GIFT OF Mrs- W. W* Campbell 3TSTRONOM7 STATICS MINCHIN Hontion MACMILLAN AND CO. PUBLISHERS TO THE UNIVERSITY OF Clarentrmt press TREATISE ON STATICS CONTAINING THE FUNDAMENTAL PRINCIPLES OF ELECTROSTATICS AND ELASTICITY BY GEORGE M. MINCHIN, M.A. DUB. PROFESSOR OF APPLIED MATHEMATICS IN THE ROYAL INDIAN ENGINEERING COLLEGE, COOPER'S HILL SECOND EDITION COEEECTED AND ENLAEGEB AT THE CLARENDON PKESS M.DCCC.LXXX [All rights reserved] DEPT; 251 MS" ft st toft S* PKEFACE TO THE SECOND EDITION. THE present edition of this work is, I venture to think, a considerable improvement on the previous one. With such a large number of examples, not only were misprints, but mistakes, more or less trifling, almost inevitable ; but, owing chiefly to the kindness of correspondents, very few of these can remain in this edition. My acknowledgments are, in the first place, due to Mr. Robert Graham, of Kingstown, who supplied me with a long list of corrections and some good suggestions. Mr. J. C. Malet kindly called my attention to some mistakes which I had overlooked in the chapter on Attractions. In some places where a better choice of language was possible for the elucidation of the subject, I have adopted alterations kindly pointed out by Colonel Chesney. In the earlier part of the book the examples have been subjected to a rearrangement, the order of relative difficulty being better kept in view ; and some of them which were of the purely mathematical and fantastic character have been expunged. Besides alterations of the above description, four others deserve special mention. Firstly, the proof of the parallelogram of forces has been based entirely on the Newtonian definition of force, and has M177047 VI PREFACE. therefore been made to follow from the composition of velocities. Secondly, the principal propositions of Graphic Statics (so far as coplanar forces are concerned) have been introduced. The subject is a small one and very simple, and I believe that in the few pages in which I have treated it (see end IT O \ of Chapter V) the student will find enough to enable him to read with ease a more elaborate and formal treatise on graphic methods. Thirdly, the portion dealing with Electrostatics has been so enlarged as to contain several propositions of importance which had been omitted in the previous edition. Fourthly, and chiefly, a Chapter on Strains and Stresses has been introduced. So far as English works on Statics, in general, are concerned, this is an innovation, and a very important one. In view of the enormous development of Mathematical Physics, and the wonderful inventions depending on the small strains and vibrations of natural solids, which have been made within the last few years, the study of the equilibrium and motion of bodies as they are, and not as they exist in abstraction,^ surely a subject of which it is impossible to exaggerate the importance. We may well ask whether in this country too much valuable time is not spent in the discussion of neat mathematical unrealities in the calculation of the behaviour of impossible bodies under impossible condi- tions, A certain amount of this is of course necessary for the study of the fundamental principles of Dynamics; but the equilibrium and motion of natural solids ought to occupy the attention of every student of Physics after he has acquired a sound and firm knowledge of the fundamental propositions concerning the action of Force. Yet Applied Mechanics, as a PKEFACE. Vll sequel to, and corrective of, Rational Dynamics, is a subject the study of which is confined almost exclusively to scientific students of Engineering. I am very far indeed from asserting or implying that the few pages on Strains and Stresses in this work supply ade- quately this deficiency in our general scientific education. They are addressed to students who have attained consider- able proficiency in pure mathematics, and have a reference much more to the Theories of Light, Magnetism, and Elec- tricity than to ordinary Applied Mechanics. For students of lower attainments a short treatise dealing first with plane elasticity and proceeding thence to strains in three dimensions would be extremely desirable. In dealing with the theory of Strains and Stresses and with the subject of Electrostatics, I have had the benefit of the invaluable advice and criticism of Mr. Fitzgerald, whose assistance was always given with the utmost zeal. In two Chapters of his Elements of Dynamic the late Professor Clifford gave a discussion of 'Strain-Steps' and 'Strain- Velocities' marked by all the elegance and simplicity of treatment which characterised everything he wrote. From these chapters I have derived considerable assistance; but their (quaternion) method is, of course, different from that which I have adopted. For the view of the theory of Friction presented to the student in this work, I am almost wholly indebted to Mr. Jellett, whose method of treating the rational theory of Friction, both in his Lectures and in his Treatise on the subject, has invested it with a completeness and precision which it had not previously attained. Our knowledge of the laws of Friction has been recently extended by the Vlll PREFACE. experiments of Professor Osborne Reynolds on rolling friction (Phil. Trans., vol. 166, pt. i), and by experiments made on an extensive scale on the London Chatham and Dover Railway by Captain Douglas Galton (Proceedings of the Institution of Mechanical Engineers, June and October, 1878). A reference to these experiments will be useful to the student. I have again to thank Mr. Eagles for his very useful and painstaking assistance in correcting the press and verifying results. Mr. Reilly's references to sources of information have been, as before, of very great value to me ; and I have to thank Professor Wolstenholme for continuing his permission to draw from the inexhaustible store accumulated in his Book of Mathematical Problems. COOPEK'S HILL, December, 1879. TABLE OF CONTENTS, CHAPTER I. Page The Composition and Resolution of Forces acting in one Plane at a Point i CHAPTER II. General Conditions of the Equilibrium of a Particle under the Action of Forces in one Plane 21 CHAPTER III. The Equilibrium of a Particle on Plane Curves 40 Section I. Smooth Curves" 40 Section II. Rough Curves 51 CHAPTER IV. The Principle of Virtual Work 66 Section I. A Single Particle 66 Section II. A System of Two Particles 77 CHAPTER V. Composition and Resolution of Forces acting in one Plane on a Rigid Body 85 CHAPTER VI. The Conditions of equilibrium of a Rigid Body under the action of Forces in one plane deduced from the Principle of Virtual Work for a single particle 118 CHAPTER VII. Applications of the Conditions of Equilibrium of a Body . . . . 133 CHAPTER VIII. Equilibrium of a System of Smooth Bodies under the Action of Forces in one Plane 160 CHAPTER IX. Equilibrium of Rough Bodies under the Influence of Forces in one Plane 180 CHAPTER X. Equilibrium of a Rigid Body under the Action of any Forces . . . 2 1 3 X TABLE OF CONTENTS. CHAPTER XI. Page Centroids [Centres of Gravity] . . . ' . * .242 Section I. Investigations which do not involve Integration * .242 Section II. Investigations requiring Integration . ... 252 CHAPTER XII. The Principle of Virtual Work applied to any System of Bodies . .287 CHAPTER XIII. Equilibrium of Flexible Strings . '-. . , 33 1 Section I. Flexible Inextensible Strings , . . . . 33 2 Section II. Flexible Extensible Strings . . . . . 3 6 4 Section III. Method of Virtual (or Potential) Work . . . -375 CHAPTER XIV. Simple Machines 3 8 3 CHAPTER XV. Attractions. Theory of the Potential . . . . , . . , .400 Section I. Solid Distributions of Matter in general .... 400 Section II. The Attraction of Ellipsoids . . . ... . 429 Section III. Superficial Distributions ... . . -437 CHAPTER XVI. Analysis of Strains and Stresses 457 Section I. Analysis of Small Strains . . . . . . .457 Section II. Analysis of Stresses . . . . - . , . . 486 Section III. Expression of Stress in terms of Strain .... 502 ERRATUM. W In page 179, line 3, for TFread . NOTE. In example 39, p. 159, omit everything after the words 'horizontal line through A ' and insert instead the words ' show that as the point C varies, the position of the beam being always the same, the magnitudes and lines of action of the pressure on the axis will be represented by lines drawn from A to a certain right line parallel to A B ; and if the position of the beam varies, while AC is always equal to AS, find the curve whose radii vectores will represent the pressure on the axis.' STATICS CHAPTER I. THE COMPOSITION AND RESOLUTION OF FORCES ACTING IN ONE PLANE AT A POINT. ARTICLE 1.] Definition of Force. Force is an action exerted, upon a body in order to change its state either of rest or of moving uniformly forward in a right line. This is the definition of Force given by Newton (see Prin- cipia, Book I, Def. IV). 2.] Divisions of the Science. The Science which treats of the action of Force on bodies is called Dynamics. Of this science there are two branches : the first treats of the laws to which forces are subject when they keep bodies at rest, and this branch is called Statics ; the second treats of the laws to which the motions of bodies are subject when these motions are produced by given forces, and this branch is called Kinetics. 3.] Matter. Matter is something which exists in space, and attests its presence by such observed qualities as extension, resistance, and impenetrability. A limited portion of matter is called a Body, and the quantity of matter contained in a body is called its Mass. A very small portion of matter is called a Particle. K 4.] Velocity. Suppose a point to move along a right line in such a way that it always takes the same time, tf, to travel over the same length, s, of the line, in whatever points of the line the extremities of this length are situated. Then we readily say that the point's ' rate of moving ' is the same all through, and this rate we measure by the quotient - The rate of moving we call the velocity of the moving point. But if the time of moving over the length s is not the same all through but depends on the * B 2 COMPOSITION AND EESOLUTION OF FORCES. [5. points of the line between which it is measured, the velocity, or rate of moving, is clearly not uniform. Nevertheless we recognise the fact that at each of its positions the moving point has a particular rate of going. How is this rate to be esti- mated? Like all rates, it must be measured by a differential coefficient. Thus, if P and Q are two extremely close positions, and if is any fixed point on the line of motion, the distance between and P being called s and the distance OQ being called s + A s, and if the point has taken the infinitesimal time A^ to get from P to Q, we shall t>e very near the truth in assuming that its rate of moving has remained uniform in the passage from P to Qj and the velocity in this interval will, as above, be the quotient - The smaller the interval PQ (and A ' therefore the smaller A* and A^) the more nearly true is the assumption of uniformity of the rate of moving from P to Q. Hence if we could find the value of the ratio when both A s A and A are indefinitely diminished, we should have the exact rate of moving at P. But the limit of this ratio is the els differential coefficient > which is easily found by the rules of dt the Differential Calculus. We have thus not only a conception of different rates of moving, but also a method of estimating these rates numerically at different points of the path. 5.] Criterion of the Action of Force. Instead of the motion of a mere mathematical point, let us consider the motion of a material particle. How can we tell whether this moving particle is acted on by force or not ? The answer is unless the particle is completely at rest, or failing this, moving ivith a uniform velocity in a right line, it is acted on by some force. Observe the two distinct characters which must be possessed by the motion of a particle which is not acted on by force the velocity must be constant in magnitude and the path must be a right line. 6.] Measure of Force. Suppose a particle to move along a right line in such a way that in any interval of time, t, there is the same addition made to its velocity, between whatever epochs of time the interval t is reckoned. Then the velocity is 7-] WAYS IN WHICH FORCE IS PRODUCED. 3 obviously increased at the same rate at every point of the path, and the particle is said to be continuously acted on by a uniform force in the line of motion. The rate at which this increase of velocity takes place is taken as the measure of the force acting on the particle ; that is, if the same particle moves along a right line in such a way that its velocity is increased at a constant rate which is double the previous rate, it will be continuously acted upon in the second motion by a force which is double the previous force. If the rate of increase (or in other words, the acceleration) of the particle's motion is not uniform, the force acting on it is not uniform, and its magnitude at any point of the particle's path is estimated by the rate of increase of the velocity of the particle at this point. Since the velocity of one and the same particle is capable of having all possible rates of increase, all forces may be compared with each other by means of their effects on a single particle. 7.] Ways in which Force is produced. One of the simplest ways in which a force can be made to act on a particle consists in attaching a string to the particle and pulling this string so as to cause the particle to move. If no other force acts on the particle, and if the string is always pulled in the same right line, the particle will continue to move in this right line ; and the rate> per unit of time, at which its velocity is being increased at any point of its path is a measure of the magnitude of the force with which the string pulls it ; so that if for any finite time we observed its velocity to remain constant, we should know that during this time the string ceased to be pulled, and that no force acted on the particle in this particular interval. There are other ways in which forces act on particles, but the manner in which they act is not in every case known to us. . For example, if the particle consists of a small piece of soft iron and we hold it near the pole of a magnet we shall see it rushing with continually increased velocity towards the magnet, and it is therefore by definition acted on by some force towards the magnet. This force can be measured, as before, at every point of the particle's path by the rate, per unit of time, at which it produces an increase of velocity in the particle; nevertheless it is quite uncertain how this force is produced whether it is an action at a distance or a stress in some intervening medium. B 7, 4 COMPOSITION AND RESOLUTION OF FORCES. [8. But whatever its cause may be, we can measure it numerically by its effect viz., rate of increase of velocity produced in a material particle. Again, since the velocities of planets towards the sun and of meteoric stones towards the earth are perpetually accelerated, the planets are acted upon by forces towards the sun, and the meteors by forces towards the earth. These forces are called forces of attraction ; but the nature or precise mode of operation of this attraction is a matter on which no certainty exists. 8.] Linear Representation of Forces. Consider a single material particle. Every velocity which it can have possesses three characteristics it must have a certain numerical magni- tude, it must take place in a certain right line, and it must take place in a certain sense (from right to left or from left to right) along this line ; or, in other words, it must have magni- tude, line of action, and sense. Now every velocity can be regarded as produced in the particle by the uniform action of a force for a definite time. Hence forces are also characterised by magnitude, line of action, and sense. Two forces acting on a particle are therefore compared by specifying the two lines and senses in which they would cause it to move if each acted separately, and also the magnitudes of the velocities which they would thus generate in it if they both acted for the same time on it. Hence any force may be completely represented by a right line drawn in the direction and sense in which it would cause a material particle to move, the length of this line representing, on any scale, the rate per unit of time at which the force would generate velocity in the particle. And all other forces may be compared with this force as to magnitude, direction, and sense by drawing right lines in the several directions in which they would produce motion, and taking the lengths of these lines to represent, on the same scale as before, the rates at which they would severally generate velocity in one and the same particle. Forces may also be compared with each other by means of their effects on different particles. For, let n perfectly equal particles be placed side by side in a row (fig. i), and let each of them be acted upon uniformly for the same time by a force which at the end of this time generates the same velocity, 9.] COMPOSITION OF VELOCITIES. 5 f> in each of them. Now if instead of being n separate particles they were all glued together so as to form a body of n times the mass of each particle, and if each of them is still acted on by the same force as before, this body will, at the end of the time considered, have the same velocity as each separate particle had, and will be acted upon by n times the force which generated this velocity in the particle. Comparing a single particle, then, with the body whose mass is n times the mass of this particle we see that to produce the same velocity in two bodies by forces acting on them for the same time, the magnitudes of the forces must be proportional to the masses to which they are applied. And hence, generally, if we define momentum as the product of mass and velocity The magnitude of a force is proportional to the rate per unit of time at which it generates momentum. The greater the mass on which the force acts, the less the rate at which it increases the velocity of this mass ; and the less the mass, the greater the rate of increase of velocity ; the product of the two being always the same for the same force, whatever le the masses to which it is applied. So that if P is a force which generates velocity at the rate -j- in a body of mass m> and if P' is a force which generates dt 7 / velocity at the rate -^- (per unit of time) in a body of mass m, we have ^_ [vnn \ 9.] Composition of Velocities. We propose to show how a particle may be moving with two velocities in two different directions at the same time. Let a board be placed on a hori- zontal table ; let a rectilinear groove, OA (fig. 2), be cut in this board, and let a particle be placed at in the groove. Suppose, for definiteness, that the unit of time is one second. Let the particle be moved along the groove with "aT uniform velocity represented by OA, and at 6 COMPOSITION AND RESOLUTION OF FORCES. [lO. the same time let the board (i.e. every point in the board) be moved along a groove cut in the table with a uniform velocity represented in magnitude and direction by OB. Over what point in the table will the particle be found at the end of one second? Before the motions begin, complete the parallelogram OACB. At the end of a second the particle must be found in the groove at the point A, and also at the end of the same time the point A of the groove must be found at the point of the table vertically under C. Hence this latter point is the position of the particle at the end of a second. Let the foot of a perpendicular dropped from the particle on the table be called the position of the particle referred to the table. How do we know that the position of the particle referred to the table has described the right line OC (or rather a line in the table vertically under OC) ? In this way if we demanded the position of the particle referred to the table at the end of any fraction or multiple of a second, we should find that the distance which it has travelled along OA is to the distance which the groove has travelled in the direction OB as OA is to AC, and therefore the positions of the particle referred to the table trace out a right line vertically under OC. Consequently the two simultaneous velocities OA and OB which were impressed on the particle have combined to give it a single velocity represented in magnitude and direction by OC. The velocity OC is called the resultant of the velocities OA and OB, and these latter are called components of the velocity OC. Hence we arrive at the proposition which is the foundation of Dynamics : If a point, 0, move with two coexistent velocities represented in magnitudes, directions, and senses by two right lines, OA and OB, it will have a resultant velocity represented in magnitude, direction, and sense by the diagonal, drawn through 0, of the parallelogram determined by the lines OA and OB. This proposition is called by the name of The Parallelogram of Velocities. 10.] Composition of Forces. From the Parallelogram of Velocities, the Parallelogram of Forces follows at once. Since two simultaneous velocities, OA and OB, of a particle result in a single velocity, OC, and since these three velocities may be 12.] EQUILIBRIUM OF THREE FORCES. 7 supposed to be produced by the separate action of three forces all acting for the same time, it follows that the effect produced on a particle by the combined action, for the same time, of two forces may be produced by the action, for the same time, of a single force which is therefore called the resultant of the other two forces. And these forces will be represented in magnitudes, lines of action, and senses by the lines OA, OS, and OC (Art. 8); hence If two forces be represented in magnitudes, lines of action, and senses by two right lines OA and OB, their resultant is represented in magnitude, line of action, and sense by the diagonal, OC, of the parallelogram OACB determined by these lines. This is the proposition of the Parallelogram of Forces. COR. The resultant of two forces acting along the same right line and in the same sense is equal to their sum ; and if they act in different senses, the resultant is equal to their difference. 11.] Equilibrium of Three Forces. In fig. 2 produce CO through to Cf so that CO OC' . Now imagine that, when the particle is started along the groove and the board along the table, the table itself is moved in a groove cut in the floor in th ; direction OC' with a velocity represented by OC '. In this case it is evident that the position of the particle with reference to the floor is fixed ; that is, the particle is at rest with regard to fixed space (the floor being supposed fixed). Consequently if three forces represented by the lines OA, OB, and OC f act together on the particle, no motion will ensue. In this case each force is equal and opposite to the resultant of the other two ; for it is obvious that OA is equal and opposite to the diagonal, through 0, of the parallelogram determined by OB and OC '; and that OB is equal and opposite to the diagonal of the parallelogram determined by OA and OC' \ 12.] Statical point of view. The primary conception of force is that of a cause of motion in a body or in a material particle, and the magnitude of any force is estimated by the rate at which it generates momentum (Art. 8). Nevertheless in Statics it is only the tendency which forces have to produce motion that is considered. Forces in this branch of Dynamics are considered as acting in such ways as to counteract each other's tendency to produce motion, or as producing a state of equilibrium in the bodies to which they are applied ; but the magnitude of each force is estimated with reference to the 8 COMPOSITION AND KESOLUTION OF FOKCES. [13. amount of momentum which it would actually generate if it were completely unfettered by the action of other forces. Forces in Statics are usually expressed as multiples of the weight of some standard body arbitrarily chosen. Thus a force is said to be a force of 1 kilogrammes if it is just capable of lifting vertically a body whose weight is equal to that of the mass of water which at a temperature of 4 C C. fills a volume of 10 cubic decimetres. But even here the Newtonian definition of force, as a cause of change of motion, is not discarded but merely kept in the background. For the weight which is called a kilogramme is merely a force which generates momentum at a certain rate in a body of certain mass ; and the vertical force which is just able to raise a body from the ground is a force which could generate momentum in the body at the same rate as its weight and in the opposite sense. For practical purposes this measurement of forces as multiples of a weight is used by engineers and others ; but in the very important branch of Dynamics which treats of Electricity and Magnetism an absolute measure of force is resorted to i. e. a measure which is one and the same all over the earth, and indeed all through the universe. The mass of a body is something which cannot con- ceivably change, whether the body is taken to different parts of the earth or to different parts of the universe; and the force which, acting uniformly on this mass for a certain time (say one sidereal second), will at the end of this time have caused it to move with a certain velocity (say one centimetre per second), must be one and the same wherever the experiment is tried. The mass selected to define the unit force is a mass equal to that of the water which, at its temperature of maximum density, fills one cubic centimetre; and this absolute unit of force is called a Dyne. Compared with even such a small force as the weight of a gramme, the dyne is exceedingly small; but in many problems of Electricity and Magnetism where the forces at play are very small, the dyne as a unit force is convenient enough. 13.] Force must act upon Matter. Although the Newtonian definition and measure of force render it clear that whenever force acts it must act on something material, it is not impossible that beginners may lose sight of this fact and suppose that a force could, for example, act on a mathematical point. We may without error speak of forces as acting at a point, but not on it, 1 6.] RESOLUTION OF FORCES. 9 if their lines of action pass through the point. Thus, in fig-. 3, two forces acting- along the lines OA and OB may be spoken of as two forces acting at the point ; but their action would be physically impossible unless it took place on some material body, such as a particle placed at 0. Wherever force is exhibited, there is evidence of the existence of matter, both acting and acted upon. 14.] Proper Representation of Forces. In representing the resultant of two forces which act together at a point, 0, the student should be careful to draw the two forces acting from the point. Q,\_ _ J Thus, if of the two forces, P and Q, * x x S x \ one, P, is represented as acting from 0, and the other towards 0, we niust produce the line QO to Q', so that OQ'=OQ', completing, then, the pa- Fig. 3. rallelogram OPRQ', its diagonal, OR, will represent in magnitude and direction the resultant of P and Q. The marking of lines representing forces with arrow- heads will serve to exhibit the senses of the forces in every case. 15.] Resolution of Forces. Having proved the principle of the Composition of Forces, the principle of the Resolution of Forces at once follows. If two forces, P and Q, are equiva- lent to a single force 0(7 = R (fig. 4), it is evident that the single force R acting along 00' can be replaced by the two forces P and Q, represented in magnitude and direction by two adjacent sides of a parallelogram of which 0(7 is the diagonal. Since an infinite number of parallelograms, of each of which 0(7 is the diagonal, can be constructed, the force R can be resolved in an infinite number of ways into two other forces. These forces are called the components of R. 16.] Theorem. It being given that the direction of the resultant of every two forces is that of the diagonal of their parallelogram, its magnitude must be represented by this dia- gonal ; and conversely, Let it be granted that the resultant of P and Q acts in the diagonal, 0(7 (fig. 4), of the parallelogram determined by P and Q. Measure backwards through a line, OR, the length of which represents the magnitude, R, of the resultant. A system of forces acting at 0, represented in magnitude and direction by P, Q, and R, will evidently be in equilibrium. Each 10 COMPOSITION AND RESOLUTION OF FORCES. [17. force is, therefore, equal and opposite to the resultant of the other two. If, then, we consider P as equal and opposite to the resultant of Q and R, OP', the production of OP, must be the diagonal of the parallelogram determined by Q and E. Now, since OQP'R is a parallelogram, OR = P'Q-, Fig> 4 * and since OP' Q (/ is a paralle- logram, P'Q = 00' ; therefore OR = 00'. Q. E. D. Again, for the converse proposition, let it be granted that OR =00*, while 00" and OR are not necessarily in one right line; and let OP' be diagonal of the parallelogram, OQP'R, determined by OQ and OR; then OP is equal in magnitude to OP*, since the resultant of Q and R has a magnitude equal to OP'. Comparing the triangles OQO'and OQP' we have 00'= QP', QO' = OP', and OQ common to both; therefore the angle QOO' = the angle OQP', therefore QP' is parallel to 00'-, but QP' is also parallel to OR, therefore OR and 0(7 are in one right line. Therefore, &c., Q. E. D. 17.] Relations between Three Forces in Equilibrium. When three forces maintain a particle in equilibrium, each force is equal in magnitude to the resultant of the other two, and acts in the sense exactly opposite to this resultant. Thus, in fig. 4, each of the lines, OP, OQ, and OR, which represent in magni- tude and direction the forces P, Q, R, is equal and opposite to the diagonal of the parallelogram determined by the two re- maining lines. This enables us to express the relative magnitudes of three forces in equilibrium by means of the three angles between them. For (fig. 4) the forces P, Q, R are equal in magnitude to the lines OP, PC/, O'O, respectively. Now, since the sides of a plane triangle are to each other as the sines of the opposite angles, we have OP:P(/:0'0=: sin PO'O : sin O'OP : sin OPO / . Denote by PQ, QR, RP, the angles between the directions of the forces P and Q, Q and R, R and P, respectively. Then, evidently, sin PO'O = sin Q0(/ = sin QOR = sin QR ; 1 8.] RELATIONS BETWEEN THREE FORCES IN EQUILIBRIUM. 11 sin O'OP = sin ROP = sin RP ; sin OP(/ = sin POQ = sin Hence we have the fundamental relations P : Q : R = sin QR : sin RP : sin It may, perhaps, assist the beginner to mark the angle opposite to each force by the corresponding small letter (fig. 5) ; and then the ratios be- tween the forces may easily be remembered in the form P : Q : R = sin p : sin q : sin r. (a) Since the sides of the triangle OPO' (fig. 4) are connected by the equation OO' 2 = OP 2 - we have evidently an equation which gives the magnitude of the resultant of two forces in terms of the magnitudes of the two forces and the angle between their directions, the forces being represented by two lines, both drawn from the point at which they act, as in Art. 14. If PQ = 0, the above equation gives R = P + Q, or the resultant of two coincident forces is equal to the sum of the forces. If PQ = TT, R = P Q ; or, the resultant of two forces which act at a point in exactly opposite senses is equal to the difference of the forces. 18.] Theorem. If any one set of forces (P, Q, R) acting in three given directions is in equilibrium, all other sets acting in equilibrium in the same directions are merely multiples of the set (P, Q, #). For, let the given directions make angles p, q, r with each other in pairs, and let the sets (P, Q, R) and (P', Q', R') acting in these directions be separate systems in equilibrium. Then we have P : Q : R = sin p : sin q : sin r and P':Q':R'= sin p : sin q\ sin r. Therefore, P : Q : R = P r : Q' : K, or ^ = ^- = ^ - Hence 12 COMPOSITION AND RESOLUTION OF FORCES. [19. the forces P', ', R' are separately proportional to P, Q, E, and therefore the former set is not essentially distinct from the latter. This theorem is equivalent to the statement when we have determined any one set offerees in equilibrium in three given directions, we have determined all such sets. Thus, if we know (see Example 1, p. 16) that three forces acting along the bisectors of the sides of a triangle drawn from the opposite angles, and proportional to the lengths of these bisectors, are in equilibrium, we know that this is the only set in equilibrium in these directions. 19.] Principle of the Transmissibility of Force. When a force acts on a particle, the force will produce the same effect if it be supposed applied at any point along a string connected with the particle, the string lying in the line of action of the force. Thus, if a force of P grammes (fig. 6) act on a particle, 0, in the direction OA, P may be supposed to act at A or B at the end of a string attached to 0. Imagine the particle to be connected with an indefinitely thin rigid mem- brane, aoc ; then any force P acting on may be supposed to be directly applied at any point of the membrane in the line of action of P. This axiom is known as the principle of the transmissibility of force ; it is one of the fundamental principles of Rational Statics, and in most treatises on the subject, it constitutes the basis of the investigation of the conditions of equilibrium. It is essen- tially necessary to observe that it holds good only for a rigid body that is, a body whose parts, under all circumstances, must maintain constant distances from each other. Thus, if we sup- pose such a body about to be acted on by any set of forces given in magnitudes and directions, we can say, before the forces are actually applied at certain points in the body, that the effect will be the same if these forces are applied at any other points in their respective lines of action. On the contrary, if the body is deformable, we can make no such assertion. Take, for example, a set of parallel rulers, ABCD (fig. 7), of which the ruler CD is fixed, and suppose a force F to act on the ruler AB at the point a. \^previom to the action of the force, it were allowable to transfer its 19.] TEANSMISSIBILITY OF FORCE. 13 point of application to b, on the fixed ruler CD, it is clear that the system would remain at rest. But we know that the force F, applied at a, will cause the ruler AB to move A until the braces AD and CB are pa- \ ~"\ rallel to the direction of F. However, ^ -- f ^ after the deformable body has taken up -p. a position of equilibrium under the action of the forces, each force may be transferred to any point in its line of action, just as in the case of an indeformable body. Several other very obvious instances of the inapplicability of this principle will doubtless present themselves to the student. It is essential to observe at the outset that in nature there are no such things as rigid bodies. For a great many practical matters there are bodies which may be treated as if they were rigid or indeformable ; but the fact that the particles of solid bodies like iron can be thrown into vibration by the application of even small impulses as is evidenced by the production of sound from bells and gongs proves that these bodies are not absolutely rigid. Bodies which most nearly approximate to the notion of rigidity are called Natural Solids. EXAMPLES. 1. Find the magnitude of the resultant of two forces of 10 kilogrammes and 8 kilogrammes which act at an angle of 105. Ans. 72 = 2^/41-10(^6- >/2) = 1 1 -06 kilogrammes. 2. Two forces, P and Q, of which P is given, act at an angle of 60 ; given the magnitude of their resultant, R, find the magnitude of Q. n Ans. Q= From this it appears that R cannot be less than --P; explain this result by a figure. 3. Two forces, P and Q, inclined at an angle of 120, have a resultant, R ; when they are inclined at an angle of 60, the resultant becomes n times as great as before ; show that R 14 COMPOSITION AND RESOLUTION OF FORCES. [20. 4. If two forces, acting at a given angle, be each multiplied by the same number, show that their resultant is also multiplied by this number and unchanged in direction. . 5. Two forces act at an angle o> ; each force becomes n times as great as before, and the angle between the forces is reduced to -; each of these latter forces again becomes n times as great as before, and the angle between them reduced to - It is observed, that in all these cases the magnitude of the resultant is unaltered. Show that o> = 4cos- 1 ( v " 4 -) 6. Two chords, OA and 0^, of a circle represent in magnitude and direction two forces acting at the point ; show that if their resultant passes through the centre of the circle, either the chords are equal or they contain a right angle. 7. Find the components of a force, P, along two directions making angles of 30 and 45 with P on opposite sides. 2P PV~2 Ans. and 1+V3 1 + -/3 8. Show that a force represented in magnitude and direction by the diameter of a circle may be resolved into two rectangular components represented by any two rectangular chords of the circle drawn from the extremity of the diameter. 9. Two rectangular forces, P and P-/3, act on a particle lying on the ground. If P makes an angle of 30 with the horizon, show that the particle will have no horizontal motion. 10. Three forces equal to P } P + Q, and P Q, act on a particle in directions mutually including an angle - ; find the magnitude and direction of their resultant. 1 20.] Theorem. Tbe following theorem is of wide application in the composition of forces If two forces acting at a point, 0, are represented in mag- nitudes and directions by OB and n . OA, their resultant is represented in magnitude and direction by (-fl) OG, the Fi g point G being taken on AB so that BG = n.AG. For, produce OA to C so that 00 '= n . OA. Then the two forces acting at 0' are represented by OC and OB. Complete the parallelogram OCRB. Then the diagonal OR is the re- sultant force. 21.] THEOREM. 15 From C draw CH parallel to AB. Then the triangles CHR and EGO are equal in all respects, therefore HR OG. Now since OC = n.OA, it follows that OH n.OG, therefore OR ( 4-1) OG, which proves the proposition for the magnitude of the resultant. CH CO Again, -^- = -= #, therefore CH=n.AG, and since (JA. CH= BG, we have BG = n.AG. As a particular case, the resultant of two forces represented by OA and OB passes through the middle point of AB> and is equal to twice the line joining to this point. If the two forces are equal to n . OA and m . OB, the resultant passes through the point G determined so that -^ = > and is AG m represented on the same scale by (m + n) . OG. For, diminishing the scale to which the forces are drawn in the ratio of m : 1, the two forces will be represented by OB and OA. It then follows, by what precedes, that the resultant M acts through a point G, such that BG = -.AG, and is equal in x WL \ magnitude to ( - + 1 J OG. If, now, we revert to the original scale, this must be multiplied by m, and we have for the resultant (n+m). OG.- Q. E. D. 21.] Graphic Representation of the Resultant. If several forces, P x , P 2 , ...act together at a point, their resultant is found thus : Take the resultant of P 1 and P 2 ; compounding this resultant with P 3 , we get a new force which is the resultant of P ly P 2 , and P 3 ; com- pounding this force with P 4 , we get the resultant of P x , P 2 , P 3 , and P 4 ; and carrying on this process until all the forces have been used, we obtain in magnitude and direction the re- sultant of the whole system. -p. Lst ff^ be the middle point of the line P t P 2 , which joins the extremities of the first two forces. Then the resultant of P l and P 2 is represented in magnitude and direction by 2.0^. Compounding the force 2.0^ with P 3 , 16 COMPOSITION AND RESOLUTION OF FORCES. [21. we get a resultant represented in magnitude and direction by 3 . Og 2 (Art. 20), where g 2 is a point on ^P 3 such that P 3 ff 2 2.g 1 t/ 2 . Again, the resultant of 3.0^ 2 and P 4 is 4.0^ 3 , where g.^ is the point on P 4 # 2 such that P 4 ^ 3 = 3 .g. 2 g y If there are n forces acting on 0, and if G is the last point determined as above, the resultant is represented in magnitude and direction by n . OG. DBF. The point G, thus determined, is called the Centroid of the points P 19 P 2 , ...P n . COR. 1. If the point 0, at which the given forces act, is the centroid of the extremities of the forces P 15 P 2 , ...P n , the resultant force vanishes, and the point is in equilibrium. COR. 2. The more advanced student will perceive that if at the points P lt P 2 , ... P n there be placed equal particles, each of mass m, and if each of these particles attracts or repels the particle with a force proportional to m and to the distances OP 15 OP 2 ,...OP W , respectively, the resultant attraction or repulsion on will be nm.OG, or M.OG, where M = the sum of the masses and G is their centre of mass. COR. 3. If the attracting or repelling particles form a con- tinuous body, of mass M, and the law of attraction or repulsion is that of the direct distance, the resultant attraction or re- pulsion will be M . OG, acting in the line OG } where G is the centre of mass of the body. This important result is, therefore, seen to be a simple con- sequence of the theorem in this Article concerning the resultant of a number of forces acting on a particle a theorem which was first given by Leibnitz. EXAMPLES. 1. Find a point inside a triangle such that, if it be acted on by forces represented by the lines joining it to the vertices, it will be in equilibrium. Ans. The intersection of the bisectors of the sides drawn from the opposite angles. 2. P v P 2 ,...P n are points which divide the circumference of a circle into n equal parts. If a particle, Q, lying on the circumference, be acted upon by forces represented by QP l} QP 2 , ... QP n > show that the magnitude of the resultant is constant wherever Q is taken on the circumference. Ans. It is n . QO, being the centre of the circle. 3. A particle placed at is acted on by forces represented in magnitudes and directions by the lines, OA l} OA^ ... OA n , which join to any fixed points, A lt A%, ... A n \ where must be placed so that the magnitude of the resultant force may be constant 1 22.] EXAMPLES. 17 Ans. If the resultant is represented by a line of length R, may TO be placed anywhere on a sphere of radius described round the centroid of the fixed points as centre. 4. Two forces are represented by two semi-conjugate diameters of an ellipse ; prove that their resultant is a maximum when the diameters are equal and so taken as to include an acute angle; and that their resultant is a minimum when they are equal and include an obtuse angle. 5. ABCD is a quadrilateral of which A and C are opposite vertices. Two forces acting at A are represented in magnitudes and directions by the sides AB and AD ; and two forces acting at C are represented in magnitudes and directions by the sides CB and <7Z). Prove that the resultant force is represented in magnitude and direction by four times the line joining the middle points of the diagonals of the quadrilateral. 6. is any point in the plane of a triangle, ABC, and D, E, F are the middle points of the sides. Show that the system of forces OA, OS, 00 is equivalent to the system OD, OE, OF. (Wolstenholme, Book of Mathematical Problems.) 7. If be the centre of the circumscribed circle of a triangle, ABC, and L the intersection of perpendiculars from the angles on the sides, prove that the resultant of forces represented by LA , LB, and LC will be represented in magnitude and direction by 2 LO. (Wolstenholme, ibid.) If r is the centroid of the triangle, the resultant is 3 .LO (Art. 21); but this, by a well-known theorem in Geometry, is 2.LO. 22.] Graphic Representation of the Resultant. There is another mode of exhibiting the resultant of a number of forces acting on a particle. When two forces, OA and OB (fig. z, p. 6) act at 0, their resultant is the diagonal of the parallelogram OACB ; or, again, it may be considered as the third side of the triangle determined by OA and AC, the latter line being drawn from the extremity of the force OA parallel to the other force, OB. Let any number of forces, OA, OB, OC, OD (fig. 10), act at 0. Then drawing oa (fig. n) parallel and equal (or proportional) to OA, and from the extremity a drawing ab parallel and equal (or proportional, on the same scale) to OB, the resultant of the forces OA and OB is represented by ob, the third side of the triangle oab. (Of course the resultant acts at 0, and is parallel to ob). Again, drawing be parallel and equal (or proportional) to OC, the resultant of ob and be is oc. Com- c 18 COMPOSITION AND RESOLUTION OF FORCES. pounding this with cd, which represents OD in the above manner, we get the resultant of the whole system represented in magnitude and direction by od, the last side of the polygon oabcd. Hence to represent the resultant of any number of forces acting at a point, Take any pointy o, and draw the sides of a polygon successively parallel and equal (or proportional) to the forces acting at ; then the last side, or that which is required to magnitude and direction the re- Fig. ii. close up the polygon, represents sultant of the system. */ 7 COR. 1 . If the last vertex, d, of the polygon of forces closed up into o, the side od would vanish, or the resultant force would vanish ; that is, the system of forces would be in equilibrium. Hence If the sides of a closed polygon marked with arrows, which all go round the polygon IN THE SAME SENSE, represent in magnitude and direction the forces which act together on a particle, these forces form a system in equilibrium. COR. 2. When only three forces act, the preceding Cor. shows that they will be in equilibrium if they are parallel and pro- portional to the sides of a triangle which are marked with arrows all going round the triangle in the same sense. This proposition is known as the Triangle of Forces. 23.] LAPLACE'S PEOOF OF THE PARALLELOGRAM OF FORCES. Among purely statical proofs of this fundamental proposition, i.e. proofs which do not depend on the consideration of velocity, Laplace's appears to be the most elegant, and as, moreover, it does not involve the principle of transmissibility, it is thought desirable to include it in the present treatise. Let two rectangular forces, P and Q, represented by the lines OA and OS (fig. 12) act at 0, and let R be the unknown magnitude, and OC the unknown direction, of their resultant. It is evident that if P and Q give a resultant equal to R acting in OC, nP and nQ will give a resultant equal to nR acting also in OC, because taking multiples of the forces is the same thing as merely altering the 'A Fig. 12. 23-] LAPLACE'S PROOF. 19 scale of magnitude to which they are referred. Conversely, whatever n may be, nit may be replaced by nP, making an angle (= CO A) and nQ, making an angle -$(= COB) with 7J the direction of R. Let n be taken = ^ and draw A' OB' per- pendicular to 00. Then, since 22 may be replaced by P in OA and Q in 0.5, P 2 P0 P -^i n 0(7 tV-inOA'; O 2 may be replaced by - in 05' and -| in 00. Hence the forces P and Q are equivalent to a force = + | i n 0(7, a force in OA', and a force in OB*. Jt -ft XL 1 But these last are equal and opposite, and therefore they destroy each other. Hence P and O are equivalent to a single force P 2 -f O 2 = -- p acting in the direction of their resultant ; therefore or R=VP 2 + Q 2 . (1) Thus we have found the magnitude of the resultant of any two rectangular forces. We now proceed to find its direction. If P and Q are equal, their resultant bisects the angle between them, and (l) therefore shows that it is represented in magnitude and direction by the diagonal of their parallelogram. Let three forces, at right angles to each other, OA, OB, and OC (fig. 13) each equal to P, act on a particle ; complete the cube as in the figure. By what precedes, the resultant of OB and OC is OF', combining this with OA, we see that the direction of the resultant lies in the plane FOA. Simi- larly, it can be proved to lie in the plane COD; hence its direction is 0(7, H the intersection of these planes, or the diagonal of the cube. Now from (l) viiag u jj.aj. UA UJJLC tJuuc. 0.1 uw ixuiii \L j /: ~ OF= PV2, and the resultant of the three forces is the same as the resultant F . z of P\/2 along OF and P along OA. By (1) the magnitude of the resultant is P \/!J, and since 20 COMPOSITION AND RESOLUTION OP FORCES. [23. 0(7 = P \/3, we have proved that the diagonal, 0(7, of the parallelogram FOA represents in magnitude and direction the resultant of two forces P and P\/2. Suppose now that OA = P, B = P\/2, and 0(7= P, and complete the parallelepiped. We have just proved that the resultant of OB ( = P*>/2) and OC (=P) is the diagonal OF (=P/v/3); and since the resultant of the three forces must lie in the planes COD and FOA, it must act in the diagonal 00". But this resultant is the resultant of PA/S along OF and P along OA, and by (l) its magnitude is P\/l, which is the magnitude of 00', the diagonal of the parallelogram FOA. By keeping OA and OC each equal to P, and giving OB the values P, P\/2, P\/3, ...P\/m, successively, we prove in this way that the parallelogram law holds for P and P */m ; hence, multiplying the forces by Vn, the law holds for P\/n and P\/mn ; or, replacing mn by , the law holds forPVn and P\//f, where ft and # are any two integers. But the numbers n and /I can be varied in such a way that A/ - shall be equal to any given quantity. Hence the parallelogram law holds for two rectangular forces which bear to each other any given ratio. From this the proposition follows easily for oblique forces. Let OA and OB (fig. 14) represent two oblique forces, P and Q ; complete the par- allelogram, draw the line mn through perpendicular to Fig> I4 ' the diagonal OC, and let fall the perpendiculars Ap, Am, Bq, and Bn, on OC and mn. By what we have proved, the force OB ( = Q) can be replaced by Oq and On, and OA ( = P) can be replaced by Op and Om. But Om is evidently equal and opposite to On, therefore OC is the line of action of the resultant, and its magnitude = Op + Oq, which = OC. This proof will be found at greater length in the first chapter of Moigno's Legons de Mecanique Analytique. CHAPTER II. GENERAL CONDITIONS OF THE EQUILIBRIUM OF A PARTICLE UNDER THE ACTION OF FORCES IN ONE PLANE. 24.] Absolute Condition of Equilibrium. One condition is necessary and sufficient for the equilibrium of a particle and that condition is, that the magnitude of the resultant force acting upon it shall be zero. In the case of a body (as distinguished from a mere particle) the student will afterwards see that this single condition is not sufficient. The vanishing of the Re- sultant may be called the absolute condition of the equilibrium of a particle. 25.] Several Forces. When several forces act upon a particle, the condition of its equilibrium may be expressed as in Cor. 1, p. 16; or as in Cor. 1, p. 18. But, in practice, these represen- tations would frequently be found clumsy, and we obtain simpler results by using the principle of the Resolution of Forces than those given by the principle of Composition. It is to be observed that forces acting on a particle are to be considered as forces whose lines of action all pass through one common point. 26.] Resolution of Forces in given Directions. It has been proved that a force can be resolved into two others along any two directions in the same plane. Simplicity is gained by taking these two directions at right angles to each other. Thus, let Ox and Oy be any two lines at right angles to each other, and P any force acting at in the plane Oxy. Then, completing the parallelogram OXPY, we find the components, OX and OY, of the force P along the axes Ox and Oy. Let OX and OY be denoted simply by X and Y. It is, then, evident that X = P cos 9, Y = P sin 6, where 6 is the angle which the direction of P makes with Ox. 22 CONDITIONS OF EQUILIBRIUM OF A PARTICLE. [27. In strictness, when we speak of the component of a given force along a certain line, it is necessary to mention the other line along which the other component acts. For example, the force P may have an infinite number of components along the same right line Ox. If the line associated with Ox be Om, and if the parallelogram OMPM' be completed, the component of P along Ox will be OM, the other component X N M Fig. 1 6. being OM'. If, again, the resolution of P be effected along Ox and On, and the parallelogram ONPN' be drawn, the com- ponent of P along Ox will be ON; and it is evident that if o> be the angle between the axes along which P is resolved, the coin- sin (o> 0) ponent along Ox will be P In what follows, unless the contrary is expressed, by the component of a force along any line we shall understand the rectangular component ; that is, the resolution is supposed to be made along this line and the line perpendicular to it. ^ It must be remembered, then, that The component of a force, P, along a right line is P. cos (angle between right line and direction of P). 27.] Equations of Equilibrium, or Analytical Conditions. If several forces, P 19 P 2 , P 3 , . . . , act at 0, each of them may be replaced by its two components, one along Ox, and the other along Oy, which is perpendicular to Ox (fig. 17). Thus, the com- ponents of P 1 are P x cos X , and P! sin X ; those of P 2 are P 2 cos 2 , and P 2 sin 2 , and these latter are mea- sured in exactly the same senses as the components of P l ; that is to say, P 2 cos 2 is the component of P 2 along Ox in the sense Ox. The component of P 2 in the figure is actually in the sense opposite to Ox, that is, in the sense 0, x ; still, x, "9 Fig. i 27.] EQUATIONS OF EQUILIBRIUM. 23 the component in the sense Ox is P 2 cos0 2 , for cos0 2 is negative. If the senses Ox and Oy are regarded as the positive senses, any components which act in the opposite senses, 0, x and 0, y, would subtract from the positive components, and must be considered negative. It will be seen that the negative sign of every component will be perfectly represented and accounted for by the general expressions, P cos d and P sin 6, for the two components. Thus, the figure shows that both components of P 3 are negative, and accordingly both of the expressions P 3 cos 3 and P 3 sin 3 are negative, since 3 is In order that the expressions P cos and P sin may always represent components in the positive senses Ox and Oy, the angle 6 must le measured from Ox towards the line of action of the force in a fixed sense that opposite to watch-hand rotation being generally chosen. With this understanding, then, we may say that the com- ponents of P 15 P 2 , P 3 in the direction Ox are P 1 cos^ 1 , P 2 cos0 2 , and P 3 cos0 3 , and those in the direction Oy are P 1 sin t , P 2 sin 2 , and P 3 sin 3 . Replacing each of the forces, P 15 P 2 , P 3 , ...,by its com- ponents, we have P 1 cos X + P 2 cos 0.a + P 3 cos 3 + ..., or 2P cos along Ox, and P l sin 0!+P 2 sin 2 -fP 3 sin 3 + ..., or SP sin along Oy. If the component, P cos 0, of a force, P, along Ox, be de- noted by X, and that along Oy by J, the whole system of forces is equivalent to the two single forces, X A + X 2 + X 3 + , . ., or 2X along Ox, and 1^+72 + 73+..., or SFalong Oy. Now, since (Art. 23, p. 20) the resultant of two forces, P and Q, at right angles is \/P 2 + Q 2 , the resultant, R, of the system of forces P 15 P 2J ... , is given by the equation (1) For the equilibrium of it is necessary and sufficient that R = 0. Hence (2) 24 CONDITIONS OF EQUILIBRIUM OF A PARTICLE. [28. Now this equation cannot be satisfied, so long as 2X and are real quantities, unless 2X=Oand2r=0. (3) These, then, are the two necessary and sufficient conditions for the equilibrium of the particle, and they are equivalent to the single condition R = 0. (See Art. 24). The equations (3) are equivalent to the following state- ment : For the equilibrium of a particle acted on by any number of forces in one plane, it is necessary and sufficient that the algebraic sum of the rectangular components of the forces, along each of two right lines at right angles to each other in the plane of the forces, should vanish. Since the directions Ox and Oy, along which the forces are resolved, may be any whatever in their plane, we may evidently vary the above statement thus the algebraic sum of the rectangular components of the forces along every right line in their plane is zero. It is merely for uniformity of notation that we have mea- sured lt a , 0,p...(fig. 17) all in the same sense that opposite to watch-hand rotation. In resolving forces along a line, Ox, it is simpler x in practice to use the acute angles made by the forces with the line, and to indicate negative components by the sign minus. Thus, if (fig. 1 8) the forces P, P', P" make acute angles 0, &, &", with Ox, the sum of the components of the forces along Ox is P cos 6-P' cos 0' P" cos , sm a 27 sin (co a) 2JC 29.] Tension of a String. When a string is employed to connect two or more particles which are acted on by given forces, the fibres of the string become subject to a certain pull, stress, or tension, which, if increased beyond a certain limit, will cause the string to break. This tension is a force which at any point of the string may be conceived as acting in either of two opposite senses, or in both of these senses at once, according to the nature of the question under discussion. Let us consider, as a simple example, the case of a string, AB (fig. 20), whose weight we may neglect, fixed at the extremity A, and attached at B to a weight W. If, now, we imagine the string to be cut at any point p, and the lower portion, pB, to be removed, it is clear that the re- maining portion, pA, will not be in the same state of stress as before unless we apply at the section p a force equal to W 9 and acting downwards. Again, let the string be cut a little above p, at q, and suppose the portion qA removed. Then the small portion, pq, will not remain in its place unless an upward force equal to -p. W is applied at the section q. The small portion of the string included between p and q is then kept at rest by two equal and opposite forces, each equal to W. Thus, then, if we consider any portion, pq, as isolated from the rest of the string, we must represent it as subject to two equal tensions directly opposed to each other. If we considered the action of the upper portion, p A, on the lower, pB, we should represent pB as acted on by an upward force applied at p ; and if we consider the 20. 26 CONDITIONS OP EQUILIBRIUM OF A PARTICLE. [30. action of the lower on the upper, we must represent pA as acted on by a downward force applied at the section of separation of pA and j.Z?. Thus, the action at B of the string on the body W is an upward force, or tension, equal to W\ while the action of W on the string consists of an equal force in the opposite direction. 30.] String passing over Smooth Pegs or Surfaces. When a string whose weight we neglect passes over a smooth peg, or over any number of smooth surfaces, we shall assume for the present that the stress of its fibres, or its tension, is the same at all of its points. Should it, however, be knotted at any of its points to the other strings, we must regard its continuity as broken, and the tension will not be the same in the two portions which start from a knot. Thus, if the string pass over two smooth surfaces. A and B (fig. 21), and if it is pulled at one extremity by a force P, it must be pulled at the other extremity with an equal force ; but if, after leaving the surface A, it is knotted at C to another string which is pulled with a force equal to R, the tensions in the portions between C and A and between C and B are no longer the same, and their relative magnitudes must be determined by equation (a) of Chap. I, Art. 17. 31.] Equilibrium of a System of Particles. When several particles are connected together and form a system, each par- ticle being acted upon by special forces in addition to the forces produced upon it by its connexion (by strings or rods) with the other particles, we can consider the equilibrium of any one particle apart from all the others, provided that we take account of all the forces which are produced on it by its connexion with the others, in addition to the special forces acting upon it. Thus, in No. 8 of the following examples, we may write down equations for the equilibrium of the particle N as if it were entirely disconnected with the other points, A, P, M, B, if we represent it as acted on by the force, W> and by the tensions, T 2 and T 3 , of the strings by which it is connected with the system. EXAMPLES. 27 EXAMPLES. 1. At the point, 0, of intersection of diagonals of a square (fig. 22), let two forces of 8 grammes, and 12 grammes, act along the diagonals, and two forces of 10 grammes, and 2 grammes, act perpendicularly to two sides ; required the magnitude and direction of their resultant. Resolving the forces along Ox, the line of action of one of them, the component of the force 10 is 10, that of the force 8 is 8 cos 45, that of 2 is zero, and that of 12 is 12 cos 45. Fig. 22. /2 8 I 2 Similarly, 27=- + 2 + -/= = Hence 8 12 -)^= 10-2/2. Therefore R = V (10 - 2 /2) 2 + (2 + 10/ 2) 2 = Again, if a be the angle made by R with Ox, tan a = 2+10/2 _ 1 + 5/2 10-2/2 ~~ 5-/2 = 2J (nearly). ?, act on a particle : find the magnitude of 2. Three forces, P, their resultant. Let the angles opposite P, Q, and R be denoted by p, q, r (fig. 5, p. n). Then resolving all the forces along the direction of P, we get for their combined component in this direction P + Q cos r + R cos q. Resolving them perpendicularly to P, the component = Q sin r R sin q. Hence the square of the resultant = (P + Q cos r + R cos Q) 2 + (Q sin r R sin qf. Remembering that p + q + r = 2ir, this is easily seen to be 3. Verify in the last question that if the three forces are in equilibrium, the expression given for the resultant vanishes. When the forces are in equilibrium, P : Q : R = sin p : sin q : sin r. Hence the expression for the square of the resultant is proportional to sin 2 p + sin 2 q + sin 2 r + 2 sin p sin q cos r + 2 sin q sin r cos p + 2 sin r sin p cos , of inclination of BC to the horizon, would do equally well ; and it is evident that, since either angle suffices, each must be capable of being expressed in terms of the other, and the given magnitudes in the question. Let AB = a, AC = b. Then, for the equilibrium of the point (7 we have, by equation (a), p. 1 1, P _ COS0 W ~ sin (0 + 0)' 30 CONDITIONS OF EQUILIBKIUM OF A PAKTICLE. [31. To this equation must be joined the relation between 6 and $ given by the geometry of the figure. We have, evidently, AC. sin ACS = AB.sm <, or b sin (0 + ) = a sin <. (1) Equation (1) gives a sin W &COS0 = P~' or bW sin < = 75 cos B, aP Expanding sin (6 + 0) in (2), and substituting these values of sin (f> and cos <, and reducing, we have the equation The student will do well to observe that the coefficients of this equation are ratios of magnitudes of the same kind. Thus, force and linear magnitude are quantities of essentially different kinds. It is true, indeed, that the magnitude of a force may be conventionally represented by the length of a line, but it is only in comparison with other forces that any one force can be so represented, and the scale of representation is arbitrary. Hence cos 6, which is a mere number, if it is expressed in terms of force, must be expressed as the ratio of one force to another ; and if it is expressed in terms of linear magnitude, it must be as the ratio of one line to another. If, for example, the Pa 3 coefficient of cos 3 6 in (3) being unity, the last term had been =- > we should have known at once that the result was wrong. For the numerator and denominator of this expression are not of the same degree in force ; neither are they of the same degree in linear magni- tude. Such a term as -=^ denotes the product of an area, > by the p reciprocal of a force, -== Similar remarks as to the homogeneity of our results will be of frequent occurrence in the sequel. By attention to considerations of this kind the student will often be able to detect an error in his work. 6. If, in the last example, the weight W, instead of being knotted to the string at (7, is suspended from a smooth ring which is at liberty to slide along the string ACB, find the position of equi- librium. Si-] EXAMPLES. 