree by the action of external forces. Actual bodies, however, undergo changes of form and size when forces are applied to them. In some cases the changes are so slight as to be of little importance. Thus, a steel beam, resting upon two supports, may bend only slightly under very heavy loads. For certain purposes this bending is unimportant, while for others it must be taken into account. The statics of non-rigid bodies is less simple than that of rigid bodies. Certain classes of problems may, how- ever, be treated without great difficulty by means of the following general principle. 135. Non-Rigid Body in Equilibrium. The conditions of equi- librium for a rigid body apply to any body, every portion of which is in equilibrium. Suppose a body, acted upon by any external forces, to be de- formed in any manner, reaching finally a condition of equilibrium. This condition being attained, the rigidity of the body is of no further consequence so far as its equilibrium is concerned. The external forces must therefore satisfy the same conditions as if the body were actually rigid. 136. Flexible and Inextensible Cord. A flexible cord has been defined as one which may be bent by the application of exter- nal forces. A perfectly flexible cord would offer no resistance to bending. The following discussions will relate to cords assumed to possess this ideal property of perfect flexibility. The results are of practical value, since many actual cords are flexible to a high degree. It will be assumed further that the cords are inextensible. The forces acting upon a cord may be concentrated or distributed. (Art. 37). These two cases will be considered separately. 137. Geometrical Conditions of Equilibrium of Cord Acted Upon by Concentrated Forces. Consider a perfectly flexible, weightless cord, suspended from two fixed points, and acted upon by EQUILIBRIUM OF FLEXIBLE CORDS. "5 any concentrated forces. Thus, let Jf and Y (Fig. 81) be the fixed points of suspension, and let forces P x , P 2 , P 3 be applied at points L, M, N. Suppose the cord to have assumed a form of equilibrium, the segments XL, LM, MN, NY being straight. Let T x , T 2i T 3 , T A denote the tensions in the successive segments of the string, taken in order from X to Y. Then there is a definite relation between any- applied force, as P x , and the tensions in the adjacent portions of the string, as T x and T 2 . By Art. 115 any portion of the string may be treated as a separate body, and the conditions of equilibrium may be applied to the forces acting upon it. For the portion of the string near L the forces are three, namely, P x and the forces due to the tensions in XL and LM, that is, a force T x in the direction LX, and a force T % in the direction LM. Similarly, for the por- Fig. 81. tion of the string near M we have three forces in equilibrium : the force P tt a force T 2 in the direction ML, and a force T s in the direction MN. A similar analysis applies to the point N. If P lt P 2 and P z are known, the tensions may be found, either geometri- cally or by writing the equations of equilibrium for each of the sets of forces named. The geometrical construction is shown in Fig. 81. The line AB is drawn to represent P x in magnitude and direc- tion ; then BO and OA drawn parallel to LM and LX represent the forces T. 2 and T x . Draw BC to represent P 2 in magnitude and direc- tion ; then for the equilibrium of the portion of the cord near M, CO must 1 epresent the force T. A . Similarly, if CD represents P 3 in magnitude and direction, DO must represent the force T K . The Il6 THEORETICAL MECHANICS. figure * thus shows at a glance the relations of magnitude and direc- tion that must subsist among the forces P ly P 3i P. 6 and the tensions -* 1 J * % I -* 3 > M 4 Algebraically, the relations between these quantities may be ex- pressed by writing two equations of equilibrium for each of the three systems of forces acting at L, M and N. 138. General Problem of Cord Acted Upon by Concentrated Forces. If a cord fixed at two points is acted upon by concentrated forces, numerous problems may arise, depending upon what quanti- ties are given and what are required. Thus, in the case represented in Fig. 8 1 , the quantities involved are the applied forces P x , P. l and P A ; the tensions T x , T. t , T 3 , T t ; and the coordinates which give the positions of the points X, L, M, N, Y. (Instead of these coordi- nates, we may use the lengths and directions of the segments XL, LM, etc.) If enough of these quantities are known the others may be determined. Many of the problems which may arise do not admit of simple treatment. If the problem is determinate, as many equa- tions may be written as there are unknown quantities. Part of these are the equations of equilibrium, written for each point at which a force is applied in the manner above indicated ; the others must be written from the geometrical relations of the figure. The following examples illustrate simple cases. Examples. 1. Let a single vertical load of /Mbs. be suspended from a cord at distances / and m from its ends ; and let the ends X and Y be fixed in such positions that the distance XY is a and the line XV makes with the horizontal an angle 6. Determine the tensions in the cord. 2. In the preceding example let / = 10 ins., m = 8 ins., a 12 ins., 6 = 30 . Ans. 0.0744 P and 0.988 P. 3. Let the ends of the cord be fixed at points in the same hori- zontal line at a distance apart equal to a, and let vertical loads /^and Q be suspended at points dividing the length of the cord into seg- ments /, m and n. Determine the relation between P and Q so that * The construction here described is of importance in Graphical .Statics. The polygon formed by the successive segments of the cord is called a funic- ular polygon. (Sometimes called also equilibrium polygon and string polygon.) EQUILIBRIUM OF FLEXIBLE CORDS. 117 the middle segment may be horizontal. Prove that, if the middle segment is horizontal, the following relation must be satisfied : (P- Q)I(P + Q) = ( a - ni(a - my. 4. If, in the preceding example, a = 12 ins. ,7=4 ins. , m = 6 ins., ?i = 6 ins., prove that P = 3.5Q. Determine the tensions, assuming Q = 10 lbs., P = 35 lbs. Ans. Tension in horizontal cord == 12.37 lbs. 139. Equations of Equilibrium for Any Part of Cord Carry- ing Distributed Vertical Load. Let it be required to determine the form of the curve assumed by a cord suspended from two fixed points and sustaining a vertical load distributed along the cord in any manner. Let the coordinates of any point of the curve ( Fig. 82) be x and y, the axis of x being horizontal and that of y vertical, and the origin being any point in the plane of the curve. Let cf> denote the angle between the tangent to the curve at any point and the axis of x, and T the tension in the cord at that point. Also let w be the load sustained by the cord per unit length (made up of its own weight and any other applied load). In gen- eral w varies continuously along the cord. Let P and Q be any two points of the cord, and let the values of x % y, T and at P be denoted by x ly y i} 7\ , (/> t ; while at Q the val- ues are x 2 , y, , T 2 , 2 7", cos (f> l = o; . . . (1) resolving vertically, T 2 sin , T x sin <^> 1 ^f=o. . . (2) A moment equation might also be written, but is not needed in the present discussion. Equation (1) expresses the fact that the horizontal component of ''*V' 4h O Fig. 82. Il8 THEORETICAL MECHANICS. the tension has the same value at all points of the cord. Let this value be represented by H, so that T x cos (/>j = T % cos (j> 2 = T cos = H = constant. . (3) Substituting in equation (2) the values T x = Hj cos lt T 2 = HI cos 2 , that equation becomes tan (f> t tan , = WjH. . . . (4) Equations (3) and (4) may be used instead of (1) and (2). If the values of T and are known at any one point of the cord, H becomes known. If also W is known for every portion of the cord, equation (4) serves to compute the slope of the curve at any point. A differential equation may, however, be deduced which is more useful. 140. Differential Equation of Curve of Loaded Cord. Let s denote the length of the curve measured from some fixed point up to the point (x, y), and let s x and s % be the values of s at P and Q. From equation (4) we have (tan & - tan fa)/ (s. z - s t ) = W/H (s. z - *,) Let the limiting value of each member of this equation be found as P and Q approach coincidence. Evidently, limit [(tan (j> 2 tan x ) / (s t s x )] = ^(tan )/ds; limit [W/(s 2 s x )] = w. Hence the equation becomes d(tan)/ds = wjH. . . . (5) To apply this equation to any particular case, the distribution of loading must be known, so that w can be expressed in terms of the coordinates of the curve. Two important cases will be considered. 141. The Common Catenary. The curve assumed by a cord whose weight per unit length is uniform, is called the common cat- enary. In this case w is constant ; let Hj w = c, so that the constant c denotes the length of cord whose weight is equal to the horizontal tension H. Writing/ for tan $, equation (5) becomes dp jds = ijc. . . . . (6) EQUILIBRIUM OF FLEXIBLE CORDS. 119 Since dp Ids = {dp / dx){dx / ds) == (dpjdx) cos , and cos = i/l/ 1 -j- / 2 , equation (6) may be written Integrating, dp/Vi + p 2 = dx/c. (7) log(/ + l/i+/*)= */'+*". To determine the constant ^T, suppose the tangent of the curve at some point to be horizontal, and let the origin of coordinates lie in a vertical line through that point, so that when x equals zero, p equals zero. Then K equals zero, and the equation may be written *x\c Solving for p, Integrating, p = dy/dx == y 2 {e xic e-*' c \ y = y 2 c(e x!c -f e-*' c ) + K'. (8) (9) . (10) If the origin of coordinates is taken at the lowest point of the curve, so that y = o when x = o, K' is equal to c. If the origin is at a distance c below the lowest point of the curve, K' = o. Adopt- ing the latter supposition, the equation becomes ;= %c(e*l* + e~ xlc \ (11) In the solution of prob- lems relating to the catenary, use is made of certain rela- tions between the following quantities : the length of the cord ; the tension at any point ; the tension at the lowest point ; the inclination of the tangent at any point ; the coordinates of any point. The following equa- tions may be deduced : (a) From equation (6), by integration, Fig. 83. s = cp = c tan (12) the constant of integration being zero if s is reckoned from the point at which (f> = o, that is, from the lowest point of the curve. 120 THEORETICAL MECHANICS. (/?) Eliminating x between equations (9) and (11), there results y = cv 1 -\- p 2 == ^sec . . . . (13) (V) Eliminating/ between (12) and (13), r-s'=c 2 (14) The use of these equations will now be illustrated. Examples. 1 . Let a cord be suspended from two points at a given distance apart in the same horizontal line, the load per unit length (w) being known, and the tension at the lowest point (//) being assigned. Required (1) the length of the curve, (2) the position of the lowest point, (3) the slope at points of suspension, (4) the tension at points of suspension. Since w and H are given, c is known, its value being 77/ w. Also, the value of x at a point of suspension is known (being half the distance between the two given points); hence from (11) the value ofjj/at a point of suspension can be computed. Subtracting cfrom this value of y gives the depth of the lowest point below the points of suspension ; which answers question (2). The length of half the curve may be found by substituting in equation (14) the value of c and the value ofy at the point of sus- pension. This answers question (1). From (13) may be found the value of at the point of suspension, which answers question (3). From equation (3), T //sec , from which may be determined the value of T at the points of sus- pension, thus answering question (4). In solving a numerical case of this example, the value of the exponential terms in equation (11) must be found by logarithms. The value of e, the base of the system of hyperbolic logarithms, is 2. 7 1 828 1 8 and its common logarithm is 0.4342945. 2. Let the points of suspension be 100 ft. apart in the same hori- zontal plane, let the load per foot be 50 lbs., and let it be required that the tension at the lowest point shall be 1000 lbs. From the given data, c = Hjw = 1000/50 = 20 ft. The coordinates of one of the points of suspension are x = 50 ft., y = 10 O 2 - 5 -J- *- 25 ) =122.6 ft. The lowest point of the curve is therefore 102.6 ft. lower than the points of suspension. EQUILIBRIUM OF FLEXIBLE CORDS. 121 For the half length of the curve, s 2 = y* c 2 = (122.6) 2 (20) 2 ; s = 1 2 1 .0 ft. For the value of (f> at the point of suspension we have cos (f> = c/y = 20/122.6 = o. 163 1 ; = 8o 37'. The value of T at the end is T== H sec <\> = 6, 130 lbs. 3. A cord of known length is suspended from two given points in the same horizontal plane ; determine the position of the lowest point of the curve, and the tension at any point. In this case the value of x at the end of the cord is known, but the corresponding value of y is unknown, as is also the value of c. To solve the problem, y and c must be determined so as to satisfy equations (n) and (14). This is best accomplished by trial when numerical data are given. 142. Load Distributed Uniformly Along the Horizontal. Let w' denote the load per unit horizontal distance. Since the load on a length ds of the curve falls upon a horizontal distance dx, wds = w'dx, or w' =5 w{ds/dx) = w sec $. . . (15) Consider now the case in which w' is constant. Writing/ for tan , and substituting w ' cos (j> for w, equation (5) (Art. 140) may be written dp jds = (w' cos <\>)IH = {cos )/r', . . (16) in which c' is written for 11/ w', and is a constant denoting the hori- zontal distance upon which the load is equal to the horizontal tension H. Again, since dp/ds = (dp/dx)(dx/ds) = (dp/dx) cos (/>, the equation becomes dp/dx = d 2 yldx* = i/c'. . . (17) Integrating, taking the axis of y through the point where the curve is horizontal, so that when x = o, dy/dx = o, there results p = dy/dx = x/c f . . . . (18) Integrating again, taking the origin on the curve, so that y = o when x = o, y=x 2 /2c' (19) This represents a parabola with vertex at the origin of coordinates and axis vertical. ds = c'V \ + p 2 - dp, 122 THEORETICAL MECHANICS. The length of the curve, measured from the origin to any point, may be determined as follows : Since cos a= l/vj -f- p 2 , equation (16) may be written ds = c'V the integral equation of which is s = y*cXpVi -\-p 2 + log (p + Vi +/)], . (20) the constant of integration being zero if s ..ass o when p = o. If it is desired to express s in terms of x, p may be eliminated by means of equation (18). Examples. 1. A cord carrying a known load with uniform horizontal distri- bution is suspended from two points on the same level at a given distance apart. The lowest point of the cord is at a known distance below the points of suspension. To determine (a) any number of points of the curve ; (b) the tension at any point ; (c) the length of the cord. (a) Since the values of x and y at the points of suspension are known, c can be determined from equation (19); after which the same equation serves to determine the coordinates of any number of points of the curve. (b) The value of w' being known, //"can be computed from the relation H = c'w'. The slope at any point may be found from the equation tan =/ = dy\dx = xlc\ and the tension at any point by the relation T = H sec a= H V\ +f. (V) The length of the cord from the lowest point up to any given point may be found from equation (20), using the above value of p. (2) A cord carrying 40 lbs. per horizontal foot is suspended at two points on the same level 40 ft. apart, so that the lowest point of the cord is 10 ft. below the point of suspension. Required the ten- sion at the lowest point ; the tension at a point of suspension ; and the length of the cord. 143. Approximate Solution of Problem of Loaded Cord. If a cord is suspended from two points on the same level whose dis- tance apart differs little from the length of the cord, the relation between the load, the sag,* the distance between points of sus- * By the sag is meant the vertical distance of the lowest point of the cord below the level of the points of support. EQUILIBRIUM OF FLEXIBLE CORDS. 1 23 pension and the length may be determined approximately as follows : If the weight per unit length is uniform, it may be assumed, with small error, that the load is uniformly distributed along the hori- zontal, and that w' = w. On these assumptions the curve is a para- bola and equations (19) and (20) of Art. 142 are applicable. An approximate value of s may be obtained in simple form. If the value of s given by equation (20) be developed in powers Oi /, the first two terms of the result are This result, expressed in terms of x, is s = c'{xlc' + x 3 [6c n ) = * (1 -f xV6c n ). Examples. 1. A wire weighing o. 1 lb. per "yard of length is suspended be- tween two points 100 ft. apart so that the sag at the middle is 2 ft. Determine the length and the greatest tension. The equation of the curve being y = x 2 /2C f , the value of c', found by substituting in this equation the coordinates of any point of the curve, is c r as x 2 l2y = 2500/4 = 625 ft. The tension at the lowest point is H = c'w' = 625/30 3= 20.8 lbs. The value of p at the support is p = x/c' = 50/625 = 0.08, and the value of the tension at th e point of s uspension is therefore T H sec = 20.8V 1 -f- 0.0064 = 2 -9 l s - The half length of the curve is s = c'(p + p s /6) = 625 (0.08 -j- 0.000512/6) = 50.053 ft. 2. Solve Ex. 1, assuming the sag equal to (a) 1 ft. and (b) 3 ft. 144. Suspension Bridge. The cables of a suspension bridge usually hang in parabolic curves. If the total load carried by such a cable has a uniform horizontal distribution, there is no tendency of the cables to depart from the parabolic form. If, however, this dis- tribution of loading is departed from, there is a tendency to change the form of the curve assumed by the cable. To counteract this tendency to distortion is the office of the "stiffening truss." For a complete discussion of the theory of suspension bridges, works on the theory of bridges must be consulted. CHAPTER IX. CENTROIDS. i. Centroid of a System of Parallel Forces. 145. Forces with Fixed Points of Application. In the discus- sion of systems of forces applied to rigid bodies, it has been assumed that a force may be applied at any point in its line of action, since its effect upon the motion (or tendency to motion) of the body is the same for all such points of application. In certain cases, however, forces are applied at definite points which remain fixed in the body, whatever displacement it may undergo. Thus, the force of gravity acting upon a body is the resultant of forces applied to every particle of the body and these points of application remain fixed in the body, however it may be turned from its original position. In the following discussion of systems of forces with points of application fixed in the body, it will be assumed that the forces are parallel, and that they remain constant in magnitude and direction, however the body may be displaced. It obviously amounts to the same thing, so far as the relations of the forces are concerned, whether their direction remains fixed while the body turns, or whether their direction changes while the body remains fixed. 146. Resultant of Two Parallel Forces with Fixed Points of Application. Let two parallel forces be applied at fixed points, A and B. Their resultant is equal to their algebraic sum, and its line of action divides the line A B into segments inversely proportional to the given forces (Art. 86). Let C (Fig. 84) be the point in which the line of action of the resultant intersects AB. If the two given forces remain parallel while their direction changes in any manner, their points of application still being A and B, the line of action of the resultant will continue to pass through the same point C ; hence C may be taken as the fixed point of application of the resultant. 147. Resultant of Any Number of Parallel Forces with Fixed Points of Application. If, in addition to the two parallel forces applied at A and 73, there is a third force parallel to them applied at CENTROIDS. 125 a point D, the resultant of the three may be found by combining the third with the resultant of the first two. Since the resultant of the first two has a fixed point of applica- tion for all directions in which the forces may act, and since a point may also be found which may be regarded as the fixed point of application of 1 y / v * - 2 the resultant of this force and the third of the given forces, it follows R that the line of action of the resultant of the three given forces will always pass through a certain point (as E, Fig. 84), which may be taken as its fixed point of application. Since this process may be continued to include any number of forces, it is seen that the resultant of any number of parallel forces with fixed points of application acts in a line which always passes through a certain fixed point, whatever the direction of the forces. 148. Centroid. The point of application of the resultant of any system of parallel forces with fixed points of application is called the centroid of the system. Methods of determining the centroid will now be considered. 149. Determination of Centroid of Coplanar Forces. Let 7^, P , . represent the given forces, and let the coordinates of their points of application with reference to a pair of rectangular axes be (x x , yd* ( x -i *y*)> - - - - Let R denote the resultant and x, y the coordinates of its point of application, that is, the coordinates of the centroid of the system. Assume the forces to act first parallel to the axis of y and then parallel to the axis of x. The magnitude of the resultant is the same in both cases, and in each case it acts in a line which contains the centroid. Since the moment of the resultant of any system of forces is equal to the algebraic sum of the moments of the several forces, we may write, for the case in which the forces act parallel to the axis of y } the equation Rx = P x x x + P t x % + \ . . ; and for the case in which they act parallel to the ^r-axis, *?-/Wr+"J&.+ 126 THEORETICAL MECHANICS. From these, since R is the algebraic sum of the given forces, the coordinates of the centroid are found to be X = (P X X X + P 2 X 2 + . . . )/(/> +/*+:.,,)== ^PX^P\ y = tp^ + i\y,+ . . . )/(/; + /y+ . . . )=2/>/sp. Here the sign 2 denotes the summation of a series of terms of sim- ilar form. 150. Non-Coplanar Parallel Forces. It will be well to extend the discussion to cases in which the forces, though parallel, are not restricted to a plane and the points of application have any positions in space. The reasoning of Arts. 146 and 147, showing that every system of parallel forces with fixed points of application has a centroid, ap- plies to this general case. To determine the position of the centroid, the following method may be employed. (a) Coplanar points of application. If the points of application are coplanar, while the direction of the forces is unrestricted, the centroid may be determined by assuming the forces to act in the plane of the points of application. If the axes of x and y are chosen in this plane, the coordinates of the centroid are given by the formulas deduced for coplanar forces. (b) Non-coplanar points of application. Let s lt z 2 , etc., denote the ordinates of the points of application of the several forces meas- ured from any assumed reference plane. Any two of the given points of application lie in a plane perpendicular to the reference plane ; hence the formula deduced for the case of coplanar points of application applies to any two forces. Or, if z' is the ordinate of the centroid of P x and P 2 , Similar reasoning applies to the two forces (P t -f- P 2 ) and P 3 ; if z" denotes the ordinate of the centroid of these forces, *' = [(/> + />,) *' + /,.*,]/[(/>, + /) + PZ = (P.ir, + P,s,+ P t *,)KP, + P % + P 3 ). By extending this reasoning, any number of forces may be included, with the result that the ordinate of the centroid is j? = (/y*, + ^ 2 + . . . )/(/>, + p 2 + . . . ) = zpz/zp. CENTROIDS. 127 Since the plane of reference may be any plane whatever, an indefi- nite number of equations of the above form may be written. Three equations, however, suffice for the determination of the centroid. If three rectangular coordinate planes are selected, and the points of application of the forces are (x ly y A , z x ), (x-i,y etc -> tne coor- dinates of the centroid are given by the three equations x = 2,Pxp.P; y = ZPy/ZP; Z = ZPs/ZP 2. Centroids of Masses, Volumes, Areas and Lines. General Method. 151. Center of Mass Defined. If parallel forces be conceived to act in the same direction upon all portions of a body, such that the resultant forces acting upon any two portions, however small, are proportional to their masses, the centroid of this system of forces is called the center of mass of the body. The forces of gravity acting upon all portions of a body form a system which is practically such as described in this definition. The centroid of the forces of gravity is called the center of gravity of the body. The terms center of mass and center of gravity are often used interchangeably, although it is to be remembered that the forces of gravity do not exactly satisfy the conditions assumed in the defini- tion of center of mass. The determination of the mass-center of a body or of a system of bodies is a special case of the determination of the centroid of a system of parallel forces. The word centroid is, in fact, often used to designate the center of mass. 152. General Method of Determining Center of Mass. Let the given body be divided into any number of parts whose masses areWp m ai . . . , and let the coordinates of their respective centers of mass be (x lf y lt z x ), (x. it y t ,z i ), .... Assuming forces such as described in the definition of mass-center (Art. 151) to act upon all portions of the body, the problem is to find the point of application of the resultant of these forces. Now, from the definition of mass- center it follows that the forces acting upon all parts of tn x may be replaced by their resultant, taken as applied at the mass-center of tn x ; and similarly with each of the other portions. Hence,'if x, y } z are the coordinates of the mass-center of the whole body, there may 128 THEORETICAL MECHANICS. be written three equations similar to those which give the cooordi- nates of the centroid of any system of parallel forces (Art. 150); the force-magnitudes being replaced by the masses m lt m. n . . . . The equations are, therefore, x = ^mx/^m ; y = 'Zmy/'Zm ; z = ^mz /2w. To apply these formulas, it is necessary to know the center of mass of each of the separate masses. If the given body can be sub- divided into finite parts whose masses and mass-centers are known, the application of the formulas is simple. If such subdivision is impossible, resort must be had to the integral calculus, if an exact determination is required. The above formulas are directly applicable if it is desired to de- termine the center of mass of any number of bodies taken together, their several masses and mass -centers being known. Examples. 1. Find the center of mass of a system of four bodies whose masses are 10 lbs., 50 lbs., 12 lbs., 40 lbs ; their several mass-centers being at the points whose rectangular coordinates are (10, 4, 3), (12, 3, 2), (4, 5, 6), (8, 3, 6). 2. Show that the mass-center of three particles of equal mass placed at the vertices of a triangle lies at the intersection of the medians. 3. Show that the mass-center of four particles of equal mass placed at the vertices of a triangular pyramid is at a distance from the base equal to one-fourth the altitude. 153. Density. A body is said to be homogeneous if the masses of any two portions, however small, are proportional to their volumes. The density of a homogeneous body is a quantity proportional directly to the mass and inversely to the volume. If the unit density is that of a body of which unit volume contains unit mass, the den- sity of any homogeneous body is equal to the mass of unit volume of the body. If p denotes the density, V the volume and M the mass, p = M/V, or M= pV, for a homogeneous body. Dimensions of unit density. The unit density above described is a derived unit (Art. 13), depending upon the units of mass and CENTROIDS. 129 length. This dependence is expressed by the dimensional equation (Art. 15) (unit density) === (unit mass)/(unit volume) = M/L 3 . If the pound is taken as the unit mass and the foot as the unit length, the unit density is that of a body of which each cubic foot of volume contains one pound mass. The density of any body is then ex- pressed as a certain number of pounds per cubic foot. Other units of mass and length give other units of density, for example a gram per cubic centimeter, or a kilogram per cubic meter. The methods of determining center of mass are independent of the particular unit of density. Average density. If a body is not homogeneous, its average density may be defined as the density of a homogeneous body whose total mass and total volume are equal to those of the given body. The average density is thus equal to the whole mass divided by the whole volume. Actual density at a point. In case of a heterogeneous body, the average densities of different portions are unequal. The concep- tion of actual density at a point may be reached as follows : Let AM be the mass and A V the volume of any portion con- taining the given point ; its average density p' is then p ; = AM/AV. If A V become smaller, approaching zero as a limit, but always con- taining the given point, AM also generally approaches zero, and p' approaches a definite limit p. This limiting value of p' is the den- sity at the point considered. That is, the required density is given by the equation p ^ dM j dVt 154. Mass-Center of Homogeneous Body or System of Bodies. For a homogeneous body, or for a system of such bodies of equal densities, volumes may be substituted for masses in the for- mulas for the coordinates of the mass-center. For if p is the den- sity of each body, and v x , v t \ . . . are the volumes correspond- ing to masses m x , ffht , we have m l = pv x , m 2 = pv 2i . . . ; and substituting these values in the formulas for x, y } and 2, the factor p cancels, leaving x = 1vx/1v ; y = ^vy/^v ; 2 = lLvz[S*v. I30 THEORETICAL MECHANICS. 155. Centroids of Volumes, Surfaces and Lines. Volumes. If parallel forces be conceived to be applied to all portions of a body, the resultant forces acting upon any two portions, however small, being proportional to their volumes, the centroid of this system of forces is called the center of volume (or volume- centroid} of the body. The centroid of a given volume evidently coincides with the mass-center of a homogeneous body occupying the volume, and may be found by the formulas already given. Surfaces. The centroid (or center of area) of any surface is the centroid of a system of parallel forces conceived to be applied to all portions of the surface, the resultant forces acting upon any two por- tions, however small, being proportional to their areas. This defini- tion applies to curved surfaces as well as to plane areas. In case of a plane surface the centroid lies in the plane ; but the centroid of a curved surface does not in general lie in the surface. The formulas for the coordinates of the centroid of a homogeneous body or of a volume may be applied in finding the centroid of an area, if v lt tf, . . . represent partial areas and (x x , y x , 2 X )> (*2 > y* 1 #*) . . . the coordinates of their centroids. Lines. The centroid (or center of length) of any line is the cen- troid of a system of parallel forces conceived to be applied to every portion of the line, the resultant forces acting upon any two portions, however small, being proportional to their lengths. If the line is divided into parts whose lengths and centroids are known, the centroid of the whole line is found by formulas identical in form with those used for finding the coordinates of the centroid of a volume or area ; but^, v 2 , . . . must represent lengths. 156. Moment with Respect to a Plane. The moment of a mass with respect to a- plane is defined as the product of the mass into the distance of its mass-center from the plane. The moment of a volume with respect to a plane is the product of the volume into the distance of its centroid from the plane. The moment of a surface (or of a line) with respect to a plane is the product of its area (or length) into the distance of its centroid from the plane. The formula for the jr-coordinate of the centroid of any mass (Art. 152) maybe written {m x -f m 2 -j- . . . )x = m x x x + m 2 x % + , , . . CENTROIDS. 131 Since the plane from which x is measured may be any plane, the equation expresses the proposition that The moment of any mass with respect to a plane is equal to the sum of the moments of any parts into which it may be divided. The same proposition obviously holds for a volume, a surface or a line. As a special case, if the given plane contains the centroid of the body, surface or line, the sum of the moments of the parts is zero. I 57 Symmetry. If a volume, a surface or a line has a plane of symmetry, the centroid lies in this plane. For the volume, sur- face or line can be divided into pairs of equal elements such that the centroid of each pair lies in the plane of symmetry. Similarly, if there is an axis of symmetry it contains the centroid ; and if there is a center of symmetry it coincides with the centroid. A similar proposition is true for a mass if homogeneous. From the principles of symmetry the centroids of many geometri- cal figures may be located, either completely or partly, by inspection. Thus : The centroid of a straight line is at its middle point. The centroid of a circular arc lies upon the line bisecting the angle subtended at the center. The centroid of a rectangular area lies upon the line bisecting two opposite sides, and is therefore at the intersection of the two lines so drawn. The centroid of the area of an ellipse is at its center of figure. The centroid of an ellipsoid is at its center. 158. Centroids of Plane Figures and of Volumes. Parallelo- gram. Let A BCD (Fig. 85) be a par- allelogram. Since the relation of the A. Lj B area to AB and its relation to the oppo- site side CD are exactly similar, the centroid is equally distant from these two lines. For a like reason it is equal- ly distant from AD and BC. It is there- j} ~M fore at the intersection of the bisectors p IG 8s LM, UK. This point of intersection is, in fact, the center of symmetry of the parallelogram. It is also the point of intersection of the diagonals. 132 THEORETICAL MECHANICS. Triangle. In the triangle ABC (Fig. 86) draw AX bisecting the side BC, and draw lines such as B' C, parallel to BC and lim- ited by AB and A C. These lines are all bisected by AX. Draw B 'b and C'c parallel to AX, thus constructing a parallelogram B ' C'cb. The centroid of every such paralelogram lies on A X, hence the cen- troid of the figure made up of all of them lies on AX. If the num- ber of the parallelograms be increased indefinitely, the width of each approaching zero, this figure approaches coincidence with the triangle ; hence the centroid of the triangle lies upon AX. Similar reasoning shows that it lies on the line drawn from C bisecting AB, and on the line drawn from B to the middle point of A C. Hence it is at the common point of intersection of these three lines. Let M ( Fig. 86) be the middle point of BC, N the middle point of A C, and G the point of intersection of A M and BN. By comparison of the similar triangles ABC and NMC, it is seen that NM is equal to half of AB. Comparing the similar triangles GAB and GMN y it is seen that GMis half of A G and GN half of BG. Hence GM is one-third of AM, and GN one-third of BN. Any plane polygon, regular or irregular, may be subdivided into triangles whose areas and centroids may be computed ; so that the centroid of the whole area can be determined by the general formulas of Art. 154. Right prism. A right prism may be generated by the motion of a plane area perpendicular to itself. The centroid of the area will describe a line which must contain the centroid of the prism. For, any elementary areas will generate volumes proportional to the areas ; and if moments be taken with respect to any plane con- taining the line generated by the centroid of the generating area, the moments of the elementary volumes will be proportional to the moments of their generating areas. But the moment of the whole area is zero for such a plane ; hence the moment of the whole volume is also zero, and its centroid therefore lies in the plane. The centroid is obviously equidistant from the two bases of the prism. CENTROIDS. 133 Oblique prism. An oblique prism may be generated by the rectilinear motion of a plane area in a direction not perpendicular to itself. The reasoning given for the case of a right prism may be applied to this case, with a like result. Triangular pyramid. The centroid of the volume of a triangular pyramid lies in the line joining the vertex with the centroid of the base. To prove this, let A be the vertex, BCD the base, and Mthe centroid of the base (Fig. 87). Let B' CD' be a plane section* par- allel to BCD, M' being its centroid ; it may be proved without difficulty that M' lies on the straight line AM. Let B" C" D" be another plane section parallel to the base, and let a prism be generated by the translation of B'C'D' parallel to AM until it falls into the plane B" C" D" . The centroid of this prism lies in AM. Let any number of plane sections parallel to BCD be taken, and let each be the base of a prism generated in the same way as the one de- scribed. Since the centroid of each prism lies upon A M, so also does the centroid of their combined volume. If the number of plane sections be increased indefinitely, their distance apart approaching zero as a limit, the volume of each elementary prism approaches zero, but their com- bined volume approaches that of the pyramid ; hence the centroid of the pyramid lies upon AM. To determine the position of the centroid, let N (Fig. 87) be the centroid of the face A CD ; then the centroid of the pyramid must lie on BN, and must therefore be the point of intersection of AM and BN. If E is the middle point of CD, M lies upon EB, and EM= EB/3; also, N lies upon EA, and EN = EA/3. The triangles EAB and ENM are therefore similar, and NM = * The sections B'C'D' and B"C"D" are not shown in Fig. 87 ; but it is to be understood that like letters denote corresponding vertices of the three triangles BCD, B'C'D', B"C"D". 134 THEORETICAL MECHANICS. AB/3. Again, comparing the similar triangles GAB, GMN, it is seen that GM=AG/ 3 = AMJ^; GN = BG/3 = BN/4. Therefore, the distance of the centroid from the base is equal to one-fourth the altitude. Any pyramid or cone. The above proof that the centroid lies upon the line drawn from the vertex to the centroid of the base applies to a pyramid whose base is any polygon. Moreover, if the base be divided into triangles, the pyramid can be divided into tri- angular pyramids whose centroids all lie in a plane parallel to the base and at a distance from it equal to one-fourth the altitude ; hence the centroid of the given pyramid also lies in that plane. If the base is bounded by any curve, this may be regarded as the limit of an inscribed polygon the number of whose sides is increased indefinitely. Hence the following proposition may be stated : The centroid of any pyramid or cone lies in the line joining the vertex with the centroid of the base, a?id at a distance from the base equal to one-fourth the altitude. Examples. 1. Find the centroid of the area of half of a regular hexagon. 2. Determine the centroid of an area composed of a square, and an equilateral triangle one of whose sides coincides with a side of the square. 3. Determine the centroid of a volume composed of a cube, and a right pyramid whose base coincides with a face of the cube and whose altitude is equal to the length of an edge of the cube. 4. Find the centroid of a volume made up of a right circular cylinder of given base and altitude, and a cone of any altitude whose base coincides with a base of the cylinder. 5. Prove that the centroid of the area of a triangle coincides with the mass- center of a system of three particles of equal mass situated at the vertices of the triangle. 6. Determine the centroid of the volume of a frustum of a cone or pyramid. 7. Prove that the centroid of the volume of a tetrahedron coin- cides with the mass-center of a system of four particles of equal mass situated at the vertices of the tetrahedron. CENTROIDS. 135 3. Determination of Centroids by Integration. 159. General Formulas. In many cases it is not possible to subdivide the body, surface or line whose centroid is required into finite portions whose centroids are known. In such cases, if an exact result is desired, resort must be had to integration. Starting with the general formula x = (m x x x + m.x., + . . : )l(m l -f m 2 + . . . ), let the number of parts into which the body is divided be increased, while their volumes and masses become smaller, approaching zero. Each term in the sum m x x y + m.,x. 2 + . . . approaches zero, while the sum in general approaches a finite limit, which has the value limit \m x Xi -j- m % x t -f . . . ] fxdM, the limits of the integration being so taken as to include the whole body. The denominator in the expression for x is equal to M (the whole mass of the body), and may be written in the form/ dM. The values of x, y and z may therefore be written x=fxdMlfdM; y =fydAf/fdM; z=fzdMjfdM. (1) If p denotes the density at the point (x, y> z) (being in the most general case a variable), we have (Art. 153) dM= P dV } and the equations may be written x=fxpdVifpdV ] y=fypdV/fpdV; z^fzpdVjfpdV. . . . (2) Homogeneous body. If the body is homogeneous, p is constant, and the formulas reduce to x =fxdVlfdV; y=fydV/fdV; z=fzdV/fdV. (3) These are also the formulas for the coordinates of the centroid of a volume. Any area. Starting with the formulas for the coordinates of the centroid of an area already given (Art. 155), let the number of partial areas be increased while their size decreases, approaching zero as a 136 THEORETICAL MECHANICS. limit. The values of x % y and z approach limiting values identical in form with those given by equations (3). These equations may there- fore be used in finding the centroid of an area if d V represents an element of area. In case of a plane area, only two of the equations are needed, the plane of the area being taken as the plane of two of the coordinate axes. Any line. The values of x, y and $ given by equations (3) may be regarded as the coordinates of the centroid of any line, straight or curved, if dV'is taken to represent an element of length. For a plane curve only two of the equations are needed, and for a straight line but one equation is needed. The centroid of a straight line is evidently at its middle point. 160. Methods of Solution. In applying the general formulas for center of mass, center of volume, center of area and center of length to particular problems, many special methods are found useful. Thus, the integration may be double, single, or triple, depending not only upon the number of dimensions of the body considered, but also upon the way in which the differential element is taken. Again, it is often convenient to employ polar coordinates, or to otherwise change the variable with respect to which the integration is to be performed. These various devices are best explained in connection with the problems in which they are employed. Several such prob- lems will now be solved. 161. Applications. I. To determine the centroid of a circular arc. Let AB (Fig. 88) be the given arc, the radius being r. By sym- metry it is seen that the centroid lies upon OX, drawn through the center bisecting the angle A OB. Let de- note the angle between OX and the radius vector drawn from to any point of the arc. If s represents the length of the arc, measured from some fixed point, ds replaces dV in the general formula for x ; that is, x =fxdslfds. Introducing as the variable, x = r cos } ds = r d0. Fig. 88. CENTROIDS. 137 If the angle A OB is a, the limits of integration are a/2 and Hence -a/2. r r 2 cos 0<^0 a/ 2 Xa/2 -a/ 2 rdd But x a/2 / , / 2 cos * = aV 3 ; f(a 2 x 2 Y !2 dx = i[x(a' x 2 )" 2 + a 2 sin" 1 (x/a)]" --= ira 2 ^; / O x = (a"/^/(7ra 2 /4.) = \alyrr. From symmetry it is evident that y = x. Double integration. Let the element of area be taken as dV = dx dy. Then x =Jfx dx dy Iff dx dy ; y =Jfy dx dy Iff dx dy ( Notice that x and y now denote the coordinates of any element dx dy, not the coordinates of the curve.) The limits of the integra- tions with respect to the two variables will depend upon which inte- gration is performed first. If the ^-integral is first taken, the limits for x are o and V a 2 y ' l ; since for any value of y these are the extreme values of x for the area in question. The integration with respect toy would then have the limits o and a. Thus, x = ff So *s o xdy dx dy dx CENTROIDS. I39 Performing the ^-integration, O x = 1 C V a *y % the axis of x, and any ordinate. 3. Find the centroid of the area bounded by the parabola, the axis of x, and any two ordi nates. 4. Find the centroid of the volume bounded by the surface of a sphere of radius a and two parallel planes distant x x and x. t from the center. Ans. *= [3^ + ;r 2 )(2tf 2 ^, 2 ^/)]/[4(3^ 2 -^i 2 x x x^ x 2 2 )\ 5. Find the centroid of the area of any zone of a sphere. 6. Determine the centroid of the area bounded by a circle and any two parallel lines. 7. From a circular area of radius r, a second circular area of radius r' is removed. The distance between the centers being c, determine the centroid of the area remaining. 8. Determine the centroid of the area common to two circles of radii r and r\ the center of the latter lying on the circumference of the former. 9. Determine the centroid of the volume of a hemisphere. Ans. At a distance from the center equal to three-eighths the radius. 140 THEORETICAL MECHANICS. 10. Determine the center of mass of a hemisphere in which the density varies directly as the distance from the center. An s. At a distance from the center equal to two-fifths the radius. 11. Determine the center of mass of a sphere made up of two hemispheres, each homogeneous, but one twice as dense as the other. 12. Determine the position of the centroid of the surface of a right cone. CHAPTER X. FORCES IN THREE DIMENSIONS. i. Concurrent Forces. 162. Resultant of Any Number of Concurrent Forces. The resultant of any number of concurrent forces, whether coplanar or not, may be found by the repeated application of the principle of the parallelogram or triangle of forces. The resultant of any two is a single force equal to their vector sum. This resultant, combined with a third force, gives as the resultant of -the three a single force equal to their vector sum ; and so on. That is, the resultant of any number of non-coplanar concurrent forces is a single force equal to their vector sum and acting at their common point of application. 163. Parallelopiped of Forces. Any three non-coplanar con- current forces and their resultant may be represented in magnitude and direction by three edges and a diagonal of a parallelopiped. 0[ D Thus, let OA, OB and OC (Fig. A.-" "A /f7\ 9 1 ) represent, in magnitude and di- V -f \ ^ ' ! \ rection, three non-coplanar forces. \ / ,' Y B The resultant of OA and OB is rep- yij*^*^ >' ' resented by OC, the diagonal of the Q ^~C parallelogram OA C'B. The result- Fig. 91. ant of OC and OC is represented by OD, the diagonal of the parallelogram OC DC. But OD is the diagonal of the parallelopiped of which OA, OB and OC are adja- cent edges. 164. Resolution of a Force into Three Components. A force may be resolved into three components acting in any three non- coplanar lines passing through a point in its line of action. For if the given force be represented in magnitude and direction by a vector, a parallelopiped may always be determined of which this vector is a diagonal, and whose edges are parallel to the three chosen lines. The resolution of a given force into three components acting in three chosen non-coplanar lines can be made in but one way ; for a parallelopiped is completely determined if a diagonal is given in 142 THEORETICAL MECHANICS. length and direction, and if the directions of the three adjacent edges are also given.* If the three components are mutually perpendicular, their mag- nitudes may be simply computed from the angles they make with the resultant. Thus, if the parallelopiped in Fig. 91 is rectangular, and if a, ft, 7 are the angles between OD and the three edges OA, OB, OC, we have OA == OB cos a ; OB = OD cos /3 ; OC = OD cos 7. 165. Computation of Resultant of Concurrent Forces. The simplest method of determining the magnitude and direction of the resultant of non-coplanar concurrent forces is usually to replace each force by three rectangular components, and then combine these components. Thus, having given any concurrent forces, of magnitudes P x , P ti . . . , choose a set of rectangular axes OX, OY and OZ (Fig. 92), and let the angles made by P x with these three axes respectively be a, , fi x , 7, , with similar notation for the other forces. The axial components of P x are P x cos c P x cos fi x , P x cos y x ; and similar expressions may be written for the compo- nents of every force. The whole system is thus reduced to three sets of collinear forces, which combine into three forces as follows : Fig. 92. P., cos a., -f- parallel to OX; " " OY; " " OZ. a force P x cos a x " " /\cos&-f /> 2 cosft + " " P x cos y x + /> cos 7, + Let R denote the magnitude of the resultant ; a, b, c its angles with OX, OY, OZ ; and X, Y, Z its axial components. Then X = R cos a = Pi cos a x -J- P 2 cos a, -f . . . ; Y = R cos b = P x cos /3 x + P, cos /8 a -f . . . ; Z = R cos c = P, cos 7j + P 2 cos y 2 -j- . . . ; R* = X* + F 2 + Z 2 ; Jf/^ ; cos = F/Jfc ; cos r = Z/7?. cos * If the three components are coplanar with the given force, the problem is indeterminate. If only two of the components are coplanar with the given force, the third component is zero. FORCES IN THREE DIMENSIONS. 1 43 166. Equilibrium. The general condition of equilibrium for any system of concurrent forces is that the resultant is zero,, From the equation R = V X 2 + Y 2 + Z 2 = o may be derived the three equations X == o Y = o, Z = 0. For if these three equations are not satisfied, R cannot equal zero unless either X 2 , Y 2 or Z 2 is negative, which would make X, Y or Z imaginary. But neither of these quantities can be imaginary for any real system of forces. Since the directions of resolution may be assumed at pleasure, it follows that If a system of concurrent forces is in equilibrium, the sum of their resolved parts in any direction must equal zero. In accordance with this principle an infinite number of equations may be written. Only three such equations can be independent as the following analysis shows : If the sum of the resolved parts in one direction is zero, the re- sultant, if not zero, must act in a plane perpendicular to the direction of resolution. If the sum of the resolved parts is zero for a second direction, the resultant, if not zero, must act in a line perpendicular to the two directions of resolution. If the sum of the resolved parts is zero for a third direction not coplanar with the other two, the result- ant must be zero. It follows that if three equations be formed by resolving in three non-coplanar directions, any equation obtained by resolving in a fourth direction may be derived from these three. Examples. 1. A particle of 45 lbs. mass is suspended by three strings, each 8 ft. long, attached at points A, B, Cm the same horizontal plane, the distances AB, BC, CA being each 4 ft. Determine the tensions in the supporting strings. Ans. 16.37 l s - m eacn string. 2. A particle of 60 lbs. mass is suspended by three strings, each 12 ft. long, attached at points A, B, C, in the same horizontal plane, the distance AB being 6 ft. and BC and CA each 8 ft. Determine the tensions in the supporting strings. 3. A particle of 75 lbs. mass is suspended by three strings, each 10 ft. long, attached at points A> B, C, in the same horizontal plane, 144 THEORETICAL MECHANICS. the distances AB, BC, CA being 4 ft., 5 ft. and 6 ft. respectively. Required the tensions in the supporting strings. Ans. 33.62 lbs., 9.04 lbs., 35.84 lbs. 4. A body of 5 kilogr. mass is suspended by a string which is knotted at C to two strings CA, CB, these being attached to fixed supports at A and B in a horizontal plane. At C is knotted a fourth string which passes over a smooth peg at D and sustains a mass of 4 kilogr. In the position of equilibrium CD is horizontal and at right angles to AB. The distances A C, BC } AB are 180 cm., 150 cm., and 120 cm. respectively. Determine the position of equilibrium and the tensions in the strings AC, BC Ans. Inclination of plane to vertical, 38 40'. Tension in AC, 1. 2 1 kilogr.; tension in BC, 5.44 kilogr. 2. Composition and Resolution of Cotdples. 167. Representation of Couple by Vector. It has been shown in Chapter IV that two couples applied to the same rigid body are equivalent if (1) they act in parallel planes, (2) their moments are equal in magnitude, and (3) their rotation-directions coincide. So long as the discussion was limited to coplanar forces, a couple could be completely represented by a quantity whose numerical magnitude specified the value of the moment and whose algebraic sign specified the direction of rotation in the plane. It is now needful to consider couples whose planes are not parallel. A couple may be completely represented by a vector in the follow- ing manner : (a) The vector is to be taken perpendicular to the plane of the couple ; (b) its length must rep- resent, to some chosen scale, the magnitude of the moment of the couple ; (c) its direction must correspond to the rotation-direction of the couple in its plane. These requirements may be further explained as follows : Let the plane of the paper be parallel to the plane of the couple ; let P denote the magnitude of each force and p the length of the arm ; and let the lines of action and direc- Fig. 93. tions be as shown in Fig. 93. Then {a) the vector which represents the couple must be perpendicular to the plane of the paper ; (Jj) its length must be Pp units (the unit length which represents one unit of moment being FORCES IN THREE DIMENSIONS. 145 chosen arbitrarily) ; (c) it must be decided which of the two oppo- site directions perpendicular to the plane of the paper represents the kind of rotation shown in the figure and which the opposite kind. It will here be assumed that the couple shown in the figure is to be represented by a vector pointing upward from the paper. The rule thus assumed may be stated generally as follows : If a right-hand screw be placed with its axis perpendicular to the plane of the couple, when its direction of rotation coincides with that of the couple its direction of advance will coincide with the direction of the vector representing the couple. 168. Resultant of Any Two Couples. The resultant of any two couples is a couple, and the vector representing it is equal to the sum of the vectors representing the two given couples. This has already been proved for the case in which the planes of the couples are parallel, the vector sum in this case reducing to the algebraic sum (Art. 93). If the planes of the two couples are not parallel, let the plane of the paper be perpendicular to their line of intersection, this line being projected at B (Fig. 94), and B M, BN being the traces of the two planes. Let G x de- note the moment of the couple in the plane BM and G. 2 the moment of the couple in the plane BN, their directions of rotation being such that they are represented by the vectors A'B\ B'C. Replace the for- mer couple by an equivalent couple of which the forces are perpendicular to the plane of the paper, P being the magnitude of each force and G l /P its arm. One of these forces may be taken as acting at B and the other at A, if AB = G X \P. Similarly, the second couple may be replaced by a couple of forces P, one act- ing at B, opposite to the force P already assumed to act at B, and the other at C, if BC = G.JP. Since the two forces applied at B 10 Fig. 94. 146 THEORETICAL MECHANICS. balance each other, the two couples are equivalent to a couple of equal and opposite forces P applied at A and C, its moment being PX AC It may be shown that this resultant couple is completely repre- sented by the vector A'C. In the triangles A'B'C, ABC, the side A' B' is perpendicular to AB and equal to P X AB ; and the side B'C is perpendicular to BC and equal to P X BC. There- fore A'C is perpendicular to AC and equal to P X AC> which is the moment of the resultant couple. It is also evident that the direc- tion of the vector A ' C represents correctly the rotation-direction of the resultant couple. Thus the proposition is proved. 169. Resultant of Any Number of Couples; Resolution of a Couple. By repeated applications of the above principle, any num- ber of non-coplanar couples may be combined ; it is seen that the resultant will always be a couple, and that the vector representing it is the sum of the vectors representing the several couples. This may be briefly expressed by saying that the resultant couple is the vector sum of the given couples. By reversing this process, a given couple may be replaced by several component couples. Since in general a vector may be resolved, in one way only,* into three components whose directions are given, a couple may be resolved (in one way only) into three component couples whose planes are given. The simplest method of computing the resultant of any number of couples is usually to replace each by its components parallel to three rectangular planes. The whole system is thus reduced to three sets of coplanar couples, each of which reduces to a single couple by algebraic addition of the several moments. IfZ, M y iVare the mo- ments of these partial resultants, the final resultant has the moment G = VL i + M' 1 + N\ If /, m, n are the angles between the vector representing the resultant couple and the vectors representing its three components, cos / = L/G ; cos m = M/G ; cos n = NjG. * If the three components are coplanar, the resolution is impossible un- less the given vector lies in the plane of the components ; in which case the resolution can be made in an infinite number of ways. FORCES IN THREE DIMENSIONS. 3. Any System of Forces. 147 170. Any System Reduced to a Force and a Couple. Any system of forces applied to a rigid body is equivalent to a single force and a couple. Let P represent the magnitude of any one of the given forces. At any point O, introduce a force equal and parallel to P and an equal and opposite force. One of these, with the given force, forms a couple ; so that the given force is equivalent to an equal force P applied at O and a couple. In a similar manner each one of the given forces may be replaced by an equal force applied at O and a couple. The whole system is therefore equivalent to a system of concurrent forces applied at O, equal in magnitude and direction to the given forces, together with a system of couples. The former combine into a single force equal to the vector sum of the given forces ; and the latter into a single couple, equal to the vector sum of the several couples. 171. Computation of the Force and Couple. The single force which results from the above process of combination may be computed as if the forces were concurrent (Art. 165); its magnitude and direction are the same, whatever point is chosen as its point of application. The couple may be computed most readily by a differ- ent method of resolution. Choosing a set of rectangular axes, OX, Fand OZ, let x, y } z be the coordinates of the point of application of any force P, and let P be replaced by its axial components X, Fand Z (Fig. 95). Each of these components may be replaced by an equal force at O and two couples, as follows : Consider the force X applied a.tA( Fig. 95). Introduce, in the line CB, a force equal to X and an equal and oppo- site force ; and in the line OX also a pair of equal and opposite forces of magnitude X. The force X at A and the equal and oppo- site force in the line CB form a couple whose plane is parallel to the plane xy and whose moment * is Xy. The remaining force X in the line CB and the equal and opposite force in the line OX * The moments of couples in planes parallel to the yz plane are called plus or minus, according as the vectors representing them point toward the plus or the minus direction of the .r-axis. A similar convention is adopted for each of the other coordinate planes. 148 THEORETICAL MECHANICS. form a couple lying in the plane zx, of moment Xz. These two couples, with the remaining force in the line OX, are equivalent to the given force X applied at A. In a similar manner it may be shown that the force Y (Fig. 95) is equivalent to an equal force applied at 0; a couple parallel to the plane yz, of moment Yz ; and a couple parallel to the plane xy, of moment Yx. Also, Z (Fig. 95) is equivalent to an equal force ap- plied at 0; a couple parallel to the plane zx, of moment Zx; and a couple parallel to the plane yz, of moment Zy. * Combining the three forces at into a single force, and the six couples into three, it is seen that the given force P, applied at A, is equivalent to an equal and parallel force applied at O, together with three couples as follows : In the ^-plane, a couple of moment Zy Yz. In the zx- plane, a couple of moment Xz Zx. In the .zy-plane, a couple of moment Yx Xy. These three couples may be represented by vectors having the directions OX, Y and OZ, respectively, and of lengths represent- ing the values of the moments. Let there be given any number of forces, P x applied at the point * These results may be obtained without actually repeating the process employed in case of the force X, by cyclic substitution of the letters xyz and XYZ. FORCES IN THREE DIMENSIONS. 1 49 C*ii y\y #i) ^2 applied at the point (x 2 , y 2 , z 2 ), etc. ; and let each be replaced by a force and three couples in the manner described. Let Lx = Z x y x Y x z x ; M x = X x z x ~ Z x x x ; N x = Y x x x - X x y x ; with similar notation for each force. The whole system is equivalent to a force R, applied at O, equal to the vector sum of the given forces ; and a couple which is the resultant of three couples as fol- lows : In the yz-p\ane, a couple of moment L x -f- Z 2 -f- . . . = L. In the gar-plane, a couple of moment M x + M 2 -f- . . . = M. In the ^/-plane, a couple of moment N x -f N 2 ~\- . . . = N. The moment of the resultant couple is G = V D + M* + N 2 ; and the direction of its plane may be found as in Art. 169. 172. Resultant of any System of Forces. Having reduced a system of forces to a force and couple as above, if it is attempted to reduce it to a still simpler system, it appears that in certain cases it may be reduced to a couple and in other cases to a single force ; but that in general the simplest system to which it can be reduced consists of two non-coplanar forces. (a) If the vector sum of the given forces is o, the resultant is evidently a couple, determined as above. (b) If the plane of the couple found by the above process is par- allel to the single force R, this force and couple are equivalent to a single force (Art. 94) which is the resultant of the given system. This resultant force is equal to the vector sum of the given forces, and its line of action can be determined as in Art. 94. (V) If the plane of the couple is not parallel to the force R, let it be replaced by an equivalent couple of which one force acts in a line intersecting the line of action of R. This force may be combined with R, thus reducing the system to two non-coplanar forces. This reduction to two forces may be made in an infinite number of ways, but in general no further reduction is possible. 4. Equilibrium. 173. Equations of Equilibrium. Let a system of forces ap- plied to the same rigid body be reduced to a force and couple as in Art. 171. Let R denote the magnitude of the force, a, b, c the I50 THEORETICAL MECHANICS. angles it makes with the coordinate axes, and X, Y, Z its axial components ; and let G denote the moment of the couple, /, m t n the angles made with the axes by the vector representing the couple, and L, M, N the moments of the three component couples parallel to the coordinate planes. In order that the system may be in equilibrium, both R and G must reduce to zero ; otherwise the force and couple may be reduced (as in Art. 172) either to a single force, to a couple, or to two non- coplanar forces. The condition R = o requires that X=o; Y=o; Z = o; . . . (1) and the condition G = o requires that L = o ; M = o ; N = o. . . . (2) For unless X, Y and Z are severally equal to zero, either X 2 , Y 2 or Z 2 must be negative, and X, Y or Z therefore imaginary. But from the manner of computing these quantities it is evident that they must be real. Similar reasoning applies to L, M and N. Equations (1) and (2) are therefore six independent equations of equilibrium. Since the position of the origin may be chosen at pleasure, equations (2) maybe written in an infinite number of ways. Also, since the directions of the axes may be chosen arbitrarily, equa- tions (1) may be written in an infinite number of ways. In order that the conditions of equilibrium may be stated in a concise form, it is convenient to generalize the definition of moment of a force in the following manner. 174. Moment of a Force About an Axis. The moment of a force with respect to an axis has been defined, for the case in which the axis is perpendicular to a plane containing the line of action of the force (Art. 70), as the product of the magnitude of the force into the perpendicular distance of its line of action from the axis. Thus, if the line of action of the force P lies in the plane of the paper (Fig. 96) and the axis is perpendicular to that plane at O, the moment of the force is equal to Pp. If the axis is not perpendicular to the force (that is, to any plane containing the line of action of the force), the moment is to be com- FORCES IN THREE DIMENSIONS. 151 puted by replacing the force by two components, one parallel and the other perpendicular to the axis, and taking the moment of the latter. 175. Conditions of Equilibrium. Referring to the process by which a system is reduced to a single force and a couple (Art. 171), it is seen that Z x y x Y x z x is equal to the moment of the force P x with respect to the axis OX; X x z x Z X X X to its moment with re- spect to O Y; and Y x x x X x y x to its moment with respect to OZ. The quantity L is therefore equal to the sum of the moments of the given forces about the axis OX ; M to the sum of their moments about O Y; and N to the sum of their moments about OZ. The meaning of the two sets of equations of equilibrium (Art. 173) may therefore be stated in words as follows : (1) The sum of the resolved parts of the given forces parallel to each of the axes is zero. (2) The sum of the moments of the given forces about each of the given axes is zero. Since any line whatever may be taken as one of the axes, it fol- lows that, for equilibrium, (a) The sum of the resolved parts of the forces in any direction is zero. (b) The sum of their moments about any axis is zero. These principles lead to an infinite number of equations ; but only six can be independent. Examples. 1. A three-legged stool rests upon a smooth horizontal floor. De- termine the pressures at the three points of support. 2. The three points of support of a three-legged stool being at any distances apart, what must be the position of its center of gravity in order that the pressures at the points of support may be equal? 3. A four-legged table rests upon a smooth horizontal floor. How much can be determined about the supporting forces ? 4. A windlass consists of a circular cylinder supported at two points in bearings which permit free revolution about its axis, a crank attached to one end of the cylinder, and a cord wound upon the cylinder and having one end free. To the free end of the cord is attached a heavy body which is to be lifted by the application of a force to the crank-handle. The axis of revolution of the cylinder is horizontal. Assuming that the force applied to the crank is perpen- dicular both to the axis of revolution and to the crank-arm, and that 152 THEORETICAL MECHANICS. the bearings are frictionless, it is required to determine all forces acting on the windlass when the weight of the lifted body is known. Let W weight of body lifted ; P = force applied to crank- handle ; r= radius of cylinder; a = length of crank-arm (z. e., radius of circle described by point of application of P) ; / = distance between centers of bearings ; b = distance from center of circle de- scribed by point of application of P to nearest bearing ; l x and /. 2 = distances from centers of bearings to suspended cord. The bearings being assumed smooth, the supporting forces exerted by them may have any directions perpendicular to the axis of the cylin- der ; let them be replaced by horizontal and vertical components, and let H x , V x denote the components of the force at the bearing near the crank and H 2 , V t the components of the force at the other bear- ing. The direction of the force P t if always perpendicular to the Fig. 99. crank-arm, will vary as the cylinder revolves. Let 6 = angle between crank-arm and the vertical. All forces acting upon the windlass act in planes perpendicular to the axis of revolution ; hence, if forces are resolved parallel to this axis, no equation results. Five independent statical equations may, however, be written as follows : Resolving horizontally, H x + H % + Pcos = o. . . . (1) Resolving vertically, V 1 + V 2 Psin 6 W= o. . . (2) Taking moments about the axis of revolution, Pa Wr = o (3) Taking moments about a vertical line through center of circle described by point of application of P, H x b + H. t {b + /) = o. . . . (4) FORCES IN THREE DIMENSIONS. 1 53 Taking- moments about a horizontal diameter of same circle, V x b + Vlb + /) - W{b + /,) = o. . . (5) These five equations contain five unknown quantities, H x , V x , H. t , V 2 , P. Their solution is simple. Thus, (3) determines P at once ; (1) and (4) then determine H x and H. z ; (2) and (5) determine V x and V,. 5. In Ex. 4, take numerical data as follows : l x = 7/3 ; b = //io ; # = 6r. Solve for = o, 90, 180, 270, 45. ^tzj. P = W/6 for any value of 6. For # = o, //, = -(u/6o)W, #, = (1/60)^, ^ = (2/3)^, V t = (1/3)^ For = 90, H x = H 2 = o, F t - (51/60) W, r 2 = (19/60)^. 6. In Ex. 4, substitute for the force P a weight Q suspended from the crank-handle. If Q is known, determine the position of equi- librium and all unknown forces. 7. Assume dimensions as in Ex. 5, and let the weight Q, applied as in Ex. 6, be equal to W. Determine position of equilibrium and all forces. Ans. = 9 36' or 170 24' ; H x = H t = o ; V x = (17/20) W; V, = (19/60) w. 8. A circular cylinder is placed with axis horizontal, one end rest- ing in a smooth cylindrical bearing and the other on an inclined timber. The longitudinal axis of the timber lies in a plane perpen- dicular to the axis of the cylinder and is inclined to the horizontal at a known angle. If its surface is so rough as to prevent sliding, can the cylinder be in equilibrium ? 9. In Ex. 8, can equilibrium be produced by hanging a weight from a string wound on the surface of the cylinder, assuming sliding to be impossible? If equilibrium is thus possible, specify the re- quired weight, and all forces acting on the cylinder. 10. Two timbers are placed with longitudinal axes parallel and with upper surfaces in the same horizontal plane. A circular cylin- der rests upon the timbers. Can the supporting timbers be tilted without destroying the equilibrium of* the cylinder, assuming the sur- faces to be sufficiently rough to prevent sliding ? 1 1. The points of support of a three-legged stool form an equi- lateral triangle ABC oi side a. The center of gravity is vertically above the centroid of the triangle. The stool is pulled horizontally by a string lying in the vertical plane containing AB and at a dis- tance // above the floor. Show that the stool will tip without sliding if the coefficient of friction is greater than a^h. 12. A uniform straight bar is placed with one end on a rough horizontal floor and the other in the right angle between two smooth vertical walls. Show that it will be in equilibrium if its inclination 154 ' THEORETICAL MECHANICS. to the vertical is less than tan~' (2/x), /j, being the coefficient of friction. 13. In Ex. 12, show that the resultant force acting upon the bar at either end acts in the vertical plane containing the bar. Deter- mine these forces. 14. A bar is placed with one end upon a rough horizontal floor and the other against a rough vertical wall. Determine the posi- tions of limiting equilibrium. CHAPTER XI. GRAVITATION. i. Attraction Between Two Particles. 176. Law of Gravitation. Every portion of matter exerts an attractive force upon every other portion. The magnitude and direc- tion of this attractive force for any two bodies may be determined in accordance with Newton's law of universal gravitation, which may be stated as follows : Every particle of matter attracts every other particle with a force which acts along the line joining the two particles, and whose magni- tude is proportional directly to the product of their masses and inversely to the square of the distance between them. The proportionality expressed in this law may be stated algebra- ically as follows : Let P denote the magnitude of the attractive force between two particles whose masses are m x , m. z , and whose distance apart is r ; and P' the force between two particles whose masses are ml, m. 2 \ and whose distance apart is r' . Then PIP' = (ni x mj r 2 )l{mlm.; I r'*). This equation may be written in the form Pr 2 lm x m 2 = P'r^/m^ml = 7. Since a similar equation may be written for any pair of particles whatever, the law of gravitation may be expressed by the equation Pr il lm x m. l = 7, or P = rpnjnjr*, in which 7 is a constant. The value of 7 is to be determined by experiment ; but having been determined for one case it is known for all cases. This formula applies strictly only to particles, but it gives, to a close approximation, the attraction between two bodies of finite size whose linear dimensions are small compared with the distance be- tween them. We shall first consider the attraction between two 156 THEORETICAL MECHANICS. particles, and shall then give a brief discussion of methods of com- puting accurately the attraction between bodies of finite size. Attraction is a mutual action between two particles or bodies ; i. e. , each exerts an attractive force upon the other, the two forces being equal in magnitude and opposite in direction. This is implied in the above statement of the law of gravitation. It is also in ac- cordance with the law of "action and reaction," Newton's third law of motion (Art. 35). 177. The Constant of Gravitation. The quantity 7 in the above formula is called the constant of gravitation. Its numerical value depends upon the units in which force, mass and distance are expressed. For any given system of units, 7 is numerically equal to the attractive force between two particles of unit mass at unit dis- tance apart ; for if m x = I, m % = 1 and r = I, the formula gives If mass, length and force be expressed in any units employed in ordinary practical problems, the numerical value of 7 is very small. Thus if mass is expressed in pounds and distance in feet, 7 is equal to the attractive force between two particles of one pound mass each, placed one foot apart. This force is too small to be detected by ordi- nary methods of measuring forces ; and if expressed in terms of any unit in common use its numerical value is very small. (See Arts. 184, 186.) 178. Gravitation Unit of Mass. Instead of choosing the units of mass, length and force independently, they may be so chosen as to give 7 any desired value. Moreover, two of these units may still be chosen arbitrarily, the third being then so taken as to satisfy the equation P = ym^mjr 2 with the desired value of 7. In order to simplify the equation, let 7 = 1. Then the units must be so chosen that two particles, each of unit mass, placed at the unit distance apart, attract each other with unit force. The unit mass which satisfies this condition, the other units having been chosen at pleasure, is called the gravitation unit. To determine the value of this unit of mass in terms of the pound, gram, or other known unit, requires the same process of experiment and reasoning which is involved in the determination of the constant 7 when all the units have been chosen. This subject will be resumed in Art. 185. GRAVITATION. Examples. 157 1. If two particles 20 ft. apart attract each other with a force of 12 lbs., with what force will they attract each other if placed 50 ft. apart ? 2. If two particles of 1 , 000 grams and 1 2, 000 grams mass respect- ively attract each other with a force equal to the weight of P grams when 30 cm. apart, with what force will two particles of 1,800 grams and 3,000 grams attract each other when 100 cm. apart? Ans. o. 0405P grams weight. 3. In Ex. 2, what is the value of the constant 7 in terms of P? Ans. 7 = 3/740,000. 4. With the data of Ex. (2), what is the gravitation unit of mass? Ans. A mass equal to 200/1/(3/*) grams. [The answers given to examples 3 and 4 imply that the centi- meter is the unit length and the weight of a gram the unit force.] 179. Dimensions of Units. The formula expressing the law of gravitation for two particles may be written Pr 2 lm 1 m 2 = 7. The constant 7 is therefore of dimensions (Art. 14) PL'/M 1 if the units of force, length and mass are all taken as fundamental, i. e., are all chosen independently (Art. 13). The law of gravitation, however, itself furnishes a means of mak- ing one of these units depend upon the other two. For the equa- tion expressing the law may be written P = Mjfftjr*, if the units be so chosen as to satisfy the condition stated in Art. 178. The relation between the units will then be expressed by the dimen- sional equation, p __ M 2 /L 2 Any two of these units being chosen arbitrarily, the dimensions of the third are given by this equation. Thus, if, as in Art. 178, the units of force and length are made fundamental, the dimensions of the unit mass are expressed by the equation M = LF 2 The relation among units which is expressed by a dimensional equation is thus, to a certain extent at least, arbitrary. This will be further illustrated when the kinetic or ' ' absolute ' ' system of units is explained. (See Art. 219.) 158 THEORETICAL MECHANICS. 2. Attractions of Spheres and of Spherical Shells. 180. Law of Gravitation Applied to Continuous Bodies. The law of attraction is stated above as applying to particles, and these are treated as bodies of finite mass but without finite size. If a body consisted of a finite number of such particles, its resultant attraction for any other body or particle would be computed by find- ing the resultant of the attractive forces due to all its individual par- ticles. If, instead of being made up of discrete particles of finite mass, a body occupies space continuously (Art. 5), so that any portion whose mass is finite is of finite volume, the resultant attraction Of one body upon another may be computed in a similar manner, but the process involves integration. Let m x and m. L denote the total masses of two continuous bodies, and Am x , Am, any small elements of these masses. If the dimensions of these elementary portions are small in comparison with their dis- tance apart, the attractive force exerted by each upon the other has approximately the value yA7n x A7n.Jr 2 , r being the distance between the elements. If the attraction of every element of m x upon every element of m. l be computed approximately in the same manner, the resultant of these forces will be an approx- imate value of the resultant attraction of m x upon m. x . The approx- imation is closer the smaller the elements into which the bodies are subdivided ; the exact value is the limit approached by the approx- imate value as the elements approach zero in size. The magnitude of every component force thus approaches zero and the number of components approaches infinity, but their resultant has in general a finite value. The computation of this value involves a process of in- tegration. In applying this method only a few simple cases will be here con- sidered. 181. Attraction of a Homogeneous Spherical Shell Upon an Interior Particle. Proposition. The resultant attraction exerted by a homogeneous spherical shell of uniform thickness upon a particle within its inner surface is zero. Let O (Fig. 98) be the position of the particle, and through O GRAVITATION. 1 59 draw any straight line intersecting the outer surface of the shell in two points A and B. On the outer surface of the shell take an ele- mentary area containing the point A ; from every point of the per- imeter of this element draw a line through 0, and prolong it to inter- sect the spherical surface at a point near B. We thus get a cone cutting from the spherical surface at B a second elementary area. Call the areas of the elements at A and B, a and b respectively, and let h be the thickness of the shell. The volumes cut from the shell by the two branches of the conical surface are approximately ha and hb respectively, and their ratio has approximately the value ha/hb = a/b, the approximation being closer as the values of h, a and b are taken smaller. It may be shown that as a and b approach o, a/b approaches OA 2 /OB\ The tangent planes to the sphere at A and B are equally inclined to the line AB ; hence in the limit, as a and b approach o, they are proportional to their orthographic pro- jections upon a plane perpendicular to AB ; that is, to the right sections at A and B of the two branches of the con- ical volume whose vertex is at O. But these right sections are directly propor- tional to OA 2 and 0B % \ hence the areas a and b, the elementary volumes ha and hb, and the masses of these elementary volumes, are proportional toOA 2 andOB'\ If m is the mass of the particle at 0, and if m' and m" are the masses of the elements of the shell at A and B respectively, the particle is attracted toward A with a force ymm'l OA 2 , and toward B with a force ymm'/OB*. But, as just shown, m'/m f = OA 2 /OB\ or m'/OA 2 =m"/OB i ; i6o THEORETICAL MECHANICS. hence the attractive forces of m and m" upon m are equal and op- posite, and their resultant is zero. The whole volume of the shell may be divided into elements which, taken in pairs, are related in the same way as the two ele- ments considered. Hence the resultant attraction of the shell upon the particle at O is zero. In the above reasoning, it was assumed that the thickness of the shell becomes small, approaching zero as a limit. If the thickness has any finite value the result still holds. For the shell may be regarded as made up of a great number of very thin shells for each of which the conclusion is true at least approximately ; and if the number of shells is increased without limit, the thickness of each approaching zero, the conclusion holds strictly for each elementary shell and therefore for the given shell. The proposition is also true if the density of the shell varies in such a way that its value is the same at all points equally distant from the center of the sphere. 182. Attraction of a Homogeneous Shell Upon an Exterior Particle. It is evident from symmetry that the resultant attraction of a homogeneous spherical shell of uniform thickness upon an ex- terior particle is di- rected along the line joining the particle with the center of the sphere. Let A (Fig. 99) be any point of the surface of the shell, C the center of the sphere, and O the position of the ex- terior particle, its mass being m. Take an element of the spherical surface containing A, and let a denote the area of this element, h the thickness and p the density of the shell, R the radius of the sphere. Let OA = r, OC = c, angle A CO = 6, angle AOC = The conical surface whose base is a and apex C cuts from the shell an element of volume whose mass is pha. The attraction of Fig. 99. GRAVITATION. l6l this mass upon the particle at O is ymphajr 2 , the resolved part of which in the direction OC is (ympAafr*) cos . Let the element of area a be taken as part of a zone of infinitesimal width, included between two circles whose planes are perpendicular to OC For every point of such a zone, r and have the same values ; hence the total attraction in the direction OC due to the portion of the shell corresponding to such a strip is ympk cos cf> dA/r 2 , if dA represents the area of the zone. The attraction due to the whole shell is therefore cos (j> dA P = ymp/i J the integration being so taken as to cover the whole surface of the sphere. We have now to express , r and dA in terms of a single variable and integrate. Let r be the variable chosen. We have dA = Rdd 2-jtR sin = 2-rrR 2 sin d dd. But r 2 = c l + R l 2cR cos ; rdr = cR sin d0 ; hence dA = {2irRjc)r dr. Again, R 2 = c 2 + r' 1 2cr cos , or cos = (V 2 + r 2 R 2 )/2cr. Substituting the values of cos and dA in the value of P, and ex- pressing the proper limits, _ irymp/iR f* + Jt ? ^ 2 + r 2 The integral expression can be separated into two parts, thus : {c 2 R 2 ) - + r. r ii 1 62 THEORETICAL MECHANICS. Taking the value between the proper limits, dr x ' + * c 2 R l h c- R r -[(c-R) - (*+ R)] + [(r+ R) - (c-R)] = 4 R. Hence P = TrymphR 4-R/c 2 = y ^rrR 2 hp m/c 2 . But 4.irR 2 hp = M = total mass of shell. Hence P=yMm/c i ) .... (2) which shows that The resultant attraction between the shell and an exterior par- ticle has the same value as if the entire mass of the shell were con- centrated at its center. The result obviously holds for a shell of any thickness whose density has the same value at all points equally distant from the center. For such a shell may be subdivided into elementary shells, each of which may be taken as thin and as nearly of uniform density as desired. Interior particle. The reasoning by which (1) is deduced ap- plies also to the case of an interior particle, except that the limits of the integration must be different. Thus, for the attraction on an interior particle, we have __ TrymphR n RJc c 2 R l + r* _ fc" h-c ~ r 2 which agrees with Art. 181. 183. Attraction of a Sphere Upon a Particle. The forego- ing results may be applied to a sphere whose density has the same value at all points equally distant from the center. Exterior particle. The resultant attraction of such a sphere upon an exterior particle has the same value as if the entire mass of the sphere were concentrated at its center. Interior particle. The resultant attraction of such a sphere upon an interior particle is the same as if the portion of the mass nearer the center than the particle were concentrated at the center ; the re- maining shell being disregarded because its resultant attraction is zero. gravitation. 1 63 Examples. 1 . Assuming the earth to be a sphere whose density is a function of the distance from the center, compare the weight of a body at the surface and at a point h feet above the surface. The ' ' weight " of a body is the force with which the earth at- tracts it.* If R is the radius of the earth, and if Pand P* are the values of the earth's attraction upon the body at the surface and at h feet above the surface respectively, we have P/P* = (R + k)*IR\ or P' = PR' 2 /(R + h)\ Hence the weight of a body at height h above the surface is the fraction R 2 /(R -\- h) 2 of its weight at the surface. If h is a small fraction of R, we have, approximately, R%R + h) 2 = (1 + k/R)- 2 = 1 2/1 1 R. 2. At what height above the surface will the weight of a body be 1 per cent less than at the surface ? [The mean radius of the earth is very nearly 6.3709 X io 8 cm. or 20,902,000 ft.] 3. If the pound-force is defined as the weight of a pound-mass at the earth's surface, how much does a change of elevation of 10,000 ft. affect the value of this unit force ? 4. Let G denote the weight of a given mass at the earth's sur- face, and G' its weight at a depth h below the surface. If the earth were a sphere of uniform density and of radius R, show that G'/G = (R h)IR = 1 k/R. 5. Assuming that the earth is a sphere and that the density is a function of the distance from the center, let p denote the mean den- sity of the whole earth and p the mean density of the outer shell of thickness h. Determine the relation between the weight of a body at the surface and its weight at a depth h below the surface. Let M be the mass of the whole earth, M ' that of the inner sphere of radius R k, m the mass of the given body, G its weight at the surface and G' its weight when at depth k below the surface. Then G is equal to the attraction between two particles of masses M and in whose distance apart is R ; and G' is equal to the attraction between two particles of masses M' and ;;/ whose distance apart is R h. That is, G = yMm/R 2 , G' = yM'mfcR h)\ Hence =*1-*1Y\ . . . (o G M\R h) But M = 4 ttR* p ; MM' = 4 7rp [R* (R hf\ * The effect of the earth's rotation upon the apparent weight is here dis- regarded. See Art. 311. 164 THEORETICAL MECHANICS. M , pjR C-(^)H-?K 6 ( M' = _gj\ l R h \* M ' * " p Substituting this value in equation (1) G' I p\( R V Po(R R (2) H'-3fe^r + ?m * Approximate simple formula. \i h is a small fraction of R, equation (3) may be reduced to the following as a first approxima- tion : ^ =i+ ( 2 -w (4) 6. Show that, if the mean density of the outer layer of the earth is less than two-thirds the mean density of the whole earth, the weight of a body increases as it is carried below the surface. This follows from equation (4) above. It will be seen that, as a body is carried below the surface of the earth, its change of weight is the resultant of two opposite effects. The total mass attracting it is less because the outer shell has on the whole no effect (Art. 181) ; while the attraction of the inner sphere is greater because the body is nearer its center (Art. 183). The latter effect is greater than the former unless the mean density of the outer shell is at least two- thirds that of the whole earth. According to the best determinations the mean density of the earth is very nearly 5.527 times that of water, while the average den- sity of rocks near the surface may be taken at about half this value. If p /p = 1/2, equation (4) becomes G'/G^i +k/2R. . . . ( 5 ) That the weight of a body increases when it is carried below the surface of the earth has been shown by experiment. 7. What would be the change in the weight of a body if carried 2,000 ft. below the earth's surface, assuming the density of the outer portion of the earth to be half the mean density ? 184. Value of the Constant of Gravitation. If the attrac- tion between two particles of known mass at a known distance apart can be determined by experiment, the value of 7, the constant of gravitation, can be determined. From the foregoing results (Arts. 182, 183), spheres of any size may be used instead of particles. The attraction between bodies of ordinary size is so small that it can be measured only by the most delicate apparatus. The relation between the constant of gravitation and the mass of the earth may be shown as follows, assuming the earth to be a sphere whose density is a function of the distance from the center, We shall GRAVITATION. 1 65 take as units those of the British gravitation system. The unit mass is the pound, the unit force the weight of a pound mass at the earth's surface (the pound-force), the unit length the foot. Let R denote the radius of the earth in feet, M its mass, p its mean density. Consider the attraction of the earth upon a body of mass m at the surface. By the general formula of Art. 176 the value of this attraction is yMm/R 2 . But expressed in pounds-force it is also equal to m. Hence m = yMm/R 2 , or yM =: R\ Since the value of R is known, this equation serves to determine either of the quantities 7 and M when the other is known. Instead of M, we may introduce the earth's mean density p, since M = ^ttR^p/t,' The equation then becomes vp = sIattR. The best determinations give as the earth's mean density about 345 pounds per cubic foot. Taking R = 20,900,000 ft., the value of the constant of gravitation * is 7 = 3/47^/0 = 3/(47T X 20,900,000 X 345) = 3-3 1 X io- u . 185. Value of Gravitation Unit of Mass. Let m pounds be the mass of each of two particles which, placed one foot apart, attract each other with one pound-force. The value of m may be found from the formula P = ym x mjr 2 } by putting P = 1, r = 1, m x = m% = m, and 7 = 3.31 X io~ u as found above. The result 1S m = i/j/7 = 173,800 pounds. If a mass equal to 173,800 pounds be taken as the unit mass, the constant 7 becomes unity, and the formula for the attraction between two particles whose masses are m x and m 2 and whose distance apart isris P=m x mjr\ Thus, the " gravitation unit of mass" (Art. 178) is a mass equal to about 173,800 pounds, if distance is expressed in feet and force in pounds-force. *The above is not given as the actual method of determining T, but merely as showing about what its value is. The above value of the earth's mean density is based upon the experimental determination of T by the meas- urement of the attraction between bodies of known mass. 1 66 THEORETICAL MECHANICS. 1 86. Values of the Constant of Gravitation and of the Grav- itation Unit of Mass in the C. G. S. System.* If the unit mass is the gram, the unit length the centimeter and the unit force the dyne (Art. 217), the value of 7 may be found as follows : The attraction of the earth for a body of m grams mass at the surface is approximately 98 \m dynes. By the law of gravitation it is also equal to yMm/R 2 . Hence 981^ = jMm/R 2 , or yM= 9S1R 2 . Here M is the mass of the earth in grams, and R is its radius in centimeters. Introducing the earth's mean density p instead of M, the equation is 7P = 3X 981/47^. Taking p = 5.527 grams per cubic centimeter and R = 6.371 X 1 o 8 centimeters, 8 7 = (3 X 98i)/(4T X 5-527 X 6.371 X io 8 ) = 6.65 X 10 Let m grams be the mass of each of two particles which, when 1 centimeter apart, attract each other with a force of 1 dyne. Then 1 = p~' moves. If the motion of P' is known, Fig. 102. the velocity at every instant is known. 200. Notation for Time-Derivatives. When a variable quan- tity is a function of the time, its derivatives are often designated by the use of dots, as follows : dx/dt = x ; d^x/dt* = dx/dt = x ; etc. This notation will often be used in the following pages. 2. Velocity-Increment and Acceleration. 201. Increment of Velocity. Let v x and v 2 be the values of the velocity at instants t x and t 2 respectively ; then v 2 v x is the velocity-increment for the interval from t x to t. 2 . This increment may be either positive or negative, according as v t is (algebraically) greater or less than v x . Geometrical representation. If, as described in Art. 199, the velocity at every instant be represented in magnitude and direction by a length O'P' laid off from a fixed point O' (Fig. 102), the point P' will move as the velocity varies. When v increases (algebraic- ally), P' moves in the plus direction ; when v decreases, P' moves in the minus direction. The displacement of P' in any interval repre- sents the velocity-increment. The velocity -increment must not be confounded with the displace- ment of the particle. There is not necessarily any relation between them. The point P' may or may not move in the same direction as the particle. 202. Uniformly Varying Velocity. If the velocity receives equal increments in all equal intervals of time, however small these may be, the velocity is uniformly variable. Recurring to the geometrical representation given above (Fig. 102), if the velocity of P changes at a uniform rate, the point P 172 THEORETICAL MECHANICS. moves uniformly. The motion of P' has the plus or the minus direction according as the velocity of P is (algebraically) increasing or decreasing. 203. Acceleration. The rate of change of the velocity of a par- ticle is called its acceleration. If, throughout a certain interval of time, the velocity changes at a uniform rate, the acceleration is proportional directly to the total velocity-increment and inversely to the duration of the interval. The numerical value of the acceleration depends upon the unit in which it is expressed. 204. Unit of Acceleration. Although the unit of acceleration may be chosen arbitrarily, it is convenient to make it depend upon the units of length and time. Hence the unit acceleration is defined as that of a particle whose velocity increases by one unit in every unit of time. Thus, with the foot and second as fundamental units, the unit acceleration is that of a particle whose velocity increases by one foot- per-second in each second. With the meter replacing the foot as the unit length, the unit acceleration is that of a particle whose velocity increases by one meter-per-second in each second. Similarly, a third unit of acceleration is derived from the centi- meter and second as fundamental units. Dimensions of unit acceleration. The above definition of the unit acceleration makes it proportional directly to the unit velocity and inversely to the unit time. Representing the unit acceleration by A, there may be written the dimensional equation A = V/T = L/T 2 . 205. Numerical Value of Acceleration. When the velocity of a particle increases at a uniform rate during any interval, the value of the acceleration in terms of the unit above denned is computed by dividing the increment of velocity by the duration of the interval. Thus, let v denote the velocity at any instant /, v x and v 2 the values of v at the instants t x and t t , and p the acceleration. Then P=(v 2 v l )/(t. 2 -t l ). This value ofp may be in feet-per -second- per-second, in meters- per-second-per-second, in centimeters-per-second-per-second, or in other units, according to the units in which t and v are expressed. MOTION IN A STRAIGHT LINE. 173 206. Algebraic Sign of Acceleration. If v 2 v x is negative, the value of p is negative. This means that the velocity is decreasing algebraically ; that is, that the velocity-increment received by the particle in any interval has the minus direction. If p is plus, the velocity is increasing algebraically, although numerically it may be decreasing. Thus, if v is positive and increasing in magnitude, p is positive ; " V ' ' negative ' decreasing ' ' " f M .*.-. \f-v " positive " decreasing M M /r" negative ; 1 ' v " negative ' ' increasing " " p " ' . 207. Non-Uniform Variation of Velocity. In case the incre- ments of velocity in equal intervals of time are unequal, acceleration may still be defined as the rate of change of the velocity, but the above method of computing its value becomes inapplicable. Acceleration must now be understood to be a variable quantity, having a definite value at any instant. Its meaning is best under- stood by a consideration of ' ' average acceleration. ' ' 208. Average Acceleration. The average acceleration of a par- ticle for an interval during which its velocity varies in any way is an acceleration which, if constant, would result in the same velocity- increment in the same interval. The formula / as (v, v,)i{t 2 - O always gives the value of the average acceleration for the interval from t x to t t . 209. Approximate Value of Acceleration at an Instant. An approximate value of the acceleration at an instant may be found by computing the average acceleration for a very short interval of time. Let v denote the velocity at an instant /, and &v the velocity- increment for a short time At. Then Av/At is an approximate value of the acceleration at the instant /. The approximation is closer the shorter the interval At. 210. Exact Value of Acceleration at an Instant. The exact value of the acceleration at the time / is the limit approached by the approximate value as At is taken smaller and smaller, approaching zero. This limit is the derivative of v with respect to / ; that is, / =3 lim \Av\kti\ =s dv/dt. 174 THEORETICAL MECHANICS. Since v = dx/dt = x, we have p = (Px\dt % dx/dt = jr. 211. Application of Formulas for Velocity and Accelera- tion. The formulas V = dx/dt, p = dvjdt *= d 2 xjdt\ are of fundamental importance in the solution of various problems in rectilinear motion. Suppose, for example, that x is known as a func- tion of / ; that is, that the position at every instant is known. In this case v and p can be determined by differentiation. Thus, if x = f(f), we have v = dxldt = df{t)jdt =f'(f), p == dvjdt = d*xjdt l = df\t)jdt =/"(*}. As another case, suppose the velocity or acceleration known at every instant ; that is, let v or p be a known function of /. In this case the value of x in terms of / can be found by integration. In general, if a relation is known between any or all of the quan- tities x, t, v and /, the motion can be completely determined either by differentiation or by integration. If the solution involves integra- tion, either one or two arbitrary constants will be brought in, the values of which cannot be determined without additional data as to the conditions of the motion. Thus, if there is given such a relation as fix, /, v, p) == o, the substitution of the general values of v and p as derivatives with respect to t gives a differential equation whose solution determines the motion. Examples. i. The position of a particle is given by the equation x = kt'\ k being a constant. Show that the acceleration is constant, and deter- mine its value in terms of k. 2. In Ex. i, if the velocity is 8 ft.-per-sec. at the instant t = 3, what are the values of k and of the acceleration ? 3. If the position of a particle is given by the equation x = kt 2 + k't + k" ', show that the acceleration has the same value as if the equation were x = kt' 1 . 4. The position of a particle is given by the equation x = 10 -J- 5^ + 2/ 2 -\- t'\ x being in feet and / in seconds. Compute the average MOTION IN A STRAIGHT LINE. 175 acceleration for o. 1 sec. after the instant / == 2, and compare it with the exact value of the acceleration at the beginning of the interval. The velocity is given by the formula v = x = 5 + 4/ -j- 3/ 2 . When / = 2, v = 25. When / = 2. 1, V -== 26.63. Hence the av- erage acceleration is Av/At = (26.63 2 5)/- 1 == x 6-3 ft. -per-sec. -per-sec. The true acceleration at any instant is / = dvjdt = x = 4. -\- 6t. When / = 2, p = 16 ft. -per-sec. -per-sec. 5. If the velocity of a particle is given by the equation v = 12 16/, determine the position at any time. The equation may be written i: = dxjdt =12 16/. Integrating with respect to /, X = 12t 8t 2 -f C. In order that the constant of integration C may be determined, the position of the particle at some given instant must be known. Thus, if t is reckoned from the instant at which the particle is at the origin, the values (x = o, / = o) must satisfy the equation. This requires that C== o. But if the " origin of time" (z. e., the instant from which t is reckoned) is chosen as the instant when x = 4, the values (x = 4, t = o) must satisfy the equation, and therefore C = 4. If jr is in feet, in what units is C? 6. If v= 12 16/, determine the acceleration at the instant t = 5, and the average acceleration for o. 1 sec. after t = 5. 7. Let the velocity be given by the equation v = 12 -f- 18/ 2 , / being in seconds and 7/ in centimeters-per-second. Determine the position, the velocity and the acceleration at the instant t =10 sec. In order that the value of x may be completely determined, what additional data must be given ? 8. With data of Ex. 7, determine the average velocity and the average acceleration during o. 5 sec. and during o. 1 sec. following the instant / = 10 sec. 9. The velocity of a particle at a certain instant is too ft. -per-sec. in the positive direction. The acceleration is constant and equal to 24 ft. -per-sec. -per-sec. in the negative direction, {a) When will the particle be at rest ? {B) What will be its velocity 4 sec. later ? Ans. (a) At the end of 4^ sec. (<) 96 ft. -per-sec. 10. The acceleration being constant, determine the position and velocity at any time. 176 THEORETICAL MECHANICS. Let f be the given constant value of p. Then the problem is to solve the differential equation d % x\dt % = dvjdt = f. Integrating with respect to t, v = dx/dt = ft + C, C being a constant of integration. To determine its value, an ' ' initial condition " must be known ; that is, the value of the velocity at some definite instant must be known. Let v denote the value of v when / = o ; then, since the last equation is true for any simultaneous values of v and /, v o / + C ; or C = v . Hence v = x =ft -\- v . Integrating the last equation with respect to /, x = \fi % + v + c. To determine the constant of integration, C\ the position at some given instant must be known. Let x denote the value of x when t = o ; then, the last equation being true for x = x and / = o, C must equal x . Hence x=\ft l -fV+iJ. 11. The velocity of a falling body is observed to increase by 32.2 ft.-per-sec. during every second of its motion. How far will it fall from its position of rest in t seconds ? [The acceleration is constant and equal to 32.2 ft.-per-sec.-per- sec. Hence this is a special case of Ex. 10.] 12. The acceleration of a particle increases in direct ratio with the time reckoned from a given instant. Determine the position and velocity at any time. Let distance be expressed in centimeters and time in seconds, and assume that the acceleration increases by 4 units in 1 sec. Assume further that the velocity is zero when the ac- celeration is zero, and that x is reckoned from the position of the particle when the acceleration is zero. That is, the four quantities /, v, x, t are all zero together. Ans. x e= f/ 3 , v = 2/ 2 , / = \t. 13. With data as in Ex. 12, where was the particle and what was its velocity 3 sec. before the instant at which the acceleration was zero? 14. If jit were reckoned from the position of the particle 3 sec. before the acceleration was zero, how would the results of Ex. 1 2 be changed ? (Make no change in the origin of time. ) Does this change in the origin of abscissas imply a different case of motion ? MOTION IN A STRAIGHT LINE. I77 3. Motion and Force. 212. Laws of Motion. In the foregoing analysis of the motion of a particle, nothing has been said of the laws in accordance with which the motions of bodies actually occur. By studying the motions of bodies in nature, it is found that the motion of any given body is influenced by its relation to other bodies. This is often expressed by saying that one body is ' ' acted upon ' ' by other bodies. Such an ' ' action ' ' of one body upon another, measured in a particular way, is called a. force. (Art. 32.) Newton's three laws of motion give a concise statement of the way in which the motion of a body is influenced by forces, i. e., by the action of other bodies. A full understanding of these laws re- quires a more general analysis of motion than has been given in this Chapter. The present discussion of the laws of motion must be limited to the case of motion in a straight line. A formal statement of Newton' s laws, with a fuller explanation of their meaning, will be given in Chapter XIV. 213. Motion of a Body Which Is Not Influenced by Other Bodies. Newton's first law asserts that a body not acted upon by force (that is, wholly uninfluenced by other bodies) will remain at rest, or else will move in a straight line with uniform velocity. The meaning of this law has already been briefly explained (Art. 31). 214. Effect of Constant Force on Body Initially at Rest. If a force of constant magnitude and direction acts, for a certain interval of time, upon a body initially at rest, the body will have at the end of the interval a velocity whose direction is that of the force, and whose magnitude is proportional directly to the force and to the duration of the interval, and inversely to the mass of the body. Let P denote the magnitude of the force, / the duration of the interval, and m the mass of the body acted upon ; then at the end of the interval the velocity of the body is proportional directly to P and to /, and inversely to m ; that is, it is proportional to Pt/m. Thus, if a force P\ acting for a time t* upon a particle of mass m initially at rest, gives it a velocity v'\ and if a force P" acting for a time t" upon a particle of mass m initially at rest gives it a velocity *'"' then v':v" = P't'lm :P"t"lm". 178 theoretical mechanics. Examples. 1. A given force acting upon a mass of 1 lb. for 1 sec. gives it a velocity of 10 ft.-per-sec. What velocity would an equal force im- part to a mass of 5 lbs. in 4 min.? 2. A force of magnitude P acting for 4 sec. upon a body of mass 20 lbs. gives it a velocity of 10 ft.-per-sec. What velocity will be imparted to a body of 50 lbs. mass in 9 sec. by a force of magnitude 3. A force of 8 lbs. (z. e., equal to the weight of 8 lbs. mass ; see Art. 47), acting upon a body of mass m lbs. for 12 sec, gives it a velocity of 45 ft. -per-sec. What velocity will be imparted to a body of mass 7m lbs. by a force of 22 lbs. acting for 9 sec. ? 4. Two bodies whose masses are 2 lbs. and 5 lbs. , starting from rest at the same instant, are observed to have equal velocities at every subsequent instant. Compare the magnitudes of the forces acting upon them. 5. A body near the surface of the earth will, if unsupported, fall toward the earth. It is observed that two bodies of different masses, starting from rest, acquire equal velocities in equal times. What inference can be drawn as to the forces acting upon them ? 6. A body falling from rest under the attraction of the earth is observed to have a velocity of 32. 2 ft.-per-sec. at the end of the first second. Assuming the attraction of the earth to be a constant force, what velocity will the body have at the end of 20 sec. ? 7. If g ft. -per-sec. is the velocity acquired by a falling body in 1 sec, what is its velocity at the end of t sec, assuming that the only force acting upon the body is the constant attraction of the earth ? 215. Effect of Constant Force on Body Not Initially at Rest. If a body moving in a straight line is acted upon by a force of con- stant magnitude whose direction coincides with that in which the body is moving, its velocity will receive, during any interval of time, an increment proportional directly to the force and to the duration of the interval, and inversely to the mass of the body. Let m be the mass of the body, P the magnitude of the force, and let the velocity have values v x and v t at any two instants t x , t y Then v. 2 v x is proportional to P (7 2 t^/m. This case obviously includes that given in the preceding Article ; for if the initial velocity is zero, the final velocity is equal to the velocity-increment. If a body is acted upon by a constant force whose direction is opposite to that in which the body is moving, the velocity is de- MOTION IN A STRAIGHT LINE. 1 79 creased, during any interval of time, by an amount proportional directly to the force and to the interval, and inversely to the mass. If the interval be long enough, the velocity will be reduced to zero, and will then be reversed in direction. This case and the preceding are both included in the general statement that A constant force acting upon any body gives it a velocity -incre- ment zvhose direction is that of the force, and whose magnitude is proportional directly to the force and to the time during which it acts, and inversely to the mass of the body. If the two directions along the line of motion are distinguished as plus and minus, the sign of the velocity-increment agrees with that of the force. It should be observed that this principle has no reference to the distance described by the body during the interval in question. It gives us a rule for estimating the change in the velocity ; the result is independent of the actual velocity at the beginning of the interval or at any other instant. The distance passed over depends upon the value of the velocity at every instant throughout the interval, not merely upon the amount by which the velocity changes. The above proposition is a statement of Newton's second law of motion, as applied to the motion of a body under the action of a single force whose line of action coincides with the line of motion. Newton's third law is the law of action and reaction (Art. 35). For a formal statement of the three laws, see Art. 259. 216. Equation of Motion for Particle Acted Upon by Constant Force. Let a particle of mass m } acted upon for a time At by a force P, receive a velocity-increment Av ; and let a particle of mass m\ acted upon by a force P' y receive in a time At' a velocity-incre- ment Av'. Then Av : Av' = PAt/m : P'At'/m ; or mAvjPAt = m'Av'/P'At'. That is, the quantity mAv/PAt has the same value whatever the particular case of motion considered ; the force P and the mass m having any values whatever, and Av being the increment of velocity received in a time At. There may therefore be written the equation m Av\PAt = k } or Av =k(PAtjm), . . . . (1) l8o THEORETICAL MECHANICS. k being a constant. This may be called the general equation of motion for a particle acted upon by a constant force directed along the line of motion. \ip is the acceleration, the equation may be written p = k(Plm) t .... (2) since p = Av/At when the velocity varies at a uniform rate. The numerical value of k depends upon the units employed in expressing the values of Av, P, m and At. These units having been chosen, k can be determined by a single experiment. The nature of the experiment must be as follows : A force of known magnitude (say P' units) is applied to a body of known mass {m' units) for a known time ( At' units), and the ve- locity produced is measured (call it Av' ft.-per-sec). Then, since these values must satisfy equation (i), k = m'Av'/P'At'. The units of force, mass, length and time may thus be chosen arbi- trarily and a value of k determined which will make equation (i) true for all cases in which the same units are employed. The unit velocity is here assumed to be derived from the units of length and of time as in Art. 191. Examples. 1. Take the unit mass as a pound, the unit force as the weight of a pound-mass {i. e. y a pound-force), the unit time as the second and the unit length as the foot. Determine the value of k. Let a body of m pounds-mass fall freely from rest under the ac- tion of gravity. Experiment shows that the velocity increases at a uniform rate. In one second the increment of velocity is about 32.2 ft.-per-sec. The force producing this effect is the weight of m pounds-mass, or in terms of the pound-force its value is m. Hence in the above value of k we may substitute m' = m, P' = m, At' = 1, Av' = 32.2; k = {m X 2t 2 - 2 )/(. m X 1) = 32.2. The equation of motion for this system of units is therefore p = Av/At = 2>2.2P/m. The value 32.2 ft. -per-sec. -per-sec. is only an approximate value of the acceleration of a body falling freely under gravity. The true value varies somewhat with the position on the earth's surface. If g denotes the value at any given locality, k = g, and the equation of motion is p = g\m). . . . . (3) MOTION IN A STRAIGHT LINE l8l Since the pound-force varies with the locality in exactly the same ratio as the value of g y the numerical value of any given force (rep- resented by P in the equation) varies inversely as^-, so that equa- tion (3) is always true if g is given its true value for the particular locality at which the pound-force is determined. 2. If the unit mass is equal to 200 lbs., the unit force equal to the weight of 10 lbs., the unit time the second, and the unit length the foot, determine the value of k. Ans. k = g/20. 3. If the pound is the unit mass, the foot the unit length, and the second the unit time, what must be the unit force in order that k may equal 1 ? Ans. A force equal to i/g times the weight of 1 lb. 4. If the pound-force is the unit force, the foot the unit length, and the second the unit time, what must be the unit mass in order that k may equal 1 ? Ans. A mass equal to g lbs. 5. Show that, whatever units are employed, the constant k is nu- merically equal to the acceleration due to the unit force acting upon the unit mass. Answer examples 1, 2, 3 and 4 by the direct appli- cation of this general result. 6. The equation of motion may be written P = k'mp, in which k' = \\k. Show that the constant k' is numerically equal to the force which will give the unit mass the unit acceleration. 217. Kinetic Systems of Units. It has been seen that, in the general equation of motion given above (Art. 216), the four quanti- ties, force, mass, velocity and time (or force, mass, length and time) may be expressed in any arbitrary units, provided the value of k is properly determined. It is also apparent that k may be given any desired value by properly choosing the units of force, mass, length and time. In order to simplify the equation of motion, let k = 1, and con- sider what restriction is thus imposed upon the choice of units. The equation of motion becomes p = AvjAt = Pirn. . . . (1) In order to satisfy this equation it is necessary to express force, mass, length and time in such units that a unit force acting for a unit time upon a unit mass will give it a unit velocity. Any system of units satisfying this requirement may be called a kinetic system. It is obvious that, even with this restriction, any three of the four units named may be chosen arbitrarily. But when three are chosen as fundamental, the fourth becomes a derived unit. 1 82 THEORETICAL MECHANICS. For the purposes of pure science the common practice is to take as fundamental the units of mass, length and time ; the unit force being derived from these in accordance with the requirement above stated. Two such systems of units may be mentioned. The centimeter -gram- second system. In this system the cent- imeter is the unit length, the gram the unit mass and the second the unit time. It is briefly called the C. G. S. system. The unit force is called a dyne. A dyne is a force which, acting for one second upon a mass of one gram, gives it a velocity of one centimeter-per-second. The C. G. S. system is almost universally employed in purely scientific work. The foot-pound-second system. In this system (called briefly the F. P. S. system) the foot is the unit length, the pound the unit mass and the second the unit time. The unit force is called a poundal. A poundal is a force which, acting for one second upon a mass of one pound, gives it a velocity of one foot-per-second. Examples. i. A body whose mass is 40 lbs. is acted upon by a constant force of 12 poundals. What is the velocity after 4 sec. (a) if in- itially at rest, (b) if the initial velocity is 20 ft.-per-sec. in the direction of the force, (V) if the initial velocity is 20 ft.-per-sec. in the direction opposite to that of the force. Ans. {a) 1.2 ft.-per-sec. (b) 21.2 ft.-per-sec. (c) 18.8 ft.-per- sec. 2. A body of 20 lbs. mass is acted upon by a constant force which in 1 2 sec. gives it a velocity of 60 ft. -per-sec. What is the mag- nitude of the force in poundals? 3. A body of 6 lbs. mass, starting from rest and falling freely under the earth' s attraction, is observed to have, after 2 sec. , a velocity of 64. 4 ft. -per-sec. If the earth' s attraction upon the body is a constant force, what is its magnitude in poundals ? 4. A body of m lbs. mass, falling vertically under the action of its own weight, receives during each second a velocity of g ft. -per-sec. What is its weight in poundals ? (That is, what attractive force is exerted upon it by the earth?) Ans. mg poundals. 5. What is the ratio between a force of one pound and a force of one poundal? 6. A body of 20 gr. mass is acted upon by a constant force of 5 dynes. Determine its velocity after 16 sec. {a) if initially at rest, MOTION IN A STRAIGHT LINE. 1 83 (b) if its initial velocity is 16 c.m.-per-sec. in the direction of the force, (c) if its initial velocity is 16 c.m.-per-sec. in the direction opposite to that of the force. Ans. (c) 12 c.m.-per-sec. 7. A body whose mass is 3 kilogr. is acted upon by a constant force which, in 5 sec, changes its velocity from 10 met. -per-sec. in one direction to 12 met. -per-sec. in the opposite direction. What is the value of the force ? Ans. 1,320,000 dynes in the direction of the final velocity. 8. A body of 2,000 gr. mass, starting from rest and falling freely under the earth's attraction, has at the end of 2 sec. a velocity of 19.6 met. -per-sec. If the earth's attraction upon the body is a con- stant force, what is its magnitude in dynes ? Ans. 1.96 X 1 o 6 dynes. 218. Engineers' Kinetic System. For the purposes of the en- gineer the most convenient unit of force is the weight of a definite mass at the earth's surface. Such a unit, though not exactly the same for all localities, is sufficiently definite for most practical pur- poses. Let the pound-force (already defined as the weight of a pound- mass) be taken as the unit force, let the foot and second be taken as units of length and time, and let the unit mass be so determined as to satisfy the equation P = m(Av/Af) = mp. The unit mass must now be defined as a mass whose velocity will increase by one foot-per-second during each second if acted tip on by a force of one pound . The ratio of this unit to the pound-mass may readily be deter- mined. A force of 1 lb. acting upon a mass of 1 lb. for 1 sec. gives it a velocity of^- ft. -per-sec. The same force acting upon a mass of g lbs. for 1 sec. will therefore give it a velocity of 1 ft. -per-sec. The required unit mass is therefore g times as great as the pound-mass. If this system of units is employed, a mass whose value is given in pounds must be reduced to the unit just defined by dividing It will hereafter be assumed, unless otherwise expressly stated, that a kinetic system of units is employed, so that the equation of motion takes the form P= m p. Examples. 1. A mass of half a ton is acted upon by a force of 50 lbs. 184 THEORETICAL MECHANICS. Write the equation of motion, using the pound as the unit force. What is the acceleration ? [1 ton = 2,000 lbs.] Ans. p = 1 .61 ft. -per-sec. -per-sec. 2. With French units, let the unit length be the meter, the unit time the second and the unit force the weight of a kilogram. Deter- mine the value of the unit mass in terms of the kilogram, so that the equation P = nip may be satisfied. 3. A mass of 200 kilograms is acted upon by a force equal to half its weight. Write the equation of motion, taking units as in Ex. 2. Determine the acceleration. Ans. p = 4.9 met. -per-sec. -per-sec. 219. Dimensions of Units in Kinetic System. The relation which must subsist among the units in order that the constant k in the general equation of motion shall be unity may be expressed by a dimensional equation. Let the units be represented by the same symbols as heretofore (Arts. 15, 191, 204). The equation to be satisfied is P = mp. The dimensional equa- tion is therefore p __ j^ But also V = L/T ; A = V/T = L/T 2 ; hence F = ML/T 2 . If, as in the C. G. S. and the F. P. S. systems (Art. 217), the fundamental units are M, L and T, this equation gives the dimen- sions of the unit force. If, however, F, L and T are made funda- mental, the equation gives the dimensions of the unit mass ; in this case it may be written M = FT 2 /L. In accordance with scientific usage, we shall generally regard M, L and T as fundamental units in the general equation of motion and all equations derived from it ; force being regarded as having the dimensions given by the above equation. 220. Effect of Two or More Constant Forces. If a body mov- ing in a straight line be acted upon by two or more forces directed along the line of motion, it receives, during any interval, a velocity- increment equal to the algebraic sum of the increments which would be produced by the forces acting separately. The same effect would be produced by a single force equal to the algebraic sum of the sev- eral forces. That is, the resultant of any number of forces acting MOTION IN A STRAIGHT LINE. 1 85 upon the same particle and directed along its line of motion is a single force equal to their algebraic sum. 221. Effect of Variable Force. If a body is acted upon by a force of variable magnitude, the velocity-increment produced in any given interval cannot be estimated so simply as in the case of a con- stant force. In this case the equation of motion becomes a differential equation. If a body of mass m receives a velocity-increment Av in a time A/, the equation P = m(Av/At) gives the magnitude of a constant force which will produce the same velocity-increment in the same time. If the interval At is long, the actual magnitude of the force may vary widely from this value. But if the time At be taken smaller and smaller, approaching the limit zero, Av also approaching zero, the value of P given by the equation approaches as a limit the actual magnitude of the force at the begin- ning of the interval. Hence, if P denotes the value of the force at any instant, P = lim [m(Av/Af)] = m{dvjdf). The determination of the change of velocity produced by the force in any finite time requires the integration of this equation. This is not possible unless the law of variation of P is known. 222. Equation of Motion of a Particle Acted Upon by Any Number of Constant or Variable Forces. It is now evident that, if a kinetic system of units be employed, the equation of motion may be written P = m{dvjdt), or P= mp, if p is the instantaneous value of the acceleration and P the instan- taneous value of the algebraic sum of all forces acting on the particle. If the position of the particle be specified by its distance x from a fixed point in the line of motion, p = dvjdt = d 2 xldt'\ and the equation of motion may be written P = m(d 2 x/dt 2 ). CHAPTER XIII. MOTION IN A STRAIGHT LINE : APPLICATIONS. i. General Method. 223. Classes of Problems. The equation P=mp . . . . (1) may always be applied in the solution of problems relating to the motion of a particle in a straight line. The problems that may con- ceivably arise are of various kinds, depending upon what is known and what unknown regarding the motion. The most important cases are the following : (1) The motion being known, it is required to determine the re- sultant force. (2) The forces being known (as functions of the position or of the time or of both), it is required to determine the motion. The second of these problems will require the solution of a dif- ferential equation, and may be called the inverse problem ; the first will be called the direct problem, since it involves no integration. 224. Direct Problem : To Determine the Resultant Force When the Motion Is Known. If the acceleration is known at every in- stant, the resultant force can be determined by substitution in the equation P = mp. If the position or the velocity is known as a function of the time, the acceleration can be found by differentiation. Examples. 1. A body of 1 lb. mass, starting from rest, moves so that its dis- tance from the starting point at every instant is given by the formula x = 16. 1 1'\ x being in feet and t in seconds. Required the mag- nitude of the resultant force acting on the body at any instant. Ans. The force is constant and equal to 32.2 poundals. 2. If the velocity of a body is constant, what is the magnitude of the resultant force acting on it ? 3. If the position of a particle of mass m is given by the formula x = a sin bt, determine the value of the resultant force acting upon it as a function of x. Ans. P = mb l x. MOTION IN A STRAIGHT LINE: APPLICATIONS. 1 87 4. A body of m lbs. mass, acted upon by no force except the earth's attraction, is observed to receive each second a velocity- increment of g ft. -per-sec. What is the magnitude of the force act- ing upon it ? Ans. mg poundals. 5. A body whose mass is 18 lbs. moves so that its position at any instant is given by the equation x = $t 2 -j- 6/ -f 8, / being in seconds and x in feet. Required the magnitude of the resultant force acting upon it at any instant. Ans. 180 poundals or 5.59 pounds-force. 6. What force will give an acceleration of 1,000 cm. -per-sec - per. -sec. to a mass of 600 gr. ? 7. What constant force acting upon a particle of m grams mass will increase its velocity by^ cm. -per-sec. in 1 sec. ? 8. What is the weight in dynes of a mass of m grams ? Ans. mg, if g is the acceleration due to gravity, expressed in cm. -per-sec. -per-sec. Its value is known to be about 981. 9. The velocity of a particle is proportional to its distance from a fixed point, and is 24 ft. -per-sec when the distance from the fixed point is 2 ft. If the mass of the particle is 4 lbs., what is the value of the resultant force acting upon it when 8 ft. from the fixed point? Also when 3 ft. from the fixed point ? Ans. 4,608 poundals. 1 ,728 poundals. 225. Inverse Problem: To Determine the Motion When the Forces Are Known. If every force is known as a function of one or more of the three quantities x, t, v, the general equation P = mp becomes a differential equation, the complete solution of which de- termines x and v as functions of t. In general two integrations will be required, thus introducing two constants. To determine the constants, some information must be given concerning the motion at one or more definite instants, or in some one or more definite positions. The information usually avail- able is the following : The velocity and position at a certain instant are completely known. Note on constants of integration. The method of determining constants of integration has already been illustrated in several par- ticular cases. A clear understanding of the general method is of importance to the student. To determine such a constant it is always necessary to have some information in addition to that which enables us to write the differential equation. Let there be given any equation containing two variables, x and y y and a constant, C. Then the value of C can be determined if one 1 88 THEORETICAL MECHANICS. pair of simultaneous values of;r and jy be known ; that is, if it be known that ' ' when x = some known value, y = some known value. ' ' Thus, let the given equation be ffcy, C) = o. If it is known that when x = a, y = b, a and b being known con- stants, there may be written /{a, b, C) = o, from which C can be determined. In a problem relating to the motion of a particle, the variables being usually some two of the quantities x, t and v, the information necessary for determining the constant is equivalent to some knowl- edge as to the position at some instant, or as to the condition of motion of the particle either at some instant or in some position. It is to be noticed that the same method will serve for determining any constant in an equation, whether introduced by integration or otherwise. 2. Motion Under Constant Force. 226. Solution of General Problem. Let a particle of mass m be acted upon by forces whose resultant n jc has the constant value P directed parallel O A. to the line of motion. Let O (Fig. 103) Fig. 103. be a fixed point in the line of motion, and let x be the distance of the particle from O at the time t. The equation of motion is then mOPxfdt*) = P = constant. . . . (1) Two integrations give midxldf) =Pt+ C x \ . . . (2) m*:==iFP'+)CS+.C M . . . . (3) To detemine C A and C. 2 , let it be known that at a certain instant t the velocity is v Q and the abscissa of the particle x . Then from (2), and from (3), C x = mv a ~P k ; C 2 = mx iP/ ' (ptv, P/ )l = mx mvj a + h_P/\ MOTION IN A STRAIGHT LINE: APPLICATIONS. 189 Since the origin of time may be chosen arbitrarily, let t be reckoned from the instant at which v = v Q and x = x ; then t = o, and C x mv ; C 2 = mx . Equations (2) and (3) now become midxjdf) = Pt -f mv ; or m{v v ) = Pt\ . (4) *&r = \Pf 4 w%/ + **#o- (5) These equations give the velocity and position at any instant. 227. Motion of a Body Falling Vertically Near the Earth. A body near the surface of the earth is attracted toward the earth's center with a force which varies as the body moves upward or down- ward. If the range of motion is small compared with the earth's radius, the force varies so little that for most purposes it may be regarded as constant. Hence if a body starts from rest, or is pro- jected vertically upward or downward, and is then left to the influence of the earth's attraction, it presents a case of rectilinear motion under a constant force. Experience shows that bodies of unequal mass, acted upon by gravity alone, are equally accelerated. Their weights are therefore proportional to their masses. Let W x and W 2 denote the weights of two bodies whose masses are m x and m 2 ; then WJIV 2 = mjm 2 . This equation is true for any units of force and mass. If these units satisfy the condition prescribed in Art. 217 (so that the unit force gives the unit mass the unit acceleration), the acceleration of the mass m x acted upon by the force W x is WJm x , and the acceleration of the mass m 2 acted upon by the force W 2 is WJm 2 . If the known value * of the acceleration due to gravity is denoted by g, *The value of g varies with the position on the earth. This variation is approximately represented by the following formula : Let

dv = kxdx. MOTION IN A STRAIGHT LINE: APPLICATIONS. 1 95 Multiplying through by 2 and integrating, v 2 = kx 2 + C. Applying condition (V), we find C = ka 2 ; hence the last equation may be written v l = k(a l x% . . . . (2) which gives the velocity in any position. To determine a relation between x and t, a second integration is required. Equation (2) may be written (dx/dt) 2 = k{a l x% or k*dt = dxjVd'x 2 . Integrating, k^t = sm~\xla) + C. From condition (b) it follows that when / = o, x = 0, and there- fore C sin -1 (o) = o. Hence the equation becomes k%t -- sin _1 (.r/tf), or x = a sin (k l/2 t) (3) Equation (3) gives the value of x at any time. [In determining the value of C , it was assumed that the angle whose sine is is o. But this is only one of many allowable values. The general value would be sin -1 (o) = mr> where n is any integer. Using this value we have whence sin_1 ( ^ = k%t + **' x\a = sin (Jfat -j- nif) = sin (k&) cos ntr -f cos (Jfct) sin nir. But sin nir = o, and cos mr = 1 ; the sign depending on whether n is even or odd. Hence x = a sin (kfit). The double sign shows that the assumed conditions are not sufficient to fully determine the motion ; it is left uncertain whether the motion has the positive or the negative direction when t = o. The former supposition will be adopted, and the plus sign used, as in equation (3)-] A discussion of equation (3) shows that x is periodic; that is, in successive equal intervals it passes repeatedly through the same series of values. The greatest value of x is a ; and it is seen that x a as often as sin (k&f) = 1 ; that is, when kfit = 73-/2, 57r/2, 97J-/2, etc. ; or when t = 7r/2k y *, 5^/2^, 971-/2^, etc. 196 THEORETICAL MECHANICS. The interval between any two successive values of t in this series is 27r//^ ; which is the interval required for the particle to return to the point at which it is at rest after once leaving it. Again, the greatest negative value of x is a ; which occurs when sin (#*/) = 1 ; that is, when k V *t = 37T/2, 77T/2, II7T/2, etC. | or when / = 371-/2$*, 777-/2$*, 1171-/2$*, etc. The interval between two successive values of / in this series is 27r/$ / 2. It is also seen that the interval between two instants at which x = a and x = a is 7r/$ /2 . Again, the interval between the in- stants when the particle is at O and at *--- a *--- a * A is seen to be 7r/2$ /2 . Hence, if A' A! 0'*~JC -4 A is the point at which x = a, it is FlG - io 5- seen that, in successive intervals of 77/2$^ sec. , the particle passes from A to 0, from to A', from A' to 0, and from O to A ; and then re- peats the cycle. Motion in accordance with this law of force is called harmonic motion, and is of great importance in mathematical physics. 230. Repulsive Force Proportional Directly to the Distance. Consider next the case in which the force is repulsive, i. e., acts always away from the fixed point, and let the magnitude of the force vary in direct ratio with the distance from the center of force. Let P' be the magnitude of the force when the particle is at the unit distance from O ; then P = P'x, and the equation of motion is m'x = P'x, or dvjdt = x = P'x/m = kx, . . (1) where k is a positive constant, and means the magnitude of the repul- sive force per unit mass when the particle is at unit distance from O. Equation (1) is identical with the differential equation for the motion when the force is attractive (equation (1), Art. 229), except that k takes the place of k. The integration may therefore be car- ried out in the same manner ; and if the same initial conditions are assumed, the final result will take the same form, with the substitution of k for k. The position would be given by the equation corres- ponding to (3) of Art. 229: x = a sin (jv k). MOTION IN A STRAIGHT LINE: APPLICATIONS. 1 97 In order to avoid this imaginary form, the integration may be effected in a different manner. It will be found necessary to change the initial conditions as the solution proceeds. Integrating as in Art. 229, (dx/dtj 2 = k(x l d), . . . (2) the constant being determined from condition (V) assumed in the pre- ceding case, that when x = a, v = o. Equation (2) may be written k*dt = dx/Vx 2 d\ Integrating, , k*t = log (x + V x 2 d) + C. If, as in the preceding case, it be assumed that x = when t = (condition (Jj) of Art. 229), the value of C is imaginary. This shows that such a condition is inconsistent with the condition already as- sumed ; if the particle is at rest when x = a, it can never pass through the origin. This is of course obvious in the case of a repulsive force ; it is, in fact, evident that x can never be less than a. Therefore, instead of assuming condition (b) of Art. 229, let / be reckoned from the instant when x = a. This gives C = log a, and therefore k*t = log [O -f- V x * d)la\ or x -f Vx 2 a? == 04**'. Solving for x } x t x = \a(e** + e~ kt ). . . . (3) Here e denotes the base of the natural system of logarithms, its value being 2. 7 18-}-. Equation (3) shows that the motion is not periodic, as in the pre- ceding case. If t = 0, x = a ; and this is the least value x can have. Again, if / be given equal positive and negative values, the corresponding values of x are equal. Hence the motion after the instant / = o is exactly the reverse of the motion before that instant. As / increases, x increases ; and the particle never returns after it be- gins to recede from its nearest position to O. Examples. 1. Let the force be attractive, its magnitude at 1 ft. from the center of force being 4 poundals per pound of mass of the attracted particle ; and let the particle be at rest at a certain instant at 10 ft. from the center. Determine the position and velocity in terms of /. I98 THEORETICAL MECHANICS. 2. In the same case, what are the position and velocity 5 sec. after the particle is at rest? Ans. x = 8.39 ft. ; v = -(-10.89 ft.-per-sec. 3. In the same case, with what velocity does the particle pass the center of force? Ans. 20 ft.-per-sec. 4. How often does the particle return to the starting point ? Ans. At intervals of 3. 14 16 sec. 5. Solve example^ 1 and 2, assuming the force to be repulsive, the remaining data being the same as before. Ans. When t = 5, x = no, 100 ft., v = 220,200 ft.-per-sec. 6. Solve the problem of the motion of a particle under a force varying directly as the distance from a fixed point in the line of mo- tion and directed away from that point, taking all data as in the above general solution with the following exceptions : Instead of the con- dition v = o when x a, assume that v = v when x o. 7. Solve with data as in Ex. 5, except that the velocity is to be 10 ft.-per-sec. when the particle passes the center of force. 8. Let the force be attractive, its magnitude at 1 met. from the center of force being 1,000 dynes per gram of mass of the attracted particle. Let the particle start from rest at 150 cm. from the center. Determine the position and velocity at any time. 9. With data as in Ex. 8, determine the velocity when the par- ticle is 100 cm. from the center of force, and determine the position and velocity 10 sec. later. Ans. When x = 100 cm., v 354 cm.-per-sec 10. Assume data as in Ex. 8, except that the force is repulsive. Determine the position and velocity 10 sec. after the particle is at rest. 231. Force Varying Inversely as the Square of the Distance. Let a particle be attracted toward a fixed point with a force whose magnitude varies inversely as the square of the distance from that point. Let P' denote the magnitude of the force when the particle is at unit distance from the center of attraction ; then the equation of motion is , 2 mx = F jx\ or x = dxjdt = /x 2 , . . . (1) in which k (= P'jm) is the magnitude of the attractive force per unit mass at unit distance from the origin. * *It is to be noticed that equation (1) applies only when the particle is on the positive side of the origin. The equation makes the acceleration negative for all values of x, while in fact the force (and therefore the acceleration) is positive when x is negative. Therefore the equation must be used for positive values of x only. For negative values of x the equation must be x = k/x 2 . MOTION IN A STRAIGHT LINE: APPLICATIONS. 1 99 Let equation (i) be integrated subject to the condition that the particle is at rest when at a distance a from the origin, and that / is reckoned from that instant. The first integration can be performed by the method described in Art. 228. Multiplying through by dx and integrating, \x> = k\x- + C. The initial conditions make the constant equal to kja, hence x 2 = (dxjdff = 2k{\\x ijd). . . (2) Equation (2) gives the velocity when the position is known. To find the relation between x and /, we have dtV 2kla = dxVx/(a x). Integrating tv 2k\a b=3 1 ax x' 1 + \& sin -1 \[ix d)ld\ -j- C . If t = when x = a (as above assumed), C = -\a sin -1 ( 1) ; hence tV 2kja = V ax x' 1 -+- \a sin _1 [(2^ a)/a] \a sin _1 (i), or tv 2k ja = V ax x' 1 -f \a cos -1 [(2^ d)ja\ . (3) 232. Motion Under the Attraction of the Earth. If the earth were a sphere of uniform density throughout, or a sphere in which the density had the same value at all points equally distant from the center, its attraction upon any body outside its surface would vary inversely as the square of the distance from the center. (Art. 183.) In the actual case the attraction is very nearly expressed by the law stated. Hence equations (1), (2) and (3) of the preceding Article may be applied with small error to the motion of a body falling vertically toward the earth from a great height, supposing gravity to be the only force acting. In Art. 227 the motion of a falling body was dis- cussed on the supposition that the attraction of the earth upon it is constant. In the present case the range of motion is supposed to be so great that a more accurate treatment is desirable. In order to apply the results of Art. 231 to the present problem, it is necessary to determine the value of the constant k. If R denotes the radius of the earth, we know that when x = R the acceleration is g. Hence from equation (1) we have -g=~k/R\ or k=gR\ 200 THEORETICAL MECHANICS. If g is in feet-per-second-per-second, R must be in feet, and k means the attraction in poundals on a pound-mass if placed one foot from the earth's center (supposing the same law of variation of the earth's attractive force to hold within the surface as without). The value of R may be taken as 20,900,000 feet. Examples. 1. A body starts from rest at a height above the surface equal to the earth's radius. Compute the velocity when the surface is reached. Putting k = gR 2 , the velocity is given by equation (2) (Art. 231 J, which may be written v 2 = 2gR\\\x i/a). In this example a = 2R ; hence for x = R we have v 2 = 2gR\i/R 1/2R) =gR. In feet-per-second, v = V 32.2 X 20,900,000 = 25,900. 2. What is the greatest velocity a body could acquire in falling from rest to the earth's surface ? [Put a = do, x = R.~\ 3. With what velocity must a body be projected upward at the earth's surface in order that it may never return? Ans. A velocity not less than 6.95 miles-per-sec. 4. Deduce a formula for the velocity acquired by a body in falling to the surface from a height h. Putting a = R + h and x R, the above general formula for v 2 becomes ^ = 2g k[RI{R + h)\ If h is small compared with R, we may put as a first approxima- tion ^/(^ + fy p- J ; ... 7 r> = 2g / L This is identical with the formula which would apply if the attraction were constant. See Ex. 1, Art. 227. In using the accurate formula, if hIR is small we may write R/(R + h) = (1 + /1/R)- 1 and therefore v 2 = 2g/i[i -wr) + WRy-(A/Ry+ . . . ]. By taking any number of terms of this series an approximate result may be obtained which is correct to any desired degree. 5. Let a body fall to the surface from a height of 5,000 ft. Compute the velocity acquired, using first the approximate formula and second the accurate formula. (Take g = 32. 2.) 6. Determine the value of fc, using C. G. S. units. MOTION IN A STRAIGHT LINE I APPLICATIONS. 201 4. Miscellaneous Problems. 233. Constant Force Combined With Central Force. Let a body be acted upon simultaneously by a constant force and a force directed toward a fixed point. Then if P l and P t are the values of the two forces at any instant, the general equation becomes **='/>+ i> (0 and we must put for P x and P 2 their values as functions of x and /. If the origin is taken at the center of force, and the central force is a function of the distance from the center, we have P x = constant; P 2 =/(*). The only case that will be discussed is the following : Body suspended by elastic string. A body is suspended from a fixed point by an elastic string, and is acted upon by no force except its weight and the supporting force exerted by the string. To de- termine the motion. We must here make use of the property that a stretched elastic string exerts a resisting force proportional directly to the amount of stretching. This law applies not only to elastic My. strings but to a bar or rod of any elastic material. The law is established by experiment, and for many substances is nearly, though perhaps in no case exactly, true. Let MO (Fig. 106) represent the unstretched string, M being the point of support. Let / denote the length MO, and x the amount of stretching at the time / ; at which instant the end of the string, originally at 0, is at A. The upward force exerted by the string upon the particle is then proportional to x, and we may put P 2 = ex in equation (1), c being the force necessary to pro- ^ duce an elongation of one unit of length. Since the force p IG Io6 producing a given elongation is greater in proportion as the unstretched length is less, c is inversely proportional to /, and we may write . P % ~ (e/l)x y e being a constant. For P x may be written mg, the weight of the body ; hence equa- tion (1) becomes m(d' 2 x/dt 2 ) = mg (e/l)x. . . . (2) 202 THEORETICAL MECHANICS. This equation can be integrated directly, but the process is simpler if the origin of coordinates is changed. Let 0' (Fig. 106) be the position of the end of the string when the two forces just balance each other ; the distance of this point from O is the value of x found by putting P x -\- P 2 = o, that is mg exjl = o, or x = mgl/e. Introducing a new variable *, such that x = z + mglje, equation (2) becomes d 2 z/dt 2 = (e/ml)z. The variable z evidently means the abscissa of the particle measured from O'. The last equation is identical in form with equation (1) of Art. 229, and the solution there given is here applicable. It is thus seen that in the case now under discussion the motion is the same as if the particle were acted upon by a single force directed toward the point O' and varying directly as the distance from that point. Examples. 1. If a force equal to 10 pounds- weight would change the length of the string from 5 ft. (its natural length) to 6 ft, and if the mass of the body is 5 lbs. , write the equation of motion. A ns. d' 2 x/dt 2 (1 2X)g, or d 2 zjdt 2 = 2gz. 2. In the same case, let the body be at rest when the string is unstretched ; determine the motion completely. Ans. z = \ cos (ty 2g). 