31 In this case, the string PBCA, which passes over a smooth surface at B, and through the smooth ring, will have its tension constant at each of its points (Art. 25), and therefore equal to P. Hence, putting T = P, and resolving forces vertically for the equilibrium of C, we have JT-2Psin0=0, or W 7. A string, whose weight is neglected, passes over three smooth pegs, A, B, (7, which are in the same horizontal line. From the extremities of the string are suspended two weights, P and P'; and to two given points in it are knotted two weights, W and W ', the first suspended between A and B, and the second between B and C. Find the position of equi- librium. In this problem the given quan- tities are the suspended weights, Fig. 25. P, W, P', and W, the distances AB and BC, and the length of the portion mBm of the string (fig. 25). Evidently the quantities which we wish to determine are the inclinations, 0, 0, ... , of the portions of the string to the horizon. Let AB == a, BO = a', and the length of mBm = 1c. Consider the equilibrium of the point m. Since the string PAm passes over a smooth peg at A, the tension in it = P throughout. If T =. tension in mBm', we have for the equilibrium of m, COS0 ~ sin (1) cos ~ sin (0 + 0) Again, for the equilibrium of m', P' COS 0' F~'~~sin(0' + 0' T cos 6' (2) Equating the two values of T, we have W cos W cos sn sn (3) These are all the equations that can be obtained from statical considerations. One more equation is required to determine the four 32 CONDITIONS OF EQUILIBRIUM OF A PARTICLE. [31. unknown quantities, 0, $, 0', and ('. This is obtained by expressing that the length of mBmf = k. Evidently a sin i *> / Bm = - , 5 and Bm a am a sin 9 . -7-. sin (0' + <')' a' sin 0M * ( 4 ) sm (0 + ') These four equations determine 0, <, 0', <', and therefore the position of equilibrium. 8. A string, BMNPA, whose weight is neglected, is suspended from two fixed points, A and B ; and from given points, M y N, P, ... , in the string, are sus- pended a series of equal par- ticles whose weight is W. Find the inclinations, X , 2 , 3 , . . . , of the successive por- tions of the string to the horizon. Consider the equilibrium of the particle M. It is acted on by three forces, viz., W acting vertically, T lt the tension of the string MB, and T z the tension of MN. Fig. 26. Resolving these forces vertically, W+ T z sin 2 - TI sin 6 1 = ; and, resolving horizontally, TI cos O t T 2 cos 2 = 0. For the equilibrium of N t resolving horizontally, (i) Hence T l cos 0j = T z cos 2 = T 3 cos 3 = ... ; or in other words, the horizontal components of the tensions in the different portions of the string are constant. Let this constant be denoted by T\ then T T TI = -^, T 9 = -, & c . COS cos 0., Substituting these values in (1), we have ) 1 = tan^ + -^. Similarly, 31.] EXAMPLES. 33 Hence the tangents of the successive inclinations form a series in Arithmetical Progression. In the figure ~W *) PP" ^ ~W 4 = 0, /. tan 3 j tan 2 = ~^- > tan 6^ = =- If the suspended weights are not equal, it is still true that the horizontal components of the tensions are all equal. The figure formed by the string BMNPA is called the Funicular Polygon. y ROQ.P n m I x Fig. 27. 9. To construct the Funicular Polygon, when the horizontal pro- jections, RQ, Qp, pn, nm, mb, . . . , of the successive portions of the chain are all of constant length, a. Let Pp c ; then, since (last example) the tangent of the incli- nation of PN = 2 . tangent of inclination of PQ, it follows that, Pn being horizontal, Nn = 2Pp 2c. Also tan of inclination of MN = 3 tan of inclination of PQ j therefore Mm = 3c. Hence, taking the middle point, 0, of the horizontal portion, RQ, as origin, and the horizontal and vertical lines through it as axes of x and y, the co-ordinates of P are (fa, c) ; those of N are (fa, c + 2c) ; those of M are (fa, c + 2c + 3c); and those of the n th vertex from Q are evidently _2n+l _n(n+l) The value of the ordinate, y, of any vertex at once enables us to determine this vertex. If we eliminate n from the two equations for x and y, we get an equation which is satisfied by all the vertices indifferently. This equation denotes, therefore, a curve passing through all the vertices of the polygon. Eliminating n, we get 2-^1 L c 4 This denotes a parabola whose axis is the vertical line Oy. The A vertex of the parabola is vertically below at a distance = - The smaller the distance RQ, Qp, pn, ... , the more nearly does the Funicular Polygon coincide with the parabolic curve. 10. To represent graphically the forces in the general case of the Funicular Polygon. 34 CONDITIONS OF EQUILIBRIUM OF A PARTICLE. [31. For convenience, let the vertices of the string or chain be denoted by the numbers 1, 2, 3,..., and let the forces P 2 , P 3 , ... act at the vertices. Let also the tension in the portion of the string (1, 2) be denoted by T 12 , &c. Fig. 29. Now, take any point, 0, and from it draw the line t l2 parallel to the string (1, 2), and proportional to the tension jP 12 . From the extremity of t lz draw the line, p z , parallel and proportional to the force P 2 . It follows, then, that since the forces T Vi , T Z3 , and P 2 form a system in equilibrium at the point (2), the third side, 23 , of the triangle t l2 , p 2 , 2 23 is parallel to T 2S , and proportional to it (Cor. 2, p. 18). In the same way, drawing ^; 3 parallel and proportional to P 3 , the side t 34 is parallel and proportional to T 34 ; and continuing this construction, the tensions in the successive portions of the string are all represented by the lines 12 , 23 , 84 , ... in the new figure (fig. 29). The figure (fig. 29) which represents by its lines, both in magnitude and in direction, all the forces of the system in fig. 28, is called by Professor J. Clerk Maxwell, a ' Force Diagram ' of the system. (Transactions of the Royal Society of Edinburgh, vol. xxvi.) When, as in example 8, all the applied forces, P 2J P 3 , ,.. are parallel, the Force Diagram of the system consists of a triangle with lines drawn from the vertex to different points in the base. Thus, taking any point, (fig. 30), and drawing ob parallel to MB (fig. 2*7), and proportional to the tension in it ; and then drawing bm vertical and proportional to the weight suspended at M, it follows that om will be parallel to MN, and proportional to the tension in it. Similarly for the rest of the figure. If all the suspended weights are equal, the lines bin, mn, np, pq, ... *' 3 ' are all equal, and fig. 30 at once shows that the tangents of the successive inclinations of the parts of the chain are in Arithmetical Progression. This figure also exhibits the con- stancy of the horizontal components of the tensions ob } om, on, ... these components being all equal to oq. 31.] EXAMPLES. 35 11. Suspension Bridge. The number of vertices of the polygon being very great, and the suspended weights all equal, the parabola which passes through all the vertices virtually coincides with the chain forming the polygon, and gives the figure of the Suspension Bridge. In this bridge the weights suspended from the successive portions of the chain are the weights of equal portions of the flooring. The weight of the chain itself and the weights of the sustaining bars are negligible in com- parison with the weight of the flooring and the Fig. 31. load which it carries. Fig. 30 may be taken to represent the Force Diagram of the Suspension Bridge, the vertical line ab, representing the weight of the flooring, being divided into as many equal parts as there are divisions of the chain. If these parts are sufficiently numerous, the lines ob, om, on, &c., are parallel to tangents to successive points of the chain. Let the span, AB, of the bridge = 2a, and let the height OH = h. Then, the equation of the parabola referred to horizontal and vertical axes of y and x, respectively, through (fig. 3 1 ) is m being a constant ; and the tangent of the inclination to the vertical of any portion _ dy __2m _ y ~ dx ~~ y ~~ 2x Hence the tangent at the point of support, B, makes with the horizon . , , . 2h an angle whose tangent is Therefore, oq (fig. 30) being parallel to the tangent at the lowest point of the bridge, and ob parallel to the tangent at the point B, tan boq = -- a Hence, since bq represents half the weight of the bridge, and ob the terminal tension of the chain at B, Terminal tension = - - = W -, > 2 sin boq 4A W being the weight of the flooring. Also, the vertical tension at B = \ W, and the constant Horizontal tension = W^ 4h 12. The entire load of a suspension bridge is 160,000 kilograms, the span is 64 metres, and the height is 5 metres ; find the tension at the points of support, and also the tension at the lowest point. Ans. Terminal tension = 268,208 kilograms. Horizontal tension = 256,000 36 CONDITIONS OF EQUILIBRIUM OP A PARTICLE. [31. 1 3. If the vertical bars which support the roadway of a suspension bridge are not at equal horizontal distances, prove that the vertices of the polygon formed by the chain will still lie on a parabola, provided that each vertical bar supports half of the adjacent portions of the roadway. This follows from the fact that the cotangent of the inclination of any chord of a parabola to the axis is proportional to the sum of the ordinates of the extremities of the chord. 14. If R is the resultant of any number of forces, P lt P 2 , P 3 , ..., acting in one plane on a particle, prove that R* = 2P 2 + 2SP X P 2 cos (PpP a ), where P 19 P 2 means the angle between P 1 and P 2 . (This result is true for non-coplanar forces). 15. If a particle is in equilibrium under the action of any forces, prove that the sum of the oblique components of the forces along any right line is zero. If 2JT and EF denote the sums of the components along two lines inclined at an angle = co, the square of the resultant is equal to (SZ) 2 + 2 (2JT) (S 7) cos o) + (2 F) 2 ; and this = (2X+2F) 2 cos 2 |+(2X-2F) 2 sm 2 |. Hence the result follows as in equations (3), p. 24. It is otherwise evident, since the resultant is the third side of a triangle, two of whose si^esare SXandSF. 16. If in example 7 the weights W and W, instead of being knotted to two given points in the string, are attached to two smooth rings which are capable of sliding freely along the string, determine the condition and position of equilibrium. Here, since the string passes freely over and under smooth surfaces, the tension is constant throughout its length. Now, the tension in Am is P, and that in CW=P'. Hence P=P'. For the equilibrium of m, we have, resolving vertically, W TF=2Psin0; .-. sin0= ^; 2 -I and for the equilibrium of m', W' TT=2Psm<9'; .-.sin^ 2x 17. A heavy particle is attached to one end of a string, the other end of which is fixed. Find the horizontal force which must be applied to the particle in order that the string may deviate by a given angle from the vertical, and find also the tension of the string. Ans. If F = the horizontal force required, T = tension of string, W = weight of particle, and 6 == angle of string's deviation, F= JFtanfl, F= TFsectf. 3I-] EXAMPLES. 37 18. A string ACB (fig. 24, example 5) has its extremities tied to two fixed points, A and B ; to a given point, C, in the string is knotted a given weight, W. Find the tensions in the portions CA and CB. Ans. Since AC and EC are given, the angles CAB and CBA are also given. If these angles are denoted by 6 and tf', and if T and T' are the tensions in CA and (72?, sn Tfcosfl 19. If (same figure) the extremities A and B are fixed, and the weight W is that of a smooth heavy ring at (7, which is capable of sliding freely along the string, find the horizontal force which must be applied to the ring C in order that the system may take a given position of equilibrium. Ans. If the angles CAB and CBA are and tf, and F = the required force, 6 & Fig. 32- 20. ABCD (fig. 32) is a system of pegs forming a square in a vertical plane ; a string attached to A and B passes through a heavy smooth ring, R, while another string is attached to C and R. The ring is kept in equilibrium half way between H, the middle point of CA, and 0, the centre of the square ; find the tensions in the strings ARB and CR. Ans. If W = weight of ring, T = tension in ARB, and T'= tension mCR, T- T'-= W. 32 16 21. In the last example if the tensions in the two strings are equal, find the point at which the ring must be placed on OH. f) 7? Ans. If = x > x is determined by the equation This equation has only two real roots, one between and 1, and the other be- tween 1 and 2. 22. A string whose weight is neglected passes over three smooth pegs, A>B,G (fig. 33), in a vertical plane, and sus- tains two equal weights, W, from its ex- tremities. Find the pressures on the pegs; and find also the magnitudes of the angles a, /3, and y when the system of pegs is least likely to break, the pegs being all equally strong. Fig. 33- 38 CONDITIONS OF EQUILIBRIUM OF A PARTICLE. [31. Ans. If P, Q, and R be the pressures on the pegs A,B, and C, respect- ively, P = 2 W cos ^ Q = 2 JFcos ^ ,72= 2 IF cos ; and since the sum 2 2-2 of a, /3, and y is given (= 27r), it follows that in the best arrangement 2 2 a = /3 = y = -'7r. For, unless each of the angles == - TT, some one of 71 the pressures must be >2 TFcos > or TT; and if the pegs are of equal 3 strength, it is best under these conditions, to have the pressures on them all equal. 23. If the string passes over any number of equally strong smooth pegs in the same vertical plane, find the best arrangement. Ans. If there are n pegs, each of the angles, a, /3, y, 8, . . . must be n 24. In example 14 calculate the pressures on the pegs A, S, C. Ans. The squares of the pressures are respectively P(2P+W), i{4P 2 + WW- 2 - JF 2 )(4P 2 - W' 2 ), P(2P+W). 25. If the strengths of the pegs, A, B, (7, in example 20, are propor- tional to Z, ra, n, find the best arrangement of the system. Ans. The angle a is given by the equation 2mncc 3 + (Z 2 + m 2 + ?i 2 ) tf-1? = 0, in which x = cos -. The angles /3 and y are at once found from a. 2 26. Let A A 1 ...A 5 (fig. 34) be any funicular polygon, with weights P lt P 2 ,P 3 ,P 4 suspended at its vertices A I} A 2 , A 3 , A^, respectively; draw any line, a 5 , meeting the verticals through A , A^, ... in the points a ot dtp d 2 ,..., and let A A 5 meet these verticals in A , DI , D 2 , . . . . Now construct a new polygon, a^cL^ a 2 ...a 5 , by taking d^ ^ = D^ A^ d 2 a 2 = ^D^A 2 - and so on, n being any number. Prove that the new poly- gon, whose fixed ends are a Q and a 5 will be kept in equilibrium by the set of forces P x , P 2 , P 3 , P 4 applied at its vertices a , a 2 , a z , a. Although this may be readily proved geometrically by principles of Graphic Statics, the student will do well to establish it by the method of example 8. He will easily prove that, if a and /3 are the incli- nations of A Q A 5 and a a 5 to the horizon, 01 , # 02 ,... the inclinations 31,] EXAMPLES. 39 of the sides A Q A 1} A 1 A t , ... ,and $ 01 , <#> 12J ... those of a Q a lt a^a^,... to the horizon, we shall have tan 01 tan/3 = -(tan 01 tan a) ; % tan $ 12 tan /3 = - (tan 12 tan a), &c. But if T denotes the constant horizontal tension in a* funicular polygon, the conditions of its equilibrium are P P tan 01 -tan 12 = -\ tan 12 tan 23 = ^; &c. These conditions are satisfied in the polygon a a v ...a s on the supposition that the horizontal tension in it = nT; and it is axiomatic that if internal forces can preserve equilibrium, they will. Of course all the ordinates (and not merely those through the vertices) of the derived polygon are proportional to the corresponding ordinates of the original. 27. Show that the last example enables us to construct for a given parallel system of forces a funicular polygon which shall pass through three given points. (A solution of this problem for any system of forces will be given in a subsequent chapter). 28. Given the base, NS (fig. 35), of a triangle NFS, and also the sum of the cosines of the base angles, SNP and NSP ; let the curve locus of P be constructed. Prove that if a particle be placed at any point of the curve and acted on by two forces, one repulsive from N and equal to , and the other attractive towards S and equal to -o2 ' ^e resultant force is, at Fig. 35. every position of the particle, directed along the tangent to the curve. N. B. This curve is called the l Magnetic Curve,' being one of those in which small iron filings would arrange themselves under the influence of a fixed magnet whose poles are N and S. It is to be observed that each little piece of iron is a magnet, having two poles at its extremities, and that it must therefore set at the point, P, where it is placed, in the direction of the resultant force on either of its poles. 29. Prove that the line of action of the resultant force of a magnet on a magnetic pole at P divides NS externally in the ratio NP Z : SP 3 . 30. Iron filings are sprinkled over a sheet of paper on which a magnet rests ; prove that all those filings which dip towards the same point on the line of the magnet lie on a circle (neglecting their mutual actions). CHAPTER III. THE EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. SECTION I. Smooth Curves. 32.] Smooth Surface. When a body is placed in contact with a surface, it is evident that, in addition to the given forces acting on the body, there is a certain force produced by the surface the force, namely , which the surface exerts to prevent the body from passing through it. This force is called the Reaction of the surface. Now, the surface being supposed to be rigid, there is evidently no limit to the magnitude of the force with which it is capable of reacting ; but the direction of the force depends on the nature of the surface itself. If the surface be perfectly Fmooth, it can react on any body in contact with it only in the direction of the normal to the surface at the point where the body is in contact with it. Thus (fig. 36), if a body, M, acted on by any given system of forces, be in contact at a point with a smooth surface, AB, the force which this surface exerts on the body takes the direction, ON, of the normal to the surface at the point of contact, 0, and its mag- Fig. 36. nitude will be such as to destroy the effect of all the other forces acting upon M. To the magnitude of the reaction, R, there is no limit ; so that if each of the other forces acting on M were increased 100 times, for example, the surface would react with a force equal to I oo R ; but the direction of R is strictly limited to that of the normal. We may therefore state that When two smooth bodies are in contact, their mutual reaction is normal to the surface of contact. 34-] NORMAL TO A CURVE. 41 s Fig- 37- 33.] Example. If P (fig. 37) is a heavy particle whose weight W> placed on a smooth spherical surface whose vertical diameter is AS, what is the position of equilibrium ? Here the forces acting on P are only two in number namely, its weight, W> and R, the reaction of the smooth surface. Now, this reaction takes place in the direction of the normal, PO, to the sphere at P; and since the particle is in equilibrium under the action of only two forces, these must be equal in magnitude, and act in opposite senses along the same right line. Hence, since Tracts vertically, PO must be a vertical line ; that is, P must be placed at A, the lowest point of the sphere, or outside the surface at J5, the highest point. Whatever be the smooth surface on which the particle is placed, it is evident that the points on it at which the particle will rest are points the normals at which are vertical lines. And, generally A particle will rest at those points of a smooth surface at which the normal coincides with the direction of the resultant of all the forces acting on the particle. 34.] Normal to a Curve. The normal to a curve at a given point is not, like the normal to a surface at a given point, a definite line, but is any line whatever in the plane perpendicular to the tangent at the point. Hence, for the equilibrium of a particle placed inside a smooth tube of any form, the resultant force on the particle need not act in a given right line, but must act in a given plane namely, the plane which is normal to the tube at the point where the ^e^^**^ B particle is placed. Thus, for example, let AB (fig. 38) be a smooth tube of any form, and let P be a particle placed inside it. If we imagine a string attached to P, coming out of the tube through an opening at P, which is not sufficiently large to allow P to come out, it is evident that we may pull at P with any force however great in the plane normal to the tube, and in all directions round P Fig. 38. 42 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [35. and the equilibrium of the particle will not be disturbed. But if we incline the string- ever so little to the normal plane at P, motion will ensue along the tube. 35.] Plane Curve. In the present chapter we shall consider only plane curves, i.e., curves which lie altogether in one plane. Moreover, when a particle is placed on a curve, and acted on by given forces, we shall suppose that all the forces act in the plane of the curve. Now, it is evident that the only effect which a curve produces on a particle placed upon it is a normal reaction of some definite magnitude. If, then, we produce upon the particle, by any other means, a force identical with this reaction, we may dispense with the curve altogether. This being so, if we call the reaction of the curve H, we may suppose the particle acted upon by all the given forces, and also by a new force equal to R> this latter acting in the direction of the normal to the curve. Thus, the case is the same as that treated in the last chapter namely, the equilibrium of a particle acted upon by any number of forces in one plane ; and in writing down the equations of equilibrium, we shall merely have to include the new force R among all the others. EXAMPLES. 1. A heavy particle is placed on a smooth inclined plane, AB (fig. 39), and is sustained by a force, F, which acts along AB in the vertical plane which is at right angles to AB ; find F, and also the pressure on the inclined plane. The only effect of the inclined plane is to produce a normal re- action, R y on the particle. Hence, if we Introduce this force, we may imagine the plane removed. Let W be the weight of the par- ticle, and i the inclination of the plane to the horizon. Resolving the forces along AB, we have F Fsin i = 0, or F = TFsin i; and, resolving perpendicularly to AB, R W cos i = 0, or R = W cos i. If, for example, the weight of the particle is 4 grammes and the inclination of the plane 30, there will be a normal pressure of grammes on the plane, and the force F will be 2 grammes. 35-] EXAMPLES. 43 2. In the previous example, if F act horizontally, find its magni- tude, and also that of R. Resolving along AB, and perpendicularly to it, we have, successively, j^cos i Wsmi = 0, or F=Wi&ni' } W and Fsini+ Wcosi R = 0, .-. R .; cost R is therefore in this case greater than it was before, as is sufficiently evident a priori. 3. If the particle is sustained by a force, F, making a given angle, 6, with the inclined plane, find the magni- tude of this force, and of the pressure, all the forces acting in the same vertical plane. Resolving along the plane, (fig. 40), ._ Wsmi t ' cos 6 ' and resolving perpendicularly to the plane, cose The student will, of course, observe that these values of ^and R could have been at once obtained, without resolution, by the equation (a), p. ii. 4. A heavy particle, whose weight is W, is sustained on a smooth inclined plane, by three forces applied to it, each W equal to ; one acts vertically, another o horizontally, and the third along the plane (fig. 41); find the inclination of the plane. Since we do not want R, the pressure on the plane, we shall resolve forces at right angles to R, that is, along the plane. Hence W . W W sin ^\ 1 cos i JFsm i = 0, 3 o o or 2suu = 1 +GOSZ, . (1) If we reject the factor cos for the present, we have 2 which determines the inclination. The student should observe that we have expelled the factor cos from equation (1), and this amounts to rejecting the solution cos - = 0. 44 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [35. Now in this, as well as in many physical and geometrical problems, such a solution ought not to be rejected, unless it is shown to be irrelevant to the question. So long as our equations are perfect interpretations of the physical or geometrical conditions of the pro- blem, no factor can furnish an irrelevant solution. It is only when an equation expresses more or less than is implied in the given con- ditions that irrelevant factors can present themselves. Instances of these factors frequently occur in the operations of Algebra and Analytic Geometry as, for example, when we rationalize an equation by the process of squaring. If, before this process, the square root of a quantity was affected with a minus sign, this sign will be indifferent in the rationalized result, and this latter, consequently, expresses more than was contained in the original equation. Hence it may happen that the result will furnish us not only with what is relevant, but, in addition, with what is wholly irrelevant. In the present instance the equation cos = would give the incli- 2 nation of the plane = 180, and the figure would then become fig. 42, in which the particle is placed under- W 180 neath the plane in such a way that equilibrium is manifestly impos- sible. Hence it appears as if the equa- tion cos - = were wholly without Fig. 42. meaning. A little reflection, however, will show that it is quite relevant. For equation (1) is merely the analytical expression of the physical con- dition that the component of the acting forces along the plane shall be zero. Now it is not enough for equilibrium that the component along some one line shall be zero ; for this, the component along some other line must vanish as well. Hence our result does not express the com- plete condition of the particle's equilibrium, but merely a part of that condition ; and each of the equations i 1 i tan-=-> and cos- = 0, expresses perfectly all the physical conditions contained in (1). For when the inclination is 180, the force which acted along the in- 3 clined plane becomes a horizontal force opposite to the given hori- W W zontal force ; and the vertical furnishes no component along the o o plane. The magnitude of R is | W. 5. A heavy particle, P (fig. 43), is placed inside a smooth parabolic tube whose axis is vertical, and is acted upon by a horizontal force, F } 35-] EXAMPLES. 45 equal p PM, PM being the ordinate of the point P ; find the position of equilibrium, Here the forces acting are TF, the weight of the particle, R, the normal reaction of the tube, and F. We shall obtain an equation between F and TF, without R, by resolving along the tan- gent at P. If 9 = angle between the tangent at P and the vertical, W cos 9 = F sin 9 = \iy . sin 9, where 2m Hence, for the position of equilibrium, retaining the factor cos 9, cos 6 ( Wpy tan 9) = 0. But if the equation of the parabola is y* = kmx, tan 9 = the equation is cos9(W-2fj.m) = 0. This equation of equilibrium can be satisfied in two ways, we can have cos 9=0, Hence (i) Firstly (2) or 6 = - } which gives the vertex of the tube as the position of equi- 2 librium. This position is a priori evident, since the particle would at the vertex be acted upon only by its weight and the reaction of the tube, the force F here being = 0. Secondly, the equation will be satisfied if TF-2//m = 0. (3) Now, this is simply a relation between the constants of the problem, and gives no value of that is, no definite position of equilibrium. In fact, if the equation (3) is satisfied, (1) will be satisfied, no matter what 9 may be. In physical language, then, the result is as follows : W if jut = - the particle will rest in all positions ; and if this relation 2w* does not hold, the vertex is the only position. It is well for the student to observe that /u, is here the quotient of a force by a line, the force being expressed in the same units as those of W, and the line in the same units as those of PM. For since we have put F = p PM, if Q is a force in the same units as those of JF, and I a line in the same units as those of PM, it is clear that the proper representation of F would be something of the form Q-j~3 therefore /LI = I I 6. A heavy particle, resting on a smooth inclined plane, is at- tached to a string which, passing over a smooth pulley, sustains another heavy particle: find the conditions and position of equi- librium. 46 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [35. Let W be the weight of the particle on the plane, P that of the hanging particle, and the inclination of the string to the inclined plane in the position of equilibrium. For the equilibrium of the particle on the plane, we have, resolving along the plane (since the tension of the string = P), W sin i P cos 6 ; W sin i .'. cos 6 = -p In order that there may be a position of equilibrium, this value of cos 6 must be < i, therefore W sin i must be < P. Explain the result when P = W. 7. Three particles, whose masses are TWj, w 2 , m 3 , are placed at three points, A, B, C (fig. 44), inside a smooth circular tube; they attract or repel each other with forces directly proportional to their masses and their distances ; find the posi- tion of equilibrium of the system. Consider the equilibrium of m^ at A. It is acted upon by two forces equal to m. 2 AB and m 3 AC, in the directions AB and AC. The resultant of these must be normal to the tube at A. But (Cor. 2, p. Fig 44> 1 6) the resultant acts towards a, the centre of mass of m 2 and w 3 , and if is the centre, OB = OC. Hence - = - ; and, bv considering the sin z m s equilibrium of -B, we have -a - = ^ Therefore sin x : sin y : sin z = m 1 : m 2 : w 3 . Also x + y + z = TT' } therefore #, y, and z are the angles of a triangle whose sides are proportional to m ly m 2 , and m 3 . These angles being known from some such equations as cos x = 2 - - - 5 &c., the relative positions of the particles are at once determined. The centre, 0, of the tube is the centre of mass of the particles. 8. Two smooth heavy rings, A and C (fig. 45), slide on two rods which are inclined to the horizon at angles i and i' ; a string con- necting A and C passes through a smooth heavy ring, B. Find the condition of equilibrium. Let the weights of A, B, (7, be P, W, P*, respectively, and let R and R' be the reactions of the rods on A and C. Construct the force- diagram of the system by drawing om from an arbitrary origin, 0, parallel and proportional to Rf, and mn parallel and proportional to P* ; then on will be parallel to BC and proportional to the tension in it. Drawing again np parallel and proportional to W, op will be 35-] EXAMPLES. 47 parallel to 13 A, and represent its tension. Finally, if pq_ represents P, oq will represent R. Since the tension in ABC is constant, on = op\ therefore a perpendicular from on mq bisects np. The p r W f P Fig. 45- length of this perpendicular is on the one hand (mn + J np) tan i', and on the other (pq + ^ np} tan i. Hence, equating these, we have tan ?= (P tent. This is a relation between the constants of the problem, and it there- fore constitutes a condition that equilibrium should be at all possible. If this condition is fulfilled, the position of equilibrium can be obtained by finding the angle, 0, which the string BC makes with the vertical. Evidently, from the force-diagram W+2P' tan e = tan i . 9. Two heavy rings, whose weights are P and P f (fig. 46), rest on the circumference of a smooth vertical circle, and are connected by a weightless string on which a heavy ring, whose weight is Q, slides freely. Find the position of equilibrium. Fig. 46. Construct the force-diagram. Let 6 and 0' be the inclinations of the radii CA and CA' to the vertical, and let (f) be the inclination of the portions of the string AB and BA' to the vertical. 48 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [35. The force-diagram then gives the statical equations n n (i) (2) & j To these must be added the geometrical equation which connects the length, /, of the string, with the radius, a, of the circle. Since the horizontal projections of the broken lines AC A' and ABA' are the same, we have a (sin B + sin ^) = I sin <. (3) Equations (1), (2), and (3) are sufficient to determine the unknown angles 0, 0', and . 10. A body, whose weight is 10 kilogrammes, is supported on a smooth inclined plane by a force of 2 kilogrammes acting along the plane and a horizontal force of 5 kilogrammes ; find the inclination of the plane. . _, /3 X Ans. sin (*)* 11. A heavy body is sustained on a smooth inclined plane (incli- nation i) by a force P acting along the plane, and a horizontal force, Q. The inclination being halved, and the forces P and Q each halved, the body is still observed to rest ; find the ratio of P to Q. Ans. 7; = 2 cos 2 - Q 4 12. Two weights, P and Q (fig. 47), rest on a smooth double-inclined plane, and are attached to the extremities of a string which passes over a smooth peg, 0, at a point vertically over the intersection of the planes, the peg and the weights being in a vertical plane. Find the position of equilibrium. Ans. If I the length of the string, and C0=7i, the position of equilibrium is defined by the equations sin a _ _ sin /3 cos cos (f> cos a cos (3 _ I sin 6 sin P, Q. Then the base angles of this triangle are those made with the vertical by the radii of the wire drawn to the rings. 18. Two weights rest on the convex side of a parabola whose axis is vertical, and are connected by a string which passes over a smooth peg at the focus; show that equilibrium is impossible unless the weights are equal. 19. Two weights, P and Q (fig. 48), rest on the concave side of a parabola whose axis is horizontal, and are connected by a string which passes over a smooth peg at the focus F. Find the position of equilibrium. Ans. Let I = length of the string ; 6 the angle which FP makes with the axis; 4ra = the latus rectum of the parabola ; then Fig. 48. * The student will afterwards see that this would be the case if the natural length of the string were so small as to be negligible in the problem. 50 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [35. 20. A particle is placed on the convex side of a smooth ellipse, and is acted upon by two forces, F and F' t towards the foci, and a force, F", towards the centre. Find the position of equilibrium. Ans. If r = the distance of the particle from the centre of the FF' curve ; b = semi-axis minor ; and n = =; ; then b 21. A heavy particle, P, is placed on the concave side of a smooth vertical circle whose lowest point is A and highest point B. If the particle is acted upon by two forces, in the directions AP and HP, equal to pBP, and pAP, respectively, find the position of equi- librium. Ans. Let W = the weight of the particle ; 6 =: the angle made with the vertical by the radius to P ; a = the radius of the circle ; then 22. A particle, P, is acted upon by two forces towards two fixed points, S and H, these forces being - and ,, respectively; prove that P will rest at all points inside a smooth tube in the form of a curve whose equation is SP. PH = & 2 , k being a constant. 23. A particle, P, is placed inside a smooth circular tube, and acted upon by two forces towards the extremities, A and B, of a fixed diameter, AB ; the forces are respectively proportional to PA and PS : prove that the particle will rest in all positions. 24. Two weights, P and Q, connected by a string rest on the convex side of a smooth cycloid. Find the position of equilibrium. Ans. If I = the length of the string, and a = radius of generating circle, the position of equilibrium is defined by the equation where 6 is the angle between the vertical and the radius to the point on the generating circle which corresponds to P. 25. Two weights, P and Q, rest on the convex side of a smooth vertical circle, and are connected by a string which passes over a smooth peg vertically over the centre of the circle; find the position of equi- librium. Ans. Let ^ = the distance between the peg, B, and the centre of the circle ; and < = the angles made with the vertical by the radii to P and Q, respectively; a and /3 = the angles made with the tangents to the circle at P and Q by the portions PB and QB of the string ; I = length of the string ; then 37.] LAWS OP FRICTION. 51 cos a cos p sin^ sin| v cos a cos /3 ' & cos (0 4- a) = a cos a, cos ( which is just half as 2o great as before. Hence, though the area over which friction acts is doubled, the intensity of pressure is halved ; and it is consistent with common sense that the friction per unit of area should be halved also. Thus, on the whole, the same total tangential force is required to set up sliding in both cases. 40.] Actual Magnitudes of Coefficients of Friction. It is well that the student should have some idea of the actual magni- tudes of coefficients of friction between bodies. For this purpose he should look at a table of these coefficients. Practically there is no observed coefficient much greater than 1. In Rankine's table the coefficient for damp clay on damp clay is given as 1 , and that for shingle on gravel is at the most 1.11. Most of the ordinary coefficients are less than J. 41.] Other Coefficients of Friction. It is found by experiment that the friction which resists the beginning of sliding is greater than that which resists its continuance. Again, the resistance which is opposed to the rolling of one surface on another is dis- tinguished by the special name of Rolling Friction, but it would more properly be called Resistance to Rolling. At present we shall limit ourselves to the consideration of the friction of the beginning of motion which is ex- pressed by the equation F= (J.R. 42.] Reaction of a Bough Curve or Surface. Let AB (fig. 49) be a rough curve or surface ; P the posi- tion of a particle on it ; and suppose the forces acting on P to be con- fined to the plane of the paper. Let R l = the normal resistance of the surface, acting in the normal, PN, and F = the force of friction, acting along the tangent, PT. 54 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [44. The resultant of R l and F is a force which we shall call the Total Resistance of the surface. It is represented in magnitude and direction by the line PR = R, which is the diagonal of the parallelogram determined by R l and F. We have seen that the total resistance of a smooth surface is normal ; but this limitation does not apply to a rough surface. The angle, $, between R and the normal is given by the equation F tan = ^- ^i Hence, < will be a maximum when the force of friction bears the greatest ratio to the normal pressure. But this greatest ratio is what we have called the coefficient of friction, p ; and this ratio is attained when the particle is just on the point of slipping along the surface. Therefore the greatest angle ly which the Total Resistance of a rough curve or surface can deviate from the normal is the angle whose tangent is the coefficient of friction for the bodies In contact ; and this deviation is attained when slipping is about to commence. 43.] Angle of Friction. The angle between the normal and the total resistance of a rough surface when slipping is about to take place is called the Angle of Friction*. We shall throughout denote it by A ; and if /ot is the coefficient of friction, tan \ = IJL. 44.] Experimental Determination of fji. Let P be the position of a heavy particle, whose weight is W 9 on a rough plane, AB, whose inclination is gradually increased until P is on the point of slipping down. Consider the equilibrium of P in these circum- 0. stances. It is acted upon by two forces, namely, its weight, W> and the total resistance, R, of the plane. For equilibrium these forces must be equal and act in opposite senses. Hence R acts in a vertical line ; and since slipping is about to take place, the angle between R and the normal, PN, to the plane must (Art. 42) be equal to A, the angle of friction. But the angle between the * Sometimes called the Angle of Repose. 45-] LIMITATION OF THE TOTAL RESISTANCE. 55 vertical and PN is also equal to the inclination of the plane to the horizon. Hence the inclination of a rough plane on which a particle, acted upon solely by its own weight } is just about to slip, is the Angle of Friction. This result might have been proved by the resolution of forces. Thus, if 2t l be the normal pressure, the force of friction acting up the plane is pR-^ since slipping is about to begin. Hence, resolving forces horizontally for the equilibrium of P, JRi sin i JJL ^ cos i = 0, i being the inclination ; or tan i = jot, therefore i = A. Morin determined the coefficient of friction between two substances by placing one on a fixed horizontal plane made of the other, and then measuring the least horizontal force which should be applied to the body resting on the plane to cause it to slide. The ratio of this force to the weight of the body is the required coefficient of friction. 45.] Limitation of the Total Resistance. As in the case of the resistance of a smooth curve or surface, there is no limit to the magnitude of the total resistance of a rough curve or surface for the surfaces with which we are at present concerned are supposed to be capable of resisting penetration to any extent the only limitation to which the total resistance is subject being one of direction, and this limitation is thus expressed : The Total Resistance of a rough curve or surface, though un- restricted in magnitude, can never make with the normal an angle greater than the angle of friction corresponding to the two bodies in contact. Within this limit, the total resistance can assume any magni- tude and direction, so that we at once deduce the following important principle : If the Total Resistance can maintain equilibrium, it will do so. Thus, let P (fig. 51) be a heavy par- ticle placed upon a rough plane whose inclination is less than A, the angle of friction. Then it is clear that, to keep P at rest, the total resistance, R, has only to be equal and opposite to W, the weight of P. But drawing PQ, making the angle of friction, A, with the 56 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [46. normal, PN, we see that the direction of R falls within the prescribed limit ; and therefore the equilibrium will subsist, no matter how great W may be, for there is no limit as to the magnitude of R. 46.] Limiting Equilibrium. A particle acted upon by any forces and placed upon a rough surface is said to be in limiting equilibrium when it is in such a position that the total resistance of the surface makes the angle of friction with the normal. In such a position if any slight change should occur in the circum- stances of the particle, in virtue of which the total resistance would be compelled to make a greater angle with the normal, equilibrium could subsist no longer ; for the total resistance can never be inclined to the normal at an angle greater than the angle of friction. Or we may put the matter thus. In every case the equilibrium of a particle restricted to a rough curve or surface is broken only by some circumstance which compels the total resistance to make with the normal an angle greater than the angle of friction. The manner in which this is supposed to happen depends on the particular problem. For example, let us enquire into the circumstances of the equilibrium of a heavy particle, whose weight is W t on a rough curve, AB (fig. 52), whose plane is vertical, the particle being acted upon by a horizontal force, F. The problem proposed for solution may be any one of the three follow- ing : (a) Determine the least horizontal force that will sustain a particle, of weight W> at a given point, P, of a given rough curve, AB. (b) Determine the point at which a 52< particle, of weight W, will be just sus- tained by a given horizontal force, F, on a given rough curve, AB. (c) Determine the least coefficient of friction that will allow a particle, of weight W, to rest at a given point, P, of a curve, AB, the particle being acted on by a given horizontal force, F. If PN be the normal at P, and PR be drawn making the angle of friction, A, with it, PR will be the direction of the total resistance, since, by supposition, the particle is about to slip down. All three problems are solved by the equation 47-] FRICTION IN NON-LIMITING EQUILIBRIUM. 57 being the inclination of the tangent at P to the horizon. But the manner in which equilibrium is supposed to be broken is not the same in each of them. If, in the first case, F< TFtan (0 A), in the second, > A + tan" 1 (-~) > and in the third, A< tan- 1 77F '' (jy) , the particle will not rest at P. Thus the equilibrium may be broken by (a) a slight change in some of the acting forces ; (b) a slight change in the position of the particle ; or, (c) a slight change in the nature of the supporting surface, i.e., a diminution of its roughness. If the particle is in limiting equilibrium (i.e., if the total resistance makes the angle of friction with the normal to the supporting surface) it is evident that equilibrium will always be broken if the third of these changes occurs ; but it may not be broken by either of the others. Take, for example, a heavy particle placed on an inclined plane whose inclination to the horizon is the angle of friction. It is evident that any change may be made, either in its weight or in its position on the plane, and equilibrium will still subsist ; for in neither case is the total resistance (equal and opposite to W) compelled to make with the normal an angle > A. In every case of equilibrium it is to be observed that the Force of Friction (Art. 37) acts in the sense opposite to that in which motion would ensue if the bodies in contact became gradually smoother. 47.] Friction in non-limiting equilibrium. The beginner is very prone to assume that, if ju, is the coefficient of friction between two bodies, in every case in which one of these bodies rests against the other the force of friction is jmP, where R is the normal pressure between them. That this is not so he will easily see by considering the case in which a heavy piece of metal rests on a horizontal plane of wood the coefficient of friction between the metal and the wood being, say, f, and no forces, other than its weight and the resistance of the plane, acting on the body. So far from the force of friction being f of the normal pressure, the force of friction is zero, and will come into 58 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [47. existence only when some horizontal force is applied to the body. The force of friction will always be equal to this horizontal force and will attain the value f R only when slipping is about to take place. The changes both in magnitude and in direction which the Total Resistance between two rough surfaces in contact undergoes while equilibrium changes from a state bordering on motion in one direction to a state bordering on motion in the opposite direction may be very simply illustrated by solving the following problem : A heavy body of weight W is held on a rough inclined plane of inclination i by a horizontal force P ; the force P being varied gradually from the value required just to sustain the body to the value required just to drag it up the plane, it is required to repre- sent graphically the different magnitudes and directions of the Total Resistance corresponding to the successive values of P. Let (fig. 53) be the position of the body, and measure off a vertical line OW to represent the magnitude of W. Then, for dif- ferent values of P, the resultant of W and P will be represented by lines drawn from and terminating on the hori- zontal line WH. The Total Resistance of the plane on the body is, of course, equal and opposite to the resultant of P and W, and it will therefore be represented by a line drawn from to a horizontal line, R^R^ drawn at the same distance above as the line WH is below it. Let ON be the normal to the plane at 0, and draw the lines OR-L and OR 2 making the angle, X, of friction with the normal at opposite sides of it. Let these lines be produced to meet the line JTHiiL the points r x and r 2 . Then for equilibrium the resultant of P and W must be repre- sented by some line intermediate between O x and Or 2 . When the resultant of P and W is Or v the Total Resistance 53. 48.] EXAMPLES. 59 of the plane is OR V and since this makes the angle of friction with the normal, the body is on the point of slipping down. When the resultant of P and W is Or 2 , the Total Resistance is OR 2) and the body is on the point of slipping up. The values of P which will just sustain the body and just drag it up are, respectively, W tan (i - X) and W tan (i + X), as appears at once from the figure or by calculation. If P has a value between these limits, the Total Resistance, OR, will be intermediate between OE l and OR 2 , and the equilibrium will not be limiting, i.e., the body will not be on the point of slipping either up or down ; and the force of friction, which is the component of R along the plane, will not be //, times the normal pressure, except in the two states bordering on motion. If P has the value ^Ftan i, which is intermediate between its extreme values, the Total Resistance will be normal to the plane, and in this state there will be no force of friction exerted between the plane and the body. 48.] Passive Resistances. The force of friction between a body and a rough surface belongs to a class of forces called Passive Resistances^ i.e., forces which come into existence only on account of the action of other forces and which always endeavour to destroy the effect of these other forces. To this class, indeed, belongs also the normal pressure between any two bodies, and also the resistance of air or any other fluid to a body moving through it. And it is an axiom with regard to all passive resistances that if they can preserve equilibrium they will. EXAMPLES. 1 1. A heavy particle is placed on a rough plane inclined to the horizon at an angle less than the angle of friction ; find the limits of the direction of the force required to drag it down. Let PN (fig. 54) be the normal to the inclined plane, and let PQ be drawn, making the angle NPQ = A, the angle of friction. Now, the necessary and sufficient condition that equilibrium should exist is, that the resultant of the weight, W, and the force applied, F, should fall within the angle NPQ. Hence, producing NP and QP to n and q, we see that no force applied to P within the angle nPq 60 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [48. will disturb the equilibrium. F must, therefore, be applied within the angle NPq, and act from P towards the left of the figure. ' 2. Two heavy particles, whose weights are P and Q, rest in limiting equilibrium on a rough double-inclined plane, and are connected by a string which passes over a smooth peg at a point, A (fig. 55), ver- tically over the intersection, B, of the two planes. Find the position of equilibrium. Let the inclinations of the planes be a and /3; let the length of the string be Fig. 54- Z, and AB = h ; and let the portions of the string make angles 6 and tf> with the planes. Suppose that P is on the point of ascendiog, and Q of descending. Then, since the motion of each body is about to ensue, the total resistances, R and S, must each make the angle of friction with the corresponding normal; and since the weight P is about to move upwards, R must act towards the left of the normal, while, since Q is about to move downwards, S must act to the right of the corresponding normal. If T is the tension of the string, we have for the equilibrium of P, T=P cos(6-\)' Again, for the equilibrium of Q, T-n .sin(/3-X) NJ / 1 , \ \ * Fig. 55- Hence, equating the values of T, p sin(q + A) _ *cos(0 A) ~ sin(/3-A) (1) This is the only statical equation connecting the given quantities. We obtain a geometrical equation by expressing that AB and the length of the string are given. This is, evidently, L/c^lL^ /LI v, _ C, sn (2) , and con- -* Equations (1) and (2) determine the values of 6 and stitute the solution of the problem. Other Solution. Instead of considering the total resistances, R and S, we may consider two normal resistances, S l and S lt and two forces of friction, ^R l and y L S l , acting respectively down the plane a and up the plane /3. In this case, considering the equilibrium of P, and 48.J EXAMPLES. 61 resolving forces along and perpendicular to the plane a, we have P sin a + fJ?! = T cos 0, ) ; j and for the equilibrium of Q, Eliminating E 19 S lt and ^from the systems (A) and (J?), we arrive at the same statical equation as before. The method of considering total resistances instead of their normal and tangential components is almost always more simple than the separate consideration of the latter forces. . 3. If in the last question P is given, what are the limits of Q con- sistent with equilibrium ? If Q be so large that it is about to drag P up, its value, Q lt will be given by equation (1), o _ sin (a + A) cos ((ft + A) t ' Bin (/3- A) cos (0- A)' and if Q be so small that P is about to descend, its value, Q z , will be Q _ i> sin (a -A) cos ((ft -A) ^ ^CdLi^A in3 A0 A' the angles 6 and (ft being connected by equation (2). ' 4. A heavy ring is placed on a rough vertical circle ; find the limits of its position consistent with equilibrium. Ans. Draw two diameters making the angle of friction with the vertical diameter. The ring will rest anywhere on the circumference between the two upper extremities, or between the two lower extremi- ties, of these diameters. : 5. A heavy body whose weight is 20 kilogrammes is just sustained on a rough inclined plane by a horizontal force of 2 kilogrammes, and a o force of 10 kilogrammes along the plane ; the coefficient of friction is - ; o find the inclination of the plane. _. ,25 X Ans. 2 tan 1 ( ) 6. A heavy particle is placed on a rough plane whose inclination to 3 the horizon is sin" 1 ( ) j and is connected by a string passing over a o smooth pulley with a particle of equal weight, which hangs freely. Supposing that motion is on the point of ensuing up the plane, find the inclination of the string to the plane, the coefficient of friction being -. Ans. By resolving forces along the inclined plane, we have, if 6 = inclination of the string to the plane, 1 /, 1 6 e * e - sin 6 = I cos 6, or - sin - cos - = sin 2 - j & 6 A f tt 62 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [48. 6 1 one solution of which is = 0, and the other is tan - = - & 7. In the second solution of the last question, exhibit the position of the string, and explain the result. 1 8. A heavy particle acted upon by a force equal in magnitude to its weight is just about to ascend a rough inclined plane under the influence of this force ; find the inclination of the force to the inclined plane. Ans. If 6 is the required inclination, A = angle of friction, and i =. inclination of the plane, 0=-^ i, and 6 = 2\ + i-^ & 6i are possible solutions. (6 is here supposed to be measured from the upper side of the inclined plane. If - > 2 A + i, the applied force will 2 act towards the under side). 9. In the first solution of the last question, what is the magnitude of the pressure on the plane ? Ans. Zero. Explain this. 10. Prove that the horizontal force which will just sustain a heavy particle on a rough inclined plane will sustain the particle on the same plane supposed smooth, if the inclination is diminished by the angle of friction. 1 1 . What is the least coefficient of friction that will allow of a heavy body's being just kept from sliding down an inclined plane of given inclination, the body (whose weight is W) being sustained by a given horizontal force, P ? Wtzni-P Explain <& priori, why we get a negative value for the coefficient of friction unless PFtan i > P. 12. It is observed that a body whose weight is known to be TFcan be just sustained on a rough inclined plane by a horizontal force P, and that it can also be just sustained on the same plane by a force Q up the plane; express the angle of friction in terms of these known forces. Ans. Angle of friction = cos"" 1 ^ , r>2 . ~W2 " 13. It is observed that a force, Q lt acting up a rough inclined plane will just sustain on it a body of weight JF, and that a force, Q. 2 , acting up the plane will just drag the same body up ; find the angle of friction. Ans. Angle of friction = sin" 1 48.] EXAMPLES. 63 14. A body is held on a rough inclined plane (i > X) by a force which acts up the plane ; this force being varied gradually from the value required just to sustain the body to the value just required to drag it up, it is required to represent graphically the different magnitudes and directions of the Total Kesistance. 15. In example 8, p. 47, if the rings A and are equally rough, find the condition that there may be a limiting equilibrium in which each is about to slip down. Ans. If A is the angle of friction, the required condition is (P' + jj tan (i'-\) = (P + ^ In this case the lines Om and Oq must be drawn making angles i' A, and i A, respectively, with the line mq. 16. In the same example, if one of the rings, C, is in a position of limiting equilibrium, find the direction of the string, the position of the other ring, A, and the direction of the total resistance at it. Ans. The position of the string is determined by the equation the + or sign being used according as C is about to slip up or down. When 6 is known, the position of A is known ; and the direction of the total resistance at A is found from the equation (^ + P) tan Oqm = (^ + P' ) tan (i' A). & & 17. A heavy body is to be dragged up a rough inclined plane : find the direction of the least force requisite. Ans. The force must make the angle of friction with the inclined plane. This follows at once either by resolution of forces, or by drawing the force-diagram. Viewed in the latter way, the problem is this : Given one force (the weight) in magnitude and direction, and the direction of another (the total resistance), when is the resultant a minimum ? Evidently when it is at right angles to the total resistance. N. B. This result is often expressed thus : The best angle of traction up a rough inclined plane is the angle of friction. 18. Two weights, P and Q, connected by a string, whose weight is neglected rest on a rough vertical circle, the string being supposed not to be anywhere in contact with the circle ; find the limits of the position of equilibrium. Ans. If 6 be the angle made by the radius to P with the vertical, I = the length of the string, and a = the radius of the circle, 6 may have any value between l and 2 , these being given by the equations 64 EQUILIBRIUM OF A PARTICLE ON PLANE CURVES. [48. tan 6 l = Q sin (- + A) + P sin A - j - ; Q cos (- + A) + P cos \ $sin(- A)- PsinA tan 2 = , Q cos ( A) + P cos A A being the angle of friction. 19. Two heavy bodies rest, at points P and Q, on any rough curve in a vertical plane, and are connected by a string, which is nowhere in contact with the curve; show that in the limiting positions of equi- librium the total resistances at P and Q intersect on the circle passing through P, Q, and the point of intersection of the normals at P and Q. 20. Two heavy particles, P and Q (fig. 56) rest, one on a rough diameter, A B, of a rough vertical circle, and the other on the convex side of the circle, the particles being con- nected by a string which passes over a smooth peg at the upper extremity, B, of the diameter. Find the position of equilibrium, the string being supposed to be nowhere in contact with any rough surface, and the coefficients of friction for P and Q being different. Ans. If a = the inclination of AB to the vertical, 6 = inclination of the radius drawn to Q to the vertical, fj. = coefficient of friction between P and AB, n' = coefficient of friction between Q and the circle, the limiting positions of equilibrium are given by the equations Q (sin 0j + 1/ cos 6^ = P (cos a /a sin a), Q (sin 2 fjf cos 2 ) = P (cos a + [JL sin a). 21. AB is the vertical diameter of a rough circular tube, of which C is the centre; P is a heavy particle placed inside the tube, and attached to three strings which, passing through a narrow slit in the inner side of the tube, pass over smooth pegs fastened at A, B, and C. Find the position of equilibrium. Ans. If the weight of the particle = W, and the weights sus- pended over the pegs, A, C, and B = P lt P 2 , and P 3 , respectively, the angle 6, which CP makes with the vertical when the particle is about to slip down, is given by the equation Fig. 56. -P 2 sin A-P 3 sin ( + A) = 0; and by changing the sign of A in this equation we obtain the position 48.] EXAMPLES. 65 in which P is about to slip up. ' Anywhere between these positions the particle will rest in non-limiting equilibrium. 22. Two heavy particles, P and Q (fig. 57), rest on two rough circular arcs which have a common vertical tangent at ; P and Q are connected by a string which passes over a smooth pulley at ; find the positions of limiting equilibrium. Ans. Let & and $ be the angles sub- tended by the arcs OP and OQ at the centres of the corresponding circles, a and 6 the radii of the circles, A and the angles of friction for P and Q, respectively, and I the length of 'the string ; then, if P is about to slip down, the equations Fig. 57- p cos(0 + A) _ Q cos (e) co S (|-0 and determine the position of equilibrium. Changing the signs of A and e, we obtain the position in which Q is about to slip down. 23. A particle rests on a rough curve whose equation is/ (x, y) = Q, and is acted on by forces the sums of whose components along the axes of x and y are X and Y ; prove that the particle will rest at all points on the curve at which 1 dx -r~ dy >cos A. 24. Two rings whose weights are P and Q are moveable on a rough rod inclined to the horizon at an angle i ; these rings are connected by a string of given length which passes through and supports a smooth heavy ring W '; find the greatest distance between P and Q. Ans. If 6 is the inclination of either portion of the string to the vertical, the greatest distance between the rings is obtained by giving tan 6 the greater of the values W+2Q W Q being the upper ring. tan (\i), W+ZP W CHAPTER IV. THE PRINCIPLE OF VIRTUAL WORK, SECTION I. A Single Particle. 