3. Can the initial conditions be such that the equation deduced above does not apply throughout the whole of the motion? If so, how could such a case be treated ? [The value of P. 2 , the upward pull of the string on the body, be- comes zero when the string shortens to its natural length, the body being then at O (Fig. 106). If the body rises above this point, P t is zero, and so long as this is the case the body moves under the single force of gravity. If the string be replaced by a rigid bar of elastic material (such as steel or wrought iron when stretched within the ' ' limit of elas- ticity"), the force P 2 will follow the same law when the bar is short- ened as when it is lengthened; that is, exjl is the value of P 2 throughout the whole motion,* even if the initial conditions are such that the length of the bar becomes less than the natural length.] * This assumes that the " modulus of elasticity " has the same value in compression as in tension. MOTION IN A STRAIGHT LINE : APPLICATIONS. 203 4. With the data of Ex. 1, let the body have a velocity of 10 ft.- per-sec. at the instant when the string is unstretched. Determine the motion completely. 234. Motion in a Resisting Medium If a body be moving through the air, or through any fluid, the fluid exerts upon it forces which depend upon the velocity. The effect of these forces is to retard the motion of the body ; in other words, the resistance of the medium is equivalent to a force whose direction is always op- posite to that in which the body is moving. Hence in the general equation of rectilinear motion, such a resistance will enter as a force whose magnitude is some function of the velocity, and whose direc- tion is opposite to that of the velocity ; that is, for such a force, care being taken that the correct sign is used Examples. 1. Write the equation of motion for a body moving vertically under gravity and the resistance of the air, the latter being assumed to vary directly as the velocity. Explain the meanings of any con- stants entering the equation. Integrate the equation completely. 2. In the same case, assume the resistance of the air to vary as the square of the velocity. Can a single equation be written for the whole motion in this case, if the body has initially an upward velocity? Integrate the equation once, thus determining the relation between velocity and time. 235. Motion of Connected Particles. If two particles are con- nected by a string which is kept tight, each is acted upon by a force due to the string. If the string is without weight and is not in con- tact with any body except the two particles mentioned, the tension sustained by it has the same value at every cross-section. Hence the string exerts equal and opposite forces upon the particles. If the string passes around smooth pegs or pulleys, the tension is still uniform throughout its length, and the forces exerted upon the particles by the string are equal though they may not be opposite. If the equation of motion be written for each particle, each equa- tion will contain the force due to the tension of the string. By com- bining the two equations this unknown force may be eliminated. This will be illustrated by a particular case. Let two particles whose masses are m y and m % be connected by a 204 THEORETICAL MECHANICS. string which passes over a smooth pulley. Let the initial conditions be such that each particle describes a vertical line, one rising while the other falls. Assume the string to be without weight, perfectly flexible and inextensible. Assume also that the mass of the pulley is so small as to be negligible, and that it can revolve without frictional resistance. The arrangement is shown in Fig. 107. Let m x be greater than m 2 , and assume the par- ticles to be at rest at a certain instant. Then m x will fall and m 2 will rise. Take the position of rest as origin for each particle, and let x denote the dis- '~\tti !~]m tance f eacn fr m its origin at the time t. If T = Fig. 107. tension in string, the resultant force acting upon m x is m x g T downward, and the resultant force act- ing upon m 2 is T m 2 g upward. The acceleration of m x is x downward, and that of m 2 is x upward. The equations of motion for the two particles are therefore m x g T = m x x; . . . . (1) T m 2 g --= m 2 x. . . . (2) The unknown quantity T may be eliminated by adding the two equations. The result is (m x m 2 )g = {m x -f m 2 )x. . . . (3) This equation is the same as would apply to the motion of a particle of mass m x -\- m t acted upon by a force {in x m.^g ; i. e. , by a force equal to the difference between the weights of the particles. It is as if the combined mass of the two particles were being pulled in opposite directions by forces m x g and m.,g. Evidently equation (3) may be treated by the methods employed in Arts. 226 and 227. The equations derived from the motion of a falling body may be made applicable to the present case by substitut- ing [(**, m 2 )/(m x + m. 2 )]g for g. Value of the tension. By eliminating x between any two of equations (1), (2) and (3), the value of T is found. It is T = 2[m x mJ(m x -f- m 2 )]g. . . . (4) Examples. 1. If the two masses are 4 lbs. and 4. 1 lbs., determine the accel- eration, the tension, the velocity acquired after 0.5 sec, and the distance fallen through in 0.5 sec. Ans. Acceleration = g/81. MOTION IN A STRAIGHT LINE : APPLICATIONS. 205 2. Two masses of 2 kilogr. each are suspended from the ends of a string which passes over a smooth pulley. The system being at rest, a mass of 10 gr. is added to one side. Determine the subse- quent motion. 3. Three particles are connected by two strings, one of which passes over a smooth pulley as in Fig. 108. De- termine the motion, and the tensions in the strings. Let m x , m 2 and m 3 denote the masses of the three particles, T x the tension in the string m x m 2 (Fig. 108), and T 2 the tension in m 2 m. A . Let x be the distance of m z below a certain fixed point ; then x is also the distance of m x above some fixed point, and of m % above another fixed point. The equations of motion for the three particles are T x m x g = m x x ; T 2 T x m 2 g = m^x ; m 2 \_Jm 3 m. g T 2 = mJc, The unknown quantities T x and T 2 may be elimi- I i ra, nated by adding the three equations. The result Fig. 108. ls (m 3 m x tn^)g = (m x -\- m 2 -f- m 3 )x. This is identical with the equation of motion for a particle of mass m x -f- m 2 -f- m. A acted upon by a force m^g in one direction and by forces m x g, m 2 g in the opposite direction. The values of T x and T 2 are T x = m x (g + x) = 2[m x m 3 /(m x + m 2 + m^g; T 2 = m,(g x) = 2 [0 1 -j- m 2 )m.J(m x + m 2 -f m^\g. 4. In Ex. 3, let m Xi m 2 and m. Ai expressed in kilograms, have values 2, 3 and 4. Determine the acceleration and the tensions. Arts. T x = 16/9 kilograms-weight, T 2 = 40/9 kilograms-weight. 5. Let m x = 4 lbs., m 2 = 3 lbs., m 3 = 2 lbs. Determine T x , T 2 and the acceleration. Ans. The acceleration of;;/, is 5^/9 downward. Miscellaneous Examples. [In the following examples the student should in every case write the differential equation of motion, even if it is not found possible to complete the solution. He should also examine what conditions are needed for the determination of the constants of integration. If this is done, the problem in Dynamics is reduced to one in mathematical analysis.] 1 . Two particles whose masses are equal are connected by an elastic string. They are projected from the same point with equal and opposite velocities. Determine the motion, assuming no forces to act except those due to the tension of the string. 206 THEORETICAL MECHANICS. Ans. Let a be the natural length of the string and k the force necessary to double its length. If A is the position of one particle when the string begins to stretch, that particle describes half of a harmonic oscillation (Art. 229) about A, returning to A after a time 7Ti/(am/2k). The two particles then approach each other with constant velocity. 2. Assume the conditions as in Ex. 1, except that the masses are unequal. 3. Two particles whose masses are m l and m 2 , connected by an inextensible string, are suspended from a fixed point by an elastic string attached to m x . If the system is held at rest with the elastic string at its natural length and then released, determine the subse- quent motion. Determine also the tension in the lower string in the lowest position. Ans. The required tension is twice the weight of m t . 4. Two particles, each of mass m, are connected by an elastic string whose natural length is /. The force necessary to double the length of the string is F. The particles repel each other with forces varying directly as their distance apart ; the repulsive force being F' when the distance is /. If both particles are held at rest at a dis- tance apart equal to /and are then released, determine the subsequent motion. Ans. \i F ^> F' , the motion of either particle is given by the equation x _ F + F' sin [tVi^F F")/ml ~~\ l~ 2 (FF') The motion is a harmonic oscillation (Art. 229) about a point distant F//2(F F') from the point midway between the particles, the time of a complete oscillation being wy/\_2m//{F F')\ If the origin be taken at the center about which the oscillation takes place, the motion is given by equation (3) of Art. 229, if k = 2(F F')jml. If F' > F y the motion reduces to a case under Art. 230. 5. Solve with data as in Ex. 4 except that the masses are un- equal. 6. In Ex. 4 let the mass of each particle be 40 gr. ; let / = 20 cm., F = 50 grams-weight, F' = 30 grams-weight. Determine the period of a complete oscillation, and the range of motion. Ans. Time of oscillation = 0.9 sec. 7. In Fig. 108 let the string m l m 2 be elastic, its unstretched length being /. Let the system be initially at rest with m l m 2 = /. Determine the subsequent motion. 8. Two particles, each of mass m, repel each other with forces varying inversely as the square of their distance apart. When the distance is / the force is F. They are held at rest in a vertical line MOTION IN A STRAIGHT LINE : APPLICATIONS. 207 at a distance apart /', and are then released. Determine their subse- quent motion, assuming the only forces acting on the particles to be gravity and their mutual repulsion. Ans. If r is the distance of the particles apart at any time, their velocity of separation drjdt is given by the equation {drfdff = UFP/m)(i// f i/r). This equation can be integrated, giving /asa function of r. The center of gravity of the two particles falls a distance \gt 2 in /sec. These two relations determine the motion. 9. A particle of mass m is attached to one end of an elastic string whose natural length is /. The other end of the string is at- tached to a fixed support. The particle is dropped from this fixed point of attachment. Determine the motion. Let the force neces- sary to stretch the string to double its natural length be F. 10. The natural length of an elastic string is /; under a pull Fit stretches to a length 2/. The ends are attached to fixed pegs whose distance apart is 37/2. A particle of mass m is attached to the string at a point distant 4// 5 from one peg, and is forcibly brought to the point midway between the pegs and then released. Determine the time of oscillation. Ans. fair/ 15))/ (14m// F). 11. A body is projected into a resisting medium which exerts a retarding force proportional to the velocity. If no other force acts upon the body, determine the motion. Let m = mass, V == initial velocity, and let the force = F when the velocity == V. Ans. Let;r' denote the distance the body would move, and t' the time it would move, before coming to rest against a constant force F. Then the distance described in time t against the actual force is x= ix' {1 e~ flt '). It is seen that x approaches the limit 2x ', but the particle never comes to rest. 12. A mass of 5 kilogr. moves in such a way that its acceleration is directly proportional to its velocity but has the opposite direction. When expressed in centimeter-second units the velocity and accelera- tion are numerically equal. In a certain position the velocity is 50 c.m.-per-sec. Determine the value of the force after the particle has moved 40 cm. from this position. Ans. 50,000 dynes. 13. In Ex. 12, determine the value of the force 1 sec. after the instant at which the velocity is 50 c. m. -per-sec. When will the par- ticle come to rest, and how far will it move ? Ans. 92,000 dynes. The distance passed over will approach 50 cm. as a limit, but the particle will never come to rest. CHAPTER XIV. MOTION IN A CURVED PATH. i. Position, Displacement and Velocity. 236. Direction. When the path of a particle is a straight line, its motion has at every instant one of two opposite directions. These two directions may be fully specified by signs plus and minus. But when the path is a curve, the direction of the motion continually changes, and cannot be specified so simply. In the foregoing analysis of the motion of a particle whose path is a straight line, definitions have been given of displacement, velocity and acceleration. These definitions must now be enlarged. Each one of these quantities is at every instant associated with a definite direction in space, and this direction is an essential element in its value. The quantities named are, in fact, vector quantities (Art. 16). 237. Position Given by Vector. If a particle is not confined to a straight line, but has any motion in space, its position at any in- stant may be specified by means of a vector of position. UP (Fig. 109) is the position of the particle at any instant, and any fixed point taken as origin of reference, the position of Pis known if the vector OP is known. For, to know the vector OP completely is to know (1) the direction from O to P and (2) the length of the line OP. Knowing these, P may be located from O. OP is called the position-vector of the particle. As the particle moves, the position-vec- tor varies either in length, or in direction, or in both length and direction. p IG IO Practically, in order to describe com- pletely the magnitude and direction of a vector, the values of certain angles and distances, measured from definite lines, planes or points regarded as fixed, must be given. For the present purpose it is not necessary to consider how the value of a vector may best be specified in practice. 238. Displacement. If the particle moves from A to B during any interval of time, the vector A B is its displacement. MOTION IN A CURVED PATH. 209 This definition is independent of the actual path followed, which may be any line joining A and B. If the path is the straight line AB (Fig. no), the vector AB is the actual displacement. If any other path ACB is followed, the vector AB is still regarded as the total or resultant dis- placement. 239. Uniform Motion. The mo- tion is said to be uniform when the particle receives equal displacements in any equal intervals of time, however these intervals be chosen. It is to be particularly noticed that, since displacement is a vector quantity, successive displacements are not equal unless they agree in direction as well as in magnitude. Hence, uniform motion as here defined has a constant direction ; i. e. , it is rectilinear. This case of motion has been considered in Chapters XII and XIII. 240. Variable Motion. If the particle receives unequal displace- ments in equal intervals of time, the motion is variable. The un- equal displacements may differ in length only, in direction only, or in both length and direction. If the successive displacements agree in direction, the path is a straight line, and the motion falls under the case already treated. The general analysis which follows includes this as a special case. 241. Velocity. In case of motion not confined to a straight line, the velocity of a particle may still be defined as its rate of dis- placement /* but the definition must be interpreted with reference to the meaning of displacement as a vector quantity. Whatever path a particle may describe, it is at every instant moving at a definite rate and in a definite direction. The velocity is a vector quantity whose direction coincides with that in which the particle is moving, and whose magnitude measures the rate of motion. The meaning of the definition of velocity and the method of esti- mating its value are best understood by considering "average" velocity in case of the unrestricted motion of a particle. 242. Average Velocity. The average velocity of a particle for an interval of time during which it receives any displacement, is the *As in rectilinear motion, Art. 190. 14 2IO THEORETICAL MECHANICS. velocity of a particle which, moving uniformly, would receive the same total displacement in the same time. The average velocity is a vector quantity, its direction being that of the total displacement for the interval. Note that this definition is inde- ^^ pendent of the path. Thus, if the particle moves from A to B (Fig. Fig. in. in) along any path ACB, the total displacement is the vector AB. If t x and t 2 are the values of t corresponding to the positions A and B, the average velocity is given by the expression (vector AB)l(t 2 t,), whatever the form of the path A CB. 243. Approximate Value of Velocity at an Instant. An ap- proximate value of the velocity of a particle at any instant may be found by determining the average velocity for a very short time. Thus, let A (Fig. in) be the position of the particle at the time t } and B its position after a short interval At ; then (vector AB)/At is the average velocity for the interval A/, and is an approximate value of the velocity in the position A. The approximation is closer the shorter the interval At. 244. Exact Value of Velocity at an Instant. The true value of the velocity at the instant /, when the particle is at A, is the limit approached by the above approximate value as At approaches zero. That is velocity at A = limit [(vector AB)'At]. This limit is a vector quantity. (1) Its direction is that of the tangent to the path at A, since the chord AB approaches the tangent as B approaches A. (2) Its magnitude is ds/dt, if s denotes the length of the path measured from some fixed point to the position of the particle. For as the point B approaches A, the chord AB and the arc AB ap- proach a ratio of equality, so that limit [(chord AB)/At] = limit [(arc AB)/At] = ds/dt. The magnitude of the velocity, considered without reference to MOTION IN A CURVED PATH. 211 direction, is called the speed. If this is denoted by v } its value is al- ways given by the equation v = dsjdt. 245. Graphical Representation of Velocity ; Hodograph. Let AB (Fig. 112) be the path of a par- ticle, described in any manner. From jp some point 0' draw O'A' to repre- ' / ^"~"~~ -^^^ sent in magnitude and direction the J^r velocity of the particle in the position ^^^^^ V At and O'B' to represent the velocity n ,^^^^^ \ in the position B. Also, for every ^""""^J r \P' intermediate position of the particle, ^*\^^ \ as P, draw a vector (9'P' represent- ^^"\^^ 1 ing the velocity of the particle when no in that position. Through the ex- Fig. II2 - tremities of all such vectors draw a curve. This is called the curve of velocities or hodograph of the motion. Examples. i . A particle describes a circle of 2 ft. radius with uniform speed, the whole circumference being described in half a second. Required (a) the speed ; (Jj) the values of the average velocity for 1/10 sec. and for 1/8 sec. ; (V) the values (both direction and magnitude) of the instantaneous velocity at two instants 1/10 sec. apart. 2. A particle describes a circle of radius r with uniform speed v. Determine the magnitude and direction of the average velocity for an interval t. Ans. Its magnitude is (2r/t) sin (vt/2r). 3. A particle describes a circle of radius r in such a way that s = af\ s being the length of the path described in time t, and a being a constant. Required (a) the speed at any time ; r b) the average velocity during the interval from t x to / 2 . 4. In Ex. 3, let r = 60 c. m. , and let the length of arc described during the first second be 1 20 c. m. Required (a) the speed at the end of 2 sec. ; (b) the average velocity during an interval of o. 1 sec. after /= 2. Ans. {a) 48oc.m.-per-sec. ifi) 478.5 cm. -per-sec. 5. Draw the hodograph for the motion described in Ex. 1. 6. Draw the hodograph for the motion described in Ex. 4. 7. A particle describes any curved path with uniform speed v. What is the form of the hodograph ? 212 THEORETICAL MECHANICS. 2. Velocity -Increment and Acceleration. 246. Increment of Velocity. Let the values of the velocity of a moving particle at two instants t x and t t be represented by the vec- tors O'A and O'B' respectively (Fig. 113). Then the vector A'B' is the velocity-increment for the interval from t x to t. 2 . For A'B' is the vector which must be added to O'A' to produce O'B'. (See Art. 22.) It will be seen that velocity-increment as thus defined is not the amount by which the speed changes. The increment of the speed is length O'B' length O'A' = v 2 v ly if v x and v % are the initial and final values of the speed. It is only when the initial and final velocities are parallel that the velocity- increment \s equal in magnitude to v 2 v x . 247. Acceleration. If a particle is moving in a curved path, the acceleration may be defined as the rate of change of the velocity \ just as in the case of rectilinear motion (Art. 203). But since "change of velocity " (or velocity-increment) is a vector quantity, acceleration is also a vector quantity ; and in computing its value both direction and magnitude must be considered. In this definition of acceleration, "rate of change of velocity" must be understood to mean ' ' increment of ve- locity per unit time." Fig. 113. Consider first the case in which the velocity is varying uniformly ; by this is meant that the increments of velocity during any different intervals of time have the same direction and are proportional to the intervals. Thus, let the vector A'B' (Fig. 113) represent the velocity-increment during an interval of time At, and let the velocity-increments during any partial intervals into which At may be divided have the direction A'B' and be proportional in magnitude to the 'partial intervals. Then the acceleration as above defined has a constant value through- out the time At, that value being (vector A'B')/At. When the velocity is not varying in this uniform manner, the MOTION IN A CURVED PATH. 213 acceleration is a variable vector quantity, having a definite mag- nitude and direction at any instant. The exact meaning of the definition of acceleration in this case, and the method of computing its value, may be best understood by a consideration of "average acceleration." 248. Average Acceleration. The average acceleration of a par- ticle for an interval of time during which the velocity varies in any man- ner is an acceleration which, if con- stant in magnitude and direction, would result in the same velocity- increment in the same time.* The value of the average acceleration is found by dividing the velocity-incre- ment received during the whole in- terval by the duration of the interval. \{ A B' (Fig. 114) represents the velocity-increment for the in- terval from t y to t 2 , the expression (vector A'B')/(t 2 t x ) or (vector A'B')/At gives the average acceleration for that interval. This expression cannot be reduced, as in the case of rectilinear motion (Art. 208), to the form in which v x and v 2 denote the initial and final values of the magnitude of the velocity ; for it is only when the two velocities are parallel that v 2 v x gives the velocity-increment. \ 249. Approximate Value of Acceleration at an Instant. The average acceleration for a short interval immediately following any instant is an approximate value of the true acceleration tf/that instant. Thus, if the velocity at the given instant is represented by O'A' * Compare Art. 208. t If, however, i\ and v 2 be regarded as vector symbols representing the initial and final values of the velocity in direction as well as in magnitude, the vector v 2 v x is the value of the velocity-increment (Art. 22) and [vector (z' 2 - v x )\l(t % - t x ) is the value of the average acceleration. 214 THEORETICAL MECHANICS. (Fig. 114), and the velocity after the short interval A/ by 0'B\ then (vector A'B') I At is an approximate value of the required acceleration. The approxi- mation is closer the shorter the interval At. 250. Exact Value of Acceleration at an Instant The true value of the acceleration at the instant t is the limit approached by the above approximate value as the interval At approaches zero. That is, acceleration = limit [(vector A'B') /At]. To determine this limiting value, the following reasoning may be employed : Let O'A' and O'B' (Fig. 1 1 5) represent the values of the velocity at the beginning and end of At respectively, so that vector A'B' de- notes the increment of velocity, the curve A'B' being the hodograph or curve of velocities (Art. 245). As At approaches o, the point B' approaches A' along the curve A'B'. The limit- ing direction of (vector A'B") I At is that of the tangent to the curve at Fig. 115. A'. Also, the limit of the magnitude of this vector quantity is the derivative of s' with respect to /, if s' is the length of the curve A'B' measured from some fixed point up to the point which is describing it ; that is limit [(length A'B')/At] = ds'/dt. Consider the curve A'B' (Fig. 115) to be described by a moving particle whose position at every instant corresponds to that of the given particle describing the path AB ; so that if P represents the position of the latter at any instant, and the vector O'P' its velocity, P' is the position of the point describing the hodograph or curve A'B'. Evidently, ds'/dt is the velocity of the point P'. Hence The acceleration of a point describing any patJi in any manner is at every instant equal {in magnitude and direction) to the velocity of the point describing the hodograph. The direction of the acceleration is always inclined toward the concave side of the path, but nothing further can in general be said MOTION IN A CURVED PATH. 215 as to its direction relative to the path. Neither can any simple expression be given for its magnitude in terms of the coordinates of the path.* 251. Sudden Change of Velocity. In what precedes it has been assumed that when the velocity changes from one value to another, it passes through a continuous series of intermediate values, and that for a finite change of velocity a finite interval of time is re- quired. It is conceivable that this condition might be violated, and that a particle might receive an instantaneous increment of velocity. In case of the motions actually occurring in nature there is no reason to believe that instantaneous changes of velocity occur. Sudden changes of velocity do, however, occur and will be treated in another place. (See Chapter XVI.) Examples. 1. A particle describes a circle of radius 2 ft. with a uniform speed such that the entire circumference is described in 1 sec. Required (a) the velocity-increment for an interval of o. 1 sec. ; (b) the average acceleration for the same interval (give its magnitude and the angle its direction makes with that of the initial velocity). Ans. {a) 7.766 ft. -per-sec. ; (&) 77.66 ft. -per-sec.-per-sec. ; angle = 108 . 2. A particle describes a circle of radius r with uniform speed v. Required {a) the magnitude and direction of the increment received by the velocity while the particle describes one-fourth of the circum- ference ; (Jb) the magnitude and direction of the average acceleration for the same interval. Ans. (b) (21/2/V) (z/ 2 /r), inclined 135 to initial velocity. 3. With data as in Ex. 2, determine the average acceleration in any finite interval At. (Give its magnitude and the direction it makes with the initial velocity.) Let AB (Fig. 116) represent the arc described in time At. The velocity at the beginning of the interval is represented by a vector O' A\ of length v, perpendicular to the radius OA ; the velocity at the end of the interval is represented by a vector 0'B\ also of length * If z' is regarded as a vector symbol representing the velocity in direction as well as in magnitude, and if A v is the increment of the vector v in the time At, the value of the acceleration may be written p = limit [(vector Av/At)~\ = vector (dv/dt). But the meaning of dv/dt is quite different from that of the same expression when v denotes the speed. 2l6 THEORETICAL MECHANICS. v, perpendicular to the radius OB. The velocity-increment is there- fore vector A 'B', and the average accel- eration is (vector A'B ')/At. The length A'B' is 2V sin (A'O'B'/i). But (expressing angles in radians), angle A'O'B' = angle A OB = (arc AB)/OA = (vAt)/r; hence length A'B' = 2V sin (v&tfzr), and the average acceleration has the magnitude FlG " II6 " ^ ''/Af = (2Z//A0 sin (vAt/zr). Its direction is perpendicular to the chord AB. 4. Taking the result reached in Ex. 3, let the interval At approach the limit zero ; determine the magnitude and direction of the limit- ing value of the average acceleration. Determine thus the exact value of the acceleration at any instant. As At approaches zero, the chord AB approaches the tangent to the circle at A, and the direction of the vector A ' B' , which is per- pendicular to the chord AB, approaches AO. Also, since the angle A OB ( A'O'B') approaches zero as At approaches zero, the value of the angle in radians may in the limit replace the sine ; that is, vAtfer may replace sin (y AtJ2r). Therefore limit [A'B' /At] = limit [(2v/At)(vAt/2r)] == v 2 /r. That is, the actual acceleration at the instant when the particle is at A is directed toward the center of the circle and has the magnitude v 2 /r. 5. What is the value of the acceleration of a particle moving as described in Ex. 1 ? Ans. 78.9 ft.-per-sec.-per-sec. toward the center. 6. Draw the path and the hodograph for the motion described in Ex. 2. Choosing any point P on the path, find the corresponding point P' on the hodograph. If P moves with the moving particle, what is the velocity of P' at any instant ? From the principle stated in Art. 250, that the velocity of P' is equal in magnitude and direc- tion to the acceleration of P, verify the result of Ex. 4. 7. A particle describes the perimeter of a regular hexagon with uniform speed v. It receives what sudden increment of velocity as it passes an angle of the polygon ? 8. What sudden increments of velocity are received by a particle describing the perimeter of a regular polygon of ;/ sides with uniform speed v ? MOTION IN A CURVED PATH. 217 3. Motion and Force. 252. Effect of a Constant Force. The effect of a constant force upon the motion of a body which initially is either at rest or moving along the line of action of the force has been considered in Arts. 214 and 215. The principle there stated is that the force produces a velocity-increment proportional directly to the force and to the time during which it acts, and inversely to the mass of the body. The case in which the body has initially a velocity not parallel to the force must now be considered. If a moving particle be acted upon by a force whose direction is inclined to that of the motion, there results a continuous change in the direction of the motion. Thus, suppose a particle to move along the straight line AB (Fig. 117) with uniform velocity, not being acted upon by any force until arriving at the point B. Let it then be acted upon by a force represented by the vector MN, which remains constant in magnitude and direction for a period A/. The particle is deflected from the original straight path AB, and describes some curve BC. The velocity thus changes in direction, and in general it changes also in magnitude. Representing the velocity at the beginning of the interval A/ by a vector O'B' (Fig. 117) parallel to AB, and the velocity at the end of the interval by a vector O'C parallel to the tangent to the path BC at C, the total change of velocity (or velocity-increment) is represented by the vector B'C. The magni- tude and direction of this velocity-increment depend upon (1) the magnitude and direction of the force, (2) the duration of the interval A/, and (3) the mass of the particle. (1) The direction of the velocity-increment agrees with that of the force, and its magnitude is proportional directly to the magnitude of the force. (2) Its magnitude is proportional directly to the duration of the interval. (3) Its magnitude is proportional inversely to the mass of the particle. 218 THEORETICAL MECHANICS. These principles may be stated in the following general proposi- tion : A constant force acting upon a particle gives it a veto city -incre- ment whose direction is that of the force, and whose magnitude is proportional directly to the force and to the time during which it acts and inversely to the mass of the particle. This proposition is identical in form with that given in Art. 215. But since the discussion was there restricted to the case of rectilinear motion, the velocity-increment was necessarily parallel to the line of motion. This restriction being removed, the proposition is still true, velocity and velocity-increment being regarded as vector quantities. The special case of rectilinear motion is included in the general state- ment. It is thus seen that the effect of a constant force upon the motion of a particle is estimated in the same way, whatever the initial ve- locity of the particle. The velocity-increment is in every case equal to the velocity which would be produced by the same force acting for the same time on the same particle initially at rest. The final velocity is the vector sum of this velocity-increment and the initial velocity. The path of the particle will depend both upon the initial velocity and upon the magnitude and direction of the force. To illustrate this, compare two cases in which the force has the same value while the initial velocity has different values. The two cases are shown in Fig. 118. In each case the initial velocity v x is represented by the vector 0'A' y the direction being the same in both cases but the MOTION IN A CURVED PATH. 2ig magnitude being much greater in the second case than in the first. The force (represented by the vector MN) has the same magnitude and direction in both cases. The mass of the particle is also the same, so that the values of the velocity-increment in a given time are equal in the two cases. The velocity-increment being represented by the vector A'B' , the final velocity v., is given by the vector O'B'. The values of v % in the two cases differ both in magnitude and in direction. The two paths are also quite different. If A and B are the initial and final positions of the particle, the tangent at A is par- allel to v 1 and the tangent at B is parallel to v v The curvature of the path is evidently greater in the first case than in the second ; while the length of the path is greater in the second case, because the speed is greater throughout the interval. Examples. i. A body is projected horizontally with a velocity of 50 ft. -per, sec. , after which it is acted upon by the constant force of gravity. Determine the magnitude and direction of its velocity at the end of o. 5 sec. , 1 sec. and 2 sec. [The velocity-increment due to gravity is to be computed as if the body fell vertically from rest. See Art. 227.] 2. A body is projected with a velocity of 50 ft.-per-sec. in a direc- tion inclined 40 upward from the horizontal. Determine the mag- nitude and direction of the velocity at the end of o. 5 sec. , 1 sec. and 2 sec. Ans. At the end of 2 sec. the velocity is 50 ft.-per-sec, directed 40 downward from the horizontal (very nearly). 3. A body falling vertically at the rate of 30 ft. -per-sec. receives a blow in a horizontal direction which, if the body had been at rest, would have given it a velocity of 100 ft.-per-sec. Determine the magnitude and direction of the velocity immediately after the blow. 4. With data of Ex. 3, determine the magnitude and direction of the velocity 1 sec. after the blow, assuming gravity to be the only force acting on the body. An s. 1 17.8 ft.-per-sec. inclined 31 56' downward from the hori- zontal. 253. Equation of Motion for Particle Acted Upon by a Con- stant Force. The general principle above stated (Art. 252) may be expressed by an equation similar to that given in Art. 216 for the case of rectilinear motion. Let the increment of velocity (vector A'B', Fig. 118) be represented by Av } and the force (vector MN, Fig. 118) by P. Then if m is the mass of the particle and A/ the 220 THEORETICAL MECHANICS. time in which the constant force P produces the velocity-increment Av, the general principle is equivalent to the equation Av = k(PAt/m). But this must be interpreted as a vector equation, expressing identity of direction as well as equality of magnitude. Taking units of force, mass, length and time as in Art. 217 or Art. 218, so that k = 1, the equation may be written P = m{AvjAt). But Av/Aty. the increment of velocity per unit time, is the accel- eration (Art. 247). Hence, if acceleration is designated by/, the equation may be written P = mp, or p = P/m. Unless otherwise stated, it will hereafter be assumed that units are so chosen that k = I. 254. Effect of a Variable Force. A particle may be acted upon by a force which varies in magnitude, in direction, or in both mag- nitude and direction. Consider the effect of such a variable force upon the motion of the particle. Let the vector O'A' (Fig. 114) represent the velocity of the par- ticle at an instant /, and the vector O' B r the velocity after an interval At. Then A ' B' is the velocity-increment received by the particle during the interval. The value of a constant force which would produce the same velocity-increment in the same time is (by Art. 253) m (vector A'B')/At. If At is very small, the actual force differs little from this value during the interval ; and the true value of the force at the beginning of the interval is the limit approached by this approximate value as At approaches zero. That is, if P is the value of the force and / that of the acceleration at the beginning of the interval, P -= limit [in (vector A 'B')[At] = mp. Here P and p must be understood as vector symbols, and the equa- tion expresses equality both in magnitude and in direction. Thus the equation of motion has the same form when the force is variable as when it is constant. MOTION IN A CURVED PATH. 221 255. Resultant of Two or More Forces Acting Simultaneously Upon the Same Particle. Any number of forces acting simultaneously upon a particle are equivalent, in their effect upon the motion, to a single force. For the acceleration has at every instant a definite value p, and this same acceleration would result from the action of a single force of magnitude mp, agreeing in direction with/. A single force which, acting alone, has the same effect as several forces acting simultaneously, is called their resultant (Art. 55). The resultant of any number of forces acting at the same time upon a particle is a force equal to their vector sum. This proposition, for the case of two forces, expresses the principle of the parallelogram of forces. Assuming its truth in case of two forces, its truth for any number of forces follows immediately. Its meaning may be explained as follows : hetp l be the acceleration due to a force P x acting alone upon a given particle, and p t the acceleration due to a force P 2 acting alone upon the same particle. Then if the particle be acted upon by P x and P., at the same time, its acceleration is the vector sum of p x and p t . If the particle be acted upon at the same time by a third force P 3 which, acting alone, would produce acceleration p^ , the actual accel- eration is the vector sum o{p l ,p 2 and/ 3 . And similarly for any number of forces. Thus, the acceleration due to the concurrent action of several forces may be computed in either of two ways, (1) by determining the acceleration due to each acting alone and taking the vector sum of these separate accelerations, or (2) by determining the acceleration due to a single force equal to the vector sum of the several forces. This principle must be regarded as a fundamental law of Dynam- ics, derived from experience and not deducible from any simpler law or laws. 256. Equation of Motion for Particle Acted Upon by Any Number of Forces. The equation of motion may be written in the same form when several forces act upon the particle as when only a single force acts. It is n & P = mp, if m is the mass of the particle, p its acceleration at any instant, and /'the vector sum of all forces acting upon the particle at that instant. 257. Remarks on General Equation of Motion. The equation P=mp . . . . . (1) 222 THEORETICAL MECHANICS. may thus be called the general equation of motion for a particle acted upon by any forces. In interpreting it, P may be taken to mean any single force and/ the acceleration due to that force ; or P may mean the resultant of several forces and p the acceleration due to their combined action. In either case it is to be understood that P and/ are vector quantities having the same direction. It must also be remembered that unless the units of force, mass, length and time are related in a certain manner (Art. 217), the equa- tionwillbe p = K p h)t or P==k 'mp. The constant k (or k') will always be unity provided the unit force is taken as that force which will give to the unit mass the unit accel- eration. Thus, equation (1) may be used if the units are the foot, pound- mass, second and poundal ; or the centimeter, gram, second and dyne. (Art 217.) Or, any three of the units involved may be chosen arbitrarily, and the other determined in such a way as to make the constant k unity. In particular, the "engineers' kinetic system" of units, de- scribed in Art. 218, may be employed. That is, in using the equa- tion P - mp, we may take the pound-force as the unit force, pro- vided the mass be expressed in units, one of which is equal to g pounds-mass. If the mass of the body is known in pounds, its value in these new units is found by dividing by g. Thus, if M denotes the mass in pounds, the equation of motion will be P = (M/g)p. 258. Absolute Unit of Force. The unit force which was usually employed in the earlier chapters of this book was the weight of a unit mass. The most precise method of determining the magnitude of a force practically consists in comparing it with this unit. Since, however, the weight of a given mass changes with its position on the earth's surface, forces determined experimentally in different localities in terms of this unit cannot be accurately compared unless the ratio between the weights of the same mass in the two localities is known. In Art. 217 another unit of force was defined, the force which, acting upon the unit mass, produces the unit acceleration. This has been called an absolute unit, since it is not dependent upon any par- ticular force such as the weight of a body, nor upon the place at which its value is determined. A force whose value is known in terms of the weight of a unit MOTION IN A CURVED PATH. 223 mass at a given place may be expressed in terms of the absolute unit if the value of g (the acceleration due to gravity) at that place is known. Thus, a force equal to the weight of m pounds-mass at a certain place is equal to mg poundals, if g is the acceleration due to gravity at that place, expressed in feet-per-second-per-second. (Art. 227.) The most accurate method of determining the value of^at any place is to determine the time of vibration of a pendulum of known length. (Arts. 309 and 425.) This definition of the absolute unit of force depends upon the ef- fect of force in accelerating mass. Any accurate definition of force as a measurable quantity must rest upon the same basis. For this reason the general proposition or ' ' law of motion ' ' stated in Art. 252 is often regarded as a definition of force rather than as a proposi- tion about forces. (See Art. 260.) 259. Newton's Laws of Motion. The fundamental laws of mo- tion, as formulated by Newton,* are as follows : Law I. Every body continues in its state of rest or of uniform motion in a straight line, except in so far as it may be compelled by force to change that state. Law II. Change of motion is proportional to impressed force, and takes place in the direction of the straight line in which the force acts. Law III. To every action there is always an equal and contrary reaction ; or, the mutual actions of any two bodies are always equal and oppositely directed. These laws include all the principles upon which the foregoing discussions have been based. Law I expresses the principle stated in Arts. 31 and 213. This is often called the law of inertia ; inertia being defined as that prop- erty of matter by virtue of which a body cannot of itself change its own state of motion. Law II includes in its meaning the principles stated in Arts. 252- 256. In order that this may be clearly understood, some explana- tion of the terms employed by Newton is needed. The word motion is to be understood as meaning what is defined as momentum (Art. 312); that is, a quantity proportional directly to the mass of a body and to its velocity at any instant. Change of * These laws are stated nearly in the language of Kelvin and Tait's "Natural Philosophy," g 244, 251, 261. 224 THEORETICAL MECHANICS. motion is therefore change of momentum, and is a vector quantity proportional to the mass and to the change of velocity or velocity- increment. Again, impressed force must be understood to mean, not simply force as the term is now understood and has been used in the foregoing discussions, but a quantity proportional to the force and to the time during which it acts. Thus, if a force of magnitude P acts for an interval of time At upon a body of mass m, and if Av repre- sents the resulting velocity-increment, the law asserts that m Av is proportional to P At. This is evidently the same result as that stated in Art. 252 for the case of a constant force, and extended to the case of a variable force in Art. 254. That Law II includes also the law of composition of forces acting simultaneously (Art. 255) is not immediately evident from the lan- guage in which it is expressed. By Newton, however, the law was doubtless intended to imply that every force acting upon a body produces its effect independently of the action of other forces upon the same body, and that these effects (accelerations) combine accord- ing to the law of vector addition. * There is some ground for the opinion that the law of composition of forces constitutes an independent principle among the fundamental laws of Dynamics. Whether this be the correct view or not, it is desirable that this law should receive explicit statement. Such a statement is given in Art. 255. Law III has been stated and explained in Art. 35. In spite of its importance and its simplicity, its true meaning has often been missed. In the explanation of this law given in Art. 35, the word action was interpreted as meaning force, thus limiting the law to a state- ment regarding the forces exerted by two bodies upon each other. By Newton the law was applied in a wider sense, as indicated by the scholium f which he added to it. This scholium is here omitted, since its full meaning cannot be explained without anticipating the results of a later discussion. The law will for the present be under- stood as referring only to forces. So long as the attention is confined to the motion of a single particle, no use is made of the third law. But it is of fundamental importance in the analysis of the motions of two or more particles which exert forces upon one another. The importance of the law in Statics has already been seen in Chapter VI. *See Kelvin and Tait's "Natural Philosophy," | 254, 257. t Thomson and Tait's ''Natural Philosophy," \ 263. MOTION IN A CURVED PATH. 225 Examples. 1. A constant force acting upon a body of mass 40 lbs. for 3 sec. changes its velocity from 20 ft.-per-sec. north to 80 ft.-per-sec. east. Required the magnitude of the force in poundals. Ans. 1099 poundals. 2. If the same force continues to act, what will be the velocity of the body after 3 more sec. ? Ans. 161. 3 ft.-per-sec. S. 82 52' E. 3. A particle of 5 lbs. mass is acted upon by two constant forces at right angles to each other; one equal to the weight of 1 lb., the other equal to 20 poundals. What single force would produce the same effect? Ans. Magnitude of force = 37.9 poundals. 4. In Ex. 3, if the particle has initially a velocity of 20 ft.-per- sec. in a direction opposite to that of the lesser force, determine its velocity after 2 sec. Ans. 17.6 ft.-per-sec. at angle of 47 2 with initial velocity. 5. A particle whose mass is 250 gr. is acted upon by two con- stant forces whose directions are inclined to each other at an angle of 6o. One force is equal to the weight of the particle, the other is equal to 50,000 dynes. If the velocity at a certain instant is 600 c.m.-per-sec. in a direction bisecting the angle between the forces, determine the velocity 3 sec. later. 6. In Ex. 5, what single force may replace the given forces? 7. If a body of 38 lbs. mass describes a circle of 40 ft. diameter at the uniform rate of 30 ft.-per-sec, what is the magnitude and direction of the resultant force acting upon it at any instant ? [Use result of Ex. 4, following Art. 251.] 8. A body of mass m describes a circle of radius r at the uniform rate of v ft. -per-sec. Required the magnitude and direction of the resultant force acting upon it at any instant. 260. Laws of Motion Regarded as Definitions of Force and Mass. In the foregoing exposition of the laws of motion, it has been tacitly assumed that force and mass are quantities each of which is measurable independently of the other ; so that either or both may be made fundamental in the system of units which we may choose to adopt. (Art. 216.) But when it is attempted to define these quantities with scientific accuracy, it appears to be impossible to avoid making use of the laws of motion themselves. It has already been shown that the second law furnishes a means of denning the unit force in an exact manner when the unit mass has been chosen (Art. 258). A closer analysis shows that exact definitions of both force and mass as measurable quantities are supplied by the laws of motion. 15 226 THEORETICAL MECHANICS. Notice first that if it is possible to subject two different bodies to the action of equal forces, the second law furnishes a definite com- parison of their masses. Let bodies whose masses are m x and m 2 be acted upon by equal forces of magnitude P, and let p x , p 2 be their accelerations. Then (Art. 253) m x p x = m 2 p 2 ; and if p x and p 2 can be measured, the ratio of the masses is at once known. If a third mass m^ be acted upon by a force equal to P, the ratio of its mass to that of m x or m t becomes known if its ac- celeration p 3 is measured. In fact, i = (A /A )i ; ^3 = (a /a)**i ; and if m x is chosen as the unit mass, the numerical values of m 2 and m 3 are known. Thus the mass of any body whatever can be expressed in terms of m x as a unit if it can be determined what ac- celeration it will have when acted upon by a force equal to P. Evidently, however, it is not necessary always to apply a force of the same magnitude P. The masses of two particles may be com- pared by comparing the accelerations produced in them by equal forces of any magnitude whatever ; and two masses may be compared indirectly, by comparing each independently with a third mass, forces of different magnitudes being used in the two comparisons. Now our conception of the laws of motion implies that all such comparisons of two masses m x} m 2 , direct and indirect, with any values whatever of the forces whose effects are measured in making the comparisons, lead to the same value of the ratio m x jm 2 . This must be true if the science based upon these laws is self- consistent. Assuming it to be true, we have a method of defining mass as a measurable magnitude in a consistent and exact manner. But is it possible to define the meaning of ' equal forces applied to different bodies ' ' without making use of the notion of mass itself? The answer to this question is that such a definition is supplied by the third law, the equality of action and reaction.* * Maxwell's answer is that the desired object may be accomplished by- conceiving the forces applied by means of an elastic string, one end of which is attached to each of two or more bodies in turn. The other end is pulled in such a way that the string elongates by a certain amount which is the same in every case and is kept constant during each experiment. Assuming the MOTION IN A CURVED PATH. 227 The acceleration of a particle at any instant is due to the influ- ence of other particles. The "action" of a particle A upon a par- ticle B which causes acceleration of B is called the ' ' force exerted by A upon B. ' ' The third law of motion asserts that A and B influ- ence each other mutually,- so that the force exerted by A upon B and the force exerted by B upon A are at every instant equal in magnitude and opposite in direction. Let the masses of the two particles be m x and m 2 . Let that part of the acceleration of A which is due to B be p x , and let that part of the acceleration of B which is due to A be p 2 . The "force" exerted by B upon A is then measured by m x p x , and the force ex- erted by A upon B by m t p 2 . By the third law, these are equal in magnitude ; that is, m x p x = m t p 2 , or mjm x = pjp 2 . This equation may be regarded as the definition of the ratio of tlie masses of the two bodies. Or, in words, The masses of any two bodies are in the inverse ratio of the ac- celerations they give each other at any instant. This conception of the meaning of mass and of force, and of the significance of the laws of motion, may be embodied in the following statements : (1) The acceleration of a particle is always due to the influence of other particles. (2) The actual acceleration of a particle is at every instant equal to the vector sum of accelerations due to the individual particles which are influencing it. (3) The accelerations which two particles experience by reason of their mutual influence are at every instant oppositely directed along the line joining them. (4) The masses of two particles are in the inverse ratio of the accelerations they give each other. This is a definition of mass. (5) The product of the mass of a particle A into that part of its acceleration which is due to another particle B is the measure of the force exerted by B upon A. This is a definition of force. elastic properties of the string to remain constant, it may be assumed that the magnitude of the force is the same in every case. ("Matter and Motion," Chapter III.) Even if no actual string can be found for which this will be true, this may perhaps still be regarded as a satisfactory method of explaining ideally what is meant by equal forces. 228 THEORETICAL MECHANICS. The first of these statements is virtually Newton's first law ; for it implies that if a particle were free from the influence of other par- ticles (that is, from the action of forces) it would have no acceleration. The second statement is the law of composition of forces, usually regarded as included in Newton's second law. The second law as applied to the effect of a single force is included in statements (4) and (5). Newton's third law is included in (3), (4) and (5).* 261. Particles and Bodies. The geometrical analysis of motion in the present Chapter and in Chapter XII referred to a particle. The laws of motion, although stated by Newton as applying to "bodies," have here been applied only to the ideal bodies called particles. The propositions at the end of Art. 260 were also stated as applying to particles. These particles have been regarded as pos- sessing finite mass while occupying no finite volume. That such a particle exists, either in isolation or associated with other particles so as to constitute one of the ' ' bodies ' ' of our experience, we are not justified in asserting. It is, then, pertinent to inquire what applica- bility this abstract theory has to the actual bodies of nature. To answer this question it is necessary to anticipate the results which are reached when the theory of the motion of a particle is extended to systems of connected particles. Assuming the accelera- tion of every particle to be in accordance with Newton's second law as above explained, and assuming the influence of the particles upon one another to be in accordance with Newton's third law, it is possible to deduce certain general laws governing the motion of con- nected systems of particles. These laws are found to describe with great accuracy the actual motions of natural bodies. Moreover, it is immaterial whether a body is regarded as made up of a finite number of particles, each of finite mass, or whether it is regarded as occupying space continuously (Art. 5). If by particle we understand a portion of matter whose volume and mass are both vanishingly small, the foregoing theory of the motion of a particle may be applied if it be assumed that the force acting upon any par- ticle becomes vanishingly small with the mass. To extend the theory * For a critical analysis of the laws of motion along the lines of the above discussion, the following works may be consulted: " Science of Mechanics," by Dr. Ernst Mach, English translation by Thomas J. McCormack. "Gram- mar of Science," by Karl Pearson. "Theoretical Mechanics," by A. E. H. Love. MOTION IN A CURVED PATH. 229 to the motion of a body of finite size involves, on this hypothesis, a process of integration, but the general laws deduced do not differ from those resulting from the other assumption. Newton's first and second laws are not intelligible as applied to a body of finite size, unless it be assumed that all parts of the body have at every instant equal and parallel velocities. This is, indeed, a possible state of motion, but its actual occurrence is very excep- tional. It is not possible to describe simply the way in which a body would move if not influenced by other bodies ; still less can a simple general statement be made describing the effect of a single external force in accelerating the various parts of a body. One of the results reached in the theory of the motion of con- nected systems of particles is that there is in every such system a certain point (the center of mass) whose motion is accurately described by Newton's first and second laws, if the whole mass be regarded as concentrated at that point and acted upon by forces equal in magnitude and direction to the external forces which are actually applied to the system. 4. Simultaneous Motions. 262. Meaning of Simultaneous Motions. The actual motion of a particle is sometimes regarded as the resultant of two or more component motions occurring simulta- neously. Thus, if the total displacement dur- ing a given interval is AB (Fig. 119), the particle may be regarded as receiv- ing simultaneously the displacements AC> CB, even though the actual dis- placement follows a path A MB not con- taining the point C. Similarly, if the actual velocity at a given instant is represented by the vec- tor A'B' (Fig. 120), the particle may be regarded as having simul- taneously two velocities represented by A'C, CB'. And if the actual acceleration at any instant is represented by the vector A" B" (Fig. 121), the particle may be regarded as having simultaneously two accelerations represented by A" C\ C" B" . 23O THEORETICAL MECHANICS. Any actual displacement, velocity or acceleration may thus be regarded as the resultant of any number of "simultaneous" com- ponents which satisfy the condition that their vector sum is equal to the given resultant. C" 263. Resolution Into Simultaneous Motions an Arbitrary Process. Obviously a particle cannot at the same time describe two different paths ; neither can it have, at the same instant, two different velocities or accelerations. The motion has at every instant a definite direction and a def- inite speed, and the actual velocity is represented by a definite vector. This may be expressed as the sum of several component vectors, but it is only by an arbitrary use of language that these com- ponent vectors can be said to represent velocities actually possessed by the particle at the same instant. 264. Effects of Simultaneous Forces May Be Estimated Sep- arately. One case in which it may be advantageous to regard the actual motion of a particle as the resultant of two or more simulta- neous motions is that in which the particle is acted upon by two or more forces such that the effect of each acting alone admits of simple determination. From the law of composition of forces (Art. 255) the actual acceleration is the vector sum of the accelerations which would result from the several forces acting singly. Hence these component accelerations may be computed separately and then com- bined. This method may be carried further, and applied to the computation of the velocity-increment and the displacement due to the simultaneous action of several forces, constant or variable, during any interval. 265. Velocity-Increment Due to Several Forces Acting Simul- taneously. Since acceleration is by definition (Art. 247) the incre- ment of velocity per unit time, and since, by the law of composition of forces, the acceleration at each instant is the vector sum of the MOTION IN A CURVED PATH. 23I accelerations which would result from the separate action of the forces, it follows immediately that the actual increment of velocity prodiiced in any interval is the vector sum of tJie velocity-increments which would result from the separate action of the forces during the same interval. If at the beginning of a certain interval the velocity is v , and if during the interval the particle is acted upon by forces P i% P %i . . . , constant or variable, which separately would produce ve- locity-increments v 1 , y t , . . . , the velocity at the end of the interval is the vector sum of v , v 1 , v 2 , 266. Displacement Due to Several Forces Acting Simultane- ously. Since the displacement of a particle during any interval depends upon the magnitude and direction of its velocity at every instant throughout that interval, the total displacement may be com- puted as the vector sum of the displacements due to any components into which the actual velocity may be resolved. Thus, suppose a particle, having at the beginning of a certain interval At some definite velocity v Q , to be acted upon during the interval by a force P. The position of the particle at the end of the interval may be determined as follows : Determine separately (a) the displacement in the time At due to the initial velocity if no force acted, and () the displacement in the time At due to the force P acting on the particle initially at rest. The vector sum of these displacements is the actual resultant displacement. This vector, drawn from the initial position of the particle, determines its position at the end of the interval. Thus, the displacement due to the Fig. 122. initial velocity v is a vector whose di- rection is that of v and amose length is v At. Represent this by AB (Fig. 122). If BC represents the displacement which the par- ticle would receive in the time At if initially at rest and acted upon by the force P, the vector A C represents the resultant displacement. The path followed by the particle from A to C is some curve AMC tangent to AB at A. If two forces P l and P 2 act upon the particle during the interval At, the total displacement may be computed as the vector sum of 232 THEORETICAL MECHANICS. three components : a component v At having the direction of the initial velocity v , a component equal to the displacement of the par- ticle if initially at rest and acted upon by the force P x alone, and a component equal to the displacement due to P 2 acting alone on the particle initially at rest. The same method may be applied, whatever the number of the forces. Examples. 1. A particle is projected horizontally with a velocity of 20 ft.- per-sec, after which it is acted upon by no force except gravity. Determine its velocity and its position after o. 1 sec, 0.2 sec, 0.5 sec, 2 sec. Ans. Let v = velocity, = angle between vsrnd the horizontal, r = distance from starting point, 6 = angle between r and the hori- zontal. At the end of 2 sec. v = 67.4 ft. -per-sec , (f> = 72 45', r= 75-8 ft., 0= 58 9'. 2. A particle is projected horizontally with a velocity v , after which it is acted upon by no force except gravity. Determine its velocity and its position after / seconds. 3. A particle is projected with a velocity of 100 ft. -per-sec. in a direction inclined 20 upward from the horizontal. Determine its velocity and its position after o. 1 sec , o. 5 sec , 2 sec. 267. Relativity of Motion. The motion of a body is always estimated with reference to some standard body which is assumed to be "fixed." Thus, in all ordinary practical problems, the earth is regarded as a fixed body, of unchanging dimensions ; the displace- ment and velocity of any terrestrial body are estimated as they appear to an observer on the earth. If a body moves from A to B f these being two points fixed upon the earth, the straight line AB is its total displacement when the earth is the standard body with refer- ence to which motions are specified. If, however, the rotation of the earth upon its axis is considered, the displacement has a very different value. Similarly, the velocity and acceleration of a par- ticle at any instant have different values if different reference bodies are selected. In the foregoing discussions, and in those that follow unless the contrary is explicitly stated, it is to be understood that only a single reference body is considered. When the motions of different bodies or particles are discussed, or when the motion of a particle is regarded as made up of components which follow certain rules of combination, MOTION IN A CURVED PATH. 233 all these motions are estimated with reference to the same standard body. 268. Motion of a Particle Referred to Two Different Bodies. In some cases the study of the motion of a particle is aided by regarding it from two different points of view, the motion being referred first to one standard body and then to another. This is illustrated by the following simple case : Let a particle P slide along a tube AB (Fig. 123), while the tube itself is in motion. Let the particle move from A to B while the tube moves from the position A'B' to the position A"B"; AB being in every position parallel to its original position A'B', and the ends of the tube 2?" describing the straight lines A'A",B'B". The total displacement of the particle is from A' to B ". The actual displace- ment may follow any curve or broken line lying between A ' B' B" and A 'A " B" . A ' **' . Fig. 123. If the motion of the particle along the tube from A to B, and the motion of the tube from A'B' to A"B", both take place uniformly, the actual displacement will be along the vector A'B"; but the same total displacement of the particle may occur in many other ways. It will be observed that in thus analyzing the motion of P two different reference bodies are employed in estimating the motion. Thus, AB (or A'B' or A"B") is the displacement of the particle with respect to the tube ; while A' A" (or B'B") is the displacement of the tube with respect to some "fixed" body not specified ; and A'B" is the displacement of the particle with respect to this same "fixed" body. Of these three displacements the third is seen to be equal to the vector sum of the first and second, and the actual displacement of the particle with respect to the " fixed " reference body is therefore regarded as the resultant of the two "simultaneous" displacements AB', B'B". Let it now be assumed that the particle P moves along the tube AB at a uniform rate, while the tube moves in the direction A' A" also at a uniform rate. If the particle is at the point A of the tube when the tube is at A'B', and if it moves from A to B while the tube moves from A'B' to A" B" , the actual path of the particle is the vector A'B". If this total motion occupies At seconds, 234 THEORETICAL MECHANICS. the actual velocity of the particle throughout the motion has the value (vector A'B")/At. This is obviously the vector sum of two components, (vector AB)/At and (vector B'B")/At, of which the former represents the velocity of the particle with re- spect to the tube, and the latter the velocity of the tube with respect to the "fixed" reference body. Hence the " actual" velocity of the particle (t. e., its velocity with respect to the assumed ' ' fixed ' ' reference body) may be regarded as the resultant of two simultaneous velocities, one of which is its velocity with respect to the tube and the other is the velocity of the tube. An analysis like the above is useful in many cases in which the displacement and velocity of a particle may easily be estimated with reference to a body which is itself in motion. Examples. i. A straight tube AB, i met. long, moves in such a way that every point has a velocity, relative to the earth, of 12 c. m. -per-sec. in a direction inclined 45 to AB. A particle slides uniformly from A to B in 3 sec. Required (a) the velocity of the particle relative to the tube ; (b) its velocity relative to the earth ; (c) its total dis- placement relative to the earth while it slides from A to B. [Give direction as well as magnitude of each of the required vector quan- tities. ] 2. A stream of water flows uniformly at the rate of 2 miles per hour. A boat is rowed in such a way that in still water its velocity would be 5 ft. -per-sec. in a straight line. If headed directly across the current, (a) what is its velocity referred to the earth? (b) If the stream is 3,000 ft. wide and the boat starts from one shore, where will it strike the opposite shore ? 3. If, in Ex. 2, the boat is headed up-stream at an angle of 30 with the shore, answer questions (a) and (b). Arts, (a) 2.86 ft. -per-sec. , 29 12' up-stream, (b) 0.317 miles up-stream. 4. In Ex. 2, how must the boat be headed in order to strike the opposite shore as near as possible to the starting point ? Ans. 35 55' up-stream. 5. Show that the current does not affect the time of crossing the stream. MOTION IN A CURVED PATH. 235 6. A straight tube AB, 4 ft. long, rotates about the end A at the uniform rate of 1 revolution per sec. A particle slides along the tube from B toward A at the rate of 10 ft.-per-sec. Determine the velocity of the particle relative to the earth when it is midway between A and B. Ans. If Cis the position of the particle, its velocity is 16.06 ft.- per-sec. in a direction making the angle 51 30' with CA. CHAPTER XV. PLANE MOTION OF A PARTICLE. i . Methods of Specifying Motion in a Plane. 269. Restriction to Plane Motion. The principles developed in Chapter XIV apply to any motion of a particle in space. In con- sidering the methods of applying these principles to the solution of particular problems, the discussion will be limited mainly to the case in which the path of the particle lies in a plane. It is needful now to consider how the motion can conveniently be specified in this restricted case. 270. Coordinates of Position. If the motion of a particle is restricted to a plane, its position at any instant may be specified by the values of two quantities. Any two quantities serving this puroose are coor- dinates of position. Of the various possible systems of coordinates, the most useful are the rect- angular system and the polar system. Rectangular coordinates. Let OX and OY (Fig. 124) be any two fixed lines at right angles to each other, called axes of coordinates, and let P be the position of the moving particle at any instant. Then the distances PN and PM, measured parallel to OX and O Y respectively, are the rectangu- lar coordinates of P, and will be denoted by x and y. As the particle moves, x and y vary. If x and y are known functions of the time, the position of P is known at every Q J[ instant. Fig. 125. Polar coordinates. Let the fixed point O and the fixed line OX (Fig. 125) be taken as pole and initial line respectively ; then the angle POX (or 0) and the length OP (or r) are the polar coordinates of the particle P. As P moves, y N X .. . TD 1 j- y M [ X Fig. 124. PLANE MOTION OF A PARTICLE. 237 its position is known at every instant if r and 6 are known functions of the time. 271. Velocity and Acceleration Each Given by Two Compo- nents. Any vector quantity lying in a given plane is known when its components in two given directions are specified. In analyzing the motion of a particle in a plane, the displacement, velocity and acceleration may each be regarded as the resultant (or vector sum) of two components, the directions of resolution being chosen at pleasure. The particular directions chosen will be determined by convenience, depending upon the nature of the problem under consideration. In nearly every case, however, the two directions of resolution will be taken at right angles to each other. The values of the two components of the velocity and of the accel- eration may be expressed in terms of the coordinates of position and their time-derivatives. 272. Resolution Parallel to Coordinate Axes. If the position of the particle is specified by its rectangular coordinates, it is usually convenient to resolve the displacement, velocity and acceleration parallel to the coordinate axes. The components in these directions may be called axial components. 273. Axial Components of Dis- placement. If the particle moves from A to B (Fig. 1 26) during any interval At, its total displacement is given by the vector AB. This total displacement may be resolved into components CD and EF, parallel to the coordinate axes. Let x, y be the coordinates of position at any instant ; let x x , y x be the coordinates of A y and x tJ y 2 those of B, Then the axial components of the dis- placement are evidently equal to x % x x and y 2 y v 274. Axial Components of Velocity. The average velocity- of the particle during the interval At is (vector AB)\At ; the exact value of the velocity at the beginning of the interval is the limit approached by this average value as At is made to approach zero (Art. 244). But since X t X\ and y., y x are always the values of the axial com- 2 3 8 THEORETICAL MECHANICS. ponents of vector AB } the axial components of limit [(vector AB)/At] are limit [(x 2 ^i)/A/] and limit [(j 2 y^/At]. The axial components of the velocity are therefore limit [{x t x^/At] = limit [A*/A/] = dx/dt = x = ;r- component ; limit [(y. z y x )/At] = limit [Ay I At] = dyjdt = y = j^-component. Let s denote the length of the path measured from some fixed point up to the position of the particle ; v the magnitude of the re- sultant velocity ; a, /3 the angles between v and the x- and j-axes respectively. Then V = ds/dt ; x = dxjdt = v cos a = (ds/'dt) cos a ; y = dy/dt = v cos ft = (ds/df) cos /3 ; tan a = cotan ft = j//ir. 275. Axial Components of Velocity-Increment. Let velocities be represented by vectors laid off from a point as in Art. 245. Let O' (Fig. 127) be this point, and from O' draw O'X* and O'V parallel re- spectively to the axes of x and y. Let vectors 0'A\ O'B' represent the values of the velocity at the begin- ning and end of an interval At. The velocity-increment for the interval is then represented by vector A'B'. This increment is completely known if we know its axial components CD' and E'F'. As found above, x and y are the values of the axial components of the velocity at any instant. Let x x , y x be the values of these compo- nents at the beginning of the interval, and x t% y., their values at the end of the interval. Then (Fig. 127) x x = 0'C\ y x = 0'E'\ x 2 = 0'D\ y 2 = Q'F\ PLANE MOTION OF A PARTICLE. 239 Hence CD' = x 2 x x ; E'F' = y 2 jt x . The axial components of the velocity-increment are therefore x 2 x x = Air = ^r-component ; j., ji = Ay = jy-component. 276. Axial Components of Acceleration. The average accel- eration for the interval A/ is equal to (vector A'B') /At. The exact value of the acceleration at the beginning of the interval is the limit approached by (vector A'B')/ At as At is made to approach zero (Art. 250). Since the axial components of vector A'B' are equal to x, x x and y 2 y v or Ax and Aj>, whatever the length of the in- terval, the axial components of limit [(vector A'B')/ At] are limit [Air/A/] and limit [A>/A/]. That is, the axial components of the acceleration are limit [Ax I At] = dx/dt = d 2 x/dt 2 = x = ^-component ; limit [Ay /At] = dyjdt = d 2 y/dt 2 = y = j/-component. Let/ denote the actual or resultant acceleration at any instant, and a , /3' the angles its direction makes with the x- and /-axes re- spectively. Then x = d' l x\dt 2 = p cos a'; y =d 2 y/dt 2 ==/cos/3'; p 2 = x 2 +y 2 ', tan a = cotan /3' z=y/x. 277. Hodograph. If the vector representing the velocity at any instant is O'P' (Fig. 128), O' being a fixed point while P' moves as the Y r value of the vector varies, the curve F' traced by P' is the hodograph (Art. 245)- The rectangular coordinates of the point P', referred to axes O'X', O'Y', parallel to the axes of x and JE' y, are x and y. If the values of x and y are known at every instant, the form of the curve can be deter- O' C M' D' X 1 mined. Fig. 128. 24O THEORETICAL MECHANICS. 278. Cases Adapted to Polar Coordinates. There are certain problems to whose treatment polar coordinates are well adapted. This is often the case when the acceleration of the particle is directed toward a fixed point. In this case, and in some others, it is conven- ient to replace the acceleration and the velocity by their resolved parts in directions parallel and perpendicular to the radius vector. The values of these components in terms of r and 6 and their deriv- atives will now be deduced. 279. Components of Velocity Parallel and Perpendicular to Radius Vector. Let a particle move from A to B along the curve AB (Fig. 129) during the interval At. Let r, 6 be its polar coor- dinates at any time /, and let the initial values of these coordinates be 1\ , l , and the final values r % , 6. 2 . Then r x = OA t r t = OB, X = A OX y 6 2 = BOX. The average velocity during the interval is (vector AB)/At, and the true velocity at the beginning of the interval is the limit of this value as At is made to approach zero. Take Fig. 129. OA' = OA = r lt and let (vector AB) be resolved into components (vector A A') and (vector A' B). Then the true velocity at the beginning of the interval is equivalent to two components limit [(vector A A') /At], limit [(vector A'B)/At]. The latter component has the direction OA and the former is per- pendicular to OA. Hence the component of velocity parallel to the radius vector is limit [A'B/At] s= limit [(r 2 r x )/At] = limit [Ar/At] dr/dt, and the component of velocity perpendicular to the radius vector is limit [AA'/At] = limit \r x (0 % Oj/At] = limit \r x Ad I At] = r(d6/dt). These results may also be deduced from the values of the axial components of velocity. Thus, let x, y be the coordinates of posi- tion referred to a pair of rectangular axes, the origin being at the pole and the ;r-axis coinciding with the initial line of the polar coordinates r, 0. (Fig. 130.) Then x = r cos 6 ; y = r sin 6. . , . (1) TH PLANE MOTION OF A PARTICLE. 2 4 I The velocity is equivalent to a component x in direction OX and a component y in direction OY. The resolved part of the velocity in any direction is equal to the sum of the resolved parts of x and y in that direction. Hence the resolved parts of the velocity along and perpendic- ular to OP are as follows : Y \ Ay , P &* y X y sin 6 -f- x cos 6 == component along OP; y cos 6 x sin 6 = component perpendicular to OP. The values of x and y may be ex- pressed in terms of r, 6 and their derivatives by differentiating equations (1) with respect to /. Thus, Fig. 130. dr a x cos v dt dr r sin 6 d6 dt d6 (2) y = sin 4- r cos 6 dt dt Substituting these values, the components of velocity parallel and perpendicular to the radius vector become dr ~dt dd dt = component in direction OP) = component perpendicular to OP. 280. Components of Acceleration Parallel and Perpendicular to Radius Vector. Since the acceleration is equivalent to two com- ponents, x in direction OX (Fig. 131) and y in direction OY, its re- solved parts parallel and perpendicu- lar to OP are as follows : r y / P ? x y X y sin 6 y cos 6 x cos 6 in direction OP ; x sin 6 perpend, to OP. Fig. 131, The values of x and y in terms of r, 6 and their derivatives may be found 16 242 THEORETICAL MECHANICS. by differentiating equations (2), Art. 279, with respect to t. The resulting values are .. ePr a drdd . a (d0\ 2 a d 2 6 . a x = cos u 2 sine/ r cos a r sin u ; dt' 2 dt dt \dt) dt 2 .. d*r . a . drdd a ldd\ 2 . d 2 6 a y = sin v -4- 2 cos*, u r | sin u 4- r cos v. f\ dt 2 dtdt \dt) dt 2 Hence ....... Q d 2 r ld0\ 2 y sin v 4- x cos a = r\ dt 2 \dt) = component of acceleration parallel to radius vector ; a .... drdO . d 2 6 y cos C7 .rsin u = 2 \- r dt dt dt 2 = component of acceleration perpendicular to radius vector. The value of the latter component may be written in another form, as follows : drdd . d 2 1 dt t d6\ 2 [- r = \r l . dtdt dt 2 rdt\ dt) This may be verified by differentiation. 281. Angular Motion of a Particle. The line joining a mov- ing particle A to a. fixed point O in general turns about as the particle moves. Let 6 denote the angle between OA and a fixed line in the plane of the motion ; then the rate of turning of the line is measured by the rate of change of 0. By the ' ' angular motion of A about (9" is meant the angular motion of the line OA. Angular velocity may be defined as the rate of angular motion. It is measured by the angle described per unit time. If the line OA turns at a uniform rate, let Ad be the increment of 6 in a time At, and let co denote the angular velocity ; then co = AOjAt. If 6 increases at a variable rate, AO/At is the average angular veloc- ity for the time At, and the instantaneous value of the angular veloc- ity is the limit of A0/At as A^ approaches o ; that is, co = dOldt. A ngular acceleration is the rate of change of angular velocity. If the angular velocity co varies at a uniform rate, let Aco be its incre- PLANE MOTION OF A PARTICLE. 243 ment during a time At, and let (f) denote the angular acceleration. Then = Ao>/A/. If co changes at a variable rate, Aco/ A/ is the average angular accel- eration for the time A/, and the instantaneous value of the angular acceleration is the limit of Aco/ At as At approaches o ; that is (j> = dco/dt. If the point A is describing a circle of radius r with center at O, and if As is the length of arc subtended by the angle A#, As = rAO; hence ds/dt == ridOuit). That is, the linear velocity is at every instant equal to the product of the radius by the angular velocity. The angle 6 must be expressed in radians in order that this relation may hold. Examples. i. A particle describes a circle in any manner. Show that its angular velocity about any point of the circumference is at every in- stant equal to half its angular velocity about the center. The angular velocities about all points in the circumference are equal. 2. A particle P describes a straight line at the uniform rate of 24 ft. -per-sec. Determine its angular velocity and angular acceleration about a point A, 6 ft. from the path, in two positions : (a) when AP is at right angles to the path, (b) when AP is inclined 45 to the path. Ans. (a) co = 4 rad. -per-sec, (f> = o. (b) co = 2 rad. -per-sec, (f> 8 rad. -per-sec. -per-sec. 3. A particle P describes a straight line in such a way that its angular velocity about a point A distant h from the path is constant and equal to 00. Determine the linear velocity and linear accelera- tion in any position. Ans. Let 6 denote the angle between AP and a perpendicular to the path, v the linear velocity and p the linear acceleration ; then v = hco sec 2 6, p = iJico 1 tan 6 sec 2 6. 282. Relation of Velocity and Acceleration to Path. The methods of resolution above considered, though useful in the analy- sis of particular cases of motion, do not show clearly the relation of velocity and acceleration to the path of the particle. This relation is best shown by resolving along the tangent and normal to the path. 244 THEORETICAL MECHANICS. The values of the components of velocity and acceleration thus de- termined are independent of any particular system of coordinates of position. 283. Tangential and Normal Components of Velocity. Since the resultant velocity, for any position of the particle, has the direc- tion of the tangent to the path, its resolved part in the direction of the tangent is the velocity itself, its value being ds/dt (Art. 244) ; while the resolved part along the normal is zero. 284. Tangential and Normal Components of Acceleration. Let A (Fig. 132) be the position of the particle at an instant t, and AB the path described during an in- terval At. The speed in the position A will be denoted by v. From any point O' lay off the vector O'A' to represent the initial velocity and O'B' to represent the final velocity ; these vectors are then parallel to the tangents to the path at A and at B respectively. Fig. 132. The velocity-increment during the interval is equal to (vector A'B') ; the average acceleration is (vector A' B' )/At; and the true accelera- tion at the beginning of the interval is the limit of this average value as At is made to approach zero. The vector A ' B ' will now be replaced by two components whose limiting directions are those of the tangent and normal to the path at A. These components are A'C, C'B', if the point C is so taken that O'C = 0'A' = v. It is evident that as At approaches zero, the limiting direction of A'C is perpendicular to O'A' and that of C'B' is parallel to O'A'; but O'A ' is parallel to the tangent to AB at A. The tangential and normal components of the acceleration at the instant t are therefore limit [(vector C'B') I At] and limit [(vector A'C) I At]. It remains to determine the magnitudes of these limiting values. Evidently C'B' is equal in magnitude to the increment received by the speed during the interval At. Hence limit [C'B' I At] = limit [Av/At] = dv/dt, PLANE MOTION OF A PARTICLE. 245 which is the value of the tangential component of the acceleration. Next, to determine the normal component. Draw normals to the path at A and B, and let O be their point of intersection. Lay off OC = OA. The triangles AOC, A'O'C are similar ; hence A'C'/A'O' = AC/AO; AC = AC(AO'/AO) = AC(v/AO); limit [AC /At] = limit [(v/AO)(AC/At)]. As At approaches zero, the point approaches as a limiting position the center of curvature of the curve AB at A, and the limiting value of AO is the radius of curvature at A. Call this R. Again, as At approaches zero, the points C and B approach coincidence ; A C and arc AB approach a ratio of equality. Hence limit [A CI At] = limit [(arc AB)/At] = limit [As /At] = ds/dt = v. Substituting these values, limit [A' CI At] = (v/R)v v 2 /R. The required components of the resultant acceleration^ are there- fore dvjdt = d' 2 s/dt 2 = tangential component ; v 2 /R = (ds/dt) 2 /R = normal component. These values are independent of any particular system of coordi- nates of position, but they may, if desired, be expressed in terms of the coordinates. Examples. 1. Determine the magnitude and direction of the resultant accel- eration of a particle describing a circle of radius r with uniform speed v. [Express the values of the tangential and normal components and determine their resultant. Compare with result of Ex. 4, follow- ing Art. 251.] 2. The path of a particle is a circle of radius r lying in a vertical plane. The speed is given by the equation v = k\/z, where z is the vertical distance below a horizontal plane through the center of the circle. Compute the tangential and normal components of the acceleration for any value of z. Ans. Tangential component = (k 2 1 2r)\/ (r 2 z 2 ) ; normal component k 2 z/r. 246 THEORETICAL MECHANICS. 3. In Ex. 2, let v = 8 ft.-per-sec. when s = 1 ft., and let the radius of the circle be 2 ft. Compute the tangential and normal com- ponents of the acceleration when the particle is in its lowest position. Ans. Tangential component o; normal comoonent = 64 ft.- per-sec-per-sec. 4. A particle describes a circle 01 50 ft. radius in such a way that s = 16/ 2 , in which s is in feet and / in seconds. Compare the true acceleration when t = 2 with the average acceleration for the ensu- ing o. 1 second. 2. Motion in a Plane Under Any Forces. 285. Conditions Under Which the Motion Will Be Confined to a Plane. If the direction of the resultant force acting upon a particle is always parallel to a fixed plane, and if the motion at any instant is parallel to that plane, the resolved part of the velocity per- pendicular to the plane will remain zero, and the path of the particle will therefore lie in the plane. 286. Two Independent Equations of Motion. The general equation of motion, P = mp, is a vector equation, expressing equality of direction as well as of magnitude between the two members. If both Pandfl be resolved in any direction, we have (resolved part of P) = m X (resolved part of f). Hence by resolving in different directions, any number of true equa- tions may be written. In the case of plane motion, however, only two of these equations can be independent. For if the resolved parts of a vector in two directions are given, the vector is completely de- termined ; hence the two equations obtained by resolving in two dif- ferent directions include all that is expressed by any equation ob- tained by resolving in a third direction. The form of the two independent equations depends upon the system of coordinates employed in specifying the position of the par- ticle, and also upon the directions of resolution. 287. Equations of Motion in Terms of Rectangular Coordi- nates. Let the position of a particle be specified by its coordinates x and j/, referred to any pair of fixed rectangular axes in the plane of the motion, and let the two equations of motion be obtained by re- solving parallel to the axes. Let the axial components of the result- ant force P be X and Y. If several forces act upon the particle, X PLANE MOTION OF A PARTICLE. 247 is the algebraic sum of their resolved parts in the ^-direction, and Y the algebraic sum of their resolved parts in the ^-direction. The axial components of the acceleration are x and y (Art. 276). The equations of motion are therefore X = m'x = m(d % xldt*) ; . . (1) Y = my == m(d 2 yldt 2 ). . . . (2) 288. Equations of Motion in Terms of Polar Coordinates. If polar coordinates are employed, the equations are obtained in the most convenient form by resolving parallel and perpendicular to the radius vector. Let P r and Pg be the resolved parts of the resultant force P in these directions respectively. The values of the resolved parts of the acceleration are given in Art. 280. Using these values, the equations of motion become P r ---= m \dti m (r^Uo (2) d 2 r idO\ dt 2 ld_l t de\ rdr Here P is the magnitude of the force. To determine the motion these equations must be integrated. For the complete integration, P must be a known function of one or more of the variables r, 6, t. In the most important case P is a function of r only. One important result may, however, be obtained for any case of motion under a central force. From equation (2), multiplying by r and integrating, we have J 6 7 r = h = constant. . . (i) dt To interpret this result, let A denote the area swept over by the radius vector, in moving from some given position. Thus, in Fig. 1 34, let N be the position of the particle at time t, ON = r, NOX = 6 ; and let MN be a portion of the path, described in the direction MN. Then A = area MON In time At let the particle move to TV'; the increments of 6 and A are Ad = angle NON\ AA = area NON'. Approximately, AA = ir 2 A0, or AA/At = lr\A6/At). In the limit, as At is made to approach zero, the equation is exact ; that is, r\d6/dt) = 2 (dA/dt). Equation (3) may therefore be written dA/dt = kJ2 = constant. . . . (4) The quantity dA/dt may be called the areal velocity of the par- ticle, being the rate at which the radius vector is describing area at any instant. The last equation therefore shows that the areal veloc- ity is constant ; and this conclusion evidently holds for all cases of motion of a particle in a plane, so long as the acceleration is directed toward a fixed point, PLANE MOTION OF A PARTICLE. 257 Integrating equation (4) between limits corresponding to instants t x and t 2 , A 2 -A, = \h(t. t - O- That is, the area described by the radius vector during any interval is proportional directly to the interval. The constant h is the double areal velocity. Its value is known in any case if the velocity in some one position is known in magni- tude and direction. Thus, at a certain instant let the distance from the center of force be a, and let the resolved part of the velocity at right angles to the radius vector be V. The areal velocity at that instant is a V/2, and therefore h = a V. As applied to the motion of a planet under the attraction of the sun, equation (4) expresses Kepler's law that the radius vector de- scribes equal areas in equal times. The law is seen to be true, not only when the force follows the law of gravitation, but in every case of central force. For use in the treatment of certain problems, equation (i) takes a more convenient form if a new variable, u = i/r, be introduced instead of r. This is accomplished as follows : From equation (3) 5 = ^ ! (5) at Also, differentiating the equation r = i/, dr _ 1 du _ 1 dudB _ ,du dt~ u 2 dt~ u 2 d6~dt~ dO (putting for dd/dt its value hu 2 ). Again, d 2 r _ _,dldu\_ d 2 udd _ 2 d 2 u dt 2 ~ dt\d6!~ ~ a 7 e 2 a 7 t~ ~ * dO 2 ' Substituting the values of dO/dt and d 2 r/dt 2 in (1), we have finally P d 2 u - = h 2 u 2 + h 2 u\ . . . (6) m dd 2 Equations (5) and (6) may be used instead of (1) and (2). To determine the law of variation of the force zvhen the path is given. If the polar equation of the path is given, d 2 ujd6 2 may be found by differentiation. By substituting its value in equation (6), the force may be determined as a function of 0> u or r. 17 258 theoretical mechanics. Examples. 1 . Show that a particle cannot describe a straight line under a force directed toward or from a fixed point not in the path. [Applying the above method of determining the law of force, it will appear that P = o. ] 2. If the orbit is a circle passing through the center of force, show that the force varies inversely as the fifth power of the distance. 3. A particle describes a conic section under the action of a force which is always directed toward a focus. Show that the force varies inversely as the square of the distance. 302. Force Varying Inversely as the Square of the Distance. The most important case of motion under a central force is that in which the force varies inversely as the square of the distance from the fixed point. In this case it is most convenient to employ equa- tions (5) and (6) of Art. 301. Let the attraction per unit mass at unit distance from the center be k ; then P/m = k/r' 1 = ku 1 , and equation (6) reduces to the form k d l u . + * (7) This, and the equation h 1 dO' 1 $ = **, .... (8) at are equivalent, in the present case, to the general differential equa- tions of motion. The path of the particle may be found by the integration of equation (7). To accomplish this, put u = z -f- k//i 2 , and there results (Pz____ d6* ~ Proceeding as in Art. 229, the first integration gives S)=-* !+c ' (9) and the second, z = C sin (Q 4 ) Here C and a are constants whose values must be found from the initial conditions. Restoring the value of g t k u== 1- Csin (0 + a). . . (10) PLANE MOTION OF A PARTICLE. 259 The form of this equation shows that the curve is a conic section, the focus being at the pole or center of force. To simplify its form, let C and a be determined from the condition that when 6 = 0, r is equal to a, and the velocity is perpendicular to the radius vector. The last condition makes drjdt, dujdt and dnjdQ all zero when = o. From (10), Hence the condition dujdO = o when 6 = o gives cos a = o ; sin a = 1 . Also from (10), applying the condition that u = i/a when 6 = o, ^. " I _ k The equation of the path is therefore h 1 \a /W li l \_ \ak I J This represents a conic section of eccentricity k 2 /a& 1 and semi- parameter k 2 /k, the focus being at the pole of the system of coor- dinates and the principal diameter lying in the initial line. Since a conic section is an ellipse, a parabola, or an hyperbola, according as the eccentricity is less than, equal to, or greater than unity, it is seen that the orbit is an ellipse if Jffak < 2 ; a parabola if H l \ak = 2 ; an hyperbola if h 2 /ak > 2. Since h depends upon the velocity in the initial position, it is seen that if the initial velocity has a certain value the orbit is a parabola ; if it is less than this value the orbit is an ellipse ; and if greater, the orbit is an hyperbola. To find this critical velocity, let V denote the velocity when r = a and 6 = o, its direction being perpendicular to the radius vector. The double areal velocity k has the value a V ; hence h l \ak = a V' z lk y and the above relations become : V' 2 < 2k/a for ellipse ; V 2 = ik\a for parabola ; V 1 > ik\a for hyperbola. 260 theoretical mechanics. Examples. i . A body moves under the action of a central force which varies inversely as the square of the distance. At a certain instant it is 10 ft. from the center of attraction and moving at right angles to the radius vector at the rate of 5 ft.-per-sec. In this position the force is 10 poundals per pound of mass. Write the equation of the path, and determine whether it is an ellipse, a parabola or an hyperbola. 2. If v is the velocity of the particle when at a distance r from the center of attraction, show that the path is an ellipse, a parabola or an hyperbola, according as v 1 is less than, equal to, or greater than 2k\r ; this result being independent of the direction of v. 3. Prove that v 2 2k/r is constant when the force varies in- versely as the square of the distance. 4. Integrate equation (6), Art. 301, for the case in which the force varies inversely as the cube of the distance. 5. Constrained Motion. 303. Meaning of Constrained Motion. The motion of a body is said to be constrained if certain conditions are imposed upon it, while the forces which must act in order that the motion may con- form to those conditions are not specified. A bead sliding on a wire of any form which is either at rest or moves in a specified manner furnishes an example of constrained motion. Another case is that of a body sliding on an inclined plane which is either at rest or is moving in a given manner. The laws of motion and the general equations deduced from them, in any of the forms given in the foregoing discussions, may be applied in the solution of problems in constrained motion. The con- straint is always produced by the action of forces. The feature which distinguishes constrained motion is that certain of the forces are not given directly, but are given indirectly by specifying their effect upon the motion. Such forces are called constraining forces. 304. General Method. In dealing with a case of constrained motion of a particle, we are to apply the general differential equa- tions of motion, the constraining forces being introduced as unknown quantities. In addition there will be one or more equations express- ing the conditions which are to be imposed upon the motion. If rectangular coordinates are used, we have (a) PLANE MOTION OF A PARTICLE. 26 1 m'x = X, . . . . . (i) my = Y, . . . . . (2) in which X and Y involve the unknown constraining forces as well as the known forces ; and (<) certain equations of condition which may involve any or all of the quantities x, y, t, x, y, x, y. The only case which will here be considered is that in which the path is assigned. The equation of condition is then of the form A*> y) = (3) Before discussing this problem in its general form, the simple case of motion in a straight line will be considered. 305. Motion on a Smooth Inclined Plane Under Gravity. Let a body of mass m pounds be placed upon a smooth plane sur- face, inclined to the horizontal at an angle a. The forces acting upon it are its weight and the pressure ex- erted by the body which it touches. The \^ former is mg poundals vertically down- \ ward, and the latter is perpendicular to \ ^ the plane but of unknown magnitude $ -J/^ (say N poundals). The force N is called <^^^ into action only to resist a tendency of ^<\ N the body to pass through the surface ; JT^<^\ , \ it cannot cause the body to leave the x ma surface, for it ceases to act as soon as Fig. 135. the bodies separate. The motion is thus " constrained" to follow the plane. It will be assumed also that the motion is confined to a vertical plane, which is taken as the plane of the figure. Referring to rectangular coordinates, let the axis of x(OX, Fig. 135) lie in the plane, the positive direction being downward along the plane. The equations of motion are mx = mg sin a; ..... (1) my = N mg cos a. . . . . (2) But since the path is the straight line y = o, so that y = o, equa- tion (2) becomes N mg cos a = o. . . . (3) Equation (3) serves to determine N, while (1) determines the motion. 262 THEORETICAL MECHANICS. Comparing equation (i) with the equation of motion for a body falling vertically, it is seen that the results deduced in Art. 227 may be applied to the present case by substituting g sin a for g. The three cases in which the initial velocity is zero, down the plane, and up the plane, may be discussed as were the corresponding cases in Art. 227. 306. Motion on a Rough Plane Under Gravity. If the plane is rough, friction must be included among the forces entering the equa- tions of motion. While the body is sliding, the force of friction is equal to jjlN, fi being the coefficient of friction (Art. 129) ; the direc- tion of this force is opposite to the sliding. Taking axes as in Fig. 136, the equations of motion are mx = mg sin a /jlN ; . . (1) my = N mg cos a. ... (2) The latter equation becomes N mg cos a = o, . . . . (3) because of the condition y == o. Equation (3) gives the value of N \ substituting this in (1), x ~ g{su\ a fi cos a). . . (4) The integration of equation (4) gives results agreeing with those of Art. 227, with the substitution o{g(s'm a /jl cos a) for g. This equation applies only to the case in which the motion is down the plane. If it is up the plane the direction of the force /jlN must be reversed. If x is still taken as positive downward along the plane, the case of motion up the plane gives the equation x = ^(sin a -j- jjl cos a). . (5) It should be noticed that, in the case of motion down the plane, the velocity will increase or decrease, according as the second member of (4) is positive or negative. Between these cases is that in which sin a jjl cos a = o, or /jl ~ tan a, which reduces equation (4) to the form x = o. PLANE MOTION OF A PARTICLE. 263 This is the case of equilibrium ; the particle will remain at rest or will move uniformly down the plane. Uniform motion up the plane is impossible. Examples. 1. A body slides down a smooth plane whose inclination to the horizon is 20 . How far does it move during the first 2 sec. after starting from rest? Ans. 22 ft. 2. A body slides down a smooth plane whose inclination to the horizon is 25 . What is the velocity when it reaches a position 12 ft. lower than its position of rest? Ans. 27.8 ft.-per-sec. 3. Solve Ex. 2 if the inclination of the plane is 40 , the data being otherwise unchanged. 4. A body starting from rest slides down a smooth plane whose inclination to the horizon is a. What is its velocity after moving a distance whose vertical projection is k? Show that the velocity ac- quired in falling to a given level is independent of the inclination of the plane. 5. A body is projected up a smooth plane inclined 30 to the horizon with a velocity of 200 c.m.-per-sec. What is its velocity after moving 40 c. m. ? When will it come to rest, and when will it return to the point from which it is projected? Ans. 28 c.m.-per-sec; after 0.41 sec. ; after 0.82 sec. 6. A body is projected with a velocity V up a smooth plane whose inclination is a. When will it come to rest, when will it return to the starting point, and how high will it rise ? Prove that it will rise to the same height as if projected vertically upward with the same speed. 7. Solve Ex. 1 assuming the plane rough, the angle of friction being 1 o. Ans. 11.4 ft. 8. A body is projected with a velocity of 20 ft. -per-sec. down a plane whose inclination is 25 , the coefficient of friction being 0.4. Determine the position and velocity after 2 sec. Ans. 43.9 ft. from starting point ; 23.9 ft. -per-sec. 9. In Ex. 8, if fi = o. 5, when will the body come to rest ? Ans. After 20. 4 sec. 10. In Ex. 8, if the angle of friction is 25 , determine the motion. 11. In Ex. 8, let the direction of the initial velocity be reversed, and let the angle of friction be 25 . Determine the motion. Ans. The body will come to rest after 0.73 sec. and will remain at rest. 1 2. A body is placed at rest on a rough plane, and the inclina- tion of the plane is increased until sliding begins. If the coefficient of friction is 0.2, what is the angle of incipient sliding? 264 THEORETICAL MECHANICS. 13. A body is projected up an inclined plane. What condition must be satisfied in order that it may slide down after coming to rest? 14. A body is projected up a plane whose inclination to the hori- zon is 30. The angle of friction is 20 . If the initial velocity is 20 ft. -per -sec. , (a) when will the body come to rest, (p) when will it return to the initial position, and (r) with what velocity will it pass through the initial position ? Ans. {a) At end of 0.76 sec. (b) 1.60 sec. after coming to rest. (<:) 9.52 ft.-per-sec. 1 5. A body is projected on a horizontal plane with a velocity of 50 ft.-per-sec. It comes to rest after 6 sec. What is the degree of roughness of the plane? 16. A stone thrown horizontally on a sheet of ice with a velocity of 1 20 ft. -per-sec. comes to rest after sliding 1,400 ft. Assuming the coefficient of friction to be independent of the velocity, determine its value. Ans. 0.16. 17. Write the equations of motion of a body sliding on a smooth inclined plane, under the action of no force except gravity and the pressure of the plane, assuming that the motion is not confined to a vertical plane. Show that the path is a parabola. 18. In the case described in Ex. 17, show that the motion is given by equations like those deduced in Art. 295 for a projectile, with the substitution of g sin a for g, a being the inclination of the plane to the horizon. 307. Motion of a Bead on a Smooth Wire The general case in which a particle is constrained to move in a given plane curve will now be considered. To fix the ideas, let the particle be a bead sliding on a smooth wire bent into any 1 I plane curve, and suppose all forces act- / ing on the bead to be known, except qq / X' the pressure exerted by the wire. This /PST _ pressure is a "passive resistance" yy .' ^\ (Art. 41); it comes into action to re- /a.. sist any tendency of the bead to leave L LJ the wire. The direction of the pres- p IG I37 sure exerted by the wire upon the bead is normal to the path of the particle. Let N represent the magnitude of this unknown normal pressure, regarded as positive if directed toward the concave side of the path. (See Fig. 137.) Let Q denote the resultant of all forces applied to the bead except the normal pressure N. Two independent equations of motion are now to be written, ob- PLANE MOTION OF A PARTICLE. 265 tained by resolving resultant force and resultant acceleration in any two directions. Two sets of equations will be given. (1) Resolution along fixed rectangular axes. Choosing any pair of rectangular axes, let 6 denote the angle between the tangent to the path and the axis of x. Then the normal pressure N makes with the ^r-axis the angle 90 + 6, and with the jj/-axis the angle (Fig. 137). The axial components of A^ are therefore N sin 6 and N cos 6. But if s means the length of the path, measured from some fixed point, sin 6 = dy/ds, cos 6 = dx/ds ; therefore N(dy/ds) = .^-component of N; N{dxjds) =jj/-component of N. Let X\ Y' denote the axial components of the known force Q t and X, Y the axial components of the resultant of all forces acting on the bead (t. e., the resultant of Q and N). Then X == X' N{dylds) ; Y= Y' + N(dx/ds). The equations of motion then become (Art. 287) m(d 2 x/dt 2 ) = X' Nidylds) ; . (1) m(dy/df 2 ) =m Y' .+ N(dx/ds). . . (2) These two equations, together with the equation of the path, serve to determine the motion. (2) Resolution along tangent and normal. Let Q t and Q n rep- resent the tangential and normal components of Q ; then if forces and acceleration be resolved along the tangent and normal to the curve at each instant (Art. 289), the equations of motion become m(dv/df)=Q t ; .... (3) m{v 2 IR) =N + Q n . . . . (4) In equation (4), R denotes the radius of curvature of the path at the position of the particle. Equations (3) and (4) are just equivalent to (1) and (2), and either set may be used, as most convenient. 308. Bead Sliding on Smooth Wire Under Gravity. Let the resultant applied force Q be constant in magnitude and direction. If 266 THEORETICAL MECHANICS. the ^r-axis is taken in the direction of this force, X' = Q, V = o. If the force Q is the weight of the particle, Q = mg, and the equa- tions become m(d*xldt*) = mg N{dylds) ; . v5) m(d 2 y!dt 2 ) = N(dx/ds). .... (6) An important result may be deduced from these equations, irre- spective of the form of the curve. Multiplying (5) by dx and (6) by dy and adding, there results ldxd 2 x dyd 2 y ,\ , m dt -f- -dt = mg dx, \dtdf dt dt 2 I f. the integration of which gives f dx\ 2 1 (dy y 1 tdyy H 1 = gx -f- constant, 2\dt) Adt) S or v 2 = 2gx -\- constant, if v is the speed. Taking the integral between limits x A and x 2 for x, and limits 7\ and v., for v, V * V* r== 2 ^ 2 ^) = 2^-//, if ^5 is the height through which the body falls while the velocity changes from v x to v t . Comparing this result with that of Ex. 1, Art. 227, it is seen that the change in the speed of the particle is the same as if it had fallen freely through the same vertical distance under the action of gravity. The further consideration of this problem will be restricted to the case in which the path of the particle is a circle. The problem then becomes identical with that of the motion of a simple pendulum. 309. Simple Pendulum. A particle suspended from a fixed point by means of a weightless rigid rod or flexible inextensible string, and acted upon by no forces except gravity and the force exerted by the rod or string, is called a simple pendulum. Such a pedulum exists only ideally. An actual pendulum con- sists of a body suspended by a bar or string possessing weight. The motion of any actual pendulum, if frictional resistances be neglected, is the same as that of some simple pendulum, so that the discussion of the ideal simple pendulum is of practical value as well as theoret- ical interest. Consider then the case of a particle suspended by a perfectly PLANE MOTION OF A PARTICLE. 267 flexible but inextensible string. So long as the string remains tense the path must be a circle with the point of suspension as center ; and the force exerted by the string upon the particle is directed toward the center. Let O (Fig. 138) be the point of suspension, A the lowest posi- tion occupied by the particle during its motion, and B the position at any in- stant. Represent the angle A OB by 0, and let OA = /. Also let m de- note the mass of the particle, and N the magnitude of the force exerted by the string, its direction being toward the center. The only other force act- ing upon the particle is its weight, a force of magnitude mg, directed down- ward. Let s denote the length of the arc AB ; then s = 16. (Both 6 and s will be negative if the particle passes to the left of A.) Equations (5) and (6) of Art. 308 are now applicable ; but it will be better to resolve along the tangent and normal to the path. The equations thus take the forms (3) and (4) (Art. 307). . Resolving in the direction of the tangent, m(d' 2 s/dt' 2 ) = mg sin 6; . . (7) and resolving in the direction BO, m(v 2 /l) = N mg cos 6. . . . (8) Since s = 1 6, (7) may be written d t 0/dt t = (g/l) sin 0. . . . (9) The integration of this equation determines the motion, while from (8) N may be determined when the motion is known. This would constitute the complete solution of the problem. The complete in- tegration of (9) will not be shown, since it involves an elliptic integral. One integration, however, may readily be performed, and the value of N determined. If the range of the motion is small in comparison with the radius (i. e. y if the angle 6 is small), a complete approximate solution may be made which is practically correct. ( 1 ) Exact partial solution. Mutiplying both members of equation (9) by dd and integrating, there results VJOidtf = (g/l) cos 6 + C. 268 THEORETICAL MECHANICS. To determine C, let a be the value of 6 when the particle is at rest in its highest position ; then C = (g/0 cos a, and the equation becomes (dOldt) 2 = 2(-//)(cos cos a). . . (io) The velocity in any position is therefore v = l(dO\dt) = V-2gl(cos 6 cos a). . (i i) The value of N may now be determined by substituting the value of v 2 in (8). That is, JV = m(v 2 /l) -f- mg cos 6 = w[2^(cos cos a) -f g cos 0], or AT = mg($ cos 6 2 cos a). . . (12) (2) Approximate complete solution. Suppose the angle 6 to re- main very small throughout the motion ; then, approximately, sin 6 = s/l, and equation (7) becomes d*s/dt* = -(gU)s (13) Multiplying by 2{dsjdt)dt and integrating, (ds/dty = (g/l)(a* - i- 2 ). . . . (14) Here the constant of integration has been determined on the as- sumption that s = a when the particle is at rest in its highest position. Equation (14) may be written dtVJfl = ds/V 'a 1 s 2 . Integrating, and determining the constant by the condition that s = o when / = o, tV g\l = sin-'O/tf) sin-^o). Taking sine of each member and reducing,* s = asm(tV / g/iy . . . (15) This result is identical in form with that found in Art. 229 for the motion of a particle in a straight line under a central attractive force varying directly as the distance. The motion of a simple pendulum when the range is small, is thus a harmonic oscillation. * It will be noticed that the conditions assumed in determining constants of integration leave it uncertain whether the plus or the minus sign should be prefixed to the second member of (15). If it be further assumed that s is plus for small positive values of t (i. e., that the particle is moving in the positive direction through A when t = o), the plus sign must be used. See Art. 229. PLANE MOTION OF A PARTICLE. 269 Time of vibration and of oscillation. If C and C (Fig. 139) are the extreme positions of the particle, the passage from C to C or iirom C to C is called a vibration, while the passage from C to C and back to C is called an oscillation. The par- ticle is at C as often as s - - a, and at C as often as s = a. While s changes from a to a, sin (tvg/l) changes from +1 to 1, and tV g\l changes by 180 or 7r. Hence the time of one vibration is T = Wlfg, . . . (16) Q h&^ c while the time of a complete oscillation is 2T. Fig. 139. Seconds pendulum. A seconds pendulum is one that vibrates once every second. Its length may be found from (16) if^- is known ; thus, if 7" = 1, / = g/ir 2 . Determination of g. By determining the time of vibration of a pendulum of known length, the value of g may be determined from equation (16). Examples. 1. What is the length of the seconds pendulum at a place where the value of^* is 32.2 ft.-per-sec.-per.-sec? 2. If g x and g 2 are the values of g at the surface of the earth and at an elevation of h ft. above the surface respectively, what is the rela- tion between g x and g 2 ? Ans. If P x and P 2 are the values of the earth's attraction upon the body in the two positions, the law of gravitation (Art. 176) gives the proportion pjp % rp _l k) 2 /R 2 . Or, since PJP 2 =gJg 2 , gJg>=(R + hyiR\ 3. If h is small compared with R, show that the result of Ex. 2 reduces approximately to the form^ 2 g x (i 2/1 /R). 4. If T x and T. 2 are the times of vibration of the same pendulum at two places of which the second is h ft. higher than the first, what is the relation between T x and T 2 ? Show that this relation is approx- imately expressed by the equation T x = T. 2 (i k/R). 5. Show how the last result may be used in determining the dif- ference of elevation between two points in the same neighborhood, the same pendulum being swung in the two places. 6. If N x is the number of vibrations of a pendulum in any time at a certain place, and N. 2 the number of vibrations of the same pendu- lum in an equal time at a place h ft higher, show that, approximately, kjR = (AT, - NMN V 270 THEORETICAL MECHA^JCS. 7. At sea-level a pendulum beats seconds. At the top of a moun- tain it beats 86, 360 times in 24 hours. What is the height of the mountain ? 8. Determine roughly a value of g by swinging a weight sus- pended by a string. 9. A body whose mass is 1 lb. is suspended from a fixed point by a string 12 ft. long. The string is swung to a position 6o from the vertical and the body released. Determine the velocity when the body is in its lowest position ; also when 2 ft. above its lowest position. 10. In Ex. 9, determine the tension in the string in each of the positions specified ; also in the initial position. Ans. Tension in lowest position is twice the weight of the body ; in highest position, half the weight of the body. 1 1. Two bodies A and B are connected by a flexible, inextensible string which passes over a smooth pulley C. The body A is verti- cally below C and rests on a horizontal plane. The body B is held on a level with C, the string being tight, and is then allowed to fall under the action of gravity. How far will it fall before A leaves the floor? Ans. Let m f m' be the masses of A and B respectively, and let 6 = A CB. Then at the instant when A is lifted, cos 6 = m/$m'. If m > 3#z', the body A will not be lifted. 310. Uniform Circular Motion. If a particle describes a circle of radius r with uniform speed v, its acceleration is at every instant directed toward the center of the circle, and is equal to v 2 /r. Therefore, from the general equation of motion, the re- sultant of all forces acting upon the par- ticle is a force mv' 2 /r directed toward the center. Let A (Fig. 140) be the particle, and suppose it to be attached to an in- extensible string A O, of which the end O is fastened to a fixed peg. If the Fig. 140. particle be projected in a direction at right angles to A O with a velocity v t and if no force acts upon it except that due to the string, the equa- tions of motion obtained by resolving along the tangent and normal to the path (Art. 289) are m(dv\df) = o, m(y*/r) = T; PLANE MOTION OF A PARTICLE. 27 1 T being the force exerted upon the particle by the string. The first equation shows that the speed v remains constant ; the second gives the value of T. The string thus exerts upon the particle a force whose direction is always toward the point O and whose magnitude is mv % \r. By the law of action and reaction, the particle exerts upon the string a force mv^fr in the direction OA. The string sustains a tension whose value is uniform throughout its length ; i. e., if it be conceived to be divided by a section at any point, the two portions exert upon each other forces which are equal and opposite (Art. 43), each being equal to mv 2 jr. The peg is pulled by the string in the direction OA with a force* equal to mv a /r. The value of the resultant force acting upon the particle may be expressed in terms of its angular velocity (Art. 281) about the center of the circle. If the angular velocity \s co, v rco, and mv a /r = mrco'\ Examples. 1. A body of 2 lbs. mass is swung in a circle of 2 ft. radius by a string which is capable of sustaining 2 lbs. against gravity. At what rate may it move without breaking the string ? Ans. About 8 ft.-per-sec. 2. A locomotive of 250,000 lbs. mass describes a curve of 2,000 ft. radius at the rate of 30 miles-per-hour. What is the resultant force acting upon it ? What horizontal pressure does it exert upon the rail? Ans. Resultant force = 7,520 lbs. toward the center. 3. A locomotive of mass m describes a curve of radius r with speed v. (a) Determine the resultant force acting on the locomo- tive, (fi) Determine the magnitude and direction of the resultant pressure between the track and the locomotive, (r) What should be the difference in elevation of the rails in order that there shall be no tendency of the locomotive to slide laterally? Ans. (a) The resultant force acting upon the locomotive is a force mv 2 !r directed toward the center of the curve. This resultant is made up of two components, a downward force mg (the weight of the body) and the supporting force exerted by the track. (6) This supporting force assumes such magnitude and direction that, when combined with the weight, it gives the resultant mv*/r toward the center. Hence its magnitude is * This force acting upon the peg is sometimes called the "centrifugal" force. It seems better, however, to reserve this term for another use, in con- nection with problems in relative motion. 272 THEORETICAL MECHANICS. V(mv*/ry + {pigf = mVfy*jr) % + g' z ; and its inclination to the vertical is tan -1 (v 2 jrg). (c) The pressure between the rails and the locomotive should be normal to the surface determined by the tops of the rails ; hence this surface should make with the horizontal the angle whose tangent is v 2 /rg. If the hori- zontal distance between centers of rails is a, the outer rail should be higher than the inner by v 2 a/gr. 4. A horizontal platform rotates about a vertical axis at the uni- form rate of 6 revolutions per min. A body of 20 lbs. mass rests upon the platform at a point 16 ft. from the axis of rotation, (a) What is the resultant force acting upon the body ? (b) Of what ac- tual forces is this resultant composed? (c) How great must be the coefficient of friction to prevent the body from sliding ? Ans. (a) 126.3 poundals toward the axis of rotation, (b) The actual forces are the weight (20^ poundals) and the pressure of the supporting platform. The normal component of this pressure is 20^ poundals upward, the tangential component is 126.3 poundals toward the axis of rotation. (c) The coefficient of friction must be at least 126.3/20^* = o. 196. 5. In Ex. 4, if the coefficient of friction is o. 5, how far from the axis may the body be placed without sliding? Ans. 40.8 ft. 6. In Ex. 4, if the body rests upon a smooth surface which is fixed to the platform, what must be the slope of the surface? 7. In Ex. 4, if the body is suspended by a string from a support fixed to the platform, what direction will the string assume, and what tension will it sustain? Ans. The lower end of the string will swing directly away from the axis of rotation until the string is inclined n 6' to the vertical. The tension will be 656 poundals. 8. A platform rotates about a vertical axis with uniform angular velocity co. A body distant x from the axis is placed on a smooth surface whose inclination is such that the body rests without sliding. Show that the surface is inclined to the horizontal at an angle whose tangent is xco 2 !g. The surface slopes directly toward the axis. 9. Using the result of Ex. 8, show that a smooth surface upon which a body would rest wherever placed must be a surface gener- ated by a parabola which rotates with the platform and whose prin- cipal diameter lies in the axis of rotation. If x, y are the coordinates of a section of the surface by a plane containing the axis of rotation, the origin being at the point in which the axis pierces the surface, the result of Ex. 8 gives dy/dx = (co 2 /g)x. Integrating, y = (co 2 /2g)x 2 . 10. Two bodies, each of 5 lbs. mass, are connected by an elastic string which passes through a straight tube 3 ft. long and by its PLANE MOTION OF A PARTICLE. 273 tension holds the bodies against the ends of the tube. The system rotates uniformly about a vertical axis perpendicular to the axis of the tube at its middle point. The natural length of the string is such that a pull equal to the weight of 2 lbs. is required to stretch it to the length of the tube. What is the pressure of each body against the tube when the system is making 10 revolutions per minute? (Take g = 32.2 ft.-per-sec.-per-sec.) Ans. 56.2 poundals or 1.75 pounds-force. 11. Let the system described in Ex. 10 rotate about a vertical axis 6 ins. from the middle point of the tube. Determine the pres- sure of each body against the tube. Ans. The body nearer the axis presses against the tube with a force of 58.9 poundals ; the other with a force of 53.4 poundals. 12. The system and axis of rotation being as described in Ex. 10, suppose the angular velocity gradually to increase. When will the bodies cease to press against the tube? Ans. When the angular velocity reaches 2.93 rad. -per-sec. 13. If the axis of rotation is located as in Ex. n, and the angular velocity gradually increases, which body will leave the tube first? When this occurs, what will be the pressure of the other body on the tube? Ans. When the angular velocity reaches 2.54 rad. -per-sec. the body 2 ft. from the axis of rotation will cease to press against the tube. The pressure of the other against the tube will then be equal to half the tension in the string or 32.2 poundals. 14. A particle suspended from a fixed point by an inextensible string and acted upon by no force except its weight and the pull of the string describes a horizontal circle. If a is the length of the string, 6 the angle it makes with the vertical, a> the angular velocity, T the tension in the string, and m the mass of the particle, show that cos 6 = g/aa?, T = maw 2 . 15. In Ex. 14 let the mass of the particle be 1 kilogr. and the length of the string 1 met. If the number of revolutions per minute is 40, determine 6 and T. (Assume " = 981 C. G. S. units.) Ans. 6 = 56 (nearly) ; T = 1.755 X io 6 dynes == 1.788 kilo- grams-weight. 311. Effect of the Earth's Rotation Upon Apparent Weights of Bodies. If by the weight of a body is meant the gravitational pull of the earth upon it, the apparent weight differs from the true weight because of the earth's rotation. Let Fig. 141 represent a meridian of the earth, NS being the polar diameter and B a point in the equator. The form of the meridian is known to be nearly elliptical, the length of the polar radius being 6.356 X io 8 cm. and that of the equatorial radius 18 274 THEORETICAL MECHANICS. 6.378 X io 8 cm., very nearly. In Fig. 141 the ellipticity is greatly exaggerated. If the earth were spherical and of uniform density, or of density varying only with the distance from the center, the attrac- tion upon a body at the surface would be directed toward the center (Art. 183). In fact the direction differs from this. In Fig. 141, let AH' be the direction of the earth's attraction upon a body at A, and let P' denote the magnitude of the attraction upon a body of mass m. To fix the ideas, suppose the body to be suspended by a string A C attached to a spring balance, and let P be the supporting force as measured by the balance. If the body were in equilibrium, P Fig. 141. would be equal and opposite to P . But since the body moves with the earth in its daily rotation, the string assumes such a direction and the supporting tension such a magnitude that the resultant of P and P' is just sufficient to maintain the actual motion of the body. Let r radius of circle described by body, co = angular velocity of earth's rotation ; then the resultant force acting upon the body has the direction AD and the magnitude Q = mrco 2 . The value of Q may thus be computed, since w is a known constant and r is known for any latitude. The relation between P, P' and Q is represented by the vector triangle EFG f in which EF is parallel to AH', FG to AC, and EG to AD. The value of P is determined by experiment ; Q may be computed as above ; and P' may then be determined by solving the triangle EFG. The force P is equal and opposite to the ' apparent ' ' weight of PLANE MOTION OF A PARTICLE. 275 the body. Its direction is normal to the earth's surface at A {i. e. y to a level surface, such as the surface of still water). The angle between A C and the plane of the equator is the latitude of the place. It will be shown presently (examples 3 and 4 below) that Q is very small in comparison with P, so that the angle EFG is also very small ; assuming this to be true, the value of the angle and of the difference between P and P' may be determined to a close approx- imation in the following simple form : angle EFG = /3 = (Q sin )/P; . . (1) P' P = Q cos cf>. . . . (2) Here ft is expressed in radians. Examples. 1. The time of one revolution of the earth is 86,164 mean solar sec. Determine the value of od, the angular velocity, in rad. -per-sec. Ans. log (o = 5.8628 10 ; co = 0.00007292. 2. Determine the linear velocity of a body at sea level at the equator, using the value of the equatorial radius given above. Ans. 4.65 X io 4 cm. -per-sec. 3. Determine the difference between the true and the apparent weight of a body of m gr. at the equator. Ans. For a body at the equator P, P' and Q are all parallel, and P' P= Q. Also, Q = mv 2 /r, in which v = 4.65 X io 4 (Ex. 2), r = 6.378 X io 8 . Therefore Q 3.391^ dynes. 4. If the value of^ at sea-level at the equator is 978.1 c. m. -per- sec. -per-sec. , what would be its value if not modified by the earth's rotation ? The value of g as experimentally determined is the ' apparent ' ' acceleration, i. e., the acceleration on the assumption that the earth is at rest. If P is the apparent weight of a body of mass m, g = Pjm. The acceleration which would be due to the true attraction is P f /m. At the equator, P'/m = Pjm + Qjm = 978. 1 + 3.4 == 981.5. 5. Show that if the earth should rotate seventeen times as rapidly as at present, its form and density being unchanged, the apparent weights of bodies at the equator would be reduced nearly to zero. 6. In latitude 40 the radius of a parallel of latitude is very nearly 4.894 X io 8 cm. The value of^-at sea-level is about 980.2 C. G. S. units. Determine the influence of the earth's rotation on the mag- nitude and direction of g. 276 THEORETICAL MECHANICS. Using C. G. S. units, the value of Q is mrooi 1 , in which r = 4.894 X io 8 , ft) = 7.292 X io -5 (Ex. 1) ; that is, Q = 2.602m dynes. From equation (2), P' P = Q cos --= 2.602m cos 40 = 1.993W. From the given data, P = 980. 2m, hence P' = 982. 2m ; and equa- tion (1) gives /3=(Q sin )/P' = (2.602 sin 4o)/982.2 == 0.001703 radians = 5' 51". The rotation of the earth thus changes the magnitude of g by 1.993 C. G. S. units, and its direction by 5' 51". 7. If the length of a degree of longitude is 78,849 meters in lati- tude 45 , and the value of g 980. 6 C. G. S. units, compute the effect of the earth's rotation on the value of g. 8. If the diminution of the apparent weight of a given body in latitude cf> is w, and if the value of w at the equator is w\ show that w = w' cos 2 (very nearly J. CHAPTER XVL MOMENTUM AND IMPULSE. i. Rectilinear Motion. 312. Momentum. The fnomentum of a particle is a quantity proportional directly to its mass and to its velocity. Momentum is sometimes called mass-velocity. It is often re- garded as a measure of the " quantity of motion " of the particle. The numerical value of the momentum of a particle of given mass moving with given velocity depends upon the unit in terms of which it is expressed. Unit of momentum. For most purposes it is convenient to choose as the unit the momentum possessed by a body of unit mass having the unit velocity. The unit thus defined depends upon the units of mass, length and time. If these are the pound, the foot and the second, respectively, the unit of momentum is that of a body of one pound mass having a velocity of one foot-per-second. Other units would be the momentum of a particle of one gram mass having a velocity of one centimeter-per-second ; and that of a kilogram mass having a velocity of one meter-per-second. In the discussion which immediately follows, it is to be under- stood that the motion is restricted to a straight line. 313. Increment of Momentum. If the velocity of a particle varies, so also does the momentum. If the mass is tn, and if the velocity has values v x and v t at the beginning and end of a given interval respectively, the increment of the momentum during the interval is , -. mv 2 mv x = m{v 2 vj. 314. Acceleration of Momentum. The rate of increase of the momentum (or increment of momentum per unit time) is called the acceleration of momentum. If the velocity varies at a uniform rate, receiving an increment Av in an interval of time At, the acceleration of momentum is equal to m(Av/Af). If the velocity changes at a variable rate, the instanta- neous value of the acceleration of momentum is m(dv/dt). In either case its value is mp, p being the acceleration. For this reason it is often called mass-acceleration. 278 THEORETICAL MECHANICS. The most convenient unit of mass-acceleration is that of a unit mass having unit acceleration. With the usual British units of mass, length and time, this unit would be the mass-acceleration of a pound- mass having an acceleration of one foot-per-second-per-second. If the gram and centimeter are taken as units of mass and length re- spectively, the unit mass-acceleration is that possessed by a particle of one gram mass having an acceleration of one centimeter-per-sec- ond-per-second. 315. Impulse of a Constant Force. The impulse of a force which remains constant in direction and magnitude is a quantity pro- portional directly to the force and to the time during which it acts. The numerical value of the impulse of a given force acting for a given time depends upon the unit in terms of which it is expressed. Unit impulse. Let the unit impulse be that of a unit force act- ing for a unit time. Then the value of any impulse is equal to the product of the force into the time. This unit depends upon the units of force and of time. In the F. P. S. system the unit impulse is that of a poundal force acting for a second. In the C. G. S. system it is the impulse of a dyne acting for a second. These may be designated as a poundal-second and a dyne-second respectively. If the engineers' system (Art. 218) is em- ployed, the unit impulse is the pound-second. Impulse is a localized vector quantity (Art. 26), its direction and line of action coinciding with those of the force. Restricting the discussion to forces acting upon a particle moving in a straight line which is the line cf action of all the forces, the only directions to be distinguished are the two opposite directions along the path ; these may be distinguished by signs + and . 316. Impulse of a Variable Force. If the magnitude of a force varies continuously throughout a certain interval, its impulse must be computed by integration. Let the value of the force at any instant t be denoted by P, and let it be required to determine the value of the impulse during the interval from / = t' to t = t" , Let this interval be divided into small intervals each equal to At ; and let P x , P 2 , . . . denote the values of P at the beginnings of the successive small intervals. Represent the required impulse by P' ; then, approximately, P' = P l At + P s *t+ (0 MOMENTUM AND IMPULSE. 279 The approximation is closer the smaller At is taken. If At is made to approach zero, the number of terms in the series (i) increases in- definitely while the value of every term approaches zero. The sum in general approaches a definite value which is the exact value of the impulse P' . That is, P' = limit [P x At + P 2 At + . . . ] = C" Pdt. . (2) When P is constant this reduces to P = Pit" - /')> agreeing with Art. 315. 317. Relation Between Impulse and Momentum. If m is the mass of a particle acted upon by a single force P directed along the line of motion, the general equation of motion is P = m(dv/dt). Multiplying through by dt and integrating between limits /' and t'\ x Pdt = m(v" v[), . . (3) v' and v" being the values of the velocity at the instants t' and t" re- spectively. The first member of equation (3) is the value of the impulse of the force P during the interval from /' to t" . In the second member, mv' is the value of the momentum at the instant /', and mv" its value at the instant t"; m(z>" v') is the increment of the momentum for the interval from t' to t" . Equation (3) therefore expresses the prop- osition that The impulse of a force acting alone upon any particle during a given interval is equal to the change of momentum of the particle during that interval. If several forces act upon the particle, its change of momentum during any interval is equal to the impulse of the resultant force ; and this is evidently equal to the algebraic sum of the impulses of the several forces. Examples. 1. A body whose mass is m pounds falls vertically under the action of gravity, (a) Compute the impulse of the force during t sec. (J?) What increment of momentum is received during t sec? Ans. (a) mgt poundal-seconds. (J?) mgt momentum-units. 28o THEORETICAL MECHANICS. 2. In Ex. i, if the body has initially an upward velocity v x , what is its final velocity? Ans. gt v^ ft.-per-sec. downward. 3. A body of 12 lbs. mass is projected downward with a velocity of 20 ft. -per-sec. Write the equation of impulse and momentum for an interval of 3 sec, and thus determine the velocity at the end of the interval. 4. A particle of mass m is acted upon by a force P = kt, t being the time reckoned from a definite instant. Compute the impulse during the interval from t = o to / = t' . Write the equation of im- pulse and momentum, assuming the velocity to be zero when / = o. Explain the meaning of k. 5. In Ex. 4, let m = 500 gr., and let P= 10,000 dynes when / = 1 sec. Write the equation of impulse and momentum and deter- mine the velocity when t = 8 sec. Ans. v" = 640 cm. -per-sec. 318. Dimensions of Impulse and Momentum. If units of force, mass, length and time are all chosen arbitrarily, the unit impulse is of dimensions FT, and the unit momentum of dimensions ML/T. But with such an arbitrary choice of units the equation of impulse and momentum must contain a constant, taking the form (impulse) = (constant) X (change of momentum), the value of the constant being determined in accordance with the particular fundamental units chosen. If, as in Art. 317, a kinetic system of units is employed, the unit force has dimensions* ML/T 2 , and the unit impulse is therefore of dimensions ML/T. The equation impulse = change of momentum thus becomes homogeneous, both members being of the same dimen- sions in terms of the fundamental units M, L and T. 319. Sudden Impulse. The impulse of a force which acts for a very short time is called a sudden impulse. If the time of action is infinitesimal, the impulse is instantaneous. The value of a sudden impulse will be small unless the force is very great. For PAt is the impulse of a constant force P acting for a time At, and this product will be very small if At is very small, unless P is very great. An instantaneous impulse cannot have a finite value unless the * See Art. 219. MOMENTUM AND IMPULSE. 28 1 force is infinite ; impulses which are strictly instantaneous therefore do not come within our experience. There are, however, cases in which a very great force acts for a very short time, and in which the variation of the force within the interval considered cannot be made the subject of observation. Such impulses may often be regarded as instantaneous without important error. Thus, when two bodies come into collision, the force which each exerts upon the other cannot usu- ally be determined, since its time of action is very short ; but the value of the impulse may sometimes be found by observing its effect upon the motion. 320. Effect of a Sudden Impulse. The effect of a sudden im- pulse is to produce a sudden change in the momentum of the body acted upon. If this change of momentum can be observed, the value of the impulse can be determined, although the force and the time both remain unknown. 321. Ordinary Forces Neglected. In considering the effect of a sudden impulse, all forces of ordinary magnitude may usually be neglected without important error. Consider the case of a ball struck by a bat. During the impulse the ball is acted upon by the force of gravity (a downward force equal to the weight of the ball). But during the short time of action of the force due to the blow, the change of momentum due to the force of gravity is exceedingly small in comparison with that due to the blow, and may be neglected with- out appreciable error. If an impulse is strictly instantaneous, any finite force whatever may be neglected in comparison with the force of the impulse, since the momentum due to a finite force during an infinitesimal time is infinitesimal. Let P be the force of an instantaneous impulse which produces a finite change of momentum, and let Q be a finite force acting on the particle at the time of the blow. Then the equation of impulse and momentum is r t" r t" Pdt + I Qdt= miro" v'\ J t < J t , If the time of action of the blow, /" t\ approaches zero, the second term of the first member of this equation approaches zero. Hence Q may be neglected in estimating the change of momentum during the blow. 282 theoretical mechanics. Examples. i. A ball weighing 5^ oz., moving horizontally at the rate of 100 ft. -per-sec. , is struck by a bat. Immediately after the blow the ball moves at the rate of 1 50 ft. -per-sec. in the direction opposite to that of its original motion. What is the value of the impulse? 2. In Ex. 1, if the time occupied by the blow is o. 1 sec, what is the average value of the force (z. e. , what constant force would pro- duce the same change of momentum in the same time) ? Ans. 820.3 poundals. 3. What would be the velocity of the ball after the blow, if initially at rest and acted upon by the same impulse as in Ex. 1 ? 4. If the same blow were applied to a body of 1 lb. mass, what would be its effect? 322. Impact or Collision of Bodies. When two bodies come into collision, each exerts upon the other at the place of contact a force whose magnitude increases from zero up to some maximum value and then decreases to zero again. The time of action of these forces is so small that their magnitudes cannot be measured ; there- fore in. dealing with cases of impact the discussion must usually be limited to a consideration of the total changes of velocity and mo- mentum produced in the bodies. Direct and indirect impact. If two bodies moving in the same straight line collide, their impact is said to be direct ; their original velocities may have the same direction or opposite directions. If the paths of the two bodies before collision are intersecting lines, the impact is indirect. The present discussion will be restricted to the case of direct impact. It will further be assumed that the bodies are of such symmetrical form that the forces acting be- tween them when in contact are directed along their common line of motion. Elastic and inelastic bodies. As the two bodies come into col- lision, each is distorted by the pressure, the amount of distortion in- creasing as the pressure increases. If the bodies are wholly inelastic, the distortion is permanent, the bodies showing no tendency to re- gain their original forms. In this case the two bodies will move on with a common velocity after the impact. If they are elastic, each tends to regain its original form, and in so doing exerts a force upon the other so that the two tend to separate. Elasticity, as thus de- fined, is possessed by bodies in various degrees. The method of specifying the degree of elasticity will be considered below. MOMENTUM AND IMPULSE. 283 323. Law of Action and Reaction Applied to Impact. HA and B are any two bodies in collision, the forces exerted by them upon each other are at every instant equal and opposite, in accord- ance with Newton's third law (Art. 259). It follows that the impulses of these forces during any time are equal and opposite. Also, since the momentum due to an impulse is numerically equal to the impulse, the two bodies receive during any part of the time of the collision equal and opposite quantities of momentum. We thus reach the fol- lowing principle : The collision of two bodies causes no change in their total mo- mentum. In the case of direct impact this principle may be stated algebra- ically as follows : Let m x , m 2 be the masses of two bodies moving in the same straight line, v x , v 2 their velocities before collision and /, V^ their velocities at any instant during or after the collision. (In any particular case algebraic signs must be given to the velocities in accordance with their directions along the line of motion.) Before the collision the total momentum of the two bodies is m l v l -f m,v 2 ; at a succeeding instant its value is m x v x -f- m 2 v 2 . By the above principle these two quantities are equal, giving the " m x v x -f- m 2 v 2 = m x v x + m 2 v 2 . . . (1) Equation (1) is not sufficient for the determination of v x and v 2 ; but if an additional relation between these quantities is known they may be determined. For example, suppose there is some instant during the collision at which the two bodies have the same velocity v ; then for that instant v x = v 2 = v, and (1) becomes m x v x -f m 2 v 2 (m x -f #z 2 )^> or v = (m x v x -f m 2 v 2 )/(m x + m 2 ). . . (2) 324. Inelastic Impact. If two bodies moving in the same line are inelastic, they will not separate after collision (Art. 322) ; hence in this case equation (2) will give their common velocity after im- pact. Examples. 1. A sphere of mass 10 lbs. moving at the rate of 20 ft.-per-sec. overtakes a sphere of mass 20 lbs. moving at the rate of 10 it. -per- 284 THEORETICAL MECHANICS. sec. If the bodies are inelastic, what is their velocity after the col- lision? Ans. 13^ ft.-per-sec. 2. In Ex. 1, what is the value of the impulse acting on each body during the collision? Taking the direction in which the bodies are moving as positive, the momentum of the first body changes from 200 to 133/^ (F. P. S. units), while the momentum of the second changes from 200 to 266^3. Hence the impulse acting upon the first body is 66^3 poundal-seconds, and that acting on the second is + 66^3 poundal- seconds. 3. A body whose mass is 5 kilogr. , moving at the rate of 700 cm. -per-sec. , meets a body whose mass is 4 kilogr. Both bodies are brought to rest by the collision. Required (a) the original velocity of the second body, and (b) the value of the impulse acting on each body during the collision. Ans. (#) 875 cm. -per-sec. (J?) Two opposite impulses each equal to 3. 5 X io 6 dyne-seconds. 4. Two bodies collide while moving in opposite directions with velocities inversely proportional to their masses. Show that their total momentum is zero at every instant during the collision. If they are inelastic, what is their common velocity at the end of the col- lision ? 5. A body of 2 lbs. strikes a body of 50 lbs. which is at rest on a smooth horizontal surface. Immediately after the collision the heavier body has a velocity of 4 ft. -per-sec. while the lighter body is at rest. What was the initial velocity of the lighter body ? What total impulse acted on each body? 6. Is the solution of Ex. 5 changed by assuming the horizontal surface to be rough? [See Art. 321.] 7. A 2-oz. bullet passes through a block of wood weighing 4 lbs., its velocity being thereby changed from 1,100 ft.-per-sec to 950 ft.-per-sec. With what velocity does the block move after the impact, if free? What is. the value of the total impulse acting on the block ? 8. A block weighing 10 lbs. is struck by a 2-oz. bullet which remains imbedded in it. After the collision the block has a velocity of 10 ft.-per-sec. What was the original velocity of the bullet? A ns. 8 1 o ft. -per-sec 325. Elastic Impact. If two elastic bodies moving in the same straight line collide, it is convenient to divide the time occupied by the impact into two parts : the time T x up to the instant of greatest distortion of the bodies, and the time T 2 after that instant. At the instant of greatest distortion the bodies have the same velocity. Before that instant the forces acting between them are resisting MOMENTUM AND IMPULSE. 