49.] Orthogonal Projection. Let Ox and AB (fig. 58) be any two right lines inclined at an angle 0. If from the extremities, A and B, of the right line AB, two perpendiculars, Aa and Bb, be let fall on Ox, the line ab is called the orthogonal projection of AB on Ox. If the lines Aa and Bb had been each drawn parallel to a given line, which is not perpendicular to Ox, ab would be an oblique projection of AB. In the case of orthogonal projection it is evident that ab = AB cos 0. 50.] Projection of a Broken Line. Let ABCD (fig. 59) be a zig-zag or broken line. Then it is evident that the projection (orthogonal or oblique) of the line AD, joining the first and last Fig. 58. points, A and D, is equal to the sum of the projections of the separate lines, AB, BC, and CD, on any line Ox. This is also true when the line Ox, on which the projection takes place, cuts any or all of the lines AB, BC, . . . between 5 1- VIRTUAL WORK. 67 the vertices, A, B, C, . . . , of the polygon formed by them, as in fig. 60. If the sides of a closed polygon taken in order be marked with arrows pointing from each vertex to the next one, and if their projections be marked with arrows flying in the same directions, then, lines measured from left to right being considered positive, and lines from right to left negative, we may evidently state this result as follows : The sum of the projections of the sides of a closed polygon on any right line, allowance being made for positive and negative projections, is zero. 51.] Virtual Displacement. Virtual Work. If a point at (fig. 61) be conceived as displaced to A, OA may be called the virtual displacement of the point. / Let OP be the direction of a /\'' force, P, and let AN"be drawn per- /'' \ pendicular to it ; then ON is the projection of the virtual displace- ment along OP, and the product of the force, P, by the projection, ON, of the virtual displacement is Fi g- 6l - called the virtual work of the force. We shall therefore say that The VIRTUAL WORK of a force is the product of the force and the projection along its direction of the Virtual Displacement of its point of application. If 6 be the angle between the force and the virtual displace- ment, The Virtual Work = P.ON= P. OA cos6 = P cos 6 . OA. Now P cos 6 is the projection of the force along the direction of displacement, and is equal to OM, if PM is perpendicular to OA. Hence we may also define the virtual work of a force as follows : The virtual work of a force is the product of the virtual displace- ment of its point of application and the projection (or component) of the force in the direction of this displacement. */ / J- This latter definition is for some purposes more convenient than the former. It is to be observed that the projection of a line, AB (fig. 58), of given length remains unaltered in magni- tude when AB is moved parallel to itself into any position. F 2 d _< 68 THE PRINCIPLE OF VIRTUAL WORK. [52. 52.] Theorem. The virtual work of a force is equal to the sum of the virtual works of its components, rectangular or oblique. Let a force R, represented by OR (fig. 62), act at 0, and let its com- ponents be P and Q, represented by OP and OQ. Let OA be the virtual displacement of 0, and let its pro- jections on R, P, and Q, be r, jo, and q, respectively. Then the virtual works of these forces are R. r, P.p, Q.q. Draw Pm and Rn perpen- dicular to OA. Then On is the pro- jection of R in the direction of the displacement, and by the end of Art. 51, -r, r\ 4 r\ R.r = OA x On. Similarly P .p = OA x Om, and Q . q = OA x mn. Hence P.p + Q.q = OA(Om + mn) = OA x On = R.r.Q.E.D. 53.] Theorem. The sum of the virtual works of any number of forces acting at a point is equal to the virtual work of the resultant. This may be proved by taking the forces two-and-two; and using the last Theorem, or by making use of the polygon of forces (see fig. n, p. 18). The sum of the virtual works of the forces is equal to the virtual displacement multiplied by the sum of the projections along it of the sides of the polygon parallel to the forces (Art. 51). But (Art. 50) the sum of these projections is equal to the projection of the remaining side of the polygon, and this side represents the resultant. Therefore, &c. It follows, then, that When a system of forces acting at a point is in equilibrium, the sum of the virtual works of the forces =. 0. For such a system will be represented by a closed polygon, and (Art. 50) the sum of the projections of the sides of the polygon along any right line = 0. 54.] Convention of Signs. If the virtual displacement, OA (fig. 63), project on the line of the force P in the sense opposite to that in which P acts, the projection ON is to be considered negative, and the virtual work is negative. In this case P will also project on the line of displacement in the sense opposite to OA. In fig. 64 the virtual displacement, OA, is such as to give 56.] GENERAL EQUATION OP VIRTUAL WORK. 69 positive projections, Or and Op, along the forces R and P, and a negative projection, Oq, along Q. And if in this case the R Fig. 63. Fig. 64. lengths of Or, Op, and Oq are denoted by r,jo, and q, the equation of virtual work will be ~ft r _ p Q ~ 55.] Nature of the Displacement. It must be carefully observed that the displacement of the particle on which the forces act is both VIRTUAL and perfectly ARBITRARY. In the motion of a particle, treated of in Kinetics, the displacement is often taken to be that which the particle actually undergoes ; but in the statical problem of the equilibrium of forces, the relation between them, expressed in an equation of virtual work, holds, whatever the displacement may be that is, it holds whether the displacement be an actual or merely an imagined one. Since with regard to the equilibrium of forces a state of absolute rest and a state of uniform motion in a right line are not essentially different, we shall see that the most useful applications of the Principle of Work are made in the case of machines moving uni- formly. The second characteristic of the displacement namely, Us arbitrariness is most important, as will presently appear. 56.] General Equation of Vir- tual Work. Let several forces, P lt P 2 , . . . (fig. 65), act in equilibrium on a particle, 0, and let OA be any conceived, or virtual, displacement of 0. Letting fall perpendiculars, the forces, the 2 , on projections Op 2 , Op 3 , and Op, are all positive, while Op l and Op 5 are negative (Art. 54). Hence the equation of virtual work is 70 THE PRINCIPLE OF VIRTUAL WORK. . [57. If the projections of the displacement be denoted by p lt p^ . . . , and if these quantities are supposed to carry their proper signs with them, this equation becomes, the number of forces being any whatever, P 1 .p 1 + P 2 .p 2 +P 3 .p,+ ... = J (1) or 2(P.j?) = 0. (2) 57.] General Displacement of a Particle. The most general displacement of a single particle is a simple motion of translation from the point, 0, which it occupies, to another point., A. It is true that in Molecular Dynamics, very small portions of matter are conceived as capable not only of translations but also of rotations about axes through themselves. Indeed every portion of matter, since it must possess extension in space, must be capable of both kinds of displacement ; but the second kind does not belong to our present purpose. 58.] Deduction of the Equations of Equilibrium from the Equation of Virtual Work. Through draw any two axes, Ox and Oy, rectangular or oblique, and let a and /3 be the projections of the virtual displacement, OA> along these axes. Replace the force PL by its components, X : and Y ly along Ox and Oy. Then (Art. 52) P 1 . A = aX 1 + Similarly, ^2-^2 = a ^2 + Hence equation (1) of Art. 56 becomes or a2X+/3Sr= 0. (1) Now a and /3 are perfectly independent of each other. For the displacement OA may be chosen so as to keep a constant while varying /3 at pleasure, or vice versa. Suppose, then, that /3' and a are the projections of a new virtual displacement, and we shall have a2X+/3'27= 0. (2) Subtracting (2) from (1), we have Now /3 /3 7 is not = 0, therefore 2 T must = ; and in the same 6o.] CASE IN WHICH VIRTUAL WORK VANISHES. 71 way 2JT = 0. Hence we arrive at the equations of resolution of forCeS 2X= 0,27=0, which were deduced in Chap. II.* 59.] Elementary Virtual Work. In the general equation of virtual work, for forces acting in equilibrium on a single particle, namely, P 1 .jo 1 + P 2 .j9 2 + P 3 .^ 3 +... = 0, or 2P.^ = 0, no limitation has been placed upon the magnitude of the virtual displacement. This equation is true, independently of its magnitude ; but it is generally more convenient to assume the virtual displacement to be infinitesimal, even in the case of the equilibrium of a single particle, and it is absolutely necessary to do so (as will presently be seen) in treating of the equilibrium of a connected system of particles. If the virtual displacement is infinitesimal, its projections, _p l9 p 2 , . . . , on the several forces acting upon the particle are all infinitesimal. We shall, therefore, denote these small projections in future by bp v bp 2 , . . . , and the equation of elementary virtual work will be P 1 .^ 1 + P 2 .6j? 2 + P 3 .5^ 3 +... = 0, or SPSj? = 0. 60.] Case in which the Virtual Work of a Force vanishes. If a force P act at a point 0, and if the virtual displacement OA is at right angles to the direction of P, it is clear that bp, the projection of OA on the direction of P, is equal to zero. Hence, when the virtual displacement is at right angles to the direction of the force, the virtual work of the force = 0, and the force will not enter into the equation of virtual work. Such a virtual dis- placement is always a convenient one to choose when we desire to get rid of some unknown force which acts upon a particle or a system. For example, let a particle, 0, of weight W, be sus- tained on a smooth inclined plane by a force, P, making an angle * These equations are, of course, implied in the proof of the principle of virtual work (Art, 53). 72 THE PRINCIPLE OF VIRTUAL WORK. [60. 6 with the plane. If we wish to find the magnitude of P in terms of W> without bringing the unknown reaction, R, into our equation, we conceive as receiving a virtual displacement, OA (the magnitude of which is, in the present case, unlimited), at right angles to R, that is, along the plane. Drawing Am and An perpendicular to W and P } respectively, the equation of virtual work is yr Qm-P. On = 0. But Om = OA . sin i y and On = OA . cos ; therefore Wsini Pcos0 = 0. As a second example, let us suppose that the plane is rough, and that the particle is on the point of being dragged up the plane. The normal resistance will then be replaced by the total resistance, 21, inclined to the normal at an angle = A, the .p. 6 angle of friction. Let the virtual displacement, OA (fig. 67), now take place perpendicularly to R. Then the equation of virtual work is W.0m-P.0n= 0. But Om = OA. sin (i + A), and On = 0.4. cos (A. 0) ; therefore W. sin (i + A) = P cos (A 0). As a third example, let us find the horizontal force which is necessary to keep a heavy particle in a given position inside a smooth circular tube (fig. 68). Let the virtual displacement, OA, be an indefinitely small one = ds, along the tube. Then since ds is infinitesimal, the pro- jection of OA on R will be zero. Also Om ds . sin 0, and On = ds.cosO; therefore the equation of virtual work is Wds .sin 0+Pds .cos 6 = , P = JFtan 0. \v Fig. 68. or If the tube is rough, and the particle in limiting equilibrium, instead of the normal reaction we must draw the total resistance. 6s.] NORMAL TO CURVES. 73 making the angle A. with the normal at the right or left hand side, according as P is the force which just sustains the particle, or the force which will just drag it up the tube, and take the virtual displacement, not along the tube, but at right angles to the total resistance. In this case we obtain P = 7Ftan(<9 + A). 61.J Condition of Equilibrium of a Particle as determined by the Principle of Virtual Work. It will now be sufficiently clear that For the equilibrium of a free particle acted on by any forces in one plane it is necessary and sufficient that the virtual work of the system of forces for every arbitrary displacement whatsoever should vanish. First, it is necessary that the virtual work should vanish for every displacement. For the sum of the virtual works of the forces is equal to the virtual work of their resultant, and if this sum did not vanish, the resultant force could not vanish, and therefore the particle could not be in equilibrium. Secondly, it is sufficient that this sum should vanish for every displacement. This sum is equal to the virtual work of the re- sultant, and if this vanishes for all possible displacements, the resultant force itself must be zero, and therefore the particle is at rest. For, if possible, let there be a resultant R, which is not zero. Then, since the virtual displacement is quite arbitrary, we may choose it so that it gives a projection = br (which is not = 0) on the direction of R. Now, since the virtual work of the system vanishes, we have Rbr = 0. But since br is not =0, R must be = 0, and the particle is, therefore, at rest. 62.] Normals to Curves. The equation of virtual work furnishes a ready method of drawing nor- mals to certain curves. For ex- ample, to draw a normal at any point, 0, of an ellipse (fig. 69) : let a particle be placed at inside a smooth elliptic tube whose foci are F and F' t and let it be kept in Fi S- 6 9- equilibrium by two forces, P and P* 9 directed towards the foci. Let OF = r, OF' = /. Then by the property of the ellipse, r -f / = a constant. ~ A 74 THE PRINCIPLE OF VIRTUAL WORK. [62. Hence, proceeding to a close point, A, we have 3r + 8/=0. (1) Now the resultant of P and P f is normal to the curve, and is destroyed by the normal reaction. Drawing Am and Am per- pendicular to P and P', the equation of virtual work is P. Om - P'. Om' = 0. But Om = br, and Om' = 8/; therefore this equation becomes P.br + P'.br' = 0. (2) Equation (l) gives 8/= br ; therefore, substituting in (2), we have P = P' or the forces towards the foci must be equal. But the result- ant of two equal forces bisects the angle between them. Hence the normal at any point of an ellipse bisects the T-,. angle between the focal radii !lg. 70- drawn to the point. Again, the ovals of Cassini are given by the equation rr' = &* 9 r and / being the distances of a point, 0, on the curve (fig. 70), from two fixed points, F and F '. If two forces, P and P', act at towards F and F', their resultant being normal to the curve, we have for a small virtual displacement along the curve P5r + P / 8/ = 0. (1) But, differentiating the equation of the curve, rbr + rbr' = 0. (2) Hence from (l) and (2) _v P 7 "" r' Now, if C is the middle point of FI" t we have / sin F sin COF * Therefore r sin F' ' sin COF' P sin COF F~ sin COF' 63.] NORMALS TO CURVES. 75 But if ON be the direction of the resultant, P_ _ sin NOF' ^ &Jr> If, P' == sin NOF' Hence NOF'= COF\ and the normal is, therefore, constructed by joining the point 0, on the curve, to the middle point of the line joining the foci, F and I", and then drawing the right line ON so that L NOF' = L COF. The line ON is the normal at 0. EXAMPLES. T I . If the equation of a curve is expressed in the form = k, k being a constant, and r, r the distances of any point on the curve from two fixed points, A, B, show that the normal to the curve divides AB externally in the ratio k 2 : 1, and that the curve is therefore a circle. 2. Prove that the normal to the curve -f = - divides AB in r n r n a n the ratio ( )* +2 - v r ' 3. Give a simple construction for the normal to a Cartesian oval, whose equation is Ir + mr = a. 4. The equation of the magnetic curve is cos o> + cos a/ = k (example 28, p. 39). If N and S are the poles, prove that the normal at a point P is constructed by measuring, on lines perpendicular to PN and PS, lengths proportional to PS 2 and PN 2 , respectively, and proceeding as in last Article. 5. The equation of any curve being / (r, r'} = 0, prove that if the normal is constructed by measuring constant lengths, Pa and Pb, from a point P on the curve, along the lines PA and PB, the curve must belong to the Cartesian ovals. [This follows at once from the integral of the equation - = &-/>; dr dr for this integral gives / = rp- = & ~, do> aco 1 df k df . df . , df tJL- r ^L, or sin o>r- = & sm 76 THE PRINCIPLE OF VIRTUAL WORK. [62. for the method of obtaining which integral see Boole's Differential Equations, p. 328]. 7. Apply the result in the last example to construct the normal to an ellipse at any point. [The equation of the ellipse is tan -tan = k.~\ & 2 The general theorem* of which these are particular cases is the following : Let the equation of any curve be expressed in flff v r r\ J V 1 > ' 2 5 ' 3 5 *) U 5 r ni denote the distances of any point, P, (fig. 71), on the curve, from a number of fixed points, A lt A 2 , A^ ... A n ; then, if on PA lf PA 2 , PA Z , ... PA n , we mea- sure off lengths P%, Pa 2 , P# 3 , ... Pa n proportional to df df df df the form where r- U/T-i ^^"2 3 and find G, the centroid of the points a l9 a 2 , a 3 ,...a n , PG will be the normal to the curve at P [/ is used for shortness instead of The proof of this theorem is exceedingly simple from a statical point of view. Suppose a number of forces, P lt P 2 , P 3 , ... P n , to act at P along the lines PA^ PA 2 , PA 3 , ... PA n \ then these forces will have a resultant normal to the curve if But d f 8 *i+^r^ 2 +-^r = 0; henceif P 1 : : P 3 : ... P n = dr the resultant acts in the direction of the normal. The rest easily follows by Leibnitz's graphic method of representing the resultant of any number of concurrent forces (see p. 15). This theorem may be extended to curves given by equations of the form 2 , o> 3 , = 0, * This theorem is, I believe, due to Tschirnhausen. The student will find another proof of this and the following theorem in Williamson's Differential Calculus, Art. 193, third edition. 63.] A SYSTEM OF TWO PARTICLES. 77 where u) 1} o) 2 , o> 3 , ...o> n are the angles which PA lt PA 2 , PA 3 , . . . PA n make with a fixed line. Let forces Q^ Q, 2 , Q 3) ... Q n , act at P perpendicularly to the lines PA V PA 2 , PA B , ...PA n . Then the virtual work of Q t for a displacement along the curve is evidently Qi^i^ !- Hence the forces will have a resultant normal to the curve if .p df df df df therefore the resultant will be normal if i df i df i df i df Q ' ' ' = * * Consequently, the rule is measure off lengths, P6 19 Pb 2 , &c., "I r!~P "I 7-/* proportional to - - -^ > -^ > &c., on lines drawn at P perpen- dicularly to PA lt PA 2 , &c., in the directions in which the angles ft>],o> 2 , ^ c v increase; find the centroid of the points, b l9 b 2) &c. ; then the line joining this point to P is the normal to the curve. SECTION II. A System of two Particles. 63.] Projection of a Displaced Line of Constant Length. Let a line, AB (fig. 72), be a right line which is displaced into any close position, A'ff, its length a remaining constant. Let 80 be the small angle between AB and A'ff, and let ab be the projection of A'ff on its original position. Then Aa, the projection of the displacement AA', is equal to JBb, the projection of the displacement BB' t if infinitesimals of a higher order than the first are neglected. For, ab = A'ff. cos (50) = A'V (l - Hence the difference between ab and A' I? (or AB) is of the order of (50) 2 ; and therefore, rejecting (80) 2 , we have AB = ab, .-. Aa, = Bb. 78 THE PRINCIPLE OF VIRTUAL WORK. [64. The result may be thus stated : the difference between AB and ab is infinitesimal compared with the greatest displacement in the figure. 64.] Projection of a Displaced String of Constant Length. Let APB be a string which passes over a peg at P, and, the length of the string remaining the same, let the extremities, A and B, be slightly dis- placed to A and If. Let Aa and Bb be the projections of the displacements AA' and BB' on the original portions of the string (fig. 73). Then Aa = Bd. For Pa = PA', cos a PA = PA', as in the last Article. Also Pb=PB'. Hence, since PA' + PB'=PA + PB ) Pa + Pb = PA + Fi g- 73- ^-^5 therefore Aa = Bb. If in the last Article I = the length of AB, and in the present, I = length of the string, both results are expressed in the equation = 0. 65.] Virtual Work of the Tension of an Inelastic String. In fig. 73 suppose the peg to be smooth. Let A and B be two particles which are acted on by any forces which keep the system in equilibrium in the position indicated by the figure. Then if we consider the equilibrium of A alone, we may replace the string by a force = T (the tension) acting in AP. Con- sidering then a virtual displacement AA , the tension would furnish the term T.Aa, or -T.br, to the equation of virtual work, the length PA being denoted by r. Similarly, considering the equilibrium of B, the tension would furnish to its equation of virtual work, for the virtual displacement BB', the term T.S6, or -T.br', / denoting the length of PB. Taking the two equations together, the term contributed by the tension will be, by addition, or -T.M 66.] TYPICAL EXPRESSION FOR VIRTUAL WORK. 79 which = 0, since the particles A and B are imagined to be simultaneously displaced in such a manner that the length of the connecting string is constant. Hence For any small virtual displacement in which the length of a string is unaltered, the virtual work of its tension = 0. In the same way, if, in fig. 72, the rod AB, connecting two particles A and B, be subject to a tension, T, in the direction of its length, the virtual work of this tension for the displace- ment A'B' will be T.(AaBb), or T.bAB, which = 0, because the length of AB is constant. Hence The virtual work of the tension of a rod connecting two points whose mutual distance is unaltered in the virtual displace- ment is zero. 66.] Typical Expression for the Virtual Work of a Force. Example. We have seen (Art. 59) that if a force, P, act on a particle, 0, whose vir- tual displacement, OA, has a projection = bp on the line of action of P in the sense in which P acts, the virtual work of P is P. bp. Fig. 74. Generally, if p denote the co-ordinate, referred to some fixed axis, of the point of application of a force, P, the virtual work of the force is P. ftp, ftp being supposed to be a positive incre- ment, and the co-ordinate being measured in the sense in which P acts. As an example, let us determine the relation between two weights, P and P 7 (fig. 74), which rest on two smooth inclined planes, of inclinations i and i'. Let y and y' denote the co- ordinates of the weights, referred to a horizontal plane through 0. Then the equation of virtual work for the system, the displacement being supposed to be along the planes, is P.8y + P-.S/=0. (1) [Here it will be observed that the normal reactions do not enter, because the virtual displacements take place at right angles to them (see Art. 60) ; and the tension does not enter, 80 THE PRINCIPLE OF VIRTUAL WORK. [67. since the virtual displacement does not alter the length of the string (see Art. 65)]. To this must be added the geometrical equation connecting y and /. If I be the length of the string, we have, clearly, sm i sin i Differentiating this equation, we have by by i v "f~ / 0. \^) sin i sin i Hence, from (l) and (2), P sin i = P' sin i' 9 an equation which is, of course, otherwise evident. If the weights are connected, as in example 1 2, p. 48, we have still the equation of virtual work, y and y' denoting the vertical distances of P and Q in the figure of that example from a horizontal plane through C. The geometrical equation connecting y and y' is, evidently, A/y 2 cosec 2 a -f 2% + h 2 + \// 2 cosec 2 /3 + 2%' -j- #* = /. (4) Differentiating (4), we have 2 cosec 2 /3 + 2% + ^ 2 Hence, from (3) and (5), we obtain p _ 2 Equations (4) and (6) are sufficient to determine y and y, on which the position of equilibrium depends. 67.] Geometrical Forces. When a particle is compelled to satisfy some geometrical condition as, for instance, to rest on a given smooth surface, or to preserve a constant distance from some other particle this condition is equivalent to the action of a certain force on the particle. If the particle is compelled to rest under given forces on a smooth inclined plane, we have seen that this condition may be removed if we produce, by any means, a force exactly equal to the normal reaction of the plane on the particle. In the same way, the connexion of the particle with another by means of a rigid rod may be severed if we 68.] CHOICE OF VIKTUAL DISPLACEMENTS. 81 produce on the particle the force which is actually impressed upon it by the rod. Forces proceeding from geometrical connexions are called Geometrical Forces, and if these forces are actually produced on the particle by other means, the conditions may be violated, and the particle considered absolutely free from constraint. 68.] Choice of Virtual Displacements. When two or more particles constituting a system are connected by rods or strings, and constrained to rest on given smooth curves or surfaces, there is an advantage, when seeking for the position of equi- librium, in choosing such virtual displacements as do not violate any of these conditions ; because, as we have seen, the tensions of the connecting rods or strings, and the reactions of the smooth curves or surfaces, will, for such virtual displacements, contribute nothing to the equation of virtual work of the system. Thus we get rid at once of all such unknown forces. Of course, any geometrical condition may be violated in a virtual displacement at the expense of bringing into the equation of virtual work the corresponding geometrical force. For example, if a particle, (fig. 75), is placed on a smooth plane whose inclination is i, and we wish to find the horizontal force, P, which will sustain it, the best displacement to choose is one along the plane, i. e., one which does not violate the geome- trical condition, because, if this is chosen, the unknown reaction, E } will not appear in the equation of virtual work. But we shall still get a valid equation if we choose a virtual displacement, OA, which does violate the condition. This equation is 72 . Qr-P. Op- W. Ow = 0, Or, Op, and Ow being the projections of OA on the directions of R, P, and W, respectively. On the other hand, if we wish to determine 72, without determining P, the best virtual displacement to choose is one at right angles to P, i.e., a vertical displacement which does violate the geometrical condition. 82 THE PRINCIPLE OF VIRTUAL WORK. [68. In the typical expression, Pdp, for the virtual work of a force, the letter 8 has been used to signify that the small displacement is any whatever ; but it is usual in the Differential Calculus to denote small increments of the co-ordinates of a point on a curve or surface by the letter d. Hence in the following examples we shall denote small displacements on the curves considered by this letter. EXAMPLES. 1. Two heavy particles, P and P f (fig. 76), rest on the concave side of a smooth vertical circle, and are connected by a string passing over a smooth peg, A, at the ex- A tremity of the vertical dia- meter. If the particles are acted upon by two horizontal forces, F and F f , proportional to the distances, PQ and P'Q'i of the particles from the vertical diameter, find the position of equilibrium by the principle of virtual work. Let 6 and tf be the angles which the radii to P and P f make with the vertical; let the weights of the particles be Wand W'\ the radius of the circle = a, the length of the string = , and the forces F and F'=\*..PQ and /. P'()'> respectively. Finally, let the distances PQ and P f Q f be x and of, and let the vertical distances of P and P f below the horizontal diameter of the circle be y and y. Then, choosing virtual displacements of P and P f along the circle in such a manner that the length of the connecting string remains unaltered, we have Wdy + W'dy' + Fdx + F'dx' = 0, w Fig. 76. or Wdy+W'dy' + ij.x.dx + v?x'.dx'=Q. (1) Now y = a cos 0, y' =. a cos tf , x a sin 0, x' = a sin tf. Hence (1) becomes 0. (2) A / Again, AP = 2a cos - , AP' = 2a cos - 68.] EXAMPLES. 83 Hence the geometrical equation is co 'f +COB | r =i- < 3 > Differentiating this, we have 6 tf sin -d0 + sin -. and sin rOm = -~c ; therefore % 0) P_ ~Q ~ Gp ' Again, if R is the resultant of P and Q, we have R Or _smnOm ~P " ~0m ~ sinnOr' R _ perp. from Bon P P ~~ perp. from B on R Now, if P and Q are parallel, R becomes parallel to P and Q, and we shall evidently have ~ = -~ ; hence (1) gives for parallel forces or, (2) Q ~~ GA ' and (2) gives, since R is parallel to P and R BA BG+GA Q, ~ + P' A similar demonstration holds when P and Q act in opposite directions. 71.] Construction for the Resultant of two Parallel Forces. If the lines AP and BQ (figs. 81 and 82), represent 88 COMPOSITION AND RESOLUTION OF FORCES. [7* in magnitudes and lines of action two parallel forces, the student will easily prove the following construction for the resultant : Fig. 81. Fig. 82. Draw BQ' equal and opposite to Q, and draw PQ', meeting AB in g. Then measure off AG = By. G is a point on the resultant. Through G draw an indefinite right line parallel to P and Q, and from A and P draw parallels to PQ' and AB, respectively. These lines will intercept on the line through G a length = P + Q resultant. 72.] Moment of a Force with respect to a Point. Let a force, P (fig. 83), act on a rigid body in the plane of the paper, and let an axis perpendicular to this plane pass through the body at any point, 0. It is clear, then, that the effect of the force will be to turn the body round this axis, (the axis being supposed to be fixed,) and the rotatory effect will depend on two things firstly, the magnitude of the force, P, and, secondly, the perpendicular distance, p, of P from 0. If P passes through 0, it is evident that no rotation of the body round can take place, whatever be the magnitude of P ; while if P vanishes, no rotation will take place, however great p may be. Hence we may regard the product p. f as a representation of the power of the force to produce rotation about ; and to this product the special name Moment has, for convenience of reference, been given by writers on Statics. When all the forces under consideration act in one plane, we may speak of the point, 0, in which the axis of Moments meets this plane, instead of the axis itself. We shall therefore define the Moment , with respect to a point, x>f a force acting on a body Fig. 83. 74-] TWO EQUAL AND OPPOSITE PAEALLEL FORCES. 89 to be the product of the force and the perpendicular let fall on its line of action from the point. The unit of force being a pound and the unit of length a foot, the unit of Moment will obviously be & foot-pound. 73.] Moments of Different Signs. If two forces tend to produce rotations of a body in opposite senses round a point, their moments with respect to this point are affected with opposite signs. Thus (fig. 84), the force P tends to turn the body round in a sense opposite to that of watch- hand rotation, while Q tends to turn it in the opposite sense. If, then, the former rotation is considered positive, the algebraic sum of the moments of P and Q round is t Fi g . 84. P.p-Q.q, p and q being the perpendiculars from on P and Q. Round the point 0' both forces would produce rotation in the same sense, and therefore the algebraic sum of their moments with respect to this point is p' and q being the perpendiculars from 0' on P and Q, re- spectively. In future we shall speak simply of the sum of the moments, instead of the algebraic sum of the moments, of forces with respect to a point, as we shall suppose the moment of each force to be affected with its proper sign, in accordance with the rule given at the beginning of this Article. 74.] Case of Two Equal and Opposite Parallel Forces. If the forces P and Q in Art. 69, fig. 79, are equal, the equation P x GA = q x GB gives GA = GB, or ' = 1, an equation which is true only when G is at infinity on AB. Also the resultant of the forces, being equal to their difference, is equal to zero. Two equal and opposite parallel forces acting on a rigid body constitute what is called a Couple. THEOREM I. Two equal and opposite parallel forces have a 90 COMPOSITION AND RESOLUTION OF FORCES. [74. constant moment with respect to all points in their plane. Let (fig. 85), be any point in the plane of two equal and opposite parallel forces, P, and let fall the perpendiculars Om and On on their lines of action. Then, if is inside the lines of action of the forces, these forces tend to produce rota- tion round in the same sense, and therefore the sum of their moments is equal to P(0m+0n), or Pxmn. If the point chosen is O 7 , the sum of the moments is evidently P ((/m0 f n), or Pxmn, which is the same as before. The perpendicular distance between the two forces of a couple is called the Arm of the couple. The Moment of a couple is the product of the arm and one of the forces. The Axis of a couple is a right line drawn anywhere perpen- dicular to the plane of the couple, and in a particular sense, its length being proportional to the moment of the couple. The sense of the axis is determined thus : imagine a watch placed in the plane in which several couples act. Then let the axes of those couples which tend to produce rotation in the direction opposed to that of the rotation of the hands be drawn upwards through the face of the watch, and the axes of those which tend to produce the contrary rotation be drawn downwards. THEOREM II. The effect of a couple on a rigid body is not altered if the arm be turned through any angle round one extremity. Let AC and BD (fig. 86) be a couple whose arm is AB, and let the arm turn round B into the position BA'. At A' introduce two equal and opposite forces, AC' and A'C", each of which is equal to one of the forces, P, of the given couple, and per- pendicular to BA'. At B introduce two equal and opposite forces, BI/ and BD", perpendicular to BA', each force being equal to AC 74-] THEORY OP COUPLES. 91 or P. The effect of the given couple is, of course, unaltered by the introduction of these forces. Now the forces BD and BD" may be replaced by their resultant, 2 P cos - > or 2 P sin - > a 2 which acts in the bisector, BO, of the angle DBD" ; and the forces AC and AC" may be replaced by their resultant, 2Pcos - - , or 2Psin - > which also acts in the line in a sense opposed to the previous resultant. Hence the forces BD, BD", AC, and AC") are a null system. There remain, then, the forces Elf and ACT which form a couple whose arm is BA. Hence the couple of forces P acting at A and may be replaced by a couple of forces P acting at the extremities of an arm of length equal to AB having one extremity common with AB. THEOREM III. The effect of a couple on a rigid body is not altered if the arm is moved parallel to itself anywhere in the plane of the couple. Let two forces, AC and BD, each equal to P (fig. 87), act with arm AB, and draw ABT equal and parallel to AB in the plane of the couple. At A and B' introduce, perpendicularly to A'J? t four forces AC, AC", fftf, and , each equal to P. This does Fig 87 not alter the effect of the given couple. Now since AB and A'B' are equal and parallel, the lines AB' and BA , being the diagonals of the parallelogram ABB' A, bisect each other in the point 0, suppose. Replace the forces BD and ACT' by their resultant, 2 P, which acts at parallel to BD ; and replace the forces AC and BflP by their resultant, 2P which also acts at in a sense opposite to the previous resultant. These two resultants Destroy each other, and there- fore the forces BD, AC, B'lf', and AC\ constituting a null system, may be removed. There remain the forces, AC' and J5'_D', which constitute a couple whose arm is A'B'. Therefore, &c. THEOREM IV. The effect of a couple on a rigid body is not altered if the couple is changed into another having the same moment, the arms of the couples being in the same line and having a common extremity. . 92 COMPOSITION AND RESOLUTION OP FORCES. [74- Let the given couple be AC and BD (fig. 88). each equal to P. Produce BA to A' so that i = -, and at A' and B in- JoA. Jr troduce equal and opposite forces AC' and A'C' ', BIf and the magnitude of each of these forces being Q. Now the forces AC and AC" give a Cl Fig. 88. C'' resultant = PQ at B (Art. 69) in the direction BD" ; and this force added to BD" gives a force = P which destroys BD. Hence there remain the forces A'(? and BD', which form a couple whose moment is equal to that of AC and BD, since (by construction) BA= Q.BA'. Therefore, &c. THEOREM V. A couple acting on a rigid body may be replaced by any other couple in the same plane if the moments of the couples are the same in magnitude and sign. Let P, P and Q, Q (fig 89), be two couples in the same plane, having the same moment, and tend- ing to produce rotation in the same sense ; then P, P may be transformed into Q, Q. For, we can first turn the arm AB round B until it is parallel to B'A' (Theorem II) ; then we can lengthen it until it becomes equal to B'A') changing, at the same time, the forces P into forces Q (Theorem IV) ; and finally, we can move it into the position B'A' (Theorem III). The sign of the moment of a couple is indicated by the sense in which the axis is drawn, as has been already explained (p. 90). Axes drawn upwards through the face of the watch are then considered positive, and axes drawn downwards are negative. From the foregoing Theorems it is clear that the addition of co-planar couples is effected by adding their axes, regard being had to the signs of the axes. THEOREM VI. A force and a couple acting in the same plane on a rigid body are equivalent to a single force. Let the force be F and the couple (P, a) that is, P is the 75-] GEOMETKICAL REPRESENTATION OF A MOMENT. 93 magnitude of each force in the couple whose arm is a. Then (Theorem IV) the couple (P, a) = the couple (F, ^) - Let this latter be moved until one of its forces acts in the same line as the given force F, but in the opposite sense. The given force F will then be destroyed, and there will remain a force F acting in the same direction as the given one and at a perpen- dicular distance = -== from it. Jl This Theorem is equivalent to the statement A force and a couple acting in the same plane cannot produce equilibrium. THEOREM VII. A force acting on a rigid body at any point A may be replaced by an equal force acting in the same direction at any other point B together with a couple whose moment is the moment of the original force about B. This important proposition is easily demonstrated. THEOREM VIII. The resultant of any member of coplanar couples is a couple whose moment is equal to the sum (with the proper signs) of the moments of the given couples. For, let the component couples have moments Z, M,N 9 ... , and let each of them be changed into a couple, having the same right line AB (whose length is #) for arm. Then (Theorem IV), the couple L will give rise to a force at A, and an equal force in x opposite sense at B. Hence at A we shall have the force ~ and an equal and opposite force at B. Thus we x have a couple whose moment is the product of this force by the arm x ; i.e., its moment is L + H+N+ ... , or the sum of the given moments. 75.] Geometrical Representation of the Moment of a Force with respect to a Point. Let the line AB (fig. 90) represent a force in magnitude and direction, and let it be required to represent its moment with respect to a point 0. If p = the perpendicular from on AB, the moment is AB x_p. Now this is double the area of the triangle AOB. Hence the moment of Flg< 9 ' a force with respect to a point is geometrically represented by double the area of the triangle whose base is the line representing 94 COMPOSITION AND RESOLUTION OF FORCES. [76. the force in magnitude and line of action, and whose vertex is the given point. Draw AO, and from the other extremity, B, of the given force draw an indefinite right line, BC, parallel to AO. Join A to any point, C, of this line. Then the area of the triangle AOB = the area of the triangle AOC, since these triangles have the same base and are between the same parallels. Consequently the moment of a force represented by AB about = the moment of a force represented by AC about 0, wherever C be taken on the indefinite line through B. 76.] Varignon's Theorem of Moments. The sum of the moments of two forces with respect to any point in their plane is equal to the moment of their resultant with respect to the point. Let AP and AQ (fig. 91) represent two forces whose resultant is AR, and let be the point about which moments are taken. Draw AO, and draw PC and QD parallel to it. By the last Article the moment of AP about = the moment of AC about 0, and the moment of AQ = the moment of AD ; therefore the sum of the moments of AP and AQ about = the sum of the moments of AC and 9I AD about = the moment of the sum of AC and AD (since AC and AD are forces acting in the same line) ; but, by equal triangles AC is evidently = DR ; therefore the sum of the moments = the moment of AR = the moment of the resultant. Q. E. D. The student will find no difficulty in considering the case in which is between AP and AQ, observing that in this case their moments are opposed, and that in the new figure AR will be equal to .10 -^ Of course it follows that the sum of the moments (with their proper signs) of any number of co-planar forces with respect to any point in their plane is equal to the moment of their resultant with respect to this point ; for the forces may be replaced in pairs by their resultants, &c. It also follows that the sum of the moments of the forces about any point on the line of action of the resultant is equal to zero. 77.] Varignon's Theorem of Moments for Parallel Forces. The sum of the moments of two parallel forces about 78 ] CENTEE OF PARALLEL FORCES. 95 any point is equal to the moment of their resultant about the Let the forces be P and Q (fig. 92), and let be the point about which moments are to be PA taken. From let fall perpendicu- lars OA, OB, and OG on the lines of action of P, Q, and their resultant, It, and let the forces be applied at the points A, B, and G, respectively. Then, moment of Fi S- 9 2 - P about = P.OA = P (OG + GA) j and moment of , Q about = Q . OB - Q (OG- GB) ; therefore, by addition, the sum of the moments = (P + QJ . OG + P.GA-Q.GB. But P. (L4 = Q . # ; therefore the sum of the moments = (P+ Q).OG = R. OG. A similiar proof holds when P and Q act in opposite directions, and also when is between the lines of action of P and Q. It follows that the sum of the moments (with their proper signs) of any number of co-planar parallel forces with respect to a point in their plane is equal to the moment of their resultant with respect to the point. 78.] Centre of Parallel Forces. Theorem. If any number of parallel forces, P 15 P 2 , P 3 , ... P n , act in one plane at points A lt A 2 , A 3 , ... A n) their resultant passes through a fixed point if all the forces are turned in the same sense round their points of application through an arbitrary but common angle. The point, g^ (fig. 93), of application of the resultant of P l and P 2 has been determined (Art. 69) by dividing the line A 1 A 2 so that on the supposition that the forces P 1 and P 2 are parallel, but no as- sumption has been made as to their common direction. Hence ff-L will be a point on their re- sultant in whatever direction they act, and the force at this point is P l -f P 2 . The point of application of the resultant of 96 COMPOSITION AND RESOLUTION OF FORCES. [79- P lt P 2 , and P 3 , is determined by joining g^ to A 3 , and dividing it in &, so that ft ft force at A z P 3 force at ft j and the force at ft is P 1 -|-P 2 + P 3 . Similarly, the point of application of the resultant of P ls P 2 , P 3 , and P 4 is a point, G, on ft ^4, such that and the force at G = We thus see that the point, G, of application of the resultant of the system is determined by dividing the lines g^A z , g^A^ ... in certain ratios which depend simply on the magnitudes, and not on the directions, of the forces at A lt A 2 , A 3 , . . . . The theorem is, therefore, evident. Of course no one point on the line of action of a force which acts on an indeformable body has a special right to be called the point of application of the force ; nevertheless, we shall speak of the point, G, as the point of application of the resultant force, since, as we have seen, it is a point through which the resultant of forces equal to P 15 P 2 , ... always passes, whatever be the common direction of these forces. The theorem of this article is true also in the case in which neither the parallel forces nor their fixed points of application lie in the same plane. Aa 79.] Centre of Mean Position. Let there be any number of points, A \>- A v ^ 3 > (% 94), in one plane, and let the line, .A^A 2 , be divided at g^ so that J? efr \ n L Fig. 94. let ^^3 be divided at ft, so ffzffi let g 2 A be divided at g z , so that and so on, until by a final construction we arrive at a point, G. 79.] CENTKE OF MEAN POSITION. 97 It is required to express the distance of G from an arbitrary line, Jv, in the plane of the points in terms of the distances, z l , z 2 , z 3 , ... of A 19 A 2 , AS, ... from this line*. Draw A^mn parallel to L. Then h ^ = --r^r(*2-*i)- But the distance of g l from L is equal to m* . . y/*j z l +g l m=z l + (^2-%) = Calling this distance z lt we have the distance of g 2 from L equal to /^ 4. m I ; /*n2 tyy* since ^^ 3 is divided at ^ 2 in the ratio - - Continuing the m^-\-m^ application of this method, we have evidently ... +m n z being the distance of G from L. This equation is generally written in the form in which 2 denotes a summation. The point G thus arrived at is called The Centre of Mean Position of the given points for the system of multiples %, m 2) m^ .... The points A 19 A^ A 3 , ... remaining the same^ and the system of multiples being altered to jpj, j0 2 , _p 3 , ... the point G arrived at would, of course, be different. The distance of the new point would be In particular, the distance, z 9 of the centre of parallel forces from any plane is given by the equation * All this holds if the points A l} A 2) ... are not in the same plane and L re- presents any plane from which their distances are measured. H 98 COMPOSITION AND RESOLUTION OF FORCES. [79. EXAMPLES. 1. The centre of mean position of three points, A, B, C, for a system of equal multiples, is the intersection of the bisectors of the sides of the triangle A BC drawn from the opposite angles. 2. The centre of mean position of three points, A, B, 0, for a system of multiples sin 2 A, sin 2 B, sin 2 C, is the centre of the circle circumscribed about the triangle ABC. 3. The sides of the triangle being a, 6, c, the centre of mean position of A, B, C, for the system of multiples a, b, c, is the centre of the inscribed circle. 4. For the system of multiples tan .4, tan I?, tan (7, the centre of mean position is the intersection of perpendiculars. The construction given in this Article for the Centre of Mean Position of the points A 19 A 2 , A 3 , ... is of course the same when the points do not all lie in one plane. In the latter case it is easily seen that if z lt z 29 z 3 , ... denote the distances of the points from an arbitrary plane, the distance, z, of the centre of mean position from this plane, for the system of multiples m-^ m 2) m 3 , . . . , is given by the equation 2mz '' ^m ' Centre of Mean Position is a generic term which comprises under it particular points which must be specially noticed. One, the Centre of Parallel Forces, has been already mentioned. Another is the Centre of Mass, called also the Centre of Inertia.. If at the points considered, A 19 A 2) A 3 ,... there be placed material particles whose masses are respectively %, m 2 , m 3 , ... and we find the centre of mean position of these points for the system of multiples m lf m 2 , m 3 , . . . we shall arrive at the Centre of Mass of this system of particles. Nothing is here assumed about the closeness of the points A lt A 2 , A 3 , ..., or the particles placed at them, and the process of arriving at the point G will be unaltered if these particles constitute a continuous body. Hence the Centre of Mass of any body is the Centre of Mean Position of all the points within it for a system of multiples pro- portional to the masses of the particles placed at these points re- spectively. A. body whose points do not suffer any relative changes of position will therefore continue to possess the same centre of mass no matter into what part of the universe the body may be 79-] CENTRE OF MEAN POSITION. 99 taken. A different arrangement of its particles, would, of course, in general alter its centre of mass. The centre of mass of a rigid body is, then, something which it possesses absolutely, or apart from all contingency of position in space or relation to other bodies. The distance of this point from any plane is given by the equation last written, in which the sign 2 is to be replaced by the integral sign f, and the element of mass at a distance z from the plane denoted by dm. Thus. _ _ fzdm ~ fdm Again, if at the points A lt A Z9 A 3 , ... there be placed particles whose weights are w^ , w 2) w S) ... these weights constituting a system of parallel forces, the centre of these parallel forces is called the Centre of Gravity of the given particles. The effect of altering the position of the body in the most general manner possible is merely to turn the forces, w l9 w 2 , w 3 , ... round their fixed points of application,^, A 2 , ... through the same angle, and by the last article we see that the resultant of the weights of the particles will, in all positions of the body, pass through a fixed point, G, in the body. The resultant of all the elementary weights is equal to their sum, and is called the weight of the body. We may, therefore, define the centre of gravity of a body thus The centre of gravity of a body is that unique point in it through which passes, in all possible positions of the body, the resultant of the system of parallel forces formed by the weights of the indefinitely great number of indefinitely small particles into which the body can be divided. The centre of gravity of a body is,, then, the centre of the particular set of parallel forces which act on its various elements in virtue of the attraction of the Earth. The existence of such a point depends on the parallelism of the forces produced by the Earth on the elements of the body, and this parallelism, again, depends on the minuteness of the volume of the body in com- parison with that of the Earth. If the body were carried to the surface of the Sun, or any other such large attracting mass, the individual weights of its elementary portions, and therefore its total weight, would be greater than they are at the Earth's surface, but the position of the centre of gravity in the body would remain the same. On the other hand, if the dimensions H 2 100 COMPOSITION AND RESOLUTION OP FORCES. [8o. of the body were comparable with those of the attracting mass, the forces of attraction on its elementary portions would not be a parallel system, and the resultant attraction would not, in general, pass through any fixed point in the body independently of the relative positions of the two masses. The term weight of a body is used to signify the resultant attraction produced on the body by the Earth, or other planet, on whose surface the body exists, and it is therefore, unlike mass, a mere contingent property of the body; and the centre of gravity is essentially distinguished from the centre of mass ; although, since weight and mass are always proportional, when the first point exists, it coincides with the second. In considering the equilibrium of a rigid heavy body we represent its weight as a single force acting vertically through its centre of 80.] Conditions of Equilibrium of a Rigid Body acted on by Forces in One Plane. 1. Let the forces be parallel. Take any point, 0, and draw through it a right line, Oy, parallel to the forces (fig. 95). At introduce two forces, P/ and P/', each equal to P 19 these new forces being directly opposed to each other along Oy. Now, P/ and P x " form a couple whose moment is PI .flit if Pi is the perpendicular from on the line A 1 P x . Introducing, in the same way, two forces, P 2 ' and P 2 ", equal to P 2 , directly opposite to each other along Oy, we have P 2 at A 2 replaced by a force P 2 " acting at along Oy' and a couple whose moment is P 2 .j) 2 , p 2 being the perpendicular from on the line A 2 P 2 . The sign is attached to this couple because the couple (P/, P 2 ) tends to produce rotation in a sense opposite to that in which the couple (P/', P 2 ) tends to produce rotation. Proceeding in this way with all the forces in the above figure, we have the whole system of forces at A 19 A 2) A S) A^ ... equiva- lent to a single force, P 1 -P 2 +P 3 -P 4 +,.., acting at in the direction Oy, and a couple, 95- 8o.] CONDITIONS OP EQUILIBRIUM. 101 tending to turn the body round in a sense opposite to that of watch-hand rotation. In general, denoting the resultant force by 7?, and the moment of the resultant couple by G, we have R = ZP, (I) Now, by Theorem VI, of Art. 74, a couple and a force in the same plane are equivalent to a single force, and cannot, there- fore, conjointly produce equilibrium. Hence, for equilibrium, the force and the couple must vanish; or 2P = 0, (3) and 2(P.j?) = 0; (4) that is to say, for the equilibrium* of a system of coplanar parallel forces acting on a body (a) The sum offerees must = 0, and (b) The sum of the moments of the forces about every point in their plane must = 0. 2. Let the forces act in any directions. Take any point whatever, 0, (fig. 96), in the plane of the forces. At introduce two opposite forces, P and P/', each equal and parallel to P v Let Pj and PJ" be considered as forming a couple. Then P l at A l is equivalent to P l acting at 0, and a couple whose mo- ment = P 1 . p v Replace P 2 at A 2 in the same way by P 2 " (or P 2 ) acting at 0, and a couple (P 2 , P/) whose moment is P 2 .j 2 . Thus the whole Fig. 96. system of forces will be re- placed by forces, P 15 P 2 , P 3 , P 4 ,..., acting at 0, and a number of couples whose moments are P 1 .jt? 1 ,P 2 .jo 2 , Pz-P^PfPi) ... (the forces acting as in the above figure). The forces acting at will have a single resultant, R, and the couples will form a * The attention of the student is particularly directed to the remark at the end of this chapter. ' < 102 COMPOSITION AND RESOLUTION OF FORCES. [8 1. single couple whose moment, G, is (Theor.VIII, Art. 74) the sum of the moments of the couples. For equilibrium it is necessary that each of these should vanish. Hence, for the equilibrium * of a body acted on by coplanar forces (a) The resultant which the forces would have if they all acted together at a point, each in the direction in which it acts on the given body, must = ; and (b) The sum of the moments of the forces round every point in their plane must = 0. The first of these conditions asserts that there must be no force in any direction ; and the second that there must be no moment round any point. Thus, the conditions of equilibrium of a rigid body embrace the condition (a) of the equilibrium of a particle (Art. 24, p. 21); and (b) a condition distinctive of the susceptibility of a body of finite extension to receive a motion of rotation. It is to be observed, then, that a system of coplanar forces acting on a body can be reduced to a single resultant force, R, acting at any arbitrary point, 0, in the plane of the forces, and a couple, G, also in this plane ; and that whatever point, 0, is chosen, the force R is constant in magnitude and direction, while the magnitude of the couple G varies with the point chosen. The force R is called the Resultant of Translation. 81.] Analytical Conditions of Equilibrium. Through any point, 0, draw two rectangular lines, Ox and Oy, and resolve the force, P l , acting at A , into two components^ X 1 and Y 19 parallel to Ox and Oy. Now (Art. 76) the moment of P l about is equal to the sum of the _. * moments of X, and Y, about 0. Fig. 97. But if rotation opposite to that of a watch-hand is considered positive, the moment of Y l about is Y lt x^ and the moment of X is Y 1 .^ l , where x^ and y^ are the co-ordinates of A referred to the axes Ox and Oy. Hence the moment of P about is * See remark at the end of this chapter. EQUATION OF THE KESULTANT. 103 Adding together the moments of P 1? P 2 , ... , we get the total moment G = S(7*-Xy). (1) If the sum of the components of the forces along Ox is denoted by 2Jf, and the sum of the components along Oy by SJ", the resultant of the forces acting at (fig. 96) is given by the equation ^ 2 = (2JT) 2 +(2r) 2 . (2) Now, since for equilibrium we must have R 0, and = 0, the conditions, analytically expressed, are 2X= 0,27=0, (3) 2 (Yx-Xy) = 0. (4) These equations are the expressions of the conditions of Art. 80. 82.] Equation of the Resultant. We have seen (Art. 80), that a system of coplanar forces is equivalent to a single force, R> acting at any arbitrary origin, together with a couple, G. The direction and magnitude of the resultant force, R, will be the same whatever origin may be chosen, but the couple will vary with the origin. Now, supposing that the resultant of the forces does not vanish, the couple and the force R can (Theorem VI, Art. 74) be replaced by a single force equal to R ; and the sum of the moments of the forces about any point on its line of action is equal to zero (Art. 76). Let (a, /3) be the co-ordinates of any point referred to rect- angular axes through an arbitrary origin, (fig. 97). Then the moment of the force, P 19 about this point, is evidently Taking the sum of the moments of all the forces about the point, we have G'= 0-aSr+0SX, (1) G' being the sum of the moments about the point (a, jB). Since, for any point on the resultant G' 0, the equation of its line of action is a27-/32X = G. Equation (1) gives at once the following result The sum of the moments of a system of coplanar forces about any point, 0, is equal to the sum of their moments about any other point , (7, plus 104 COMPOSITION AND RESOLUTION OF FORCES. [83. the moment about of their resultant of translation supposed acting at a. 83.] Force Polygon and Funicular Polygon. Let there be any system offerees, P 15 P 2 , P 3 , P 4 , P 55 (fig. 98) acting in one Fig. 98. plane on a body. Starting with any point, 01, draw lines, (01, 12), (12, 23), (23, 34), (34, 45), (45, 56), parallel to the lines of action of the forces and respectively proportional to them. The figure formed by these lines, (01, 12), (12, 23), ... , is called the Force Polygon of the given system of forces. Now take any point, 0, and from it draw lines, 001, 012, 023,..., to the vertices of the force polygon. From any point, J\ , on the line of action of P x draw two lines, fif Q and^yj > parallel to the lines 84.] THEOREM. 105 001 and 012; from the point / 2 in which /i/ 2 meets P 2 , draw /2 /s parallel to 023 and meeting P 3 in/ 3 ; from/ 3 draw/ 3 / 4 parallel to 34 ; and so on. The system of lines / / / 2 / 3 / 4 / 5 / 6 parallel to the radii drawn to the vertices of the foree polygon from any point, 0, is called a Funicular Polygon of the given system of forces. The point the radii from which to the vertices of the force polygon determine the funicular is called the Pole corresponding to the funicular. Let any other pole, O 7 , be chosen, and from an arbitrary point,//, on P 15 let///,' and//// be drawn parallel to O'Ol and 0'12, respectively; and let a new funicular, ////...//, be constructed. Then the sides (such as/ 2 / 3 and////) of these polygons which reach between the lines of action of the same two forces are called corresponding sides. Since the point / may be taken anywhere on P x it is clear that for a given pole, 0, we may construct an infinite number of funiculars of the system,, but the corresponding sides of them are of course parallel. If the force at each vertex of a funicular of the system is resolved into two components directed along the two sides of the funicular which meet at this vertex, the components at the extremities of each side of the funicular are equal and opposite. For, suppose P 3 resolved into two compo- nents in/ 3 / 2 and/ 3 / 4 ; ^ en these components are represented by the lines 23 Oand 034; also if P 2 is resolved into components in / 2 / 3 and/ 2 / 15 these will be represented by 023 and 120, respectively ; thus the components in the side/ 2 / 3 are equal and opposite. 84.] Theorem. The corresponding sides of any two funiculars of a given system of forces intersect on a right line> which is parallel to that joining the poles of the two funiculars. At the points / 2 and // let two equal forces (each P 2 ) be applied in opposite senses along the line / 2 //; suppose them to act away from both of these points, as P 2 is represented in fig. 98. Considered as acting on a rigid body, these forces are in equilibrium. Now let P 2 at/ 2 be resolved into its com- ponents along / 2 / x and / 2 / 3 . These components will be re- presented in magnitudes and senses by 012 and 230, respectively. Similarly, resolve P 2 at// along//// and////; and these 106 COMPOSITION AND RESOLUTION OF FORCES. [85. components will be represented by 12 0' and 0' 23. These four components are therefore in equilibrium. Take the sum of their moments about the point of intersection of the lines/ / and ///'. Then, since this sum is zero, it follows that the resultant of the two components (012 and 120') in the lines / 2 / and // /' must pass through the point of intersection of / / 3 and // / 3 ' ; but it also passes through the point of intersec- tion of/g/! and////; therefore its line of action is the line joining these two intersections. Now this line of action is parallel to the line 00'; for, two forces represented by 012 and 120' give a resultant represented by 00" in magnitude and sense. Hence the corresponding sides// and///',/ /s and/'// intersect on a line parallel to 0(7; similarly the sides / 2 / and / 2 '/ 3 ', /.A and/'// intersect on a line parallel to 0(/, which, of course, must be the same line as before. This line is LM in the figure. 85.] Problem. Given one funicular of a given system of coplanar forces^ to construct all funiculars of the system. Let the given funicular be/ /i / / 3 Draw any line LM in the plane of the forces ; produce the sides, / /, / / 2 , . .. , of the given funicular to meet LM ; from the point of intersection of LM and/ / draw the arbitrary line///', which meets P 1 in /'; join /"to the point of intersection of LM and//; this joining line will meet P 2 in//, which is the second vertex of the new funicular ; join // to the point of intersection of LM and //; this will give//; and so on. Hence a new funicular is formed, and since the lines LM and//// were drawn at random, an infinite number of funiculars of the system can be described in this way. 86.] Problem. To construct the resultant of a given system of coplanar forces. On any scale construct a force polygon 01, 12, 23, ...of the given system ; then the line of action of the resultant must be parallel to the side (01, 56) which closes the force polygon. Take any pole, 0, and construct a funicular / / / 2 . . . , of the system. Then the resultant must pass through the point of intersection of the extreme sides, /Q/ and/ 5 / 6 , of the funicular. For, by resolving each force into components along the two 87.] GRAPHIC CONDITIONS OF EQUILIBRIUM. 