285 their approach, after that instant the forces are urging their sepa- ration. Let the whole impulse which either body exerts upon the other be divided into two parts ; the part acting during T x being called the impulse of compression, and the part acting during T 2 the impulse of restitution. Let the ratio of the latter impulse to the former be called the coefficient of restitution, and be represented by e. Let m x and m 2 be the masses of the bodies ; v x and v 2 their ve- locities before collision ; v their common velocity at the instant of greatest compression ; v Xi v 2 their velocities after the collision. Let Q denote the impulse acting upon m x during compression, and em x the impulse acting on the same body during restitution. The im- pulses acting on m 2 are then Q and eQ. Consider either body, as that whose mass is m x . Before the col- lision its momentum is m x v x ; at the instant of greatest compression it is tn x v ; at the end of the collision it is m x v x . Hence (increase of momentum due to impulse Q) = m x (v v x ) ; ( " " " " " eQ) = m x (y x ' v). Or, since the impulse is numerically equal to the momentum it produces, Q = m x {v V 1 );. . . . (3) eQ = m x (v x v) (4) Eliminating Q and solving for v x \ v x = (1 -\~ e)v ev x . But by equation (2) (Art. 323), v = (m x v x + m 2 v 2 )/(m x -f m 2 ) ; which substituted in the value of v x gives, after reducing, 111 vi = v x - -a (1 + )(-, - ,) (5) m x -f- m 2 In like manner, by writing equations similar to (3) and (4) for the other body, we may find < = *. =t-(i +"')(, .) (6) m x -f- m. Equations (5) and (6) determine the velocities of the bodies after they separate. The value of the impulse Q and of the total impulse 286 THEORETICAL MECHANICS. (i -j- e)Q may be determined from the equation of impulse and mo- mentum. Thus, the total change of momentum of the body m x is m x (y{ v x ). Equating this to the total impulse, and using the value of v x given by equation (5), (1 +e)Q = - m < m < (i + *)0, -v 2 ); . (7) Q = 'tivr v,). . . . (8) Equation (7) gives the whole impulse acting upon the body m l . Assuming v x greater than v 2 , the impulse is negative. Equation (8) states that the impulse up to the instant of greatest compression is independent of e, that is, independent of the degree of elasticity. The value of e may lie anywhere between o and 1 . For inelastic impact e = o, and the value of v x is v x = {in x v x -f m 2 v 2 )j(m x -f m 2 ) } . . (9) as in Art. 324. For perfect restitution, e = 1, and ITU v; = v x -*&, v 2 )\ . . (10) 2nt \ r \ r \ v 2 = v % - (y 2 v x ). . .(11) m x -\- m t Examples. 1. A man weighing 160 lbs. leaped into a boat whose mass was 100 lbs., thereby causing it to move from rest with a velocity of 10 ft. -per-sec. With what velocity did the man leap ? An s. 16^ ft. -per-sec. 2. A boat weighing 200 lbs. is at rest in still water when a person weighing 150 lbs. starts to move from one end to the other. If he moves 10 ft. relatively to the boat, how far does he move relatively to the water, and how far does the boat move ? By Newton's third law, the force urging the man forward is at every instant equal to the force urging the boat in the opposite di- rection. Let P be the magnitude of each of these forces in poundals ; then at any instant the acceleration of the boat is P/200 and that of the man P/i^o in the opposite direction. The velocities due to these accelerations in any time are in the same ratio as the accelera- tions ; and the distances described by the two bodies in the same time are proportional to their velocities, and therefore to their accel- erations. The boat thus moves three-fourths as far as the man. MOMENTUM AND IMPULSE. 287 3. Two perfectly elastic bodies moving along the same line col- lide. Prove that they exchange velocities. 4. A body whose mass is 5 kilogr. , moving at the rate of 5 met. -per-sec. , collides with a body whose mass is 4 kilogr. , moving in the same direction at the rate of 4 met. -per-sec. The coefficient of restitution is 0.5. Determine the velocities of the bodies after they separate. Ans. 4*/ 3 met. -per-sec. and 4% met. -per-sec. 5. In Ex. 4, compute the value of the impulse acting during com- pression, and of the total impulse during the collision. Ans. (2/9) X 1 o 6 dyne-seconds; (1/3) X io 6 dyne-seconds. 6. A body weighing 10 lbs., moving at the rate of 12 ft. -per-sec. , overtakes a body weighing 18 lbs., moving at the rate of 5 ft. -per- sec. After the collision the velocity of the former body is 6 ft. -per- sec. Required (a) the final velocity of the second body ; (b) the coefficient of restitution. Ans. (a) 8^ ft. -per-sec. (b) }i. 7. Prove that, in any case of direct collision, the ratio of the rel- ative velocity of the two bodies after impact to their relative velocity before impact is numerically equal to the coefficient of restitution. 8. A rifle weighing 3 lbs. is discharged while lying on a smooth horizontal plane. The weight of the bullet is 2 oz. and it leaves the barrel with a velocity of 1,400 ft. -per-sec. What will be the velocity of the " kick "? What is the impulse of the kick ? 9. If the discharge of a rifle were strictly instantaneous, what would be the force of the kick ? 10. A man weighing 165 lbs. leaps from a boat weighing 124 lbs. into a boat weighing 150 lbs. If the boats are initially at rest, compare their velocities after the leap. 2. Motion in Any Path. 326. Momentum a Vector Quantity. The momentum of a moving particle has been defined (Art. 312) as a quantity propor- tional directly to its mass and to its velocity. When motion in a curved path is studied, it must be remembered that direction as well as magnitude is an essential part of the value of velocity, and there- fore of momentum. Momentum is, in fact, a vector quantity, its direction coinciding at every instant with that of the velocity of the particle. 327. Moment of Momentum or Angular Momentum. The momentum of a particle is at every instant associated with a certain position. It is, in fact, a localized vector quantity (Art. 26), its posi- tion-line at any instant being the straight line along which the motion 288 THEORETICAL MECHANICS. is directed. The position-line is the tangent to the path at the instantaneous position of the particle. The moment of momentum of a particle with respect to any point in the plane of the motion is the product of the momentum into the perpendicular distance of its position-line from the origin of moments. Moment of momentum is also called angular momentum. 328. Increment of Momentum. If the velocity of a particle varies (either in direction or in magnitude or in both) so also does the momentum. The increment of momentum for any interval may be found in the manner already de- scribed for finding increment of velocity (Art. 246) or of any variable vector (Art. 22). Thus, let AB (Fig. 142) be the path described by a particle during any interval, v x being the magnitude of the velocity in the position A and v 2 its Fig. T 42. magnitude in the position B. At the beginning of the interval the momentum has the value mi\ , directed along the tangent to the path at A ; at the end of the interval its value is mv % , directed along the tangent at B. Representing these values by vectors OA' and OB\ the vector A'B' represents the increment of momentum for the given interval. 329. Acceleration of Momentum a Vector Quantity. The ac- celeration of momentum (or mass -acceleration) of a particle is a vector quantity proportional directly to the mass and to the accel- eration. Its direction is that of the acceleration ; it does not, in general, coincide with that of the tangent to the path at the position of the particle. Acceleration of momentum bears the same relation to momentum that acceleration bears to velocity. It may be defined concisely as momentum-increment per unit time; just as acceleration is defined as velocity-increment per unit time (Art. 247). Examples. 1. If the velocity of a particle of mass 18 lbs. changes from 12 ft.-per-sec. in a certain direction to 20 ft.-per-sec. in the opposite direction, what is the momentum-increment? Ans. 576 F. P. S, units. MOMENTUM AND IMPULSE. 289 2. If the velocity of a particle of 18 lbs. mass changes from 12 ft.-per-sec. in a horizontal direction to 20 ft. -per-sec. vertically down- ward, what are the magnitude and direction of the momentum-incre- ment? Ans. 419.9 F. P. S. units ; 30 58' from vertical. 3. A particle of 60 lbs. mass describes a circle of 6 ft. radius at the uniform rate of 120 revolutions per minute. Determine the magnitude and direction of the increment of momentum received by the particle while making (a) one-sixth of a revolution, (b) one-fourth of a revolution, and (c) one-half a revolution. Ans. (a) 4,524 F. P. S. units. 4. A particle of 20 gr. mass describes a circle of 90 cm. radius with a uniform speed of 250 cm. -per-sec. Determine completely (in magnitude and direction) f he increment of momentum in one- hundredth of a second. Ans. Its magnitude is 139 C. G. S. units, nearly. 5. With the data of Ex. 4, compute the moment of the momentum {a) about the center of the circle ; {b) about a point in the circum- ference, (c) Compute the greatest and least values of the moment of the momentum about a point 40 c m. from the center. 330. Impulse of a Force Whose Direction Varies. If the direction of a force varies, this variation must be taken into account in computing the impulse. Impulse must now be regarded as a vector quantity, its direction being determined by that of the force ; and the impulses for successive intervals of time are to be combined by vector addition. Thus, if P x ' and P 2 f denote the values of the impulse for two successive intervals At x and At. i} the impulse for the entire interval At x -f At 2 is the vector sum of P x and P 2 f . If the force has a constant direction during the interval At x , and a constant but different direction during the interval At.,, P x and P. 2 ' may each be computed as in Art. 315 or Art. 316. But if the direction of the force varies continuously, a different method must be employed. Let it be required to compute the impulse of any variable force P during the interval from t' to t" . Let the interval be subdivided into small parts At x , At 2 , . . . , and let the values of P at the beginnings of these intervals be P x , P., , . . . . Approximate values of the impulses for the successive small intervals are P x At x in direction of P lf P.At 2 in direction of P s , and the required total impulse P' is approximately equal to the *9 29O THEORETICAL MECHANICS. vector sum of /^A/, , P 2 At, , . . . . The approximation may be made as close as desired by taking the intervals sufficiently small. The exact value of the total impulse is the limit approached by the approximate value as A/, , A/ 2 ... all approach zero. That is, P' = limit IP Ah ~h PA*t + ] . it being understood that the -f- sign denotes vector addition.* It will be noticed that the above process of computing im- pulse is independent of the path of the particle upon which the force acts. So far as its magnitude and direction are concerned, impulse depends only upon force and time. 331. Vector Equation of Impulse and Momentum. The prin- ciple of impulse and momentum, already stated for the case of rectilinear motion, may now be extended to the case of a particle de- scribing any path under the action of a force varying in any manner. The impulse of t lie force during any interval is equal in magni- tude and direction to the change of momentum of the particle during that interval. This statement includes the proposition given in Art. 317. The equation . , n impulse = momentum-increment must now be interpreted as a vector equation ; in the special case of rectilinear motion it reduces to an algebraic equation. 332. Algebraic Computation of Impulse of Variable Force. Usually the simplest method of computing, the value of the impulse of a force whose direction varies is to resolve the force into compo- nents parallel to fixed rectangular axes and determine the impulse of each axial component separately. Restricting the discussion to the case in which the path of the particle and the line of action of the force lie in a plane, let P be the value of the force at time /, and let the axial components of P be X and Y. Let P' denote the impulse of Pfor the interval from t' to /", X' and Y' being the axial com- ponents of P' . Then X' = f Xdt; V = C Ydt. . * The above reasoning may be expressed in the language of differentials, as follows : Let /'be a vector symbol denoting the value of the force at the instant I, and P / a vector symbol denoting the value of the required impulse. Then dP / = P dt. This is a differential equation, both members of which are vectors. MOMENTUM AND IMPULSE. 29 1 From the values of X ! and V the magnitude and direction of P' may be determined. 333. Algebraic Equations of Impulse and Momentum. In the case of plane motion the vector equation of impulse and mo- mentum is equivalent to two algebraic equations. These may be deduced immediately from the differential equations of motion (Art. 287). These equations are X = mXy Y =z= my. Multiplying each through by dt and integrating between limits f and /", the values of x and y at the limits being x ', x" and y' , y" respectively, \ Xdt= m(x" x')\ f Vdt = m(y"y' The first members of these equations are the axial components of the impulse of the resultant force acting upon the particle (Art. 332); the second members are the axial components of the increment of momentum of the particle. Together the equations express the proposition that the resultant impulse is equal in magnitude and direction to the total increment of momentum. The general principle of Art. 331 is thus an immediate consequence of the fundamental equations of motion of a particle. 334. Impulse and Momentum as Localized Vector Quanti- ties. In the foregoing discussion no reference has been made to the path of the particle nor to the position in space of the line of action of the force. These must, however, be taken into account in a com- plete explanation of the principle of impulse and momentum. The momentum of a particle is at every instant directed along a definite line in space. This may be regarded as its position-line, momentum being thus a localized vector quantity (Art. 26). A force acting upon a particle has at every instant a definite line of action. The impulse of a force whose line of action remains constant may therefore be regarded as a localized vector quantity whose position-line is the line of action of the force. If the line of action of the force varies, the elementary impulse Pdt correspond- ing to an elementary time dt may be regarded as a localized vector quantity whose position-line coincides with the instantaneous position of the line of action of the force. 292 THEORETICAL MECHANICS. 335. Angular Impulse. Let P denote the value of a force at any instant /, and P' the impulse during the time from t' to /". Choosing any origin of moments (Fig. 143), let a be the perpen- dicular distance from O to the line of ac- tion P. Then Pa is the moment of the force with respect to 0. During a small interval At let the particle to which the force is applied move a short distance AB. The change in P will be small ; the impulse during the small interval will be approxi- mately equal to PAt ; and the moment of q ' the impulse will differ little from Pa At. Fig. 143. Let the whole time from t' to t" be subdivided into small parts A^, , A/ 2 , . . . , and let the values of P at the beginnings of these small intervals be P lt P 2i . . . , the corresponding values of the arm a being a x , a . . . . The sum P x a x At x + P 2 a 2 At, -f . . . may be regarded as an approximate value of the moment of the resultant impulse P . Taking the intervals smaller and smaller, approaching zero, the exact value of the moment of the impulse is limit \\P x a x At x + P,a 2 At, -f . . . ] = f Padt. If the moment of the force with respect to O is represented by G, G = Pa, and the moment of the impulse is equal to x Gdt. The moment of the impulse of a force about a given point is also called its angular impulse about that point. 336. Position-Line of Resultant Impulse. It has been pointed out that the impulse of a force during an elementary time dt may be regarded as having a definite position-line, even when the line of action of the force is variable. The resultant impulse for an interval of any length may be regarded as localized in a line whose position P'a' = \ Padt. MOMENTUM AND IMPULSE. 293 is determined by the value of the moment of the impulse about any given point. Its distance a! from the origin of moments must satisfy the equation f It will be observed that the method of determining the magni- tude, direction and position-line of the resultant impulse for any interval is essentially the same as that of finding the magnitude, direction and line of action of the resultant of any system of forces acting in the same plane on a rigid body (Art. 97). In the case of forces, the resultant is equal to the vector sum of the components ; and the moment of the resultant is equal to the algebraic sum of the moments of the components. In the case of the impulse, the resultant impulse for any interval is equal to the vector sum of the impulses for the elementary intervals ; and the moment of the re- sultant impulse is equal to the algebraic sum of the moments of the elementary impulses. The similarity of these relations may be fur- ther illustrated. Let AB (Fig. 144) represent the path described by the particle during the time from t' to /". Let the whole time be subdivided into small intervals A^ , A/ 2 , . . . ; let A, a, b, . . . be the positions of the particle at the beginnings of these intervals ; and let P x , P 2 , . . . be the corre- sponding values of the force, their di- rections being as shown in the figure The impulses during A/ x , A/, , . . . have approximately the values P x At, along the line AA\ P 2 At 2 along the line aa\ Let these be combined as if they were forces ; their resultant is equal to their vector sum and acts along some determinate line, as M'N' ; and this resultant is an approximate value of the resultant impulse P'. The exact value is the limit approached by the approximate value as A/, , A/ 2 , . . . are all made to approach zero. The line M'N' approaches a limiting position MN, which is the position-line of the resultant impulse P'. 294 THEORETICAL MECHANICS. Fig. 145. Examples. 1. A particle is projected horizontally with a given velocity V t after which it is acted upon by no force except gravity. Determine the position-line of the resultant impulse of the force of gravity during any time. Let the particle be projected from the point A (Fig. 145) in the direction AB', and let AB be the path described during the time T. Let this time be subdivided into a number of small intervals, each equal to At, and let A, a, b, . . . be the positions of the particle at the beginnings of these intervals. If m is the mass, the force is equal to mg and is constant in direction. Applying the method above described, the required impulse is approximately equal to the resultant of a series of equal impulses mgAt acting in lines AA' , aa' , bb', . The successive distances be- tween these lines are equal, since the horizontal velocity is constant ; hence the position-line of the resultant is midway between the extreme lines AA', gg '. The limiting position of this line, as At is made to approach zero, bisects AB'. The magnitude of the resultant im- pulse is obviously mgT. 2. In Ex. 1, compute the moment of the resultant impulse by integration, and thus verify the above result. 337. Equation of Angular Impulse and Angular Momentum. The impulse of a force and the increment of momentum produced by it, during any element of time, are not only equal in magnitude and direction but have the same position-line. From this it follows that their moments about any origin are equal. And since this equality holds for every element of time, it holds for any interval whatever. That is, The angular impulse of a force for any interval of time is equal to the angular momentum produced by the force during that time. This equation and the two given in Art. 332 are the complete algebraic expression of the principle of impulse and momentum as applied to the motion of a particle in a plane. To summarize : The proposition that impulse is equal to change of momentum, in its full meaning, must be understood to express equality of magni- tude, agreement in direction, and identity of position-lines. In case of plane motion, this is completely expressed by three algebraic MOMENTUM AND IMPULSE. 295 equations. Two of these may be obtained by resolving in any two directions, the third by taking moments about any point. 338. Geometrical Illustration of General Equation of Impulse and Momentum. Let the curve AB (Fig. 146) be the path de- scribed by a particle during the interval from t' to t" . Let v' and v" be the in- itial and final values of the velocity. From any point draw vectors OA', OB', representing the initial and final values of the momentum. These vec- tors are parallel to the tangents to the path at A and at B respectively, and *OA' = mv' , OB' = mv" . The incre- ment of momentum m Av is then rep- resented by the vector A ' B' . To apply the principle of impulse and momentum, the position-line of the momentum-increment must be determined. The position-line of the initial momentum is the tangent to AB at A ; that of the final momentum is the tangent to AB at B ; the position-line of the momentum-increment therefore passes through C, the intersection of these tangents. Let P* be the resultant impulse applied to the particle during the interval ; then P' = m A?', and its position -line is CD, parallel to A'B'. In the foregoing explanation of the general principle of impulse and momentum, it has been assumed that each of these quantities is subject to the laws of composition and resolution which apply to forces acting upon a rigid body. As applied to impulse, these laws have been explicitly stated in Art. 336. As applied to momentum, it is assumed that the increments of momentum received by a par- ticle during successive intervals are to be combined as if they were forces acting on a rigid body in order to produce the momentum- increment for the entire interval. These are arbitrary assumptions, amounting in reality to defini- tions of resultant impulse and resultant momentum. They are ex- tremely useful because they make it possible to state very concisely the full meaning of the equation of impulse and momentum. These definitions are especially valuable when the general principle is ex- tended to systems of particles and to rigid bodies. 296 THEORETICAL MECHANICS. Examples. 1. In Ex. 1, Art. 336, show that the tangent to the path at B bi- sects AB' . [Apply the general principle of impulse and momentum, using the result already found in the solution of the example. ] 2. If A and B are any two points in the path of a projectile, show by the principle of impulse and momentum that the tangents at A and B and the vertical line bisecting the chord AB intersect in a point. [The position-line of the resultant impulse may be located as in Ex. 1, Art. 336.] 3. A particle describes a circle at a uniform rate. Show that the resultant impulse applied to it during any interval passes through the center. Show also that its position-line bisects the path described during the interval. 339. Sudden Impulse. If a particle receives a sudden impulse whose direction is inclined to that in which the particle is mov- ing, the direction of motion is suddenly changed. The relation between the in- itial and final velocities and the impulse is expressed by the vector equation of impulse and momentum, irrespective of the time occupied by the impulse. During the time of action of a sud- den impulse the particle moves over a very short distance. The direction of the motion may, however, change by any angle. In general, therefore, a sud- den impulse causes a rapid curvature of the path. Let a particle describing the straight path LA (Fig. 147) receive a blow whose total impulse is rep- resented by the vector A'B'. If OA' represents the initial momentum, OB' will represent the momentum after the blow. The path after the blow will be some line BM, parallel to OB'. The path during the impulse will be some curve AB. The shorter the time occupied by the blow, the shorter the path AB. If the impulse were instantaneous, A and B would fall together, Fig. 148. 147. as in momentum and impulse. 297 Examples. 1. A ball whose mass is 5^ oz. is moving horizontally at the rate of ioo ft.-per-sec. when it is struck by a bat in such a way that immediately after the blow it has a velocity of 150 ft.-per-sec. in a direction making an angle of 30 upward from the horizontal. Re- quired the value of the impulse. Ans. 79.34 poundal-seconds, inclined i84 upward from the horizontal, assuming the horizontal velocity to be reversed in direc- tion by the blow. 2. In Ex. 1, suppose the ball to receive the same blow when moving at an angle of io downward from the horizontal. How will it move immediately after the blow ? 3. In the same case, if the blow occupies 0.01 sec, what is the average value of the force? What will be the effect of gravity on the motion of the ball during the blow ? 340. Collision of Bodies Moving in Different Paths. The collision of bodies moving in different paths presents a less simple problem than that treated in Arts. 322-325. No discussion of this problem will be given here, except to point out the application of the law of action and reaction. As a consequence of this law, the total impulses acting on two bodies during their collision are equal and opposite. It follows that their momentum-increments are equal and opposite. The vector sum of their momenta therefore has the same value after collision as before. Its value is, in fact, the same at every instant during the collision. Examples. 1. A body of 5 lbs. mass strikes a fixed surface which deflects it 20 and changes its speed from 25 ft.-per-sec. to 10 ft.-per-sec. What is the total impulse exerted upon the body during the collision? Ans. 79.8 poundal-seconds, directed at angle 167 38' with the original motion. 2. A ball A of 6 lbs. mass is at rest on a horizontal plane when it is struck by a ball B of 4 lbs. mass moving at the rate of 12 ft.- per-sec. After the collision A is moving at the rate of 4 ft. -per-sec. in a direction inclined 6o to the original velocity of B. How does B move after the collision ? Ans. With velocity 10. 4 ft. -per-sec. at right angles to motion of A. 3. Two bodies moving in opposite directions with velocities in- versely proportional to their masses collide in such a way as to be deflected from the original directions of motion. Show that after the collision they move in opposite directions with velocities inversely proportional to their masses. CHAPTER XVII. WORK AND ENERGY. i. Work in Case of Rectilinear Motion. 341. When a Force Does Work. A force is said to do work upon the body to which it is applied when its point of application moves in the direction toward which the force acts. 342. Quantity of Work Done by Constant Force. The amount of work done by a constant force is a quantity proportional directly to the magnitude of the force and to the distance moved by its point of application in the direction of action of the force. The numerical value of the work depends upon the unit in which it is expressed. The unit of work usually chosen is the quantity of work done by the unit force when its point of application moves the unit distance in the direction of action of the force. If the pound is taken as the unit force and the foot as the unit distance, the unit work is the quantity of work done by a force of one pound when its point of application moves one foot in the direction of action of the force. This unit is called a foot-pound, and is the unit commonly employed by engineers. In the C. G. S. system the unit of work is the quan- tity of work done by a force of one dyne while its point of application moves one centimeter ; the name erg has been given to this unit.* In the F. P. S. system the unit of work is the foot-poundal. When the point of application of a force moves a certain distance in the direction of action of the force, the force is said to act through that distance. The unit work may therefore be defined as the quan- tity of work done by the unit force acting through the unit distance. * The erg is very small in comparison with the foot-poundal or the foot- pound. Thus, 1 foot-poundal = 421,390 ergs; 1 foot-pound 1.356 X io 7 ergs. (The last result assumes^ = 981 C. G. S. units.) For practical use, especially in electrical engineering applications, an- other unit of work, the joule, has been introduced, defined as equal to io 7 ergs. WORK AND ENERGY. 299 343. Positive and Negative Work. If the point of application of a force moves in a direction opposite to that of the force, the force is said to do negative work upon the body. If P is the algebraic value of a constant force, and Ax that of the displacement of its point of application, the positive direction being the same for both, the value of the work is given with proper sign by the product p ^ x Examples. 1. A body weighing 10 lbs. is thrown upward against gravity. Compute the work done upon it by its weight (a) while it rises 10 ft. and () while it falls 10 ft. Ans. (a) ioo^foot-poundals. (b) -\-ioog foot-poundals. 2. If the resistance of the air amounts to a constant force of 2 lbs., compute the work done by it in both cases of Ex. 1. Ans. 20 foot-pounds in each case. 344. Work Done by Variable Force. If the value of a force varies during the displacement of its point of application, the work must be computed by integration. Let OB (Fig. 149) be the path described by the point of appli- cation (also the line of action of the force). Let P be the value of the force when the point of application is at distance x from 0, and let it be required to compute the work done by P while the point of applica- j- Xt-- h tion moves from A to B. ,-- x\ -." ""~ ~;j Divide AB into small parts A a O A a b c B -= Ax lt afr = Ax 2 , &C = Ax., y . . . Fig. 149. and let P lt P t1 P,, . . . be the values of P when the point of application is at A, a, b, . . . Let W represent the required work ; then, approximately, W= P x Ax x + P 2 Ax t + PA** + .... By taking A^Tj, Ax. 2 , . . . small enough, the approximation may be made as close as desired. The exact value of the work is the limit approached by the approximate value as Ax ly Ax. i} . . . are made to approach zero. That is, if x i = OA and x 2 = OB, W = limit [/> Ax x + P, Ax t + . . . ] = f Pdx. *i A 3OO THEORETICAL MECHANICS. 345. Graphical Representation of Work Done by Variable Force. At every point of the line AB representing the displace- ment of the point of application (Fig. 150), erect an ordinate whose length is equal to the value of P for that position ; the extremities of these or- dinates lie on a curve A'B'. The j{ B~ area bounded by the curve, the line Fig. 150. AB, and the ordinates AA', BB\ is equal to the work done by P during the displacement AB. For, the expression above found for the work, Pdx, L is also the value of the area. Let the work-diagram be drawn in each of the following examples. Examples. 1. A body is suspended by an elastic string whose unstretched length is 4 ft. Under a pull of 10 lbs. the string stretches to a length of 5 ft. Required the work done on the body by the tension of the string while its length changes from 6 ft. to 4 ft. Ans. 20 foot-pounds. * 2. A body whose mass is m falls vertically to the earth's surface from a height equal to the radius R. Compute the work done by the earth's attraction during the fall. Ans. \mgR kinetic units. 3. In Ex. 2, let m = 100 lbs., P == 21,000,000 ft. Give the value of the work in foot-pounds. Ans. 1.05 X io 9 . 4. A particle moves in a straight line under the action of a force directed toward a fixed point in the line of motion and varying di- rectly as the distance from that point. Compute the work done on the particle during a given displacement. Ans. Let P' be the value of the force when the particle is at the unit distance from the fixed point, x x and x t the initial and final dis- tances from that point. Then the required work is \P'{x* x*). 346. Dimensions of Unit Work. If the units of force and length are both chosen arbitrarily, the unit work has dimensions FL. But in the kinetic system (Art. 219), F = ML/T 2 , which reduces the unit work to the dimensions ML 2 /T 2 . WORK AND ENERGY. 3OI 2. Work in Any Motion of a Particle. 347. General Definition of Work. When the point of applica- tion of a force receives a displacement which is not along the line of action, the definition of work must be enlarged. Let a force P remain constant in magnitude and direction while its point of application describes any straight line AB (Fig. 151). The quantity of work done depends upon the value of AB' , the orthographic projection of AB upon the line of action of P. This projection may be called the effective displacement. The quantity of work done by a constant force while its point of application receives any rectilinear displacement is equal to the prod- uct of the force into the effective displacement. If the direction of the effective displacement coincides with that of the force, the work is positive ; if the effective displacement is op- posite to the force the work is negative. If the actual displacement Fig. 151. Fig. 152. Fig. 153. is at right angles to the direction of the force, the effective displace- ment is zero. These three cases are illustrated in Figs. 151, 152, 153- The cases of positive and negative work may be illustrated by a body sliding along an inclined plane. Let P be the weight of the body. In computing the work done by the force P, the effective displacement is the vertical projection of the distance moved by the body. If the body descends a vertical distance //, the work is -Ph. If it rises a vertical distance h, the work is Ph. If the plane is horizontal, the work is zero. 348. Work Done by Constant Force Whose Point of Applica- tion Receives Any Displacement. The above rule for computing the work of a constant force may be extended to the case in which the displacement follows any curve AB. It may be shown that the quantity of work done by the force is equal to the product of the 302 THEORETICAL MECHANICS. magnitude of the force into the projection of AB upon aline parallel to the force. For, suppose the path to be replaced by a broken line AabB (Fig. 154), a and b being any points of the curve. The work done during the displacement along this broken line is equal to P{Aa' + a'b' + b'B') = PX AB'. It is evident that the same value of the work results if the displace- ment follows any broken line whatever joining A and B. If the d number of points such as a and b is taken greater and greater, the broken line ap- proaches the curve as a limit ; hence the work done during the motion along the curve has the value P X AB' . l? 7 ^ * It follows that the work done by a p IG I54 force which remains constant in magni- tude and direction while its point of ap- plication moves from A to B has the same value whatever the path described between A and B. An example of this is the work done by gravity during the movement of bodies near the earth's surface. 349. General Algebraic Formula for Work of Constant Force. The general rule for computing the work done by a force P which remains constant in magnitude and direction while its point of appli- cation moves from A to B along any path may be stated algebraic- ally in the form n .. , A D a . J P X (AB cos 0), in which is the angle between the straight line AB and the direc- tion of P. The work is positive or negative, according as the angle is less or greater than 90 (it may always be so measured as not to exceed 180). Since P X (AB cos 6) = (P cos 6) X AB, it is evident that the work done is equal to the product of tlie displacement into the resolved part of the force in the direction of the displacement. It is useful to remember the rule in both forms. 350. Work Done by Variable Force During Any Displace- ment of Its Point of Application. Let a force P vary in magni- tude and direction while the particle to which it is applied describes any curved path AB (Fig. 155). On AB choose any number of points a, b, c, . . . , dividing the curve into small parts. Let P iy P 2 , . be the values of the magnitude Pwhen the par- WORK AND ENERGY. 303 tide is at A, a, . . . ; and let { = angle between P x and chord Aa> 9., = angle between P 2 and chord ab, etc. Then an approximate value of the work done by the force is P x cos 6 X {A a) + P % cos d, (ab) -f P, cos 3 (&)'+'.'... The smaller the parts ^4#, #$, &. Here represents the angle between the tangent to the curve at any point and the corresponding direction of P, s is the length of the curve from some fixed point to the position of the particle, and the limits of the inte- gration must be so taken as to include the entire displacement AB. If s ' and s " are the values of s at A and at B respectively, W= C(P cos 0)ds. 351. Work of Central Force. If the force is always directed toward a fixed point, the above expression for the work may be re- duced to a convenient form as follows: Let O (Fig. 156) be the center of force, and AB the path. Let r be the radius vector of the particle measured from O, r f and r" being the values of r at A and B. Let s be measured positively in the direction AB. Drawing MN tangent to the path, is the angle OMN. Evidently, dr ds cos (w 0) = ds cos 6. The above value of the work therefore re- duces to w-^= f r Pdr If P is a known function of /', IV may be found by direct integration. 304 THEORETICAL MECHANICS. This formula shows that if a particle moves under the action of a central force which is any function of the distance from the center, the work done by the force during any motion depends only upon the initial and final distances from the center. Examples. 1. Compute the work done on a particle by an attractive force varying according to the law of gravitation. Let P be the magnitude of the force at any distance r from the center of attraction, and let P x be the value of P when r = r x . Ac- cording to the law of gravitation we have P : p = r : r' l \ or P = PirJ/r* = kjr\ k being a constant. If the body moves so that the initial and final values of r are r' and r", the work done is w rp d r=-C**=k[r--i\ 2. In Ex. 1, let the earth be the attracting body. In this case, if G denotes the weight of the body at the earth's surface, and R the radius of the earth, we have k = GR 2 ', hence W= GR 2 (i/r" i/r'). 3. If a body weighs 1,000 lbs. at the earth's surface, how much work will be done upon it by the earth's attraction while it falls to the surface from an elevation of 10,000,000 ft.? (Take R = 21,000,000 ft.) 4. A particle moves under the action of a central force varying directly as the distance from the center. Compute the work done by the force during any motion. [Since the work done depends only upon the initial and final dis- tances from the center of attraction, it is evident that it has the same value as in Ex. 4, Art. 345.] 5. A body describes a straight line while acted upon by a force of constant magnitude P lbs. directed always toward a fixed point distant h ft. from the line. Determine the work done by the force while the body passes between any two given positions. 6. In Ex. 5, let h = 10 ft., P= 14 lbs. Compute the work done by the force while the body moves 40 ft. , starting from its near- est position to the fixed point. Ans. 437 foot-pounds. 352. Work Done by Resultant of Any Number of Forces. The quantity of work done by the resultant of any number of con- current forces during any displacement of their point of application WORK AND ENERGY. 305 is equal to the algebraic sum of the quantities of work done by the several forces duri?ig tJie same displaceme?it. Let P be the resultant ; P u P 2 , . . . , the several forces ; t d lt 0. it . . . , the angles between P, P x , P 2) . . . and the tangent to the path of the particle. Since the resolved part of P in any direction is equal to the algebraic sum of the resolved parts of P x , P 2 , . . . in that direction, P cos = P x cos 6 X + P, cos *,+ >.../'., Hence (Pcos d)ds = (P x cos d x )ds + (/> cos d 2 )ds + . . . ; and C (P cos d)ds = f (P x cos 6 x )ds -f f (/, cos 0. 2 )ds + . . . . The first member of this equation represents the work done by P t while the successive terms of the second member represent the quan- tities of work done by P lf P 2i . . . , respectively (Art. 350). Hence the proposition is proved. Examples. v 1. A body weighing 50 lbs. slides a distance of 8 ft. down a plane inclined 20 to the horizontal, against a constant retarding force of 4 lbs. due to friction. Compute the total work done upon the body by its weight and the friction. Ans. 104.8 foot-pounds. 2. Prove that the total work done by gravity upon a system of particles during any motion has the same value as if the entire mass were concentrated at the center of gravity. Let the system consist of particles whose weights are W x , W 2 , . and whose elevations above a certain horizontal plane are initially g lt s t , . . . ; and let their elevations above the same plane in the final positions be z x \ gJ, .... The total work done by gravity upon all the particles is W x (z x - O + W,{z 2 gi) + . . . ^(w x z x +w % z,+ . . o-( r i^+;w+ . ). But if the elevation of the center of gravity changes from 2 to z\ W x z x +W 2 z 2 + . . . =(IV X +W 2 + . . . )2; W x z x ' + w 2 z.; + . . . ==(&V+W r ,+ . . . Yz'\ and the above value of the work is therefore equal to (j#V+ IV 2 + . . . )(2-2'), which is the same as the work which would be done by gravity on the whole mass if concentrated at the center of gravity. Evidently the same proposition holds for any body or system of bodies. 306 THEORETICAL MECHANICS. 353. Total Work Done on a Particle Expressed in Terms of Its Initial and Final Velocities and Its Mass. Let a particle describe any path under the action of any forces. Let P be the value of the resultant of all the forces at any instant and 6 the angle between P and the tangent to the path. Let m be the mass of the particle and v its velocity. The tangential component of the acceleration is dv/dt, and the tangential component of the resultant force is P cos 6. Hence there may be written the equation of motion (Art. 289) FlG ' I57 ' P cos == m(dv/dt). But dvjdt = (dv/ds)(ds/dt) == v{dv\ds)\ hence P cos 6 = tnv(dvlds), or (P cos 0)ds = mv dv. Let s' and s" be the values of s for any two positions A and B, v' and v" being the corresponding values of v. Then JCPcos 6)ds = f tmtdv = \m(z>" 2 v' 2 ). t *> V' The first member of this equation is the total work done on the par- ticle by all forces during the motion AB. The last member depends only upon the values of the velocity of the particle at A and at B. Vis viva. The quantity mv' 1 is often called the vis viva of the particle. Using this term, the above equation expresses the propo- sition that the increase in the vis viva of a particle is equal to twice the total work done upon it. This is often called the principle of vis viva. Since the development of the theory of energy, the term vis viva has largely given place to another, half the vis viva being called the " kinetic energy." The significance of this name will be explained below. The above principle should be applied in the solution of the fol- lowing examples. Examples. v 1. A body whose mass is 20 lbs. falls freely under the action of gravity. How far does it fall while its velocity changes from 40 ft. - per-sec. to 50 ft.-per-sec? Ans. About 14 ft. 2. A body is projected horizontally with a velocity of 100 ft. -per- WORK AND ENERGY. 307 sec. If acted upon by gravity alone, at what rate will it be moving when 20 ft. lower than the point of projection ? A us. 1 06. 2 ft. -per-sec. 3. A body falls vertically from a height of 5,000,000 ft. above the earth's surface. With what velocity will it reach the earth, neglecting the resistance of the air? [Use the result of Ex. 2, Art. 351.] A n s. 1 6, 1 00 ft. -per. -sec. 3. Energy of a Particle. 354. Meaning of Work Done by a Body. When a body moves against a resisting force, it is said to do work against that force. If the point of application of the force is displaced in a direction exactly opposite to that of the force, the quantity of work done by the body is equal to the product of the magnitude of the force into the dis- placement. If the displacement has some other direction, the effec- tive displacement (or projection of the actual displacement on a line parallel to the force) must be used in computing the work. Thus, a body thrown vertically upward does work against the force of gravity while rising, the quantity of work being equal to the product of the weight of the body into the distance it ascends. If it moves upward along an inclined plane, the effective displacement is the projection of the actual displacement upon a vertical line. A body moving through the air in any direction moves against the resisting force due to the air, and thus does work against that force. A body projected along a horizontal plane moves against the force of fric- tion ; in moving any distance the body does an amount of work equal to the product of the frictional force into the distance the body moves. & Let P be the magnitude of a constant force : \- f\ applied to a body at a points (Fig. 158), j \ \ and let A receive the displacement AB, mak- B' A P ing the angle 6 with the direction of P. The Fig. 158. effective displacement of A against the force is AB' AB cos 6. Hence the work done by the body against the force is - P X AB cos 0. This value is positive when 6 lies between 90 and 180 . The conception of work done by a body against a force is, however, to be so extended as to include negative values ; although it is only when 308 THEORETICAL MECHANICS. the effective displacement and the force have opposite directions that the force can properly be said to ' ' resist ' ' the motion of the body. A body moving - downward is thus regarded as doing negative work against gravity ; in algebraic language, the force of gravity is a neg- ative resistance. 355. Work Done by a Body Always the Negative of Work Done by a Force. From the foregoing definitions it will be seen that whenever positive work is done by a body against a force, neg- ative work is done by the force upon the body ; and vice versa. The two statements are merely different ways of expressing the same fact. Whatever displacement the point of application of a force may re- ceive, the work may be computed as done either by the force upon the body or by the body against the force. The two quantities are equal in magnitude but opposite in sign. All the results above found in the discussion of the work done by forces upon a particle may be extended to work done by a particle against applied forces, by merely changing the sign of the work in every case. 356. Energy of a Body Defined. When the condition of a body is such that it can do work against a force or forces that may be applied to it, the body is said to possess energy. The quantity of energy possessed by a body is the amount of work it can do in passing from its~ present condition to some standard condition. The unit of energy is the same as that of work. 357. Kinetic Energy of a Particle. If a particle is in motion, it will continue to move for a time, however great a resisting force may be applied to it. It cannot be brought to rest until it has moved over some distance. By reason of being in motion, therefore, a par- ticle can do work against any force or forces which resist the motion ; that is, it possesses energy. Since this energy is possessed by the particle by reason of its motion, it is called energy of motion or kinetic energy. A single particle can possess no energy except kinetic energy. When the definition of energy is extended to systems of particles, another form of energy will be recognized. In defining the quantity of energy of a body (Art. 356), reference was made toa " standard condition. ' ' The full meaning of ' ' con- WORK AND ENERGY. 309 dition ' ' cannot be explained until the theory of energy is extended to systems of particles and bodies in general. In case of a single particle,, the only condition affecting its capacity to do work is its velocity ; the ' ' standard condition " is a standard velocity. So long as a particle possesses any velocity, however small, it can move against resisting forces ; in estimating the energy of a particle, there- fore, the standard condition will be taken as a condition of rest. 358. To Determine the Quantity of Kinetic Energy of a Par- ticle. It will now be shown that a particle of given mass moving with a given velocity possesses a definite quantity of kinetic energy. In order to determine the quantity of energy possessed by a par- ticle of mass tn and velocity v, it is necessary to determine how much work will be done by the particle in coming to rest. Let it follow any path whatever and be acted upon by any forces. By the principle proved in Art. 353, the total work done upon the particle while its velocity changes from v' to v" is \m{v"' l v"\ The total work done by the particle is the negative of this quantity, In coming to rest from a velocity v the particle therefore does an amount of work \mv 2 . That is, The kinetic energy of a particle is equal to half the product of its mass into tJie square of its velocity. 359. Principle of Work and Energy for a Particle. The prin- ciple expressed by the equations total work done upon particle = \rniv'" 1 v' 2 ) } total work done by particle == \m(v' 2 ^" 2 )> may now be called the principle of ivork and energy for a particle. Expressed in words, During any motion of a particle its kinetic energy increases by an amount equal to the total work done upon the particle by all forces ; its kinetic energy decreases by an amount equal to the total work done by the particle against all forces. These two statements are identical in meaning, if the words in- crease and decrease are understood in their algebraic sense. This principle is extremely useful in solving certain problems. 3IO THEORETICAL MECHANICS. The equation of work and energy can often be written immediately, the work being expressed in terms of the forces acting and the dis- tances through which they act, and the change of kinetic energy in terms of the mass of the particle and its initial and final velocities. It must be remembered that a kinetic system of units (Arts. 217, 218) must be employed in writing the equation, since it is deduced from the fundamental equation of motion, P = mp. Dimensions of kinetic energy. From the formula mv i J2 i it is evident that kinetic energy has dimensions ML 2 /T 2 . This agrees with the dimensions of work when kinetic units are employed (Art. 346), as should be the case, since the equation work = change of kinetic energy must be homogeneous. The following examples should be solved by applying the prin- ciple of work and energy. The value of g (the acceleration due to gravity at the surface of the earth) is taken as 32.2 foot-second units. Examples. 1 . Compute the kinetic energy of a body of 20 lbs. mass moving at the rate of 150 ft.-per-sec. Ans. 6,990 foot-pounds. 2. If the direction of motion of the body in Ex. 1 is vertically upward and if no force acts upon the body except its own weight, how high will it rise ? (That is, how far must it move against gravity in order to do an amount of work equal to the known decrease of kinetic energy?) v 3. If the resistance of the air is constantly equal to 4 lbs., how high will the body rise? Ans. 291 ft. v 4. If a body of 10 lbs. mass is projected horizontally on a rough plane with a velocity of 50 ft.-per-sec, how far will it move before its velocity is reduced to 20 ft. -per-sec. , the retarding force due to fric- tion being constantly 5 lbs. ? Ans. 65.2 ft. 5. A body weighing 10 lbs. falls vertically under gravity against a constant force of 1 lb. due to the resistance of the air. How far must it fall in order that its velocity may change {a) from zero to 20 ft.-per-sec; {b) from 10 ft.-per-sec to 20 ft.-per-sec. ? Ans. (a) 6.9 ft. {b) 5.2 ft. 6. If a body is at rest at a distance from the earth's surface equal to the radius (taken as 21,000,000 ft.), and falls under the action of its own weight, what will be its velocity on reaching the surface ? Ans. 26,000 ft.-per-sec 7. With what velocity must a body be projected from the surface WORK AND ENERGY. 3H of the earth in order that it may never return, no force except the earth's attraction being supposed to act? Does the direction of pro- jection affect this result ? Ans. 36,700 ft. -per-sec. = 6.96 miles-per-sec. 8. A particle whose mass is 500 gr. is attached to one end of an elastic string the other end of which is fixed. The ' ' natural length ' ' of the string is 50 cm. ; under a pull equal to the weight of 500 gr. its length becomes 70 cm. The string being stretched to double its natural length, the particle is held at rest and then released. If no force acts upon the particle except that due to the string, what great- est velocity will it acquire? (Take^ = 981 C. G. S. units.) Ans. 350 cm. -per-sec. 9. Solve Ex. 8, taking account of the force of gravity, the string being assumed to hang vertically downward. Ans. 210 cm. -per-sec 4. Energy of a System. 360. Material System May Possess Energy Not Due to Mo- tion. The definition of "energy of a body " given in Art. 356 has thus far been applied only to a single particle. When the definition is applied to an actual body or to an aggregation of particles re- garded as a system, our conception of energy must be enlarged. Besides the kinetic energy which may be possessed by the individual particles, the system as a whole may possess energy of another form. In Part III will be developed the theory of the motion of a system of particles, including the general theory of energy for any material system.* We shall here give an elementary explanation of the meaning of energy of a system, depending only upon principles de- veloped in the foregoing chapters. 361. Energy of a Deformed Elastic Body. An example of a body which can do work even if initially at rest is furnished by an elastic string. Such a string, if not acted upon by external forces, assumes a certain ' ' natural ' ' length. By the application of equal and opposite forces at the ends it may be stretched, the amount of elongation depending upon the magnitude of the applied forces. If the forces are diminished, it shortens, and unless the stretching has been too great it nearly or quite resumes its original ' natural ' ' length when the forces cease to act. * See Chapter XXIII. 312 THEORETICAL MECHANICS. While in the stretched condition, the string possesses energy in accordance with the definition (Art. 356) ; its condition is such that it can do work against applied forces. It will, in fact, do work against the forces applied to the ends if these are gradually dimin- ished so as to permit the string to shorten. This energy is not kinetic ; it is not due to the motion of the body as a whole nor of its parts. It is due to the relations of the parts of the body to one another. Just what these relations are we do not know, since the ultimate structure of the body is unknown. It can only be said that the particles tend to assume certain relative positions, that any departure from these positions calls into action resisting forces, and that these forces will, if the external forces cease to act or decrease in magnitude, bring the particles back to- ward these positions. Any elastic body, while deformed and tending to resume its 1 ' natural ' ' shape, possesses energy irrespective of its condition of motion. Such energy may be called "energy of position," since it is due to the tendency of the particles of the body to assume def- inite relative positions. 362. Energy Due to Gravity. A body near the earth whose position is such that it can move to a lower level can do work against forces resisting its descent. The condition of the body is therefore such as to satisfy the definition of energy (Art. 356). This energy is, however, possessed by the system consisting of the body and the earth rather than by the body alone. It is the weight of the body that enables it to move against the forces which resist its downward motion. The amount of work which can be done against the resist- ing forces is just equal to the amount of work done on the body by the earth's attraction during the descent. This energy, like that of a deformed elastic body, may be called energy of position, since it depends upon the relative position of the bodies constituting the system and upon the action of internal forces tending to cause a definite change in their relative position. Potential energy is the name usually given to energy of position. After the foregoing illustrations of the meaning of potential en- ergy, it will be well to restate the definition of energy with explicit reference to a system of bodies or of particles. 363. Energy of a System Defined. A system of bodies or of particles is said to possess energy when its condition is such that WORK AND ENERGY. 313 it can do work against external forces which may be applied to it. The quantity of energy possessed by a system is the amount of work it will do against external forces in passing from its present condition to some standard condition. It is important to notice the word "external" in this definition. The distinction between external and internal forces has been ex- plained in Art. 119. What shall constitute a ' ' system ' ' of bodies or of particles is a matter of arbitrary selection. Thus, we may regard a single body as forming a system, or we way regard the body and the earth as form- ing a system. The body itself possesses kinetic energy if in motion ; the body and the earth regarded as a system possess energy because of their relative position. Similarly, the particles of a deformed elastic body individually possess no energy unless they are in motion ; collectively they pos- sess energy because of the internal forces which tend to cause the body to assume a definite shape. These cases illustrate the general principle that potential energy is due to the action of internal forces. 364. Choice of " Standard Condition." It is now clear that the words "standard condition" in the definition of energy refer to some definite set of velocities and positions of the members of the system. The actual value of the energy depends upon what is chosen as the standard condition. This choice is governed solely by convenience. In estimating kinetic energy it is convenient to assume rest as the standard condition for each particle, since the algebraic expression for the energy is thereby simplified. In computing potential energy it is often a matter of indifference what set of positions constitutes the standard condition. In particular applications we are concerned with the difference between the values of the energy in different con- ditions rather than with absolute values ; this difference is the same whatever be assumed as the standard condition. 365. Value of Potential Energy Due to Gravity. Potential energy due to the weights of bodies near the earth is usually briefly referred to as energy possessed by the bodies themselves ; although as above stated it is strictly possessed by the system of which the earth and the bodies are members. In computing its value, a hori- 314 THEORETICAL MECHANICS. zontal surface or ' ' datum plane ' ' must be chosen as specifying the "standard" positions of the bodies. It may be shown that the ' ' gravity ' ' potential energy possessed by a particle is equal to the product of its weight into its height above the assumed reference plane ; and that the total potential en- ergy of a system of particles is the same as if the entire mass were concentrated at the center of gravity. Let W be the weight of a particle whose height above datum is Z. Its potential energy is equal to the work it can do against resist- ing forces in descending to the standard position, i. e., to the datum plane. If the particle is at rest in the initial position and comes to rest in the standard position, the work it has done against all forces except its weight is equal to the work done upon it by its weight (Art. 359). The value of this work is Wz ; which is there- fore the value of the potential energy of the particle (strictly, of the earth and the particle regarded as a system) when it is at a height z above its standard position. Again, let a system consist of particles whose weights are W\ , W 2 , . . . , at heights z x , z % , \ . . above the plane of ref- erence, and let z be the height of the center of gravity of the system above the same plane. Then ( w, + m + . . . )* = w x z x + w,z. 2 + . . . . The second member of this equation is the total potential energy of all the particles in the supposed positions, and the first member is the potential energy of the system on the assumption that all the par- ticles are concentrated at the center of gravity. Potential energy due to the weights of bodies and to their posi- tions on the earth is one of the most important forms of energy with which we are concerned practically. In the case of streams of water, this kind of energy is available for the performance of useful work. The amount of available energy depends upon two factors, weight of water and available fall. 366. Principle of Work and Energy for a Material System. During any change of condition of a material system, its total en- ergy (kinetic and potential} increases by an amount equal to the total work done upon the members of the system by external forces ; its total energy deceases by an amount equal to the total work done by its members against external forces. WORK AND ENERGY. 315 This principle is in accordance with the definition of energy of a system (Art. 363). Thus, consider the system consisting of the earth and a particle whose mass is M pounds. If the height of the particle above its standard position is z feet and if its velocity is v feet per second, the total energy of the system is Mz -f- Mv 2 J2g foot-pounds. Suppose the elevation and velocity of the particle change, their initial values being g lt v x , and their final values z 2 , v t . If no external force (/'. e. , no force except its weight) acts upon the particle during the change, the principle of Art. 359 shows that Aft* - Bt) = mpf - v*)lig % or M{z y + vlJ2g) = M(z, + vflTg) ; that is, the final value of the total energy of the system is equal to its initial value. But if the motion is resisted by an external force against which the particle does W foot-pounds of work, the principle of Art. 359 gives the equation M(Si - *0 - W = M(v* - v^\2g, or M{z x + V*l2g) W == MQtt + vil2g). That is, the system has lost W foot-pounds of energy. As another illustration, consider a deformed elastic body. Let its ' ' natural ' ' shape be taken as the standard condition for estimating its potential energy, and suppose it to be so deformed that in return- ing to the natural shape it can do E units of work against external forces ; then (assuming its particles to possess no energy of motion) E is its total energy. Now suppose the deformation to decrease gradually until the body has done W units of work against the ex- ternal forces ; it can still do E W units of work in coming to the standard condition, so that its energy is now E IV. That is, its energy has decreased by an amount equal to the work it has done against external forces. * * The reader may notice that the principle of work and energy for a mate- rial system, in fact the very supposition that potential energy has any definite value, involves the assumption that the total work done by the internal forces during any change of condition has a definite value which is independ- ent of the way in which the change of condition takes place. For a rigorous discussion of this assumption, see Chapter XXIII. 3l6 THEORETICAL MECHANICS. 367. Machine. A machine has been defined (Art. in) as a device for the application of force. In most cases, however, the pro- duction of motion is a part of the object of a machine. Stated com- pletely, the function of a machine is to do work against external forces. Thus, a system of pulleys used to lift a heavy body against gravity does work against the force which the body exerts upon the system by reason of its weight. A steam-engine is designed to cause certain bodies connected with it to move against resisting forces ; in accomplishing this object the moving parts of the engine do work against the forces which these bodies exert upon them. In doing work a machine continually gives up energy (Art. 366), and it cannot continue to operate unless energy is supplied to it. The work done upon it by one set of external forces must on the whole be equal to the work done by it against another set. Thus, the ex- ternal force which is doing work upon a steam-engine is the pressure of the steam upon the piston. The external forces against which the engine is doing work are {a) the tension in the belt which connects it with the driven machinery, and (J?) the frictional resistances to the motion of the parts of the engine. The external work done by the engine is thus only in part utilized, that done against frictional resist- ances being lost.* Efficiency. The efficiency of a machine is the ratio of the useful work done by it to the energy supplied to it. The efficiency is always less than unity. 368. Power or Activity. The rate at which a machine does work is called its power or activity. It would be natural to take as the unit of power that correspond- ing to a unit of work done per unit time ; as a foot-pound per second, or a meter-kilogram per second, or an erg per second. Such a unit is too small to be convenient for ordinary use. The ordinary British unit is the horse-power, equal to 550 foot- pounds per second. The French horse-power, or force de cheval, is equal to 75 meter-kilograms per second. *The energy represented by the work done against friction is "lost" only in the sense of being unavailable for any useful purpose. In reality this energy is transformed, principally into heat-energy. See Chapter XXIII. WORK AND ENERGY. 317 The unit of activity commonly employed in electrical engineering is the watt, or the kilozvatt. A watt is defined as io 7 ergs per sec- ond, a kilowatt being therefore io 10 ergs per second. Examples. v 1. A stream of water is 20 ft. wide, its average depth is 3 ft., and the average velocity in the cross-section is 3 miles per hour. If there is an available fall of 200 ft. , how much available potential energy is possessed by one minute' s supply of water (or strictly, by the system consisting of the water and the earth) ? Assume the density of water to be 62.5 lbs. per cu. ft. Ans. One minute's supply of water is 15,840 cu. ft. or 990,000 lbs. In falling 200 ft. this water gives up 198,000,000 foot-pounds of potential energy. 2. In Ex. 1, compute the kinetic energy of one minute's supply of water. Ans. 297,600 foot-pounds. 3. What available H. P. is represented by the stream described in Ex. 1? An s. 6,000 H. P., neglecting kinetic energy. 4. Water is to be lifted 150 ft. at the rate of 5 cu. -ft. -per-sec. What effective H. P. must be realized by the engine and pump ? Ans. 85.2 H. P. V 5. A nozzle discharges a stream 1 in. in diameter with a velocity of 80 ft. -per-sec. (a) How much kinetic energy is possessed by the amount of water which flows out in 1 min.? (fr) If this energy could all be utilized by a water-wheel, what would be its power ? Ans. (a) 162,800 foot-pounds. () 4.93 H.'P. \ 6. In Ex. 5, suppose the jet to drive a water-wheel connected with a pump which lifts water 20 ft. If the efficiency of the whole appara- tus is 0.48, how much water is lifted per minute? Ans. 62.5 cu. ft. 7. A well 4 ft. square and 85 ft. deep is excavated in earth weigh- ing 180 lbs.-per-cu.-ft. How much work is done in lifting the earth to the surface ? 8. Show that the relative values of the units of activity above de- lined are as follows : 1 horse-power = 746 watts = 1. 01 385 force de cheval. 1 kilowatt =1.34 horse-power =1.36 force de cheval. 9. In Ex. 5, let the energy of the jet be employed in producing an electric current by means of a water-wheel driving a generator. If the efficiency of the water-wheel is 0.75 and that of the generator 0,85, what power in kilowatts is represented by the current? 318 THEORETICAL MECHANICS. 5. Virtual Work. 369. Work-Condition of Equilibrium. Let a particle, acted upon by any number of forces, receive any displacement. The work done by the resultant of the system is equal to the algebraic sum of the quantities of work done by the several forces. (Art. 352.) If the forces are in equilibrium, their resultant is zero and the work done by it is zero. Therefore, If a system, of con air rent forces is in equilibrium, the algebraic sum of the quantities of work done by them during any displacement of their point of application is zero. 370. Virtual Displacement of a Particle. Whatever forces may be acting upon a particle, it may at any instant be at rest, or it may be moving in any direction. Its displacement during a short interval may therefore have any direction ; its actual direction being determined by previous conditions. The principle stated in the last Article is not restricted to the actual displacement, but is true for any arbitrary hypothetical displacement. Any hypothetical displacement of a particle is called a virtual displacement. It may or may not coincide with the actual displace- ment. 371. Principle of Virtual Work. The work done by a force during a supposed (or virtual) displacement of its point of application is called its virtual work. The principle of Art. 369 may therefore be stated as follows : If a system of concurrent forces is in equilibrium, tJie algebraic sum of their virtual works is zero for any possible displacement. In applying this principle, it is common to assume the displace- ment to be infinitesimal. This is not necessary, however, if the forces are assumed to remain constant in magnitude and direction through- out the displacement. 372. Conditions of Equilibrium. The conditions of equilibrium for concurrent forces may be deduced from the principle of virtual work. Let A (Fig. 1 59) be the point of application of several forces in equilibrium, and let it receive a displacement AB = //. The work done by a force of magnitude P, acting in a direction inclined at angle 6 to AB, is Ph cos 6. For any number of forces, P lt P %t WORK AND ENERGY. 319 . . . , acting at angles X , 6. n . . . with AB, the total work is (P x cos 0, + P t cos t + : . . )/*. If the forces are in equilibrium, this work is zero ; therefore P x cos 0, + P., cos $ % + . . . = o. This is identical with the equation obtained by resolving forces par- allel to AB. By taking virtual displacements in different directions, any number of equations of the above form may be obtained, just as by resolving forces in different directions. It is evident that, for coplanar forces, only two of these possible equations can be independ- ent. For if the total virtual work is zero for any displacement, the only possible direction for the resultant force is perpendicular to that dis- placement ; and if the total virtual work is zero for each of two displacements which are not parallel, the resultant must be zero. 373. Forces in Three Dimensions. The principle of virtual work is readily seen to hold for forces in three dimensions. For the proposition upon which it depends, that the work done by the resultant of any forces is equal to the sum of the quantities of work done by the several forces, is true without restricting the forces to a plane. This is evident from Art. 352. It may also be seen by applying the principle first to two forces and their resultant (which are coplanar), then to this resultant and a third force, and so on. 374. Application of Principle of Virtual Work. Any prob- lem in equilibrium of concurrent forces may be solved by the principle of virtual work. The method has no advantage over the usual method of resolving forces, so long as only a single system of con current forces is involved. It is, however, useful in the solution of many problems relating to systems of particles, and systems of rigid bodies. (See Chapter XXIII.) In applying the principle of virtual work to a particle, any force may be eliminated from the equation of work by choosing the dis- placement in such direction that the virtual work of that force is zero ; that is, at right angles to the direction of the force. 320 THEORETICAL MECHANICS. Examples. t. A particle rests on a smooth inclined plane under the action of a horizontal force. Required the relation between the supporting force and the weight of the particle. Let a = inclination of plane to hori- zontal, W = weight of body, P = hori- zontal force, N = normal reaction of plane (Fig. 1 60). In order to eliminate the unknown force N y assume a displacement along the plane. Let its direction be up the plane and its length AB = h. The virtual displacement of P is + h cos a ; that of W is h sin a ; and the equation of virtual work is Ph cos a Wh sin a = o, from which P = W tan a. The value of N may be found by taking a displacement at right angles to P. 2. A bead whose weight is W is free to slide on a smooth circu- lar wire in a vertical plane. A string attached to the bead passes over a smooth peg at the highest point of the circle and sustains a weight P. Determine the position of equilibrium. In Fig. 161, AB represents the vertical diameter of the circle, C the bead, O the center of the circle. Let = angle CA O, 20 = angle COB. The bead A is acted upon by three forces : W vertically downward, P in direction CA, normal pres- sure N in direction CO. Take a displace- ment h upward along the circle. The equa- tion of virtual work is Ph sin 6 Wh sin 2d = o. Solving, cos = P/2 W. This and the preceding problem both ex- emplify the statement above made, that the equation of virtual work possesses no advan- tage over the equation obtained by resolving forces. After canceling the displacement, Fig. 161. which is a common factor in every term of the work equation, this becomes identical with the equation obtained by resolving forces in the direction of the displacement. Part III. MOTION OF SYSTEMS OF PARTICLES AND OF RIGID BODIES. CHAPTER XVIII. MOTION OF ANY SYSTEM OF PARTICLES. i . Motions of Individual Particles of a System. 