107 sides of the funicular which start from the vertex at which the force may be supposed to act, these components will be mutually destroyed, with the exception of those in the extreme sides, f Q f and f & f Q . Hence the whole system of forces is equivalent to two forces acting in these sides, and represented in magnitudes on the scale adopted by the lines 001 and 056. The line of action of the resultant therefore passes through the intersection of the extreme sides and is parallel to the line joining 01 to 56, and the magnitude is represented by the length of this joining line, its sense being of course from 01 to 56. COR. 1. Whatever be the path described by the pole, the point of intersection of the extreme sides of the funicular describes a fixed right line. This is the line of action of the resultant of the given system of forces. COR. 2. The point of intersection of any two sides of a funicular describes a fixed right line, when the pole varies in any manner. Thus the sides f^f^ and/^^ will always intersect on the line of action of the resultant of the forces P 2 , P 3 , P 4 . 87.] Graphic Conditions of Equilibrium. When a system of coplanar forces acting on a rigid body is in equilibrium, the forces when compounded two and two must finally reduce to two equal forces of opposite senses acting in the same right line. Since the resultant is proportional to the line required to close the force polygon, this line must be zero; hence the force polygon of the system must close up of itself. Again, since the system is finally reducible to two forces acting in the first and last sides, ff^ and^y^, of any funicular, these sides must coincide ; or, in other words, the funicular must be closed. Hence the conditions of equilibrium are 1. The Force Polygon of the system must be closed. 2. Any Funicular Polygon of the system must be closed. COR. 1. If any one funicular of the system is closed, every funicular of the system is closed. COR. 2. If the system is equivalent to a couple, the force polygon is closed, and the first and last sides of all funiculars are parallel. 1:08 COMPOSITON AKD RESOLUTION OF FORCES. [88. 88.] Problem. To represent the moment of a force about a point. Let it be required to repre- sent the magnitude of the moment of a force P about a point (fig. 99). Draw ab parallel to P and representing it on any scale. Let o be a point taken at a unit distance from ab ; draw oa and ob. Assume any point, Q, on the line of action of P, and draw QM and QL parallel to oa and od, respectively. From draw a line, I/M, parallel to P. Then the length LM represents the moment of P about 0. For, the triangles oab and Q ML are similiar ; therefore if p is the length of the perpendicular from Q, on LM 9 we have = > therefore LM = P . p , since ab represents P. Hence LM is the moment on the scale adopted. If the pole o is at a distance k units from ab } we shall have Fig. 99. 89.] Problem. To represent the sum of the moments of any system of coplanar forces about a point. Let A (fig. 98) be the point about which the sum of the moments of the forces is required. The sum of their moments = the moment of their resultant about the point. Let this resultant be constructed by Art. 86, and let the moment of the resultant be constructed by last Art. Now the resultant is represented by the line joining 01 to 56 (fig. 98), and if is a pole assumed at any distance, , from this line, we are to draw from any point on the resultant, two lines parallel to 001 and 056, and through A a line parallel to the resultant, R. Now the extreme sides, f$fi andj^jfg, of the funicular intersect in a point on R, and are parallel to the lines 01 and 056. Hence the intercept made by the extreme sides of the funicular on a line drawn through the given point A parallel to the resultant will represent the sum of the moments of the forces about the point. 90.] PROPERTY OF PERSPECTIVE TRIANGLES. 109 This intercept multiplied by k will be the sum of moments. 90.] Property of Perspective Triangles. Two triangles, ABC and A'B'C', are said to be in perspective when their vertices can be joined in pairs by three right lines which meet in a point. If the lines joining A to A', B to B\ and C to Cf meet in a point, A and A are called corresponding vertices, as are also B and B', C and C' ; and the sides, AB and A'B', &c., which join corre- sponding vertices in the triangles are called corresponding sides. The fundamental property of triangles in perspective is that the points of intersection of corresponding sides lie in one right line. To prove this projective property it is sufficient to prove it for the simplest figure into which the two triangles can be projected. Let the line CC' be projected to infinity. Then AA' and BB" will become parallel lines ; also the sides AC and BC of the first triangle will become parallel, as will A'C' and ffC' of the second. For the simple figure thus obtained there is no difficulty in proving the proposition. To construct a triangle whose three sides shall pass each through a given point, and whose three vertices shall each lie on one of three concurrent lines. Let it be required to construct a triangle whose vertices, A, B, C, shall lie on three concurrent lines, AO, BO, CO, and whose sides shall pass through the points a, 6, c, (fig. 100). Suppose it done, and let ABC be the triangle. Take any point, C, on CO, and draw C'a and C'b meeting BO and AO in 1? and A respectively. Then the triangles ABC and A'B'C* are in perspective, therefore the sides AB and A' I? intersect in P, a point on j^ I00 . the line ab. Hence P is known, since it is the intersection of ab with the line AB' which is con- structed by arbitrarily assuming C f . P being known, join it to c, and the vertices A and B are determined, and C follows at once. Q. E. F. NOTE. In Art. 88 if the unit of force is CT and the unit of length 7 A, the moment of the force P about will be LM x -or x 7* For db p A will obviously be A. 110 COMPOSITION AND RESOLUTION OF FORCES. [90. EXAMPLES. 1. A heavy rod, or beam, is supported horizontally on two smooth props at its extremities, and loaded with given weights at given points in its length ; find the pressure on the props. Suppose the line a a 5 (fig. 34, p. 38) to be horizontal and to represent the loaded beam, the loads, P lt P 2 , ... (including its weight among them) being applied at the points, d 1} d 2 , ..., and let the pressures at the props and a 5 be P and P 5 . Starting from any point 01 draw a vertical downward line to represent on any scale the force P lt and let this line terminate at the point 12 ; from 12 draw a vertical downward line representing P 2 on the same scale, and let this line terminate at the point 23 ; from this point draw a vertical down- ward line to the point 34 to represent P 3 ; from 34 draw a vertical downward line to the point 45 to represent P 4 . Then from 34 we must draw a vertical upward line to represent the pressure P 6 , and this line will terminate at the point 56, which, however, is at present unknown. The pressure P Q will, of course, be represented by the upward line between 56 and 01. To determine 56, assume any pole, 0, and join this pole to the points 01, 12, ... . Across the lines of action of the forces acting on the beam draw the lines A A^ ) A 1 A 2) ... parallel to the lines 001, 012, ... , and draw the closing line, A A 6 , of the funicular polygon. Then the line through parallel to this closing line is that joining to the required point, 56. 2. A beam is supported horizontally at its extremities on two vertical props and loaded with given weights at given points in its length ; it is required to represent the Sending Moment at any point of the beam. Def. When a beam is in equilibrium under the action of any forces, the Bending Moment at any point means the sum (with their proper signs) of the moments about this point of all those forces which act at one side (either side will do) of the point. Suppose a Q a 5 (fig. 34, p. 38) to represent the beam, as in last ex- ample, and let P be the point about which the leading moment is required. The pressure on the prop a being P Q , the bending moment at P is the sum of the moments of P , P lt and P 2 ; and if we con- struct any funicular of the system this moment will, by Art. 89, be the intercept on a vertical line through P made by the extreme sides of the funicular of the forces P Q , P l , and P 2 . But these extreme sides are obviously A Q A 5 and A 2 A S . Hence the bending moment at any point P is represented by the vertical ordinate, mn, drawn through P, of any funicular polygon of the system. Of course, if k is the distance of the pole of the assumed funicular from the vertical line which serves as the force diagram, the bending iy moment will be mn x k X -r (See Note, p. 109.) 3. To construct for any system of coplanar forces a funicular polygon which shall pass through three assigned points. 90.] EXAMPLES. Ill Let the given system of forces beP 1} P 2 , P 3 , P 4 , P 5 (fig. 98, p. 104), and let it be required to construct a funicular polygon which shall pass through the points D, E, F. Consider the triangle formed by the sides, / / x , / 2 / 3 , and f 5 / 6 , of the funicular which pass through the three given points. The vertex formed by the intersection of f / x and / 2 / 3 lies on a given line, P 12 , (not drawn in figure) which is the resultant of P x and P 2 (Cor. 2, Art. 86); the vertex formed by the intersection of / 2 / 3 and /g/ 6 lies on a given line, P 346 , which is the resultant of P 3 , P 4 , and P 5 ; and the vertex formed by the intersection of / / r and / 5 / 6 lies on a given line, ^ 12345 , which is the resultant of P 19 P 2 , P 3 , P 4 , and P 5 . Moreover the three lines P 12 , P 345 , and P 12345 obviously meet in a point ; for the resultant of P lf ... P 5 may, if we please, be constructed by first finding the resultant of P 15 P 2 , and then finding the resultant ofP 3 ,P 4 ,P 5 . Hence the triangle formed by the sides of the funicular which are to pass through the assigned points is one whose vertices lie on three concurrent lines and whose sides pass each through a fixed point. Let this triangle be constructed by Art. 90. Then knowing the force diagram of the forces and drawing two lines, 01 and 230 say, parallel to the two sides / /j and / 2 / 3 , the pole is known, and thence the whole figure. 4. Construct a funicular polygon which shall pass through three given points, two of which lie on one side of the polygon. Ans. This side of the polygon is known, and it intersects the side passing through the remaining point in a point lying on a given line. Hence the side passing through the remaining point is known, and hence the pole of the funicular. 5. For a given system of vertical downward forces, P lt P 2 , ... P n _ 1? equilibrated by two extreme vertical upward forces, P , P w , let any funicular polygon be constructed. Prove that the area of this C polygon = 5 where C is constant and k the distance of its pole from k/ the vertical line which is the force diagram of the forces. (The value of C is obtained by multiplying each force of the system by half the product of the distances between its line of action and the lines of action of the extreme forces, and adding all such products together, and multiplying the result by - See Note, p. 109.) 6. A uniform beam is supported at its extremities on two vertical props ; find the bending moment at any point in it. Ans. If y is the distance of the point from one extremity, the bending moment is W ^ ' where W is the weight of the beam. 2 ' 7. In the last example, what is the curve of bending moment 1 Ans. A parabola passing through the ends of the beam, its vertex lying on the vertical line through the middle of the beam at a distance 112 COMPOSITION AND RESOLUTION OF FORCES. [91. - from the beam. (The bending moment at any point is the product o of W and the vertical distance of the point from the parabola.) 91.] Astatic Equilibrium. When any number of forces, P 1} P 2 ,..., acting at points, A 19 A^ ... , in a body keep this body in equilibrium, these forces will not, in general, continue to preserve equilibrium when the body is displaced in any manner, each force still retaining its magnitude, direction, and point of application in the body. If for all displacements of the body the forces continue to preserve equilibrium, the body is said to be in astatic equilibrium. The simplest example of astatic equilibrium is furnished by a heavy body suspended by a vertical string attached at its centre of gravity. Here the system of forces consists of the weights of the particles of the body and the tension of the string ; and however the body may be displaced about its centre of gravity, all these forces will retain their individual magnitudes, direc- tions, and points of application, and the body will remain at rest. Again, a system of two equal reversed magnets rigidly con- nected by an axis through their centres is astatic for displace- ments round this axis. When a system of forces applied to a body is not in equi- librium, it happens that in certain cases this system can be astatically equilibrated by a single applied force ; i. e., in all displacements which the body can receive, each force acting on it with invariable magnitude, direction, and point of application, it may be possible to equilibrate the system by one force of constant magnitude, direction, and point of application. It is evident that this is always the case for a system of parallel forces. A single force equal and opposite to their re- sultant, applied at their centre, will astatically equilibrate them. Into the general discussion of astatic equilibrium we do not enter *. Suffice it to say that a system of (non-coplanar) forces must in general be astatically equilibrated by three forces ; and if the forces are all parallel to one plane, by two. When (as in the present chapter) the forces are all coplanar we shall prove * On this subject the student may consult Moigno's Statique (Dixieme Le9on), a memoir by M. Darboux (sur ISEquilibre Astatique), and a paper by the author in the Proceedings of the London Mathematical Society (vol. ix). 92.] CENTRE OP A SYSTEM OF COPLANAR FORCES. 113 that for displacements of their points of application in their plane, the system can be astatically equilibrated by a single force. In this case it is clear that instead of considering the body to which they are applied as displaced, we may consider the body fixed and each force rotated in a fixed sense round its point of application through a constant angle a motion of translation of the body or points having obviously no effect on the system of forces. We shall now prove that if all the forces in a coplanar system are rotated in the same sense, through the same angle, in their plane, round their points of application, their resultant (unaltered in magnitude, of course) passes through a fixed point in the body. Let two forces, P and Q, act at two fixed points, A and B, (fig. 101) in the directions OA and OB, being the point of intersection of their lines of action; and let the forces be turned in the same sense round A and B through the same angle, so that the point of intersection of their new lines of action is (7. Now, since LOACt ' = /.OB(y, a circle described through A, B, and will pass through (/, and the angle AC/B, between P and Q when they are turned round, is equal to the original angle, AOBj between them. Also, the forces being unaltered in magnitude, it follows that the angles which the resultant at (/ makes with them are the same as the angles which it makes with P and Q at 0. If, then, OC is the direction of the resultant at 0, (JC must be the direction of this resultant at (7. Hence, the resultant of P and Q passes through the fixed point C. In exactly the same way it is proved that the resultant of three forces passes through a fixed point when the forces are turned round their fixed points of application through a constant angle ; and so on for any number of forces. This point may be called the astatic centre of the system of forces*. 3 92.] To find the Astatic Centre of a System of Coplanar Forces. Taking an arbitrary origin and arbitrary axes, the * Of course it is understood throughout this discussion and in the examples at the end of this chapter that the displacements of the body or forces are always supposed to take place in the plane of the forces. I 114 COMPOSITION AND RESOLUTION OF FORCES. [92. point required lies on the resultant whose equation is (Art. 82) a2r-/32J- = 0, (1) (a, (3) being the running co-ordinates. Now, if the force P 15 acting at the point (# 1? y^) is turned round in the plane of xy through an angle co, X l becomes P l cos (0 1 + co), where 6 l is the original angle made with the axis of x by P x , or X 1 cos co Y l sin co ; Y l becomes X l sin co + Y l cos o> ; and YI&I X\y\ becomes (I^ X\y\) cos co + (Zj x^ + Y\y\) sin co. Hence, 2 X becomes cosco.SX sin co.27, \ 2F sin co. 2X+ cos co. 2 7, ( (A) G G cos co -j- T sin co, ) where F=2 (Is +I*. The equation of the new resultant is, therefore, (a2r-2X-)cosco + (a2X+2r-r)smco = 0, (2) and the astatic centre of the system of forces is the intersection of the lines given by equations (1) and (2). This point may evidently be determined by (l) and by the equation a2X+/32r-r= 0. (3) Hence for the co-ordinates of the astatic centre we have T2X+ V If the astatic centre were the origin, a and ft would be each = 0., and G would = 0, since the point is on the resultant (Art. 76). Hence for the centre of the forces we have = 0, r=0. (5) If the co-ordinates of A, the point of application of a force, P (fig. 102), with respect to rectangular axes, Ox and Oy, are x and y, the quantity Xx + Yy is equal to ~M x P (x cos 0+y sin 0), being the angle which P makes with Ox. Now if OM is Fig. 102. Xj and AM is y, it is evident that x cos 6 +y sin 6 = AN, N being the foot of the perpendicular from on the line of action of P. Denoting AN by q, we have, then, for the Virial * This quantity is called by Clausius the Virial of the forces. 94-] REMARK ON THE CONDITIONS OF EQUILIBRIUM. 115 Hence, if 'any number of coplanar forces be turned each round a fixed point of application through an arbitrary but common angle, there exists a point in the plane of the forces such that both the Virial and the sum of the moments of the forces about it t continue to vanish for all displacements. It is easy to see that if AN be in the sense in which P acts, the sign of the product Pq will be changed. The value of F with respect to axes through a point (a, /3) parallel to Ox and Oy is evidently 2{X(# a)-f Y(y /3)}, or F a2Z /32Z. Hence the locus of points for which this quantity = is given by equation (3), which denotes a right line passing through the astatic centre, and evidently perpendicular to the resultant. 93.] Theorem. If any number of coplanar forces are in equi- librium, and if the forces be turned, each round a fixed point, in the same sense through any common angle, the new system is equivalent to a couple. For, from equations (A) Art. 92, it appears that if 2X= and 2 T = before the rotation, they will = after it ; hence the new system has no resultant of translation, and it must, there- fore, be a couple. Now, since by hypothesis (r=0, the axis of the new couple is, by equations (A), equal to F sin CD. We see, then, that the system of forces will remain in equi- librium, whatever be the angle through which they are turned, if T= 0. 94.] Remark on the Conditions of Equilibrium. It must be carefully borne in mind that the conditions of equilibrium given in Arts. 80 and 81 are sufficient only in the case of in- deformable bodies. For, having reduced a system of forces to a resultant of translation, R, acting at an arbitrary point, together with a couple of moment G, the logical conclusion is that If R and G = 0, those motions of the system which would be produced by R and G respectively are thereby destroyed. Now by a fundamental principle of Kinetics, which we antici- pate, if R = there is no resultant linear momentum of the system in any direction, or in other words its centre of mass is at rest ; and if, in addition, G = 0, there is no resultant angular momentum about the centre of mass of the system. i a 116 COMPOSITION AND RESOLUTION OF FORCES. [94. These two things we can conclude from the equations R = 0, G for all systems, whether they are gases, liquids, deform - able frameworks, natural solids, or rigid bodies. Now the destruction of resultant linear and angular mo- mentum will, except in the case of rigid bodies, be quite con- sistent with the existence of motions of parts of the system among themselves, negative momenta cancelling positive. Hence, whenever a system is capable of altering the relative positions of its parts, the complete equilibrium of the system will require more than the vanishing of the resultants of translation and rotation of the forces applied to it. In fact, its internal forces will have to be taken into account. In rigid bodies the destruction of the above-mentioned motions will necessitate the destruction of all motion, and the conditions R = 0, G = are both necessary and sufficient. In these bodies there is no restriction placed on the internal forces, so that they are always capable of assuming such magnitudes and directions as will enable them to destroy the action of the external forces. On the contrary, in deformable bodies, there are restrictions placed on the internal forces so that they are not capable of preserving equilibrium against all systems of external forces. For example, in a freely jointed framework, the action between bar and bar must consist of a single force restricted to passing through the joint. This is the reason why two equal forces applied in opposite senses in the same line to two opposite sides of a set of parallel rulers will not hold them in equilibrium, unless the rulers are placed in a certain configuration ; and it is also the reason why two equal and directly opposite forces applied to the ends of a string, elastic or inelastic will not hold it in equilibrium until it has assumed a certain state. Hence also the necessity for considering the internal forces (pressures) in Hydrostatics. We shall afterwards enunciate a single principle,* or condition, of equilibrium which will embrace all systems indiscriminately. These observations are recommended to the most careful consideration of the student. * The Principle of Virtual Work. 94-] EXAMPLES. 117 EXAMPLES. 1. If the sums of the moments of any number of coplanar forces round three points which are not in a right line are each = 0, the forces are in equilibrium. 2. If the sums of the moments round three points not in a right line are equal, the forces are either in equilibrium or equivalent to a couple. 3. If the sum of the moments of a system of coplanar forces round three given points are /, m, and n, and if the sides and angles of the triangle formed by the points are a, b, c, A, B, C, show that the resultant force is equal to (I 2 a 2 + m? 6 2 + n 2 c 2 2 Imab cos C-^ 2 mnbc cos A 2 nlca cos B)? 2A where A is the area of the triangle ABC. 4. If a system of coplanar forces applied at fixed points is in equilibrium, the co-ordinates of the astatic centre become indeterminate. Explain this. Ans. In this case the system must be astatically equilibrated by two equal and opposite forces (couple). 5. In the last case show how to find an astatically equilibrating couple for the system. Ans. Take the astatic centre of any number of the forces, and also the astatic centre of the remaining forces. These will be the points of application of the forces of the required couple (whose moment, of course, varies with the displacement of the body or forces), and the forces of the couple are equal to the resultants of the two partial sets. 6. Three forces are applied at the middle points of the sides of a triangle, ABC, perpendicular to these sides and respectively propor- tional to them ; find a couple which will astatically equilibrate them. Ans. A couple one of whose forces is applied at the middle point of any one side, AB, and the other applied at the point of intersection of a parallel to AB drawn through C with the perpen- dicular to AB at its middle point. 7. When a system of coplanar forces in equilibrium continues in equilibrium for all displacements in the plane of the forces, show that the astatic centre of any number of them must be coincident with that of the remainder. CHAPTER VI. THE CONDITIONS OF EQUILIBRIUM OF A RIGID BODY UNDER THE ACTION OF FORCES IN ONE PLANE DEDUCED FROM THE PRINCIPLE OF VIRTUAL WORK FOR A SINGLE PARTICLE. 95.] Theorem. If a particle in equilibrium under the action of any forces be constrained to maintain a fixed distance from a given fixed point, the force due to the constraint (if any) is directed towards the fixed point. Let B be the particle, and A the fixed point. Then the string or rigid rod which connects B with A may be removed if we enclose the particle in a smooth circular tube whose centre is A\ for evidently the preservation of the constancy of the distance AB receives sufficient expression in this manner. Now, in order that B may be in equilibrium inside the tube, it is necessary that the resultant of the forces acting upon it should be normal to the tube, i. e., directed towards A. COR. 1. If A and B be two particles in equilibrium, con- nected by a rigid rod whose weight is neglected, the reactions of A and B on the rod are two forces equal in magnitude and opposite in direction. COR. 2. If any body be in equilibrium under the action of two forces only, these forces must be equal and opposite in the same right line. COR. 3. If a particle in equilibrium under the action of any forces is constrained to maintain a fixed distance from each of a number of other particles or points, the forces corresponding to these constraints are directed in the right lines joining the particle to each of the other particles or points. This is evidently true whether the invariable distances are maintained by straight rigid bars or by crooked bars. 96.] System of Particles Rigidly Connected. Let there be any number of particles, %, m^ m 3 , ... (fig. 103), each acted on by any forces, and connected with the others in such a way that the figure of the system is invariable. 96.] SYSTEM OF PARTICLES RIGIDLY CONNECTED. 119 Then, by the last Article, the force proceeding from the connection of m 1 and m 2 is in the line % m 2 , which we may Fig. 103. imagine to be a rigid bar. Let this force be denoted by T 12 . Similarly, let the forces in the bars m 2 m 3 and m 3 m 1 be denoted by 7^3 and T 31 , respectively. These internal forces may tend either to increase the distances between the particles or to diminish them. In the figure we have supposed the latter to be the case, but the result will be the same if the former sup- position is made. Imagine that the system is slightly displaced so as to occupy the position abc. Now, it has been already proved (Art. 65, p. 78) that the equation of virtual work for two particles rigidly connected will not involve the force due to the connection ; but, for clearness, we reproduce the proof here. Let fall the perpendiculars aa 2 and aa 3 on the lines m 1 m% and m l m 2 ; 6b t and bb% , on m 2 m z and m l m 2 ; cc^ and cc 2 on m. 2 m 3 and %^ 3 . Let the sum of the virtual works of the external forces (not including T 12 and T ls ) acting on % be denoted by 2P8/J, and let 2 Qbq and 272Sr denote similar quantities for m. 2 and m 3 . Then the equation of virtual work for m 1 is evidently = 0; (1) that for m 9 is = 0, (2) and that for m 3 is T^.m.c.-T^.m.^ = 0. (3) Now (Art. 63, p. 77) m l a s = % 3 ; % 2 = w 3 c 2 ; m z ^ = m 3 " r 120 EQUILIBRIUM OF A RIGID BODY. [97. Hence, by addition, the internal forces disappear, and the equation of virtual work for the whole system is r = 0. or 2(Pbp + Qbq + Rbr) = 0. (4) The same result is evidently true, whatever he the number of particles forming the system ; and it is well to note Chat we have been enabled to obtain equation (4) connecting the external forces acting on the system, by choosing a virtual displacement compatible with the geometrical conditions of the system, that is, in the present case, a virtual displacement which allows the mutual distances of the particles to remain unaltered ; or, again, such a virtual displacement as might be an actual one; for the system could actually occupy the position abc. 97.] Elimination of the Internal Forces of a System. By the Internal Forces of a system it is already sufficiently clear that we mean forces proceeding from the internal connections of the parts of the system among themselves. Such forces are directed from particle to particle, and will contribute nothing to the equation of virtual work of the system, if in the virtual displacement the distance between every two particles re- mains the same as before. It is evident that if the virtual displacement violates any geometrical condition of the system, the corresponding internal force will appear in the equation of virtual work. Thus, if in fig. 103, the distance ab is not equal to the distance between m l and m^ we shall have by addition the term or T 12 . 6 (m-L m 2 ), where 8 (m l m 2 ) denotes the change or variation of the distance between % and m 2 . And, generally, if any internal force, F 9 tend to vary any internal function,/ 1 , in a system, this force will contribute to the equation of virtual work of the system the term F.SS, so that if in the supposed displacement of the system, the function f is actually altered, the force F will appear in the equation, but will not appear if/" is unaltered. 98.] VIRTUAL WORK OF FORCES ACTING IN ONE PLANE. 121 98.] General Equation of Virtual Work for Forces Acting in One Plane on a Rigid Body*. If the particles m 1 ,m. 2 ,m 3) ... form a continuous body, on which, forces P lt P 2 , P 3 ,... act in one plane at different points A L , A^ A^, .., of the system (fig. 104), Fig. 104. the condition necessary and sufficient for tlie equilibrium of the system is that the sum of the virtual works of the forces is equal to zero for any and every virtual displacement which violates none of the geometrical conditions of the system. For we have seen (Art. 61, p. 73) that the condition necessary and sufficient for the equilibrium of any one particle of the system is the vanishing- of the virtual work of all the forces acting upon it, the internal forces proceeding from its connection with the other particles of the system being, of course, included, as in equations (l), (2), (3) of Art. 96. Expressing thus the conditions for the equilibrium of all particles of the system, and adding the results, there remains for the condition of equilibrium the equation ... = 0, (1) into which no internal force enters. Conversely, if the sum of the virtual works of the forces * We formally confine the discussion for the present to Kigid Bodies, although it is clear from last Article that what follows is applicable to systems such as freely articulated bars which, without being rigid systems, satisfy certain geo- metrical conditions that are not violated in the virtual displacement ; and it is equally clear that these conditions may be violated if we include in our equations the work of internal forces. 122 COMPOSITION AND RESOLUTION OF FORCES. [99. vanishes for every virtual displacement, the system is in equi- librium. For, if it is not, it will take a determinate motion, each point of the system describing a certain line in virtue of its con- nections with the other points. Now, this motion will be in no way interfered with if we introduce new connections which render it the only motion possible for the system. Under the new circumstances it is clear that if we prevent the motion of any one point, we prevent the motion of the system. Suppose the motion of the point A to be stopped by the application of a force, F, in the direction A' A, A being" the point to which A moves. Now, equilibrium exists under the action of (a) the given external forces, (/3) the newly-introduced geometrical connections, and (y) the force F\ hence the sum of the virtual works of these forces for every displacement. Choose that displacement which the system is supposed actually to undergo when the force F is not applied at A. Now, by the last Article, since none of the geometrical conditions (/3) are violated by this displacement, the forces proceeding from them will do no work. Hence the equation of work is 2Pb, at 0. Let 0! be the angle between OA l and the direction of P r Then, evidently, the projection of A 1 a^ on the direction of P l is Jj^.sin r o y But A l a 1 = co. OA^; therefore the Fig. 107. virtual work of P l is a)P l . OA sin 0J. If p t ne perpendicular, Oq^ from on the line of action of P 13 this is evidently 126 COMPOSITION AND RESOLUTION OF FORCES. [105. Similarly, the virtual work of P 2 is o>P 2 .p 2) and that of P 3 is &>P 3 .jt? 3 . Hence the equation of virtual work is or 2Pp = 0. But the product of a force, P, and the perpendicular, p, let fall upon it from the point 0, is the moment of the force with respect to the point 0, or rather with respect to an axis through O perpendicular to the plane of the figure. Hence, equation (2) asserts that for equilibrium the sum (with their proper signs) of the moments of the forces with respect to any point in their plane is zero. As regards the signs to be given to the moments, P^p^ P 2 p 2 , ... of the forces, we see that Those forces which tend to rotate the body in the same sense round the point give virtual work of the same sign, and therefore have moments of the same sign with respect to 0. Thus, in fig. 107, the forces P l and P 2 tend to turn the >ody round 0, in a sense opposite to that of watch-hand rotation, while P 3 tends to turn it in the opposite sense. Hence, in the Equation of Moments, as the equation 2Pp= is called, P l p and P 2 p 2 have the same sign, and P 3 p 3 has an opposite sign. 105.] Absolute Conditions for the Equilibrium of a Rigid Body Acted on by Forces in One Plane. It is now clear that, as all possible displacements of a rigid body are exhausted in a motion of translation common to all its parts, and a motion of rotation common to all its parts, all possible conditions of its equilibrium under the action of forces acting in one plane are exhausted in the conditions of Articles 103 and 104, namely 1. The sum of the components of the acting forces along every arbitrary line in their plane = 0. 2. The sum of the moments of the forces with regard to every arbitrary point in their plane = 0. These are the conditions which were deduced in the last chapter ; and it is clear that since all possible displacements of a deformable system are by no means exhausted in motions of translation and rotation common to all its parts, the equation of virtual work for such a system does not lead to the above con- ditions as sufficient. io6.] DISPLACEMENT OF A RIGID BODY. 127 106.] Analytical Expression for the Displacement of a Rigid Body. We have seen (Art. 100) that the displacement of a rigid body is known from the displacement of any fixed triangle in it; and that the displacement of such a triangle consists of a motion of translation common to all its parts, and a motion of rotation common to all its parts. The displacement of translation may be that which moves each side of the triangle parallel to itself until the vertex m l (fig. 105) comes into the position a ; or it may be that which moves each side parallel to itself until the vertex m 2 comes into the position b; or, again, that which moves the system until the vertex m 3 comes into the position c. In the first case the magnitude of the motion of translation is m 1 a t in the second, m< 2 b, and the third, m 3 c. Now these three quantities are all of different magnitudes. But after any one of these motions of translation has taken place, the motion of rotation is constant, since the angles between the sides of the triangle are invariable. Hence If a rigid body occupying the position (A) is displaced by a motion parallel to one plane into the position (B), the body may be brought from the position (A) to the position (B) by : (a) a variable motion of translation common to all its parts, whereby any one point, P, of the body is brought directly from its old to its new position, ; and (ft) a subsequent motion of rotation round an axis through per- pendicular to the plane of motion, the angular magnitude of the rotation being a constant quantity for all such axes. We shall investigate the changes produced in the co-ordinates of a point by given small motions of translation and rotation. Let the motion of translation first take place. Then draw any two rectangular axes, Ox and Oy, through (fig. 108) the new position of a point 0^ Let the motion of translation 0^0, common to all parts of the body, be resolved in two components, a and 6, parallel to Ox and Oy. Then, if x and y denote the co- ordinates of a point Qi i n the body with reference to fixed axes drawn through 1 parallel to Ox and Oy } these quantities will be increased by a and b, respectively, by the motion of translation. To find Fi S- lo8 - how much they will be subsequently altered by an angular 128 COMPOSITION AND RESOLUTION OF FORCES. [107. rotation = o> round 0, let Q describe a small arc of a circle, Qq f round 0. Let fall the perpendiculars Q M and qm on Ox, and Qp on qm. It is evident that OM = x and QM = y. Then the increase of y produced by the rotation = qp, and the increase in x ' = QP. Now Qp = Qq.sin QOx = u.OQ.sin QOx = co.Qlf = coy; and g7? = Qq.cos QOx = co. OQ.cos QOx = co. OM = &x. Hence, if bx and 6y denote the changes produced in x and y by the two motions combined, bx = fl o>y, (1) 5y = + , and then adding a and # to the results.) 107.] Analytical Conditions of Equilibrium. If any forces, P l9 P 2 , P 3 , ... , act on a rigid body in one plane, the condition necessary and sufficient for equilibrium is (Art. 98) P l a/> 1 + P 2 ty 2 + P 8 ty 2 +... = 0. (1) Let X l and Y 1 be components of P 1 along two rectangular axes, Ox and Oy t and let ^ and y^ be the co-ordinates of the point at which P l acts. Then (Art. 52, p. 68) PI*PI = XI*XI+Y^. (2) Making similar substitutions for P 2 bp 2 , P 3 8/? 3 , . . . , equation (1) becomes J 1 8a? 1 +J 1 8y 1 + Z 2 8a? a + r 2 6y 2 + ... = 0, (3) or 2(X8ar+rty) = 0. (4) Substituting in (4) the values of 8a? and fy given in the last Article, we have 2 {X(0_ft,y) + Y(b + u>x)} = 0, or .2JT+^.2FH-a).2(a?7-^X)= 0, ... (5) since #, ^, and to are common to all points of the body, and may be taken outside the sign of summation. Now the displacements a, b, and o> are completely independent of each other, and therefore equation (5) requires that io8.] VARIGNON'S THEOREM OF MOMENTS. 129 o,sr= different. Then we have aSX + 1>2 Y+ o>' 2 (xY yX] = 0. (7) Subtracting (7) from (5), ( ') 2(^7 J/.J) = 0. But since a> a/ is not = 0, this equation requires that 2(aJ-.yJ) = 0. Similarly, by making a alone variable, we prove that 2JT = 0, and by making b alone variable, 2F = 0. The three equations (6) constitute the analytical conditions of equilibrium of the body, and they are the expressions of the two absolute conditions of Art. 105. The first two of the equations (6) are called the equations of translation, and the last is called the equation of moments or rotation. 108.] Varignon's Theorem of Moments. The moment of the resultant of two forces with respect to any point in their plane is equal to the sum of the moments of the forces with respect to this point. Let R (fig. 109) be the resultant of two forces, P and Q, applied at a point A, and let be any point in their plane. Then the virtual work of R for any displacement of A = the virtual work of P + the virtual work of Q. Let the virtual displace- ment of A be one of rotation round 0, through a small angle = o>. Then, as in Art. 104, the Fig. 109. virtual work of R is co . R . OA . sin OAR; but this = o> . R x the perpendicular from on R to x the moment of R with respect to 0. Similarly, the virtual work of P = co x moment of P with respect to ; and virtual work of Q co x moment of Q with respect to 0. Therefore, &c. Q. E. D. In precisely the same way, the moment of the resultant of any number of forces is proved to be equal to the sum of the moments of the forces separately. 130 COMPOSITION AND RESOLUTION OF FORCES. [109. 109.] Particular Case in which the Resultant of Transla- tion Vanishes. When forces applied to a particle have no resultant of translation, their whole effect is null. It is not necessarily so, however, if they are applied to a body of finite dimensions. For example If the forces acting upon a rigid body form ly their magnitudes and lines of action the sides of a closed polygon taken in order, their resultant of translation vanishes, and they have a constant moment / / with respect to all points in their plane. Let forces P lt P 2 , P 3 , ... (fig. no) act at points A 19 A 2 , A B , ... in one plane, in a body and let these forces be represented in mag- nitudes and lines of action by the sides of the polygon formed by their points of application. Now since (Art. 50) the sum of the projections of the sides of this polygon on any arbitrary line = 0, the condition of Art. 103 is fulfilled, and the forces have no resultant of translation. Let be any point inside the polygon, and take the sum of the moments of the forces round it. If the perpendiculars from on the sides A^A^ A 2 A 3 ... be jt? l5 j 2 , ... the sum of the moments will be Fig. no. And since P 15 P 2 , ... are equal to the sides of the polygon, G is evidently = 2 . area of polygon. This is a constant for all points inside the polygon. Now if we take the sum of the moments round any external point, (7, we shall have since P 4 turns the body round (/ in a sense opposite to that in which the other forces turn it. But this sum is equal to + A and this is again equal to 2 . area of polygon. Hence for all points in the plane, the sum of the moments, is constant. III. PROBLEM. 131 110.] Theorem. If a number of forces acting in one plane upon a rigid body have a constant moment with respect to all points in the plane, they can have no resultant force, and must be reducible to a couple. For, suppose that they have a resultant = R, then if p is the perpendicular let fall on E from any point, 0, the sum of the moments of the forces = E .p (Art. 108). Hence by varying the position of 0, the sum of the moments varies, which is contrary to hypothesis. They are reducible to two equal, parallel, and opposite forces. For their resultant is zero ; hence, compound- ing them in pairs, they must reduce to two parallel, equal, and opposite forces forming a couple, or to two such forces directly opposite to each other in a right line. But in the latter case the sum of their moments about any point would be zero ; therefore if this moment is not zero, the forces must reduce to a couple. 111.] Problem. To find the resultant of two parallel forces, P and Q, acting in the same sense. Let AB (fig. in) be the shortest distance between P and Q, and let the forces be supposed to act at A and B. Also let the reversed resultant, JR, act at some point, 0, in AB. Since the forces are in equilibrium, their virtual work = for every virtual displacement (Art. 98). Choose first a virtual displace- ment of translation along AB. For this displacement the vir- tual work of the forces P and Fig. in. s Q = 0, therefore the virtual work of E = 0, therefore E parallel to P and Q. Again, choose a virtual displacement of rotation about through an angle = o>. The virtual work of P is then P.uOA, and that of Q is - Q . w OB, while that of E is zero. Hence P. OA Q . OB = 0, (1) OA_Q L '* OB~~P' Finally, to find the magnitude of E, take a virtual displacement of translation parallel to the forces. This evidently gives (2) 132 COMPOSITION AND RESOLUTION OF FORCES. [ill. Therefore the resultant of two parallel forces acting in the same sense is a force parallel to them in the same sense, equal to their sum, and dividing the line joining their points of application in the inverse ratio of the forces. Equation (1) asserts that the moments of two parallel forces with respect to any point on their resultant are equal and opposite a result which is, of course, con- ^ tained in equation (1) of Art. 104. If P and Q act in opposite senses (fig. 112), the resultant is obtained in P magnitude and direction by simply chang- ing the sign of Q. B Thus (l) becomes Fig. 112. OA _ Q OB~~P* which shows that is on the production of AB at the side of the greater force ; and (2) gives R = P-Q. (4) In illustration of this chapter some of the examples in the next are solved by the Principle of Virtual Work. CHAPTER VII. APPLICATIONS OF THE CONDITIONS OF EQUILIBRIUM OF A BODY. 112.] Condition of Equilibrium of a Body under the Action of Two Forces in a Plane. If two forces maintain a body in equilibrium, they must be equal and opposite in the same right line. For, take moments round any point on the line of action of one of them, P. The sum of the moments must (Art. 104) be = 0. Hence the other force, Q, must pass through the assumed point. Again, take any other point on P } and take moments round it. The sum must be = 0, and Q must, therefore, pass through this point. Hence P and Q act in the same line. Now their sum must = (Art. 103). Therefore P and Q are equal and opposite. Q. E. D. 113.] Condition of Equilibrium of a Body under the Action of Three Forces in One Plane. If three forces maintain a body in equilibrium, their lines of action must meet in a point, or be parallel. For, take moments round the point of intersection of two of them, P and Q. The sum must (Art. 103) = ; therefore, either the third force, R, is zero, or it passes through the intersection of P and Q. If R is not = 0, it must pass through this point. The three forces may then be supposed to act at this point, and to keep it at rest. Hence, each force must be equal and opposite to the resultant of the other two ; and if the angles between them in pairs be JK>, , r, the forces must satisfy the conditions T> ^ -n - ta\ P : Q : E = sin p : sin q : sin r. (/3) If two of them are parallel, the third must be parallel to them and equal and directly opposed to their resultant. 134 APPLICATION OF THE CONDITIONS OF EQUILIBRIUM. [l 13. EXAMPLES. ^1. Three forces, P, Q, R (fig. 113) act at the middle points of the sides of a triangular plate, each force being perpendicular and pro- portional to the side at which it acts. If the forces all act inwards, or all outwards, they are in equilibrium. For (a) they satisfy the first conditions of equilibrium of three forces, namely, that of meeting in a point (Art. 113); and (/3) they are pro- portional to the sines of the angles between them in pairs, since P :Q : R = a:b:c = sin -4 : sin B : sin G = sin QOR : sin ROP : sin POQ. They, therefore, satisfy both of the conditions of Art. 113. In exactly the same way it is proved that if three forces act perpen- dicularly to the sides of a triangle, and be proportional to them, they will be in equilibrium, provided that they pass through any common point, and all act outwards or all inwards. v 2 . Three forces acting along the perpendiculars of a triangle keep it at rest ; find the relations between them. They satisfy the first condition of equilibrium, namely, that of meeting in a point. Then if the forces perpendicular to the sides a, b, c, be P, Q, R, respectively, the relations (/3) of Art. 113 give P : Q : R = sin A : sin B : sin C = a : b : c, as might have been concluded from the remark at the end of the last example. 3. Three forces acting along the bisectors of the angles of a triangle, all either from or to wards the angles, keep it at rest ; find the relations between them. The forces evidently satisfy the condition of meeting in a point. Let P } Q, R, be the forces in the bisectors of A, B, C, respectively. A _i_ 7? Then the angle between P and Q is easily seen to be TT - Hence P : Q : R = cos : cos : cos 222 4. Three forces acting in the bisectors of the sides of a triangle drawn from the opposite angles maintain equilibrium ; find the rela- tions between them. They satisfy the first condition. Let the lengths of the bisectors of the sides a, 6, c (fig. 114) be /3 1} /3 2 , and /3 3 , and let p and q be the perpendiculars from C on P and Q. Take moments round C for the equilibrium of the forces. Then PP = Q* (i) "3-3 EXAMPLES. 135 (The moments of P and Q with respect to C have opposite signs, since Q tends to turn the body round G in the sense of watch-hand rotation, while P tends to turn it in the opposite sense). Again, pft = q(3 2) (2) each side of this equation heing the area of the triangle. Divide the sides of (1) by the cor- responding sides of (2). Then Hence A or the forces are proportional to the bisectors. 5. At the middle points of the sides of any indeformable polygon (fig. 115) forces act perpendicularly to the sides, each force being proportional to the side at whic! tt acts. If the forces all act inwards outwards, they form a system equilibrium. For (example 1) the resultant of P, and P z is a force acting at the middle point of AC, perpendicular and pro- portional to AC. Again, this force and P 3 may be replaced by a force acting at the middle point of AD, perpendicular and proportional to AD. Replacing the given forces in this manner, the result follows by ex- ample 1. Fig. 115. 6. If from any point perpendiculars be drawn to the sides of a polygon, and forces act along these perpendiculars, either all inwards or all outwards, each force being proportional to the side to which it is perpendicular, the system is in equilibrium. This follows, exactly as in the last example, by dividing the polygon into triangles, and attending to the remark at the end of example 1. 7. From any point, 0, inside (or outside) a triangle, ABC (fig. 116), are let fall perpendiculars, Oa, 0/3, Oy, on the three sides. At the points a, /3, y, are applied forces P, Q, R, each of which is proportional and perpendicular to the side at which it acts. The forces are then all turned round their points of application in the same sense, so as to make equal angles with the perpendiculars Oa, 0/3, and Oy. Show that in this latter case the resultant of the system of forces is 136 APPLICATION OF THE CONDITIONS OF EQUILIBRIUM. [ 1 13. a couple whose moment is proportional to the square root of the area of the triangle A'B'C', enclosed by their lines of action. (The forces act all outwards or all inwards). Let the sides of ABC be a, b, c, and let P = ka, Q = kb, R = kc, k being a constant coefficient. Let B be the angle, OaB', between P and the perpen- QV x \ dicular Oa. Then Replace P by two compo- nents, one along BC and the other perpendicular to it. Similarly, replace Q and R. Then the perpendicular com- ponents are ka cos 6, kb cos B, and kc cos 6 ; and since they meet in a point, 0, and are proportional to the sides at which they act, they are in equilibrium (example 1 ). Hence the forces are equi- valent to three, ka sin B, kb sin B, and kc sin B, acting along the sides of ABC in cyclical order, and therefore, by Art. 109, their equivalent is a couple = 2&A sin 6, A denoting the area of the triangle ABC. (See also Art. 93, p. 115.) Now the triangle A'B'C' is similar to ABC. For, since the angles OaB and OyB are right, and the angles OaB' and OyB' are equal, a circle will go round the points OB'aBy. Hence Z yOa = Z yB'a ; therefore their supplements, B and B' are equal. Similarly, A A', and C = C'. Again, the side A'B' = AB . sin B. For in the circle round yOB'aB, yB' is a chord making an angle B with a chord yO, and an angle - with the perpendicular chord, yB. Therefore 2 yB' =y0.cos6 + yB. sin B. (1) Similarly, in the circle round yA'OftA , we have yA' = yO.cosQ yA .sintf. (2) Subtracting (2) from (1) we have A'B' = (yB + yA) . sin 6 AB . sin B. Now if A' be the area of A'B'C', sin 6 = A / -r- and therefore the moment of the forces = 2&\ // AA'. EXAMPLES. 137 C' 8. If the triangle be replaced by a polygon of any number of sides, prove that the equivalent of the forces is a couple whose moment is proportional to the square root of the area of the (similar) polygon enclosed by their lines of action. 9. A triangular plate, ABO (fig. 117), is acted upon at each angle by forces, along the two sides containing it, r represented in magnitudes and lines of action by the distances between the angle and the feet of the perpendiculars let fall from the other two angles on these sides. Find the line of action of the resultant force. Let the perpendiculars let fall on the three sides, a, b, c, from any point, P, . on the resultant be x, y, z, respectively, and let A', B', C f be the feet of the perpendiculars. Then the force in AB in the sense AB is AC'BC', or b cos A a cos B. Hence the moment of this force about P is z (b cos A a cos B), and since the sum of the moments of all the forces (estimated in cyclical order) round P is = (Art. 76), we have Now, one set of values of x, y, and z, which will satisfy this equa- tion, is, evidently, a, b, c. Hence the resultant passes through the point the perpendiculars from which on the sides are proportional to a, b, c. This point is thus found : Let G be the centre of gravity of the triangle ; from A draw a line, AG', which makes Z.CAG'= /.BAG, and from B draw a line, BG', which makes Z CBG' = /.ABG. These lines intersect in G', the required point. Again, another set of values of x, y, z, which will satisfy (1), is cos A, cos j?, cos C ; and the resultant passes through the point whose perpendiculars on the sides are proportional to these quantities. This point is the centre of the circumscribed circle. Hence the line of action of the resultant is known. 10. Show that the resultant of the system of forces in the last example is 4A /i Ii r abc where A is the area of the triangle. 11. Forces P, Q, R act along the sides of a triangle, ABC, and their resultant passes through the centres of the inscribed and circum- scribed circles : prove that P = Q = R cosB cosC cosCcosA cosAcosB (Wolstenholme's Book of Mathematical Problems). 12. A heavy beam, AB (fig. 118), rests against a smooth horizontal plane, CA, and a smooth vertical wall, C, the lower extremity, A, 138 APPLICATION OF THE CONDITIONS OF EQUILIBRIUM. [113. being attached to a rope which passes over a smooth pulley at (7, and sustains a given weight, P. Find the position of equilibrium, and the pressures on the plane and wall. Let 6 be the inclination of the beam to the horizon in the posi- tion of equilibrium; let W= weight of the beam : and let the centre of gravity, G, divide the beam into two portions, A G = a, and BG = b. Now, the reactions, R and S, Fig. 1 1 8. of the wall and plane are nor- mals to these surfaces ; and since they are both unknown, we shall obtain an equation for 6 which will contain neither of them, by taking moments about 0, their point of intersection. Hence, since the force P acts on the beam along A C, and tends to turn it in a sense opposite to that in which W tends to turn it round 0, we have P(a + 6) sin 6 Wa cos = 0, Wa /. tan#= - (1) P(a + b) Again, resolving forces vertically, we have R = W. (2) And resolving horizontally, S P. (3) Solution by Virtual Work. Imagine a displacement in which the ends A and B remain in contact with the planes. Then the virtual works of R and S are both zero, and the equation of virtual work is (if y is the height of G above the horizontal plane) -W.dy-P.d(AO) = 0. (4) Now y = a sin B, A C = (a 4- b) cos B ; .'. dy = a cos OdO, d (A 0) (a + 6) sin Od9 ; and (4) gives Wa cos = P (a + b) sin 0, which gives the same value of as (1). 13. If the beam rest, as in the last example, against a smooth vertical and a smooth horizontal plane, and a rope be attached firmly to the point (7, and to a point in the beam, find the limit to the position of this latter point consistent with equilibrium. Let fig. 119 represent the beam in any position, and let m be the middle point of the beam. Suppose the rope attached to (7, and to a point, n, in the upper half of the beam. Then the forces acting on the beam are IF, T (the tension of the rope nC), R, and S. Let EXAMPLES. 139 be the point of intersection of W and T. Now, the resultant of W and T must, for equilibrium, be equal and opposite to the re- sultant of R and S ; hence the resultant of It and S must act in the line Op ; but this line is not between the lines of action of T and W, that is, inside the angle WpC ; therefore the resultant of K and S cannot be equal and opposite to that of W and T with such a position of the rope, and, therefore, equili- brium is impossible, no matter _. what the inclination of the beam may be. Hence, in order that equi- librium may be possible, the rope must be attached to some point, such as P, between A and m. 14. In the last example, given the point of attachment of the rope, find the tension in it. It is easy to see that if P, the point of attachment, be given, and also I, the length of the rope, CP, the position of the beam is given. For, if = L BAG, we have sin 2 0, = Pk an equation which determines 0. The angle PC A is also known. Denote it by <. To determine T, the tension of the rope, without bringing R and S into our equation, take moments round 0, their intersection. Hence, a and b being the segments of the beam made by the centre of gravity, we have = T.OCsmOCP= T. (a + 5) sin (0-0), To obtain T by the principle of Virtual Work. Choosing a virtual displacement which keeps A and B in contact with the planes, the equation of work is -Wdy-Td(PC)=0, (1) y denoting the height of G above the horizontal plane. Now PC 2 = BP 2 cos 2 + AP' 2 sin 2 0, and this equation will also hold in the displaced position. Hence we may differentiate it, and then we obtain PC . d (PC) = -(PB*- PA 2 ) sin 6 cos dQ sin<9 _PA ss i?4;rZY3-p PAz. - Also y = a sin 0, therefore dy = a cos 6dB] and substituting these values of dy and d(PC) in (1), we obtain the same value of T as before. 140 APPLICATION OF THE CONDITIONS OF EQUILIBRIUM. [l 14. Note. If = $, T= co. In % this case the rope is attached to m, the middle point of the beam, and therefore its direction always passes through 0, the intersection of R and S. Now, it is easy to see that in this case the conditions of equilibrium are theoretically satis- fied, because the resultant of T and W acts along T, whose direction passes through 0. But if $ > B, no value of T can even theoretically satisfy the conditions (see last example). 15. ABC is any triangle, of which C is the vertex. It is acted on by the forces CA, CB, and AB. Prove that it will be kept in equi- librium by a force equal to 2BC, acting parallel to BC, at the middle point of AB. 16. In example 12, it is clear that two positions of equilibrium of the beam are a vertical and a horizontal position ; explain why these positions are not given by the equation (1) which determines the position of equilibrium. 17. Explain why the proof in example 5 would not hold for a polygon formed of bars freely jointed together and therefore capable of turning about the joints. 114.] Action of a Hinge or Joint. Among the internal forces of a system, the action of a joint is one of frequent occurrence. If the joint be smooth, the re- action between two bars or beams connected by it consists of a single force. For, let PQS (fig. 120) represent a section of the joint connecting two beams : then, since their surfaces are in contact, either throughout the whole of the circumference or a part of it, there will be (since the joint is smooth) normal reactions at the points of contact, since all these pass through the centre of the circle, they have a single resultant. Consequently, the action in this case consists of a single force. But, if the joint be rough, the reactions at the points of con- tact will not be normal, that is, their lines of action will not meet in a point, and, therefore, they may reduce to a couple, or to a single force. When slipping is about to ensue at the joint, it is easy to see that the total resistances at the points of contact envelop a circle (or rather a cylinder). For, at any point, P, of contact (fig. 121), draw PR, making the angle of friction, A, I2I< with the normal, PC, to the surface of con- tact. The perpendicular from (7, the centre of the joint, is equal ,Q P,Q,.... Now, II4-] ACTION OF A HINGE OR JOINT. 141 to PC', sin A, and is, therefore, constant. Hence, PR envelops a circle whose radius = PC. sin A. If PC = a, and ds is the element of the surface of contact at P, it is evident that the sum of the moments of the reactions about C is (R being the reaction per unit of surface) a sin Kf Rds. As an example, let us consider the equilibrium of two equal beams which are connected by a joint, C, and rests on a perfectly smooth cylinder, in a vertical plane at right angles to the axis of the cylinder. Firstly, let the joint be rough, and suppose the contact to be complete all over its surface : then it is clear that such a position as that represented in fig. 122 is a possible position of equilibrium if the joint is suf- ficiently rough. Let fig. 120 represent an enlarged view of the circle which is enveloped by the total resistances at the various points of the surface of w Fig. 122. W S A contact at the hinge, C. Then, if the total resistances at the lower portion of the joint be considerably greater than those at the upper portion, it is possible that the resultant of the whole set may be a horizontal force, R, acting through a point, P, below the joint. In the position of equilibrium of the beams represented in fig. 122 the weight, W t of the beam CD 19 and the normal reaction, S, of the smooth cy- linder, meet in a point A 19 through which point the force produced by the action of the other beam must pass. In the same way the action of the beam CDi on CD 2 must pass through the point A 2 . Hence the resultant action of each beam on the other must be directed in the line A 1 A 2 ; and we have seen that if the contact along the joint extend Fig. 123. 142 APPLICATION OF THE CONDITIONS OF EQUILIBRIUM. [114. over its surface, this is a possible line of action, though it does not intersect the joint. Secondly, let the joint be rough, and let the contact take place at only one point, -^(fig. 124). Suppose the joint to consist of a pin, UN, which forms part of the beam CD 2 (fig. 119), and let this fit loosely into the beam CD, , It is clear, then, that the action between the beams consists of a single force, J2, acting at N, and making the angle of friction, A, with the radius CN, if slipping is Fig. 124. about to take place. As before, this force must pass through the points A lt A 2 . In this case, then, the point of contact of the beams is con- structed by drawing a radius, CN, of the cylindrical axis consti- tuting the joint, inclined to the horizon (since A l A. 2 is horizontal) at the angle of friction. Thirdly, let the joint be smooth. In this case the beams must assume such a position that the line A 1 A 2 passes through the centre of the joint ; and this position is practically the same as that in the last case, because since the dimensions of the joint are negligible compared with those of the beams, the line of resist- ance RN (fig. 1 24) may be supposed to pass through the centre, C, of the joint. A similiar explanation is to be given in the case of two equal beams rigidly connected, and form- ing one piece, the system resting, as in the previous example, on a smooth cylinder. In this case the beams can take only one position, which must be a position of equili- brium, and the action between them must accommodate itself to the geometrical necessity of the figure. (In the following figure the cylinder is not drawn.) If we consider the Fig. 125. equilibrium of one of the beams, CD (fig. 125), by itself, we shall have to supply to it whatever force is actually produced upon it by the other beam. Now, if C is the section along which the 1 15.] GEOMETR1CO-STATICAL PEOBLEMS. 143 system is considered as divided by the removal of the second beam, it is clear that the internal forces in the neighbourhood of B tend to tear the beams apart, if A is below the section C, while those about C tend to press the beams more closely together. Hence the action of the second beam on CD consists of a number of forces whose horizontal components near B act from left to right, as the force JBF, and whose horizontal com- ponents near C act from right to left, as the force CF'. If, therefore, the forces near B are greater than those near C, the resultant of the whole system will consist of a horizontal force, AR, acting outside the section CB> so as to pass through the point, A) of intersection of the weight and the normal reaction of the cylinder. In this case, then, the action, over a section BC, between two rigidly connected pieces consists of a force outside the section ; which force may, of course, be replaced by one at any point in the section, together with an accompanying couple (see Art. 74). In all cases in which contact over a finite surface takes place between two bodies, the student must be careful to examine the nature of the forces exerted between them at the individual points of contact with a view to ascertaining whether the resultant action of one on the other consists of a single force at all ; or, if so, whether it can be assumed to act at any point in the surface of contact or must be assumed to act wholly outside it. 115.] Geometrico - statical Problems. In many statical problems which relate to the positions of equilibrium of bodies the result is independent of the magnitude of some given force, and such independence can be perceived a priori. Thus, suppose the question to be What is the limiting inclination to the horizon of a heavy uniform beam which rests against a rough vertical and a rough horizontal plane ? In this problem we may, if we please, assume W> the weight of the beam, and 2a, its length ; but it is evident a priori that the result cannot involve either of these quantities. For, if the angle which the beam makes with the ground be 0, the position of equilibrium will be denned by some of the trigonometrical functions of 0, such as sin 6 or tan 6. Now, the trigonometrical function of an angle are mere numbers, or ratios of quantities of the same kind. Hence, if the expression for tan 6 (suppose) involve force, it must involve the ratio of one force to another force, and if there is only one 144 APPLICATION OF THE CONDITIONS OF EQUILIBRIUM. [ll6. force given in the problem, we have no other force to combine with it in the form of a ratio or a mere number. Consequently, the weight of the beam can in no way influence its limiting in- clination. Precisely similiar remarks hold with regard to the only linear magnitude in the question, viz., the length of the beam. There is no other quantity of the same kind with which to compare it. Therefore, we are enabled to state a priori that the inclination of the beam to the horizon in its limiting position of equilibrium depends simply on the coefficients of friction for the beam and the two rough planes, or that 0=/0*,M'). H and // being these coefficients, and f denoting some (as yet) unknown function. Again, suppose the question to be What force applied to one of the handles of a table drawer will pull the drawer out ? * It is evident that the answer must be either no force, however great, will pull it out, or any force, however small, will pull it out. And the result will depend simply upon the relation between the coefficient of friction for the drawer and the table, and the ratio of the side of the drawer to the distance between the handles. This is evident, because there is no given force in terms of which the required force could be expressed. Numerous examples of this class of questions will be given in the sequel. Such problems, then, in which the result is in- dependent of a force magnitude, we shall classify as Geometrico- statical Problems, because, though they involve conceptions concerning the directions of forces, they do not involve their magnitudes. In all such problems, once the requisite theorems concerning the directions of forces are made use of, the result follows at once from the geometry of the figure ; and a solution by the method of resolving forces and taking moments is, in reality, an illogical process. 116.] Useful Trigonometrical Theorem. In connexion with the class of geometrico-statical problems, the following theorem in Plane Trigonometry will be found extremely useful : If a right line, CP (fig. 126), drawn from the vertex of a triangle, divide the base into two segments m and n, or segments which are to each other in the ratio of m to n t * The friction of the bottom is neglected. IT 6.] EXAMPLES. (m + n) cotO = mcotan cotjS, a and ft being the angles which CP makes with the sides AC and HC, and 6 the angle which CP makes with the base. For, if AP = m, and BP = n, Fig. 126. Also, sm A sm (0 a) , . . ., = m . = m 7 = m (sm 6 cot a cos 0). sin a sin a CP = sm sin (3 = n (sin Scot ,3 + eos 6). Hence ^ (sin e cot a cos 0) = n (sin cot ft + cos 0), from which (l) follows at once. We have also the equation (m + n) cot 6 = ncotAm cot J9. For, sin A sin A m (2) CP = sin a sin (0A) sin cot A cos Similarly, CP = : - j-. ; sinOcot-B + cosB ' therefore, &c. Q. E. D. EXAMPLES. 1. A heavy beam rests on two smooth inclined planes whose inter- section is a horizontal line, the beam lying in a vertical plane perpendicular to this line of intersection ; find the position of equilibrium and the pres- sures on the planes. Let a and b be the segments, ACr and S6f, of the beam, made by its centre of gravity, G ; 6 the inclination of the beam to the horizon, a and /3 the inclinations of the planes, R and R' the pressures on these planes, re- spectively, and W the weight of the beam. Fig. 127. Then, since the beam is in equilibrium under the action of only three forces, they must meet in a point, 0. L 146 APPLICATION OF THE CONDITIONS OF EQUILIBRIUM. [116. Now the angles GO A and GOB are equal to a and /3, respectively, and BGO = ~^-0. Hence (a + 6) cot BGO a cot GO A -b cot GOB, or (a + b) tan0 = a cot a b cot/3, (1) which determines the position of equilibrium. Again, by the relations between three forces in equilibrium, (2) sm(a sina Hence, if T = - 3? the beam will rest in a horizontal position. b tan ^3 Suppose that a cot a &cot/3 is positive, and that (a + 6) tan /3 < a cot a 6 cot /3. Then, ct fortiori (a + b) tan < a cot ab cot /3, since 0, the angle made with the horizon by the beam in any such position as AB, is necessarily a + b, it will be impossible ; that is, if a - - ' >a + b ; or a tan /3 (cot a tan (3) > b ; or a cot a b cot /3 > a tan /3. But, by supposition a cot a b cot /3 is positive and > (a + b) tan fi, therefore AP>AB, which is manifestly impossible. Hence the only position of equilibrium in this case is one of continuous contact with the plane (a). [We have supposed all through that the end A of the beam is to rest on the plane (a).] The least inclination of the plane (a) which will allow of a position of continuous contact with (/3) is found by drawing at B a perpendicular to the plane (/3) and joining its point of intersection with the vertical through G with A. The joining line is the normal to the plane of least inclination (a). 1 1 6.] EXAMPLES. 147 2. A uniform heavy beam, AB (fig. 129), rests with one extremity, A, against the internal surface of a smooth fixed hemispherical bowl, while it is supported at some point in its length by the rim of the bowl; find the position of equilibrium. It is a priori evident that the result must be independent of force, since the weight of the beam is the only force that may be supposed to be given ; and it is also evident that the result depends on the only two linear Fig. 129. magnitudes which may be supposed to be given viz., the length of the beam, 2 a, and the radius, r } of the bowl. Draw the three forces which keep the beam in equilibrium. They are the weight, a reaction at A perpendicular to the surface of contact, and therefore perpendicular to the bowl, and a reaction at C which for the same reason is perpendicular to the beam. These must meet in a point, 0. Let 6 = the inclination of the beam to the horizon = LACD. Let the line OG meet the semicircle DAC in the point Q. Then AQ is a horizontal line. Also LQAG = LDGA = 6, therefore LOAQ =26. Hence AQ = AO cos 20, and also AQ = AQ cos 0', therefore 2 r cos 2 6 = a cos 0, or 4r cos 2 acosO 2r = 0. This equation gives two values of cos 0, one of which supposes the hemisphere to be completed into a sphere, the end A of the beam to rest against the upper portion of the sphere, and the action of the sphere on A to consist of a pull. The student will have no difficulty in representing this position, or in proving that the reaction at a .-yy.yvJL ot.'l^ZoC *'**-( the extremity C of the string being fastened to the point occupied by B when the door is horizontal. Given the length of the string, find the magnitude and direction of the pressure on the hinge line, and the tension of the string. Produce the line of the string to meet the line of action of the weight in a point, 0. Then, since the door is in equilibrium under the influence of only three forces, they must meet in a point. Hence the pressure on the hinge-line must pass through 0, and since the plane of the tension, T, and the weight, TF, intersects the hinge-line at A, the pressure, J?, must act through A (the hinge being smooth). Fig. 131. EXAMPLES. 149 To determine T take moments about A . Then, if p = the per- pendicular from A on BC, ^.p = W .AD. (1) Let the angle SAO = 2 a, and let AB = 2 a. Then p = 2 a cos a, AD = a cos 2 a, therefore cos 2 a ~ * cos a Again, by the triangle of forces we have and substituting the above value of T, this gives E = J W A/ 4 sin 2 a + sec 2 a. The values of T and R can be at once found in terms of the lengths AB and BC. Denoting the latter by 21, we have sin a = 5 there- fore, &c. 6. If in the last example the string, instead of being attached to (7, pass over a smooth pulley at that point, and sustain a given weight, find the position of equilibrium, and the pressure on the hinge-line. Let P be the suspended weight, and 6 = Z.OA B ; then the position of equilibrium is defined by the equation B 1 o-o = 0> (!) 2 P COS"* and 6 (2) Equation (1) gives two positions of equilibrium, and since it shows /i that one of the values of cos - is negative, one position corresponds to a value of 6 greater than 180. Such a position, of course, supposes the door capable of revolving freely about its hinge-line through four right angles. The student will have no difficulty in representing the position cf the door in this case, or in explaining why no linear magnitude enters into the equations. 7. A uniform heavy beam, AB, rests against a smooth peg, P, and against a smooth vertical wall, AD ; find the position of equilibrium and the pressures on the wall and peg. This, so far as it relates simply to the position of equilibrium, is another geo- metrico-statical problem. We have merely to draw AB in such a manner that the vertical through G and the perpendiculars at A and P to the wall and beam shall intersect in a common point, 0. w Fig. 133. Let 2 a = the length of the beam, and c = the perpendicular distance 150 APPLICATION OF THE CONDITIONS OF EQUILIBRIUM. [116. of the peg from the wall. Then the position must evidently be expressed as a function of - Let = the inclination of the beam to the vertical. a Then AP = > and AO = -7^-3 But AO = AG.ain ; therefore sin sin v CO / _- = a sin 0, sm 2 ' sin0 = (^ Resolving vertically, S. sin = W, Resolving horizontally, S cos = R, T> TT7" Ji rr r (1) (2) (3) 8. A triangular board, BC A (fig. 134), of uniform thickness, rests on two smooth pegs, P and Q, at a given distance from each other, in the same horizontal line. Find A its position of equilibrium. The position of equilibrium will evidently be known if the inclination of AB to the horizon is known. Let this inclination be ; let the angles of the triangle be denoted by A, B, C ; let a = LAMC, which the bisector, CM, of the base makes with the base ; let CM = I, and let PQ = k. Then, since no force is given except the weight of the board, will depend simply on A, B, C, I, and k, and the problem is geometrical. The reactions of the pegs P and Q are perpendicular to A C and BC, respectively, and they must meet the weight of the board actmg through its centre of gravity, G, in a point 0. The geometry which gives the solution will express that slOoa CO. sin COV= therefore or 4sin<9sin-=cos(--0), n ^n * tan = - cot - 3 n This equation determines the position of equilibrium. The pressure at A is evidently equal to cot - > W being the weight of the board. rt_ The external angle of the polygon being equal to > the incli- nations of the successive sides to the vertical are n n n and if p m be the perpendicular distance of the wall, counting B as the first, we have p m = a or n mir vertex from the (m I)TI sin Pm = n . m 2 mir. ( 2COS - 77 COS - ) v ' sm- 27T ' fcfi 11. A heavy plane body, ABO (fig. 138), of any shape, is suspended from a smooth peg, fixed in a vertical wall, by means of a string of given length, the extremities of which are attached to two fixed vU-. e-arj 9 4. n6.] EXAMPLES. 153 points, F and librium. in the body. Determine the positions of equi- Let the ellipse P l P 2 P 3 be described with foci F and JF", and axis major equal to the length of the string, where on this ellipse, suppose at P, pended from the peg, it is kept in equilibrium by its own weight acting vertically through the centre of gravity, and the two tensions in P 2 F and P 2 F". But since the peg is smooth, these tensions are equal, and their resultant must bisect the angle FP 2 F; its line of action is, therefore, normal to the ellipse. And if G is the centre of gravity of the body, the resultant tension The peg will then be some- Now, when the body is sus- 138- must pass through G, and be equal and opposite to the weight of the body. Hence the problem is solved by drawing normals from G to the ellipse, and then hanging the figure from the peg in such a manner that any one of these normals is vertical. Now, if G is inside the evolute, four normals can be drawn to the ellipse ; but it is easy to see that only three are relevant to the solution if G is inside the lower half of the evolute (as in fig. 135), or only one if G is inside the upper half. For the tangents drawn to the lower half of the evolute belong to the upper half of the ellipse ; and in order that the strings should be stretched, it is necessary that the peg should lie somewhere in the upper half of the ellipse. If GP 1} GP 2 , and GP 3 , are the position in normals drawn from G, the figure must be placed in a which any one of these lines is vertical. 12. A beam, whose centre of gravity divides it into two segments &~ + a and 6, is placed inside a smooth sphere ; find the position of equi- vo^/3 librium. Ans. Let 6 be the inclination of the beam to the horizon, and 2 a the angle subtended by the beam at the centre of the sphere ; then a b tan 6 = =- tan a. a + b 13. A heavy carriage wheel is to be dragged over an obstacle on a horizontal plane by a horizontal force applied to the centre of the wheel ; find the magnitude of the required force. Ans. Let W be the weight and r the radius of the wheel, h the height of the obstacle, and F the requisite force: then F=W h 14. If it be attempted to drag the wheel over a smooth obstacle by means of a force whose line of action does not pass through the centre, what happens? Is the result in last example modified if there is friction between the wheel and the obstacle 1 154 APPLICATION OF THE CONDITIONS OP EQUILIBRIUM. [116, 15. A heavy uniform beam, moveable in a vertical plane about a smooth hinge fixed at one extremity, is to be sustained in a given position by means of a rope attached to the other extremity ; find, geometrically, the least value of the pressure on the hinge, and the corresponding direction of the rope. Ans. The least pressure on the hinge = JTFsina, W being the weight of the beam and a its inclination to the vertical. Also if 6 is the angle made by the rope with the vertical when the pressure is least > cot = 2 cot a + tan a. 16. A vertical post, loosely fitted into the ground, is exposed to a uniform gale of wind ; a rope of given length is to be attached to the post and to the ground ; find how the attachment is to be made, in order that the rope may be least likely to break. Ans. If n is the height of the post and if the length of the rope is < h Ti -s/2, the rope must be attached to the top of the post. (See example 4.) 17. A heavy beam rests with one extremity placed at the line of intersection of a smooth horizontal and a smooth inclined plane, the other extremity being attached to a rope which, passing over a smooth pulley at a given point in the inclined plane, sustains a given weight ; find the position of equilibrium. Ans, Let 6 be the inclination of the beam, a the inclination of the plane, and the inclination of the rope, to the horizon ; a the distance of the centre of gravity of beam, b the distance of the pulley, from the line of intersection of the planes ; and I the length of the beam. Then the position of equilibrium is defined by the equations Wa cos 6 = Pb sin (a <), b sin (a <) = Z sin (6 + <). 18. A heavy uniform beam, AB, rests with one end, B, against a smooth inclined plane, while the other end, A, is connected with a rope which passes over a pulley and supports a given weight ; find the position of equilibrium. Ans. If a, 0, and <, are the inclinations of the plane, beam, and rope to the horizon, W and P the weight of the beam and the suspended weight, respectively, the position of equilibrium is defined by the equations . x , . Pcos ( a) = TFsm a, 2 tan 6 = tan < cot a. The student will easily explain why no linear magnitude enters into the result. 19. A rectangular board, A BCD, of uniform thickness, is moveable in a vertical plane about a smooth hinge, P, in the side AD ; the side AB is to rest, at a given inclination to the horizon, against a smooth peg, Q : find the position of this peg when the pressure on the hinge is equal to the weight of the board. 1 1 6.] EXAMPLES. 155 Ans. Let be the point of meeting of the forces which keep the board in equilibrium, and the centre of gravity of the board. Then QO must bisect the angle POG. Hence from P draw a line, PO, making the same angle with the side AB as AB makes with the vertical ; and from the point, 0, of intersection of this line with the vertical through G draw a perpendicular, OQ, on AS. This deter- mines Q. 20. A heavy body of any form is moveable round a smooth axis perpendicular to the vertical plane passing through the centre of gravity, and is sustained in a given position by a rope whose weight may be neglected. If the pressure on the axis bears a constant ratio to the weight of the body, prove that the direction of the rope must be a tangent to a conic whose directrix is the vertical line through the centre of gravity, and focus the point in which the axis of suspension cuts the above-mentioned vertical plane. If, in the last example, QO be the direction of the rope, the ratio 7^7777 is given, and the envelope of QO, as the direction PO varies, is a conic whose focus is P, directrix GO, and eccentricity the given ratio. 21. In example 19, if the hinge is at the corner A, and the position of the peg is given, find the magnitude of the pressure on the hinge. Ans. Let c half the length of the diagonal, a angle between the diagonal and the side AB, x = the distance of peg from A, ft = inclination of AB to the vertical ; then the pressure on the hinge is W 2cx sin/3 sin (a + /3) + c 2 sin 2 (a + ff) x 22. In the last example, find the position of the peg when the pressure on the hinge is a minimum, and the minimum value. Ans. At the point in AB vertically under the centre of gravity of the board. The minimum pressure = TFcos /3. 23. A rectangular board of uniform thickness rests in a vertical plane, with two of its adjacent sides in contact with two smooth pegs in the same horizontal line ; find the position of equilibrium. Ans. If P and Q (see fig. 134) be the two pegs, CA and CB the sides in contact with P and Q, re- spectively, a the angle made by the diagonal CD with CB, 9 the inclination of this diagonal to the horizon, c half the length of the diagonal, and I the distance PQ, the position of equilibrium is given by the equation Fig. 139- 156 APPLICATION OF THE CONDITIONS OF EQUILIBRIUM. [ll6. 24. A triangular board, ABC (fig. 139), of uniform thickness, is placed with its base on a smooth inclined plane, its vertex being con- nected with a string which passes over a smooth pulley and sustains a weight. Find the conditions of equilibrium. Ans. Assuming the inclination of the plane to be fixed, the string must take such a direction that the perpendicular let fall on the plane from the point of intersection of the string with the vertical line, Gfm through the centre of gravity of the board, falls inside the base. Hence, if Bp be the perpendicular at the extreme point of the base, and if the string cannot cross the surface of the board, all possible directions of the string are included between Cm and Cp. Again, supposing the string to have a direction, On, consistent with the possibility of equilibrium, the weight P and the reaction of the plane are thus found : From n let fall a perpendicular on AB, meeting it in a point, q, suppose. Then qn is the line of action of the reaction on the plane : and, resolving along the plane, we have W sin i = P cos 0, i being the inclination of the plane, and the angle which the string Cn makes with the plane. This equation determines the magnitude of P corresponding to the direction, Cn, of the string. If P is a little greater than the value thus found, the board will begin to slip up, and if P is less than this value, the board will begin to slip down the plane. 25. If, in the last example, the string is parallel to the plane, find the greatest inclination of the plane consistent with equilibrium. Ans. Tan-^i cot A + cot B). 26. If, in the same example, the string, instead of passing over a pulley and sustaining a weight, is knotted to a fixed peg, how are the previous conditions of equilibrium modified 1 Ans. The only condition to be satisfied is that which has reference to the direction of the string. This direction must be somewhere between Cm and Cp. 27. A rectangular board is sustained on a smooth inclined plane by a string attached to its upper corner ; the string passes over a smooth pulley and sustains a weight. Find the magnitude of this weight corresponding to a given direction of the string, and find also the pressure on the plane. Ans. Let i be the inclination of the plane, the angle made by the string with the plane, W the weight of the board, P the suspended weight, and R the pressure ; then COS cos 28. Show that a rectangular board cannot be sustained on a smooth Tl6.] EXAMPLES. 157 inclined plane by a string attached to its upper corner, if the in- clination of the plane is greater than the angle made by the diagonal of the board with one of the sides perpendicular to the plane. 29. If a rectangular picture be hung from a smooth peg by means of a string, of length 2 a, attached to two points symmetrically placed at a distance 2c from each other on the upper side of the frame, show that the only position of equilibrium is one in which this side is horizontal if the adjacent side of the frame is greater than 2c 2 V^^?' 30. A rod whose centre of gravity is not its middle point is hung from a smooth peg by means of a string attached to its extremities ; find the positions of equilibrium. Ans. There are two positions in which the rod hangs vertically, and there is a third thus defined : Let F be the extremity of the rod remote from the centre of gravity, k the distance of the centre of gravity from the middle point of the rod, 2 a the length of the string, and 2 c the length of the rod j then measure on the string a length FP from F equal to a ( 1 + ) > and place the point P over the peg. This will define a third position of equilibrium. 31. A smooth hemisphere is fixed on a horizontal plane, with its convex side turned upwards arid its base lying in the plane. A heavy uniform beam, AS, rests against the hemisphere, its extremity A being just out of contact with the horizontal plane. Supposing that A is attached to a rope which, passing over a smooth pulley placed vertically over the centre of the hemisphere, sustains a weight, find the position of equilibrium of the beam, and the requisite magnitude of the suspended weight. Ans. Let W be the weight of the beam, 2 a its length, P the suspended weight, r the radius of the hemisphere, h the height of the pulley above the plane, 6 and (f) the inclinations of the beam and rope to the horizon ; then the position of equilibrium is defined by the equations r cosec = h cot , (1) r cosec 2 = a (tan $ + cot 0), (2) which give the single equation for 0, r ( r _ a s in 6 cos 6) = ah sin 3 0. ( 3) (4) -, COB ($6) r z 32. If, in the last example, the position and magnitude of the beam be given, find the locus of the pulley. Ans. A right line joining A to the point of intersection of the reaction of the hemisphere and W. 158 APPLICATION OF THE CONDITIONS Or EQUILIBRIUM. [116. 33. If, in the same example, the extremity, A, of the beam rest against the plane, state how the nature of the problem is modified, and find the position of equilibrium. Ans. The suspended weight must be given, instead of being a result of calculation. Equation (1) still holds, but not (2); and the position of equilibrium is defined by the equation 34. If the fixed hemisphere be replaced by a fixed sphere or cylinder resting on the plane, and the extremity of the beam rest on the ground, find the position of equilibrium. Ans. If h denote the vertical height of the pulley above the point of contact of the sphere or cylinder with the plane, we have /i r cot - = h cot (/>, /i Pr ( 1 + cot -cot 6) cos < = Wa cos 0. 2 35. A heavy regular polygon of any number of sides is attached to a smooth vertical wall by a string which is fastened to the middle point of one of its sides ; the plane of the polygon is vertical and perpendicular to the wall, and one end of the side to which the string is attached rests against the wall. For a given position of the poly- gon, find the requisite direction of the string, and show that in all positions of equilibrium the tension of the string and the pressure on the wall are constant. Ans. Let A be the vertex of the polygon in contact with the wall, G the centre of gravity, the point in which the weight and the reaction of the wall meet, and M the middle point of the side to which the string is attached. Then the direction of the string is OM, and, the quadrilateral GO MA being inscribable in a circle, the angle between the string and the vertical is constant and equal to half the angle of the polygon. 36. A square board rests with one corner against a smooth vertical wall, the adjacent corner being attached to the wall by a string whose length is equal to the side of the board ; prove geometrically that the distances of the corners from the wall are proportional to 1, 3, and 4. 37. One end, A, of a heavy uniform beam rests against a smooth horizontal plane, and the other end, B, rests against a smooth inclined plane; a rope attached to B passes over a smooth pulley situated in the inclined plane, and sustains a given weight ; find the position of equilibrium. Let 6 be the inclination of the beam to the horizon, a the in- clination of the inclined plane, W the weight of the beam, and P the suspended weight ; then the position of equilibrium is defined by the equation cos0(TFsma-2P) = 0. (1) II 6.] EXAMPLES. 159 Hence we draw two conclusions : (a) If the given quantities satisfy the equation W sin a 2 P = 0, the beam will rest in all positions. (6) There is one position of equilibrium, namely, that in which the beam is vertical. This position requires that both planes be conceived as prolonged through their line of intersection. 38. Discuss the second position of equilibrium in the last example, and show that its possibility will depend on the length of the beam, and also on the inequality W > or < P cosec a. (N. B. In accounting for this position, the impossible supposition that the reaction of a plane can consist of a pull must be rejected.) 39. A uniform beam, AB, moveable in a vertical plane about a smooth horizontal axis fixed at one extremity, A, is attached by means of a rope BC, whose weight is negligible, to a fixed point C in the horizontal line through A, such that AB = AC ; find the pressure on the axis. Ans. If = Z.CAB, W= weight of beam, the reaction is CHAPTER VIII. EQUILIBRIUM OF A SYSTEM OF SMOOTH BODIES UNDER THE ACTION OF FORCES IN ONE PLANE. 117.] Action and Reaction. If in any system of bodies, connected in any manner, A and B are two bodies in contact between which an action of some kind is exercised; then, what- ever be the forces with which the body A acts upon the body B, the very same forces, reversed in directions, will constitute the action of B on A. Let the whole system of forces acting on A, excluding those produced by B> be denoted by (P), and let the forces constituting the action of B on A be denoted by (R) ; then we may sever the connexion between A and B, provided that we have other means of producing on A the system of forces (R). In the same way, if (Q) denote the whole system of forces acting on B, those constituting the action of A on it being excluded, the body B may be severed from A provided that we have the means of producing a system of forces ( R) on B, (R) denoting a system of forces obtained by reversing the direction and preserving the magnitude of every force in (R). For example, the beam CD (fig. 125) may be severed from the other beam along any section, CB, provided that there be introduced on CD either the single force R acting through A t or the complex system of tensile and compressive forces which act at the section CB. This equality of magnitude and oppo- siteness of direction of the forces existing between two distinct bodies in contact, or between ideally severed portions of the same body, is sometimes spoken of as the principle of the equality of Action and Reaction ; but it cannot be too strongly impressed on the student that it is by no means the whole of the Newtonian principle called by this name; for Newton specifies several senses in which the terms Action and Reaction can be taken, and in discussing one of them he has explicitly anticipated, in great part, the principle of the Con- servation of Energy as has been pointed out by Thomson and Tait. Il8.] EXAMPLES OF INTERNAL ACTION. 161 118.] Examples of Internal Action. The cases which we shall consider in this chapter are those in which the action between two portions of a system ideally severed consists of a single force. The simplest example of such action occurs when a single point of one body rests against the surface of another, the bodies being either rough or smooth. If the bodies are smooth, the action between them consists of a single force which is normal to the surface of contact (see p. 40) ; and if rough, the action is still a single force which is not necessarily normal to this surface. In all cases in which smooth spherical joints or hinges are concerned, the action exercised on bodies connected by them consists of a single force passing through the centre of the joint. When rough joints are used, the action will generally consist of a single force acting somewhere outside the joint; or of a force and a couple acting at the joint; or, possibly, of a couple alone. The tension of a string is also an instance of internal action, and its nature has been already explained in Chapter II. Again, if we ideally separate into two portions, by an arbi- trary surface, a mass of a perfect fluid in equilibrium, the action of one portion on the other over a small area of the ideally separating surface will consist of a single force acting normally on the area. And we may always treat as a separate body any portion whatever of a fluid in equilibrium*, provided that we produce along the surface of this ideally separated portion all the forces which are actually produced on it ly the fluid with which it was surrounded. It is by such separate consideration of portions of a fluid that we arrive at a knowledge of its internal forces or pressures. For example, if a heavy fluid, whether compressible or incompressible, of uniform or varying density, be contained in a vessel, we can prove that the pressure is the same at all points, P, Q, in the same horizontal plane. For, isolate in imagination a horizontal cylindrical column of the fluid, having small vertical and equal areas at P and Q for extremities, from the rest of the fluid. Then, we may treat the cylinder of fluid PQ as a sepa- rate body, provided that, in addition to the external force (gravity) acting on it, we introduce the forces which it actually * It is usually said that we may, under the above condition, imagine any portion of the fluid to become solidified ; but this imagined solidification is not only wholly unnecessary but misleading to the student. M 162 EQUILIBRIUM OF A SYSTEM OF SMOOTH BODIES. [u8. experienced from the surrounding fluid. Now these forces consist of normal pressures, p and g, on the areas at P and Q, together with normal pressures all over its curved surface, these latter being all at right angles to the axis PQ. If now we resolve horizontally all the forces acting on the cylinder, we get p q = 0, or p = q. This demonstration shows, moreover, that in the case of a heavy viscous or imperfect fluid, the pressures are not necessarily equal at all points in the same horizontal plane. For, in this case, the action of the rest of the fluid on PQ does not necessarily consist of forces normal to its surface, but of oblique forces. Hence the horizontal component of the pressure at P is not equal to the horizontal component at Q ; the difference between them is equal to the sum of the hori- zontal components of the oblique forces. The importance of keeping such considerations in view may be illustrated by the following example from Hydrostatics. A conical vessel is filled with water through an aperture at the vertex. From Hydrostatical principles it follows that the pressure on the base of the cone is equal to the weight of a cylindrical column of water, standing on the base, and having a height equal to that of the cone ; that is, the pressure on the base is much greater than the weight of the water contained in the cone. Now if we imagine the water to become solidified., the curved surface of the cone may be removed, and the pressure on the base will be equal to the weight of the ice, that is, the weight of the water in the cone. An apparent discrepancy is the result. But if we attend to the proviso that in the separate consideration of the equilibrium of any portion of a system, solid or fluid, we must produce upon the isolated portion all the forces which were originally produced upon it by the neighbouring portions of the solid or fluid, the difficulty disappears. In the fluid state the liquid in contact with the curved surface of the cone was pressed normally by a system of varying forces, and the circumstances of the solidified body will not be the same as those of the fluid, unless its surface is pressed in precisely the same way. These pressures have a total vertical component, which must be added to the weight of the block of ice in order that we may obtain the true pressure on the base. The action between two portions of a perfect fluid ideally Up.] EXAMPLES. 163 separated by a plane surface of any area always consists of a single force which is normal to the area ; but the action between two portions of an elastic solid along a plane section is by no means so simple; this latter is not generally reducible to a single force. 119.] Equilibrium of Several Bodies Forming a System. It will now be clear that when a system is composed of several bodies in contact with each other, we can consider the whole set as forming a single body in equilibrium under the action of given external forces; or we may consider the separate equi- librium of any one body under the action of given external forces, and the reactions of the other bodies with which it is in contact. A few examples of such systems have already been given ; but it is proposed to devote the present chapter more especially to the consideration of such questions. EXAMPLES. 1. Two uniform beams, connected at a common extremity by a smooth joint, are placed in a vertical plane, their other extremities, which rest on a smooth horizontal plane, being connected by a light rope ; find the tension of the rope and the reaction at the joint. Let AC and CB (fig. 140) be the beams, W and W their weights, a and of their inclinations to the horizon, R and Sf the reactions of the horizontal plane at A and B, and T the tension of the rope. If, then, we consider the two beams v * as forming one system, the mutual reaction at and the tension of the rope will be internal forces of the system, and will therefore disappear from, the equations of equilibrium. , , ' The forces on this system are simply W, W, R and R'. .Resolving vertically for the equilibrium of the system, R + K= W+W. (1) Again, considering the equilibrium of the beam AC, the forces acting on it are W, R, T, and the unknown reaction at C. This latter will be eliminated by taking moments about C. Thus we get 272cosa = 2^sina+Tf cos a, (the length of the beam dividing out), or R=Titma + \W. (2) Similarly, taking moments about C for the equilibrium of BC, JB'srZ'tana' + .JTr. (3) M 2, 164 EQUILIBRIUM OF A SYSTEM OF SMOOTH BODIES. [119. By adding (2) and (3), and making use of (1), we get T= (4) 2 (tan a + tan a') Again, let X and T be the horizontal and vertical components of the reaction at the joint. Then, for the equilibrium of the beam AC, Hence W+ YR = 0. W+W ~ 2 (tan a + tan of) ' TTtana-TTtana' and d (-4J5) in (5), we get the same 2 (tan a + tan a") If we wish to determine T by the principle of virtual work, let y be the height of the middle point of either beam, and we have _(JF+ W')dy-Td(AB) = (5) for an imagined displacement in which the beams are drawn out, while A and B remain on the ground. If A G = 2 a, BG = 2 a', y = a sin a, ^5 = 2 a cos a + 2 of cos a'. Therefore dy = acosada, d(AB) = 2a sinada 2a' sina'da' sin (a + a') . / . / = 2 a ? - da (since from the equation a sin a = a sma we cos a have a cos ada = a' cos a'da'). Substituting these values of value of T as before. 2. Two equal smooth spheres are placed inside a hollow cylinder, open at both ends, which rests on a horizontal plane j find the least weight of the cylinder in order that it may not be upset. Let figure 141 represent a vertical section of the system through the centres of the spheres. Let P be the weight of the cylinder, a its radius, W and r the weight and radius of each sphere, R and R' the reactions between the cylinder and the spheres whose centres are and 0', respectively. Then, the only motion possible for the cylinder is one of tilting over its edge at the point A, in which the vertical plane containing the forces meets it. For, con- sider the equilibrium of the lower sphere which rests against the ground at D. This sphere is in equilibrium under the influence of ff (reversed in figure), the reaction of the upper sphere, S, acting in the line 00', its weight, TF, and the reaction of the ground at D. Now, since three of these forces pass through 0', the reaction of the ground, whether the latter is rough or smooth, must also pass through (/. Hence, if 9 be the angle which 00' makes with the horizon, we have for the equi- librium of the lower sphere, resolving horizontally, (i) Fig. 1-41. EXAMPLES. 165 The upper sphere is in equilibrium under the action of R (reversed in figure), W, and S. Hence for its equilibrium we have in the same way, .# = #0080; (2) /. R = R'. (3) Again, the cylinder is in equilibrium under the action of R, R', P, and the reaction of the ground. Resolving horizontally for its equi- librium, we have the horizontal component of the reaction of the ground = R Rf = 0. Hence, even if the ground is rough, there is no tendency to slip, and the only way in which equilibrium can be broken is by turning round A. Taking moments, then, about A, the point at which the reaction of the ground acts, we have for the equilibrium of the cylinder (4) (5) (6) or ra - Again, for the equilibrium of the upper sphere, we have =**. -fu~.+-w Substituting this value of R in (4), we have Pa = 2 Wr cos 0. cos# = ar But evidently therefore, finally, t* 3. A heavy beam is moveable in a vertical plane round a smooth hinge fixed at one extremity; a heavy sphere is attached to the hinge by a string ; the two bodies rest in contact ; find the position of equi- librium and the internal reactions, there being no friction between the bodies. Let (fig. 142) be the hinge, OA the string by which the sphere is attached, the inclination of the string to the vertical, Cm, ; the geometrical equation is sin COB = - l + r or (2) (1) and (2) determine 6 and <, and therefore the position of equili- brium. If R is the mutual reaction of the sphere and the beam, we have, by considering the equilibrium of the sphere alone, Again, if the string is attached to the hinge but not to the beam, and if X and T are the horizontal and vertical components of the pressure of the beam on the hinge, we have for the equilibrium of the beam Hence, if S is the resultant of X and F, sin sin COS cos (0 + 0) cos 2 (4) Evidently acts in the line OD, which joins the hinge to the point of intersection of P and R. If the string is attached to the beam, X and T are the components of the resultant of the tension of the string and the pressure on the hinge. 4. Two heavy uniform rods are freely jointed at a common ex- tremity, and are connected at their other extremities with two smooth hinges in the same hori- zontal line. Kequired the mag- nitudes and directions of the pressures on the hinges, and the mutual reaction between the rods. Let AC and CB (fig. 143) be the rods; TFand IF 7 their weights, -y^ _,. acting through their middle points, / and g ; a and a their inclina- tions to the horizon ; R the mutual reaction at C] S and S' the 1 19.] EXAMPLES. 167 pressures on the hinges A and B, G the centre of gravity of the system of two rods ; and 6 the inclination of R to the horizon. Consider the equilibrium of AC alone. It is acted on by three forces W, R, and S; and since we have drawn the line OC to represent the direction of R, the direction of S must be Aq, q being the point of intersection of W and R. By taking moments about A for the equili- brium of AC, we shall express R in terms of TF, a, and 6 ; and by taking moments about B for the equilibrium of BC, we shall express R in terms of W, a?, and 9 ; equating the two values of R thus obtained, we get a value for tan 6 which is obtained by dividing the value of Y by that of X in example 1 . Considering the two rods as one system, this system is acted on by the three external forces, S, &, and W+ W acting vertically through G. Hence these must meet in a point, Q. It is evident that this problem is the same as that in example 1, and that if the reactions -S and S' are resolved each into a vertical and a horizontal component, the horizontal components will be equal and opposite (by considering the two rods as one body and resolving horizontally). These horizontal components have each the value of the tension of the rope in example 1, and the vertical components are the values of R and R '. Thus the problem might be completely solved analytically. Geometrical Solution*, The direction of the resistance at the joint C can be easily determined as follows : from A and B draw two lines to any point, Z>, on the line QG ; let AD meet qf\n E, and let BD meet rg in H. Then the line EH will meet AB in 0, the point through which the line of resistance at C passes. For, the triangles qrQ and EHD are such that the lines, Eq t DQ, Hr, joining corresponding vertices meet in a point (are parallel), therefore, by the well known property of triangles in perspective (which has been given at p. 109), the intersections, A, B, 0, of corresponding sides must lie on a right line. Hence is determined, and therefore OC, the line of resistance. The direction of R can also be found thus geometrically : Since qrO is a transversal cutting the sides of a triangle A QB, we have AO Aq Or Am np Am np _ _ _ _ v> ^ _ v^ _ . _ y L OB ~ qQ rB mn pB pB mn Am gG AC cos a W = ~ X = '' But A0 = ^ g ,and OB = BC. therefore sin (a + 0) _ cos a JF sin(a / -0)~^cW''Tr' from which we get the same value of tan as before. * This elegant solution was suggested to me by Mr. Henry Reilly. 168 EQUILIBRIUM OF A SYSTEM OF SMOOTH BODIES. [119: 5. A sphere and a cone, each resting on a smooth inclined plane, are placed in contact ; find the position of equilibrium of the system, and the reactions of the planes. Let the sphere rest on the plane OA (fig. 144) whose in- clination to the horizon is a, and the cone on OB who^e inclina- tion is a'; let W and W be the weights of the sphere and cone, R the mutual reaction between them, S the reaction of the plane OA on the sphere, T the re- action of OB on the cone, and let y be the semivertical angle of the cone. For the equilibrium of the sphere we have r 1t>4\ R=W J 44- COS , x a-y) and for the equilibrium of the cone R = W ., sm a -y- cos y From (1) and (2) we have W cos (a + a 7 y) cosy (2) (3) an equation which, instead of giving a position of equilibrium, gives a condition to be satisfied in order that equilibrium may be at all possible. It is evident that (3) is the only statical equation that can be obtained without involving the unknown reactions. Hence, if it is satisfied, every position in which the bodies are placed is one of equili- brium; and if it is not satisfied, the problem must be radically changed, and one or other of the two bodies must rest in contact with both planes. Suppose the cone in contact with both planes. Here there are only three forces acting on the sphere, and there are four forces acting on the cone, viz., W 9 R, T, and F, the reaction of the plane OA, which is perpendicular to OA. R must now be determined from the equilibrium of the sphere. Thus R= W sin a cos(a + a / y) To determine F, consider the equilibrium of the cone, and resolve along OB. Then 19.] EXAMPLES. 169 sin a cos y -] cosec (a + a'). To determine the magnitude of T, resolve the forces on the cone in the direction OA. Then sin a sin (a + a') The point 'N at which T acts is obtained by taking moments about for the equilibrium of the cone. We thus get T.ON = W'h (tan y cos a' - - sin a') + Rr cot (- r being the radius of the sphere, and h the height of the cone. ON is obtained by substituting in this equation the values of T and R given above, and it is geometrically evident that the point N lies between the foot of the perpendicular from P on OB and the foot of the perpendicular from the intersection of F and W on OB. If the sphere is in contact with both planes, the discussion proceeds in a similar manner. R is then determined from the equilibrium of the cone, T acts in the perpendicular from P on OB, and the re- actions of the planes on the sphere are easily calculated. If the weight of the sphere be greater than the value "> S\ c _ sn a cos. y given by (3), it is sufficiently clear that the sphere will descend to contact with the plane OB ; whereas if it is less than this value, the cone will descend. If the condition ( 3) is satisfied, the reaction T of the plane OB on the cone is easily found. For, let the directions of W and R meet in P ; then T must act in the per- pendicular, PQ, from P on OB, and T = r. cos(a '~ y ). cosy Similarly S may be Fig. 146. found. 6. Two blocks, AC and BO (fig. 146), rest against two fixed 170 EQUILIBRIUM OF A SYSTEM OF SMOOTH BODIES. [l2O. supports at A and B, and against each other at C\ each is acted on by a given force (in addition to its weight) ; find the lines of resistance at A, B, C. Ans. Let the resultant of the weight of the block AC and the force applied to it be the force P ; let the resultant of the weight of BC and the force applied to it be Q ; and let the resultant of P and Q be R. Draw the line AB ; take any point, h, on R, and draw Ah and Bh, meeting P and Q in / and g, respectively. Then the line fg will intersect AB in 0, the point through which the line of resistance at passes. Draw OC, and let it meet P in F and Q in 6r. Then AF and BCr are the lines of resistance at A and B. (See example 4.) 120.] System of Jointed Bars. When a system consists of a number of rods or bars articulated, or connected together by smooth joints, there will be exerted at the extremities of each rod cer- tain forces, or stresses, which are produced by the con- necting joints, and the cal- culation of the directions and magnitudes of these stresses forms an important part of Statics as applied to the construction of framework. The joint connecting any two bars may be either a portion of one of the bars or a hinge-pin distinct from both bars, and the directions of the stresses at the extremities of a bar will depend on the manner in which, the external forces are applied. Let us suppose that the joints at B and C (fig. 147), which connect the bar BC with the neighbouring bars, are distinct from BC itself, and that the forces applied to the system act at and on the joints. Then the stresses produced at B and C on the bar BC act along this bar. For, the only forces* acting on the bar are the re- actions of the joints B and C, and when two forces keep a body in equilibrium, they must be equal and opposite. Hence the stresses must act along BC. Suppose, however, that the forces, still applied at the joints, act on the extremities of the bar BC itself, and let fig. 148 represent the bar apart from the joints. Let the forces applied to it be P and Q. Now the smooth joints must produce reactions which act on the bar through the centres * The weight of the bar is supposed to be neglected. THEOREM. 171 of the joints (see p. 140). Hence equilibrium by forces acting at its extremities, and therefore the resultant of the forces at B must be a force acting in the direction BC or CB, and the resultant of the forces at C must be a force acting in the direction CB or BC. Hence the stresses pro- duced by the joints cannot act BC is again kept in Fig. 148. along the bar, but must assume some such directions as R and S. Thus, in any system of articulated bars, when the external forces are applied at the joints } the stresses will le in the directions of the oars only when the external forces act at the joints on pins which are distinct from the bars which they connect. 121.] Theorem. When a system of articulated bars is in equilibrium under the action of external forces applied at given points in the bars, the statical condition of the system may be determined by resolving the force applied to each bar into any two components acting on the joints at its extremities, and then representing each joint as in equilibrium under the action of the components transferred to it together with stresses acting on it along the directions of the bars which it connects. Let fig. 149 represent one of the bars detached from the T P IF Fig. 149. Fig. 150. joints at its extremities, and let fig. 150 represent the joint which connected the bars AB and BC (fig. 147). If a force F is applied to BC, it is, of course, allowable to break it up into any two components, P and Q, acting on the bar. Let P and 172 EQUILIBRIUM OF A SYSTEM OF SMOOTH BODIES. [l22. Q act on the bar at its extremities, and let R be the reaction of the joint at B on the bar, and S that of the joint at C. The bar is then kept in equilibrium by the forces P and R at B, and the forces Q and S at C. Hence the resultant of P and R must be a force, T, along the bar; that is to say, if the forces P and R act at any point, they produce a resultant T\ or again, if we reverse the directions of R and jT(as in fig. 150), the forces P and T are equivalent to R. Now the joint was kept in equi- librium by the equal and opposite reactions, R and R' (fig. 150) of the bars BC and AB. But we have just shown that R is equivalent to the transferred component P of the force F and the stress T, acting along CB. In the same way, R' may be replaced by a component of the force K (fig. 147) acting on AB and a stress acting along AB. We may, then, replace the external forces, K, F... (fig. 147) which act on the bars by any system of components passing through the centres of the joints, and represent two equal and opposite stresses as acting at the extremities and in the direction of each bar of the system. But it must be remembered that the stresses thus calculated (sueh as T, fig. 149) are not the total stresses at the joints. The stress in each bar, thus calculated, is the resultant of the total stress at the joint and the component of the force acting on the bar which has been transferred to the joint. For example, the stress along the bar AB is the resultant of the total stress, R, and the component of K which has been transferred to the joint B. The external forces, F, K, . . . may be each broken up into two components passing through the centres of the corresponding joints in an infinite number of ways. In the calculation of stresses in framework it is usual to break each of them up into two parallel forces. 122.] Method of Separation of the Bars. Another method, which is not really distinct from the preceding, but which is sometimes convenient in practice, consists in representing the bars as disjointed from each other, and replacing the stresses by two rectangular components at their extremities. A single example will suffice. Four equal uniform bars, AB, BC, CD, and DE (fig. 151) are 122.] EXAMPLES. 173 connected by smooth joints at B, C, and D, and the extremities A, E are fixed in a horizontal line by smooth joints; it is required to find the position of equilibrium. Let a be the common inclination of AB and ED to the horizon, and ft that of CB and CD. Let fig. 152 represent the bars AB and BC separated ; X 2 the stress at C, which is evidently horizontal ; X 1 and F x the compo- nents of the stress at B. These components act on AB in directions opposite to those in which they act on BC. Finally, let W be the weight of each bar. Kesolving vertically for the equilibrium of BC } ii = r. (i) Taking moments about C for the equilibrium of BC, 2 X 1 sin ft + TFcos ft = 2T l cos /3, or -Xi = J ^ cot ft. (2) Taking moments about A for the equilibrium of AB> (W+ 2 Ti) cos a = 2X X sin a, or, substituting the values of X and Jj from (2) and (1), tana = 3tan/3. (3) With this equation must be combined the geometrical equa- tion which expresses that AE is equal to the sum of the hori- zontal projections of the bars. If the length of each bar is , and the distance AE = c, we have c = 2 a (cos a + cos /3). (4) Equations (3) and (4) determine a and /3, and therefore the position of equilibrium* 174 EQUILIBRIUM OF A SYSTEM OF SMOOTH BODIES. EXAMPLES. 1. A triangular system of bars, AB, SO, and GA, freely jointed at their extremities, is kept in equilibrium by three forces acting on the joints ; determine the stress in each bar. Since the forces are applied directly to the joints, the stresses will act along the bars. Let P, Q, R denote the forces applied at A t B, C, respectively ; let the stresses in the sides BG, GA,AB be denoted by T lt T 2 , T s ; and let the applied forces meet in a point 0. Then for the equilibrium of the joint G, we have Tiian AGO a.OA .smAOG b.OB.smBOO' Therefore a, b, c, being the sides of the triangle. But P : Q : R = sin BOG : sin GO A : sin AOB. T l _a.OA.Q a.OAb.OB.c.OG ^1-^2^3- p Q g~ If is the centroid of the triangle, we know (p. 135) that therefore T :T = or the stresses are proportional to the sides. If is the orthocentre (or intersection of perpendiculars), therefore = OA : OB : OG. 2. A number of bars are jointed together at their extremities and form a polygon ; each bar is acted upon perpendicularly by a force proportional to its length, and all these forces emanate from a fixed point. Find the magnitudes and directions of the stresses at the joints. [This problem and the following elegant method of solution are due to Professor "Wolstenholme.] Let AB and EC (fig. 153) be any two adjacent bars of the polygon, and let P be the point from which emanate the forces, Pp, Pq, ... , acting on the bars. Then the stresses at the joints A and B, acting 753. on AB, must meet in a point, /;, on the line of action of the force Pp. Draw AQ and BQ perpendicular to the stresses in the 122.] EXAMPLES. 175 directions Ap and Bp. Now since the sides of the triangle A QB are perpendicular to three forces which are in equilibrium, and since the side AB is proportional to the force to which it is perpendicular, the sides AQ and BQ are proportional to the forces to which they are perpendicular, that is, to the stresses at A and B, respectively. Let q be the point in which Bp intersects Pq. Then the forces acting on the bar BG must act in the directions qB, Pq, and qC. Draw CQ. In the triangle BQG the sides BQ and BO are perpendicular and proportional to two of three forces in equilibrium ; therefore CQ is perpendicular and proportional to the third, that is, to the stress at 0. In the same way it can be shown that the stress at any joint is perpendicular and proportional to the line joining the joint to Q. This point Q is, therefore, a centre of stress for the system. It may be shown that the polygon of bars must be inscribable in a circle. For, since the angles at A and B are right, the quadrilateral ApBQ is inscribable in a circle whose diameter is pQ. If at the middle point of AB a perpendicular be drawn to AB, it will pass through the centre of the circle, and will, therefore, bisect Qp. But this perpen- dicular is parallel to Pp j therefore it bisects PQ in 0. Also, since the stresses at A and B are proportional to QA and QB, the same point Q must be determined by considering BC and the next bar, as was determined from the bars AB and BG ; consequently the point must be the same ; and since it is evident that OB = 00 . . . , must be equally distant from all the vertices of the polygon, that is, the polygon must be inscribable in a circle. The centre of stress is therefore constructed by joining P to the centre of the circumscribing circle, and producing PO to Q so that PO = OQ. 3. The preceding construction can be extended to the case in which the forces acting on the polygon are equally inclined, but not perpen- dicular, to the sides. Let AB, BC, ... be sides of the polygon, and let forces propor- tional to the sides act in the lines Pb, PC, ... so that LPbB LPcG = ... . It is required to prove that for equilibrium the poly- gon must be inscribable in a circle, and to find the centre of stress. The stresses at A and B must meet in a point on the force in Pb. If, then, we draw at A and B lines, QA and QB, making with the directions of the stresses angles equal to LPbB, we shall have a triangle, QAB, the sides of which are each equally inclined to the corresponding force; and, since AB is proportional to -p. the force in Pb, it follows that QA and QB are proportional to the stresses at A and B. It is easy 176 EQUILIBRIUM OF A SYSTEM OP SMOOTH BODIES. [l22. to prove that if through A and B any two lines, Ap and Bp, be drawn, meeting in a point on the right line Pb ; and at A and B lines, AQ and BQ, be drawn making with Ap and Bp, respectively, angles equal to PbB, the locus of Q is a right line, ma, making Aa = Bb, and Z maB = L mbA . Drawing the line Qd, in like manner, by making Cd = Be and Z QdB = PcC, we obtain the point Q which is the centre of stress. Now, since LPcG = Z PbB, it follows that Z 6Pc is the supplement of Z B ; and since Z $a.4 = Z $cB, it also follows that Z a$w = IT B. Hence the quadrilateral mPnQ is inscribable in a circle, and this circle must pass through 0, the point of intersection of the perpen- diculars to AB and BG drawn at their middle points, since LmOn is also the supplement of B. Hence also = -ncC, and QO = OP. Again, the stresses at A and B being proportional to QA and QB, the same point Q must be determined when BC and the next bar are considered. Hence the point is the same. But OA = OB = OG = . . . ; therefore the polygon is inscribable in a circle. The point P being given, if the angle which the forces through it make with the corresponding bars varies, the locus of the centre of stress, Q, is a circle concentric with that round the polygon, its radius being OP. To construct the centre of stress, then, we describe a circle round as centre, having radius OP, and draw PQ making the Z OPQ = the complement of the angle which the forces make with the bars. 4. A system of heavy bars, freely articulated, is suspended from two fixed points, P and Q (fig. 155); determine the magni- _ __ tudes and directions of the stresses at the joints. Fi S- J 5S- Let the bars be denoted by the numbers 1, 2, 3,..., and let their weights be W lt W 2 , TF 3 , ... . Then , transfer J W 1 and J W 2 to the joint connecting 1 and 2, which we shall denote by (1, 2). Transfer JTF 2 and TF 3 to the joint (2, 3); \W Z and JTF 4 to (3, 4), &c. Thus all the forces act at the joints. Let T lt T z , T s , ... be the tensions acting along the bars 1, 2, 3, ... on the joints, and let 12 , $ 23 , /S^, ... be the total stresses at the joints (1, 2), (2, 3), (3, 4), .... For sim- plicity suppose the bar 2 to be horizontal. Now, con- struct a force-diagram (fig. 156), by drawing a vertical line, AD, and measuring off 4 B - W *+ W * F 2 +F- TT 3 +TT 4 Fig. 156. ~T~ ~~2~ ~2~ Also take BO parallel to the bar 2 and equal to the tension T 2 , which is the constant horizontal component of each of the tensions. 122.] EXAMPLES. 