375. Material System Defined. In Part II we have been con- cerned mainly with the motion of a single particle. An important part of the discussion has related to the problem of determining the motion of a particle when the forces applied to it are known. In the definition of force (Arts. 32 and 212) the fact was emphasized that a force is an action exerted by one particle upon another ; so that the dynamical equation (force = mass X acceleration) for any particle always implies that the motion of that particle is influenced by other particles. But the attention has usually been directed particularly to a single particle. It is often desirable to consider the motions of several particles collectively instead of individually. Any number of particles thus treated as a group may be called a material system. It is found that by applying the fundamental laws of motion already explained, certain important general principles can be de- duced relating to the motion of any system of particles. The following analysis will be restricted mainly to the case in which the motion is confined to a fixed plane. It will be pointed out at the end of the chapter that most of the principles deduced are true without this restriction. 376. Coordinates of Position. Two coordinates suffice to specify the position of any particle in a given plane. Rectangular coordi- nates will be employed in the following discussion. Let the system consist of any number of particles, their masses being ; 1( m. 2i . . . , and their coordinates of position {x xy y^) t 21 322 THEORETICAL MECHANICS. C^ 2 ,j^ 2 ), .... The motion is completely known if the coor- dinates of every particle are known functions of the time. 377. Motion of Mass-Center. The center of mass plays an important part in the theory of the motion of a system of particles. If its coordinates are (x, y), its position is given by the equations (?n x + m 2 -{- . . . )x = m x x x + m 2 x 2 + , . (1) (m x -f m 2 + . . . )y = m x y x + m 2 y 2 + . . . , . (2) as in Art. 152. If the coordinates of every particle are known functions of the time, x and y are also known functions of the time ; that is, the mo- tion of the mass-center is known from equations (1) and (2) as soon as the motion of every particle is known. By differentiating these equations, the velocity of the mass-center may be determined. Its x- and j-components are dxjdt, dyjdt, given by the equations {m x + m 2 -f- . . . )(dx/dt) = m x X x + m 2 x 2 + . . . , (3) {m x + m 2 + )(dyldf) = m x y x -f- m^y, -f .... (4) The axial components of the acceleration of the mass-center are d 2 xldt 2 y d 2 y/dt'\ given by the equations (m, -f m 2 + . . . )(d' 2 x/dt 2 )= m x x x + m 2 x 2 + . . . , (5) {m x + *,+_.;.. )(d 2 y/dt 2 ) = m x y x -f ;;/ 2 > 2 + .... (6) 378. Equations of Motion for Individual Particles. If /^de- notes the resultant of all the forces acting upon a particle of mass m f and/> the resultant acceleration, the general equation of motion for the particle (Art. 256) is P = mp. Such an equation may be written for every particle of a system. Let P x denote the resultant of all forces acting upon the particle whose mass is m x , and p x its resultant acceleration ; let P 2 denote the re- sultant force and p 2 the resultant acceleration for the particle of mass m 2 ; with similar notation for every particle of the system ; then there may be written the equations P x = m x p x , P 2 = m 2 p 2 , .... . (1) Each of these is a vector equation, expressing identity of direction as well as equality of magnitude, MOTION OF ANY SYSTEM OF PARTICLES. 323 If the motion is restricted to a plane, each of equations ( 1 ) yields two independent equations, obtained by resolving in two directions. If these directions are those of the rectangular axes of x and y, the equations may be written as follows : X x = m y 'x x , Vi = wiil'i X, = m t x tl Y, = m,y 2 ; > (2) In these equations X x and Y 1 are the axial components of P x ; x x and y x the axial components oip x ; etc. The complete determination of the motion of every particle re- quires the solution of this system of simultaneous differential equa- tions. In order that this may be possible, the forces acting upon every particle must be known functions of the coordinates and the time. The forces acting upon any particle are in general part external (exerted by particles not belonging to the system) and part internal (exerted by other particles belonging to the system). The values of P x , P 2 , . . . and of their axial components include all forces, both external and internal. In many cases it is possible to express the values of these forces in terms of the coordinates. But only in exceptional cases are the resulting equations sufficiently simple to admit of complete integration. It is possible, however, by properly combining equations (2), to derive certain general equa- tions which throw light upon the motion of the system as a whole. 379. Elimination of Internal Forces. If the equations obtained by resolving along the ^r-axis be added, the result is X x +X % + . . . = m l x 1 + m t x s + (3) Similarly, from the /-equations, v i + Y t -[-...= mj x + m 2 y 2 + (4) The first members of these equations include the axial components of all forces, external and internal, acting upon every particle of the system. By Newton's third law, the two forces which any two particles exert upon each other are equal and opposite ; their components in any direction are therefore equal and opposite, and the sum of these components is zero. In the sum 324 THEORETICAL MECHANICS. x, + x, + . . . the .r-components of the internal forces may therefore be omitted, their sum for the whole system being zero. The same is true of the /-components. Therefore, if X and Fare the axial components of the vector sum of the external forces applied to all particles of the system, equations (3) and (4) may be written X = m 1 x l -f f**x% +'* (5) Y = m x y x -f m t y t -f 380. Equations of Motion of Mass-Center. Using the values of the axial components of the acceleration of the center of mass, given in Art. 377, the above equations may be written X=(m l + m a + . . . )(d 2 x/dt' 2 ) Y=(m 1 + m. 2 ^ . . . ){d*j/A*). m Equations (6) show that the acceleration of the mass-center of the system is equal to that of a particle of mass (m l + m 2 -f- . . . ) acted upon by forces whose resultant has axial components X and Y. In other words, If a particle of mass equal to the total mass of the system were acted upon by forces equal and parallel to the external forces applied to the system, its acceleration would be equal to the actual accelera- tion of the mass-center of the system. In order to determine the motion of the center of mass, it is therefore needful only to know the external forces. Examples. 1. Two particles, of masses 50 lbs. and 40 lbs., are acted upon at a certain instant by parallel forces of 75 poundals and 60 poundals respectively, whose lines of action are 4 ft. apart and perpendicular to the line joining the particles. Determine (a) the position of the mass-center and (b) its acceleration at the instant named. Ans. (b) 1.5 ft.-per-sec.-per-sec, if the forces have the same direction. 2. If the two particles of Ex. 1 attract each other with forces of 40 poundals, the remaining data being as before, compute (a) the acceleration of each particle and (b) the acceleration of the mass- center. Ans. (b) 1 . 5 ft. -per-sec. -per-sec. 3. Two bodies of masses 12 lbs. and 8 lbs. respectively, con- nected by an elastic string, are projected in any manner and left to MOTION OF ANY SYSTEM OF PARTICLES. 325 the action of gravity. At a certain instant the bodies are moving horizontally in opposite directions, each at the rate of 20 ft.-per-sec. Determine the subsequent motion of the mass-center. Ans. Its motion will be that of a projectile having initially a hori- zontal velocity of 4 ft. -per-sec. 4. Take data as in Ex. 3, except that the velocity of the mass of 12 lbs. is 20 ft.-per-sec. and that of the mass of 8 lbs. is 30 ft.-per- sec. in the opposite direction, (a) Determine the motion of the mass-center, (b) If the velocities are vertical instead of horizontal, how will the solution be changed ? Ans. (a) Its motion will be that of a body falling from rest. 5. Show that the center of mass of any system of particles will move uniformly in a straight line if no external force acts upon any particle, or if the vector sum of all the external forces is zero. 2. Angular Motion of a Material System. 381. Angular Motions of Individual Particles. From the first pair of equations (2) (Art. 378) may be obtained the following : Y x x x X x y x = m x (x x y x y x x x ). . . (1) The second member of this equation may be written in the equiv- alent form ~\m x (x x y x y x x x )\ at as may be verified by differentiation. This expression has a simple meaning. Since m x x x and M x y l are the axial components of the momentum of the particle, the moment of this momentum with re- spect to the origin of coordinates is m x (x x y x y^x)- Representing this by H Xi the second member of equation (1) is equal to dHJdt. Since X A and Y x are the axial components of the resultant of all forces acting upon the particle, the first member of equation (1) is equal to the sum of the moments of these forces with respect to the origin of coordinates. Representing this by L x , the equation may be written in the form L x = dHJdt. Let a similar equation be written for every particle ; then by addition, 1 + Z : + : . . = dHJdt + dHJdt + .... (2) 326 THEORETICAL MECHANICS. 382. Elimination of Internal Forces. The first member of equation (2) is the sum of the moments of all forces, external and internal, acting upon the particles of the system. But the sum of the moments of the internal forces is zero, because the two forces acting between two particles are not only equal and opposite but are collinear, so that their moments are equal in magnitude and opposite in sign. If, therefore, the sum of the moments of the external forces is denoted by L, L = L X + L 2 + . . . . 383. Equation of Angular Motion for the System. Let the total moment of momentum of the system (i. e., the sum of the mo- ments of the momenta of the individual particles) be represented by H. Then ////,-.// . . . , and equation (2) reduces to the form L = dHldt .... (3) In deducing this equation, the origin of moments was taken at the origin of coordinates ; but this may be any point in the plane of the motion. 384. Principle of Angular Momentum. The sum of the mo- ments of the momenta of all the particles is called the angular momentum of the system (Art. 327). Equation (3) expresses the proposition that The rate of change of the angular momentum of a system about any point is equal to the sum of the moments of the external forces about that point. The following are important special cases of this general prin- ciple : If no external force acts upon any particle, the angular momen- tum of the system about every point remains 'constant. If the resultant of the external forces acts always through a fixed point, the angular momentum with respect to that point remains constant. The above equation of angular motion, together with the equa- tions of motion of the center of mass, express all that can be deter- mined regarding the motion of a material system from the external forces alone. MOTION OF ANY SYSTEM OF PARTICLES. 327 3. Effective Forces ; D' Alemberf s Principle. 385. Effective Forces Defined. The resultant of all forces (both external and internal) acting upon any particle is called the effective force for that particle. Such resultants for all the particles of a system make up the effective forces for the system. For a particle of mass tn and acceleration p t the effective force has the magnitude mp, its direction being that of the acceleration p. Therefore, if the position, mass and acceleration of every particle of the system are known, the effective' forces are completely known. They are, in fact, forces ^,A, f*%p%\ **tAi v 1 having directions identical with those of p A , p 2 , p 3 , . . . , and points of application coincident with the positions of the particles. The magnitude, direction and line of action of the effective force acting upon any moving particle are in general continuously changing. 386. Resultant Effective Force. If all the effective forces for a system be combined by the methods used in combining forces in the statics of a rigid body, the resultant may be called the resultant effective force for the system. For coplanar forces this resultant is a single force or a couple. 387. Relation Between Effective Forces and External Forces. Since the effective force for any particle is the resultant of the exter- nal and internal forces acting upon that particle, the resultant effective force for the system may be found by combining all external and internal forces acting upon all particles of the system. But the re- sultant of all the internal forces, combined in the prescribed manner, is zero, since the entire system of internal forces is made up of stresses. It follows that The resultant of the effective forces is equal in all respects to the resultant of the external forces. It should be observed that the word resultant is here used in an arbitrary sense. To say that a force is the resultant of other forces means strictly that it is equivalent to them in effect. Two or more forces applied to different free particles are not equivalent to any single force. It is, however, convenient to use the term resultant whenever it is desired to express the fact that several forces are 328 THEORETICAL MECHANICS. combined according to the same rules as in the statics of a rigid body. The meaning of the proposition stated above is that the system of effective forces and the system of external forces satisfy exactly the same conditions which are satisfied by two eqtiivalent systems of forces applied to the same rigid body. This principle leads immediately to algebraic equations of two forms: resolution equations and moment equations. Thus, (a) The sum of the resolved parts of the external forces in any direction is equal to the sum >f the resolved parts of the effective forces in the same direction. {b) The sum of the moments of the external forces about any axis is equal to the sum of the moments of the effective forces about that axis. If the motion is restricted to a plane, three independent equa- tions may be written, of which at least one must be a moment equa- tion. For motion in three dimensions six independent equations may be written.* 388. Equations of Linear and Angular Motion. The prin- ciple that the external forces and the effective forces form equivalent systems leads immediately to the equations of motion of the mass- center given in Art. 380, and to the equation of angular motion given in Art. 383. The former are obtained by resolving along the coordi- nate axes, the latter by taking moments about the origin of coor- dinates. The axial components of the effective force for the particle m x are obviously m i x 1 and m x y Xi and the sums of the axial components for all the particles are m l x l -j- tn % x t -}-...= (m x + m 2 -\- . . . )(^ 2 ^/<^/ 2 ), m \y\ ~\~ m %y> +'...= (*#! -f- m., -f- . . . ){d 2 yldt 2 ). Equating these respectively to X and Y, the axial components of the resultant external force, the resulting equations are those before found for the motion of the mass- center. Again, the moment of the effective force for the particle m x is m x {x x y x y x x^ t * These statements may be proved by reasoning similar to that employed in Arts. 104 and 173. MOTION OF ANY SYSTEM OF PARTICLES. 329 which, as shown in Art. 381, is equal to dHJdt, if H x is the angular momentum of vi x about the origin ; and the sum of the moments of all the effective forces is therefore dHjdt, if H is the angular mo- mentum of the system. Equating this to Z, the moment of the resultant external force, the result is the equation of angular motion given in Art. 383. 389. System of Particles Rigidly Connected. In general the three independent equations, free from the internal forces, which may be written in accordance with the general principle of Art. 387, are insufficient to determine the motion completely. But if the par- ticles are so connected that their motions are compelled to satisfy certain geometrical conditions, these conditions, together with the three dynamical equations, may suffice to determine the motion of every particle. An important case is that in which the particles are rigidly con- nected, so that the distance between any two particles remains in- variable. It will be seen in the following chapters that the motion of such a system is completely determined by the three dynamical equa- tions, so that the motion of a rigid system may be determined from the external forces alone. 390. D'Alembert's Principle. The principle of Art. 387, when applied to a rigid body, is known as D'Alembert's principle. It is the basis of all rules for the solution of problems relating to the mo- tions of such bodies. The foregoing discussion makes it clear that the principle is equally true for non-rigid bodies or systems. But it is only in the case of a rigid system that the principle suffices for the complete determination of the motion. 391. Motion in Three Dimensions. Although much of the foregoing discussion has referred to two-dimensional motion, most of the results may easily be extended to the case of motion in three dimensions. The general principle of the equivalence of the external forces and the effective forces (Art. 387) is obviously unrestricted. From this follow at once the equations of motion for the mass-center, which will be three in number in the general case, and which are equiva- lent to the general proposition stated at the end of Art. 380. The equation of angular motion is also true for three-dimensional motion, and may be written in the same form as in Art. 383 ; but moments 330 THEORETICAL MECHANICS. must be taken about an axis instead of a point, and by taking mo- ments about three axes three independent equations may be obtained. Thus, for three-dimensional motion three independent equations of resolution may be written, and also three independent moment- equations. In the ease of a rigid body these six independent equa- tions suffice to completely determine the motion. 392. Moment of Inertia. In the theory of the motion of a rigid body a certain quantity called the moment of inertia plays an important part. The following chapter will therefore include a dis- cussion of this quantity as a necessary preliminary to the study of the motion of a rigid body. CHAPTER XIX. MOMENT OF INERTIA. i. Moment of Inertia of a Rigid Body. 393. Definition. The moment of inertia of a body with respect to any axis is the sum of the products obtained by multiplying every elementary mass by the square of its distance from the axis. The value of the quantity thus denned is seen to depend upon the size and shape of the body, the distribution of its mass, and the position of the axis. For a given body the moment of inertia with respect to a line fixed in the body is a definite constant ; for different axes, the moment of inertia will have different values. 394. Moment of Inertia of a System of Discrete Particles. For a system of discrete particles of finite mass, the value of the moment of inertia may be computed by the formula /= m x rf + m 2 r* -(-...= 2/^r 2 , . (1) in which m x , m t , . . . denote the masses of the particles, r x , r,i . . . their respective distances from the assumed axis, and / the moment of inertia with respect to that axis. 395. Moment of Inertia of Continuous Mass. In computing moments of inertia, bodies are assumed to consist of matter distrib- uted continuously throughout space, so that any finite mass occupies a finite volume, and the moment of inertia is in general found by integration. An approximate value of the moment of inertia of a continuous mass may be obtained as follows : Let M denote the whole mass of the body, and let it be divided into small portions whose masses are AM~ l , A^f 2 , . Let r lt r ti . . . denote the distances from the assumed axis to any definite points of the elementary masses. The value of the mo- ment of inertia is approximately f=Tr*AM l + r*AM t + . . . , the approximation being closer the smaller the elements of mass are taken. 332 THEORETICAL MECHANICS. An exact value of the moment of inertia is found by determining the limit of the approximate value as all the elementary masses are made to approach zero, their number being increased without limit. If the variable r denotes the distance of the differential element of mass dM from the assumed axis, the exact value of the moment of inertia is I=fr 2 dM, . . . . (2) the limits of the integration being so assigned as to include the entire body. Choice of element of mass. The differential element of mass may be of the first, second or third order ; and the integration indicated in equation (2) may be either single, double or triple, depending upon the order of the differential element. The element must be so taken that all points in it have a common value of r. 396. Radius of Gyration. The radius of gyration of a body with respect to any axis is the distance from the axis at which a single particle, of mass equal to that of the body, must be located in order that its moment of inertia may be equal to that of the body. If M denotes the total mass of the body, / its moment of inertia with respect to any axis, and k its radius of gyration with respect to the same axis, we have I=Mk 2 ) k l = I\M. from which k can be computed when / is known. If the whole mass M be subdivided into parts M x , M 2 , . . . , whose radii of gyration are k x , k., , . . . , the mo- ment of inertia of M is I=zM x k? + M& 4- .... It may be possible to choose a differential element of M which has one or two finite dimensions, but whose radius of gyration about the given axis is known. If dM is such an element and r' its radius of gyration, I=fr'*dM. .... (3) This agrees in form with equation (2), and in fact reduces to (2) when all points of dM are equally distant from the axis, since in that case r' = r. 397. Units Involved in Moment of Inertia. Each term in the series whose sum is equal to the moment of inertia is the product MOMENT OF INERTIA. 333 of a mass into the square of a length. The unit in which moment of inertia is expressed is therefore of dimensions ML*. In using the value of the moment of inertia in the formulas of dynamics, care must be observed that proper units of mass and length are used in com- puting it. The proper units to be used in any case may be seen from the nature of the particular application. In the present dis- cussion of general principles and of methods of computing the moment of inertia, any units of mass and length may be understood. It is usually an aid in keeping the units clearly in mind to express the moment of inertia as the product of the mass into the square of the radius of gyration. The latter quantity is obviously a simple linear magnitude. 398. Body of Uniform Density. Let V denote the whole vol- ume and M the whole mass of a body, and let p be the density at any point. Then dM= pdV. If the density is uniform throughout the body, equation (2) becomes I^pffdV. . . . . (4) The method of applying this formula is shown in the solution of some of the following examples. Examples. 1 . Determine the moment of inertia and radius of gyration of a homogeneous right circular cylinder with respect to its axis of figure. Let / denote the length of the cylinder, a the radius of the base, p the density. Taking any right section, let r, be the polar coor- dinates of any point in the cross-section, the pole of the system of coordinates being at the center. If dA is a differential element of the area of the cross-section, we may take as the element of mass a differential prism of cross-section dA and length /. The general formula for the moment of inertia becomes I=fr*dM=fr*(pldA) = ptfr'dA. Expressing dA in terms of the coordinates r, 6 y its value is rdOdr\ whence ^ 2 tz r a l=pl\ I r*d6dr. / o Jo Either integration may be performed first, since the limits for each are constants. The result of the double integration between the designated limits is /= ^dlpfe = Ma 2 /2, since the total mass M is equal to ird l lp. 334 THEORETICAL MECHANICS. If k is the radius of gyration, 1= Mk 2 = Ma 2 / 2 ; k 2 = a 2 / 2 . The value of the radius of gyration is thus independent of the density. The differential element may be so chosen that only a single in- tegration is needed. Thus, if the element is taken as the mass included between two cylindrical surfaces of radii r and r -\- dr, dM = 2/pir r dr, and I=C r 2 dM= 27r/pf r A dr = iraHp^, as before. * 2. Determine the moment of inertia and radius of gyration of a homogeneous parallelopiped with respect to an axis of symmetry. Ans. If a, b are the lengths of the edges perpendicular to the axis, k 2 = O 2 + 0*)/l2. 3. What is the radius of gyration of a homogeneous square prism whose dimensions in inches are 4X4X8, with respect to an axis through the centroids of the square sections ? How does the value of the radius of gyration depend upon the density ? How upon the length? Ans. k= 1.6331ns. 4. Determine the moment of inertia and radius of gyration of a body composed of two coaxial homogeneous cylinders, the diameters of the circular sections being 4 ins. and 8 ins. respectively, and the masses 8 lbs. and 6 lbs. Ans. k 2 = 32/7, /= 64, the inch and pound being the units of length and mass. 5. A body is made up of two portions of masses J^ and M 2 , their radii of gyration with respect to a certain axis being k x and k 2 . Determine the moment of inertia and the radius of gyration of the whole body with respect to the same axis. Ans. I = M x k 2 + M,k 2 . k l = (M x k 2 + M,k 2 )j{,M x -f M 2 ). 6. Determine the moment of inertia and radius of gyration of a homogeneous sphere with respect to an axis through the center. Let a be the radius of the sphere and p its density, and let v denote the radius of a circular section perpendicular to the axis and distant z from the center. Let dM be the mass of an element in- cluded between this section and one parallel to it at distance dz. The moment of inertia of dM is (by Ex. 1) \v 2 dM= \v\irv 2 pdz) = \-rrpv'dz, and the moment of inertia of the sphere is found by integrating this differential expression between limits z = a and z = -\- a. That is, 7 = f iirpv*ds = \irp f (a 2 z 2 ) 2 dz = 87^715. J a J a MOMENT OF INERTIA. 335 Since M, the mass of the sphere, is ^Trpa 3 /^, the value of the moment of inertia is 2Ma 2 /s> and therefore k 2 = 2^/5. 7. Determine the moment of inertia and radius of gyration of a homogeneous right circular cone with respect to its geometrical axis. Let a be the radius of the base, Ji the altitude, M the mass and p the density. Let v be the radius of the circular section distant z from the vertex, and let dM be the elementary mass included be- tween this section and another distant dz from it. The moment of inertia of this element is (by Ex. i) dl= iv 2 dM= \v\irv 2 dz)p === \irprfdz. Integrating, v*dz =53 \irp I (a^/k^dz = ira'/ip/ 10. O 'o Since M = ira^hpl^, we may write / = (y?/io)M; . . k 2 = 3^/10. 8. Determine approximately the moment of inertia and radius of gyration of a homogeneous circular lamina of small uniform thickness with respect to an axis through the centroid parallel to the circular faces. Taking a circular section of the lamina through the inertia-axis, let r, 6 be the polar coordinates of any point in this section, the pole being at the center and the initial line coinciding with the inertia- axis. Let a be the radius of the circle, h the thickness of the lamina and p its density. Take as element of mass a differential prism of altitude h and cross-section rdQdr\ the moment of inertia of this element is approximately (r sin Q)\phrdQ dr). The required mo- ment of inertia is therefore, approximately, / == f 27r Cphr* sin 2 dOdr= irrfAp/4. V o J o If k is the radius of gyration, k 1 = Hird l hp = a 2 1 4 ; . . k = a/ 2. 9. Determine the radius of gyration of a spherical shell of uni- form small thickness and uniform density with respect to a diameter. Ans. If a is the radius, k 2 = \d 2 . 10. Determine the radius of gyration of a spherical shell of uni- form density and of any thickness with respect to a diameter. Ans. If the outer and inner radii are a and b, k 2 2O 5 ^)/ 5 ( 3 P). 399. Body of Variable Density. If the density is not uniform throughout the body, the factor p cannot be placed before the sign of integration. The formula for / takes the form 336 THEORETICAL MECHANICS. I = fr' i dM = fr 2 pdV. . . . (5) The integration cannot be effected unless the law of variation of p throughout the body is known. Examples. 1. Compute the radius of gyration of a right circular cylinder with respect to its axis of figure, assuming the density to be given by the equation p = p -\- cr, r being the distance from the axis and c a constant. Proceeding as in Ex. 1, Art. 398, we may write at once 1=1) ( pr' dOdr = I J ) (p + cr)r z dddr. Integrating and reducing, /= 7r/(5/v* 4 + 4^ 5 )/io = iraH(y\p x -f f? )/io, p x being the density at the outer surface. The value of M may be found by integration : M=7rd z /(2 Pl + /> )/3. Hence # = I/M = [(4^ + Po )/(2 Pl + pjfop/io). If p = p { = p, these values reduce to those found for the case of uniform density. 2. Determine the moment of inertia of a sphere with respect to a diameter, assuming the density to vary directly as the distance from the center, being zero at the center and p x at the surface. Ans. M = vc^p x \ k 2 = 4a 2 lg. 400. Relation Between Moments of Inertia With Respect to Parallel Axes. The moment of inertia of a body with respect to any axis is equal to its moment of inertia with respect to a parallel axis through the center of mass plus the product of the whole mass into the square of the distance between -.^^ the two axes. c n ;> Let Fig. 162 represent a section of y \ the body by a plane perpendicular to j ) the assumed axes, the point rep- 0' i O X resenting the central axis and O' an \^ . J axis distant a from the central axis. Fig. 162. Let 0' OX be the trace of a plane con- taining both axes ; OY and O' Y' the traces of planes perpendicular to O' OX. Let x, y be the perpen- dicular distances of any point of the body from the planes OY MOMENT OF INERTIA. 337 and OX respectively, the distances of the same point from O'Y' and O ' X being therefore x + a and y. Let / denote the moment of inertia of the body with respect to the axis 0, and /' the moment of inertia with respect to the axis O'. Then / = /(* + y 2 )dM; /' =f[(x + a) 2 + f]dMr=f{x 2 +y> + a 2 + 2**^CAf = /(* 2 + >*M#f + tffdM + 2afxdM, the limits of integration in every case being so taken as to include the whole body. In the final expression for T\ the first term is equal to / and the second to a 2 M. The third term is equal to zero, since the plane OY (from which x is measured) contains the center of mass. (See Art. 159.) Hence I' = I+Ma 2 y . . . . (1) and the proposition is proved. Relation between radii of gyration. If k and k' are the radii of gyration of the body with respect to the axes O and ' respect- ively, / = Mk\ F = Mk"\ and therefore k' 2 = k 2 + a\ . . . . (2) Examples. 1. Determine the moment of inertia and radius of gyration of a homogeneous right circular cylinder with respect to an axis coin- ciding with one of the straight elements of the surface. Ans. If a = radius of circular section, k 2 = 3a 2 / 2. 2. Determine the moment of inertia and radius of gyration of a homogeneous square prism with respect to an axis coinciding with one of the edges. Apply the result to a prism whose mass is 1 2 lbs. and whose linear dimensions in inches are 4X4X8. Ans. If the axis coincides with one of ' the longer edges, k = (8v / 6)/ 3 . 3. Determine the moment of inertia and radius of gyration of a homogeneous right circular cone whose altitude is 6 ft. and the radius of whose base is 2 ft. , with respect to an axis parallel to the geo- metrical axis and 2 ft. from it ; the whole mass being M lbs Ans. k 2 = 5.2 ft. 2 4. Determine the moment of inertia and radius of gyration of a homogeneous right circular cylinder with respect to an axis through the center of mass coinciding with the diameter of a right section. 338 THEORETICAL MECHANICS. Let / be the length of the cylinder, a the radius of the base, and p the density. Let dM be the mass included between two circular sections distant x and x -\- dx from the center of mass. The radius of gyration of dM with respect to a diameter is aJ2 ; with respect to a parallel axis through the centroid of the cylinder the square of its radius of gyration is therefore x 2 -\- a 2 / '4, and its moment of inertia is dl= (x 2 + a 2 /i)dM= (x 2 + a 2 /4.)(7ra 2 pdx). Integrating between limits x = I/2 and x = //2, I=ira 2 lp{2>a 2 + / 2 )/i2; k 2 =^a 2 + / 2 )/i2. 5. Determine the moment of inertia and radius of gyration of a homogeneous right circular cylinder with respect to an axis coin- ciding with a diameter of the base. Ans. k 2 = (3a 2 -+- 4/ 2 )/i2. 6. Determine the moment of inertia and radius of gyration of a homogeneous right circular cone with respect to an axis through the vertex perpendicular to the geometrical axis. Ans. If h is the altitude and a the radius of the base, k 2 = 3 (/z 2 + 74)/5. 401. Product of Inertia. The sum of the products obtained by multiplying every elementary mass of a body by the product of its distances from two planes perpendicular to each other is called the product of inertia of the body with respect to those planes. For a system of discrete particles, let m lt m. 2f . . . be the masses of the particles, x iy x 2 . . . their distances from one plane and y x , y 2 , . their distances from the other ; and let H denote the product of inertia. Then N== m x x x y x + m 2 x 2 y 2 + . . ' . = Imxy. . (1) For a continuous mass, let x and y denote the distances of an element of mass dM from the two planes ; then H^fxydM. .... (2) Examples. 1. Compute the product of inertia of a homogeneous rectangular parallelopiped with respect to planes containing two intersecting faces. The three linear dimensions of the body being a, b and c, let the product of inertia be found with respect to two intersecting faces parallel to the dimension c. The traces of these planes are repre- sented by OX and (9F(Fig. 163). The density being p, the ele- ment of mass dM may be taken equal to pcdxdy, and MOMENT OF INERTIA. 339 j pcxydxdy == pcftfq^ = Mabj 4. o ^0 dM X H = 2. Compute the product of inertia of a homogeneous right cir- cular cylinder with respect to a plane containing one of the bases and a plane tangent to the cylindrical surface. An s. If a = radius of base and / = altitude, H = Malh. 3. Compute the product of inertia of a homogeneous right circular cylinder with respect to a plane containing the geometrical axis and any plane perpen- q~ dicular to that axis. Ans. H = o. Fig. 163. 402. Product of Inertia of a Body Having a Plane of Sym- metry. The product of inertia of a body of uniform density with respect to two planes, one of which is a plane of symmetry, is zero. For, let x be the distance of any point from the plane of symmetry and y its distance from the other plane. Corresponding to any ele- ment of mass for which x = a, y = b, there is an equal element for which x = a, y = b. The sum of the products of inertia of such a pair of elements is zero, hence the product of inertia of the whole mass is zero. 2. Moment of Inertia of a Plane Area. 403. Definition. In the theory of the strength and elasticity of beams, columns and shafts, there is involved a quantity whose value is given by an expression analogous in form to the expression for the moment of inertia of a solid body. The importance of this quantity, and its analogy to the moment of inertia as above denned, make it desirable to give it attention here. It may be defined as follows : The moment of inertia of a plane area with respect to any axis is the sum of the products obtained by multiplying every elementary area by the square of its distance from the axis. In the most important applications the axis is taken either in the plane of the given area or perpendicular to it. The moment of in- ertia of a plane area with respect to an axis perpendicular to its plane is called a polar moment of inertia. The term radius of gyration is also used in connection with areas, being defined by the equation MP = L 34 THEORETICAL MECHANICS. in which / is the moment of inertia of the area, k its radius of gyra- tion and M the total area. It is obvious from the definition that the moment of inertia of a plane area may be computed by a method similar in every respect to that used in case of a mass, except that elements of area must be used instead of elements of mass. It may be noticed also that the proposition of Art. 400 regarding the relation between moments of inertia with respect to parallel axes holds, using total area instead of total mass. Examples. 1 . Determine the moment of inertia and radius of gyration of a circular area with respect to an axis through the center perpendicular to the plane. [Follow the method employed in solving Ex. 1, Art. 398, but use the element of area rd6 dr instead of the element of mass plrdd dr. The radius of gyration of a circular section of the cylinder has the same value as that of the cylinder. ] 2. Show that the radius of gyration of a circular area with respect to a diameter is half the radius of the circle. 3. If b and h are the sides of a rectangle, prove that its moment of inertia with respect to an axis containing the centroid and parallel to the side b is b/1 3 / 12. From this result and the proposition of Art. 400 deduce the value of the moment of inertia with respect to an axis coinciding with a side of the rectangle. Ans. k 2 = // 2 /3- 4. Compute the moment of inertia of a triangular area of base b and altitude h with respect to an axis containing the vertex and par- allel to the base. Ans. I = bfffa ; k 1 = k 2 /2. 5. From the result of Ex. 4 and the proposition of Art. 400, de- termine the moment of inertia and radius of gyration of a triangular area with respect to an axis parallel to the base and containing the centroid ; also with respect to the base. Ans. With respect to central axis, / = b/i 6 /^, k 1 = /i*/i8. With respect to base, /= b/i*/ 12, k 1 = /i 2 /6. 6. Prove that the radius of gyration of an elliptic area of semi- axes a and b, with respect to the diameter whose length is 2a, is b/2. 404. Product of Inertia of Plane Area. The sum of the prod- ucts obtained by multiplying every element of a plane area by the product of its distances* from a pair of rectangular planes is called the product of inertia of the whole area with respect to those planes. The most important case is that in which the two planes are per- pendicular to the plane of the given area. In this case the attention MOMENT OF INERTIA. 341 may be confined to the lines in which the two assumed planes intersect the plane of the given area, and the definition may be stated as fol- lows : The product of inertia of a plane area with respect to a pair of rectangular axes lying in its plane is the sum of the products obtained by multiplying every element of area by the product of its distances from the two axes. Reasoning as in Art. 402, it is seen that if one of the axes is an axis of symmetry of the given area, the product of inertia is equal to zero. It will be shown presently that, for any area whatever, there is a pair of axes through every point with respect to which the prod- uct of inertia is zero. 405. Products of Inertia With Respect to Different Pairs of Axes Intersecting in the Same Point. Let OX and OY(Fig. 164) be any pair of rectangular axes lying in the plane of a given area, and let B represent the moment of inertia with respect to OX, A the moment of inertia with respect to OY, and H the product of inertia with respect to OX and Y. Take a second pair of rectangular axes OX', OY', the an- gles XOX' and YOY f being each equal to 6. Let H' denote the product of inertia with respect to this new pair of axes. If x, y are the coordinates of any point referred to OX and Y, and x ', y' its coordinates referred to OX' and OY' , Fig. 164. x' = x cos 6 -J- y sin 6, y' = x'y' = x 2 sin 6 cos 6 -\- xy(cos 2 6 x sin -\- y cos 6 ; sin 2 6) -\- y 2 sin 6 cos 6 = xy cos 26 -j- \{y 2 .r 2 ) sin 20. Hence H' =fx'y'dM = cos 26 -fxydM + i sin 2 <9 [ffdM- = H cos 26 + i(B A) sin 2(9. fx 2 dM] If the moments of inertia and the product of inertia with respect to OX and Y are known, H' can thus be computed for any value oid. 342 THEORETICAL MECHANICS. 406. Product of Inertia Zero for Certain Axes. The axes OX ' and O V can be so chosen that H' is zero. For, equating the above value of H' to zero gives tan 2d = 2HKA B). This equation gives a real value of for any real values of A, B and H ; a pair of axes can therefore always be found for which the product of inertia is zero. Moreover, there is but one pair of axes satisfying this condition : for any two values of 26 having the same tangent differ by some multiple of 180 , so that any two values of 6 satisfying the above equation differ by some multiple of 90 . 407. Moments of Inertia With Respect to Different Axes Through the Same Point. The moment of inertia of a plane area with respect to each of two rectangular axes through a given point, and the product of inertia with respect to the same axes, being known, the moment of inertia with respect to any other axis through that point may be determined. Thus, referring to Fig. 164, let I be the moment of inertia with respect to OX'. Its value is I=fy' 2 dM=f(xs'm 6 -\-y cos 0) 2 dM = sin 2 6 -fx 2 dM + cos 2 6 -ffdM 2 sin 6 cos 6 -fxydM = A sin 2 6 + B cos 2 6 2 H sin 6 cos 6. . . . (1) If OX and (9 Fare the pair of axes for which the product of in- ertia is zero, I=A sin 2 + B cos 2 0. . (2) Examples. 1. Find the moment of inertia and radius of gyration of a rec- tangular area of sides a and b with respect to a central axis inclined 30 to the side a. Ans. k 2 = O 2 + 3# 2 )/48. 2. Prove that the moment of inertia of a square area has the same value for all axes through the centroid. 3. Determine the moment of inertia and radius of gyration of a square, the length of whose side is 4 ins., with respect to an axis through the intersection of two sides and inclined at an angle of 6o to one side. [Take the axes of x and y coincident with two sides of the square and compute A, B, H. Since His not zero, equation (1) must be used. ] MOMENT OF INERTIA. 343 4. Show that the sum of the moments of inertia of a plane area with respect to two rectangular axes lying in its plane is equal to the moment of inertia with respect to an axis perpendicular to the plane of the area and containing the point of intersection of the rectangular axes. It follows that this sum has the same value for every pair of rectangular axes drawn through the same point. 408. Principal Axes and Principal Moments of Inertia. If the values of the moment of inertia for all axes through a given point be compared, the greatest and least values are found for the axes with reference to which the product of inertia is zero. Thus, the value of / given by equation (2) of Art. 407 may be expressed in the form I=(A-B)sm*0 + B. ... (3) Suppose A greater than B. As 6 increases from o to 90, the value of /increases from B to A. As 6 increases from 90 to 180 , the value of / decreases from A to B. Hence A is the greatest and B the least value of /. The maximum and minimum values of the moment of inertia for axes through a given point are called principal moments of inertia for that point. The corresponding axes are called principal axes for that point. If the two principal moments of inertia for axes passing through a given point are equal, the moments of inertia for all axes through that point are equal. This is obvious from equation (2). 409. Axis of Symmetry a Principal Axis at Every Point. If an area possesses an axis of symmetry, it is a principal axis at every point. That is, the axis of symmetry and any line perpendicular to it in the plane of the area are principal axes at their point of inter- section. For the product of inertia for such a pair of axes is zero, as may be proved by the method used in Art. 402. Examples. 1 . Determine the principal moments of inertia of a rectangle of sides 4 ins. and 6 ins. for axes through the middle point of the longer side. 2. Determine the principal moments of inertia of an isosceles tri- angle with respect to axes through the centroid. 3. Determine the principal moments of inertia of an area con- sisting of a square and an isosceles triangle whose base coincides with 344 THEORETICAL MECHANICS. a side of the square, with respect to axes passing through the vertex of the triangle. 4. Determine the principal moments of inertia of the area de- scribed in Ex. 3 with respect to central axes. 5. For what axis, passing through the intersection of two sides of a rectangle, is the moment of inertia a maximum? Taking axes of x and y coinciding with the two sides of the rectangle, determine the values of A, B and H to be used in the formula of Art. 406 for determining 0. The result is tan 26 = 2>a6/2(a 2 b 2 ). 6. Determine the greatest and least moments of inertia of a rect- angle with respect to axes passing through the intersection of two sides. 410. Radius of Gyration With Respect to Any Axis Through a Given Point. Let <9X and OY (Fig. 165) be principal axes of a given plane area, A being the moment of inertia with respect to Y and B the moment of inertia with respect to OX, a and b the corresponding radii of gyra- tion, and M the total area, so that A = Ma 2 , B = Mb 2 . Let / be the moment of inertia and k the radius of gyration with respect to an axis OP, inclined to Then (Art. 407) I = A sin 2 + B cos 2 d; . . k 2 = a 2 sin 2 + b 2 cos 2 6. (4) 411. Inertia-Ellipse. If in Fig. 165 the point P be so taken that the distance OP depends in some assumed manner upon the value of k, the point P will describe a curve as 6 varies. If this curve is known, the value of k for any axis can be determined. The most convenient representation results from the assumption that OP is inversely proportional to k. Let the distance OP be represented by r, and assume r = ab/k, r being thus a linear magnitude. From equation (4), Fig. 165. OX at an angle 6. a 2 b 2 = a 2 sW6 + b 2 cos 2 d } (5) which is the polar equation of the curve described by the point P. \i x andjj/ are the rectangular coordinates of P, MOMENT OF INERTIA. 345 r sin 6 = y ; r cos 6 = x ; and the rectangular equation of the curve is - 2 +-=., (6) P tf which represents an ellipse of semi-axes # and b. The most convenient method of using the ellipse depends upon the following property : If a tangent be drawn to the ellipse, inclined at angle 6 to OX, the perpendicular distance from the origin to this tangent (repre- sented by p) is given by the equation * / = a 2 sm 2 d + b 2 cos'd. Therefore, comparing with equations (4) and (5), p = abjr = k. That is, the radius of gyration with respect to any axis through O is equal to the perpendicular distance between this axis and the parallel tangent to the ellipse. This ellipse is called the ellipse of inertia for the point O. It is seen that its principal diameters lie in the principal axes of inertia (Art. 408) ; the principal semi-axis lying in OX is the radius of gyra- tion with respect to OY; and the principal semi-axis lying in OY is the radius of gyration with respect to OX. Central ellipse. For every point in the plane of the area, there is a definite inertia-ellipse whose center is at that point. The ellipse whose center is at the centroid of the area is called the central ellipse. Examples. 1. Determine the central ellipse for a rectangular area. 2. Show that the central ellipse for the area of an ellipse is a sim- ilar ellipse. What are its semi-axes? Ans. a/ 2 and 6/2. 3. Determine the central ellipse for an isosceles triangle. 4. Show that the central ellipse for an equilateral triangle is a circle, and determine its radius. Ans. (#i/6)/i2, if a is the side of the triangle. 5. Determine the central ellipse for a regular hexagon. Ans. A circle of radius equal to tfl/5/24. * Smith's "Conic Sections," Art. 115. CHAPTER XX. MOTION OF A RIGID BODY : TRANSLATION J ROTATION ABOUT A FIXED AXIS. i. Simple Motions of a Rigid Body. 412. Coordinates of Position Any quantities whose values serve to specify the position of every particle of a body are called its coordinates of position. It may be shown that six coordinates are sufficient to specify the position of a rigid body which is free to move in three dimensions of space. These six quantities may be chosen in various ways. To show that six coordinates are sufficient, notice that if three points in the body, not lying in a right line, are fixed, every point is fixed. Let A, B, Cbe three such points. The position of any one, as A, is given by its three rectangular coordinates x x , y x , z x with re- spect to an assumed set of fixed axes. The position of B is then determined by the values of any two of its three rectangular coor- dinates x. z , y % , z % , since its distance from A is fixed. When the posi- tions of A and B are given, that of C is determined by the value of one of its three rectangular coordinates, since its distances from A and B are fixed. Thus, to completely specify the positions of A, B and C (and therefore of all points of the body) it is sufficient to assign values to six of the nine quantities x x , y lt z lf x 2 , y 2 , z 2 , 413. Degrees of Freedom. The six coordinates whose values specify the position of a rigid body may be varied independently of one another ; for example, one may be varied while the other five remain constant. Such a variation causes the body to move in a particular way. This is often expressed by the statement that the body possesses one degree of freedom for every coordinate. A rigid body whose motion is unrestricted possesses six degrees of freedom. By imposing restrictions on the motion of a body the number of degrees of freedom may be artificially reduced. The following dis- cussion will deal mainly with motions so restricted as greatly to simplify the application of dynamical principles. SIMPLE MOTIONS OF A RIGID BODY. 347 414. Motion of Translation. If the velocities of all particles of a body are at every instant equal in magnitude and direction, the body is said to have a motion of tra?islation. In a motion of translation a particle of the body may describe any path in space. If no additional restriction is imposed, the body possesses three degrees of freedom. The coordinates of position of the body may, for example, be the rectangular coordinates of any one particle. The paths of all particles are alike in all respects. 415. Rotation About a Fixed Axis. If all particles of a rigid body describe circles whose centers lie in a certain fixed line, the motion is a rotation. The fixed line containing the centers of the circles described by the particles is the axis of rotation. If the motion is restricted to a rotation about a fixed axis, the body possesses only one degree of freedom. The position of the body is completely specified by the value of one coordinate, which may, for example, be the angle between two planes containing the axis of rotation, one of which is fixed in space and the other fixed in the body. Let Fig. 166 represent a section of the body by a plane perpen- dicular to the axis of rotation, and let OX and OA be the traces of planes containing the axis, OX being fixed while OA turns with the body. The position of the body at any instant is completely specified by the value of the angle A OX. Call this 6. If is known as a function of the time the motion is completely determined. Angular motion. All planes con- taining the axis of rotation or parallel to Fig. 166. . it turn through equal angles during any definite time ; hence the angular motion of any one of them as OA may be called the angular motion of the body. This angular motion may be specified as in Art. 281. The angle turned through in any definite time is the angidar displacement. The angular displacement per unit time is the angular velocity. The increment of the angular velocity per unit time is the angular acceleration. The value of the angular displacement of the body during any time At is 348 THEORETICAL MECHANICS. if l and 2 are the initial and final values of 6. If ft) represents the angular velocity of the body and its angular acceleration at any instant, their values may be expressed as in Art. ft) == d0/dt; (j> = dco/dt == d 2 0/dt\ 416. Plane Motion. Plane motion of a rigid body is a motion in which all particles move in parallel planes. This is also called uniplanar motion. A body whose motion is thus restrict- ed has three degrees of freedom. This is evident by considering that the posi- tion may be completely specified by the values of three coordinates, as follows: Let OX, OY (Fig. 167) be a pair of fixed rectangular axes lying in a plane q ~X parallel to the motion. Let A and B be p IG# !6 7- the positions of any two particles of the body lying in the plane XO V ; and let x, y denote the rectangular coordinates of A and the angle be- tween AB and OX. The position of the body is completely specified if values are assigned to x, y and 6. The general case of plane motion will be considered in Chapter XXI. 417. Determination of Motion of Mass-Center. It has been shown (Art. 380) that the motion of the center of mass of any system of particles depends only upon the external forces. To determine this motion, the entire mass of the system is conceived to be concen- trated at the center of mass and to be acted upon by forces which at every instant are equal in magnitude and direction to the external forces acting upon the particles of the system. As an example of the application of this principle, consider the motion of a projectile. If a body is thrown in any way and is then acted upon by no force except its weight, the motion of the center of mass may be determined by the results of Art. 295, which were de- duced for the case of a particle. For the resultant of the external forces is constant in magnitude and direction during the motion. The same will be true of two or more bodies connected by cords or otherwise ; or of any set of bodies whatever, regarded as a system. ROTATION ABOUT A FIXED AXIS. 349 418. Condition That Motion May Be a Translation If the motion of a rigid system throughout any interval is known to be a translation, the equations of motion of the mass-center suffice to determine .the motion of every particle, since the paths of all particles are alike in every respect. It remains to consider under what con- ditions the motion will be a translation. This question may be answered by applying D'Alembert's principle (Art. 390) that the external forces and the effective forces form equivalent systems* Resultant effective force. In a motion of translation, the accel- erations of all particles of the body are at every instant equal in magnitude and direction. The effective forces (each being equal to the product of the mass of a particle into its acceleration) form a system of parallel forces whose magnitudes are proportional to the masses of the particles. The resultant of this system evidently acts in a line passing through the mass-center, and its value is Mp if M is the total mass and/ the acceleration. Since the resultant external force and the resultant effective force are at every instant equal in all respects (magnitude, direction and line of action), it follows that the line of action of the resultant of the external forces must pass through the mass-center of the system in order that the motion may continue translatory. The converse of this proposition is not necessarily true. But if, at any instant, all particles have equal and parallel velocities, the motion will continue to be translatory if the resultant of the external forces acts in a line containing the center of mass. 2. Rotation About a Fixed Axis. 419. Rotation Under Any Forces. If a body is constrained to rotate about a fixed axis, it possesses but one degree of freedom, and one dynamical equation is sufficient to determine the motion if all external forces are known. Generally, however, certain of the ex- ternal forces are unknown, and additional equations are needed in order to determine them. The method by which, practically, the motion of a body can be restricted to rotation about a fixed axis, is by means of a hinge joint (Art. 42) or something equivalent. The force or forces exerted upon * The reader should bear in mind the remarks made in Art. 387 as to the meaning of this principle. 350 THEORETICAL MECHANICS. the body by the hinge are in general unknown, and are to be deter- mined by means of the dynamical equations. D' Alembert's principle yields three independent equations for plane motion (Art. 388), and these suffice to determine the motion and the restraining - forces in case of rotation about a fixed axis.* At least one of the three equa- tions must be an equation of moments. For the other two, equations of resolution will usually be preferred. 420. Moments of Effective Forces About Axis of Rotation. Let D' Alembert's principle be applied, taking moments about the fixed axis of rotation. The sum of the moments of the effective forces about this axis may be computed as follows : Effective force for a particle. Every particle of a body rotating about a fixed axis describes a circle with center in that axis. For any particle the resultant acceleration may be specified by its com- ponents along the tangent and normal to the circle. If r is the distance of the particle from the center of rotation and v its velocity, the two components are as follows (Art. 284): Tangential component, dvjdt\ Normal component, v 2 jr } directed toward the center. If co is the angular velocity and the angular acceleration, v is equal to rco, and since r is constant, dv\dt = r{dco\df) = rep ; v 2 /r == r 2 co 2 /r == rco 2 . If the mass of the particle is m, the effective force is equivalent to the two components mr, directed along the tangent to the circle ; mrco\ directed toward the center. * If the applied forces tend to give the body motion parallel to the axis of rotation, the method of constraint must be such as to counterbalance this tendency, and the restraining forces will have components parallel to the axis. Again, the inertia of the particles will in general tend to cause a departure from the prescribed plane motion, which tendency must be resisted by the constraining forces. In the most general case, six dynamical equations are needed in order to completely determine the constraining forces. The fol- lowing discussion is restricted to the case in which the applied forces and the distribution oi mass are such that there is no tendency to depart from a condition of plane motion, so that the three dynamical equations for plane motion are sufficient to determine the constraining forces as well as the motion. ROTATION ABOUT A FIXED AXIS. 35 1 Since the moment of the latter component about the axis of rotation is zero, the moment of the effective force is equal to mr 2 . Total moment of effective forces. If the masses of the particles of the system are m x> m %i . . . , and their distances from the axis of rotation r x , r t , . . . , the sum of the moments of the effective forces has the value (m,r 2 -f- m 2 r 2 + . . . )< = I(j>, I being the moment of inertia of the system with respect to the axis of rotation. 421. Equation of Angular Motion. The equation of angular motion is obtained by equating the sum of the moments of the ef- fective forces to the sum of the moments of the external forces. Let L denote the algebraic sum of the moments of the external forces with respect to the axis of rotation ; then L = I. If 6 denotes the angle between some line in the system and a fixed reference line, the second member of the equation may be writ- ten in either of three forms, as follows : L = f(t> = I(dco/dt) = /(d'^/dt 2 ). . . (I) If L is known, or can be expressed in terms of one or both of the variables and /, the integration of equation (i) serves to deter- mine the motion completely, provided sufficient initial conditions are known for the determination of the constants of integration. 422. Hinge Reaction. The rotation of a system about a fixed point is a case of constrained motion (Art. 303); a certain condition is im- posed upon the motion without specifi- cation of the forces necessary to maintain that condition. In the case supposed, the constraint must be effected by some- thing equivalent to a hinge at the center of rotation. Thus, if O (Fig. 168) is the center of rotation, any tendency of the body to depart from the stated mo- Fig. 168. tion is resisted by forces equivalent to a single force applied at O. This force may be called the hinge reaction. 352 THEORETICAL MECHANICS. Suppose the hinge to consist of a fixed cylindrical pin fitting into a hole in the rotating body. If the surfaces were smooth at the point of contact, the pressure of the pin upon the body would act in a line passing through the axis of rotation. In reality it will depart from this direction because of the friction. Let F be the frictional force and R the normal force ; then the moment of R with respect to the axis of rotation is zero, so that R does not affect the value of L. The moment of F must be included in L, its sign being always oppo- site to that of the existing rotation of the body. In the following discussion the friction will either be neglected or will be assumed to be included among the known external forces whose moment is L. The hinge-reaction will be regarded as acting always in a line which intersects the axis of rotation. 423. Equations of Motion of Mass-Center. The value of the hinge-reaction may be determined by writing the equations of motion of the center of mass. If P is the vector sum of all external forces (including the hinge-reaction), M the mass of the body, and/ the acceleration of the mass-center, these quantities satisfy the vector equation P = Mp. This has been proved (Art. 380) for any system of particles, whether rigid or not. In the case of a body rotating about a fixed axis, p may be ex- pressed in terms of the angular veloc- ity, angular acceleration, and distance of the center of mass from the axis of rotation. Let Fig. 169 repre- sent a section of the body by a plane perpendicular to the axis of rotation, containing the mass-center G ; OG = a = perpendicular from mass-center on axis of rotation ; = angle between OG and a fixed plane OA. Then/ is equivalent to the two components a(d 2 6/dt 2 ) perpendicular to OG, a(d6/dt) 2 in direction GO. Let all external forces be resolved into components respectively perpendicular and parallel to GO. Let T, N be the components of the hinge-reaction ; P t , P the components of all other external forces. The positive directions for the resolution must agree with those used in resolving/ ; they are shown by the arrows in Fig. 169. The figure shows P and P t as if applied at G, but it is not to be Fig. 169. ROTATION ABOUT A FIXED AXIS. 353 inferred that this is necessarily the fact ; we are not here concerned with their lines of action. The equations of motion of the mass-center may then take the forms P t + T= Ma(d 2 6/dt 2 ); . . . (2) P n + N= Ma(deidt)\ . . . (3) 424. Complete Solution of Problem of Rotation. The com- plete solution of the problem is contained in equations (1), (2) and (3), which for convenience will be written together here : L = I(d- 1 6ldt 2 ); . . . (1) P t + T = Ma(d*ejdt*)\ . . . (2) P n + N = Ma{dd\df)\ . . . (3) Equation (1) does not involve the hinge-reaction, and if all other external forces are known, the motion is determined by integrating this equation. Equations (2) and (3) serve to determine T and iV, the components of the hinge-reaction, after the motion is known. It will be observed that the complete integration of (1) is not neces- sary in order that T and N may be determined from (2) and (3). Examples. 1. A body of mass M rotates about a fixed axis distant a from the mass-center. If no force acts upon the body except the hinge-reac- tion, and if its angular velocity at a certain instant is co l , determine (a) the subsequent motion and (&) the hinge reaction. (a) In equation (1), L = o, hence d'djdt 2 = dcaldt = o ; ddldt = to = constant = co 1 ; = ay J -f $ 1 ; 6 l being the value of 6 when t = o. The angular velocity thus re- mains constant. (b) The resultant acceleration of the mass-center is aco\ directed toward the center of rotation ; the resultant effective force is there- fore Maco 2 in that direction. Since the hinge-reaction R is the only external force, it must be equal in all respects to the resultant effective force. It is evident that this result is given by equations (2) and (3). Since P t and P n are both zero, (2) becomes T = 0, and (3) reduces to N = Maw 2 . 354 THEORETICAL MECHANICS. 2. A body of mass ioo lbs. rotates about a smooth hinge 2 ft. from the mass-center, being acted upon by no external force except the hinge-reaction. If at a certain instant it is rotating at the rate of 2 rev. -per-sec. , what is the subsequent motion, and what is the value of the hinge-reaction ? Ans. Hinge-reaction = N = 3,200 ir 2 /g pounds- weight. 3. A body of mass 200 lbs. , acted upon by gravity, rotates about a smooth horizontal axis passing through the mass-center. If at a certain instant the rate of rotation is 3 rev. -per-sec. , determine the subsequent motion and the hinge-reaction. [Notice that if the axis of rotation contains the mass-center, a = o, and equations (2) and (3) simplify.] 4. A homogeneous cylinder of mass 200 lbs. and diameter 2 ft. rotates about its axis of figure (horizontal) under the action of a con- stant pull of 5 lbs. applied to the free end of a string which is wrapped around the cylinder. At a certain instant it is rotating at the rate of 200 rev.-per-min. against the pull. Determine the subsequent mo- tion. When will the body come to rest ? Determine the hinge- reaction, assuming the pull to be vertically downward. Ans. It will come to rest in 13 sec. Hinge-reaction = 205 lbs. 5. In the preceding example, let the tension in the string be due to a suspended weight of 5 lbs. , the remaining data being as before. Solve the problem completely. [Write the equation of angular motion of the cylinder and the equation of linear motion of the suspended weight ; then eliminate the tension in the string. ] Ans. It will come to rest in 13.7 sec. Hinge-reaction = 204.1 lbs. 6. A wheel-and-axle of total mass 60 lbs. is set rotating by a con- stant tension of 7 lbs. -force in the rope which unwinds from the axle. The radius of gyration of the body with respect to the axis of rotation is 10 ins. and the radius of the axle is 6 ins. Required (a) the angular velocity after 2 sec. , (d) the angle turned through in 2 sec. , and (Y) the pressure on the hinge. (Assume no friction. ) Ans. (a) o. 1 68^ rad. -per-sec. (fi) o. 1 68g* rad. (c) 6jg poundals. 7. In the preceding example, instead of a tension of 7 lbs., as- sume a tension due to the weight of a suspended body of 7 lbs. mass ; solve the problem completely. 8. In example 6, what weight suspended from the rope would produce a tension of 7 lbs. -force? Ans. 7.31 lbs. 425. Compound Pendulum. A rigid body rotating about a horizontal axis under the action of no external force except gravity and the hinge reaction, is called a compound pendulum. In Fig. 170, let the axis of rotation, perpendicular to the plane of ROTATION ABOUT A FIXED AXIS. 355 the figure, be projected at O, and let A be the center of mass. Let B be the lowest point reached by A, and A the highest point. Let angle AOB = 0, A OB = O , AO = tf, M= mass of body, k = its radius of gyration with respect to the axis of rotation, / = Mk 2 = moment of in- ertia with respect to that axis. The only external force, except the hinge reaction, is the weight of the body, its value in kinetic units being Mg and its direction vertically downward. The mo- ment of this force about the axis of rota- tion is L Mga sin 0, the negative sign being used because, in writing the equation of angular motion, the positive direction for L must agree with that for 0. Equation (i) therefore reduces to the form d 2 0/dt 2 = (ga/&) sin 0. (4) The integration of this equation will determine the motion. Equivalent simple pendulum. The equation of motion of a simple pendulum (a particle suspended freely from a fixed point by a flexible string without weight) was found in Art. 309 to be d 2 0/dt 2 = (g/l) sin 0, I being the distance of the particle from the point of suspension. This is identical with the equation just derived for the compound pendulum if / = &/a. The motion of the compound pendulum is therefore the same as that of a simple pendulum of length k 2 \a. Let OA (Fig. 170) be produced to a point 0' so taken that 00' = I = k l ja. The point O is called the center of suspension and O' the center of oscillation. The following considerations show that as O approaches the mass-center O' recedes from it. Let k 9 be the radius of gyration with respect to an axis through the mass-center A parallel to the axis of rotation ; then k 2 = k 2 + a 2 ; / = P/a = a + k 2 ja ; ,\ (/ d)a = k 2 , or OA X 0'A= k*. 356 THEORETICAL MECHANICS. That is, the product of the distances of and ' from the mass- center is constant. If the center of suspension is very near the mass-center, the value of / becomes very great ; if is very distant from the mass- center, / differs little from a. From the above relation between OA and O'A it is seen that if O' be made the center of suspension, O becomes the center of oscil- lation. For the method of integrating equation (4), and of determining the time of a small oscillation, reference may be made to Art. 309. All the results there found for a simple pendulum may obviously be applied to an equivalent compound pendulum. The first integration of equation (4) gives (d6jdt) 2 = (2tf//P)(C0S cos O ). . . (5) The solution of this equation involves an elliptic integral. But for the case of a pendulum oscillating through a small angle, an ap- proximate solution is obtained by putting 6 for sin 6. With this substitution, the integration of equation (4), subject to the condi- ditions d0/dt = o when 6 = , 6 = o when / = o, gives 6 = e sin (/ \ / ~ga\). . . . (6) Examples. 1. If the center of suspension of a homogeneous circular disc of uniform thickness is in the circumference, determine the center of oscillation (a) if the axis of suspension lies in the plane of the disc and (b) if it is perpendicular to that plane. Ans. (a) 00' = 5^/4; (b) 00' = 30/2 ; a being radius of disc. 2. The center of suspension of a homogeneous rectangular plate of sides a and b is distant r from the center of mass ; required the center of oscillation, the axis being perpendicular to the plane of the rectangle. Ans. {a 1 + ^ 2 )/i2r = distance from center of mass. 3. If, in the preceding example, a = 40 ins. , b = 2 ins. and r= 18 ins., determine the center of oscillation and the time of a small oscillation. Ans. Time of oscillation = 0.81 sec, nearly. 4. A homogeneous straight bar, whose cross-section is very small in comparison with its length, is suspended at any point. Determine the position of the center of oscillation. 5. A homogeneous bar of small uniform cross -section, 6 ft. long, is suspended so as to rotate without retardation by friction. Deter- mine the length of the equivalent simple pendulum when the distance ROTATION ABOUT A FIXED AXIS. 357 of the point of suspension from the end has each of the following values : o, i ft., 2 ft., 3 ft. For what position of the point of sus- pension will the length of the equivalent simple pendulum be 20 ft? Ans. In first and third cases, / = 4 ft. 6. Determine the hinge-reactions in case of a compound pendu- lum. Substitute in equations (2) and (3) (Art. 424) and solve for T and N. The only external force aside from the hinge-reaction is the weight of the body ; its components are P t = Mg sin 6, P n == Mg cos 6. Equation (4) gives the value of d 2 0/dt 2 , and (5) the value of {dO/dt) 2 . Hence T = Mg(i a 2 jk 1 ) sin ; . . . (7) N = Mg[(2a 2 /& 2 ) = dco/dt = d 2 6jdt 2 the angular acceleration, the value of the resultant effective force is as follows : Its magnitude is Mp ; its direction is that of p ; its moment with respect to the axis of rotation (Art. 420) is Mk 2 , hence its line of action is at a distance k 2 /p from the axis of rotation. This result may conveniently be put in another form, as follows : 358 THEORETICAL MECHANICS. In Fig. 171, let O represent the axis of rotation, A the mass- center, and O' the point in which the line of action of the resultant effective force Mp intersects OA. Let OA = a, 00' = h. Let Mp be replaced by two components acting at O r , one acting in the direction O'O, the other perpendicular to that direction. The components of p in these directions are axo 2 and a ; hence the components of Mp are Maco* and Ma. The sum of the moments of these com- ponents about O is Ma/16, Fig. 171. which must equal the moment of the resultant effective force. But (Art. 420) this is also equal to Mk l $ ; hence ah = 2 , or h = k 2 /a. The point O' thus coincides with the point called the center of oscillation in case of a compound pendulum. Resultant couple. If the axis of rotation contains the center of mass,/ = o, hence the vector sum of the effective forces is zero. Their moment Mk 2 is not in general zero. In this case the re- sultant effective force is a couple. Resultant zero. If the axis of rotation contains the mass-center and the angular velocity is constant, the ipoment of the resultant couple is zero and the resultant is therefore zero. Motion under no external forces. By D'Alembert's principle, the resultant effective force is zero if the resultant external force is zero. Hence if the external forces are balanced, the body can have no motion except a uniform rotation about an axis containing the center of mass. In this case the hinge-reaction is equal and opposite to the resultant of all other external forces, and is zero if this result- ant is zero. 427. Unsymmetrical Body. In the above analysis it has been assumed that the effective forces are coplanar. While it is true that, in the case of rotation about a fixed axis, the effective forces are in ROTATION ABOUT A FIXED AXIS. 359 parallel planes, it is not true that they are coplanar. In certain cases, however, they may be seen to be equivalent to a coplanar system. This will be true if the body is of uniform density and has a plane of symmetry perpendicular to the axis of rotation. For if the mass be regarded as made up of prismatic elements whose axes are parallel to the axis of rotation, the resultant effective force for every element will lie in the plane of symmetry. Even if there is no plane of sym- metry, the mass may be so distributed with reference to a certain plane that the resultant of the effective forces lies in this plane. The conditions which must be satisfied in order that this may be true cannot here be discussed. In the general case of an unsymmetrical body, the six dynamical equations for forces in three dimensions are needed for the complete determination of the relations among the external forces. The motion may, however, in all cases be determined from equation (i) of Art. 421. CHAPTER XXL ANY PLANE MOTION OF A RIGID BODY. i. Nature of Plane Motion. 