177 The lines OA, OC, OD, ... will then be parallel to the bars 1, 3, 4, ... and equal to the tensions in them. Hence if a be the in- clination of 3 to the horizon, tana, and in the same way the inclinations of the other bars may be ex- pressed in terms of the inclination a. W Again (Art. 121), the stress S 12 is the resultant of 7\ and -* W Hence, taking Aa = -, Oa will be equal and parallel to S 1Z . a W. W Similarly, taking Bb = 2 , and Cc = -> the lines Ob and Oc will be equal and parallel to the stresses /S^ and 34 . The tangent of the angle made by S 23 with the horizon = = - ; ^ tan a. J3\J rf o "l ^'3 Similarly for the directions of the other stresses. If the weights of the bars are all equal, the tangents of the inclina- tions of the successive bars are tan a, 2 tan a, 3 tan a, ... , and the tangents of the inclinations of the stresses are J tan a, f tan a, f tan a, , . . . 5. Six equal uniform bars, freely articulated at their extremities, form a hexagon ABCDEF (fig. 157). The bar ED is fixed in a horizontal position, and its middle point is connected by a string with the middle point of the lowest bar, AB, in such a manner that the bars hang in the form of a regular hexagon. Find, by a force- diagram, the tension of the string and the magnitudes and directions of the stresses at B and C. Am. If W is the weight of each bar, the tension of the string = 3 W] the stress at C is W /To horizontal, and = ; the stress at B = W A / > and makes 2VV / Y 12 with the horizon an angle whose tangent = 2 \/3. 6. Prove that the centre of stress for the bar BC is the intersection of a perpendicular to BC at C with the line joining the middle points of AB and BG. 7. Three bars, freely articulated, form a triangle ABC, the centre of whose inscribed circle is 0. Each bar is acted on by a force passing through 0, proportional to the sine of half the angle subtended by the bar at 0, and bisecting this angle. Prove that the stress at A makes with OA an angle whose tangent is N 178 EQUILIBRIUM OF A SYSTEM OF SMOOTH BODIES. sin COS -y- COS (This is a direct example of the Theorem of Art. 121.) 8. AB (fig. 158) is a rigid bar whose weight is neglected fixed at one extremity, A, by a smooth joint ; CD is A O another such bar fixed at C by a smooth j oint, which is vertically below A, and jointed to AB at D. From B a given weight, P, is suspended ; find the magnitudes and direc- tions of the stresses at the joints. Ans. The stresses at C and D are AB.CD Fig. 158. along CD, and each = P. the AC. AD' stress at A is in A 0, being the intersection of CD produced with the vertical through B, and AC. AD 9. In example 5, if the bars BC and CD, AF and FE, are replaced by any bars all equally inclined to the horizon, show that the stresses at C and F will still be horizontal. [One simple proof of this is obtained by taking moments about B for the equilibrium of BC, and about D for the equilibrium of CD. It follows then that the perpendiculars from B and D on the line of action of stress at C are equal.] 10. Two uniform heavy bars are freely jointed at a common ex- tremity, and are fixed at their other extremities to two smooth joints in a vertical line; find the stresses at the joints. Ans. Let G (fig. 159) be the centre of gravity of the bars, m and n their middle points. It follows, by taking moments about A and C for the equi- librium of the bars separately, that the segments of AC made by the line of action of stress a B are pro- portional to the weights of the bars. Hence, taking ng = mG, the stress acts in the line gB. The stresses at A and C act, therefore, in Ag and Cg. If W is the weight of AB, the stress at B \ W > and the stress at aA 9 n A = \W- Hence the stresses at A, B, and C are proportional to gA, gB, and gC. 11. The regular hexagon of bars in example 5 rests in a vertical plane, the bar AB being fixed in a horizontal position, and the joints F and C are connected by a string ; find the tension of the string, and the stresses acting on the bar FE at its extremities. Fig. 159- 122.] EXAMPLES. 179 Ans. The tension = TF^/3 ( W being the weight of each bar) . W /7 /s' the stress at E = A / ' an( * it makes with FE sin" 1 \ A / - ; 2 V V 7 V31 / o 5 and it makes with -F^ sin" 1 * A / 3 : 7 V 31 12. Four equal uniform heavy bars, freely jointed together at their extremities, form a square, ABGD] the joint A is fixed, while the diagonally opposite joints B and D are connected by a string, and the whole system rests in a vertical plane, the string being horizontal ; find the tension of the string and the magnitudes and directions of the stresses on the bars at A, B, and G. Ans. The tension =2 IT; the stress at G is horizontal and = \ W j the stress on the bar BG at B makes with the vertical tan" 1 J, and= W-', the stress on AB at B makes with the vertical tan" 1 !, and= W ; and the stress on AB at A intersects the line BD at 2 a distance | BD from B, and is equal to -f W. 13. Six equal uniform bars, freely jointed at their extremities, form a regular hexagon, ABGDEF] the joint D is connected by strings with the joints F, A, and , and the system hangs in a vertical plane, the joint D being fixed ; find the tensions of the strings and the stresses at the joints. Ans. If W = weight of each bar, the tensions in the strings DB and DF are each JFV's, and the tension in DA = 2W. Also, supposing the strings to be connected with pins distinct from the bars, the stresses at C and E are vertical and equal to J W, the stresses at B and F, on the bars AB and AF, are horizontal and equal to \ WV%, and the stresses at A, on the bars AB and AF, are each equal to \ WVl. These latter stresses act in the lines drawn from A to the middle points of the two vertical bars, BG and FE, respectively. 14. Two uniform heavy bars, AB and 0, connected by a smooth joint at B, rest each on a smooth vertical prop, the props being of the same height ; find the position of equilibrium, ABC being horizontal. Ans. If IF and 2 a are the weight and length of AB, W and 26 the weight and length of BG, c the distance between the props ; then x, the distance of the middle point of AB from the corresponding prop, is given by the equation N 2 CHAPTER IX. EQUILIBRIUM OF ROUGH BODIES UNDER THE INFLUENCE OF FORCES IN ONE PLANE. 123.] Criterion of the Existence of Friction. We have already learned to regard Friction as a passive resistance ; and every passive resistance comes into existence for the purpose of stopping some motion. Thus, the normal reaction of a surface on a body in contact with it comes into existence for the pur- pose of preventing the body from penetrating the surface at the point of contact; and if the circumstances of the case were so arranged that there was no tendency to this penetration, the magnitude of the force (normal resistance) required to prevent this motion would be zero. Friction comes into existence for the purpose of preventing a certain motion motion in the tangent plane of a body resting against a rough surface. If the circumstances in any case of two rough bodies in contact are such that there is no tendency to slip at their point of contact, the force required to prevent this motion (friction) will not come into existence. Generally, in the case of all passive resistances, if there is no tendency to the displacement which a passive resistance is required to prevent, this force will not come into play. Hence in many cases of contact between rough bodies the conditions and circumstances are exactly the same as if the bodies were smooth ; and to find whether in the contact of two bodies friction acts or not imagine that the bodies were smooth at their point of contact, and if no displacement would result from this supposition, friction does not come into play at that point. In illustration of this consider the problem in example 21, p. 155. How would the circumstances be altered if the peg Q were rough ? The peg being rough, let it be imagined to become smooth, and what motion occurs ? Clearly none, supposing the board to I24-] TH E CONE OF FRICTION. 181 be rigid. Hence as there is no tendency of the side AS to slip over the peg, there is no friction called into play, and the case is the same as if the peg were smooth. But if the board is not rigid, the forces acting can bend its fibres and elongate or contract them ; and if we imagine the peg to become smooth, it is possible that (even a very slight) slipping might ensue at the peg, and as this slipping is prevented by the roughness, the force of friction really acts in the case, and the pressure on the hinge is modified by the assumption of smoothness at the peg. However, even when the board is elastic, it is possible that no friction is called into play, as will be explained in Art. 130. Rankine's hint that friction is analogous to shearing stress has been already pointed out. 124.] The Cone of Friction. The essential characteristic of a smooth surface is that it is capable of resisting in a normal direction only. If two rough surfaces are in contact, their mutual reaction is not constrained to assume a direction normal to the sur- face of contact. Each surface is capable of offering resistance to the other in any direction which does not make with the normal to the surface of contact an angle exceeding a certain magnitude. Thus (fig. 160), let two rough bodies, A and , be in contact at any point, P, and let PN be the normal to the surface of contact. Let A. denote the greatest angle that the total resistance at P can make with PN, or, in other words, the greatest obliquity of the mutual reaction ; then, describing round PN a right cone, CQD, whose semivertical angle, NPJ), is equal to A, this cone is called the cone of friction, and the total resistance at P can act in any direction whatever included within this cone. This angle A. is what we have called in Chap. Ill the angle of friction, and its tangent is the coefficient of friction for the two surfaces considered. For, if R-^ denote the normal pressure between them at P, and F the force of friction (which acts in the common tangent plane), it is clear that when the resultant of R! and F acts along any generator, PD, of the cone, we have 182 EQUILIBRIUM OF ROUGH BODIES. [126. so that tan X is the greatest ratio of the force of friction to the normal pressure. This quantity we have called //. If a rigid weightless rod, M (p. 40), be pressed against a rough surface at 0, the greatest angle that the rod can make with the normal is the angle of friction. For, since the rod is acted on by only two forces, viz., the applied pressure and the total resistance at 0, these must be equal and opposite, or along the rod. Hence the greatest obliquity of the rod to the normal is A. If the resistance to slipping is not the same in different azimuths, i. e., if it is different in different planes through the normal, the value of A. will not be the same in all these planes, and the cone of friction will not be a right circular cone. 125.] Axiomatic Law of Friction. We have said that the total resistance of a rigid surface is a force which can assume any magnitude. This force will in any given case be exerted by the surface to such an extent as is necessary to preserve equi- librium, but to no greater extent. It is in its nature a passive resistance , i.e., one which can be exerted to any extent, but which will not be exerted beyond the bare requirements of the case. Within certain limits, also, as we have seen, it can assume any direction, and in any given case it will, if possible, assume such a direction as will preserve equilibrium. In fact, in virtue of its passive nature, we must regard the resistance of a rough surface as an opposition called into existence by the action of external forces ; and it seems clear that these forces will call into play only that amount of opposing force, exact both in magnitude and in direction, which will just counteract their own action. The amount of assumption contained in this principle is enunciated in the following axiom : The total resistance which acts at any point of a rough surface will, if possible, assume such a magnitude and direction as will preserve equilibrium at that point. This axiom is sometimes expressed thus : If passive resistances can give equilibrium, they will. 126.] Remarks on this Axiom. Two important observations must be made on the principles contained in this axiom. Firstly, it is important to understand the circumstances which may render it impossible for the resistances of rough surfaces to 126.] REMARKS ON AXIOMATIC LAW. 183 preserve the equilibrium of a system in any given position. Suppose that a body, acted on by given external forces, is in contact with a rough surface at a single point, P. Then, for equilibrium, it is necessary that the resultant of the given external forces should pass through P, and that the total re- sistance at P should be equal and opposite to this resultant. But if the direction of the resultant makes with the normal to the surface of contact at P an angle > A, it is impossible that the total resistance could take the required direction, and equi- librium cannot subsist. Again, take the case in which a heavy beam, AB (fig. 161), rests against a rough horizontal and an equally rough vertical plane. Describe round the normals to the planes at A and B the cones of friction, and let the sections of these cones by the plane of the figure be rAs and pBs. Let G be the centre of gravity of the beam, and Fthe vertical line through it. Then the beam, if in equilibrium, is so under the action of three forces, namely, the weight through G and the total resistances at A and B. These three forces must meet in a point, and if it be possible to find a point in which they can meet, the resistances will assume proper values. Now, in the figure it is impossible to find any point on GF, the line of action of the weight, the lines drawn from which to A and B could be directions of possible resistance at 'both A and B. For the portion of GF which is inside the cone of friction at B is outside the cone of friction at A, and vice versa. Hence, for equilibrium, there must be some portion of the line GF included in the space, pqrs, common to both cones of friction. Unless this condition is satisfied, it is not possible for the total resistances to give equilibrium,, whatever their magnitudes may be. A possible position of equilibrium is represented in fig. 162. For, if from any point on the portion, mn, of GF which is included in the space common to both cones of friction, lines be drawn to A and B, these lines are possible directions of total resistance at A and B ; and in this case the actual magni- tudes and directions of the resistances at A and B cannot be determined by what is called Rational Statics. 184 EQUILIBRIUM OF ROUGH BODIES. [126. If it be proposed to find the position of limiting equilibrium, that is, the position in which the beam is bordering on motion, we must make the vertical through G pass through r, as in fig. 163. In this case there is only one point on GV which is inside both cones of friction, viz., the point r. Hence the total re- A Fig. 162. Fig. 163. sistances act in rA and rJ5, and each makes the limiting angle (A) with the corresponding normal. Moreover, both resistances are now determinate. If 6 be the angle made by the beam with the horizon, we have, from the triangle ArB, 2cotrGB = CQlArGeoiBrG, or 2 tan 6 = cot A tan A, which defines the position of limiting equilibrium. It may, therefore, in certain cases be impossible for the total resistance at one or more points to preserve equilibrium ; and this impossibility is always due to something in the arrangement of the figure or the external forces which requires the direction of the resistance to make with the normal to the surface of contact an angle > the angle of friction. Again, in the axiom is contained the following important proposition : If a body rests against a rough surface at a pointy and if the equilibrium is about to be broken by some change in the acting forces } equilibrium at that point will, if possible, be broken by a rolling instead of a sliding motion. B W 126.] EEMAEKS ON AXIOMATIC LAW. 185" For, in this case, the point of the body actually in contact with the surface would be kept at rest. This part of the axiom is sometimes stated thus If a body can roll, it will roll, in preference to slipping. Exactly the same considerations as before determine the possibility or impossibility of the rolling motion. Such a motion will always take place if it does not require the total resistance to make with the normal to the surface of con- tact an angle > A. For example, let us discuss the following problem : A heavy cubical block rests on a rough horizontal plane, and a string, attached to \ the middle of one of the upper edges passes over a smooth pulley, and sustains a weight which is gradually increased. Find the nature of the initial motion of the block, the string and the vertical through the centre of gravity of the block being in Fig. 164. the same vertical plane. Let ABC (fig. 164) be the vertical plane in which all the forces act ; CO the line of the string, intersecting the vertical through the centre of gravity of the block in ; P the suspended weight, and JFthe weight of the block. (Since the length of the string is immaterial, no linear magnitude can enter into the result, therefore the side of the block need not be known.) Now in all such cases as this, it is necessary to observe the following rules : 1. Write down the motions of the system which are geo- metrically possible. 2. Exclude those which would obviously violate any of the fundamental rules of Statics. 3. If there remain possible cases of slipping and rolling (or turning over), solve the problem on the supposition that equi- librium is broken in the latter way, and if this does not require too great a value of the angle of friction, equilibrium will be broken in this way. In the present case, the following motions are geometrically possible : (a) The block may be lifted vertically off the plane. (/3) It may turn round the edge A. 186 EQUILIBRIUM OF ROUGH BODIES. [126. (y) It may slide in the direction AS. (b) It may turn round the edge B. Now (a) is obviously excluded, because if the block is just out of contact with the horizontal plane, it is acted on by only two forces, namely, its own weight and the tension of the string. But since these cannot be equal and opposite, equilibrium cannot be broken in this way. Suppose (/3) to happen. Then the total resistance of the plane passes through A and through 0. But it is impossible that three forces acting in the directions of AO, OC, and OW could be in equilibrium. Hence (/3) is excluded. The cases (y) and (8) remain. Now in virtue of the principle, if (5) is possible, it will happen. Solve, then, on the supposition that the block turns round B. It is then kept in equilibrium by its weight, the tension, and the total resistance which must act in BO. If the L CBO is less than X, the angle of friction, the block will turn round B ; but if CBO > X, this motion is impossible, and slipping must take place in the direction AB. To express this analytically, let be the angle made with the horizon by the string OC, and let fall from a perpendicular on BC meeting BC in p. Then " Bp BC-Op. tan 2 -tan Hence if //, (or tan X) be > - > the block can turn round JB, and will do so if P is gradually increased. The magnitude of P which will just cause the tilting of the block is found by taking moments about B. We evidently obtain P= ITT sec 0. Suppose that CBO > X, or that p < - Then the in- crease of P will produce a sliding motion, and we can easily find the magnitude and point of application of the total resistance of the plane. Now since CBO > X, the point, M, of application of the total resistance of the plane, is found by drawing from a line OH making with the normal to the plane an angle = X. The point M lies between B and the point in which the vertical through cuts AB. P can then be determined either by taking 12;.] LIMITING POSITIONS OF EQUILIBRIUM. 187 moments about JHf, or by resolving vertically and horizontally. Resolving vertically, we have resolving horizontally, E sin A = P cos ; Pcos0 _ ' ^ TT-. \ = M, or P = The direction of the string might be so modified as to render possible either a sliding in the direction BA or a tilting over A. Thus, in fig. 165, if the line of the string intersect the line of action of the weight in a point, 0, below the horizontal plane, the two motions possible are evidently one of slipping in the direction AB and one of tilting over the edge A. The latter will take place if it can. If it does, the total resistance must act in the line OA, and for this the angle DAE must be < A. But if DAE is < A, the block will slip in the direction AB, since the horizontal /j component of the tension acts in this ^ ^ | sense. The condition for tilting over A is now evidently 1 M > tan0-2* The values of P corresponding to both ( kinds of motion are calculated as before. 127.] Limiting Positions of Equi- $ librium. When a body rests in contact with any number of rough surfaces at several points, the equi- librium is said to be limiting if a slight alteration of a definite kind in the circumstances of the body would cause the equi- librium to be broken. The slight alteration referred to depends on the nature of the particular problem of equilibrium. As has been explained in Art. 46, p. 56, every statical problem relating to the equilibrium of a body is always one or other of the three following : (#) What is the least force that will sustain a body in a given position on given surfaces, or the greatest force that will allow it to rest in such a position ? (b) With given forces and given supporting surfaces, what is the position of equilibrium such that if this position be slightly altered, the body will not rest ? 188 EQUILIBRIUM OF ROUGH BODIES. [128. ( because, if it be less than this, the beam will slip. 128.] Comparative Safety of Equilibrium of a System at Different Points. When in a system in equilibrium the direc- tions of the total resistances at the various points of contact with rough surfaces are known, we are enabled to say at which of the points slipping is most likely to happen in case some of the circumstances should be altered. This will be rendered clear by the following examples, taken from Jellett's " Theory of Friction," p. 61 : Two uniform beams, AC and BC, connected at C by a smooth hinge, are placed, in a vertical plane, with their lower ex- tremities, A and B, resting on a rough horizontal plane. If equilibrium be on the point of being broken, determine how this will happen. Fig. 143, example 4, p. 166, will represent the beams if the hinges at A and B are conceived to be removed and these points rest on the ground. Then, exactly as in that example, the direction of the mutual resistance at C is determined. Supposing AC to be the longer beam, it is clear that the angle which the total resistance, AQ, at A makes with the normal to the surface of contact (i.e., to the ground) is greater than the angle which the total resistance BQ makes with the normal at B. 128.] For LIMITING POSITIONS OF EQUILIBRIUM, tan A Qn An 189 Now An = Am -\-rnn ; and if 2 a, 2b, 2c, are the sides BC, CA, AB, we have ac a (b cos a + a cos ft) Am = o cos a, mn fG = - r = ^ - - - ' , a + b a + b An a (b cos a -f a cos /3) = b cos a + * - ~ - '- a + b Similarly therefore (a 2 + 2 ab] cos 8 + b z cos a = i - '- - - - a + b Bn = - 7- (cos a cos /3). But since AC > BC, cos a > cos /3, therefore An > _Zfo. Hence the angle AQn >BQn ; that is, the total resistance at A makes with the normal at A an angle greater than that made by the total resistance at B with the normal at B. Consequently, if any circumstance should continually diminish the angle of friction (which is supposed to be the same for both beams) the total resistance at A would be the first to attain its limiting obliquity to the normal, and slipping would then take place at A in the direction BA, while the beam BC would turn round B. We might inquire which of the beams will first slip if they are drawn out so as to increase the angle C, and the same result will follow, since for any given position of the beams the direc- tions of all the resistances are determinate. In each case the angle AQn must be the first to reach the value \, and therefore the longer beam, AC, must slip first. The result may also be expressed thus in any given position of rest, equilibrium is more safe at B than at A. There are also cases in which the comparative safety of equi- librium can be determined, although the directions of total resistance are not completely determinate at all the points at contact. For example two unequal cylinders rest on the ground at given points, A Fig. 1 66. 190 EQUILIBRIUM OF ROUGH BODIES. [128. and B (fig. 166), while a third cylinder rests on them at points p and q. Supposing either that there is a gradual diminution of the coefficient of friction (which is the same at all the points of contact), or that the lower cylinders are gradually drawn asunder, determine the nature of the initial motion of the system. Denote the cylinders by the letters at their centres. Then the cylinder D is kept in equilibrium by three forces namely, 1st, its weight, which acts through A ; 2nd, the total resistance of the ground, which also acts through A ; and 3rd, the total resistance of the cylinder C at p. Now, since the first two forces act through A, the third must also pass through this point. Hence the total resistance at p acts in the \me_pA, and therefore the total resistance of the ground at A must take some inter- mediate (but unknown) direction, AR. In the same way,, the total resistance at q is proved to act in the line qB, and the total resistance of the ground at S must act in some direction, BS t intermediate to BE and Bq. The resistances in Ap and Bq at p and q meet in a point, P, on the circumference of the upper cylinder. Now the comparative safety of equilibrium at the different points of contact, A } B, p, q, will depend on the angles made by the total resistances at these points with the normals to the surfaces of contact ; and it is manifest that since the angle DAp > DAR and Dp A = DAp, the total resistance at p makes a greater angle with the normal, DC, to the surface of contact than that which the total resistance at A makes with the normal AD. Hence equilibrium is safer at A than atjo. For a similar reason, equilibrium is safer at B than at q. Consequently the final comparison is to be made between the points p and q. Now the line pq can be proved by geometry to pass through the point in which ED intersects B A ; and supposing the radius BE > AD, this point will be at the left-hand side of the figure. Let a be the acute angle which pq makes with the ground. Then, since in the triangle/* Cq the base angles at p and q are equal, it is easy to see that LqCW-LpCW^ 2 a, or qCW>pCW. But the angle which the total resistance at q makes with the normal qC is \qCW y and the angle which the total resistance at p makes with the normal pC is \pCW\ therefore if the friction were gradually and uniformly diminished everywhere, or the cylinders drawn out, the 130.] FRICTION DEPENDENT ON INITIAL ARRANGEMENTS. 191 resistance at q would reach its limiting obliquity before that at p. Hence the initial motion will be a slipping of the cylinders C and E at the point ^, and a motion of rotation at the other points of contact. 129.] Virtual "Work of the Total Besistance. Suppose one rough body to roll on another through a small angle whose magnitude is regarded as an infinitesimal of the first order. Then, neglecting infinitesimals of a higher order, the point of the rolling surface in contact with the other surface is at rest during the displacement that is, the virtual displacement of the point of application of the total resistance between the two bodies is zero. Hence for a virtual displacement which consists of a small rolling motion of one rough body on another, the total resistance will not enter into the equation of virtual work of either body. Of course in no case can the mutual action of two rigid bodies in contact enter into an equation of virtual work for both bodies. It is a principle in Kinetics that in a motion of pure rolling of a body on a rough fixed surface no work is done between any two positions by the total resistance a principle which the student will have no difficulty in comprehending, since for each small motion the work done by this force is imfinitesimal com- pared with the work done by other forces acting on the body. 130.] Friction as Dependent on Initial Arrangements. In dealing with natural solids, and not with strictly rigid or indeformable bodies, the existence or nonexistence of friction sometimes depends on the way in which a body or system has been placed in the position which we are considering. This will be made clear by the following example. A heavy trap door (or a beam) AJ3, fig. 132, p. 148, moveable about a fixed horizontal axis at A, has a rope attached at JB, and this rope is also attached to any fixed point C ; determine the pressure on the axis A. The line of action of the pressure must, of course, go through 0, the point of meeting of the other two forces, but beyond this we know nothing about it until we know the nature of the axis. If the axis is smooth^or if it is rough but so worn that the contact of the door with it takes place along a single line, the action between the door and the axis will consist of a force passing through the axis, as has been amply explained in Art. 114. But if the axis is rough and contact takes place all round it, the 192 EQUILIBRIUM OF ROUGH BODIES. line of action of the resultant force is not generally determinate. However, even in this case this resultant force may pass through the axis. The axis being rough, let us imagine it to become smooth, and what motion results ? The rope, being slightly extensible, would yield a little, and slipping would take place over a small surface at the axis ; so that the supposition of smoothness alters the circumstances of the case. But suppose that (the axis being still rough) the rope has been stretched, when the door is placed in position, to such an extent that the moment of its tension about the axis is equal to the moment of the weight of the door ; then clearly if we imagine the axis to become smooth, no motion will result no slipping at the axis; and since the displacement which friction is required to prevent does not take place, friction does not act, and the case is the same as if the axis were smooth. The resultant in this case is therefore determinate. 131.] Friction of a Pivot. Let a cylindrical pivot, ABCD (fig. 167), on the top of which a given force is applied, revolve -iH B r 1 - -~G Fig. 167. Fig. 168. in a closely fitting bearing, EFGH, and let it be required to calculate the moment of the friction on the base, AB, about the axis of the pivot. Suppose fig. 168 to represent the base of the pivot, and let P = the whole normal pressure on the base, which we shall suppose to be uniformly distributed over the base. Divide the area AS into a number of narrow circular strips, of which one is represented in the figure. Let Oa = #, Ob = x + dx, OB = r, [JL = coefficient of friction. Then since the whole pressure is uniformly distributed, the pressure on the strip whose P ZPxdx area is = 2 nxdx is ~ - 2-nxdx, or 5 r 2 Hence the sum of the forces of friction, acting in the directions of the tangents to the WEARING AWAY OF THE STEP. ZfjiPad* 193 strip, is ^ But since the tangents to the strip are all at the same distance from the centre, the moment of friction on the strip is equal to the sum of the forces of friction multiplied by the radius, SB, of the strip. Hence the moment of friction over the whole surface is f. 2 > or - (1) If the base, instead of being a full circle, is a ring, or collar, whose internal and external radii are r and / 2 , the friction per unit of surface is j and the moment of friction is 2 or - (2) 132.] Wearing Away of the Step. The piece which supports a pivot, and in which it revolves, is called a step. When the pivot revolves, the friction against the step wears away its own surface and that of the step. The amount of wear at any point of the step depends on the magnitude of the force of friction and the relative velocity of the rubbing surfaces at this point. Thus, suppose that ABC (fig. 169) represents a section of the step through the axis, 13P, of the pivot, and that Q is any point of contact of the pivot and step. If f is the magnitude of the force of friction at Q, the wearing at Q in the direction of the normal will be propor- tional to f and also to the amount of rubbing surface which passes over Q in a unit of time. Supposing the pivot to revolve round its axis with an angu- lar velocity co, the point of the pivot in contact with Q moves in a horizontal circle with a velocity = CD . QM, or o> .y ; QM, or y, being the perpendicular from Q on the axis of the pivot. But the amount of rubbing surface which passes over Q in a unit of time is evidently proportional to the velocity at Q. Hence the normal wearing of the surface at Q is proportional to B Fig. 169. 194 EQUILIBRIUM OF ROUGH BODIES. Fig. 170. If n be the magnitude of the normal pressure per unit of surface at Q, and // the coefficient of friction, we have/"= ^n. Hence the normal wearing of the surface at Q is proportional to a>pny. (a) 133.] Friction of a Conical Pivot. Let ABC (fig. 170) represent a section of a conical step by a plane through the axis, BP, of the pivot, APC being the surface at which the pivot enters the step. Supposing that the pressure on the top of the pivot is uniformly dis- tributed, it will evidently be uniformly distributed over the area APC; that is, there will be a constant normal pressure, ^, per unit of area on APC. Now it is impossible to determine by elementary methods the law of distribution of the pressure on the step. The following investigation proceeds on the assumption that the normal pres- sure per unit of area, or as it is properly called, the normal intensify of pressure, is constant over the surface of contact. Let n be the constant pressure per unit of surface of the step. If ds is a small element of the line BC at Q, the distance of which from BP is y, the corresponding elementary strip of conical surface is Znyds, and the moment round BP of the friction on this strip is or Putting ds = . . > sm0 and integrating over the surface of the step from y to y = PC = r, we have the moment of the whole friction equal to 3sin0 * p If P = the whole pressure on the top of the pivot, n = ^ , itr hence the moment of friction 2/z 3sin0 .Pr. Comparing this with the result in Art. 131, we see that the 1 34.] THE TRACTORY OR ANTI-FRICTION CURVE. 195 moment of friction in the case of a conical is greater than in the case of a cylindrical pivot of equal radius. 134.] The Tractory, or Anti-Friction Curve. In the case of a conical pivot the wearing away of the step is not uniform at all points. Hence after a sufficient time the pivot will not be in perfect contact with its step. If, however, the step has such a form that the vertical wear is the same at all points, the pivot will simply sink into the piece which supports it, and remain always in contact throughout its surface with the step. We propose to investigate the form of the step in which the vertical wear will be the same at all points. Let fig. 171 represent a section of the step through the axis of the pivot, and let CQ' be the vertical wear at C, and Q Q' the ver- tical wear at Q. Then CCf = QQ' t Q being any point on the curve BC. Hence the new curve, BQ'C', is simply the old curve J2QC moved through a vertical distance GG' h, suppose. Now (Art. 132) the normal wear at Q per unit of surface is Hence, if Qg is normal to the step at Q 3 n being the normal pressure per unit of surface on APC, which we also take to be the normal pressure per unit of surface on the step. But = the curve at Q. Hence the to " or ds = a constant, ' dy or QT = a constant. Therefore the curve J5 is such that the length of the tangent terminated by PB, or the axis of a, is constant at all points. This curve is known as the Tractory. If t = the constant o 2, 196 EQUILIBRIUM OF ROUGH BODIES. ['34- length of the tangent, and PC the axis of y, we have ds or 9 V^Sr-* or the minus sign being given to the square root, because MQ diminishes as x increases. Integrating this last equation (by assuming y = t sin $) we have for the equation of the tractory The curve approaches PB asymptotically, and the step is formed by the revolution of the curve round PB. This pivot is known as Schiele's Anti-friction Pivot. EXAMPLES *. 1. A uniform rectangular board, A BCD (fig. 172), rests in a vertical plane against two equally rough pegs, P and Q, in the same horizontal line, two adjacent sides of the board being each in contact with a peg. Find the position of equilibrium. Let A be the angle of friction, the inclination of the side AS to the horizon in the position of limiting equilibrium, G the centre of gravity of the board, PQ = a, and AG = c. Then if the board is on the point of slipping down at Q and up at P, the total resistances at P and Q will act in the directions PO and Fig. 172. QO, which are inclined at the angle A to the normals at P and Q to the sides AB and AD, respectively, if 0' (not represented in figure) be the point of meeting of the normals at P and Q, it is clear that a circle will pass through the points APO'OQ; and therefore LOAO' = A. And since A 0' = PQ = a, we have AO = a cos A, (1) * Many of the following examples are due to Mr. Jellett, in whose Theory of Friction will be found several other instructive examples which want of space compels me to omit. 1 34.] EXAMPLES. 197 Again, since LO'QP = 0, we Lave LQOG = --(A + 0), and evidently Z$0^ =0, therefore Z^0(7 = - -(A + 20). If /-GAB = a, it is clear that Z.AGN = - (0 + a). Now the position of equilibrium is found by the equation AO, KmAOG = AG. sinAGN. Substituting in this equation the value of AO from (1), we have a cos A. cos (A + 20) = c . cos (a + 0), which defines the position of equilibrium. 2. A heavy uniform beam rests against a rough horizontal plane and against a rough vertical wall, the vertical plane through the beam being at right angles to the wall and the ground; determine the greatest weight that can be affixed to it at a given point, so that equilibrium may be preserved. (a) If the beam be inclined to the vertical at an angle less than the angle of friction for the beam and the ground, equilibrium cannot be broken by attaching a weight, however great, to any point of the beam. Let AB (fig. 173) be the beam, its inclination to the horizon, W its weight, 2 a its length, P the weight suspended from the point Q in the beam, BQ = x, \ and A' the angles of friction at A and B, re- spectively. Draw the lines A and BO, making the angles A and A' with the normals, An and Bm, at A and B. Then when the resultant of IF and P passes through 0, equilibrium will be at its limit. For, if this resultant acts in a line to the left of V, the >* ^ vertical through 0, it will be possible to find an infinite number of points on it such that when joined to A and B the joining lines will be possible directions of total resistance at A and B (see Art. 126). If the resultant of W and P acts in a line to the right of OF, there will be no point on it inside both Fig. 173. cones of friction, and therefore equi- librium will be impossible. Hence for limiting equilibrium, we have by taking moments about O, W.GV=P.QV, G being the centre of gravity of the beam. The lengths GV and QV are easily obtained from the data. We may observe that if the point Q lies between G and F, equilibrium can never be broken, however great P may be. For it will then be impossible by increasing P to bring the resultant of P and W to the right of V. 198 EQUILIBRIUM OF ROUGH BODIES. 034- These results follow also from the usual mode of solution of such a problem. Let R and S be the normal reactions at A and B, and /ut and f/ the coefficients of friction at these points. Then, resolving horizontally, S = nR', (2) resolving vertically, R + p'S = P + W ', (3) taking moments about B t 2 aR (cos 6 ft sin 9) = (P# + Wa) cos 0. (4) P + JF From (2) and (3) we have R = - (5) and by substituting this value of R in (4), we get P=Wa see that BO = 2a COB / ( ? + ^-. and 57= Now it is easy to ^ v^** ~^ w , ,. . 5- ; therefore B V 2a j-? > and (5) may be written COS U '" P-w a ~ BV > BV-x from which it appears that if x = B V, the required force is infinite ; and if x> BV, it is negative, or equilibrium can never be broken by any downward force. The second part of the problem follows from (5), because if p. tan > 1, or, in other words, if the angle nAB < A, the denominator will be negative. That it is impossible to break equilibrium in this case is evident from fig. 174. For the point is now at the right of the vertical wall, and at whatever point along AB the resultant of P and W acts, it is possible to find points on it which are within both cones of friction. A\ A Fig. 174. 175- 3. Two unequal uniform beams, connected by a light rope attached to their middle points, rest in a vertical plane, an extremity of each EXAMPLES. 199 beam resting on a rough horizontal plane. If the coefficient of friction is gradually diminished, which beam will slip first ? Let the beams be AB and A'B' (fig. 175) and let G and C" be their centres, and AB>A'B'. Now the beam AB is in equilibrium under the influence of three forces, viz., its weight, the tension of the rope CC", and the total resistance at A ; and since the first two meet in (7, the third must also pass through this point, that is, the resistance at A acts along the beam. In the same way the resistance at A' acts along A'B'\ and by considering the equilibrium of the system, we see that the vertical through G, the common centre of gravity, must pass through 0, the point of intersection of the resistances. Now the angles which these resistances make with the normals at A and B are equal to mOA and mOA', respectively; and the comparative safety of the equilibrium at A and B depends on the magnitudes of these angles. Now mOA'>mOA. For, draw C'q horizontal and Cq vertical ; then, since CG < C'G, qn < nC\ and & fortiori, pn < nC'. Am tanmO-4 _ .. Therefore Am < mA : but Tf = 7-77; therefore, mOA >mOA, mA tan mOA and if the friction were gradually diminished, the total resistance at A' would reach its limiting inclination before that at A. Hence the short beam will slip first. 4. A cylinder is supported on a rough inclined plane by a string coiled round it in a direction perpendicular to its axis, the string passing over a smooth pulley and sustaining a weight. Find the limits to the direction of the string. Bound A, the point of contact of the cylinder and plane, describe the cone of friction, the section of which by the plane of the figure is nAm, the angles nAC and CAm being each = A. Let OB be any direction of the string, intersecting the vertical through the centre of the cylinder in 0. Then, so long as is between the points m and n, equilibrium is possible, because AO is a possible direction of total resistance at A. There is, of course, a particular magnitude of the sus- pended weight, P, corresponding to the direction OB of the string, and this magni- tude is found by taking moments about A. If is the angle made by the string, OB with the inclined plane, we have i being the inclination of the inclined plane. If, the direction of the string being OB, P have a value greater or less than this, W Fig. 176. the cylinder will roll up or roll down the plane. Drawing from m two tangents, mt l and mt 2 , to the cylinder, we 200 EQUILIBRIUM OF ROUGH BODIES. have the extreme directions of the string ; that is, the point at which the string leaves the cylinder must lie between the points of contact of mt^ and mt z , on the upper portion of the cylinder ; for it is evident that if the string leaves the cylinder at any point outside these limits, the point in which its line intersects that of W will be vertically above m, that is, outside the cone of friction. 5. A heavy sphere is placed on a rough inclined plane at a point P (fig. 177), and is kept in position by a heavy rough beam, AB, which is move- able about a fixed extremity, B, the coefficient of friction for the sphere and the beam being the same as that for the sphere and plane. Supposing that the friction is gradually diminished at both points of contact, P and Q, or that the sphere is pushed further up between the plane and beam, determine the nature of the initial motion. The total resistances at P and Q must meet in some point, 0, on the vertical through C, the centre of gravity of the sphere. Beyond this, however, their directions cannot be determined. The comparative safety of equilibrium at P and Q will depend on the relative magni- tudes of the angles, GPO and CQO, which the resistances at these points make with the corresponding normals. Now it is easy. to show that CQO > CPO ; for sin CPO = ^ sin COP, and sin CQO = ^ Fig. 177. sin COR, therefore sin CPO sin COP but COR > COP, therefore sin CQO sin COR ' CQO > CPO, and if from any cause the friction is diminished, or the sphere pushed higher up, slipping must take place at Q and rolling at P. 6. A cylinder is placed on a rough inclined plane, and a light rope is coiled round it in a plane perpendicular to its axis and containing its centre of gravity ; this rope, after passing round the cylinder, is attached to the middle point, H (fig. 178), of an edge of a cubical block whose height is equal to the diameter of the cylinder. Supposing the incli- nation of the plane to be gradually increased, determine the manner in which equili- brium will be broken, the co- efficient of friction being the same for the cylinder and plane as for the cube and plane. Fig. i7 8 - The motions which are here geometrically possible are I34-] EXAMPLES. 201 (1) The cylinder may roll and the cube may turn over the edge C. (2) The cylinder may roll and the cube may slip. (3) The cylinder may slip and the cube may slip. (4) The cylinder may slip and the cube may turn over. Now if is the point of intersection of the vertical through the centre of gravity of the cylinder with the rope, it is evident that the total resistance at A acts in the line AO. In the same way if 0' is the point of intersection of the vertical through G, the centre of gravity of the cube, with the line of the rope, the total resistance of the plane on the cube must pass through 0', and if D is the point in which the line of action of the weight of the cube intersects its base, the total resistance must evidently pass through some point between C and D. Now this total resistance, wherever it acts, makes with the normal to the plane an angle greater than BA ; for tan BA = \ tan i, i being the inclination of the plane, and the angle which O'D makes with the normal to the plane = i ; hence the angle made with this normal by a line joining 0' to any point between G and D is > i, and, & fortiori, > BA 0. Consequently the cylinder can never slip before the cube, and cases 3 and 4 are to be rejected. The choice then is to be made between 1 and 2 ; and (see Art. 126) if the cube can turn over, it will do so. Hence we solve on the supposition that the cube turns over C, and if this does not require too great a value of the coefficient of friction, the cube will turn over. The problem is to be solved by equating the values of the tension of the rope derived from the consideration of the equilibrium of the cylinder and that of the cube. For the equilibrium of the cylinder take moments about A, and we have T JJFsint, (1) T being the tension of the rope and W the weight of the cylinder. Again, since by supposition the cube is about to turn round C, the total resistance of the plane acts through this point. Taking moments about C for the cube, T.CH= W. CG sin (?- 1) , or T lJT(cosi sint). (2) Equating the values of T in (1) and (2), we have _ _ But in order that CO' may be a possible direction of total resistance, the angle HC(/ must be < A, or tan HCO' < \JL. Now, it is easy to see that TT+2TT * { } 202 EQUILIBRIUM OF ROUGH BODIES. W+2W' Hence if J , < JJL, equilibrium will be broken by a rolling of the cylinder and turning over of the cube. If p is less than the quantity in (4) the cylinder will roll and the cube will slip, and there is no difficulty in determining the inclination of the plane when this happens. We may either draw from 0' a line making the angle of friction, X, with the normal to the plane, and then determine T by the triangle of forces, or resolve along and perpendicular to the plane for the equilibrium of the cube. If R is the normal reaction of the plane on the cube, we find in the latter way R = JTcos f, jjiR= W'smi+T; therefore T = W (p cos i - sin i). Equating this to the value given by (1), we have i&ni =wSw" which gives the inclination at which the cube slips. 7. Two equal carriage wheels whose centres are connected by a smooth bar are placed on a rough inclined plane ; determine whether the equilibrium of the system will be best preserved by locking the hind or the fore wheel. Let C and D (fig. 1*79) be the centres of the wheels, and first sup- pose the hind wheel to be locked. Fig. 179. Since there is no friction between the bar CD and the axle at C, the action of the bar on the lower wheel consists of a force through C (see p. 140). The weight of this wheel also acts through (7, and therefore the total resistance at A, which is the third force keeping the wheel in equilibrium, must also act through C. Let G be the centre of gravity of the two wheels, and consider the equilibrium of the system formed by them. There are three forces acting on the sys- tem, viz., its weight through Gf, the total resistance at A (which has been proved to act in a line AC), and the total resistance at B. If, then, is the point of intersection of CA and the vertical through G, the total resistance at B must act in the line OB. We shall now determine the inclination at which equilibrium is broken. Since the hind wheel slips, the angle DBn = A ; also let r = the radius of each wheel, CD = 2 a, and i = the inclination of the plane. T 34-] Then EXAMPLES. tan COG _ CG tan COn ~ ~&n '' tan i a 203 or since Dn = r tan DBn = joir. The inclination of the plane when equi- librium is broken is therefore given by the equation tan i = (I) 2a + [JLr Again, suppose the fore wheel alone to be locked. In this case the total resistance at B acts in the line BD, and that at A acts in AO', 0' being the intersection of BD with OG. If i' is the new inclination at which equilibrium is broken, we have, since LCAO' A, or Dm tan i' = 2a (2) Now it is clear that i' is greater than i, and that, consequently, equilibrium will be safer when the fore wheel is locked than when the hind wheel is locked. 8. A cylinder is supported on a rough inclined plane by a light rope coiled round it in a plane perpendicular to its axis passing through its centre of gravity, the rope being attached to a fixed point. Find the direction of the rope in order that the inclination of the plane may be the greatest possible. Let O'B' (fig. i So) be the line of the rope, and CO* the vertical through the centre of gravity of the cylinder. Then evidently the total resistance at A, the point of contact with the plane, must act in the direction A 0'. If the rope took the direction OB, which is hori- zontal, the direction of the total resistance would be A 0, and evidently the angle CAO < CAO'\ or, in other words, the equilibrium of the Fig. 1 80, Fig. i Si. cylinder will be farther from its limit when the rope is horizontal than when it takes any other direction. For a given inclination, i, of the 204 EQUILIBRIUM OF ROUGH BODIES. plane the angle CAO = > and it is clear that when CAO is equal to the angle, A, of friction, the inclination of the plane will be at its greatest. Hence the greatest inclination of the plane = 2 A. If the coefficient of friction be > 1, the greatest inclination of the plane will be > 2 j and the figure of limiting equilibrium will be that represented in fig. 181, in which the angle CAO (= A) is > But whether the cylinder will stay in this position or not depends on the initial arrangement. Unless the rope is pulled with such a force as to cause the resultant of this force and W to act in the line OA, equi- librium cannot be preserved by the resistance of the plane. In fact, unless this requisite tension of the rope is produced by pressing and scraping the cylinder against the plane, it would be possible for the cylinder to take a motion of and round its centre C which would keep its surface out of actual contact with the plane ; and in this case the plane would not exert any resistance. 9. If in the preceding problem the rope, instead of being attached to a fixed point, is attached to a weight which hangs freely over a smooth pulley, find the conditions of equilibrium. Let O'B' (fig. 1 80) be the direction of the rope, P the suspended weight, W the weight of the cylinder, i the inclination of the plane, A the angle of friction, the angle which the rope makes with the inclined plane. Then for equilibrium it is necessary that A (/ should be the direction of total resistance at A, and that the moments of P and W about A should be equal and opposite. Hence we must have angle AO'= or < A, (1) and 2 cos- (2) the second condition being equivalent to that obtained by the triangle of forces for equilibrium at 0'. If the angle CA(/< A, and P is slightly increased above the value in (2), the initial motion will evidently be a rolling up, since moment of P about A > moment of W about A ; but if P is slightly diminished the rolling will be down. 10. A heavy uniform beam, AB (fig. 182), is to be sustained in a horizontal position, one end, B, resting on a rough inclined plane, while the other end, A, is attached to a light rope which passes over a smooth pulley and sustains a weight. Find (a) The limits to the direction of the rope, and the corresponding limiting values of the suspended weight. (b) The least weight that will sustain the beam, Fig. 182. EXAMPLES. 205 Let W be the weight of the beam, P the suspended weight, and BN the normal to the inclined plane at B. Then if AO be the line of the rope, intersecting the vertical through the centre of gravity of the beam in 0, BO must be the direction of the total resistance at B ; and in order that this may be a possible direction of total resistance, the angle NBO must be < A, the angle of friction. Hence the limiting directions of the rope are obtained by drawing BO and BO' making the angle A with BN on opposite sides. If the rope takes the direction AO' the beam must be on the point of slipping up, since the force of friction acts down the inclined plane ; and if the direction of the rope is AO, the beam is on the point of slipping down. The corresponding magnitudes of P are easily determined by taking moments about B. Let p l and p^ be the perpendiculars from B on AO and AO', respectively, a half the length of the beam, and P l and P 2 the corresponding values of P. Then PI The values of p l and p 2 can, of course, be easily expressed in terms of a, A, and i, the inclination of the plane. If the rope takes a direction intermediate to A and A 0', and if p is the length of the perpendicular from B on its direction, we have P Hence, if P is a minimum, p must be a maximum, since Wa is given. Now p will be a maximum when it is equal to AB, that is, when the rope is vertical. In this case the total resistance at B should also be vertical ; but if the inclination of the plane > A, this is impossible. Hence when i > A, p is a maximum (consistently with the conditions of the, problem) when the direction of the rope is AO ; and therefore in this case P l is the least value of P. If i < A, the vertical at B is a possible direction of total resistance, and therefore AB is an admissible value of p. The corresponding value of P is therefore J TF. The student will easily see that if the angle of friction is greater than the complement of the inclination of the plane, there can be no limiting equilibrium in which the beam is about to slip up. 11. A cylinder is laid on a rough horizontal plane, and is in contact with a rough vertical wall ; a string coiled round it at right angles to the axis passes over a smooth pulley and sustains a weight which is gradually increased till equilibrium is broken. Determine the nature of the initial motion. (Jellett's Theory of Friction, example 21, p. 214.) Let W be the weight of the cylinder, P the suspended weight, the angle made by the string with the horizon, A and A' the angles of friction at A and B, the points of contact of the cylinder with the 206 EQUILIBRIUM OP ROUGH BODIES. Fig. 183. vertical and horizontal planes, and the point in which the line of the string intersects the vertical through (7, the centre of gravity of the cylinder. Now, in accordance with Article 126, we first consider what motions are geo- metrically possible. These are (1) Boiling round A up the vertical plane. (2) Slipping forward at B while con- tact ceases at A . (3) Slipping at A and .B simultaneously. If (1) can happen it will (see Art. 126); let us suppose, therefore, that the cylinder is on the point of turning round A and coming out of contact at B. In this case there are only three forces keeping the cylinder in equilibrium, namely, W, P, and a total re- sistance at A. This last force should, for equilibrium, pass through and act in the direction OA. Now whether the angle OAC is less or greater than A, this is not a possible line of action of total resistance, because the plane cannot pull. Hence (1) is physically impossible. Suppose that (2) happens. Then, as before, there are only three forces keeping the cylinder in equilibrium, namely, W, P, and the resistance at B. This last must pass through 0, and must therefore act vertically. But it is obvious that such a force could not equi- librate TFand P\ therefore (2) is impossible. There remains the third case, which alone is possible. To deter- mine the value of P corresponding to limiting equilibrium, draw the lines A 0' and BO' making with the normals at A and B the angles, A and A.', of friction for the cylinder and planes. Then by taking moments about 0' we easily obtain the value of P, which may also be obtained by the ordinary equations of resolution of forces. Thus, let R and R f be the normal pressures, and therefore pR and p'R' the forces of friction, at A and B. Taking moments about B, we have (l+f*) = P(l-COS*). (1) Taking moments about A t R'(l-lJ.') = W-P(l+sm0). (2) Resolving horizontally, fj.'R'-R = Pcose. (3) Substituting in (3) the values of R and R f given in (1) and (2), we obtain the value of P corresponding to limiting equilibrium. It will be a useful exercise for the student to vary the position of the pulley in such a way as to render possible a case of limiting equi- librium in which the cylinder is about to ascend the vertical plane by turning round A. 12. A heavy right cone is placed with its base on a rough inclined plane, the inclination of which is gradually increased; determine 134.] EXAMPLES. 207 whether the initial motion of the cone will be one of sliding or tumbling over. Let ABC (fig. 184) be the vertical section of the cone through its axis, CH, and let G be the centre of gravity of the cone. (GH is J- CH, as will appear in a subsequent Chapter.) Then, in accordance with rule 3 of Art. 126, if it is possible for the cone to turn over the point A, the cone will do so. Solve, there- fore, on the supposition that equilibrium is broken by turning round A. In this case, the two forces acting on the cone are its weight and the total resistance of the plane, which, of course, passes through A ; and these forces must be .equal and opposite, i. e., ^g. 184. the total resistance must act in the vertical line AG. Now this will be possible only if AG makes with the normal to the plane an angle less thau the angle of friction, A. Hence for a tumbling motion AGH 4 tan a, the initial motion of the cone will be tumbling, and if ft < 4 tan a, the initial motion will be sliding, and this sliding will evidently occur when the inclination of the plane reaches the value A. 13. A heavy straight rod rests on a rough horizontal plane, and at one end, perpendicularly to its length and in the horizontal plane, a force is applied with gradually increasing magnitude. Find the point about which the rod begins to turn. (Price's Infinitesimal Calculus, vol. iii, p. 162.) Let I be its length and suppose it to turn round a point at a distance z from the other extremity. Then we must equate the moment of the applied force about this point to the sum of the moments of the forces of friction acting on the different elements of the rod. Take an elementary portion of length dx at a distance x from the point round which the rod turns. The weight of this portion is dx, and the force of friction on it is ft W -=- This acts I L at right angles to the rod. Hence, taking the sum of the moments for all points at both sides of the turning point, we have * But P is evidently equal to the sum of the frictions at the end adjacent to it minus the sum of those at the other end : i. e.. 7 _ ft,?; P = ft W = Hence we have * In this simple case integration is evidently not necessary. 208 EQUILIBRIUM OF EOUGH BODIES. [134. 0, /. *s(l--L)Zj or the turning point is at a distance -=. from the end at which the force is applied. * 14. A rectangular block is placed, with one of its edges horizontal, on a rough plane, the inclination of which to the horizon is gradually increased; determine whether the equilibrium of the block will be broken by a motion of sliding or one of tumbling. Ans. If a and b are the lengths of the edges which are not horizontal, b being the length of the edge which is perpendicular to the inclined plane, the initial motion will be one of tumbling if f/ > -r- > and of sliding if JJL < -? 15. A cylinder the section of which perpendicular to the axis is any given curve is to be placed, with the axis horizontal, on a rough inclined plane ; how must it be placed so that it shall be least likely to slip, the cylinder being in contact with the plane along a single line? 16. An elliptic cylinder is placed, with its axis horizontal, on a rough plane inclined to the horizon at an angle less than the angle of friction ; prove that the cylinder cannot rest if the eccentricity of the / 2 sin i section perpendicular to the axis is less than A / ; r i being the inclination of the plane. V 17. A uniform beam rests with its extremities on two rough in- clined planes whose line of intersection is horizontal, the vertical plane through the beam being perpendicular to this line ; find the limiting position of equilibrium. Ans. If i, i' be the inclinations of the planes, A, \' the angles of friction between the beam and the planes, respectively, and the limiting inclination of the beam to the horizon, 2tan0 = cot (i + X)-cot(i'-\'). Another limiting position will be got by changing the signs of A and A'. 18. A heavy uniform rod rests with its extremities on the interior of a rough vertical circle ; find the limiting position of equilibrium. Ans. If 2 a is the angle subtended at the centre by the rod, and A the angle of friction, the limiting inclination of the rod to the horizon is given by the equation Q s * n 2 ^ cos 2 A + cos 2 a 19. A solid triangular prism is placed, with its axis horizontal, on a rough inclined plane, the inclination of which is gradually increased ; determine the nature of the initial motion of the prism. 134.] EXAMPLES. 209 Ans. If the triangle AEG is the section perpendicular to the axis, and the side AB is in contact with the plane, A being the lower vertex, the initial motion will be one of tumbling if the sides of the triangle being a, b, c, and its area A. If /m is less than this value, the initial motion will be one of slipping. 20. A frustum of a solid right cone is placed with its base on a rough inclined plane, the inclination of which is gradually increased ; determine the nature of the initial motion of the body. Ans. If the radii of the larger and smaller sections are R and r, and h is the height of the frustum, the initial motion will be one of tumbling or slipping according as r* ^ ^ T h i 21. An elliptic cylinder rests in limiting equilibrium between a rough vertical and an equally rough horizontal plane, the axis of the cylinder being horizontal, and the major axis of the ellipse inclined to the horizon at an angle of 45, Find the coefficient of friction. An8 ' e being the eccentricity of the ellipse. (Employ the Theorem of Art. 116.) 22. The circumstances of the preceding problem remaining the same, except that the vertical plane is smooth, show that the coefficient of friction is ^ e 2 (Walton's Mechanical Problems, p. 82). If the horizontal plane alone is smooth, is it possible for the cylinder to rest in any position ? 23. A uniform beam, of which one end rests against a rough vertical wall, is supported by a light rope attached to the other end, and to a given point in the wall ; find the limiting positions of equili- brium (Walton, p. 81). Ans. If the length of the rope be n times the length of the beam, the inclination of the latter to the wall is given by the equation ( w a__ ju 2 - l)tan 2 + 4 /A tan + n 2 _4 = 0. 24. In order that both limiting positions may be real, what must be the limits of n ? Ans. 2n 2 must be > /** + 5 V (^+ 1) 0* 2 + 9), and 25. If n is 2, show that there is but one limiting position ; and prove geometrically that if in this case the angle of friction is 60, the limiting position is horizontal. P 210 EQUILIBRIUM -OF ROUGH BODIES. [134. 26. A heavy uniform beam rests with one end against a rough horizontal and the other end against an equally rough vertical plane ; find the least coefficient of friction that will allow the beam to rest in all positions. Ans. Unity. 27. In the previous question let the centre of gravity of the beam divide it into two segments, a and b, the latter segment being in contact with the vertical well ; given the coefficient of friction, /u, between the beam and the ground, find the least coefficient of friction between the beam and the wall which will allow the beam to rest in all positions. a Ans. -= pb 28. Two equal beams, AC and CB, are connected by a smooth hinge at C, and are placed in a vertical plane with their lower extremities, A and B, resting on a rough horizontal plane ; from observing the greatest value of the angle A CB for which equilibrium is possible, determine the coefficient of friction for the beams and the plane (Walton's Mechanical Problems, p. 96, second ed.) Ans. If the greatest value of LACB is /3, 29. Two uniform beams are placed with their lower extremities resting on a rough horizontal plane, their upper extremities resting against each other. Show how to cut a plane face from the upper extremity of one of the beams, in order that slipping may be about to ensue at their point of contact. Ans. Determine the line of action of their mutual resistance as in p. 167 ; then cut a face inclined to this line at the complement of the angle of friction. 30. A cylinder is placed on a rough horizontal plane, and a uniform plank rests with one end on the ground and the other against the cylinder (the plank being at right angles to the axis of the cylinder). If the plank is gradually lowered until equilibrium is about to be broken, show that if the weight of the cylinder exceed that of the plank the latter will always slip, whatever be the dimensions of the plank and cylinder. For any position of the plank find the direction of the reaction of the ground on the cylinder. Ans. If is the angle made by the plank with the ground, P = weight of plank, W = weight of cylinder, r = radius of cylinder, 2 a = length of plank, \//- = angle made with the vertical by the reaction of the ground on the cylinder, 31. A cylinder placed on a rough plane has a string coiled round it in a plane at right angles to its axis ; the string after passing round 1 34.] EXAMPLES. 211 the cylinder is attached to a heavy particle which also rests on the plane. If the plane is gradually tilted up, determine the nature of the initial motion. Ans. The cylinder will roll and the particle slip if both are equally rough ; and if i is the inclination of the plane when this happens, . " where W and P are the weights of the cylinder and the particle, //. the coefficient of friction, and 2 a the angle between the string and the inclined plane. 32. A heavy cylinder is laid on a rough inclined plane, its axis being horizontal ; a heavy uniform plank rests on the cylinder and against the inclined plane, the plank being horizontal at right angles to the axis of the cylinder, and touching the cylinder at its highest point. Supposing the inclination of the plane to be gradually increased, the horizontality of the plank being always perserved, determine the nature of the initial motion of the system and the inclination of the plane at which equilibrium is broken. Ans. The plank will slip at its point of contact with the plane, a rolling motion taking place at the other points of contact in the system ; and the inclination (t) is given by the equation where r = radius of cylinder, 2 a = length of plank, W = weight of cylinder, P = weight of plank, and A. = angle of friction. 33. Two particles A and B, whose weights are denoted by A and J5, are connected by a string fully stretched, and placed on a rough horizontal plane, the coefficient of friction for each particle being )u. A force -P, which is < fx. (A + J5), is applied to A in the direction BA, and its direction is gradually turned round through an angle 6 in the plane. Find the nature of the initial motion of the system. Ans. If P < fJL 4/A 2 + JB 2 and > pA, the particle A alone will slip, and this happens when sin = If P > //, */A 2 + B 2 , both J"+tiP--J3 will slip when cos 6 = - - 34. A heavy rod is placed in any manner resting on two points A and of a, rough horizontal curve, and a string attached to a point G of the cord AB is pulled in any direction in the plane of the curve so that the rod is on the point of motion. Prove that the locus of the intersection of the lines of action of the frictions at A and B is an arc of a circle and a part of a straight line ; except when G is the centre of gravity of the rod, in which case the directions of the frictions will be always parallel to the string, 212 EQUILIBRIUM OF ROUGH BODIES. [134. 35. A triangular prism, whose section by a vertical plane through its centre of gravity perpendicular to its edges is ABC, rests with its base AB on a rough horizontal plane ; a rope is attached to the middle point, 0, of its upper edge, and, passing over a fixed pulley in the horizontal line parallel to, and in the sense of, SA t is pulled with a gradually increasing force. Find the nature of the initial motion. Ans. If A B = c, AC b, and the height of the prism = h, the prism will tilt over the edge through A if c + b cos A M> - ^r - ; otherwise it will slide. 36. A cubical block is placed on a rough inclined plane and sus- tained by a rope, parallel to the inclined plane, attached to the middle point of the upper edge (which is horizontal) ; the rope lies in the vertical plane which contains the centre of the cube and is perpen- dicular to the inclined plane. Show that the greatest inclination of the plane is ' 37. Two rough inclined planes slope in the same direction and intersect in a horizontal line. A cylinder placed at their intersection and touching both all along its length has a rope coiled round it in a plane through its centre of gravity perpendicular to its axis ; this rope passes over a fixed pulley and is pulled with gradually increasing force. Discuss the ways in which equilibrium may be broken by varying the tension of the rope, finding (with a given position of the rope) (a) The condition that must be satisfied in order that equilibrium should be possible at all ; (b) The condition that the initial motion should be one of slipping on both planes ; (c) The value of the tension of the rope when this slipping takes place. 38. A heavy uniform circular wheel rests, in a vertical plane, against the ground at A and is in contact at B with an obstacle of given height ; the wheel is to be pulled over the obstacle by means of a rope (of given direction) attached at a given point to the wheel ; find (a) The condition that the initial motion of the wheel shall be a rolling over the obstacle ; (b) The condition that the initial motion may be slipping at A and B. (c) What ultimately happens when the initial motion is slipping at A and B. CHAPTER X. EQUILIBRIUM OF A RIGID BODY UNDER THE ACTION OF ANY FORCES. 135.] Resultant of any Number of Forces Applied to a Material Particle. Let a force, P, represented in magnitude and direction by 00' (fig. 13, p. 19), act on a particle at 0; let Ox, Oy, and Oz, be any three rectangular axes drawn through ; and let the angles, O'Ox, O'Oy, and (/Oz, which the direction of P makes with the axes of reference be denoted by a, /3, and y, respectively. From 0' let fall perpendiculars, OF, OH, OD, on the planes, yz, zx, and xy, respectively, and let the parallelepiped be completed as in figure. Then the force 00' may be replaced by the forces OD and OC, by the parallelogram of forces ; and OD can again be replaced by OA and OB. Hence the force P is equivalent to the three forces P cos a along Ox, Pcos/3 Oy, and P cos y Oz. The converse proposition is also evidently true namely, that any three forces, OA, OB, OC, along Ox, Oy, Oz (whether these are mutually rectangular directions or not), give a resultant represented in magnitude and direction by the diagonal, OO ', of the parallelepiped determined by the forces. If several forces, P lt P 2 , ... P n , act at and make angles (i0iyi)> ( a 2>/32;7 2 )> (>, y), with the axes, let. each of them be replaced by its three components along Ox, Oy, Oz ; and if 2Z, 27, *2Z denote the sums of the components along the axes, we shall have ... -f P n cosa n , ... +P n cos/3 n , [ (i) = P 1 cosy 1 + P 2 cosy 2 + ... +P w cosy, 214 EQUILIBRIUM OF A E1GID BODY. [136. and the whole system of forces will be replaced by the three forces, 2JT, 2F, and 2Z along the axes of #, y, and z. But the resultant of three forces in these directions is the diagonal of the parallelepiped determined by them. Hence, E being the magnitude of this resultant, E = V'(2J) 2 + (2J) 2 + (2) 2 , (2) and if 0, $, \jf, be the direction-angles of E, T 1C 5! V ^LF cos = -g- 9 cos 4 = -g- , cos ^ = -^- (3) 136.] Grapliic Representations of the Resultant. Since the resultant of any two forces, OA and OB, acting at is obtained by drawing from A a line, Ad, parallel and equal to OB, and joining to d, it follows that if a particle is acted on by n forces, OA 1 , OJ 2 , 0^ 3 , ... OJ n , the resultant is obtained in magnitude and direction by drawing A 1 a 2 parallel and equal to OA 2 , # 2 # 3 parallel and equal to OA 3 , ... a n ^ a n parallel and equal to OA n , and joining to a n ; or, in other words, the side Oa n which closes the polygon OAa 2 a^... a n represents the resultant in magnitude and direction. "When the sides of the polygon are not all coplanar, the figure is called a gauche polygon. Thus the second graphic representation of the resultant of a system of coplanar forces, which has been given in p. 18, is equally ap- plicable to non-coplanar forces. Hence, of course, it follows that a particle acted on by any set of forces which are parallel and proportional to the sides of a gauche polygon taken in order is at rest. Again, since by the parallelogram of forces, the resultant of OA l and OA 2 is 2 . Qg , where ff 1 is the middle point of A l A 2 ; and since the resultant of 20^ and OA 3 is 30^ 2 , where g 2 is determined exactly as in p. 15, it follows that Leibnitz's graphic representation of the resultant is applicable to non-coplanar forces. This result follows also analytically; for if (^ 15 y 1 , z^, (#25 Hi* z z)> " (n>yn> ^n) ^> e ^ ne co-ordinates of the extremities A l , A. 2 , ... A n of the forces acting on the particle, it is clear that ... -f x n = 2# = n . x, where x, y, z are the co-ordinates of G, the centre of mass of I37-] TRANSFORMATION OF COUPLES. 215 equal masses placed at the extremities of the forces. Hence by equations ( 1 ) of Art . 1 3 5, ^ "> an d (R> #) But it is manifest that the figure Bprq is merely the figure BPRQ turned round in its own plane through a right angle. Hence Br is the diagonal of the parallelogram determined by the axes of the component couples. Conversely, any couple may be resolved into two couples whose axes are determined from the axis of the given couple by the * According to the convention (/3) the couples in this figure are both negative, and the axes Bp and Bq should be drawn downwards. This inaccuracy in the figure was detected too late for correction. 138.] THEOREM. 217 parallelogram law ; and as in the case of forces acting at a point, any couple may be resolved into three couples whose axes are determined from the axis of the given couple by the parallele- piped law. All this follows as in Art. 135. It is well to remark that the axis of a couple represents the moment of the forces of the couple round a line in space parallel to the axis. (5) To find the resultant of any number of couples acting in any planes on a rigid body. Let the axes of the couples be all drawn, each in its proper sense according to the rule (/3), at the same point, (fig. 13), and let each axis be resolved into three components along rectangular axes Ox, Oy, Oz, drawn through 0. Let L = the sum of the axes in the direction Ox ; then L is the axis of the component of the resultant couple in the plane yz. Let M and N be the sums of the axes in the directions Oy and Oz, re- spectively. Then, if G is the resultant axis, *, (1) and if A, ju, v are the direction angles of G, L M N = , COS !>=-, equations which are exactly analogous to (2) and (3) of Art. 135. The axes of couples are, therefore, compounded and resolved in the same manner as forces. There is this difference between forces and couples, that, while any number of couples in any planes whatever always result in a single couple, any number of forces cannot, in general, be replaced by a single force, and this difference results from the veetorial nature of the axis of a couple. (e) A force and a couple acting on a rigid body cannot produce equilibrium. For, let the couple be so transferred that one of its forces, P, acts at a point on the line of action of the force, R. Then R and P at this point compound a single force which, in general, does not intersect the other force of the couple. Therefore, &c. A force and a couple acting in the same plane are, of course, equivalent to a single force. 138.] Theorem, A force acting on a rigid body in a given 218 EQUIL1BKIUM OF A RIGID BODY. ['38. Fig. 187. right line can always be replaced by an equal force acting at any chosen point together with a couple. Let a force P (fig. 187) act at a point A, and let be the chosen point. At introduce two forces, P and P 7 , opposite to each other and each equal and parallel to P. Then P at A and P' at may be taken to constitute a couple whose mo- ment is Pp, p being the perpendicular from on the line of action of P at A. There remains, then, the force P at 0; and this force together with the couple may replace P at A. Let the axis of this couple be drawn at ; let x> y, z be the co-ordinates of A with respect to a rectangular system of axes drawn through ; and let a, (3, y, be the angles which the direction of P makes with the axes of #, y, z t respectively. The direction cosines of OA are - > - > - > where OA = r, and r r r it is easy to prove that the direction cosines of the axis of the couple (which is at once at right angles to OA and to P) are y cos y z cos /3 z cos a x cosy x cos (3 y cos a P P P Hence, the axis of the couple being equal to Pjp, the projections of the axis on the axes of #, y, and z are P(^cosy 2 cos /3), P(zcos a x cosy), P (OB cos/3 y cos a) ; but it is clear from (y), Art. 137, that these are the axes of the component couples in the planes yz, zx> and xy> into which the couple Pp can be resolved. Putting P cos a = X, P cos /3 = Y, P cos y = , we see that the three couples are Zy-Yz, Xz-Zx, Yx-Xy. (l) The force P at may also be replaced by its three components, X, T, Z. (2) There is another way in which the reduction is sometimes effected. Let P at A he resolved into its three components, JT, Y, Z^ and let the line of Z meet the plane (ocy) in N, and let Z at .4 be trans- ferred to N. Let fall Nn perpendicular to Ox; at n introduce two opposite forces Z" and Z'" , each equal and parallel to Z ; and at introduce two opposite forces, Z and Z' ', each equal and parallel to Z. Now the senses of positive rotation in the planes xy, yz, zx being r - I39-] PARALLEL FORCES. 219 those indicated by the arrows, the forces Z at N and Z'" at n form a couple whose moment is Zy parallel to plane yz ; and the forces Z? at and Z" at n form a couple whose moment is Zx parallel to the plane zx ; and in addition to these there is the force X Similarly, the force X at A can be re- placed by X at together with two couples, Xz and -Xy, parallel to the planes zee and an/, respectively; and the force Y at A can be replaced by T at together with the couples Tx and Yz parallel to the planes xy and yz. Hence P at A is replaced by the forces \ Fig. 1 88. F, F, Z at and the couples ZyYz, XzZx, and Fa? JTy, parallel to the planes 3/2, zx, and #?/, respectively. 139.] Parallel Forces. Suppose a rigid body to be acted on by any number of parallel forces applied at given points in the body. Take any origin, 0, of co-ordinates, and through it draw three rectangular axes, that of z being parallel to the common direction of the forces. Then the force P, acting at (#i,yu *J may be replaced, as in last Art., by P l at along Oz, and the couples P\y\ an( l ~P\ X \ parallel to the planes yz and zx. Replacing each force in this manner, the whole system will be equivalent to a force P l4 P 2 + ... + P B , or 2P at 0, together with the couple + P nVn, or parallel to the plane yz, and the couple -P^-P^ ... P n x w or -2P0, parallel to the plane zx. These two couples compound a single couple whose axis is found by drawing OL = 2Py and OM (in the negative sense of the axis of y) = Fig. 189. and completing the parallelogram OLGM (fig. 189). If OG, the diagonal is denoted by G, 220 EQUILIBRIUM OF A RIGID BODY. [140. and R = 2P, R being- the resultant force. 140.] Centre of Parallel Forces. Since the resultant of two parallel forces, P x and P 2 , acting at the points A^ and A 2 divides A a P the line A^A^ in a point g such that -~- -^> and since by elementary geometry (see p. 97) the distance of g from any plane P x -\-P x ~y W~^ J wnere #1 an( l #2 are the distances of A and A 2 "" from this plane, it follows, by repeating this construction that the distances, #, y, z, of the centre of parallel forces from the planes yz, zx, and xy are given by the equations 141.] Conditions of Equilibrium of a System of Parallel Forces. A system of parallel forces has been reduced (Art. 139) to a single force, R, and a single couple, G. Now since these cannot in combination produce equilibrium (e, Art. 137), we must have ^ = 0, and G = 0, separately. Since G cannot be = unless 2P# = and 2Py = 0, the con- ditions of equilibrium are RQ ( 1 ) 2P# = 0, ^Py = 0. (2) DEF. The moment of a force with respect to a plane to which it is parallel is the product of the force by its perpen- dicular distance from the plane. Hence for the equilibrium of parallel forces The sum of the forces must vanish, and the sum of their moments with respect to every plane parallel to them must also vanish. EXAMPLES. 1. A heavy triangular table, ABC, is supported horizontally on three vertical props at the vertices ; prove that the pressures on the props are equal. Let P, Q, R be the pressures at A, S, C, and let a vertical plane through A and the centre of gravity of the table cut the side BG in a, its middle point. For equilibrium the sum of the moments of the forces P, Q, R, and W (the weight of the table) with respect to this 141.] EXAMPLES. 221 plane must = 0. But the moments of P and W are each = 0, since these forces lie in the plane. Hence the moments of Q and R are equal and opposite. Now the distance of Q from the plane = Ba . sinAaB, and the distance of R = Ca. sinAaC; and since Ba = Ca, these distances are equal. Therefore Q = R ; and similarly it can be shown that R = P ; therefore, &c. 2. A heavy triangular plate hangs in a horizontal plane by means of three vertical strings attached to its vertices ; at what point in its area must a given weight be placed so that the system of strings may be least likely to break 1 At the centre of gravity of the board. For if W = the weight of the board and P the sustained weight, we have or the sum of the tensions is constant, wherever P is placed. Hence if any one is less than J(JF+P), some other must be greater than this value. It is evident, therefore, that the best arrangement makes each tension = ^ ( W+ P) ; but this happens (as proved in last example) when P is placed at the centre of gravity. 3. A heavy elliptic cylinder is sustained in a vertical position by three props applied at three points on the circumference of its base ; how should the props be placed in order that the cylinder may be least likely to be upset 1 Let the base of the cylinder have any form, ABC (fig. 190), and let Cf be the projection of its centre of gravity on the plane of the base. Then, if the props are applied at A, B, and (?, the perpendiculars from G on the sides of the triangle ABC must be all equal when the equilibrium is most stable. For, suppose that the cylinder is about to be upset round the line AB; then the moment of the force re- quired to upset it is W.p, where W is the weight of the cylinder and p the perpendicular from G on AB. Again, suppose that the cylinder is about to be upset about AC', then the moment of the force required to upset it is W. q, where q is the perpendicular from G on AC. Hence if p and q are unequal, ad- vantage will be gained by increasing the lesser of them, even though the greater must be consequently diminished ; and it follows that the maximum advantage is gained when p arid q are equal. In the same way it can be shown that the perpendicular from G on BG must, in the most advantageous case, be equal to that from G on AB ; and therefore the perpendiculars from G on the sides ABG must be all equal. Hence the problem amounts to inscribing in a given curve a triangle on the sides of which the perpendiculars from a given point shall be equal. In the particular case in which the base is an ellipse, we have to find a circle concentric with the ellipse, such that a triangle circumscribed to the circle shall be inscribed in the ellipse, 222 EQUILIBRIUM OF A RIGID BODY. [141. Now (Salmon's Conic Sections, p. 330, 5th edition), let the ellipse a 2 w 2 have for equation ^ + -7^ 1 =0, and the circle # 2 + w 2 r 2 = ; or b* x z y z then the discriminant of &(# 2 + 2/ 2 -r 2 ) + + f 1 = is & -1* ) & + - * ' k+ *' and tlie re< i uired condition being 2 = 4 A . 0', we have two values for r, namely, r x = -- - and r. = - r The first value alone is admissible, because - - > b, ab a o and the circle in this case either cuts the ellipse or entirely encloses it. Since an infinite number of triangles can be inscribed in the ellipse ab and circumscribed to the circle of radius - - (Salmon, ibid.), the a -f~ o problem is capable of an infinite number of solutions. It is easy to see that in the cases in which it is possible to have a real system of in- and circum- scribed triangles for the ellipse and the circle of radius - -) the centre of the ellipse is outside the area of the triangle. a b This circle is, therefore, irrelevant to our question, 4. A heavy square board, A BCD, of uniform thickness, is hung by three vertical strings, one of which is attached to a corner, A, of the board. The plane of the board being horizontal, find the points, E and F, in the area to which the other two strings should be attached in order that it may be most difficult to overturn the board by placing a weight anywhere on it. It is evident that advantage is gained by taking the points E and F on the edges of the board. Assume AE to be the direction of the line joining the points of application of two of the strings, and suppose that a weight, P, placed somewhere in the area A BE is on the point of overturning the board about the line AE. Then the tension of the string at F = ; and if W is the weight of the board, acting at G, the weight P required to upset it is distance of G from AE fyy ^s _ . distance of P from AE Hence the greater the distance of P from AE, the less the requisite value of P, or, in other words, the more easily will the board be upset. It is evident, therefore, that the applied weight should be placed at B ; and in the same way, if the board is to be upset round the lines AF and FE, the applied weights should be placed at the corners D and C, respectively. Again, in the arrangement of greatest advantage, the board is upset with equal ease round each of the lines AE, AF, and FE. For, if it be more easily upset round one of these lines than round another, advantage will be gained by making it a little more stable with regard to the first. Hence, since the weights placed at B, D, and C 141.] EXAMPLES. 223 are all equal, we have distance of G from AE _ distance of G from AF _ distance of G from EF distance of B from AE ~~ distance of D from AF~~ distance of C from EF' The angles EAB and FAD are, therefore, equal, and each = tan" 1 (/2-1). 5. A heavy elliptic table is supported on three vertical props ; how must they be placed so that it may be most difficult to upset the table by placing a weight on it ? Ans. The props must be placed at three points, A, B, C, on the circumference of the ellipse ; and if y is the eccentric angle of C, that of B is f TT + y, and that of A is ^ir + y. The weight which, most advantageously applied, will then just upset the table is half its own weight. This may be seen as follows. Assuming any line in the area as the line joining two props, the least weight that will be required to upset the table must be placed at the point of contact of a tangent parallel to the assumed line, since the weight will have maximum leverage at this point. Also, it must be equally easy to upset the table round the three lines AB, BO, CA ; that is, the requisite weights placed at C", A', B', the points of contact of the tangents, must be all equal. If, then, x, y, z, be the perpendiculars from the centre on the lines BC, CA, AB, and P, Q, R the perpendiculars on the parallel tangents, we must have . */ P-x Q-y R-z> or, if a, /3, y, be the eccentric angles of A, B, C, a 3 fiy ay cos - cos c - - cos - a /3 /3 y ay 1 cos - 1 cos 1 + cos a negative sign being used in the last, since (y, /3, a being in ascending order of magnitude)^- is evidently >- Hence /3 = -7r + y, 2 6 O a = - TT + y ; and the weight required to upset the table = W-^ or J . W Any one position of C is, therefore, as good as any other ; and if C is made the extremity of either axis, the line AB is parallel to the other at a distance equal to of the first axis from it. 6. A rectangular board is held with its plane horizontal by three vertical strings attached to three of its corners ; find the point in its area at which a weight must be placed so that the tensions of the strings shall be given multiples of the weight of the board. Ans. Let W be the weight of the board ; let the strings be applied at the corners A, B, C ; let AC = 2 a, AB 2b; and let the tensions of the strings at A } B, C be IW, mW, nW, respectively. 224 EQUILIBRIUM OF A RIGID BODY. [142. Then the weight must be placed at a point whose distances from AB and A C are 2nl 2m 1 , . a and _ - . o. 1 The magnitude of the weight is, of course, (l + m + n 1) W. 7. A uniform circular lamina is placed with its centre upon a prop; find at what points on its circumference three weights, w lt w 2J w s , must be placed that it may remain at rest in a horizontal position (Walton's Mechanical Problems, p. 7 3). Ans. The angles which the weights subtend in pairs at the centre of the lamina are the supplements of the angles of a triangle whose sides are proportional to the weights. - 142.] Reduction of a System of Forces acting in any manner on a Rigid Body. Let any origin, (fig. 188), be assumed arbitrarily, and let any system of rectangular axes. Ox, Oy, and Oz, be drawn through it. If, then, forces P l , P 2 , P 3 , ... act on the body at points whose co-ordinates are ( lt y l , %), (#? 2 , y 2 , z 2 ) t (% 3 , y 5 , z 3 ), ... each force can be replaced by three components acting at along the axes, together with three couples whose axes coincide with the co-ordinate axes. The force P lf for example, is equivalent to X l9 Y 19 Z^ at and three couples, ^1^1^1 ^i, X l z 1 Z l x l , and Y 1 as i X l y 1 . Adding the components of the forces, and also the axes of the couples, in the directions Ox, Oy, and Oz, the whole system of forces is equiva- lent to the force 2X along Ox, 2Y Oy, and 2Z Oz; and the system of couples is equivalent to the couple S (Zy Yz), or L, in the plane yz, *S,(XzZx\ or Mj zx, and ^(Yx-Xy\ or N, osy, (Of course the axes of L, H, N are drawn along the axes of %, y, and z, respectively). Hence if E be the magnitude of the resultant of translation, R= and if G be the magnitude of the resultant couple, POINSOT'S CENTEAL AXIS. 225 y ~v y v y 7 The direction-cosines of R are -=- > -~- > and ^~ ; and those T K/T -\T R R R P n L M , N or Cr are ~ > - r > and -^ Cr Cr IT Thus, ^ system of forces acting on a rigid body can be replaced by a single resultant force acting at an arbitrary origin^ the magni- tude and direction of this force being the same for all origins^ and a single resultant couple the magnitude and direction of whose axis are both dependent on the origin chosen. It has been already remarked (Art. 137) that G is not only the axis of the resultant couple (corresponding to a resultant force acting at 0), but also the sum of the moments of the forces about a line at drawn in the direction of G ; and since the axes of couples have been proved to follow the parallelepiped and parallelogram laws, it follows that the sum of the moments of the forces about this line is greater than the sum of their moments about any other line at ; and also that the sum of the moments of the forces about any other line through is the resolved part of G in the direction of this line. Remark. The magnitude and direction of G are constant at all points along the same right line parallel to R. For R may be supposed to act at any point in this line, and the vector G may be moved parallel to itself to the point at which R is supposed to act. 143.] Poinsot's Central Axis. Any system of forces acting on a rigid body has been proved to be equivalent to a single resultant force, R, acting at an arbitrary origin, 0, and a single resultant couple G. Let (j) be the angle between R and G, and resolve G into two compo- nents, OK and On (fig. 191) along and perpendicular to R, respectively. On is the axis of a couple in the plane Jo"" R0%, perpendicular to On. Now let Fig. 191. each force of this couple be made equal _ to R, and the arm, OP*, is consequently equal to -=- ; that is, OP=^*. (,) * The point P should be represented on the production of the line xO through 0, according to the convention of Art. 137. The inaccuracy in the figure waa detected too late for correction. 226 EQUILIBRIUM OF A RIGID BODY. [l44- One of these forces may be applied at to destroy the re- sultant, R, at this point, and there finally remains a resultant force, R, at P along PT (parallel to OK), together with a couple whose axis is OK, or G cos <. Denoting OK by K, we have then K=Gcos. (2) The axis OK may, of course, be drawn at P along PT [(a), Art. 137]. Hence the whole system of forces is equivalent to a resultant force equal to R acting along the line PT and a couple in a plane per- pendicular to this line. The line PT thus determined is called Poinsot's Central Axis. To construct Poinsot's Central Axis for any system of forces Reduce the forces to a resultant force, OR, acting at any origin, 0, and a couple whose axis is OG ; then on a line perpendicular to the plane of OR and OG measure off a length, OP*, equal to P ? where is the angle between OR and OG. A line through the point P parallel to OR is the required Central Axis. 144.] Theorem. A given system of forces has but one Central Axis. This, which is sufficiently evident from the preceding con- struction, may be proved ex absurdo thus : Whatever origin be chosen, the resultant force acting at it is constant both in magnitude and in direction. Now, if it be possible, let the system of forces be equivalent to a resultant R acting at and a couple whose axis is OK, and also to a re- sultant force R acting at 0' and a couple whose axis is O'K', the lines OK and O'K' being, of course, in the direction of R. Now it is evident that the force R at and the couple OK should equilibrate the reversed force R and reversed couple O'K' at (7. But the couples give a single couple, OK <+, O'K', and the forces give also a couple which, being in a plane perpendicular to the first couple, cannot with it produce equilibrium. Therefore, &c. Since this axis is unique, equation (2) of the last Article shows that for all origins the quantity G cos 0, or the projection of the axis of the resultant couple along the direction of the resultant force is constant. 145.] Theorem. The sum of the moments of the forces round Poinsot's Axis is less than the sum of their moments * The sense of OP is determined by the convention of Art. 137. I 4 6.] PROBLEM. 227 round any other axis of principal moment. (Since for any origin, 0, the sum of the moments round OG is greater than the sum of the moments round any other line through (Art. 142), OG is called the Axis of Principal Moment at 0.) Let Oz (fig. 192) be Poinsot's Axis and OK (= K) the moment of the forces round it. Let (7 be any point in the body, and let it be proposed to find the principal moment at this point. (70 is a line drawn through 0' perpendicular to Poinsot's Axis. At 0' introduce two equal and opposite forces, O'R and O'R', each = R. Then OR and O'R' form a K.' couple, whose axis, O'n is perpendicular Q to the plane ROO'R' and equal to / R . 0(7. Transfer the axis OK to &K' (Art. 137), and draw the diagonal, O'G, of the rectangle determined by O'n and &K'. Then &G(=G) is the axis of principal moment at (7, and it is evidently > O'K'. Hence Poinsot's is the least principal moment. 146.] Problem. To find the surface traced out by the axes of principal moment at points taken along a right line inter- secting Poinsot's Axis perpendicularly. Let Ox be the assumed line, and let it be taken as axis of #, Poinsot's Axis being that of z. Let 00 7 = #, and let y and z be the co-ordinates of any point on (7(r. Then if = the angle GO'K', we have z Gn K = cot = Fig. 192. R . K or an equation which denotes a hyperbolic paraboloid. As the point (7 moves out from along Ox, the axes revolve towards the right ; as 0' moves in towards 0, they revolve towards 'the left; and after coincidence with Poinsot's Axis at 0, they revolve towards the left. At an infinite distance from the axis of principal moment is at right angles to Poinsot's Axis. Let it be proposed to find the surface traced out by the axes of principal moment at points taken all along an arbitrary curve in a plane perpendicular to Poinsot's Axis. qa 228 EQUILIBRIUM OF A RIGID BODY. [147. Let Q be any point on the curve whose equation in the plane xy is/ (#, y) = 0, and let (a, (3) be the co-ordinates of Q, and the point in which Poinsot's Axis meets the plane of xy. Then the axis of principal moment at Q is constructed by drawing QN, in the plane xy, perpendicular to OQ, taking on QN a length = R. OQ, drawing at Q a perpendicular to the plane xy equal to K, and constructing the diagonal of the rectangle determined by these two latter lines. Sup- pose P to be any point on the axis of principal moment at Q, and let N be the projection of P on the plane xy. The co-ordinates of P being x } y, z, it is clear that If is the angle made by QN with the axis of x, a = x + QN cos0 R . OQ cos ==aH or a = x+~p. (1) Similarly, ft = y - - . a. (2) Solving these equations for a and p, we have S R x+-Tfy* y--f7 xz fi Hence, since / (a, /3) = 0, we have E R which is the equation of the surface traced out. 147.] Theorem. A system of forces can be reduced to two forces in an infinite number of ways. For they can be reduced to a resultant force, R, acting at any point, together with a couple. Now the forces of the couple can be made of any mag- nitude by varying its arm ; and one of them can be combined with R. There will then remain the resultant of R and this force together with the remaining force of the couple. There- fore, &c. I49-] SYMMETRICAL REDUCTION OF A SYSTEM OF FORCES. 229 148.] Theorem. When a system of forces is reduced to a pair of forces represented in magnitudes and lines of action by two right lines, B Q\ the volume of the tetrahedron formed by these lines is constant, however the re- duction is made. Q Let the system of forces be reduced to P and Q, and let these be supposed to i act at the extremities, A and B, of the \<$ shortest distance between them. Now to -^ get the force and couple corresponding to the origin A, introduce at this point two opposite forces, AQ and AQ', each equal and parallel to Q. Compounding P and Q we get the resultant force, R ; and taking the forces Q at B and Q' at A we get a couple whose axis, AG, is at right angles to the plane QBAQ* and equal to Q . AB. Since AB is perpendicular to both P and Q, it is clear that AG is in the plane QAP and at right angles to AQ. Now since (Art. 143) G cos = K, we have Q . AB . sin Q AR = K. p But sin Q AR = -=- . sin PAQ. Hence P.Q.AB.smPAQ=K.R. Now the volume of the tetrahedron formed by the lines AP and BQ = J area ABQ X perpendicular from P on the plane ABQ ; = BQ.ABx AP. sin PAQ ; = $P. Q .AB . smPAQ. Hence if A denotes the volume of the tetrahedron, o * * This theorem has been proved in various ways. For an elegant demonstration by Mobius, see Crelle's Journal, vol. iv, p. 179, or Jullien's ProUemes de Mecanique RationneZle, vol. i, p. 71. / 149.] Symmetrical Reduction of a System of Forces. A system of forces can be reduced to two forces equal in magni- tude, equally inclined at opposite sides to Poinsot's Axis, and equally distant from this axis. This is what Thomson and Tait call the Symmetrical Case. 230 EQUILIBRIUM OF A RIGID BODY. [150. Suppose the forces replaced by R acting along Poinsot's Axis, Oz, and a couple, K. Take any point, 0' (fig. 193); draw (70 perpendicular to Oz and produce it to 0" so that (70 = 0(7'. Let R acting at be replaced by \R acting at (7 and \R acting at OF'. Also let the forces of the couple act at 0' and (7' ; for TP~ this purpose these forces must each be made = > a? being 1 00' . K 2x Now the resultant of \R and at 0' is a force 2% acting towards the right, and the resultant of \R and at 0" is a force of the same magnitude acting towards the left of the figure. If o> is the angle made with Poinsot's Axis by these new forces at (7 and (7', .^ jtL tan o> = ^- - - .=^f tx If we choose % so that = \/3 R, each of the two symmetrical forces is equal to R, and they are inclined at an angle of 60 to Poinsot's Axis. 150.] Analytical Condition for a Single Resultant. We have just seen that a system of forces acting on a rigid body is, in general, equivalent to two forces. Let the forces be replaced by a single resultant force, R, acting at an arbitrary origin, 0, and a couple G. Now the direction cosines of R referred to axes Ox, Oy, and Oz, are (Art. 142), v V "^7 , and ; R and those of G are M . N -' and -' Hence, if $ is the angle beeween G and R, Now if the resultant couple is in a plane containing R, one of its forces can be made to destroy R, and there will remain a single force ; but if G and R are not at right angles to each 151.] THEOREM. 231 other, the system of forces cannot be equivalent to a single force. The required condition is, therefore, cos = 0, or L2X + M2Y+1\ 7 2Z= 0, 60 (2) provided that 2X, 27, and 2.Z do not all vanish ; for if they do, R will also vanish, and < will be illusory. In fact, in this case, since L, M, and N alone exist, the system of forces is equivalent to a couple. 151.] Theorem. The quantity L^ X+ M^Y+N^Z has the same value for all systems of rectangular axes assumed anywhere in space. From (1) of the last Article it = R . G cos $, or R . K, where A" is Poinsot's moment (Art. 143). Hence, if this quantity vanishes for any one set of axes, the force and the axis of the accompanying couple corresponding to any origin are at right angles. The value of this quantity can be exhibited in another form which also shows that it is independent, of any particular set of axes. Substituting for L, M, and N the values (Art. 142), ^(Zy Yz), &c., the expression becomes or, substituting for X 19 T 19 Z 13 ... in terms of the forces P l) -.. and their direction-cosines, [ p i (fi cos n*i cos &).+ P i (y-2 cos Xa "2 cos ft) + ] (P x cos a x + P 2 cos a 2 + . . .) -f &c ..... It is clear at once that the terms Pj 2 , P 2 2 , ... disappear, and the products Pj P 2 , PI P 3 , ... alone remain. Collecting the coefficient of P x P 2 as a typical term, we have p i P 2 [(*i 2 ) ( cos A cos y 2 -~ cos y\ cos ft) 4 (y\y^ (cos y a cos a 2 cos a x cos y 2 ) 4 (^1^2) ( cos a i cos ft cos ft cos a 2 )]. Now (see Salmon's Geometry of Three Dimensions, p. 31, third edition, or Frost's Solid Geometry -, p. 39) if (Pj , P 2 ) denotes the angle between the directions of the forces Pj and P 2 , the 232 EQUILIBRIUM OF A EIGID BODY. quantity in brackets = d 12 . sin (P 19 P 2 ), d l2 being the shortest distance between the lines of action of the forces. Hence J J 2X + M2Y+NSZ=2P 1 P 2 .d 12 .sm(P l ,P 2 ). (l) Again (Art. 148) the term involving PjP 2 on the right side of (l) denotes six times the tetrahedron formed by Pj and P 2 ; therefore the quantity on the left side is equal to six times the sum (with their proper signs) of the - ' tetrahedra which can be formed out of the pairs of lines representing the n forces P I , P 2 , . . . P rt . This sum has, of course, no reference to any set of axes, and hence the necessarily invariant nature of Z2JT-J- M2Y+N2Z. With regard to the sign to be given to any tetrahedron of the system, we define that The moment of a force with regard to a line is the component of the force perpendicular to the line multiplied oy the shortest distance between the force and the line. Hence P a . d n . sin (P 15 P 2 ) is the moment of P x about the line of action of P 2 . Now to determine the sign which must be given to any tetrahedron, let a watch be placed so that the direction in which either force acts passes perpendicularly from the back up through the face of the watch. If then the other force tends to produce rotation in the sense in which the hands rotate, the tetrahedron is to receive a negative sign, and if the rotation is the other way, a positive sign. 152.] Conditions of Equilibrium of a Body Acted on by any Forces. The forces having been reduced to a resultant of translation, R, acting at any point, together with a corresponding couple, G; since a force and a couple cannot conjointly produce equilibrium ( (c), Art. 137) it is necessary that R = and G = 0. Substituting the values of R and G given in Art. 142, we see that these two are equivalent to the following six conditions: = 0, 2T=0, which are the analytical expressions of the fact that the forces must have no component along any line and no moment about any axis. 152.] EXAMPLES. 233 EXAMPLES. 1. "When three forces keep a rigid body in equilibrium, they must be coplanar and concurrent or parallel. Let the forces be P, Q, and R. Then the sum of their moments about every axis is zero. Take any point, p, on P and from it draw a line meeting Q in the point q, suppose. Then, since two of the forces have zero moments about this line, the moment of the third force, R, about it must = ; that is, this line intersects R, in the point r, suppose. Let another line be drawn through p meeting Q in q f . Then, as before, this line must meet R in a point, r'. Now since two points on each of the lines Q and R lie in the plane of the lines pqr and pq'r ', the lines Q and R must both lie in this- plane. Again, drawing any two lines across Q and R, each of these lines must intersect P j that is, P has two of its points in the plane of Q and R, and P, therefore, lies in this plane. Finally, taking moments about the intersection of Q and R, we see that P must pass through this point ; but if any two are parallel, the third must be parallel to them. 2. A rigid body is acted on by forces represented in magnitudes and lines of action by the sides of a gauche polygon taken in order ; prove that the forces are equivalent to a couple, and that the sum of their moments about any line is represented by double the area of the projection of the polygon on a plane perpendicular t the line. Let the forces be represented by the lines AR, BC, OD, ... (fig. 194), and let OQ be any axis. On the axis take any point, 0, and reduce the forces to a resultant, R, of translation at this point, together with a couple, G (Art. 142). This is done by in- troducing at two forces parallel and equal to AB in opposed directions, two equal and opposite to EC, &c. Now (Art. 136) the resultant of translation vanishes, and the component couples are represented by double the areas of the triangles OAB, OBG, &c. If the axes of these couples are drawn at 0, the sum of the moments of the forces about OQ will be represented by the sum of the components of the axes along OQ ; but this is the same as double the sum of the projections of the areas of the triangles on a, plane perpendicular to OQ ; that is, the moment about OQ is represented by double the area of the projection of the polygon on a plane perpendicular to OQ. Again, since G is the greatest moment round any axis through (Art. 142), it follows that the axis of the resultant couple is the line perpendicular to the plane on which the projected area of the polygon is a maximum. 234 EQUILIBRIUM OF A RIGID BODY. [152. 3. When the resultant of translation vanishes, the forces will be in complete equilibrium if the sums of their moments round any three non-coplanar axes are separately equal to nothing. For if L be the moment round the axis of x the moment If round a parallel axis through the point (a, /3, y) is Z + y2 Yfi^Z. Hence L' = L, M'= M, N'=. N ; and since the moment round an axis through (a, /3, y) making angles A, /JL, v with the axes of co-ordinates is L' cos A + M' cos p + N' cos v, it follows that the moments round all parallel axes are equal. For the three axes of moments we may take, therefore, three lines through the origin making angles (\ lt /^, v,), (^2 M-2J ^2)1 anc * (A 3 , jx 3 , v 9 ) with the axes of co-ordinates. Suppose then that L cos Aj + M cos ^ + N cos v l = 0, L cos A 2 + M cos fji z -f W cos r 2 = 0, and L cos A s + N cos p z + N cos ^ 3 = 0. These require either that L = M N = 0, or cosAj, cosA 2 , cos/z 2 , cosz> =0. COSAg, COSjUg, The latter condition requires that the three axes of moments be in one plane. If they are not coplanar, we must have L = M = N = 0, i.e. the forces are in equilibrium. 4. A tetrahedron is acted on by forces applied perpendicularly to the faces at their respective centroids. If the force applied to each face is proportional to the area of that face, prove that the tetrahedron is in equilibrium, the forces being supposed to act all inwards or all outwards. Let A, B, C, D, be the vertices of the tetrahedron and denote the areas of the faces opposite these vertices by^, B 1 ,C- L , D I} respectively. Denote also the angle between the faces A 1 and B 1 by A l B r Then evidently A A A l = B l cos A l B l + C : cos A l C + Z> x cos A 1 D l ; or if the forces perpendicular to the faces are denoted by P, Q, R, S, P-Q. cos PQ-R. cos P*R-S.cos PS = 0, which shows that there is no resultant force in a direction perpen- dicular to the face A l ; similiarly there is no resultant force in direc- tions perpendicular to the other faces ; therefore the resultant of translation vanishes. To show that there is no resultant couple, let each force be replaced by three equal forces acting at the angles of the corresponding face. Thus the force P is to be replaced by three forces each equal to ^ P acting at the points B, C, D perpendicularly to the face BOD. Let us 152.] EXAMPLES. 235 calculate the sum of the moments of the forces about the edge EG. For this purpose, let the forces \ Q and \ R at D be each resolved in the direction of the force ^ P at this point, i. e. perpendicularly to the face BCD. Supposing the forces to act outwards, the components of J Q and ^ R are ^ Q . cos PQ and ^ R . cos PR ; therefore the sum of the moments of the forces at D about EC is proportional to (A l BI . cos A l l C l . cos ^ (7j) p', r A or, again, D^ . j>, p' being the perpendicular from D on JBC, and p the perpendicular from D on the base ABC. But this last expression is three times the volume of the tetrahedron. In the same way, the sum of the moments of the forces at A is represented by three times the volume of the tetrahedron, and as these moments are in opposite senses, the forces have no moment round the edge C, and similarly no moment round any of the edges. Hence by the last example they are in equilibrium. For another simple method of proof see Collignon's Statique, p. 354. 5. Prove that a solid body of any shape is in equilibrium if it is acted on throughout its surface by normal forces, each force being proportional to the superficial element on which it acts. One very simple method of proof consists in imagining a surface precisely equal and similiar to that of the given body to be traced out in a weightless fluid which is subject to any pressure. 6. If a curved surface whose edge is a plane curve is acted on all over its surface by normal forces, each proportional to the element of surface on which it acts, prove that these forces have a single resultant if they all act towards the same side of the surface. 7. Forces perpendicular and proportional to the areas of the faces act at the centres of the circles circumscribing the faces of a tetra- hedron ; prove that they are in equilibrium, if they all act inwards or outwards. They meet in the centre of the circumscribed sphere. The proposi- tion is evidently true also for any polyhedron bounded by triangular faces. Taking the results of this example and example 4 together, we see that forces proportional to the areas and perpendicular to them are in equilibrium if they act at the orthocentres of the triangular faces of any polyhedron. 8. Find the force necessary to keep a heavy door in a given position, the hinge line being inclined to the vertical and the hinges smooth. Let i be the inclination of the hinge line to the vertical, and a the given inclination of the plane of the door to the vertical plane con- taining the hinge line. Then if W is the weight of the door, a the 236 EQUILIBRIUM OF A RIGID BODY. [-5*. distance of its centre of gravity from the hinge line, and the angle between the normal to the plane of the door and the vertical, the moment of the weight about the hinge line is Wa cos 0. This is the moment of the required force. To find 0, let lines parallel to the hinge line and the vertical be drawn through any point, 0, and through this point let a plane be drawn parallel to the plane of the door. Round let any sphere be described; let V and L (fig. 195) be the points where these lines meet the sphere ; DL the circle in which the plane of the door in- tersects the sphere, and N the point in which the normal, ON, to the door intersects it. Then VL = i, /. DL V = a, and NV = 0, and we have from the spherical triangle VDL sin VD = sin i sin a, or cos = sin i sin a, since N is the pole of DL. Hence the moment of the required force is Wa sin i sin a, and when its point of application and direction are known, its magni- tude is therefore known. 9. A beam can turn in every direction about one end which is fixed; the other end rests on a rough inclined plane. Find the limiting position of equilibrium. (See Walton's Mechanical Problems, p. 191, third edition.) Let AB (fig. 196) be the beam, A the fixed end, DPH the rough inclined plane, PH the intersection of this plane with a horizontal plane through A , A PD the vertical plane through A perpendicular to the inclined plane, BD a line parallel to PH, AO a perpendicular from A on the inclined plane, DQ a perpendicular on the horizontal plane, i the inclination of the plane, a the angle, A SO, between the beam and this plane, and jut the coefficient of friction. Now suppose first that the beam is per- fectly inelastic. Then the end B describes on the inclined plane a circle whose centre is 0, and if it is about to slip, the force of friction assumes a direction perpen- dicular to OB in the inclined plane. The extreme position of the beam will be denoted by the angle, or DOB, between the plane, A OB, through the beam normal to the inclined plane and the vertical plane, AOD. The forces acting on the beam are its weight, the reaction of the smooth joint at A, and the total resistance of the inclined plane at B. This last force we shall consider as composed of a normal reaction, R, Fig. 196. 152.] EXAMPLES. , 237 and a force of friction, pR, acting perpendicularly to BO. For the equilibrium of the beam take moments about a vertical axis through A. The moment of the normal reaction at B is R sin i x BD, or R sin i . BO sin 0, or again R sin . AB cos a sin 0. To find the moment of pR, resolve it into p,Rcos0 along BD and fjiRsin0 parallel to OD ; and resolve this latter again into a horizontal component, pR sin0 cosi, and a vertical component, pR sin# sin. The moment of jut/? is then equal to the sum of the moments of [JiR cos and pR sin cos i ; that is, it is equal to Hence the equation of moments is 72 (sin ip cos sin 0)j8Z> = pR cos 0.AQ. AO But ^O = ^P + P0 == -T ; + (OD-OP)cosi sin * 4 B . sin a _ t * cos ^ cos a cos QAB sin a cot i cos ^ smt = .42? (sin i sin a + cos i cos a cos 9) ; therefore (sin ft cos i sin 0) cos a sin = fj, cos (sin i sin a + cos i cos a cos 0), or sin i cos a sin = p cos i cos a + ju sin sin a cos 0, or tan i tan = ju yl + tan 2 + ft tan i tan a, or finally, (tan 2 -fi a ) tan 2 0-2/x tan 2 tan a tan + p 2 ( tan 2 i tan 2 a- 1) = 0. (1) If there is no horizontal plane through A obstructing the beam, it will be possible for the end B to describe a complete circle round 0. Let us inquire the condition that the beam should rest in all possible positions. For this there must be no limiting position of equilibrium, or, in other words, the value of in (1) must be imaginary. The required condition is, then, tan 2 (l+/x 2 tan 2 a) VAN ; for the vertical line is included within the cone, and through this line no plane can be drawn to touch the cone. There can, therefore, be no limiting equilibrium at either end in any position of the beam. Fig. 198. 152.] EXAMPLES. 239 11. A particle is acted on by any number of given forces, P 1? P 2 , ... ; prove that if R is their resultant, R* = 2(P') + 22(P 1 . P 2 cos P> 2 ), where P a P 2 denotes the angle between the directions of P x and P 2 . 12. Prove that a system of forces acting on a rigid body may be replaced by two equal forces whose lines of action are perpendicular to each other, and each inclined at an angle of 45 to Poinsot's Axis : the forces act at the ends of a line bisected by this axis ; the length 2K R of this line is -=r- > and each force is > R being the resultant of fi v 2 translation, and K Poinsot's moment. 13. Prove that the distance between the lines of action of the two forces which equivalently replace a given system of forces is a minimum when the forces are equal and their directions perpen- dicular. 14. Prove that the central axis of two forces divide the shortest distance between them into two parts which are inversely proportional to' the components of the two forces along the direction of their resultant. 15. A BCD is a tetrahedron; forces P, Q, R act along the edges BC, CA, AB in order, and forces P', Q', R' act along AD, ED, CD ; prove that the condition for a single resultant is rr , QQf , ssr _ AB. CD ~ 16. A rough heavy body, bounded by a curved surface, rests upon two others which themselves rest on a rough horizontal plane ; show that the three centres of gravity and the four points of contact lie in one plane. 17. A heavy beam rests on two smooth inclined planes ; show that their line of intersection must be perpendicular to the beam and parallel to the horizon. 1 8. Prove that the moment of a force represented by the right line PQ about a right line AB is six times the tetrahedron ABPQ divided l>y AB. 19. Three equal heavy spheres hang in contact from a fixed point by strings of equal length ; find the weight of a sphere of given radius which when placed upon the other three will just cause them to separate. Ans. If W and a be the weight and radius of each of the three spheres, W' and r the weight and radius of the superincumbent sphere, and I the length of each string, W 20. Three spheres are placed in contact on a rough horizontal 240 EQUILIBRIUM OF A RIGID BODY. [153. plane, and a fourth sphere is placed upon them, there being no friction between the spheres themselves. Show that equilibrium is impossible. 21. Three equal spheres are placed in contact on a rough horizontal plane, and a fourth sphere is placed upon them, there being friction between the spheres themselves. Find the least coefficient of friction between the spheres which will allow of equilibrium. Ans. If a is the radius of each of the equal spheres and r that of the superincumbent sphere, the least value of A, the angle of friction, is given by the equation sin 2A = =. --- -v/3 (The total resistance between the upper sphere and any one of the lower spheres must be capable of acting through the point of contact of the latter and the ground.) 22. Three forces whose lines of -action are given, but not their magnitudes, have a single resultant. Prove that the surface traced out by the line of action of the resultant is a hyperboloid of one sheet. (Draw any three lines across the given lines of action. Then the line of action of the resultant must always intersect these three.) 23. A heavy triangular plate of uniform thickness is suspended from a fixed point by means of three strings attached to the point and to the vertices of the plate ; prove that the tension in each string is proportional to the length of the string. (Let be the fixed point, A, B, C the vertices of the plate, and G its centre of gravity. Then G must lie vertically under 0. Take OG to represent the weight of the plate. Then, by Leibnitz's graphic repre- sentation [Art. 136], the force OG may be resolved into the forces OA, OB, OG. But a given force can have only one set of components along three given concurrent lines. Therefore, &c.) 24. At points on any right line the axes of principal moment of a given system of forces are drawn ; prove that their extremities trace out another right line. (Wolstenholme's Problems, p. 387, 2nd edition.) (At any point on the given line draw R and G. Take as axes of x, y, and z the given line, the line OG, and a line at perpendicular to R and the given line. Then at any point P on the given line at a distance x from if the axis of principal moment be drawn the co-ordinates of its extremity will be x, G, and Rx sin a, where a is the angle which R makes with the given line. Hence the extremities lie on the line y = G, z = Rx sin a.) 25. Prove that the axes of principal moment at points along any right line whatever trace out a hyperbolic paraboloid. (With the same axes as in last example, the surface has for G . equation mi = - : %.) J ' 152.] EXAMPLES. 241 25. Find the condition that a given right line should intersect Poinsot's axis. Ans. If the equations of the line are x =.mz +p, y = nz + q, the required condition is R[mL + nM+N-\-q(X-mZ}-p(Y-nZ}'\ = K(mX+nY+Z), where X is used for 2 JT, &c. (It will be found that the equations of Poinsot's axis can be put into the forms X KY-MR Y KX-LR, the origin being anywhere.) 26. The first case considered in example 9 is, equally with the second, a geometrico-statical problem. Solve it without any mention of force. [Express the condition that the vertical through the extremity A of AB is intersected by a line inclined at angle A. to the normal at B, this line lying in the plane of the normal and a perpendicular to CHAPTER XL CENTROIDS [CENTRES OF GRAVITY]. 153.] Centre of Mass. Imagine a body broken up into an indefinitely great number of infinitesimal elements of mass (without altering the relative positions of these elements) and find the mean centre of all the points at which these elements are placed, the multiple associated with each point being pro- portional to the element of mass at the point. Then if the distances of the elements dm l9 dm. 2) dm z , ... from any plane are ^ 15 z 2) 3 , ... , the distance of the mean centre from the plane is Z-L dm l -f z z dm* + ... fzdm - - j or ^ dm 1 + dm 2 + ... Jam The point thus arrived at is called the Centre of Mass of the body ; it is also often called the Centre of Inertia ; and the term centroid has lately come into use to designate it. The distance of the centre of mass from any plane is the mean distance of the body from the plane. If each element of mass is acted on by a force proportional to the mass of the element, and these forces form a parallel system ; and if w is the magnitude of the force per unit of mass, the distance of the centre of this parallel system of forces from the plane is fwzdm fzdm 7 T * Or jr^j > Jwam Jam since w is a constant. Thus the centre of the parallel system coincides with the centre of mass. The earth produces such a parallel system of forces on the elements of a body, and therefore the point thus arrived at has, until very recently, been universally called the Centre of Gravity of the body. It is only when we consider the action of such a parallel system of forces on the body as the attraction of the earth supplies that the point in question should bear the particular epithet of Centre of Gravity. 1 54.] THEOREM OF MOMENTS. 243 In numerous questions relating to the body in which the action of gravity is not considered the centre of mass plays a most important part ; and it is a point possessed by the body quite independently of any force whatever acting upon it. Hence the latter term is the one most strictly appropriate to the point determined as above ; and, except when the weight of the body is concerned, we shall use the terms centroid and centre of mass instead of centre of gravity. 