428. Coordinates of Position. Plane motion of a rigid body has been denned in Art. 416. It was there pointed out that in the general case of plane motion three coordinates serve to specify the position of the body completely. In analyzing plane motion it is sufficient to consider a single plane section parallel to the motion, since the position of every particle is determined if the positions of the particles in this plane section are known. In the following analysis little use is made of coordinates of position, but the geometrical relations are studied directly. When coordinates are employed, they will generally be chosen (as in Art. 416) as the rectangular coordinates of some particle A, and the angle between some fixed line and the line joining two particles. The point A may be chosen arbitrarily, but will usually be taken at the mass-center ; the plane of the coordinate axes being so chosen as to contain that point. 429. Displacement. The displacement of every particle of a rigid body (restricted to plane motion) is determined if the displace- ments of two particles are known. The simplest displacements of a body are translation and rotation. The displacement is a translation if all particles receive equal and parallel displacements. The direction of the straight line joining any two points is left unchanged by a translation. The displacement is a rotation if every particle describes the arc of a circle ; the centers of the circles all lying upon a straight line called the axis of rotation. In the following discussion, the atten- tion being directed to a section of the body by a plane parallel to the motion, the axis of rotation is represented by the point in which it pierces this plane. This point is called the center of rotation ; and the terms center and axis are often used interchangeably. The total displacement during any interval is called a translation or a rotation if the change from the initial to the final position can PLANE MOTION OF A RIGID BODY. 36 1 be produced in the manner specified in the foregoing definitions, whatever the series of intermediate positions through which the par- ticles actually pass. 430. General Plane Displacement of a Rigid Body. Every possible plane displacement of a rigid body is equivalent either to a rotation or to a translation. The displacement is completely known if the displacements of two particles are known. Let two par- ticles initially at A and B (Fig. 172) be displaced into the positions A' and B' . Bisect AA' at C and BB' at B, and draw from C a line perpendicular to A A' and from D a line perpendicular to BB'. If these lines are not parallel, let O be their point of intersection. The triangles OAB and OA'B' are equal in all their parts, since by construction OA = 0A\ OB = OB', AB = A'B'. Therefore the angles A OB and A' OB' are equal, and the angle AOA' is equal to the angle BOB'. It is evident that the triangle OAB may be brought into co- incidence with OA'B' by rotation about O through an angle AOA'. ~g Suppose next that A A' and BB' are par- Fig. 173. allel. In this case either A A' and BB' are equal (Fig. 173) and the total displacement is a translation, or else AB and A'B' are re- lated as shown in Fig. 174, and AB may be brought to coincide with A'B' by a ro- tation about their point of intersection 0. The proposition is thus true for all cases. A translation may be regarded as a rotation about an infinitely distant axis. Thus, Fig. 173 represents a limiting case of \ / Fig. 172, the lines CO and DO becoming v parallel and the point falling at infinity. IG ' I74 ' 431. Instantaneous Translation and Instantaneous Rotation. The motion may at any instant be translatory, even though it does A* 4 i i B 1 i I i 362 THEORETICAL MECHANICS. not continue to be so for any finite time. Again, the body may at any instant be rotating about a certain axis, even though this condi- tion does not continue for a finite time. Instantaneous translator)' motion. The instantaneous motion is translatory if the velocities of all particles are equal in magnitude and direction at the instant considered. Instantaneous rotation. The instantaneous motion is rotational if, at the instant considered, all particles in a certain axis are at rest. This axis is called the axis of instantaneous rotation, or simply the instantaneous axis. Every particle must be moving at right angles to the perpendicular drawn from it to the instantaneous axis, and the velocities of different particles must be proportional to their distances from the axis. 432. Nature of Instantaneous Motion in General Case. Any possible plane motion of a rigid body is at every instant either a translation or a rotation. Whatever the motion of the body, let MM' and NN' (Fig. 175) be the paths described by two particles. Let A, B be the positions of the particles at a certain instant, and A\ B' their positions after a short interval. In general the total displacement is equivalent to a rotation about a center 0, determined by the intersection of the perpendicu- lar bisectors of A A' and BE' (Art. 430). Draw circular arcs AA', BE', /\\ v N N |\ \ with common center O. If the inter- val be made to approach zero, these > circular arcs approach coincidence V with the chords AA\ BE', which themselves approach coincidence with 2T the tangents to the paths at A and Fig. 175. B ; and the point O approaches a limiting position Q, determined by the intersection of the normals to the paths at A and B. The in- stantaneous motion is therefore a rotation about Q. This point is the instantaneous center ; the line through Q perpendicular to the plane of motion is the instantaneous axis. In particular cases the tangents to the paths at A and B may be parallel ; if the normals are not coincident, the point Q is at infinity. In such a case the velocities of A and B are equal and parallel and PLANE MOTION OF A RIGID BODY. 363 the instantaneous motion is translatory. Translation is thus a special case of rotation, the instantaneous axis being at infinity. If the normals to the paths at A and B are coincident, their point of intersection is indeterminate. This case may occur if the motion is translatory and A Bis perpendicular to the direction of the instan- taneous motion ; or if the motion is a rotation about a center lying in AB. This case may be avoided by replacing one of the points B by another C such that AC is not parallel to AB. (Compare the third case of Art. 430. ) As the body moves, the position of the instantaneous center Q (Fig. 175) in general varies continuously, both in the body and in space. The curve described by it is called a centrode ; the curve traced in space is the space centrode, and the curve traced in the moving body is the body centrode. It is evident from the above discussion that the position of the instantaneous center can be determined if the directions of motion of two particles are known, unless both are moving at right angles to the line joining them. 433. Angular Motion of a Body. Angular Displacement. In any plane motion of a rigid body, all lines fixed in the body and parallel to the plane of the motion turn through equal angles in any interval of time. The angle turned through by any such line is called the angular displacement of the body. If 6 is the angle between a line fixed in the body and a line fixed in space (as in Art. 416), the angular displacement is equal to 6" 0', 6' being the initial and 6" the final value of 6. Angular velocity. The angular velocity is the angular displace- ment per unit time. Representing its value by co f it is given by the formula } its value is = dco/dt = d>0/dt\ These formulas for the angular motion have the same form in the general case of plane motion as in the particular case of rotation about a fixed axis (Art. 415). 434. Motions of Individual Particles. If at any instant the angular velocity and the instantaneous axis are known, the velocity 364 THEORETICAL MECHANICS. of every particle is known. If r is the perpendicular distance of the particle from the instantaneous axis and &> the angular velocity, the particle has a velocity rco at right angles to r. If v x is the velocity of a particle A and v 2 that of a particle B, the resolved parts of v x and v 2 in the direction AB are equal at every instant. For otherwise the distance AB would not remain constant. Examples. 1. The ends of a straight bar slide along intersecting lines at right angles to each other. Determine the instantaneous center at any instant. 2. If, in Ex. 1, the length of the bar is a and its angular velocity a), determine the velocity of each end and of the middle point at the instant when the length of the bar coincides with one of the guiding lines. 3. Let the length of the bar be 1 2 ins. and its angular velocity constantly 500 per sec. Determine the velocity of each end and of the middle point when the bar makes an angle of 6o with one of the guiding lines. Am. Velocity of middle point = 4.36 ft. -per-sec. , directed at angle 30 with bar. 4. One end of a bar describes a circle, the other moves along a straight line containing a diameter of the circle. Determine the in- stantaneous center in each of the two positions in which the bar coincides with the directing straight line. 5. In Ex. 4, let a = length of bar, r = radius of circle, v = ve- locity of the point describing the circle. Determine the angular velocity of the bar when it makes an angle with the guiding straight line. Ans. ft) = (V 2 a % sin 2 0) 1A v/ar cos 6. 6. In Ex. 5, let A be the point describing the straight line, B the point describing the circle, C the center of the circle. Determine the velocity of the point A when A CB is a right angle. Determine the velocity of the middle point of the bar at the same instant. 7. Where is the instantaneous center, and what is the angular velocity, when the bar is in the position described in Ex. 6 ? 2. Composition and Resolution of Plane Motions. 435. Component and Resultant Motions of a System of Par- ticles. Just as it is often useful, in the analysis of the motion of a single particle, to regard its actual velocity as made up of compo- nents, so in studying the motion of a system of particles it may be PLANE MOTION OF A RIGID BODY. 365 useful to regard any actual instantaneous motion as the resultant of two or more component motions. The meaning of component and resultant motions of a system may be defined as follows : Any definite instantaneous motion of a system of particles implies definite velocities of all the particles. Conceive two such motions M\ M" ; let v Xi vj, v 3 ', ... be the velocities of the individ- ual particles in the motion M' , and tr", p* t v./, . . . their velocities in the motion M" . Let v x be the vector sum of v x and v{ % v 2 the vector sum of v 2 and v 2 ", v. 6 the vector sum of v^ and v 3 ", . . . ; and designate by Ma. motion of the system in which the individual particles have velocities v u v 2 , v $t .... Then the motion M may be regarded as the resultant of the motions M' and M" . 436. Component and Resultant Motions of a Rigid Body. The velocities of the individual particles of a rigid body are not in- dependent of one another. In plane motion the velocity of every particle is determined if the velocities of two particles are assigned, and even these two velocities cannot be assigned arbitrarily, but must satisfy the condition that their resolved parts along the line joining them are equal (Art. 434). Using the same notation as above, let M' and M" be two possible instantaneous motions of a rigid body ; then it may be shown that the resultant motion M is a possible motion of the rigid body. In order to prove this, it is sufficient to show that the particles may possess velocities V xt v %9 V %i . . . without changing their distances apart ; in other words, that the resolved parts of any two velocities v t , v 2 along the line joining the two particles are equal. That this condition is satisfied by the velocities v x , v 2 follows at once from the fact that it is satisfied by v/, v 2 \ and also by v" % v 2 ; and the fact that v x is the vector sum of v x \ v x \ and v 2 the vector sum of v 2 , v 2 . In general, therefore, The resultant of any two instantaneous motions of a rigid body is a motion consistent zvit/i rigidity. Any actual motion of a rigid body may therefore be regarded as the resultant of two or more simultaneous motions. It remains to consider the principles governing the composition and resolution of rigid-body motions. 437. Resultant of Two Instantaneous Motions. It has been shown that any possible plane motion of a rigid body is at every 366 THEORETICAL MECHANICS. instant a rotation about a definite axis (reducing in a particular case to a translation). The resultant of two given motions may be deter- mined as follows : Let the component motions be rotations about instantaneous centers C and C" , with angular velocities co' and co"; and let the resultant motion be a rotation about an instantaneous center C with angular velocity co. It is required to determine the position of C and the value of co. Consider separately the cases in which co' and to" have (a) the same and (^) opposite angular directions. (a) The resultant velocity of a particle at the instantaneous center C is zero ; its two components due to the two given rotations are therefore equal and opposite. But .-. /*~^v /*~"\ two com P onent velocities are not f C ) f C ) ( fr* \ parallel for any particle not in the line \J-^ Z n\^^-"-^--\^_y C C" \ hence C must lie in that line. Fig. 176. Again, if co and co" have the same angular direction (Fig. 176), it is only for a particle between C and C" that the two component velocities have opposite directions ; C therefore lies between C and C" . Let .a', z" denote the distances C C, C" C respectively. A par- ticle at C has a velocity z'co' due to the rotation about C, and an opposite velocity z"co" due to the rotation about C" \ and since its resultant velocity is zero, z'co' = z"co" or z'/z" = co' '/co' '. The value of co, the angular velocity of the resultant motion, may be found by determining the velocity of any particle not at C. The two components of velocity of a particle at C are o and {z' -\- z")co"; its resultant velocity is therefore {z' -\- z")co", and the angular ve- locity of the line CC is co =co"(z' + z")/z'; or, substituting the above value of z'/z", co = co' -\- co". (b) In case the angular directions of co' and co" are opposite, let C and C" (Fig. 177) represent the two instantaneous centers, and let it be assumed that co is greater than co". PLANE MOTION OF A RIGID BODY. 367 As in the former case, it is seen that C must lie on the line C C'\ its position being such that the components of velocity of a particle at C due to the two rotations shall be equal and opposite. The com- ponents will not have opposite directions for particles between C and C"\ and since ft)' is greater than ft)", the two components of velocity can be equal only if CC is less than CC". The point (7 there- fore lies without the length C C in *%1 J* * the direction of C. Its position must f C_\ f ff 1 [ y \ be such that Vj^/ \J~^4' v!"*^/ z'co' = z"(o", or *'/#* = o)"/c'. p IG x To determine the angular velocity of the resultant motion, notice that the resultant velocity of C is the sum of components o and (z" z')(o" ; the angular velocity of the line CC is therefore ft) = (o"(z" -O/^' === m> w Case of two equal and opposite angular velocities. If, in the case in which ft)' and ft)" have opposite directions, their magnitudes are made to approach equality, the point C recedes from C, and in the limit, when ft)' = ft)", C passes to infinity. In this case ft) = o ; that is, the system has no angular velocity, and the motion is a transla- tion. The translational velocity is equal to the velocity of any particle as C, that is, it is perpendicular to C ' C" and has the value O" *')*" = (CC)co". 438. Analogy to Parallel Forces. The results of the above discussion show a close analogy between the composition of rotations and the composition of parallel forces. Thus, if C and C" ( Fig. 176 or Fig. 177) are the points of application of parallel forces of magnitudes ft)' and ft)", their resultant is a force of magnitude ft) ap- plied at C; and ft) is equal to ft)' -f- ft)" or to ft)' ft)", according as the two forces have the same direction or opposite directions. The case of two equal and opposite rotations is analogous to that of a couple ; thus a translation may be regarded mathematically as a rotation couple. The analogy may without difficulty be shown to hold for any number of component motions. The process of finding the angular velocity and instantaneous axis of the resultant is identical with the 368 THEORETICAL MECHANICS. process of finding the magnitude, direction and line of action of a system of parallel forces. The lines of action of the component forces coincide with the instantaneous axes of the component rotations ; the magnitude of any one of the forces is equal to the angular velocity of the corresponding motion ; and opposite rotation-directions are rep- resented by opposite force-directions. 439. Resolution of a Rotation Into Components. By reversing the above process, the actual motion of a body at any instant may be resolved into components. Examples. 1 . What single motion of a rigid body is equivalent to rotational velocities of 10 rad. -per-sec. and 4 rev. -per-sec. in the same direc- tion, about instantaneous centers 5 ft. apart? 2. Two rotations of 5 rad. -per-sec. and 8 rad. -per-sec. in oppo- site directions, about instantaneous centers 4 ft. apart, are equivalent to what single rotation ? 3. What is the resultant of two rotations of 10 rev.-per-min. in opposite directions, about centers 2 ft. apart? 4. A rotation of 4 rev. -per-sec. about a center C is equivalent to what two rotations about centers distant 1 ft. and 2 ft. from C on op- posite sides ? 5. A rotation of 5 rev. -per-sec. about a center C is equivalent to what two rotations about centers distant 3 ft. and 2 ft. from C in the same direction? 6. A translational velocity of 20 ft. -per-sec. is equivalent to what angular velocities about two points 8 ft. apart? What must be the direction of the line joining the proposed centers in order that the resolution may be possible? 7. Show how to resolve a given rotation into three components whose centers are given and are not collinear. 440. Translation and Rotation. The above processes of com- bining or resolving rotations include the case in which one of the components is a translation. An independent treatment of this case is, however, desirable. Composition. Let the resultant motion of a rigid body at a certain instant be made up of two components, one of which is a rotation about the instantaneous center C (Fig. 178) with angular velocity ft), and the other a translation of velocity v in the direction CM. Since this translational velocity is the resultant velocity of C, PLANE MOTION OF A RIGID BODY. 369 the instantaneous center C of the resultant motion must lie on a line through C perpendicular to CM. If z de- notes the distance CC, we must have z(a = v, or z = v/tOy since the resultant velocity of a particle at C must be zero.* Resolution. By reversing the above process, any given motion may be resolved into a translation and a rotation. As a special case, any motion may be ~Fig 173 resolved into a translation and a rotation about an instantaneous center chosen at pleasure. To accomplish this, it is only necessary to select as the translational component of velocity for every particle the velocity of the point which is to be made the instantaneous center. This will be illustrated in treating the dynamics of plane motion of a rigid body. Examples. 1. A rotation of ioo rev. -per-min. is to be resolved into two components, one of which shall be a translation equal to the actual velocity of a certain particle 4 ft. from the center of rotation. De- termine the two components completely. 2. A rotation of 100 rev. -per-min. about a center C and a trans- lation of 86 ft. -per-min. in a certain direction are equivalent to what resultant motion ? 3. Resolve a rotation of 50 rad. -per-sec. into two components, one of which shall be a translation and the other a rotation about a center 6 ft. from the center of the given rotation. Ans. The translational velocity is 300 ft. -per-sec. at right angles to the line joining the two centers of rotation. 441. Composition and Resolution of Accelerations. If, in two motions M ', M" of a rigid body, a certain particle has accelerations p\ p" respectively, its acceleration in the resultant motion is the vec- tor sum of p' and/". This follows from Art. 436, since acceleration is change of velocity per unit time. A particular case of the resolu- tion of accelerations is important in the application of D'Alembert's principle to the general case of plane motion (Art. 443). * It will be noticed that the process of combining a rotation and a trans- lation is exactly similar to the process of combining a force and a couple, explained in Art. 94. 24 370 THEORETICAL MECHANICS. If A and B are any two particles of a rigid body having any plane motion, and if the acceleration of B is resolved into two com- ponents, one of which is equal to the acceleration of A, the other component is equal to the acceleration which B would have if the body rotated about a fixed axis through A with its actual angular motion. Let v be the velocity of A' and p its acceleration ; then the truth of the proposition becomes evident by considering that if, with the actual motion, there be compounded a translation with velocity v and acceleration p, the particle A is reduced to rest while the an- gular motion is unchanged. 3. Dynamics of Plane Motion. 442. Application of D'Alembert's Principle. The way in which the condition of motion of a body is changing at any instant depends upon the shape and size of the body, the distribution of its mass, and the forces acting upon every particle. The influence of these several elements may be expressed by algebraic equations, derivable from the general principle known as "D'Alembert's principle."* The application of this principle to the general plane motion of a rigid body must now be considered. 443. Effective Forces. In order to determine the effective forces, let the actual motion be resolved into two component motions in accordance with the principle stated in Art. 441, and let the effective forces for the component motions be computed separately. One of the component motions is a 171V0O* rotation about a fixed axis with the ac- tual angular velocity and angular acceler- ation of the body ; the other is a transla- tion with velocity and acceleration equal P to those of a particle lying in the assumed Fig. 179. fixed axis. The position of the fixed axis may be chosen at pleasure ; let it be at A (Fig. 179), and let v be the velocity and^> the acceleration of a par- *See Arts. 387, 390. ->. PLANE MOTION OF A RIGID BODY. 37 1 tide of the body which instantaneously coincides with A. Let be the angular coordinate of the body (Art. 433), co = dOjdt its angu- lar velocity, and cj> = dco/dt = d' 2 6/dt 2 its angular acceleration. The two components into which the actual motion of the body is resolved are then (a) a translation with velocity v and acceleration p ; (b) a rotation about a fixed axis through A with angular velocity co and angular acceleration . Consider the effective force on any particle due to each of these component motions. Let B (Fig. 179) repre- sent the position of a particle of mass m } and let AB = r. {a) Corresponding to the translation, the effective force is equal in magnitude and direction to mp. {b) Corresponding to the rotation, the effective force is made up of a component mrco 2 in direction BA and a component mrc\> at right angles to BA. Since the translational effective forces on different particles are proportional to their masses and have a common direction, their re- sultant is a force Mp applied at the mass-center, M being the whole mass of the body. The resultant of the rotational effective forces may be determined as in Art. 426. The moment of this resultant about the axis of rota- tion is I if / is the moment of inertia with respect to the central axis. The results of the discussion may be summarized as follows: In any plane motion of a rigid body, the system of effective forces is equivalent to a force Mp applied at the mass-center and a couple of moment 7(f) ; where M is the total mass of the body, / its moment of inertia with respect to a central axis perpendicular to the plane of the motion,/ the acceleration of the mass-center, and cp the angular acceleration of the body. 444, Determination of Angular Motion. By D'Alembert's principle, the sum of the moments of the external forces about any axis is equal to the sum of the moments of the effective forces about that axis. Let the axis of moments contain the center of mass ; then the moment of the resultant of the translational effective forces is 372 THEORETICAL MECHANICS. zero, and the sum of the moments of the effective forces is equal to 1(f), if / is the moment of inertia with respect to the central axis. Let L denote the sum of the moments of the external forces about this axis ; then L = /. This is identical with the equation of angular motion about a fixed axis containing the mass -center, 445. Independence of Translation and Rotation. Any actual motion of a body being regarded as made up of a rotation about an axis containing the mass-center, together with a translation equal to that of the mass-center, we have now proved that these two motions take place independently. The meaning of this statement may be expressed definitely as follows : ( 1 ) The acceleration of the mass-center is the same as if the whole mass were concentrated at that point and acted upon by forces equal in magnitude and direction to the actual external forces. (2) The angular acceleration is the same as if the mass-center were fixed and the actual external forces applied. 446. Complete Determination of the Motion. In the general case of plane motion, three independent dynamical equations may be written. Two of these may be the equations of motion of the mass- center, the third the equation of angular motion. In any determinate problem these three equations, together with the initial conditions for determining constants of integration, and the geometrical relations (if the motion is constrained) serve to determine the motion and the unknown forces. Let the coordinates of the body be taken as in Art. 428. Choosing a pair of rectangular axes lying in a plane parallel to the motion and containing the mass-center, let x, y = coordinates of mass-center ; 6 = angular coordinate of body ; M = total mass of body ; /== Mk 2 = moment of inertia with respect to central axis; P = vector sum of external forces ; X, Y = axial components of P ; L = sum of moments of external forces about central axis. PLANE MOTION OF A RIGID BODY. 373 The three equations of motion take the form M(d 2 x/dt 2 ) = X ; .... (i) M{ a]t\ . . . (5) This determines, the motion. If the origin is so taken that x = o when 6 = o, x= ad = igsin $ t\ . . . (6) To complete the solution the values of N and T must be found. From (i), T = M(g sin j3 x) = iMg sin /3. . . (7) From (2), N = Mg cos /3. . : . . . (8) II. A homogeneous circular cylinder is placed with its axis ver- tical on a smooth horizontal plane, and is set in motion by a con- stant tension applied to the free end * 1 of a flexible cord wound upon its eK surface. Determine the motion. 0\\ Solution. Let M = mass of cyl- 7 X inder ; a = its radius ; ^ = its radius y of gyration with respect to its geo- ~^T metrical axis ; (/P = \a 2 ;) / = Mk 2 == Fig. 181. \Ma 2 ; T= tension in string; x, j/, 6 = coordinates of position taken in the usual way, the axes being as shown in Fig. 181. Since the only external force acting on the cylinder in the plane of the motion is T, the acceleration of the mass-center has the direc- tion of T. The dynamical equations are M(d 2 x/dt 2 ) = T\ . . . . (1) \Ma\d 2 6ldt 2 ) = Ta (2) If T is a known constant, the motion can be determined by the inte- gration of these two equations. If T is unknown, the motion cannot be completely determined ; but in any case the elimination of T shows that the angular acceleration and the linear acceleration bear a constant ratio to each other ; that is, d 2 x/dt 2 = \a{d 2 0jdt 2 ). Examples. 1. A homogeneous cylinder 2 ft. in diameter rolls without sliding down a plane inclined 30 to the horizon, (a) Determine the dis- PLANE MOTION OF A RIGID BODY. 375 tance moved by the mass-center in 2 sec. , starting from rest, (b) If the mass of the cylinder is 40 lbs., determine the normal and tangen- tial components of the reaction of the plane on the cylinder. Ans. (a) 21.5 ft, nearly. (&) T = one-sixth the weight of the body. 2. Solve the problem of the motion of a cylinder rolling on an inclined plane, assuming that the body has initially a motion up the plane. Apply the results to the case described in Ex. 1, assum- ing an initial velocity of the mass-center of 20 ft-per-sec. up the plane. Ans. The body will come to rest at the end of bolg sec, and will then descend as in Ex. 1. 3. Determine the motion of a homogeneous sphere which rolls without sliding down a rough inclined plane, and apply the results to a sphere of 10 lbs. mass and 2 ins. diameter rolling from rest down a plane inclined 5 to the horizon. Ans. The acceleration of the mass-center is 5/7 that of a body sliding on the plane without friction. 4. A homogeneous sphere rolls, without sliding, on the inner sur- face of a hemisphere. If it moves from rest under the action of no force except gravity and the reaction of the surface, determine the motion. Let A be the center of the hemisphere (Fig. 182), A that of the sphere, B the lowest position of A, AE that radius of the sphere which is vertical when A A' is vertical. Let a = radius of sphere, a ' = radius of hemisphere, = angle AA'B, 6 = angle between AE and vertical. In writing the equations of motion of the mass-center, let the acceleration be resolved along the tangent and nor- mal to the circle described by A. The components are {a' a)(d 2 /dt 2 ) and (a' a)(d<\>\d?j l . Let the pressure on Fig. 182. the sphere at the point of contact D be resolved into a normal component N and a tangential component T, both unknown. The only other external force is the weight Mg. The equations of motion are T Mg sin = M{a ' a)(d 2 (f>ldt 2 ); . (0 N Mg cos $ = M(a' d){d$ldf? ; . (2) Ta = (2Ma 2 / 5 Xd 2 6/dt 2 ). (3) The angles and are related in a simple manner. Since the arcs ED and CD are equal, and angle DAE = -\- (/>, a' = a{6 + ). . (4) 376 THEORETICAL MECHANICS. From equation (4), a{d 2 0ldt 2 ') = (' d)(d 2 4>ldt*). Substituting this value in (3), eliminating T between (3) and (1), and reducing, dj = k_ sin . . . ( 5 ) dt 2 7 = < , (?) '= -^-, (cos *- cos *> (6) \dtj y{a f a) From (1) and (5), T=(2Mgs\n$)h. . . . (7) From (2) and (6), N = Mg(\i cos 10 cos 4>o)l7- - ( 8 ) 5. In Ex. 4, how great must the coefficient of friction be in order that the assumed rolling without sliding may be possible ? The coefficient of friction must be not less than the greatest value of T/N found in the above solution. Let /x = TIN = (2 sin <)/( 1 7 cos c/> 10 cos < ). The value of /x increases with (f>, and has its greatest value when (f> = = angle between vertical and line joining centers ; then the spheres separate when cos = 10/17 ; (j> 53 58'. 8. In Ex. 7, if the spheres are smooth, show that they will sep- arate when cos = 2/3. 9. A circular cylinder whose mass-center is not in its geometrical axis rolls, without sliding, upon a horizontal plane. Let a = radius of circular section, c = distance of mass-center from geometrical axis, k = radius of gyration with respect to axis through mass-center, = angular displacement from position of stable equilibrium at time/. Show that ^ ^ a (d6/dty](a 2 + & + k 2 2ac cos 0) remains constant during the motion. PLANE MOTION OF A RIGID BODY. 377 10. A circular cylinder whose mass-center is not in its geomet- rical axis is placed on a smooth horizontal plane. With notation as in Ex. 9, show that (*+ C* sin 2 6)(d0idt) 2 2gc cos d remains constant during the motion. 1 1. How great must the coefficient of friction be in order that a homogeneous cylinder placed on an inclined plane with axis hori- zontal shall roll without sliding ? 448. General Method of Writing Equations of Motion. The three equations of motion are not necessarily written as in Art. 446, although this method furnishes the simplest solution of many prob- lems. The equations derivable from D'Alembert's principle are of two kinds, equations of resolution and equations of moments. All such equations are included in the two following statements : (a) The sum of the resolved parts of the effective forces is equal to the sum of the resolved parts of the external forces, for any direc- tion of resolution. (J?) The sum of the moments of the effective forces is equal to the sum of the moments of the external forces, for any axis. By reasoning analagous to that used in Art. 104 in discussing the conditions of equilibrium of coplanar forces, it may be shown that only three independent dynamical equations can be obtained by expressing the equivalence of the effective forces and the external forces. The remarks made in Art. 387 as to the meaning of D'Alem- bert's principle must here be kept in mind. The statement that the system of effective forces is equivalent to the system of external forces is merely a concise expression of the fact that the two sys- tems satisfy exactly the same conditions which are satisfied by two systems of external forces which would be equivalent in effect if ap- plied to a rigid body. This relation of equivalence between the two systems is completely expressed by three equations falling under the above general forms {a) and (b). The three independent equations may therefore be written in three different ways, as follows : (1) By resolving in each of two directions and taking moments about any axis. (2) By taking moments about each of two axes and resolving in a direction not perpendicular to the plane containing the axes. 378 THEORETICAL MECHANICS. (3) By taking moments about each of three axes not lying in one plane.* 449. Equation of Moments for Axis Not Containing Mass- Center. The simplest moment equation, so far as regards the effect- ive forces, is obtained by taking the axis through the mass-center ; the translational effective forces are thus eliminated, since their re- sultant is applied at the mass-center. But it may be advantageous to take another moment axis for the purpose of eliminating certain of the external forces. In writing such an equation, the translational effective force Mp, applied at the mass-center, must not be omitted. To illustrate the general method of writing an equation of mo- ments, consider the problem of a cylinder rolling on an inclined plane (Art. 447, Problem I). If moments are taken about the ele- ment of contact of the cylinder and plane, the forces N and T are eliminated. The only external force whose moment is not zero is the weight of the body, and its moment is Mga sin /3. The sum of the moments of the effective forces is Ma(d 2 x/dt' z ) + Mk 2 (d 2 0/dt 2 ) = (3^72) (;/ 2 0/^ 2 ). The equation of moments is therefore ^Ma 2 /2)(d 2 0/dt 2 ) = Mga sin 0, which agrees with the result given in Art. 447. Examples. 1. A uniform bar is placed in a smooth hemispherical bowl in such a position that a vertical plane through the axis of the bar contains the center of the sphere. Determine a differential equation for the motion. If a is the length of the bar and c its distance from the center of the sphere, show that its motion will be the same as that of a simple pendulum of length c -J- a' 2 / 12c. Interpret the lim- iting cases c = o, a = o. 2. A straight bar is placed at rest with one end on a smooth hor- izontal plane and the other against a smooth vertical plane which is at right angles to the vertical plane containing the axis of the bar. Write three independent equations for the motion, one of which shall be independent of the unknown reactions. * It is of course understood that the axes of moments are all perpen- dicular to the plane of the motion, and the directions of resolution parallel to that plane. PLANE MOTION OF A RIGID BODY. 379 4. Statics of a Rigid Body. 450. Balanced Forces. A set of external forces which, applied to a rigid body, produces no effect upon its motion, constitutes a balanced system or a system in equilibrium (Art. 57). The condi- tions which such a system must satisfy have been considered in Part I under the head of Statics. It may now be shown that the conditions of equilibrium are deducible from the general equations of motion of a rigid body. 451. Equilibrium of Rigid Body. If all the external forces acting upon a rigid body constitute a balanced system, the body is said to be in equilibrium. Equilibrium is not synonymous with rest ; but a body in equilib- rium can have only such motion as it could have if acted upon by no external force. A single particle acted upon by no force or by balanced forces is either at rest or moving uniformly in a straight line. The possible motion of a rigid body acted upon by balanced forces will be consid- ered below. It will be shown that the individual particles of such a body are not necessarily in equilibrium. 452. Equivalent Systems of Forces. The equations of motion of a rigid body show that two systems of external forces are equivalent in their effect upon the motion if they satisfy the following conditions : (1) Their resolved parts* in any direction are equal. (2) Their moments about an axis containing the mass-center are equal. For, the second members of equations (1 ), (2) and (3) of Art. 446 obviously have identical values for two systems satisfying these con- ditions. Equivalent forces. These conditions of equivalence are satisfied by two forces which are equal in magnitude and direction and have the same line of action, even if applied at different points in that line. This is the principle assumed in Art. 82 as one of the fundamental laws of Statics. *By the "resolved part of the system" is meant the sum of the resolved parts of the several forces ; by the "moment of the system " is meant the sum of the moments of the several forces. 380 THEORETICAL MECHANICS. 453. Resultant Force or Resultant Couple. A system of coplanar forces is in general equivalent to a single force. For, what- ever may be the values of X, Y and L for the system, a single force can generally be found having the same values of these three quanti- ties, and thus giving the same equations of motion. Moreover, the values of X, Y and L completely determine the magnitude, direction and line of action of the force. If the vector sum of the given forces is zero, X and Y are zero whatever the directions of the axes. In such a case there is no single force equivalent to the given system. The simplest equivalent system is then a couple whose moment is equal to the value of L for the system. 454. Conditions of Equilibrium. In order that a body acted upon by any external forces may move as if acted upon by no force, the values of X, Y and L in the equations of motion must be zero. That is, the conditions of equilibrium for a rigid body are (1) that the sum of the resolved parts of the external forces is zero for each of two rectangular directions, and (2) the sum of the moments is zero for an axis containing the mass-center. It may be shown as in Art. 104 that if these conditions are satisfied, ( 1 ) The sum of the resolved parts in any direction is zero, and (2) The sum of the moments about any axis is zero. From these general principles may be derived all the special conditions of equilibrium deduced in Chapter V. 455. Motion Under Balanced Forces. The nature of the pos- sible motion of a rigid body when the external forces are balanced may be seen from equations (1), (2) and (3), Art. 446. These become d*x/dt* = o, . . . (1) dy/dt* = o, . . . . (2) d 2 0ldt 2 = o (3) Equations (1) and (2) express the fact that the acceleration of the mass-center is zero, and its velocity therefore constant. Equation (3) shows that the angular acceleration is zero and the angular velocity therefore constant. It is thus seen that if every particle of the body is initially at rest, it will remain at rest. If initially there is a motion of translation, this motion will continue with unvarying velocity. But in general the PLANE MOTION OF A RIGID BODY. 381 motion will be composed of a translation in which the mass-center moves with unvarying velocity, together with a rotation with uniform angular velocity. The motion of any individual particle is the resultant of the com- ponents due to the translation and to the rotation. In general the velocity of a particle not in a line perpendicular to the plane of the motion and containing the center of mass is variable both in magnitude and in direction. Its two components of acceleration may be determined as in Arts. 441, 443. With the notation there used, the translational component/ is zero. Of the two components due to the rotation the one perpendicular to r is zero, because the angu- lar acceleration is zero ; the other component is r) are the axial components of the increment of momentum, and AH is the increment of the angular momentum. If the line of action of an instantaneous impulse is known, L can be expressed in terms of X' and Y'. 462. Linear Momentum. It has been shown that the axial components of the total momentum have the same values as if the entire mass were moving with the mass-center. This is true for any system of particles, whether rigid or not. The effect of an impulse on the motion of the mass-center may therefore be determined as if the system were a single particle. The momentum, computed as if the whole mass were concentrated at the mass-center, may be called the linear momentum of the system. The angular momentum has not in general the same value as if the mass were concentrated at the mass -center. Examples. 1. A body of mass 20 lbs. falls from rest under the action of gravity. Determine, by the principle of impulse and momentum, the velocity of the mass-center after 5 sec. 2. A body whose shape is that of a right circular cylinder, resting with one end upon a smooth horizontal plane, is set in motion by the tension in a flexible cord unwinding from its surface. If the mass of the body is 12 lbs. and the tension is constantly equal to 2 lbs. weight, what will be the velocity of the mass-center after 5 sec? Does the result depend upon the position of the mass-center in the body? 3. A body of 40 lbs. mass and of any shape, initially at rest, receives a blow equivalent to a force of 100 lbs. acting for y 2 sec. What can be determined as to the motion of the mass-center imme- diately after the blow? 4. Two bodies of masses 10 lbs. and 18 lbs. respectively are con- nected by a string. Both being initially at rest, the mass of 10 lbs. receives an impulse equivalent to a force of 50 lbs. acting for 1 sec. (a) What is the velocity of the mass-center of the system immediately after the impulse ? (b) If, immediately after the impulse, the mass THE PRINCIPLE OF IMPULSE AND MOMENTUM. 387 of 18 lbs. is at rest, what is the velocity of the mass-center of the other body ? (V) If, at a certain instant after the impulse, the mass- center of the mass of 18 lbs. has a velocity of 10 ft.-per-sec. in the direction opposite to that of the impulse, what velocity has the mass- center of the other body ? Ans. (c) 18 -f- 5g = 179 ft.-per-sec. 5. Two particles whose masses are 5 lbs. and 1 lb. respectively, connected by an inextensible weightless cord 10 ft. long, are initially at the same point. The lighter body is projected vertically upward with a velocity of 40 ft. -per-sec. (a) What will be the velocity of their common mass-center immediately after the second particle begins to rise? (<) How long will the mass-center continue to rise? An s. (a) 5. 1 5 ft. -per-sec. () 0.16 sec. 2. Rigid Body Having Motion of Translation or of Rotation About a Fixed Axis. 463. Angular Momentum in Case of Translation. If the motion of a body is a translation, the motion of every particle is known when that of the mass-center is known. The equation of angular impulse and angular momentum therefore is not needed for the determination of the motion. The value of the angular mo- mentum in case of translation will, however, be found useful. If v is the translational velocity, the momenta of the several par- ticles are parallel and equal to m x v y m 2 v y .... These momenta form a system analagous to a system of parallel forces applied to the particles and proportional to their masses. The sum of their moments is therefore the same as that of a momentum Mv at the mass-center. That is, in computing the angular momentum of a body whose motion is translatory, the whole mass may be assumed to be concentrated at the center of mass. 464. Linear Momentum of Rotating Body. Let a rigid body of mass M be rotating, with angular velocity co, about a fixed axis whose distance from the center of mass is a. Let A (Fig. 183) be the mass- WM>U> center and O the projection of the fixed \ M yVzaccf axis. The velocity of the mass-center is aco perpendicular to OA ; hence the linear momentum is Maco perpendicular & to OA. Fig. 183. 388 THEORETICAL MECHANICS. 465. Angular Momentum of Rotating Body. The angular momentum about the fixed axis of rotation may be computed as fol- lows : Let B (Fig. 183) be the position of a particle of mass m x distant r x from the axis of rotation. Its momentum is m^co perpendicular to OB, and the moment of this momentum about O is m x r x (o. The total moment of momentum about O is therefore H == {m x r x -j- m 2 r* + )(o = Iw, if / is the moment of inertia of the body with respect to the axis of rotation. 466. External Impulse Acting on Rotating Body. If a body is constrained to rotate about a fixed axis, it is convenient to classify the external forces into applied forces and constraining forces ; the latter being usually wholly unknown, except as the solution of the equations of motion serves to determine them. Similarly, the total external impulse is made up of the applied impulse and the con- straining impulse. The latter is the impulse of the hinge-reaction (Art. 422); it is unknown in magnitude and direction, but it acts through the axis of rotation. 467. Equations of Impulse and Momentum for Rotating Body Acted Upon by Instantaneous Impulse. Let momentum and im- pulse be resolved along the line joining the mass-center A to the center of rotation O (Fig. 184). Let P' be the applied impulse, and R ' the impulse of the hinge-reaction. Let the components of P' be P' n in direction A O and Pt perpendicular to A O ; and let the cor- responding components of R' be N' fT' and T'. (Fig. 184 represents P' n and <^^^ / V P' t as if applied at A ; this is not nec- ** O" "'---.._# / essarily true, but the figure shows P^*^/ correctly the directions of the compo- A. nent impulses.) Fig. 184. Assuming the impulse to be instan- taneous, the position of the mass-center A is the same immediately before and immediately after the impulse. The direction of motion of A is therefore not changed by the im- pulse, but its velocity is changed by aAco if A (4) This is identical in form with equation (3) of Art. 467, which ap- plies to the case of rotation about any fixed axis ; but in that case / is the moment of inertia about the axis of rotation. 478. Effect of Instantaneous Impulse on Free Body Initially at Rest. Let a body, initially at rest but free to move in a plane, receive an instantaneous impulse of known magnitude, direction and line of action. Since the resultant momentum is zero before the impulse, (momentum-increment) = (resultant momentum after impulse). Hence the motion just after the impulse must be such that the re- sultant momentum has the same magnitude, direction and position- line as the impulse. Let P' be the magnitude of the impulse, b the distance of its line of action from the mass-center, M = mass of body, Mk 2 = / = its moment of inertia with respect to the axis through the mass center, v = velocity of mass-center, and co an- gular velocity just after the impulse. Then by Art. 475 the resultant momentum is equal in magnitude to Mv, its direction is that of v } and its moment about the mass-center is Mk 2 co. Hence the direction of v coincides with that of P', and P = Mv, P'b - Mk 2 co; or v = P/M, co = P'blMk 1 . Instantaneous axis. Let A (Fig. 187) be the mass-center, and draw AB perpendicular to the line of action of P. Since the velocity THE PRINCIPLE OF IMPULSE AND MOMENTUM. 397 of A after the impulse is parallel to P\ the instantaneous motion is a rotation about some axis C intersecting BA produced. Let A C = z ; then v = zco, or z = v/m = P/6. The points C and B are thus related in the same way as the centers of suspension and oscillation of a com- pound pendulum (Art. 425). r _/-~~' *^> 1 The determination of the instanta- ) u^ z _ j~J neous axis and of the angular velocity ( q 4 "_l"_j ( gives a clear idea of the state of motion \ *M \ immediately after the impulse. If the "~"^ v J s-^ body is acted upon by no forces sub- Fig. 187. sequently, the angular velocity remains constant, and the velocity of the mass-center remains constant in magnitude and direction ; the instantaneous axis therefore continues at the same distance from the mass-center, and its locus is a plane parallel to the path of the center of mass. The surface traced in the body by the instantaneous axis is a circular cylinder of radius z whose geometrical axis contains the mass-center. Application of equations. To show the application of the equa- tions of Art. 477, let the axis of x have the direction of the impulse. Then X' = P\ Y' = 0] and if moments are taken about the axis through the mass-center, L = P'b. Equations (1), (2) and (4) therefore become MAx = P, MAy = o, Mk 2 Av = P'b. But Ax and Ay are the components of v, and since Ay = o, v = Ax= P'jM. Also, Aw becomes in this case a), hence co = P'bjMk\ as before. To illustrate still further the general method, let moments be taken about a point in the line of action P' . The initial angular momentum about this point is zero, the moment of P' is zero, hence 398 THEORETICAL MECHANICS. the angular momentum about this point after the impulse is zero. But by Art. 476 its value is Mtfto Mvb ; equating this to zero, v/e> = & 2 /b, which agrees with the result already found. This moment equation may take the place of one of the equations before used. Examples. 1. A straight bar of uniform small cross-section and uniform density, free to take up any plane motion, receives a blow at a given point directed at right angles to its length. Determine the instan- taneous axis of rotation. An s. With the above notation, z = I 2 1 12b, I being the length of the bar. 2. In Ex. 1, let the mass of the bar be 20 lbs., its length 2 ft., and let the impulse be equivalent to that of a force of 1,000 lbs.- weight acting for o. 1 sec. If b = IJ2, determine v and co after the impulse. Ans, v= $g ft. -per-sec ; co = 1 $g rad. -per-sec. 3. A straight bar of uniform cross-section and uniform density, 2 ft. long, receives a blow in a direction perpendicular to its length at a point 6 ins. from one end. Determine the instantaneous axis. Ans. 4 ins. from end. 4. In Ex. 1, let the direction of the impulse be inclined at angle 6 to the length of the bar. Determine the instantaneous axis. 5. If, in Ex. 3, the impulse is applied at the same point but in a direction inclined 30 to the bar, determine the instantaneous axis. An s. 4 ins. from middle point of bar. 6. A square plate of uniform small thickness and uniform density is suspended by strings attached to two corners so that two edges are vertical. It receives a blow perpendicular to its plane, applied at the middle point of the lowest edge. Determine the instantaneous axis. Ans. If a = side of square, z = a/6. 7. A circular plate of uniform density and uniform small thick- ness is suspended by a string attached at a point in the circumference. How must it be struck in order that it shall begin to rotate about a vertical tangent ? Ans. At a point bisecting a horizontal radius. 8. If the plate suspended as in Ex. 7 receives a blow in a direc- tion perpendicular to its plane and at a distance from the center equal to one-third the radius, determine the instantaneous axis. Ans. z = 3*2/2, if a = radius. 9. A homogeneous parallelopiped of mass 10 kilogr. and di- mensions (in cm.) 20 X 30 X 40, receives a blow whose direction is THE PRINCIPLE OF IMPULSE AND MOMENTUM. 399 perpendicular to one of the largest faces and whose point of application is the middle point of its shorter edge. Determine the instantaneous axis. If the velocity of the mass-center after the impulse is 1 met. - per-sec. , determine the angular velocity and the value of the impulse. Ans. z = 8y3 cm. ; co = 12 rad. -per-sec. ; P'= io 6 dyne-seconds. 10. In the above general problem, discuss the case b = o. 11. Discuss the case in which b is infinite while P'b is finite. 12. A horizontal blow must be applied at what point of a billiard ball in order that it shall begin to roll without sliding ? 479. Effect of Impulsive Couple on Free Body Initially at Rest. If a body is subjected to two instantaneous impulses which are equal in magnitude but opposite in direction and not collinear, the motion of the body immediately after the impulse must be such that the resultant momentum is a couple. This requires that the instantaneous axis shall pass through the mass-center. If L is the moment of the impulsive couple, the angular velocity after the impulse is co = L'IMk\ Examples. 1. A uniform straight bar of length /receives simultaneously two blows applied close to the ends, at right angles to the length, and in opposite directions. Either blow alone, applied at the middle point, would give the bar a translational velocity V. Determine the motion after the impulse. Ans. A rotation about a central axis, with angular velocity 12 Vjl. 2. A uniform straight bar of length / receives simultaneously two impulses P' and 2P' at right angles to the length and in the same direction, the former applied at the middle, the latter at one end. The impulse P' alone would give the bar a translational velocity V. Determine the instantaneous motion. Ans. A rotation with angular velocity 12 Vjl about an axis distant lj\ from the middle point. 3. A uniform bar 2 ft. long whose mass is 10 lbs., constrained to rotate about one end, receives an impulse at a point 0.25 ft. from the free end which gives it an instantaneous angular velocity of 360 per sec. An equal impulse is applied to an exactly similar bar which is wholly free. Determine the instantaneous motion. Ans. An angular velocity of 247J-/7 rad. -per-sec. about an axis 5^ ins. from the middle point. 4. A body is free to rotate about a fixed axis through the mass- center. Show that an impulsive couple will cause no hinge-impulse. 400 THEORETICAL MECHANICS. 480. Effect of Instantaneous Impulse on Free Body Not In- itially at Rest. The momentum-ijicrement due to a given impulse has the same value, whatever the initial motion of the body. The motion immediately after the impulse is found by combining the motion due to the impulse with the motion previously existing. Thus if x, y are the axial components of the velocity of the mass- center and ft) the angular velocity, and if values just before the impulse are denoted by a single accent and values just after the impulse by a double accent, x" = x -J- Ax, y" = y> + A>, &)" = ft)' -j- Aft) ; in which Air, Ay and Aft) are to be determined from equations (1), (2) and (4) of Art. 477. The velocity of the mass-center after the impulse is equal to the vector sum of its velocity before the impulse and that due to the im- pulse. The angular velocity after the impulse is the algebraic sum of the angular velocity before the impulse and that due to the impulse. Change of instantaneous axis. The position of the instantaneous axis of rotation immediately after the impulse is less simply deter- mined in case the body is in motion before the impulse than when it is initially at rest. It may, however, be determined, by combining the change of motion due to the impulse with the initial motion. Let the change of motion due to the impulse, computed as if the body were initially at rest, be a rotation about an instantaneous center C (Fig. 188) with angular veloc- %~^ J*C J^\ *ty ^ > * et *ke motion just before the f c\ f C"A f C '\ impulse be a rotation about an instan- \^^y \ J V_^/ taneous center C with angular velocity Fig. 188. ft>' ; and let the motion immediately after the impulse be a rotation about an instantaneous center C with angular velocity ft) ". By Art. 437 C" lies upon the line CC at such a point that CC" X ft> = C'C" X (o' y and ft) " = ft) -f- ft)'. If ft)' and ft) have like signs, C" lies between C and C; if they have opposite signs, C" lies on C C produced. THE PRINCIPLE OF IMPULSE AND MOMENTUM. 4OI If &)' and ft) are equal and opposite, C" is at infinity on CC '. In this case ft)" is zero, and the resultant motion is a translation. Examples. 1. A free circular disc of uniform density and thickness is rotat- ing about a central axis perpendicular to its plane when it receives a blow directed along a tangent. What is the angular velocity after the impulse if the velocity of the mass-center is v"l Let r = radius of disc, ft)' its original angular velocity, P' the impulse. From the given data, P' = Mv\ and the moment of P' about the mass-center is U = Mv"r in magnitude, but may be either positive or negative. If positive, the equation of moments about the mass-center is Mv"/r=iMr 2 Aco, whence Aft) = 2v"/r; ft)" = ft)' -f- Aft) = ' + 2v"jr. If L is negative, ft)" = go' 2v"/r. 2. A rotating wheel falls vertically in its own plane and strikes a horizontal plane which is perfectly rough and inelastic. Determine the subsequent motion. [" Perfectly rough" means that no sliding occurs ; "perfectly inelastic" means that the wheel does not rebound but remains in contact with the plane.] This problem may be solved by a single equation obtained by taking moments about an axis through the point of contact of the wheel with the plane. The moment of the impulse about this axis is zero, hence the angular momentum has the same value before and after the impulse. Letting v" denote the velocity of the mass-center after the impulse, and taking other notation as usual, the angular momentum before the impulse is Mk 2 co'. After the impulse it is MkW + Mv"r y if r is the radius of the wheel. Therefore MkW = Mk 2 co" + Mv"r. But also, v" = rco" ; hence ft)" = w'k 2 l(k 2 + r 2 ) ; v" = rv&Kk 2 + r 2 ). 3. A circular disc of uniform thickness and density, of radius 2 ft. and mass 50 lbs., rotating at the rate of 100 revolutions per minute, falls in its own vertical plane and strikes a horizontal plane surface. At the instant of striking its center has a horizontal velocity of 20 ft.-per-sec. and a vertical velocity of 30 ft.-per-sec. If the plane is so rough that there is no sliding and so inelastic that there 26 402 THEORETICAL MECHANICS. is no rebound, determine (a) the subsequent motion, (b) the value of the impulse. Ans. (a) v" = 20.32 ft. -per-sec. or 6.36 ft. -per-sec. , depend- ing upon the direction of v'. (b) The horizontal impulse is either 16 poundal-seconds or 1,318 poundal-seconds; the vertical impulse is 1 , 500 poundal-seconds. 4. A homogeneous right circular cylinder, resting with one end upon a smooth horizontal plane, is set in motion by a jerk applied to the free end of a string wrapped on its surface. Determine the in- stantaneous axis of rotation. An s. At a distance from the geometrical axis equal to half the radius. 5. In Ex. 4, let the radius of the cylinder be 0.5 met. and its mass 5 kilogr. If the angular velocity just after the impulse is 2 rev. -per-sec. , what is the value of the impulse ? What is the velocity of the mass-center? Ans. Impulse = 250,000^ dyne-sec. Velocity of mass-center = 507T cm. -per-sec. 6. A body has any plane motion, when a certain line in the body, perpendicular to the plane of the motion, suddenly becomes fixed. Determine the subsequent motion. The impulse by whose action the stated change in the motion is produced acts in some line passing through the axis which becomes fixed. The angular momentum with respect to this axis is therefore unchanged. The motion after the impulse may be determined by equating the values of the angular momentum before and after the impulse. 7. A homogeneous right circular cylinder is rotating freely about its axis of figure when an element of the cylindrical surface suddenly becomes fixed. Determine the subsequent motion. Let M be the mass of the cylinder, a its radius, co its original angular velocity, co" its angular velocity after the impulse. The orig- inal value of the angular momentum about an axis instantaneously coinciding with an element of the surface is Mcfco' ' I2. The angular momentum about this axis after the impulse is 3Ma 2 co"J2. Hence co" = ^co'. 8. In Ex. 7, determine the value of the impulse. The linear velocity of the mass-center changes from o to aco" ; hence the value of the impulse is Maco", its direction being perpen- dicular to the plane containing the fixed axis and the geometrical axis. 9. A uniform straight bar of mass 10 lbs. and length 2 ft., rotat- ing about its mass-center at the rate of 5 rev. -per-sec. , receives a blow equivalent to a force of 100 lbs. acting for 0.5 sec. at a point 3 ins. from one end, in a direction perpendicular to the length of THE PRINCIPLE OF IMPULSE AND MOMENTUM. 403 the bar. Determine the instantaneous center and angular velocity just after the blow. Ans. Instantaneous center is 7.10 ins. from middle point of bar. 10. With the initial motion described in Ex. 9, {a) what impulse will reduce the motion to a translation of 10 ft. -per-sec. ? (b) What impulse will leave the body at rest? Ans. (a) An impulse of 100 poundal-seconds, whose line of ac- tion is ^7r ft. from the mass-center. 11. Show that an impulsive couple, applied to a body rotating about an axis containing the mass-center, does not change the axis of rotation. 12. A body whose mass is 10 kilogr. and least radius of gyration 15 cm. has at a certain instant a translational velocity of 2 met. -per- sec. and an angular velocity of 2 rev. -per-sec. Two opposite im- pulses are simultaneously applied in lines 10 cm. apart, each equiva- lent to a force of 5,000 dynes acting for 10 sec. Determine the subsequent motion. Ans. The linear velocity is unchanged ; the angular velocity is increased or decreased by 2/9 rad. -per-sec 481. Connected Bodies. The principle that the external impulse is equal to the change of momentum it produces may be applied to any number of bodies regarded as forming a system. If the bodies press against one another, or are connected by strings or by hinges, the forces they exert upon one another are internal, and the impulses of these forces may be omitted from the equations of impulse and momentum for the system. For plane motion three independent equations can therefore be written which do not involve the impulses of these internal forces or "reactions." These are not generally sufficient to determine the motion. To complete the solution it is necessary to write equations for single bodies of the system. For any given body it is often possible to form one or more such equa- tions not containing the unknown reactions. The method of procedure will be illustrated by examples. Examples. 1. Two uniform straight bars AB, BC, connected by a frictionless hinge at B, rest on a horizontal plane so that ABC is a straight line. The bar AB receives at a certain point a horizontal blow at right angles to its length. Determine the motion just after the impulse. Let P' denote the impulse, applied at distance a from B (Fig. 189). Let ;;/, m' be the masses and /, /' the lengths of AB and irrespectively. Since the velocity of the mass-center of the system 404 THEORETICAL MECHANICS. just after the impulse will be parallel to P' , it is obvious that every point of both bars will instantaneously move parallel to P\ Let v be the velocity of the mass-center of AB and co its A angular velocity, v\ co' being the velocity of the mass- , center of i> 7 and its angular velocity. Then v* can be expressed in terms of v, co, co', as follows : If u is the velocity of B, a u = v \lco ; v' = 11 \l'co' = v \{lco -f- /'ft)'). Equating the impulse P to the linear momentum of the system after the impulse, Fig. 189. P' = mv -{- triv' = (m + m')v \m{lco -f- /'co'). (1) Taking moments about B for the body AB, thus eliminating the hinge-impulse, P'a = ml 2 co/ 12 -f- tnvl/2. . . . (2) Taking moments about B for the body BC, o s= m'l'' 2 (o'/ 12 m'v'1' I2 = (3/ft) -f- 4.1'co' 6v)m'l' 1 12. (3) Equations (1), (2) and (3) serve to determine v, co and co' when P' is known. Without knowing P\ the ratios of ft), ft)' and v can be determined. 2. If the two bars are equal in all respects and the impulse is applied at the middle point of AB, show that co = co' = 6v/jl. 3. At what point must the impulse be applied in order that the bar BC shall be instantaneously at rest? Will it remain at rest? 4. In Ex. 1, suppose the two bars equal in all respects and the impulse applied at A. Show that co 0/3 = 6^/5/. 5. Two bodies A, B of masses m, m! ', are connected by a string which coincides with the line joining their mass-centers. At a certain instant they are moving with equal velocities at right angles to the string, when A receives an impulse in the direction BA which, MA were free, would have deflected it 45 . Determine the motion just after the impulse, assuming that the string does not stretch nor break. 6. The initial conditions being as in Ex. 5, let A receive an im- pulse directed at an angle of 30 from BA produced, of such magnitude that if A were free is would be deflected 15 . Determine the motion just after the impulse. Ans. If v is the original velocity, the bodies have equal velocities \o.^i6()mj(m -\- m'y\v in the direction BA after the impulse; the velocity of B perpendicular to BA is unchanged, while the final ve- locity of A perpendicular to BA is 1.183Z/. CHAPTER XXIII. THEORY OF ENERGY. i. External and Internal Work. 482. Work of a Stress. The forces which any two particles exert upon each other are equal and opposite, constituting a stress (Art. 36). The total work done by the two forces of a stress during any displacements of the particles upon which they act may be com- puted as follows : Let A and B (Fig. 190) be the two particles, r their distance apart at any instant, P the magnitude of the force which each exerts upon the other. Assume that the force acting upon A has the direction BA ; then the force acting upon B has the opposite direction AB. Let the particles receive any innni- Fig. 190. tesimal displacements AA' } BB', such that the angle turned through by AB is also infinitesimal, and let A A" \ BB" be the orthographic projections of these displacements upon AB. Then PX A A' = work done by force P acting on A ; PX BB"= " " " " P " " B; P{_BB" A A") = total work done by stress. But BB" AA" = A"B" AB = AB' cos (dd) AB } if dO is the angle between AB' and AB. And since cos (dd) = 1 \{dOy + . . . , AB' cos (dd) differs from AB' by an infinitesimal of the second order, and AB' cos (dO) AB = AB' AB = dr. The total work done by the stress during the supposed infinitesimal displacement is therefore Pdr ; and during any finite displacement the work is equal to 406 THEORETICAL MECHANICS. I r" Pdr, . (i) r* and r" being the initial and final values of r. The same result may be reached by another method, as follows : Let (jr, , y x , z x ) and (x. 2 , y % , z 2 ) be the rectangular coordinates of the particles A and B respectively, and let a, /3, 7 be the angles between AB and the axes. By Art. 352, the work done by either of the forces P is equal to the sum of the works done by its axial com- ponents. Hence the force P acting upon A does an amount of work fP cos a dx x f P cos /3 dy x fP cos 7 dz % , and the force P acting upon B does an amount of work f P cos a - dx. t -f- f P cos $ dy., + fP cos 7 dz 2 ; the limits of the integrations being so taken as to correspond to the initial and final positions of the particles. The total work done by both the forces of the stress is therefore fP[(dx 2 dx x ) cos a -\- (dy 2 dy x ) cos /3 -f- (dz 2 dz x ) cos 7]. From the equation r> = (x 2 -x t y + {,, -y,f + (*, - z,f, there results, by differentiation, rdr = Car, x x )(dx 2 dx x ) + (7, y^{dy x ^) -f fe ^i)(^2 dzd ; from which, since (x 2 *i)/r = cos a, ( jj/ 2 jO/r = cos A (^ s x )jr = cos 7, we have dr = {dx t dx x ) cos a -\- (dy. 2 dy x ) cos ft -\~ (dz % dz x ) cos 7. The above value of the total work done by the two forces of the stress therefore reduces to Pdr, r as before. Two particles rigidly connected. If the two particles A and B are rigidly connected, or if they move in such a way that the distance between them remains constant, the work done by the stress is zero ; for if r is constant, the definite integral (1) is zero, whatever the value of P, so long as it is finite. THEORY OF ENERGY. 407 The above reasoning obviously holds for any two equal and oppo- site forces applied to two particles and directed along the line joining them, whether the two forces constitute a stress, or whether they are exerted by other bodies or particles. 483. External and Internal Work Defined. Work done upon a system of particles is external if done by an external force (a force exerted by a particle not belonging to the system) ; it is internal if done by an internal force. 484. Configuration. A definite set of relative positions of the particles of a system is called a configuration. If all the distances between the particles remain constant, the configuration remains constant. If the configuration does not change, the system either remains at rest or moves as if the particles were rigidly connected. 485. Internal Work Done Only During Change of Configura- tion. Every internal force is one of the forces of an internal stress. The work done by a stress is zero unless the particles between which the stress acts change their distance apart (Art. 482). The total work done by the internal forces upon the particles of a system is therefore zero unless the configuration changes. 2. Energy of Any System of Particles. 486. Energy of a System Defined.* When the condition of a system is such that it can do work against external forces, it is said to possess energy. The quantity of energy of a system is the quantity of external work it will do in passing from its present condition to some standard condition. The meaning of ' ' condition ' ' will be made clear by the following discussion of the method of estimating the quantity of energy. 487. Kinetic Energy of a System of Particles. It has been shown (Art. 358) that a particle of mass m and velocity v possesses energy of motion or kinetic energy to the amount imv 2 , the "stand- ard condition ' ' being one of rest. A system of particles of masses * The meaning of energy of a system has already been explained in an elementary manner in Chapter XVII (Arts. 360-366). Some of the defini- tions and principles there given are here repeated in order to make the following more rigorous discussion complete. 4 THEORY OF ENERGY. 413 the forces are equal, each being equal to h, and their direction being such that A does positive work and B therefore does negative work. The body A loses energy to the amount Ph, and the body B gains an equal amount. The action between the two bodies has therefore resulted in no change in their total amount of energy ; one has gained exactly as much as the other has lost.* Again, consider two rough bodies A and B in contact (Fig. 192), exerting upon each other forces which have components both normal and tangential to the surface of contact. The normal components are equal and opposite forces JV, and the tangential components are equal and opposite forces T. Let both bodies be displaced parallel to the forces T ; let h ' be the displace- ment of A and h " the displacement of B, their directions being the same, and ^ h' being greater than h" . No work is \tA N- h' T 7\ \A .T N B "> N N ^ done by or against either of the normal I h' forces. The forces T act in such di- Fig. 192. rections as to resist the sliding of one body over the other ; that is, the direction of the force T acting upon A is opposite to the direction of the displacement of its point of application, while the force T acting upon B acts in the direction of the displacement of its point of application. The body A does work equal to -f- Th ', and loses energy equal to + Th'. The body B does work equal to Th" , and gains energy equal to -f Th". The body A therefore loses more energy than B gains. The quantity of energy T{Ji h") has apparently disappeared. If, in the last case, h' = h", so that no sliding occurs, the energy gained by B is equal to that lost by A. If h" = o, B gains no en- ergy, and there is an apparent loss of all the energy given up by A. Another example of an apparent loss of energy is furnished by a moving body suddenly brought to rest, as a falling body striking the earth. The kinetic energy possessed by the body just before strik- ing is wholly lost, without any apparent gain by other bodies. * It should be noticed that nothing is here said of the relation of A or of B to other bodies. The energy received by B from A may be continually transferred, in whole or in part, to other bodies. 414 THEORETICAL MECHANICS. An example of the transfer of energy from one system to another is furnished by a body moving upward against the force of gravity. A body of mass m, moving upward with velocity v, possesses kinetic energy equal to \mv 2 . As it rises, its kinetic energy decreases and finally is wholly lost. Is this loss accompanied by a gain on the part of other bodies or systems ? The other body concerned is the earth ; obviously the system consisting of the earth and the body has gained an amount of potential energy equal to mgh, if h is the vertical distance ascended by the body. If, during this ascent, the velocity of the body changes from v x to v 2 , its loss of kinetic energy is \m(y? vf). But mgh and im(v* v 2 l ) are equal (Art. 227); that is, the gain of potential energy by the system of the earth and the body is exactly equal to the loss of kinetic energy by the body. The last example is an illustration of what occurs whenever a conservative system suffers a change of configuration without doing external work. The individual members of the system may gain or lose kinetic energy, but the system at the same time loses or gains an equal amount of potential energy. If, in the supposed case, the earth and the body be regarded as a system, their energy does not vary in total amount, but merely changes, in part from kinetic to potential. To summarize: When a body or a system loses energy, there may be (a) an equal gain by other bodies or systems ; (b) a gain by other systems apparently less than the loss by the given system ; or (c) no apparent gain by other systems. In some cases, then, there seems to be a disappearance of energy. 496. Equivalence of Energy and Heat. In many cases in which energy disappears, heat is generated, and it is found that there is a definite relation between the quantity of energy that disappears and the quantity of heat that is generated. Taking as the unit quan- tity of heat the amount which will raise the temperature of a pound of water one degree Fahr. (the British thermal unit), the disappear- ance of 778 foot-pounds of energy is accompanied by the generation of 1 unit of heat. * *To make the unit heat definite, the temperature of the water should be specified, since the quantity of heat required to raise the temperature of a pound of water one degree is not exactly the same at all temperatures. The numerical value of the energy-equivalent of the unit heat, expressed in foot- THEORY OF ENERGY. 415 In other cases, as in the steam-engine, energy comes into exist- ence while heat disappears ; one unit of heat being consumed for every 778 foot-pounds of energy generated. From these facts it is natural to conclude that heat and energy are equivalent, and that a given quantity of one may be transformed into an equivalent quantity of the other. 497. Forms of Energy. Modern science goes further and re- gards heat not merely as equivalent to energy but as being actually a form of energy. Moreover, heat is but one of several forms of en- ergy into which the ordinary kinetic and potential energy possessed by bodies and systems may be converted, either directly or indirectly. The meaning of the word energy is thus enlarged ; and we recog- nize mechanical energy, heat energy, chemical energy, electrical energy. Mechanical energy. Energy which depends upon the config- uration of a material system or upon the motions of its constituent particles is mechanical energy. The discussion included in Arts. 486-494 refers solely to this kind of energy. Heat energy. The reasons for regarding heat as a form of en- ergy are given above. Chemical energy. The chemical combination of two bodies is often accompanied by the generation of heat. In this case there is found to be a definite relation between the amount of chemical change and the quantity of heat generated. Thus, if one pound of water is produced by the combination of hydrogen and oxygen, the quantity of heat generated is 6,900 British thermal units. Again, by the ex- penditure of this amount of heat (or of its equivalent in other forms of energy), a pound of water may be decomposed into hydrogen and oxygen. In such a case as this the two bodies when uncombined may be said to possess chemical energy. Electrical energy. Another important form of energy is pos- sessed by bodies by reason of their electrical condition. If a body pounds, varies with the value of the pound-force, which varies with the local- ity. The value 778 foot-pounds is sufficiently correct for most purposes. Expressed in British absolute units of work (foot-poundals) the energy- equivalent of the unit heat is approximately 25,000. Using French units, the quantity of heat required to raise the tempera- ture of a kilogram of water a degree Cent, is equivalent to about 427 meter- kilograms of energy. 41 6 THEORETICAL MECHANICS. is charged with electricity, and if the conditions are such that there can be a flow of electricity accompanied by a fall of potential,* the body possesses electrical energy. If the flow actually takes place, the electrical energy is transformed into energy of some other form. In the case of an electric current, there is a continuous flow of electricity and a fall of potential in the direction of the flow. A part of the electrical energy is transformed into heat which manifests itself in the heating of the conductor carrying the current. A part may be transformed into mechanical energy by means of a motor. Or the current may be employed in decomposing a chemically com- pound substance as water into its elements, a part of the electrical energy being then transformed into chemical energy. Nature of different forms of energy. The nature of each of the above forms of energy is unknown ; any hypothesis concerning it must involve a theory as to the ultimate constitution of matter. A plausible supposition is that heat and chemical energy are dependent upon the motions and relative positions of the ultimate particles of which bodies are made up. On this supposition, these forms of en- ergy do not differ in nature from the kinetic and potential energy possessed by any system by virtue of its configuration and of the visible motions of its parts. Thus, the accepted theory regarding the nature of heat is that it is kinetic energy possessed by the mo- lecules of bodies ; and a possible explanation of the development of heat during the chemical combination of two substances is that there is a transformation of mechanical energy from potential into kinetic. The atoms of hydrogen and those of oxygen, when brought into certain relations, may exert upon each other forces, under the action of which the particles approach one another or assume definite rela- tive positions ; during this process work may be done upon the particles by these forces, and their kinetic energy may thus be in- creased, the molecular kinetic energy being heat. Before the com- bination the two bodies, regarded as a system, possess energy by virtue of the relative positions of their particles and of the forces exerted between them. If this is the true explanation, the energy, both before and after the combination takes place, is of the same * No explanation of the meaning of electrical potential can be given here. The above brief statement regarding electrical energy presupposes an ele- mentary knowledge of the theory of electricity. THEORY OF ENERGY. 4J7 nature as ordinary mechanical energy ; being at first potential, and then transformed into kinetic. Attempts have been made to explain electrical energy as me- chanical energy possessed by the hypothetical "ether." The supposition that all energy is really mechanical may or may not become more probable with the advance of knowledge ; the fact that the different forms of energy are equivalent in the sense that a definite quantity of one can be converted into a definite quantity of another is independent of any hypothesis as to their nature. 498. Conservation of Energy. One of the most important re- sults of modern science is the establishment of the principle that no system of bodies gains or loses energy wihout an equivalent loss or gain by other bodies or systems. 499. Conservation of Energy in Machines. A machine is a structure designed to do work against external forces. In doing external work, a machine loses an amount of energy equal to the work done. Energy must therefore be supplied to it if it is to continue to do work indefinitely. The operation of a machine thus involves (a) the transfer of energy to the machine from other bodies, and (Jj) the transfer of energy from the machine to other bodies. In general, these two processes go on simultaneously at nearly the same rate. If they go on at unequal rates, the quantity of energy possessed by the machine fluctuates. If energy is received by the machine and given off by it at the same rate, the machine is a body or system in equilibrium. In this case the relations which must hold among the forces acting upon it may be determined by the principles of Statics. Machines have been considered from this point of view in Part I. The theory of energy furnishes a means of dealing with machines in a more general manner. In the light of the principle of the conservation of energy, a machine may be defined as a structure designed to transfer energy from one body or system to another. 500. Utilized Energy and Lost Energy. That part of the en- ergy given up by a machine which is equivalent to the useful work done may be called utilized energy. The energy given up by the machine can never be wholly utilized. In order that the parts of the machine may move in the desired manner, they must be guided by other bodies ; these exert frictional forces, against which work must 41 8 THEORETICAL MECHANICS. be done. In doing work against the frictional forces, energy is trans- formed into heat. Such energy may be called lost, since no useful effect can be obtained from it. The external forces acting upon a machine may be classified (as in Art. in) as efforts, resistances (useful and prejudicial), and constraints. In terms of work and energy these may be defined as follows : An effort is a force which does positive work upon the machine. The machine receives a quantity of energy equal to the work done. A resistance is a force which does negative work upon the ma- chine (or against which the machine does positive work). The machine gives up a quantity of energy equal to the work done. Energy given up in doing work against useful resistances is utilized ; it is lost if given up in doing work against prejudicial resistances. A constraint is a force which does no work, positive or negative, upon the machine. The office of a constraint is to guide the motion of some machine part. No energy is gained or lost by the machine by reason of a constraint. 501. Equation of Energy for a Machine. The transformations of energy with which a machine is concerned during a given time may be expressed algebraically, in the the most general case, as follows : Let W = work done upon the machine by the efforts = energy received by the machine ; W' = work done by the machine against useful resistances energy utilized ; W" =a work done by the machine against prejudicial resistances = energy lost ; E = increase in the quantity of energy possessed by the machine. Then W = W + W" + E. If the energy possessed by the machine remains constant, the equation is W = W + W". The same equation holds if the interval is so chosen that the energy possessed by the machine has the same value at the end as at the beginning of the interval, even though it has fluctuated during the interval. THEORY OF ENERGY. 419 502. Efficiency. For the efficient working of a machine, the energy lost is to be made as small as possible. The efficiency of a machine is the ratio of the energy utilized to the energy received by the machine, the energy stored in the machine being supposed to remain constant (or the interval of time to be so taken that the total energy lost by the machine has equaled the total energy received). Consider an interval such that the energy stored in the machine has the same value at the beginning and at the end of the interval, so that E = o. The efficiency may be expressed by the equation e= W/W= W/(W + W"). If W" could be made zero, the efficiency would be i ; its actual value is always less than i. Examples. 1. The fuel used in running a steam-engine is coal of such com- position that the combustion of i lb. produces heat sufficient to raise the temperature of 12,000 lbs. of water i Fahr. It is found that 3^ lbs. of fuel are consumed per horse-power per hour. What is the efficiency of the entire apparatus ? Ans. 0.061 nearly. 2. A steam-engine uses coal of such composition that the com- bustion of 1 lb. generates 10,000 British thermal units. If 40 lbs. of coal are used per hour, and if the efficiency is 0.08, what horse-power is realized ? 3. A dynamo is driven by an engine working as described in Ex. 2. If its efficiency is 0.78, what " activity" in kilowatts is repre- sented by the current generated ? 4. Principle of Work and Energy Applied to Rigid System. 503. Potential Energy of Rigid System. It was shown in Art. 485 that internal work is done only when the configuration of the system changes. In any possible displacement of a rigid system, therefore, the internal work is zero. It follows that the potential energy of a rigid system is zero. 504. Kinetic Energy of Rigid System. When the condition of motion of a rigid system at any instant is known, the value of the kinetic energy can be expressed in a simple manner. (a) Translation. In case of translation the velocities of all par- ticles are equal in magnitude. The kinetic energy is therefore 420 THEORETICAL MECHANICS. K = \m x v? + \m { o? + . . . = \Mv\ . (i) if M is the total mass of the system and v the velocity of every particle. (J?) Rotation about fixed axis. Let r lt r %i . . . be the distances from the axis of rotation of particles whose masses are m lt m %y . . . ; / the moment of inertia of the system with re- spect to the axis of rotation ; (o the angular velocity at any instant. The velocities of the several particles have values r x (o, r z co } . . . , and the kinetic energy is therefore K = \m x {r x f -f- imr, is constant the whole change in K is equal to the change in the term ^Mv 2 . Equation (4) may therefore be applied to any case in which the re- sultant of the external forces is a single force whose line of action passes through the mass-center of the body ; since the angular veloc- ity is constant when this condition is fulfilled (Art. 444). In the case of translation, the work done by any force may be computed as if the force were applied at the mass-center, since its actual point of application receives, in any interval, a displacement equal and parallel to that of the mass-center. In the case of uniform angular velocity, the actual displacement of the point of application of each force must be used * in comput- ing W. 507. Equation of Work and Energy in Case of Rotation. If a body rotates about a fixed axis, equation (2) gives the value of the kinetic energy at any instant. If o>' and on" are the initial and final values of the angular velocity, the equation of work and energy may be written in the form w=\ii = angle between OA and P; 6 angle between OA and some fixed line in the plane of motion. *See, however, Art. 509. 422 THEORETICAL MECHANICS. If the body receives an infinitesimal angular displacement d6, the displacement of A is rdO, and its direction is perpendicular to OA. The work done by the force is therefore dW= Prsm<}>-dd. But L = Pr sin ; hence dW=LdO. That is, for an infinitesimal displacement, the work done is equal to the product of the Fig. 193. moment of the force into the angular dis- placement. For a finite displacement, let 6' and 6" be the initial and final values of 6 ; then /.: W=\ Ld6. J 0' If L remains constant during the displacement, W=L(d" - 6'). 509. Equation of Work and Energy in Any Plane Motion. Let a rigid system, restricted to plane motion, be acted upon by any forces, and receive any displacement during a certain interval. Let M, /, /, v, (o, , 6 have the same meanings as in Arts. 443, 446 ; let P denote the vector sum of the applied forces, and L the sum of their moments about the mass-center. The equation of motion of the mass-center is (Art. 445) P=Mp, being identical with the equation of motion of a particle of mass M acted upon by a force P. Resolving P and p along the tangent to the path described by the mass-center, and integrating the resulting equation exactly as in Art. 353, there results the equation W x = \M{y"*-v'% . . . (6) in which v' is the initial and v" the final value of v y and W l is the total work done by the applied forces on the assumption that they act at the mass-center. The equation of angular motion is (Art. 446) L = /(dcofdt). THEORY OF ENERGY. 423 Multiplying this equation member by member by the equation dd = codt, . LdQ === lot dco. a* Integrating, f Ldd = \I{w" 2 " are the initial and final values of a>, and 0', 6" the initial and final values of 6. The first member of this equation is equal to the work which would be done by the actual forces if the rotation occurred about a fixed axis through the mass-center (Art. 508). Calling this quantity W W 2 = il("> -<*>'*). . . . ( 7 ) Referring to the value of the total kinetic energy of a rigid sys- tem, given in equation (3), it is seen that the values of W x and W t given by equations (6) and (7) are equal respectively to the incre- ments of the two portions of the kinetic energy which have been called energy of translation and energy of rotation. Hence the following important principle may be stated : In any plane motion of a rigid system^ the increment of the energy of translation is equal to the work done by the external forces com- puted as if they were applied at the mass-center ; the increment of the rotational energy is equal to the work done by the external forces computed as if the rotation occurred about a fixed axis through the mass-center. Examples. 1. A homogeneous cylinder of mass 20 lbs. and radius 6 ins. rotates about its axis of figure at the rate of 300 rev. -per-min. Re- quired the value of the kinetic energy in foot-pounds. 2. The mass of a wheel-and-axle is 50 lbs., its radius of gyration with respect to the axis of rotation is 10 ins., and the radius of the axle is 6 ins. It is set rotating by a constant tension of 5 pounds- force in the rope which unwinds from the axle. Required the angular velocity of the body after making 4 rev. , starting from rest. Ans. 1.72 rev.-per-sec. 3. Take data as in Ex. 2, except that the tension is due to a weight of 5 lbs. suspended from the rope, {a) Determine the angular velocity of the body after making 4 rev. (b) Determine the tension in the rope. [Neglect the weight of the rope in both cases. ] Ans. (a) 1.69 rev.-per-sec. (b) 4.91 lbs. 424 THEORETICAL MECHANICS. 4. A body of any shape is projected upward against gravity ; determine the height to which the mass-center will rise, if no external force acts except gravity. Solve by the principle of work and energy. [Notice that the reasoning by which equation (6) was deduced is valid without restricting the motion to a plane.] 5. Apply the equation of work and energy to a compound pen- dulum, and deduce equation (5) of Art. 425. 6. Solve Ex. 9, Art. 447, by the principle of work and energy. 7. Solve Ex. 10, Art. 447, by the principle of work and energy. 5. The Principle of Virtual Work. 510. Equation of Virtual Work for a System of Particles. It has been shown (Art. 369) that if a particle in equilibrium be assumed to receive any arbitrary small displacement, the total work done upon it by all forces is equal to zero. By simple addition this principle may be extended to any system of particles. That is, If every particle of a system is in equilibrium, the total work done upon all the particles during any arbitrary small displacements is zero. In general the ' ' equation of virtual work ' ' obtained by applying this principle involves both internal and external forces. In many important cases, however, the displacements may be so taken as to eliminate certain of the internal forces. 511. Equation of Virtual Work for Rigid Body. Since the total internal work is zero for a displacement which leaves the con- figuration unchanged, the internal forces may be omitted from the equation of virtual work for a rigid body. That is, If a rigid body is in equilibrium, the total work done by the ex- ternal forces during any arbitrary displacement is zero. In applying this principle, it is generally necessary to take the displacement infinitely small, since a finite displacement will so change the relation among the lines of action of the forces that they will no longer be in equilibrium. Since the negative of any one of a number of forces in equilibrium is the resultant of all the rest, and since reversing a force reverses the sign of the work done by it without otherwise changing the value, it follows that the total quantity of work done by any number of ex- ternal forces acting upon a rigid body during any displacement is equal to the work done by their resultant. THEORY OF ENERGY. 425 512. Equations of Equilibrium for a Rigid Body. The gen- eral equations of equilibrium for a rigid body, already deduced in Part I, may be obtained by applying the principle of virtual work. Thus, if a body be conceived to receive an arbitrary small trans- lation in a given direction, the virtual work of any force is equal to the product of the displacement into the resolved part of the force in the direction of the displacement ; and the total virtual work is equal to the product of the displacement into the sum of the re- solved parts of the forces in the direction of the displacement. Since this total work is zero, whatever the direction of the displacement, it follows that, for equilibrium, the algebraic sum of the resolved parts of the forces in any direction is zero. Again, let the virtual displacement be a rotation about a fixed axis M. The work done by any force is equal to the product of the angular displacement into the moment of the force about M (Art. 508) ; and the total virtual work is equal to the angular displacement into the sum of the moments of the forces about M. Since this total work is zero, whatever the position of the axis of rotation, it follows that, for equilibrium, the sum of the moments of the forces about any axis is equal to zero. These two general forms include all the equations of equilibrium for forces applied to a rigid body, whether in two dimensions or in three, as already given in Chapter X. 513. Connected Bodies. If the principle of virtual work is ap- plied to a system of rigid bodies connected in any manner, as by hinges or strings, or pressing against one another, the equation of work must in general include {a) all external forces acting upon any of the bodies and (b) the forces exerted by the bodies upon one another by reason of their connections. In certain important cases, however, forces due to connections may be omitted from the equation because it can be seen that their work vanishes. Thus, any two adjacent portions of a tense string exert upon each other forces constituting a tensile stress. If the distance between the two portions remains constant, the total work of the stress is zero (Art. 482). But if the length of the string changes, the total work of the internal forces will not be zero. If a system includes two bodies which are connected by a string, the tension in this string will not enter the equation of virtual work for the system if the dis- 426 THEORETICAL MECHANICS. placement is such that the string remains tight and of unchanged length. The same is true of the contact-stress between two bodies touch- ing each other, if the surfaces of contact are smooth and the dis- placement is such as not to destroy the contact. Thus, the action and reaction between two bodies connected by a smooth hinge joint may be omitted from the equation of virtual work. But if the sur- faces are rough and the displacement is such that sliding occurs, the frictional work must enter the equation. 514. Solution of Problems in Equilibrium. The principle of virtual work furnishes a simple solution of many problems relating to the equilibrium of rigid bodies, or of connected systems of bodies. The method of applying the principle is to assume an arbitrary in- finitesimal displacement, compute the work done by all the applied forces, and equate the total work to zero. By choosing different displacements, different equations may be obtained. In general, for a rigid body acted upon by coplanar forces, three independent equa- tions may be formed. Any force may be eliminated by taking the displacement of its point of application perpendicular to the direction of the force. By this means two external forces acting on a rigid body may be elim- inated, since the directions of displacement of two points may be chosen arbitrarily. 515. Simple Machines Treated by Principle of Work. The principle of work often furnishes the most convenient method of de- termining the relation between effort and load in the case of a simple machine or a combination of simple machines. System of pulleys. To determine the relation between effort and load in case of a system of pulleys arranged as in Fig. 1 94. The system consists of two sets of pulleys, each set comprising three pulleys mounted freely upon a common axis and free to rotate independently of one another ; and a cord passing continuously around the pulleys as shown, one end being attached at A, and the effort P being applied at the other end. The axis of the set A is firmly attached to a fixed support C, while the load W is applied to B. This * ' load ' ' may be the weight of a heavy body, or it may be a force applied horizontally or in any other direction. Assuming the cord to be perfectly flexible and inextensible, and THEORY OF ENERGY. 427 neglecting axle friction, the principle of work may be applied as follows : Let the system be displaced in such a way as to shorten the dis- tance AB by an amount h, the cord remaining tense. The straight portions of the cord being practically parallel to AB, the end of the cord moves a distance 6k. The effort P and the resistance W are the only forces entering the equation of work for the system composed of the two sets of pulleys and the cord, and we have work done by effort = 6P/1 ; " " load = Wh. The equation of work is therefore 6Pk Wh = o; .-. P W/6. Train of toothed wheels. The principle of work may be applied without difficulty to any connected series of toothed wheels, one of which is acted upon by a force tending to produce rotation, and another by a counterbalancing force. Fig. 195 represents three pairs of wheels, freely mounted upon fixed axes at A, B and C, so connected by teeth that an angular displacement of one must be accompanied by a proportional angular displacement of each of the others. A force P, applied as shown, tends to pro- duce a displacement in one direction, while Q tends to produce an opposite dis- placement. The relation between the displace- ments of any two points of the system may readily be determined from the di- mensions and manner of connection. Neglecting axle friction and friction at the surfaces of contact of the teeth, the equation of work for the system shown will contain no forces except P and Q. If a and b are the distances moved by the points of application of P and Q respectively, the equation of work is Pa Qb = o. Pulleys and belts. Let motion be transmitted from one set of machinery to another through a pair of pulleys, rigidly connected Fig. 195. 428 THEORETICAL MECHANICS. and mounted freely upon a common axis, each connected by a belt with another pulley from which it receives (or to which it imparts) motion by means of the belt. (Fig. 196.) Assuming the belts to be perfectly flexible and inextensible, and neglecting *"* axle friction, the equation of work may be written' as follows : "** Let a and b be the radii of the two Fig. 196. pulleys, T and T' the tensions in the two portions of one belt, S and S' the tensions in the two portions of the other. Then, for any angular displacement A6, the equation of work is TaAd T'aAd SbAO + S'bAO = o; or (T T')a=(S S')b. Screw. A screw, used as a machine, is arranged to work in a fixed nut, and is usually provided with an arm rigidly attached to the screw and at right angles to its length. The effort is applied at the end of this arm and perpendicularly to it and to the axis of the screw, while the load is applied at the end of the screw, parallel to its length. The load may be the weight of a heavy body which is to be lifted, the fixed nut being supported upon a solid foundation in the earth. Neglecting friction, the relation between load and effort may be found as follows : Let P = effort, applied at distance a from the axis of rotation of the screw, in a direction perpendicular to the arm and to the axis of the screw ; W = load, applied in a direction parallel to the axis of rotation ; b = pitch of screw (distance between centers of two con- secutive threads, measured parallel to the axis of rotation). Let the screw and arm rotate at a uniform rate, so that the external forces are in equilibrium, and let the work done by all forces acting upon the body be computed for one revolution. The work done by P is 2iraP ; that done by W is Wb. The forces exerted by the fixed body do no work, if friction is neglected. The total work done upon the body is 2iraP Wb = o ; whence P == \Vb\21ra. THEORY OF ENERGY. 429 By making b small, W may be made very great in comparison with P. The following examples are designed to illustrate the application of the principle of virtual work. They are all problems in Statics, and may also be solved by the methods of Part I. Examples. 1. Determine the position of equilibrium of a bar resting with both ends against the inner surface of a smooth hemispherical bowl. If the bar be displaced from the position of equilibrium in such a way that the ends slide along the surface of the bowl, the work done by the reactions is zero ; hence the work done by gravity (the only other external force) must be zero. This requires that the displace- ment of the center of gravity shall be horizontal. In order that this requirement shall be fulfilled for every possible sliding displacement, the center of gravity must be vertically below the center of the bowl. 2. Find the position of equilibrium of a heavy bar resting with its ends upon two smooth inclined planes. Let AB (Fig. 197) represent the bar, C being its center of gravity, and let it receive a displacement such that the ends slide upon the planes. Reasoning as in Ex. 1, it is seen that, if the bar is displaced from the position of equilibrium, the displacement of C must be hori- zontal if the ends slide along the planes, since the work done by the weight of the bar must then be zero. From this condition and the geometry of the figure, the position of equilibrium may be determined. Let 6 be the inclination of the bar to the horizontal, and y the vertical dis- _ _ jSz^^/-@. tance of the center of gravity above a p IG# I9 y # horizontal plane through the line of in- tersection of the supporting planes. Let a and /3 be the inclinations of the planes to the horizontal, A C = a, CB = b. Let y be ex- pressed as a function of 6 ; the result is y = {a + b) 21? sin (fi 0) + a sin 6. sin (a fi) As the bar passes through the position of equilibrium while sliding on the planes, dy/dd = o. Differentiating and reducing, (a -f- b) tan 6 = a cotan a b cotan ft. 3. A bar 7 ins. long, whose mass-center is 3 ins. from one end, rests within a smooth hemispherical bowl 1 2 ins. in diameter. What is the inclination of the bar to the vertical when in equilibrium ? Ans. 84 9. 43o THEORETICAL MECHANICS. 4. A bar 1 2 ins. long, whose mass-center is 7 ins. from one end, rests upon two planes inclined 30 to the horizontal. Determine the position of equilibrium. Ans. 6 = 16 6'. 5. A straight bar rests with one end upon a smooth inclined plane and leans against a smooth peg. Determine the position of equilibrium. 6. A bar leans against a smooth peg, the lower end resting upon a smooth surface. Determine the form of the surface in order that the bar may be in equilibrium wherever placed. An s. A section of the surface by a vertical plane is represented by the polar equation r = a -\- c sec 6. 7. A uniform bar AB, of weight G, can turn freely in a vertical plane about a hinge at A . To B is attached a string which passes over a smooth peg at C and sustains a weight P. Determine the position of equilibrium. 8. A screw of % in. pitch is used to raise a weight of 5,000 lbs. Neglecting friction, what force must be applied at the end of an arm 3 ft. long? 9. Determine, by the principle of work (neg- lecting friction), the relation between the effort (P) and the load ( W) in case of the differential wheel- and-axle (Fig. 198). Ans. P/IV= (d' d)/2a. 10. In Ex. 9, let the radius of the wheel be 2 ft. and the radii of the two portions of the axle 6 ins. and 4 ins. If the load is 1,000 lbs., what must be the value of the effort ? 11. A string AB is attached to a uniform bar BC of which the end C rests against a smooth ver- Find the position of equilibrium. Ans. If the inclinations of the string and bar to the vertical are 6 and respectively, tan is the *A concrete idea of what is meant by the motion with respect to the platform may be obtained by considering how the motion will appear to an observer stationed upon the platform, whose view is so obstructed that the earth and the bodies fixed upon it are invisible to him. To his experience the platform is a " fixed " body, and is the only base available for estimating the motion of the ball. On the other hand, if a person stationed upon the earth can observe the ball, he may take the earth as his base for estimating its motion. The path described by the ball will appear quite different to the two observers, and its velocity and acceleration as estimated by them will have quite different values. 434 THEORETICAL MECHANICS. angular velocity of the platform and r the distance of the body from the axis of rotation, it has an acceleration with respect to the earth equal to o>V directed toward the axis. The general equation force = mass X acceleration therefore shows that the resultant of all forces acting upon the body is a force of magnitude mroo? directed toward the center of the cir- cular path; m being the mass of the ball. As another case, suppose the path of the ball upon the platform is a straight line passing through the axis of rotation, and that the velocity estimated with respect to the platform is constant. If the laws of motion hold when the platform is base, the resultant force acting upon the ball must be zero, since its acceleration is zero. But with respect to the earth the body describes a spiral with varying speed ; hence if the laws of motion be applied with the earth as base, the inference is that the resultant of all forces acting upon the ball varies both in magnitude and in direction. The actual forces acting upon the body are, however, the same whatever be the base of reference. Our conception of a force is that it is an action of one body upon another ; * and the actions of other bodies upon the ball at any given instant cannot be supposed to depend upon whether the motion is estimated with respect to the earth or to the platform. It is thus evident that if the laws of motion are true, and if our definition of force (which forms an essential element in the interpre- tation of those laws) is to be retained, it is not permissible to refer motions to a base chosen arbitrarily in applying the laws. It is necessary, in fact, to raise the fundamental question whether there is any base for which the laws of motion are true, f 519. Ultimate Base. It is obvious that, unless there is a base for which the laws of motion are true, these laws are unintelligible. % *See Art. 212. tit may be well to remark that this is not the same as the question whether such a base can be exactly determined by experiment. % Maxwell, in his discussion of Newton's first law ( "Matter and Motion," Chapter III), argued that no alternative law can be stated which is intelligible "unless we admit the possibility of defining absolute rest and absolute veloc- ity." And denying this possibility, he appeared to regard this as a priori evidence of the validity of the law. It would seem, however, that the argu- RELATIVE MOTION. 435 That such a base is a reality must in fact be regarded as an essential part of the laws themselves. Definition. A body or frame of reference for which the laws of motion are true may be called an ultimate base. 520. Time. In order to specify motion it is necessary to adopt some scale for estimating time. That is, some means must be adopted for comparing different intervals of time. The first essential is to be able to specify successive equal inter- vals of time ; and for this purpose the only method practically avail- able is to observe some body which is supposed to move uniformly, or to perform a definite series of motions in uniformly recurring cycles. Thus, we may take as equal the successive intervals occu- pied by complete revolutions of the earth upon its axis ; or the inter- vals occupied by successive vibrations of a pendulum. Now the time-scale specified by the motion of one body will not in general agree exactly with that specified by the motion of another, and the question arises which, if any, of the scales thus denned is the "true" time-scale. So far as a mere description of the motions of bodies is concerned, any time-scale will serve. The actual values of the velocity and acceleration of a particle will depend upon what scale is adopted, but a time-scale having been chosen, all motions can be described in terms of it. Some writers, in fact, assert that one time-scale gives as correct a description of motions as another, and that the only reason for preferring one to another is a gain of simplicity. It is obvious, however, that the laws of motion cannot be true independently of the time-scale. Thus, the first law asserts that a body not acted upon by force (i. e. , not influenced by other bodies) would move uniformly in a straight line. But motion which is uni- form when one time-scale is used may not be uniform when another is used. The laws of motion are, in fact, wholly unintelligible unless we ment which Maxwell applied to the negation of the law can be applied with equal validity to its affirmation. If the reasoning is valid, the conclusion appears to be that the laws of motion as stated by Newton are unintelligible. That Newton's conception of absolute motion corresponds to a reality can hardly be denied, unless we are prepared to discard the whole New- tonian system. Whether the term "absolute" is properly applied is another question. 436 THEORETICAL MECHANICS. admit the reality of a time-scale for which they are true ; and this reality must be accepted as a part of the laws themselves. * 521. Absolute Space and Time. The reality of an ultimate base for specifying motion, and of an ultimate scale for estimating time, was explicitly affirmed by Newton. He in fact postulated absolute space and absolute time as the space and time implied in his statement of the laws of motion. Many modern writers have found the Newtonian conception of absolute space and time so repugnant that they have discarded it. The attempt to destroy the force of Newton's arguments has not, however, been successful, and as already stated, there appears to be as much reason for accepting his conception of absolute motion as corresponding to reality as for accepting the Newtonian laws as a sound basis for the science of Mechanics. 522. Absolute and Relative Motion. The motion of a body with respect to an ultimate base may be defined as its absolute mo- tion ; and its motion referred to any other than an ultimate base may be called its relative motion, f In certain cases it is necessary to compare the motions of a par- ticle as estimated with respect to different bases. In such cases it is convenient to call one of the motions absolute, even though the base of reference is not an ultimate one. Thus, there are many practical problems in which the motion of a particle with reference to the earth has to be compared with its motion referred to some body which is itself in motion with respect to the earth. The theory of turbine water-motors presents such a case, it being needful to esti- mate the motion of the particles of water both with respect to the rotating wheel and with respect to the earth. The latter may be called the absolute motion to distinguish it from the former. In such practical problems the earth is, in fact, the ultimate base to which we refer motions, although it is known not to be a true ulti- mate base as above defined. The error which results from regard- ing the earth as an ultimate base is for many purposes unimportant. \ *It may be well to remark that to affirm the reality of a time-scale which is independent of the motions of actual bodies is not to affirm that we can practically attain to such a scale. t However objectionable the word ' ' absolute ' ' may be in itself, its em- ployment seems to be justified by usage. X A more definite discussion of the earth as a base is given in Art. 531. RELATIVE MOTION. 437 2. Transformation from One Base to Another. 523. Change of Base. In the foregoing discussion the impor- tance of the choice of a base for specifying the motion of a particle has been shown by simple illustrations, but the influence of the choice of base upon the values of the velocity and acceleration has not been explained in an exact manner. A mathematical discussion will now be given, showing how the values of the quantities which specify the motion are changed by changing from one base of refer- ence to another having a given motion with respect to the first. To discuss the problem in its full generality would be beyond the scope of this book, since the treatment already given of the mo- tion of a particle and of a rigid body does not go beyond plane mo- tion and motion of translation. The following discussion is there- fore restricted in the same manner. The case of translatory motion will be considered first ; then a restricted case of plane motion (that in which the relative motion of the bases is a uniform rotation) ; and finally the general case of plane motion. Of the two bases or frames to which the discussion refers, one may be called the primary, the other the secondary ; and the motion referred to the former will often be called the absolute motion, while that referred to the latter is called the relative motion. 524. Case in which the Relative Motion of the Bases is Translatory. Let the motion of a particle be referred to a base M\ and also to a base M which itself has any motion of translation with respect to M' . Let x ', y\ z' be the coordinates of the particle referred to a set of rectangular axes 0'X\ O'Y', O'Z', fixed in M' ; x, y, z its coordinates with respect to axes OX, O Y, OZ, fixed in M; and x , y , z Q the coordinates of the origin O referred to the axes OX', O'Y', O'Z'. The position, velocity and acceleration of the particle with respect to M' are given by the values of x , y , z , and their first and second derivatives with respect to the time ; the position, velocity and acceleration with respect to M by the values of x, y, z, and their first and second derivatives. Also, x , y , z , and their derivatives specify the position, velocity and acceleration of .^f relative to M' . 438 THEORETICAL MECHANICS. The relation between these different motions is expressed by the following equations : x* = x % + x, \ y' =io + y> > . . . . (i) z' = z -f z; ) CO x' = x + X, y = >o + >> z = z + z ; x' == -^o ~i X, y = yo + j ; > r = 4 -f- * ; (3) Equations (2) express the proposition that The velocity of the particle with respect to the base M' is equal to the vector sum of its velocity with respect to M and the velocity of M with respect to M '. Similarly, equations (3) state that The acceleration of a particle with respect to the base M' is equal to the vector sum of its acceleration with respect to M and the accel- eration ofM with respect to M '. Special case. If M has a uniform rectilinear motion with respect to M ', x , y , and z are all zero. In this case the acceleration of a particle with respect to M' is equal to its acceleration with respect to M. Examples. 1. A railway train moves uniformly on a straight track at the rate of 25 miles per hour. A passenger throws a stone so that to him it appears to move at right angles to the track with a velocity r 20 ft. -per-sec. What is its velocity with respect to the earth ? 2. In Ex. 1, how must the stone be thrown in order that its velocity referred to the earth shall be 20 ft. -per-sec. at right angles to the track ? 3. A railway train moves along a straight track in such a way that in haif a minute its speed increases uniformly from 10 miles per hour to 35 miles per hour. A stone dropped from a car window has what acceleration relative to the earth while falling to the ground ? What acceleration has it relative to the train ? 4. In Ex. 3, compute the velocity of the stone relative to the train after it has fallen 4 ft. vertically. RELATIVE MOTION. 439 525. Case in which the Relative Motion of the Bases is a Uni- form Rotation.* Let the motion of a particle be referred to a rigid body or base M', and also to a body M which rotates uniformly about an axis fixed in M' . It will be convenient to speak of M' as a ' ' fixed ' ' body, the motion of the particle relative to it being called the "absolute" motion, and that relative to M the "relative" motion. Let r denote the distance of the particle from the axis of rotation at time t ; ft) the angular velocity of the rotation of M relative to M' ; v y p the relative velocity and acceleration ; v' f p' the abso- lute velocity and acceleration. It is required to find (a) the relation between v and v\ and (^) the relation between p and p' . It will at first be assumed that the particle moves in a plane perpendicular to the axis of rotation. (a) Relation between velocities. Let Fig. 199 represent the plane of the motion, O being the point in which this plane is pierced by the axis of rotation. Let AB be the path traced by the particle on the rotating body during a short interval At. During that time this body turns through an angle coAt, and AB is carried into a position A'B' such that the angles AOA ' and BOB' are each equal to coAt. The path traced by the particle upon the fixed body M' is s6me curve AB' . The velocity of the particle relative to the rotating body at the beginning of the interval At has the direction of the tangent to AB at Ay and the magnitude V lim [(vector AB)/At]. The absolute velocity at the same instant has the direction of the tangent to AB' at A, and the magnitude v* = lim [(vector AB')IAt]. But vector AB' = vector AA ' + vector A'B' ; hence v' = lim [(vector A'B')/ At] + lim [(vector AA')/At]. * This case is a special case of that treated in Art. 526. The geomet- rical discussion here given has the advantage of showing more vividly the relations of the vector quantities involved. 44Q THEORETICAL MECHANICS. If u denotes the velocity of the point A of the body M due to the rotation, lim [(vector A A ')/At] = u. (The direction of u is perpendicular to AO and its magnitude is rco.) Also, in passing to the limit {At approaching o), vector AB may replace vector A'B'. Hence vector v' = vector v -\- vector u. That is, The absolute velocity of the particle is equal to the vector sum of its relative velocity and the velocity of that point of the rotating body which is the instantaneous position of the particle. Fig. 199. ib) Relation between accelerations. In order to compare the ab- solute and relative accelerations, let the value of each be computed in the manner explained in Art. 250. Let Av denote the increment of the relative velocity v during the time At, and Av' the increment of v' during that time. Then / = lim {AvjAt) ; p' = lim {Av'/At). We may first compare Av and Av '. RELATIVE MOTION. 44 1 The value of Av is found by drawing from some point, Q, vectors to represent the initial and final values of v. The former is parallel to the tangent to AB at A and is represented by QA" ( Fig. 199 ) ; the latter is parallel to the tangent to AB at B* and is represented by QB " ; so that vector A" B" = Av. To find the value of Av ' we must in like manner draw from some point vectors representing the initial and final values ofv'. These are parallel to the tangents to the absolute path AB ' at A and at B ' respectively. Having drawn QA" to represent the initial value of v, that of v' is found by combining with QA" a vector RQ of magni- tude rco directed at right angles to OA (representing the velocity u of the point A of the rotating body). This gives RA" as the vec- tor value of the velocity v ' at the beginning of At. Similarly, the final value is found as the sum of two components representing values of v and u at the end of At. We have already drawn QB" to repre- sent the final value of v ; but since we are now referring velocities to the fixed body, the direction of this vector must be made parallel to the tangent to A'B' at B'. That is, we take a vector QT equal in magnitude to QB" but making with it an angle co At (the angle turned through in time At by the tangent to AB at B ). The other component of the final value of v ' is perpendicular to OB ' and of magnitude OB'- co or r 1 co. Represent this latter component by RS, and draw SB'" equal to vector QT; then RB'" represents the final value of v ', and vector A" B'" = Av'. It remains to compare Av and Av '. The following vector equation may be written: Av' = Av + B"T + TB'" = Av + B"T + QS. Also QS may be expressed as the sum of two component vectors QH, HSy the point H being located as follows : Make RH equal in length to RQ and angle QRH = coAt ; then RH is perpen- dicular to OA '. This construction makes the triangle RHS similar * Although the path AB is at A'B' at the end of the interval, QB" is drawn parallel to the tangent at B in the original position, because when the rotating body is taken as base the relation between initial and final velocities is not affected by the rotation. The angle between these velocities is the angle between the tangents to AB at A and B. 442 THEORETICAL MECHANICS. to OA' B' , the corresponding sides being perpendicular each to each. Replacing QS by QH -j- HS in the last equation, dividing through by At and passing to the limit, At approaching o, p' =/ + lim[CT)/A*] + lim [(QH)/At] + Um [GffS)/A/]. (2) It is easy to evaluate the last three terms. Consider them in order, (i) As At approaches o, the angle B"QT approaches o and QB "T approaches a right angle, while QB " approaches the direction QA". Hence the limiting direction of the vector B"T\s perpen- dicular to QA", that is, perpendicular to the relative velocity v at the time /. The limiting magnitude of (B"T)jAt may be found by considering the circular arc B'T whose center is Q. Thus lim [(chord B"T)/At] = lim [(arc B"T)\At\ = lim [(^o) At)/At] = lim \t.\<*>\ = vco. [ v x is written for the value of v at the end of At. ] (2) In a precisely similar manner the limiting direction and mag- nitude of (vector QH)/At may be found from the triangle QRH. Its limiting direction is perpendicular to RQ, i. e., it is AO. Its limiting magnitude is RQa rooco = rco 2 . (3) The vector US is perpendicular to the chord A'B'. As At approaches o, A'B' approaches AB, while chord AB ap- proaches tangent to curve AB at A. Hence the limiting direction of HS is perpendicular to the relative velocity v. The limiting magnitude of (HS)/At is found by comparing the similar triangles RHSy OA'B'. Each side of the former triangle is equal to a) times the corresponding side of the latter. That is, lim [(//5)/A/] = lim [(v A'B')/ At] = - > sin ^J , (5) RELATIVE MOTION. 445 y ,f>. ( x sin 6 -f- y cos ) ( -jj ) + O cos \^/ ') g + 2 [> cos 6 y sin 61 ) ^~| + f^ s i n $ + j; cos 0~| (6) O' Let z; and / represent the velocity and acceleration of the particle referred to the base M, and v\ p' the velocity and acceleration re- ferred to the base M' . Then v is equivalent to components x, y in directions OX, O Y; and / to components x, y in these directions. V Also v' is equivalent to components x', y' in directions O'X', O'Y'; and/' to compo- nents x\ y in these directions. If v and v' were equal in magnitude and direction, the value of x' would equal the sum of the components of x and y in the direction 0'X\ being therefore given by the last two terms of equation (3); while the value of y' would be given by the last two terms of equation (4). The remaining terms in the values of x ' and y ' as given by these equations there- fore represent the difference between v and v' \ that is, they repre- sent the components in directions O'X' and O'Y' of the vector which, added to v, gives v' . Similiarly, the component of p in the direction O'X' is given by the last two terms in equation (5), while its component in the direc- tion O'Y' is given by the last two terms of (6). The remaining terms in the second members of these equations, therefore, represent the difference between/ and/'; that is, they represent the compon- ents in the directions O'X' and O'Y' of the vector which added to /gives/'. It remains to interpret these results. Interpretation of result in case of velocities. The first three terms in the second member of equation (3) give the values ofir' for a particle fixed to the base M, as is seen by the fact that they are Fig. 200. 446 THEORETICAL MECHANICS. the terms which remain when x and y are constant. Similarly for the first three terms in the value of y . Hence it is seen that the vector which represents the difference between v and v ' is equal to the velocity with respect to the base M' of that point of M which momentarily coincides with the particle. We have therefore the general result that The velocity of a particle referred to any base M' is equal to the vector sum of its velocity referred to another base M and the velocity (referred to M') of that point of M which momentarily coincides with the particle. Interpretation of result in case of acceleration. The result is less simple in the case of accelerations. Notice first that the acceleration, with respect to the primary base M\ of that point of the secondary base M which is momentar- ily the position of the particle, is given by the terms remaining in the values of x' and j/' when x and y are constant; that is, the first five terms in the second member of (5) represent its component in the direction O'X', and the corresponding terms in (6) its compon- ent in the direction O'Y'. Again, the last two terms in (5), and the corresponding terms in (6), represent the resolved parts of p in the directions OX', O'Y' respectively. There remain to be interpreted two terms in each of the two equations. In other words, we have to consider what vector it is whose resolved parts are 7/3 2 -7- (i: sin 6 -\- y cos ^ in direction O'X', ) i (7) _|_ 2 ^(^cos 6y sin 0) " " O'Y'.) dt Now ircos 6 y sin = resolved part of v in direction OX', xsmd + ycos6= " " " v " " O'Y'; hence the quantities (7) are the components oi a vector per- pendicular to v, of magnitude 2v{dQ\df) = 2Vco, if co denotes the angular velocity of the base M with respect to M '. It is thus seen that equations (5) and (6) are equivalent to the following proposition : RELATIVE MOTION. 447 The acceleration of a particle relative to any base M' is equal to the vector sum of three components: (/) its acceleration relative to another base M having any plane motion with respect to M\ (2) the acceleration relative to M' of that point of M which momentarily coin- cides with the particle, and (j) an acceleration directed at right angles to the velocity of the particle relative to M y of magnitude equal to twice the product of that velocity into the angular velocity of M with respect to M' . The direction of this third component is that assumed by the vector representing the velocity relative to M if turned through go in the direction in which M rotates relatively to M'. The propositions just proved may be concisely expressed by vector equations. Thus, if u denotes the velocity relative to M' of that point of M which instantaneously coincides with the particle, the relation between velocities is v' = v + u. . . . . (1'") And if the acceleration components numbered (2) and (3) in the above statement be represented by vector symbols p" and/'" respec- tively, the relation between accelerations is >'.-/+>"+/". (3'") These are seen to include equations (1") and (3") of Art. 525. Examples. 1. Show that the above general relation between accelerations reduces to that given in Art. 524 when the relative motion of the two bases is a translation with uniform velocity. 2. Apply the general result to the case in which the relative motion of the two bases is a uniform rotation about an axis fixed in both, and compare with the result reached in Art. 525. 3. In what cases does the component/'" vanish ? 4. A wheel of radius a rolls uniformly along the ground, the velocity of its center being V. Determine the velocity and acceler- ation relative to the earth of a point in the circumference {a) when in its highest position, (J?) when in its lowest position. 5. In Ex. 4, suppose a particle to move uniformly along a diam- eter of the wheel, and that its relative path is vertical when it passes through the center. Determine its velocity and acceleration relative to the earth at that instant. 448 THEORETICAL MECHANICS. 527. Three-Dimensional Motion of a Particle. The foregoing discussion assumes the particle to move in a plane parallel to the relative motion of the bases. In order to remove this restriction, let the point A (Fig. 200) represent the orthographic projection of the position of the particle upon a plane parallel to the relative motion of the bases. Reasoning as in Art. 525, it is seen that When the motion of the particle is unrestricted, the relation be- tween velocities is the same as in the case of plane motion; while tlie relation between accelerations is changed by the substitution for the velocity relative to M of the resolved part of that velocity parallel to the plane of the relative motion of the bases. This change affects only the component p" ' . Example. Show that the above proposition holds when the relative motion of the bases is equivalent to a plane motion combined with any translatory motion. 3. Dynamics of Relative Motion. 528. Equation of Motion for Any Base. The importance of the choice of a base of reference for the purpose of applying the laws of motion was shown in Art. 518. This is made more definite by the foregoing analysis. If the equation force == mass X acceleration holds when the motion of a particle is referred to a base M' , it will not hold when the motion is referred to another base M, unless the acceleration has the same value for the two bases ; and this will be true only if the motion of M relative to M' is a translation with uniform rectilinear velocity. It is, however, easy to write a dynamical equation which will hold for any base which itself moves in a known manner with respect to an ultimate base. Thus, applying the result of Art. 526, let the base M' be an ultimate base, and let P' denote the resultant of all forces acting upon the particle, the remaining notation being as before. The general equation of motion is P' = mp' 5 . (1) or, by equation (3'") of Art. 526, we may write the vector equation P' = mp -j- mp" -f- rnp'" . . . . (2) RELATIVE MOTION. 449 This equation serves to determine/ when the forces are known, if the motion of M relative to M' is known so that / " and p '" can be computed. 529. Fictitious Forces The last equation may be written P' mp" mp'" = mp. . . . (3) Or, if P is written for the vector P' mp" mp"\ P=mp (4) The equation of motion referred to the secondary base M thus has the same form as that referred to an ultimate base, P taking the place of the ' ' resultant force ' ' acting upon the particle. But the forces of which P is the resultant include, besides the actual forces (whose resultant has been called P'), two "fictitious" forces whose values ( mp" and mp'") depend upon the motion of the second- ary base with respect to an ultimate base. Equations of equilibrium. If the particle is at rest relative to the base M, the acceleration/'" is o, and the only fictitious force to be introduced into the equation of equilibrium is the force mp ". Examples. [In these examples it may be assumed that the earth is an ulti- mate base. The error involved in this assumption will be considered in Art. 531.] 1. A ball is thrown back and forth by two persons on the deck of a steamer which moves with uniform speed in a constant direction. How does the motion of the boat affect the game ? What is the effect of a change in the speed of the steamer ? 2. An elevator platform starts from rest with an upward accelera- tion gj<\, then moves upward with constant velocity, then has down- ward acceleration g/6. Taking the platform as base, write the equation of motion for a body resting upon the platform, for each of the three periods of the motion. 3. An elevator platform, starting from rest, moves with a down- ward acceleration which increases uniformly from o to 2g in a period of 4 seconds, (a) Write the equation of motion for a body resting on the platform, taking the platform as base, (b) Determine com- pletely the motion of the body during 4 seconds. (c) How far will it be from the platform at the end of that time ? 45 THEORETICAL MECHANICS. 4. A horizontal platform rotates uniformly about a fixed vertical axis. Write the equations of motion for a body at rest upon the platform, taking the latter as base. Ans. The actual forces acting upon the body are (1) its weight vertically downward, and (2) the supporting force exerted by the platform. Of the fictitious forces, one vanishes, since the velocity relative to the platform is zero ; the other, mp" y is a force directed at right angles to the axis of rotation and away from it. If r is the distance of the particle from the axis of rotation and co the angular velocity of the platform, p * = o>V. 5. In Ex. 2. of Art. 525, assuming the earth to be an ultimate base, (a) determine the resultant of the actual forces acting upon the par- ticle when distant r from the axis, {b) What fictitious forces must be assumed to act if the platform is taken as base ? 6. A particle slides in a smooth straight tube which rotates uni- formly about a vertical axis at right angles to its length. The actual forces acting upon the particle being its weight and the pres- sure of the tube, (a) write the equation of motion, taking the tube as base, (b) If the particle passes through the axis of rotation with velocity V relative to the tube, what will be its velocity when distant r from this position ? Ans. (5) v 2 V 2 + ri( ~, w being the angular velocity. 7. Solve Ex. 6 assuming the tube to be inclined at angle a to the horizontal, the axis still being vertical. Ans. (&) v* = V' 1 + r 2 co* cos 2 a -f 2gr sin a. 530. Centrifugal Force. When the secondary base rotates uni- formly about a fixed axis, the fictitious force mp " is directed away from the axis of rotation, and its magnitude is wft>V, co being the angular velocity and r the distance of the particle from the axis. This is called the centrifugal force due to the rotation of the base to which the motion is referred.* Whatever motion the secondary base may have with respect to an ultimate base, the component of the fictitious force mp " directed away from the instantaneous axis may be called the centrifugal force. 531. The Earth as Base. The earth is usually the most con- venient base for estimating the motions of terrestrial bodies. The fictitious forces which must be introduced into the equations of mo- tion in order to take account of the known motions of the earth are for many purposes negligible in comparison with the actual forces. * This is the proper use of the term centrifugal force, to which allusion was made in the foot-note on p. 271. RELATIVE MOTION. 45 1 The most important of these motions of the earth is the diurnal ro- tation. Correction for earth's rotation. Let a frame of reference be taken consisting of three rectangular axes intersecting at the earth' s center, one coinciding with the axis of rotation, a second lying in the plane through that axis and a fixed star, the third perpendicular to this plane. The motion of the earth with respect to this frame is a rotation with uniform angular velocity a> = 0.00007292 rad. per sec. (Ex. 1, p. 275). The effect of this rotation upon the motion of a particle relative to the earth is to give it two components of accel- eration which are the negatives of p" and/'" as defined in Art. 526. In other words, the fictitious forces to be introduced are mp" and mp'" . The former of these is the ''centrifugal" force whose magnitude is mco l r y directed away from the axis of rotation ; the latter is a force of magnitude 2nmv sin a, a being the angle between V and the axis of rotation. This latter force is always perpendicular to the axis of rotation and to the direction of the relative motion, in such a way that the particle is deflected toward the right in the northern hemisphere and toward the left'in the southern, as seen by a person standing upon the earth and facing in the direction of the relative motion of the particle. The greatest value of the acceleration p " occurs at the equator, being 3. 391 C. G. S. units* or 1/289 f g- Practically, the acceler- ation p " cannot be separated from the acceleration due to gravity except by computation. The value of^* as determined experiment- ally at any place is the resultant of the component p " and the ac- celeration due to the attraction of the earth. In practice, therefore, the centrifugal force due to the earth's rotation is always taken account of when the weight of the body is one of the forces entering the equation of motion. The acceleration p '" is small unless the velocity of the particle relative to the earth is very great. If v sin a = 100 meters per second,/'" = 1.458 C. G. S. units. The deflection of a particle toward the right in the northern hemisphere and toward the left in the southern has an important effect upon the circulation of the atmosphere. It is this which de- termines the direction of rotation of cyclonic disturbances. *See Ex. 3, p. 275. 452 THEORETICAL MECHANICS. Correction for motions of the earth about the sun and the moon. The origin of the frame of reference described at the beginning of this Article is at the center of the earth, and the frame has therefore a motion of translation due to the orbital motion of the earth's center of mass. By reason of the attraction of the sun, the center of mass of the earth describes an approximately circular orbit about the com- mon center of mass of sun and earth ; and by reason of the moon's attraction it describes an approximately circular orbit about the common center of mass of moon and earth. These are most simply considered separately. The fictitious forces which must be assumed to act upon any terrestrial particle because of these motions, if the frame of reference above described be taken as base, are easily computed from the general principle of Art. 524. The most im- portant application in which these corrections need to be made is in the theory of the tides. INDEX. Acceleration, 172, 173, 212, 214. average, 213. axial components of, 239. formula for, 173. polar resolution of, 241. tangential and normal compo- nents of, 244. Action at a distance, 19. Activity, 316. Angular impulse, 292, 383. Angular momentum, 287, 382. principle of, 294, 326. value of, 387, 388, 395. Angular motion of material system, 325, 326. of particle, 281. of rigid body, 363, 372. Attraction between particles, 155. between spheres, 164. of sphere upon particle, 162. of spherical shell upon exterior particle, 160. of spherical shell upon interior particle, 158, 162. Balance, 86. Base for specifying motion, 432, 433. ultimate, 434. Body, 2, 228. Catenary, 118. Center of mass, 127, 129. Central force, motion under, 193- 198, 253, 260. varying as distance, 253. varying inversely as square of distance, 258. Centrifugal force, 271, 450. Centroid, 124-140. determined by integration, 135- 140. of parallel forces, 125, 126. of plane area, 131-134. of volume, surface, line, 130. Circular motion, uniform, 270. Collision of bodies, 282, 340. Concurrent forces, composition and resolution of, 27-35, 14 1 142. equilibrium of, 36-48, 143. Configuration defined, 407. energy of, 408. Connected particles, motion of, 203. Conservation of energy, 417. Conservative system, 410. law of force in, 410. principle of work and energy for, 412. Constant force, motion under, 188, 249, 252. Constant of gravitation, 164, 166. Constrained motion, 260-276. Continuity of matter, 3, 228. Coordinates of position of particle, 236, 321. of rigid body, 345, 360. Coplanar forces, composition and resolution of, 60-64. equilibrium of, 65-91. Cord, flexible, 22, 1 14-123. Couples, composition of, 59, 145, 146. coplanar, 56-60. impulsive, 399. in three dimensions, 144-146. representation of by vector, 144. Curved path, motion in, 208-236. D'Alembert's principle, 327, 329, 37i. Degrees of freedom, 346. Density, 128. Dimensional equations, 7. Dimensions of derived units, 6, 157, 184, 280, 300, 310. Displacement, 167. given by vector, 208. of rigid body, 360-361. 454 INDEX. Dynamics, 4. of relative motion, 448. Earth as base for specifying motion, 45o. Effective forces, 327, 370. moments of, 350. resultant of, 357 . Efficiency, 419. Energy, 308, 312, 407. a definite quantity, 409. and heat, equivalence of, 414. forms of, 415. of a particle, 307-311. of a material system, 311-317, 407-412. of a rigid system, 419. transfer of, 412. utilized and lost, 417. Equilibrium, conditions of for con- current forces, 36-38, 143. for coplanar forces, 65-68. for non-coplanar forces, 149- 151. definition of, 26. of flexible cords, 1 14-123. of parts of bodies, 92. of system of bodies, 99. work-condition of, 318, 424. Flexible cords, equilibrium of, 114- 123. Force, conception of, 2, 16-18, 177. defined by laws of motion, 225- 228. effect of constant, 177, 184, 217. effect of variable, 185, 220. gravitation unit of, 23. kinetic or absolute unit of, 222. Forces, classes of, 18-22. equal, defined by law of action and reacti , 227. external and internal, 92. fictitious, 449. Freedom, degrees of, 346. Friction, 106-113. angle of, 108. coefficient of, 108. Friction, in machines, 11 1. laws of, 107. Gram, 24. Gravitation, constant of, 156-164, 166, law of, 155, 158. Gravitation system of units, 22-25. unit of mass, 156, 165, 166. Gravity, motion under, 189, 199, 249. Harmonic motion,- 196. Heat equivalent to energy, 414. Hodograph, 211, 239. Homogeneous body, 128, 129. Impact, 282-284. Impulse, 278, 289, 291, 383. and momentum, equations of, for material system, 384, 392. for particle, 279, 290, 291. for rigid body, 388, 395. angular, 392. instantaneous, 396, 400. resultant, 292. sudden, 280, 281, 296, 385. Impulsive couple, 399. Inertia-ellipse, 344. Instantaneous motion of rigid body, 362. Jointed frame, 95, 96. Kinematics, 4. Kinetics, 5. Kinetic energy of material system, 407. of particle, 308, 309. Kinetic systems of units, 181-184. Lever, 83. Machine, 82, 316, 426. Mass, 2. defined by laws of motion, 225, 226. Mass-center, motion of, 322, 324, 348, 352. Material body, 2. system, 321. Mechanics defined, 1. subdivisions of, 4. INDEX. 455 Metric system, 24. Moment with respect to an axis, 35, 150. with respect to a plane, 130. with respect to a point, 13, 35. Moment of inertia of plane area, 339-345- of rigid body, 33i"339- Moments, theorem of, 14, 54-56. Momentum, 277, 287, 291, 382. acceleration of, 277, 288. angular, 287. linear, 386, 387. resultant, 394, 395. Motion, absolute and relative, 436. laws of, 177, 223. Newton's first law, 16, 177. third law, 17. three laws, 223. Parallel forces, resultant of two, 5i Parallelogram of forces, 28, 221. Parallelopiped of forces, 141. Particle, 2, 228. Passive resistances, 19. Pendulum, compound, 354. simple, 266. Percussion, center of, 390. Plane motion of particle, 236- 276. of rigid body, 360-381. Plane motions of rigid body, com- position and resolution of, 364-370. Polygon of forces, 30. Position, 3. in straight line, 167. given by vector, 208. Potential energy, 312, 408. Pound force, 23. Pound mass, 23 Power, 316. Product of inertia, 33S, 341. Projectile, 249. Pulley, 83, 84, in. Pulleys, system of, 85, 430. Quantities, numerical representa- tion of, 5. Quantity, fundamental kinds of, 4. Radius of gyration, 332, 344. Relativity of motion, 232. Resisting medium, motion in, 203. Resolution of forces, 26, 31, 50, 141. Restitution, coefficient of, 285. Resultant, definition of, 26. of concurrent forces, 27-30, 142. of coplanar forces, 60, 64. of couples, 59, 145, 146. of non -coplanar forces, 142, 149. Rigid body, 3, 49. motion of, 346-381 . Rigid system, energy of, 419. Rotating body, equations of motion of, 353- Rotation, 347, 349-359. of earth, its influence on ap- parent weight, 273. Scalar, 8. Simultaneous motions, 229-235. Smooth surface, 21 Space, 436. Standard condition for estimating energy, 313, 408. Statics, 5. Statics a case of kinetics, 252, 379- 381. Strain, 104. Stress defined, 17. work of, 405. Stresses, external and internal, 102- 104. Symmetry, 131. Time, 435, 436. Translation, 347. Triangle of forces, 28. Units, fundamental and derived, 6. Vector, 7. addition, 8. increment of variable, 1 1 . localized, 13. resolution of, 12. subtraction, 9. 456 INDEX. Velocity, 167, 209. average, 169, 209. axial components of, 238. formula for, 170, 210. polar resolution of, 240. sudden change of, 215. tangential and normal com- ponents of, 244. Velocity-increment, 171, 212. Virtual work, principle of, 318-320, 424, 425. Vis viva, 306. Weight, 23. influence of earth's rotation on, 273- Work and energy, principle of, 309, 314, 412, 420. Work and vis viva, equation of, 306. Work done by body, 307. done by force, 298, 299, 301. external and internal, 407. of central force, 303. of stress, 405. virtual, 318, 424. OF THE ^>> \ UNIVERSITY OF V UNIVERSITY OF CALIFORNIA LIBRARY . THIS BOOK IS DUB ON THE LAST DATE STAMPED BELOW '* J I3 4 I91t> vtti I'd Vfl* ^25 1 916 1916 taS AUG 4 1919 DEC 20 1519 NOV S3 19 21 w. :- JUL ? m sE fSMS24 MAJNS4 -3 S& , 301-6'14 *ira -- . 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