154.] Theorem of Moments. If any number of masses be multiplied each by the distance of its centre of mass from any plane, the sum of the products thus obtained is equal to the total mass multiplied by the distance of its centre of mass from the plane. The centre of mass of any number of finite masses is obtained in precisely the same manner as the centre of mass of a number of particles. Thus, if m l and m 2 are the masses of two bodies of any magnitudes, their centre of mass is obtained by dividing the line joining their respective centres of mass in the ratio m 1 : m. 2 , just as if two particles of masses m 1 and m. 2 were placed at these points. Hence the distance, #, of the centre of mass of any number of finite masses from any plane (that of yz) is given by the equation _ '' or H. x 2mx, and the theorem at the head of this Article is merely the expression of this equation. It is obvious that the formulae which have been given for the co-ordinates of the centre of mass hold whether the axes le rectangular or oblique. For in Art. 79, p. 96, on which our formulae are founded, the distances of the points A 19 A 2 , ... from the line (or plane) L may be assumed to be measured in any common direction. It follows that if any plane be drawn through the centre of mass of a system of masses, the sum of the products obtained by multiplying each mass by the distance of its centre of mass from the plane is zero. If the plane be that of (yz\ and if SB' be the distance of the centre of mass of the mass m from the plane, this result is expressed by the equation ^mx' = 0. Given the centres of mass, g^ and # 2 , of two masses, m 1 and m 2 , 24:4: CENTROIDS [CENTRES or GRAVITY]. [155. the centre of mass of the two as one system is a point, (?, on the line fa g% dividing it in the ratio -~ = Cr^ 2 % Given the centre of mass, G, of a mass M, and also the centre of mass, fa, of a portion, m lt of the mass, the centre of mass, ^ 2 , of the remainder is a point on the line gG produced through G, r \ 3 err ff^lA^^^ when dv (and therefore dm) is indefinitely diminished, is the density of the body at the point considered. 156.] Centre of Mass of a Triangular Lamina of Uniform Thickness and Density. Let ABC be any triangular lamina of uniform thickness and density, and let it be divided by an indefinitely great number of lines parallel to the base BC into an indefinitely great number of strips. Then the centre of mass of each strip is its middle point ; and the middle points of all the strips lie on the line joining A to the - middle point of BC. Hence the centre of mass of the lamina lies on this line. Similarly, the centre of mass lies on the line joining B to the middle point of CA. It is therefore the intersection of the Usectors of the sides drawn from the opposite angles. Again, the centre of mass of a uniform triangular lamina coin- cides with the centre of mass of three equal particles placed at its vertices. For, the centre of mass of the two equal particles at B and C is the middle point of BC, and the centre of mass of the three I57-] CENTRE OF MASS OF A PYRAMID. 245 ^ , 2 , and # 3 , the distance of its lies on the line joining this point to A. Similarly, it lies on the line joining B to the middle point of CA. Therefore, &c. If the mass of each particle is m, the centre of mass divides the line joining A to the middle of BC in the ratio 2 m : m, or 2:1. Hence the centre of mass of a triangular lamina of uniform thickness and density lies on the bisector of any side drawn from the opposite angle at the point of trisection (nearest to the side) of the bisector. COR. If the distances (rectangular or oblique) of the "vertices of a triangle from any plane are i centre of mass from this plane is 3 <3 157.] Centre of Mass of a Triangular Pyramid of Uniform Density. Let ABCD (fig. 199) be a triangular pyramid. Now if any vertex, D } be joined to the centroid, N, of the oppo- site face, the joining line passes through the centroids of all triangles in which the pyramid is cut by planes parallel to this face. For, let abc be a section of the pyramid parallel to the base ABC. Draw the plane CND containing the lines CD and DN; this plane bisects the base AB in H, since (Art. 156) CN bisects AB. Let the plane CND intersect the face ABD in the right line HhD, h being the point in which this line meets ab. Then since in the triangle ABD, ad is parallel to AB, and DH bisects AB, h is the middle point of ab. Again, if the line DN meets the plane abc in n, the points h, n, and c are in a right line. For these are evidently points common to the planes CND and abc, and since two planes intersect in a right line, the points h, n, c are in a right line that is to say, n is a point on the bisector of the side ab drawn through c. Similarly, n is a point on the bisector of be drawn through a ; therefore n is the centroid of the triangle abc (Art. 156). To find the centre of mass of the pyramid, let it be divided by planes parallel to ABC into an indefinitely great number of triangular laminae. Now we have just proved that the centres of mass of all these laminae lie on the line, DN, joining the Fig. 199. 24:6 CENTROIDS [CENTRES OF GRAVITY]. [158. vertex D to the centroid of the opposite base. Similarly, the centre of mass of the pyramid lies on the line joining the vertex A to the centroid of the face BCD. It is, therefore, the point, G, of intersection of lines drawn from any two vertices to the centroids of the opposite faces. But this is exactly the con- struction for the centre of mass of a system of four equal particles placed at the vertices of the pyramid. Hence The centre of mass of a triangular pyramid coincides with the centre of mass of four equal particles placed at its vertices. Also The centre of mass of a triangular pyramid is one-fourth of the way up the line joining the centroid of any face to the opposite vertex. For, if at the vertices there be placed four equal particles, each of mass m, their centre of mass is found by joining D to N and taking-^ = -- = J, therefore GN^GD, or Cr JJ 3 in NG = \ND. COE. 1 . The perpendicular distance of the centre of mass of a triangular pyramid from the base is equal to \ height of pyramid. COR. 2. If the distances (rectangular or oblique) of the e . , r V 7 1.1. J' vertices of a pyramid from any plane are se lt x^x^ # 4 , the dis- tance of the centre of mass from the plane is 158.] Centre of Mass of a Cone of Uniform Density having any Plane Base. Consider a pyramid whose base is a polygon of any number of sides. Then, by dividing the base into triangles we can consider the whole pyramid as composed of a number of triangular pyramids. Now (Art. 157) the centre of mass of each of these pyramids lies in a plane whose distance from the base is one-fourth of the height of the pyramid ; there- fore the centre of mass of the whole pyramid lies in this plane that is, its perpendicular distance from the base is one-fourth of the height of the pyramid. Again, dividing the pyramid into an indefinitely great number of laminae, as in last Art., the centres of mass of these laminae all lie on the right line joining the vertex to the centroid of the base. Hence the centre of mass of the whole pyramid lies on this line ; and by what we have just proved, it must be I59-] THEOREM. 247 one-fourth of the way up this line. There is no limit to the number of sides of the polygon ; hence they may form a con- tinuous curve. Therefore The centre of mass of a cone whose base is any plane curve what- ever is found by joining the centroid of the base to the vertex, and taking a point one-fourth of the way up this line. '" 159.] Theorem. If 'the mass of each of a system of bodies be multiplied by the square of the distance of its centre of mass from a given point : , the sum of the products thus obtained is least when the given point is the centre of mass of the system of bodies. This theorem, which is well known in elementary geometry, admits of a very simple analytical proof. Let (#, y, z) be the co-ordinates of the centre of mass, G, of the system with reference to rectangular axes through any point, 0, and let (#1,^1 j^i), (#2'^2> z z)> ^ e ^ ne co-ordinates of the centres of mass, A 13 A z ^ ... , of the bodies whose masses are %, %, .... Then GA* = (x-xtf+(y-ytf+(z-z^ (i) Similarly, GA 2 = (*-* 2 ) 2 -f (?-y 2 ) 2 + (z-%) 2 , (2) Multiplying these equations by %,%,..., and adding, we have 2 (m . GA 2 ) = (x 2 +y* + z 2 ) . ^m-2x . ^mx - 2y . 2 my ^z.Zmz + Zm^+f + z 2 ). (3) Now (Art. 154), ^mx = #.2#&, ^my = y . 2?#, Hence (3) becomes or 2 (m . GA 2 ) = 2 (m . OA 2 ) -OG 2 . 2m, (4) from which equation it appears that 2 (m . GA 2 ) is always less than 2 (m . OA 2 ) by the quantity OG 2 . ^m. It can be shown that, if r l2 denote the distance between the centres of mass of the masses % and m 2 , and M the sum of all the masses, M.2(m. GA 2 ) = 2 (% m 2 r 12 2 ). For, let the centre of mass, G, be taken as origin. Then, denoting the co-ordinates of the points A lt A 2 ,... with reference to G by (a?/, y/, z^), 248 CENTKOIDS [CENTRES OF GRAVITY]. [159. m . GA^ = m l ( % + m 2 + . !+.... (5) Also (Art. 154) = = Squaring each of these last three equations, adding the results together, and subtracting their sum from (5), we have Q 9. 9. M.2(m.GA 2 ) = t Hence, from (4), OG = M M* under which form Lagrange expresses the distance of the centre of mass of a system of bodies from a given point (see Mecanique Analytique^ p. 61). Equation (4) can be employed to prove the well-known ex- pression for the distance between the centres of the inscribed and circumscribed circles of a plane triangle, viz. D being the distance between the centres, and r and E being their radii, respectively. (Suppose a system of particles at the vertices, the mass of each being proportional to the opposite side. Their centre of mass is the centre of the inscribed circle. The remainder is left to the student as an exercise.) EXAMPLES. ^ 1. To find the position of the centre of mass of the frustum of a pyramid. Let the frustum be formed by the removal of the pyramid abcD {%. 199) from the whole pyramid ABCD ; let h and ZTbe the per- pendicular heights of these pyramids, respectively ; and let m and M denote their masses. Now if the perpendicular distances of the centres of 'mass of the pyramid ABCD, the pyramid abcD, and the frustum, from the base ABC be denoted by z 1} 2 , and z, respectively, we have (Art. 154) Mz l = mz z + (M^m)K. (I) ' Af->ti. M 159.] EXAMPLES. 249 TT ~L But z l = > 2 = - +H h = H f h. Also the masses of the pyramids are to each other as the cubes of their heights ; therefore (1) gives or 4 (# 3 -7i 3 ) z = H*-4.Hh 3 + 3/i* - ff ~ h ~~ 2 z Instead of the heights we can use the square roots of the areas of the bases, to which the heights are proportional. If these areas are denoted by A and a, we have _Hh ^ , . ~ ' The centre of mass, G' ', of the frustum obviously lies on the line Nn (fig. 199) between N and G; and (3) evidently gives It is clear that the position of the centre of mass of the frustum of a cone standing on any plane base is also given by these equations. 2. To find the centre of mass of a board of uniform thickness and density whose figure is that of a quadrilateral. Let A BCD be the quadrilateral ; draw the line AC, which divides the quadrilateral into two triangles ; let L and M be the centroids of the triangles ABC and ADC, respectively ; and let the line LM meet .' ACinN. Then the centroid of the quadrilateral is a point, G, on LM such MG area ABC area ALC perp. from L on A C LN ^ ta LG = area ADC ~ area A M C = perp. from MonAC ~ ~MN ' *i. r LN therefore -7-^ = -^-^ , or MC = LN. LM LM The centre of mass is therefore found by taking a point, Gf, on LM, such that MG = LN. Another construction. The student will find little difficulty in proving the following construction. Draw the diagonals AC and BD, meeting in the point 0. On AC take a point C', such that AC'=CO, and on BD take a point B', such that DB'=BO. Then the centroid of the quadrilateral is the centroid of the triangle &OQ'. - 3. From a triangular board of uniform thickness and density the portion constituting the area of the inscribed circle is removed ; prove 250 CENTROIDS [CENTRES OF GRAVITY]. [159. that the distance of the centre of mass of the remainder from any side (a) is A 2s 3 -37raA 3 as s 2 TT A A being the area, and s half the sum of the sides, of the board. 4. If a tetrahedron be formed by the centres of mass of any four masses, prove that each mass is proportional to the tetrahedron standing on the opposite face and having for vertex the common centre of mass of the masses. 5. If at the vertices of a triangle there be placed three masses each of which is proportional to the opposite side of the triangle, prove that their centre of mass is the centre of the circle inscribed in the triangle. 6. Prove that the centre of mass of a system of uniform bars forming a triangle is the centre of the circle inscribed in the triangle formed by the middle points of the bars. 7. A figure is formed by a right-angled triangle whose sides are a, b, and c, and the squares constructed on these sides ; find the distance of the centroid of this figure from the greatest side (c). ab Ans. 3c 8. Prove that the centroid of a trapezium divides the line joining the middle points of the two parallel sides in the ratio - j the lengths of these sides being a and b. Prove also the following construction for the centroid : The vertices, in order, being A,B, C,D, and the parallel sides AB and CD, produce BA to A', and AB to B\ so that AA'= BB'= CD ; also produce DC to C", and CD to D', so that CC f DD' = AB ; then the point of intersection of A'C' and B'D' is the required centroid. 9. A right line passing through a fixed point intersects two fixed right lines ; find the locus of the centroid of the triangle formed by the variable line and the two fixed lines. Ans. If the co-ordinates of the fixed point with reference to the two fixed lines as axes are a and b, the locus is the hyperbola >-b) = ab. 10. If the right line in the last example, instead of passing through a fixed point, cut off a triangle of constant area, find the locus of the centroid of the triangle. Ans. If o> is the angle between the fixed lines, and k 2 the constant area, the locus is the hyperbola Qxysin to = 2k 2 . 11. From a sphere of radius R is removed a sphere of radius r, the distance between their centres being c; find the centre of mass of the remainder. I59-] EXAMPLES. 251 Ans. It is on the line joining their centres, and at a distance -r- 5 from the centre. 12. Every body has one and only one centre of mass. Hence show that the lines joining the middle points of the opposite sides of a quadrilateral bisect each other. (Consider four equal particles at the vertices.) 13. From the vertices of a given triangle let perpendiculars be drawn to the opposite sides. Find the distances of the centroid of the triangle formed by the feet of these perpendiculars from the sides of the given triangle. Ans. The distance from the side a is ^ a sin A cos (B C). 14. A thin uniform wire is bent into the form of a triangle ABO, and particles, of weights, P, Q, K, are placed at the angular points A, B, C, respectively; prove that if the centre of mass of the particles coincides with that of the wire, (Wolstenholme's Book of Mathematical Problems.) 15. Find the centroid of the triangle formed by the points in which the bisectors of the angles of a given triangle meet the opposite sides. Ans. If A denote the area of the given triangle, whose sides are a, b, c, the distance of the centroid from the side a is B 2a+b+c (a + 6)(a-f-c) ' 16. A uniform wire of given length is formed into a triangle of which one angle is given ; find the locus of the centre of mass of the wire referred to the sides containing the given angle as axes. Ans. If C is the given angle, and 41 the length of the wire, the locus is the ellipse C C (lxyY+2(lccy)(2lxy)$\tf + 4^ sin 4 = 0. 2 u 17. If particles be placed at the angular points of a tetrahedron, proportional respectively to the areas of the opposite faces, their centre of mass will be the centre of the sphere inscribed in the tetra- hedron. (Wolstenholme's Book of Mathematical Problems.) 18. Prove that the centroid of the surface of a tetrahedron is the centre of the sphere inscribed in the tetrahedron formed by joining the centroids of the faces. 252 CENTROIDS [CENTRES OF GRAVITY]. [160. SECTION II. Investigations requiring Integration. 160.] Rule. The general formulae, such as that in Art. 153, for the co-ordinates of the centre of mass of a quantity of matter arranged in any manner assume particular forms according as the matter is arranged in the form of a wire of any shape, an area or thin lamina of any shape, or a solid. Then, again, they assume particular forms in each of these cases according to the manner in which the matter is supposed to be divided into elementary portions. Many students are in the habit of remembering a special formula for each of these numerous cases ; such a habit, how- ever, is not only useless but injurious. It is much better to consider the formula of Art. 153, or the method of p. 97, as furnishing the following Hule which covers all possible cases : Divide the given quantity of matter, in any way> into elementary /:/*/ y //' / portions ; find the position of the centre of mass of each of these portions ; then multiply the mass of each portion by the co-ordinate* of its centre of mass> and take the integral of this product ; and finally divide this integral by the whole quantity of matter. The result is the co-ordinate of the centre of mass required. 161.] Centre of Mass of the Arc of a Curve. If the matter whose centre of mass we desire to find is arranged in the shape of the arc of any curve, the co-ordinates of its centre of mass are obtained from the formula of Art. 153, in which dm now denotes the mass of an ele- mentary length of the curve. Let ds denote the length of an elementary portion of the curve contained between two points, P Fig. 200. an( i Q ( n g"' 200) ; let Tc denote the mean area of a normal section of the curve between P and Q ; and let p denote the density of the matter in the neighbourhood of P and Q. Then, since the quantity of matter in any space is equal to the product of the volume and the density, the quantity of matter between P and Q is kpds. * The co-ordinates are supposed to be such as are measured parallel to a given line. The rule would not hold if by co-ordinate were understood polar co-ordinate, for instance. l6l.] EXAMPLES. 253 Again, the centre of mass of this element is evidently the middle point of PQ. And since to obtain G, the centre of mass of the whole mass, the co-ordinates of this middle point must be multiplied by the infinitesimal kpds, the co-ordinates of the centre of mass of PQ may be taken to be the same as those of P. Replacing dm in the general formulae by the linear element Jcpds, we obtain for the position of the centre of mass of matter arranged in the form of any curve the equations __ _ fkpocds = fkpds ' _ fkpyds " fkpds ' fkpzds ~~ fkpds ' The quantities k and p must be given as functions of the position of the point P before the integrations can be per- formed. EXAMPLES. 1. To find the position of the centroid of a circular arc of uniform thickness and density. Let AB be the arc, M its middle point, and the centre of the circle. Then it is manifest from symmetry that the centroid must lie on the line OM. Take OM as axis of x. Then since k and p are constant, we have /Ws x = ' ri > fds x being the co-ordinate of any point, P, in the arc. Let 6 be the angle POM and a the radius of the circle. Then x = a cos 0, and ds = adO. fcos6de Hence x a- r la > fdO the integration to be extended over the whole arc. Now if the angle BQA = 2 a, the integration must be taken from 6 = a to = a. Therefore sin a I x = a a Hence the distance of the centroid of the arc of a circle from the centre is the product of the radius and the chord of the arc divided by the length of the arc. The distance of the centroid of a semicircle from the centre is TT 254 CENTROIDS [CENTRES OF GRAVITY], [161. 2. Find the centre of mass of a circular arc of uniform section, the density varying as the length of the arc measured from one extremity. Let AB be the arc ; let the density at any point P = JJL. AP, and let OA be taken as axis of x. Then if 2.AOB = a, and AP = s, we have Ode o a sin a + cos a 1 , , - fsyd* h'* Similarly, y = J " a j sin a a cos a = 2a - - -- a 2 3. One extremity, A, of the arc, AB, of a curve being fixed, while the other extremity, B, varies, it is required to construct at any point the tangent to the locus of the centroid of the variable arc AB. Let AB be a portion of the arc of any curve, and let G be the centroid of AB. Then if B' be a point on the given curve very close to B, the centroid of the whole arc AB' is obtained by joining the centroid, G, of AB to the centroid of BB', and dividing the joining line inversely as the lengths of AB and BB*. But the centroid of BB' is its middle point. Hence the centroid of AB' lies on the line joining G to the middle point of BB'. In the limit, therefore, the line joining G to its next consecutive position is the line GB, which is, then, the tangent at G to the locus of G. 4. Find the position of the centroid of the arc of a semi-cardioid. A ns. The equation of the curve being r = a (I +cos 0), the co- ordinates of its centroid referred to the axis of the curve and a per- pendicular line through the cusp as axes of x and y are 4 * = y = -a. 5. Find the equation of the line joining the centroid of the arc of half a loop of a leinniscate to the double point. Ans. The axes of x and y being the axis of the curve and a perpendicular line, the equation of the required line is 6. Find the centroid of the arc of a semi cycloid. Ans. The axis of x being a tangent at the vertex, and a the radius of the generating circle, / 4 s 2 x=(v--)a, y = -a. 162.] EXAMPLES. 255 7. Find the distance of the centroid of the catenary X X from the axis of x, the curve being divided into two equal portions by the axis of y. Ans. If 2 Hs the length of the curve and k the ordinate of its extremity, the centroid lies on the axis of y at a distance from the axis of x. 21 8. Find a law of density of a wire of uniform section bent into the shape of a cycloid so that its centre of mass shall be half way up its axis. Ans. If the density varies as. the length of the arc measured from the vertex, the result will follow. 9. If the density of a cycloidal arc varies as the ?i th power of the arc measured from the vertex, find the position of the centre of mass of the curve. n-\- 1 Ans. On the axis at a distance 2 - a from the vertex, a n+3 being the radius of the generating circle. 10. One extremity of a circular arc is fixed while the other varies along the circle ; trace the locus of the centroids of the varying arcs, and prove that the algebraic sum of the intercepts of the locus on the diameter perpendicular to that passing through the fixed extremity of the arcs is equal to half the radius. 162.] Centroid of a Plane Area. Let APQB (fig. 201) be any curve whose equation is given, and let it be required to find the centroid of the area, CABD, of a lamina included between a given portion, AB, of the curve, two extreme ordi- nates, AC and BD, and the axis of x, the lamina being supposed of uniform thickness and density. In accordance with the rule of Art. 160, we break up the area into elementary portions. Suppose that this is done by taking rectangular strips, such as PQNM, included between two very close ordinates, PM and QN, and let g be the centre of mass of this strip. Let the co-ordinates of P be (x, y) and those of Q (sc + dx, y -f dy) ; let p be the density and k the thickness of the lamina. 256 CENTROIDS [CENTRES OF GRAVITY]. [162. Then the mass dm, of the rectangular strip is kpydx. Also the co-ordinates of g are (# + e, 3 + *')' an< ^ / being extremely small quantities of the same order of magnitude as dx and dy. Following the rule of Art. 160, to obtain the abscissa of G, the centroid of the area, we shall have to take the integral of the product Jcpy(x + e) dx. Now fdx is an infinitesimal of the second order, and is there- fore to be neglected in the integral. Hence if x and y are the co-ordinates of G, we have evidently, since Jc and p are constants, _ _ fxydx _ , ffdsc the integrations extending over the whole area CABD. EXAMPLES. 1. Find the centroid of the area of a semi-cycloid. Taking the line joining the extremities of the arc of the whole curve as axis of x, and a perpendicular through the vertex as axis of y, the curve is given by the equations y /\ Hence ydx = 4a 2 cos 4 - d0, and we have f"(e + sin 6) cos* -de /""cos 6 -d Jo 2 Jo ^ * = - -p o ~~' y = a r " / eos*! and that of the middle II -\- Z point of the second strip is k ^ 2 Hence if y and y' denote the ordinates of the centroids of the portions of U and U' cut off by PQ and P'Q', respectively, -,_T/* a (y'-*V* - 2 fk(y-z)dx k.y. Let PQ cut off in all positions a constant area from U; then it is evident that P'Q' cuts off a constant area from U'. Suppose, more- over, that in this case the locus of the centroid of the portion of U is a curve whose equation is f( x } = o then clearly the locus of the centroid of the corresponding portion of U' of constant area cut off by a right line is the curve A*, f) = o. If the lines PQ and P'Q' are replaced by two curves the second of which is deduced from the first as U' was from U t the same results evidently follow. 3. Find the centroid of a quadrant of an ellipse. 4a 46 Ans. *=-=' 258 CENTEOIDS [CENTEES OF GRAVITY]. [162. 4. A right line cuts off a constant area from an ellipse j find the locus of the centroid of the portion cut off. Ans. An ellipse concentric and coaxal with the given one. 2 2 5. Find the centroid of a quadrant of the curve (-) 3 +(T) = ! 2.4.6.8 2a 2. 4. 6. 8 26 Am ' X = 3-57779' V 5 V = 37577^* V (Assume x = a cos 3 , y = b sin 3 .) 6. Find the centroid of any segment of a parabola cut off by a right line. Ans. On the diameter conjugate to the given line at a distance from the curve equal to f of the portion of the diameter intercepted by the given line. 7. Through a given point, 0, is drawn a fixed right line meeting a curve in A ; through is also drawn another right line meeting the curve in P. It is .required to construct at any point the tangent to the locus described by the centroid of the area AOP as the line OP varies. Ans. Let G be the centroid of AOP, and take a point Q on OP such that OQ = %OP. Then GQ is the tangent to the locus at G. (See Example 3, p. 254.) 8. Find the centroid of a semi-ellipse cut off by any diameter. Ans. It is on the diameter conjugate to the given one and at a 4 a' distance from the centre, 2 y = x = a cos 4 (f>, y b sin 4 <.) The particular manner in which it is advisable to break up the area whose centroid is required varies with the nature of the , x \ v ^ (The equation of the parabola i s (-) + () = 1. Assume p/- \p* area itself. Thus, let the area be that included between the axis of o /A E\ ""*" x an< ^ ^ W0 curves > AC and EC (fig. 202) whose equations are Fig. 202. v . 4 ^ . given. In this case the area may be broken up into thin strips, such as PQP'Q', parallel 163.] EXAMPLES. 259 to the axis of x. Let (x, y) be the co-ordinates of P and (af 3 y) those of P'. Then the area of the strip is (of x) dy^ and the co-ordinates of its centroid are \(x' '+#) and y. Hence if no portion of the area considered is above a parallel to Ox drawn through C, the co-ordinates of its centroid are given by the equations . f(x'-x)dy " f(x'-x)dy in which the limits of y are and the ordinate of C. The values of x f and x are of course given in terms of y from the equations of the two curves. For example, let it be required to find the centroid of the area in- cluded between a parabola and a circle described with the vertex of the parabola as centre and a radius equal to f of its latus rectum. The centroid is on the axis of the parabola. Let the equation of the parabola be y 2 = 4 ma?; then the equation of the circle is and the ordinate of C, their point of intersection, is m \/2. mjZ = m 2 16 + 27 sin o as the student will find without much difficulty. EXAMPLES. 1. Find the centroid of the area included between the arc of a semi-cycloid, the circumference of the generating circle, and the line joining the extremities of the cycloid. Ans. The common tangent to the circle and cycloid at the vertex of the latter being taken as axis of x, the vertex being origin, and a the radius of the generating circle, 37T 2 -8 5 2. Find the locus of the centroid of the area of a parabola cut off by a variable right line drawn through the vertex. Ans. If 4m is the latus rectum of the parabola, the locus is another parabola whose equation is y z = - mx. 260 CENTROIDS [CENTRES OF GRAVITY]. (The student may verify the construction of Example 7, p. 258, for the tangent to this locus.) 3. Find the centroid of the portion of an ellipse cut off by a line joining the extremities of the major and minor axes. Ans. x =- 3 7T-2' y = o 3 7T-2 163.] Graphic Construction of the Centroid of a Plane Area. The following method of determining the centroid of any plane area is taken from Collignon's 8 fatigue, p. 315. Let APJ3Q be any plane area, and let Ox be any line in its plane. Then if the distances of the - S- - -' centroid from Ox and any other line in the plane are known, the position of the point is known. Draw any line, 0V, parallel to Ox (axis of x] in the plane of the curve, and let the perpendicular distance between Ox and 0V be a. Let the area be broken up into narrow rect- angular strips, such as PP'Q'Q, by lines parallel to the axis of x. Then if PQ = z 9 the area of the strip = zdy, the distance of PQ from Ox being y. Hence the distance, y, of the centroid of the area from Ox is given by the equation /, \ o Fig. 203. A L being the area of the figure, and the values of y running from the ordinate of A to that of R, at which points the tangents are parallel to Ox. Now take any point, 0, on Ox; draw OQ, and draw P0 / parallel to OQ. Let the line 00' meet PQ in R. Then by similar triangles QR RP OR QR OR ' PQ ~ 00 / or, / denoting the length QR, az'=yz. (2) Let the locus of R corresponding to all strips of the given area 164.] EXAMPLES. 261 be constructed. It will be a curve, ARB 3 passing through the points A and B. Substituting the value of yz from (2) in (l), we have a f z'dy y=^A* A in which the limits of y are the same as before. But/V% is the area, A 2 , between the curves ARB and AQB. Hence The distance of the centroid from Ox is therefore known. Similarly its distance from any other line can be found, and therefore the position of the point is determined. If a point S is deduced from R in the same way as that in which R was deduced from P, and if QS = /', we shall have as before p z , Jo 3 77 which = 7=. - - - Therefore 2 A/9 2-4 2 _ " :65.] EXAMPLES. 263 EXAMPLES. 1. To find the centroid of a given sector of a circle. Ans. It is on the diameter bisecting the arc, at a distance from the centre equal to f of the product of the radius and the chord of the arc divided by the length of the arc. 2. Find the centroid of a portion of an equiangular spiral included by the initial line and a given radius vector. Ans. The initial line being taken as axis of #, the equation of the spiral being r = ae ke , and a being the angle of the given radius vector, cos a = e a cos a e a sn a 3. "When a = in the preceding question, find the values of x and y, and explain the result. 4. Find the centroid of the portion of a parabolic area included between the axis and a radius vector drawn through the focus. Ans. If 4m is the latus rectum, and t the tangent of half the angle between the given radius vector and the axis, _2m 1 i\ . _2m t+t 3 : " =: " ' 165.] Double Integration. When the density of the lamina varies from point to point it may be necessary to divide it into infinitesimal portions of the second order instead of strips (triangular or rectangular) whose areas are infinitesimals of the first order. Thus, suppose that the lamina AOB (fig. 303) is not of uniform density. Then if we break it up into triangular strips, such as POQ, the element of mass will be no longer proportional to the area POQ, or JrW; and, moreover, the centre of mass of the strip will not be f r distant from 0. Let a series of circles be described round as centre, the distance between two successive circles of the series being dr. These circles will divide the strip POQ into an indefinitely great number of rectangular elements ; and if one of these is included between the circles of radii / and / + d/ 9 its area will be r' y ~ rp rr I / kpr'dr'dO / / ^a^O /a^O Let it be required, for example, to find the centroid of the area of a cardioid in which the density at a point varies as the n^ power of the distance of the point from the cusp. 165.] EXAMPLES. 265 Here p =. /u/ n , and k is constant; therefore, the abscissa being the same for the whole curve as for the half above the axis, r o Jo Integrating first with regard to /, we have n + 2 ^ r7r r n+2 d6 r o s\ /\ But r=2acos 2 -- Substituting this value and putting - = (/), we have n+3 + 2) / r I Jo These definite integrals are well known. Dividing the numerator . 1.3.5... 2n+3 77 and denominator by - - - - 5 we have 2.4.6 ... 2^ + 4 2 ' 2f+.6 ) The centroid evidently lies on the axis of symmetry, or y = 0. EXAMPLES. 1 . Find the centre of mass of a circular sector in which the density varies as the w th power of the distance from the centre. ns. n+2 ac A } where a is the radius of the circle, I the length vi -\- 3 I of the arc, and c the length of the chord, of the sector. 2. Find the position of the centre of mass of a circular lamina in which the density at any point varies as the n^ power of the distance from a given point on the circumference. Ans. It is on the diameter passing through the given point at a distance from this point equal to - a, a being the radius. 266 CENTROIDS [CENTRES OF GRAVITY]. [165. Methods of double integration are also often employed when the elements of area are expressed in Cartesian co-ordinates. In this case, let the element of area at a point P, whose co- ordinates are (#', y), be a small rectangle included between two very close lines parallel to the axis of x and two very close lines parallel to the axis of y. Then the element of area will be dx'dy'\ and if p and k are the density and thickness of the lamina at the element, the element of mass, dm = kpdafdy'. Also the co-ordinates of the centre of mass of this element are ultimately x' and if. Hence _. _ ' ffkpdx'dy' ' y ' ' A single example will suffice to illustrate this method. Let it be required to find the centre of mass of a quadrant of an ellipse included by the semi-axes, the density at any point being pro- portional to the product of the co-ordinates of this point. Here p = p . x'y', and since k is supposed constant, ~. _ y' _ ._ ffx'y'*dx'dy' ~ ffx'y'dx'dy' ' - ffx'ydx'dy'' Let the integrations be performed first over a strip parallel to the axis of y. Then we integrate with respect to y', regarding x as constant, from y'= to y' '= y, the ordinate of a point on the ellipse. fx'ydx' Hence x = J / , . J x yd x Here we must substitute the value of y in terms of x', and thus we s et _ _ - in which summations the abscissa x' is to receive all values from to a. 8 8 We easily obtain - a and - b for the co-ordinates of the centre of mass. Examples may occur in which, although the density of the lamina varies from point to point, the process of double integra- tion can be avoided by the judicious selection of an element of area. Let it be required to find the centre of mass of a quadrant of an ellipse in which the density at any point varies as the distance of the point from the axis major. l66.] CENTROID OF A SUBFACE OF REVOLUTION. 267 Here by dividing the area into rectangular strips parallel to the axis major, we obtain infinitesimal elements of the first order throughout each of which the density is constant. Hence our equations are / O 7 - = ifaPyfy. = 2 ' ' Making the usual eccentric angle substitutions for x and ^, we filld 3 377 . **-+ y = -b. 166.] Centroid of a Surface of Revolution. Let a plane curve AB (fig. 201) revolve round a line Ox (taken as axis of #) and generate a surface. Then the revolution of the elementary arc PQ(=ds) generates a portion of surface whose area is 2Tryds; and if p is the density of the matter in this zone and k its thickness, the element of mass is Znkpyds. Also the centre of mass of the zone is ultimately the point M, whose abscissa is x. Hence the centroid of the surface generated (which obviously lies on the axis of revolution) is at a distance from given by the equation _ fk pxy ds fkpyds the integrations being extended over the whole length of the generating curve. For example, to find the centroid of the surface of a semi-ellipsoid of revolution round the minor axis, the density of any zone being proportional to its distance from the equatoreal plane, and the thick- ness being constant : The area of a zone at a distance y from the equatoreal plane being 2 Ttxds, the position of the centroid is given by the equation fxy^ds V~ fxyds ' the integration extending over the arc of a quadrant of the generating ellipse. Using the eccentric angle, we have x = a cos <, y = b sin (/>, ds = *Ja a and b being the semi-axes of the ellipse. Hence I cos <> sn <>v a sn <> + cos cos (f> sin y a 2 sin 2 < + 6 2 cos 2 < . d o To find the integral in the numerator, put t for sin <, and it becomes 268 CENTROIDS [CENTRES OF GRAVITY]. [167. where a 2 & 2 = c 2 . This, again, is equal to o Jo which =-i r^ + tft^dt-^ r^ + C JQ C JQ and this, by making the first integral depend on the second, is easily proved to be _ 4c 2 Jo The integral in this expression is one of the elementary forms in the Integral Calculus. Hence the numerator is The integral in the denominator is evidently rf , _ : _ 4 / v0 2 + c 2 sin 2 (b . d sin 2 cf>, Jo which is equal to -(a 3 6 3 ). 3 c , Therefore 8 c(a*-b 3 ) For a sphere of radius a the value of y is easily proved by direct calculation to be fa; and the student may exercise himself in the evaluation of indeterminate forms by deducing this from the value of y given above. (For this purpose it will be advisable to put log - into the form \ log and expand.) o GJ c 167.] Centroid of any Portion of a Spherical Surface. Let dS denote any portion of a spherical surface, and let dl, denote its projection on any plane passing through, the centre of the sphere. Then, if this plane be taken as that of xy, and if z denote the distance of the centroid of the element dS from the plane, the distance of the centroid of any portion of the spherical surface from the plane is given by the equation fzd8 m ^ fdS ' the integration being extended over the whole portion of the spherical surface considered. 1 68.] CENTROID OF ANY SURFACE. 269 Now if r is the radius of the sphere, the cosine of the angle between the tangent plane to the sphere at the element dS and the plane of xy is - ; therefore d* = *-d8* (2) Hence fzdS rfdl, = /2, 2 denoting the projection of the whole spherical area considered ; and making this substitution in (1), we have s where S is the area of that portion of the sphere whose centroid is required. Equation (1) gives, of course, the distance of the centroid of any surface whose element is dS from the plane of xy ; and it is clear that if the surface is generated by the motion of a sphere of constant radius whose centre moves along any curve in the plane of asy, the cosine of the angle between the tangent plane at the element dS and the plane of xy will still be - > since the given surface and the generating sphere have the same tangent plane. Hence equation (2) holds in this case and therefore also equation (3). 168.] Centroid of any Surface. Let dS denote an element of any surface, d^ the projection of this element on the plane of acy, and y the angle between the plane of cry and the tangent plane to the surface at the element dS. Then if z is the distance of the centroid of dS from the plane of xy> we have fzdS f z sec y . It is not unusual to suppose the element dS cut off from the surface in the following manner. Let m (fig. 205) be a point in the plane xy whose co-ordinates are #?', y'\ let mn be drawn parallel to the axis of as and equal to dx' m , let mq be parallel to the axis of y and equal to dnf\ and complete the rectangle mn_pq. On the base mnpq describe a prism whose edges, Mm, Nn } Pp, Qq are parallel to the axis of z. This prism will intercept on the given surface an element, 270 CENTROIDS [CENTRES OF GRAVITY]. [168. MNPQ, which is dS. The rectangular projection, ^2, is then mnpq whose area is dx'dy'. Substituting this value in the above equation, we have ff z sec y dx dy f the integrations being extended over the whole projection of the given surface on the plane xy. It easily follows that the centroid of the projection (orthogonal or ob- lique) of any plane area on any plane is the projection of the centroid of the area. Take the plane on which the given area is projected as the plane of xy ; let co be the angle between this plane and the plane of the area, and let #, y be co-ordinates of the centroid of the given area. Then _ __ fxdS _ fx sec o> . d^ = ~ 205> since CD is the same for all elements. But the co-ordinate of the centroid of the projection is evidently given by this equation. Therefore, &c. ; and a similar proof obviously holds for an oblique projection, because at all points of the given area the ratio of dS to d^ is constant. EXAMPLES. 1. A section of a sphere is made by any two parallel planes ; prove that the centroid of the spherical surface included is midway between them. This is very easily proved either by direct calculation or by the application of the result of last Article. Collignon (Statique, p. 295) gives an elegant geometrical demonstration which depends on the fact that if a cylinder is circumscribed to a sphere along any one of its great circles, the portion of the area of the cylinder included between any two planes at right angles to its axis is equal to the portion of the area of the sphere included by these planes. By taking in- definitely close planes it follows that the spherical area may be transferred to the cylinder, and the centroid of any portion of a 1 68.] EXAMPLES. 271 cylindrical area cut off by planes perpendicular to the axis is evidently midway between these planes. Con. The centroid of the surface of a hemisphere is at a distance equal to half the radius from the centre. 2. To find the centroid of a spherical triangle. Let ABC be any spherical triangle, and the centre of the sphere. Produce the sides AC and AB until they become quadrants, AE and AD, and draw the arc DE of a great circle. "We shall find the distance of the centroid from the plane EOD, which is perpendicular to the line OA. The projection of the area ABC on this plane is evidently the same as the projection of the sector, COB. Now if p l is the per- pendicular arc from A on the side BC, the F j g 20 g angle between the planes COB and EOD is 90 jOjj also the area of the sector COB is \ar, a being the length of the side BC and r the radius of the sphere. Hence if 2 denote the projection of the area of the triangle on the plane EOD, 2 = \ar smp l ; and if A, B, C denote the circular measures of the angles of the triangle, and S its area, $ = ^ (A _|_ 2) + OTT). Hence, by (3) of last Article, if x denote the distance of the centroid from the plane, _ 1 a sin p l It is evident that OB is the distance from of the projection of the centroid on the line OA. Its projections on the lines OB and 00 are obtained by writing b and p 2 , c and ^; 3 instead of a and p 1 in this equation. 3. To find the centroid of the surface of a nearly spherical semi- ellipsoid cut off by the plane of the two greater axes. Let the axes in order of magnitude be a, 6, c, and let Now if doc'dy' is the projection on the plane xy (which is the base of the semi-ellipsoid) of an element of surface, dS, we have pz p being the perpendicular from the centre on the tangent plane at the element, and z the distance of the element from the plane of xy. Hence, S denoting the surface of the semi-ellipsoid, we have 272 CENTEOIDS [CENTRES OP GRAVITY]. [168. Therefore, rejecting all powers of k and k' beyond the second, Integrating from x'=oc to x'=oc, the co-ordinates of a point on the circumference of the base being so, y, we have Expressing x and y in terms of the eccentric angle, and integrating over the entire circumference, we have S.z=irabc(l o Now (Williamson's Integral Calculus), r / \ I sin Odd sin 2 9 + c 2 cos 2 0)i (6 2 sin 2 + c 2 cos 2 0)i which is easily proved to be 27rc 2 { 1 + Hence finally, F= - 4. A parabola revolves round its axis; find the centroid of a portion of the surface between the vertex and a plane perpendicular to the axis at a distance from the vertex equal to f of the latus rectum. 29 Ans. Its distance from the vertex = y^ (latus rectum). 5. Find the centroid of the surface generated by the revolution of a cycloid round its axis. 2 (15 8^ Ans. It is on the axis at a distance , ^ --a from the 15(37r 4) vertex, a being the radius of the generating circle. 6. Prove that the centroid of the lateral surface of the frustum of a right cone or pyramid lies in a plane whose distance from the base is . ^TT h, where p and p' are the perimeters of the base and upper section, and h the height of the frustum. 169.] Centre of Mass of a Solid of Revolution. If the curve AB (fig. 20 1) revolve round Ox, the rectangular area PQNM will generate a cylindrical volume equal to IT . PM 2 .MN, or Tty 2 dx. Hence if the density of the solid is uniform, we have for the position of its centre of mass (which obviously lies on Ox) _ j xy clx w ~^ o ^ " J 169.] CENTRE OF MASS OF A SOLID OF REVOLUTION. 273 the integrations being extended over the whole of the area, CABD, of the revolving curve. If the density varies, the element of mass may require to be taken differently. If the density is a function of x alone, i.e., if it is the same all over the rectangular strip PQNM, the volume may be broken up as above, and the element of mass Hence we shall have, in this case, _ ~ Suppose the density to vary as y alone. Then if we take a small rectangular area, dx dy' , at a point whose co-ordinates are #', y'> this area will generate an element of volume equal to 2iri/' doc' dy' -, therefore the element of mass = 2ttpif d&f d^ and we have ffpsc'y'dx'dy' ~ ' ~~ The integrations are to be performed first from y = to/= ^, the ordinate of a point P on the bounding curve ; and then from x'= OC to af= OD. As an example, let the curve AS be a quadrant of a circle of which OA and OB are diameters, and let it be required to find the centre of mass of the solid hemisphere generated by the revolution of this quadrant round OB (taken as axis of x), firstly when the density is uniform ; secondly when it is constant over a section perpendicular to OB and proportional to the distance of this section from the centre ; and thirdly when it is the same at the same distance from OB, and proportional to this distance. Firstly, we have x = 2 , Putting x = r cos 0, y = r sin , t/ */ where r is the radius of the circle, and integrating between = aud o x = -r. (1) / X*' Secondly, since p = px, we have x = ^ ^-j ' which easily gives J xy dx r. (2) Thirdly, p = py' y therefore - ffy'*dx'dy' - ffdx and the previous substitutions for x and y give 274 CENTROIDS [CENTRES OP GRAVITY]. [169. In this case the double integration might have been avoided by breaking the area up into rectangles parallel to the axis of x. The student will do well in such examples as this to check his results as much as possible by a common-sense view of the question. Thus, having proved that the distance of the centre of mass of a homogeneous hemisphere from the centre is f r, it is clear that when the density of a section is directly proportional to its distance from the centre, the centre of mass of the hemisphere must be at a distance from the centre > f r, since the matter is most dense in the space remote from the centre ; while in the third case above, since the ordinates of the portion of the curve near A are greater than those of the portion near B, and since the density increases with the ordinate, it is evident that the centre of mass must be nearer to the centre than in the homogeneous hemisphere. The most advantageous method of breaking up a mass of varying density into elements depends entirely on the law of variation of the density, and while all these methods are em- braced in the rule of Art. 160, it would be impossible to give formulae suited to all cases. Laplace, by assuming the change of the pressure from stratum to stratum of the earth to be proportional to the change in the square of the density, proves that if the strata of uniform density are spherical, the density of a stratum of radius x is given by the equation a being the radius of the earth, p Q the density of the centre, and JJL a constant number. Let it be required to find the centre of mass of a hemisphere whose density follows this law. Here the element of mass of uniform density is the stratum in- cluded between the hemispheres of radii x and x + doc. Hence dm = X . fJLX - sm dx. Also the distance of the centre of mass of this stratum from the /yi centre is- (Example 1, p. 270). Hence, the axis of x being the 2 diameter perpendicular to the base of the hemisphere, the distance of the centre of mass from the centre is given by the equation 1 70.] CENTKE OP MASS OF ANY SOLID. 275 I J v x j x sin ax o a (2 u 2 ) cos u + 2u sin JJL 2 == a . -~ - - -. -- 2ju (sinfi / as will be easily found. "When ju, = the hemisphere is of uniform density, and the student will see that this value of x becomes f , in accordance with our previous result. EXAMPLES. 1. Find the centre of mass of a hemisphere in which the density is proportional to the nfo power of the distance from the centre. Ans. It is at a distance = from the centre, a being the n-\-4i 2 radius of the hemisphere. 2. Find the centre of mass of a portion of a paraboloid of revolu- tion cut off by a plane perpendicular to its axis. Ans. If h is the distance of the plane of section from the vertex, ff = f h. 3. Find the centre of mass of a semi-ellipsoid of revolution round the minor axis, the density at any point being proportional to its distance from the base which is the plane perpendicular to the axis of revolution. o Ans. y = -b, where b is the semi-minor axis. 15 4. An ellipsoid of revolution round the minor axis is cut by a plane passing through this axis ; find the centre of mass of the portion included between one semi-ellipsoid thus cut off and the concentric hemisphere whose diameter is the minor axis. Ans. If a and b are the axes major and minor of the generating ellipse, the required centre of mass is on the major axis at a distance 3 a* + ab + b 2 equal to-- = trom the centre. 8 a + b Verify this result in two obvious cases. 170.] Centre of Mass of any Solid. In the solid take any point, P, whose co-ordinates are a?, y^ z, and also a close point, Q, whose co-ordinates are OB -f dx, y + dy, 2-+ dz. Then evidently the volume of the parallelepiped whose diagonal is PQ and whose edges are parallel to the axes of co-ordinates is dxdydz ; 276 CENTROIDS [CENTRES OF GRAVITY]. [170. and if p is the density of the body at P the element of mass at P is pdxdydz. Hence the co-ordinates of the centre of mass of the solid are given by the equations fffpxdxdydz _ _ fffpydxdydz . fffpzdxdydz = fff pdxdydz ' y := fff pdxdydz ' '' fff pdxdydz ' the integrations being extended over the whole solid. It may not be necessary to take infinitesimal elements of volume of the third order. From what has preceded, the student will have learned that the best mode of breaking up the given mass into elements depends entirely on the law of density which prevails. In many cases the symmetry of the solid enables us to simplify the problem by choosing elements of volume which are in- finitesimals of ihe first order only. The various elements of volume which it may be necessary to take are exemplified in the fol- lowing problems. Find the centre of mass of the eighth part of an ellip- soid, ABC (fig. 207) included between its three principal planes (1) When the density at any point is simply a function of its distance from the prin- cipal plane BC (plane Q yz). (2) When the density at any point is a function of its distances from the two principal planes AC and BC (planes of xz ana yz). (3) When the density at any point is a function of its distances from the three principal planes. In the first case, the density will be constant over a section DH perpendicular to OA. Hence, taking two such sections, DH and EF, at a distance dx from each other, the density of the solid between them may be considered uniform, and this portion of the solid may be taken as the element of mass. In the second case, the density will be constant throughout a 207. 170.] EXAMPLES. 277 portion of the body in which x and y are constant ; that is, along a perpendicular to the plane AB ; and the element of mass may be taken as the prism NQnq, the area of whose base is dxdy> and which intersects the bounding surface in the area NMQP. In the third case, the density is not the same at any two points, and the element of mass must be taken a small rect- angular prism, sir, whose volume is dxdydz. EXAMPLES. 1. In the problem just discussed find the centre of mass when the density at any point is proportional to its distance from the plane BG. Here p px ; also, the equation of the ellipsoid being the ellipse DH satisfies the equation 9- + which shows that the axes GH and GD are - and respectively. Hence, IG being = dx, the element of mass is a? ( 1 -- ^ and since the centre of mass of this element is ultimately a point whose co-ordinates are 46 / x 2 x > *~ and (see Ex. 3, p. 257), we have and Sir /" , 15ir /" ,, / *(!-- 278 CENTROIDS [CENTRES OF GRAVITY]. [170. the value of z being, of course, - 2. If the density at any point of the ellipsoid is pxy, find the centre of mass. Taking a prismatic element of volume NQnq, the element of mass is z being the height, M. m, of the prism. The co-ordinates of M being x } y, z, those of the centre of mass of this prism are evidently x, y, - Hence ffx^yzdxdy ffxy z zdxdy _ ffxyz^dxdy sp y .^ x !L_ >/ 7^-; _^, - - i : therefore ms r sin d. Again, the co-ordinates of the centre of mass of this element are ultimately the same as those of s ; therefore they are 171.] EXAMPLES. 281 r sin cos <, r sin 9 sin $, and y cos 6 ; and for the centre of mass of any finite portion of the solid we have - _ fffp /3 s ^ n2 cos fff P ^ si - - cos the limits of integration being- determined by the figure of portion of the solid considered. The angles and are sometimes called the colatitude and longitude, respectively. EXAMPLES. 1. Find the centre of mass of a portion of a solid sphere contained in a right cone whose vertex is the centre of the sphere, the density of the solid varying as the w th power of the distance from the centre. Take the axis of the cone as that of z, and any plane through it as that from which longitude is measured. Then it is clear that # = ?/ = 0, and we have _ //A n+3 sin 6 cos dr dS d(j> fffr n+ * sin Performing the integration first with respect to r, considering and constant, from r = to r = a, the radius of the sphere, we have -_ ~ Performing the integration now with respect to , the longitude, which runs from to 2-7T, we have _ _ 7i+3 /sin cos Odd ~ a If a i= the semi-vertical angle of the cone, the limits of are and a. Therefore n + 3 a 2. Find the centre of mass of a prism whose base is a given spherical 282 CENTROIDS [CENTRES OF GRAVITY]. [172. triangle and whose vertex is the centre of the sphere on which the triangle is described. Let (fig. 206) be the centre of the sphere, and take 00 as axis of z. From draw the perpendicular p a to the base AB, and let R be the radius of the sphere. The value of z given as a triple integral may be modified in the present case. Let dS denote any small element of area at any point on OP ', then the volume of a cone whose base is this element and vertex the centre of the sphere is J R dS, and the distance of its centre of mass from the plane of xy is (Art. 158) R cos 0. Hence _ 3 fcos0dS ~4 fdS Now cos Q . dS is the projection of the element dS on the plane of xy ; therefore the numerator is the projection of the whole area ABC on this plane, which, as in Example 2, p. 271, is \cR sm^; 3 . Hence, _ 3 ~ 8 3. A cardioid revolves round its axis ; find the centre of mass of the solid generated. Ans. It is at a distance from the cusp equal to T \ (axis). 172.] Theorems of Pappus. If a plane area revolve through any angle round a line in its plane, the volume generated is equal to the area of the revolving figure multiplied by the length of the path described by its centroid. Let AS (fig. 209) be the revolving figure, and Ox the line about which it revolves. Let the area be broken up into an indefinitely great number of rectangular strips, such as PQqp, by lines perpendicular to Ox. Then the volume generated by this strip in revolving through an angle co is evidently equal to 27T or denoting PM, pM, and MN by ^ 2 , y lt and due. denote the whole volume generated, Hence if V 1 73.] EXAMPLES. 283 Now the distance of the centroid of the strip from Ox is , and the area of the strip is (y^^i) d* Hence, denoting these quantities by y and dA respectively, V = = <*>A.y, A denoting* the whole revolving 1 area and y the ordinate of its centroid. Now in revolving through the angle o>, the centroid of the area describes a circular arc whose length is coy. Hence the theorem is proved. If the axis Ox intersects the revolving figure, the theorem still applies with the convention that the volumes generated by the portions of the figure at opposite sides of Ox are affected with opposite signs. Again if tJie arc of any plane curve revolve through any angle round a line in its plane, the area of the surface generated is equal to the length of the revolving arc multiplied by the length of the path described ly its centroid. For, the surface generated is wj yds, or co-Z/.y, L being the whole length of the revolving arc and y the ordinate of its centroid. As before, wy is the length of the circular arc described by the centroid of the revolving arc, and the theorem is evidently proved. If the revolving arc intersects the line Ox, the theorem is true, with the previous convention of signs. 173.] Extension of the Theorems of Pappus. The previous theorems can be easily extended to the case in which the plane of the revolving figure, instead of revolving round a fixed line, rolls without sliding on any developable surface, and the first theorem will then become If the plane of any plane area rolls without sliding on a develop- able surface, the volume generated by the area in moving from one position to another will be equal to the area of the revolving figure multiplied by the length of the path described by its centroid. A similar enunciation gives the second theorem. These propositions are evidently true, because in an indefinitely 284 CENTROIDS [CENTRES OF GRAVITY]. [174. small motion the figure is revolving round a generating line of the developable, and for such a small motion the theorem of Pappus gives the volume generated equal to the area x small space described by its centroid. Taking the sum of all such elements of volume from one position of the figure to another, we have the theorem of this Article. It is clear also that the theorems hold in the case of a plane area which moves in such a manner as to be always normal to the path described by its centroid. For the area may at any instant be considered as revolving round the line of intersection of two consecutive normal planes of the curve which the centroid describes, and the theorems are then directly applicable. 174.] Volume of a Truncated Cylinder or Prism. Let A and B denote the sections of a cylinder or prism made by any two planes. Through any line L passing through the centroid, , of B draw any plane, B ', inclined at an indefinitely small angle to B. Then G is the centroid of the section B', since this section is the projection of B made by lines parallel to the generators of the cylinder or edges of the prism, and since (Art. 168) the centroid of the projection of any plane surface is the projection of its centroid. Also the area of the section B' differs from that of B by an infinitestimal of the second order. Hence the theorems of Pappus apply, and we may consider that the area B has revolved round the line L through a small angle. But the space described by its centroid is zero ; therefore the volume between the sections B and B' on one side of the line L = the volume between them on the other side ; in other words, infinitesimals of the second order being neglected, the volume of the prism or cylinder contained between the sections A and B is equal to that contained between the sections A and B'. Allowing B' to revolve again about L through a small angle, the same reasoning applies, and we see, finally, that for the sections A and B may be substituted any two passing through their respective centroids, and the included volume will be unaltered. Let two parallel sections each perpendicular to the axis of the prism or cylinder be substituted, and the included volume will be &.>&, where H is the area of either normal section and h the distance between them. 175.] Equilibrium of a Heavy Body on a Horizontal Plane. 175.] EXAMPLES. 285 When an indeformable body rests on a horizontal plane, the contact taking place at several points, either continuous or not, it is kept in equilibrium by two forces namely, its own weight and the reaction of the plane. The condition necessary and sufficient for the equilibrium of such a body is that these two forces must be equal and opposite. Now this will be impossible unless the points of contact of the body with the plane can be so connected by right lines as to form a polygon within the area of which the vertical through the centre of gravity of the body intersects the plane. For, whether the plane be rough or smooth, resolve all the reactions at the points of contact vertically. Then it is evident that the resultant of the system of parallel vertical forces at the points of contact must necessarily fall within some polygon whose vertices are these points ; therefore, &c. The student must be careful to observe that this condition, though necessary in the case of a deformable system, is not sufficient (see Article 94, p. 115). Thus, in Example 14, p. 179, it is not true that the deformable system of two bars, AB and 13C, will rest in any position in which their common centre of gravity falls between the props. EXAMPLES. 1 . To find the volume and surface of a tore. (A tore is a surface generated by the revolution of a circle round a line in its plane.) Let r be the radius of the circle, and c the distance of its centre from the axis of revolution. Then the volume of the tore is evidently TT r 2 x 2-7TC, or 27i 2 cr 2 ; and the surface is 2?rr x 2 ire, or 47i 2 cr. 2. A triangle revolves round a line in its plane; find the volume generated. Ans. If the distances of the vertices from the lines are x lt o? 2 , # 3 , and A the area of the triangle, the volume = (x i -{-x 2 + x 3 ). 3 3. From the Theorems of Pappus deduce the volume and surface of a frustum of a right cone. (Consider a trapezium one side of which is perpendicular to the two parallel sides.) 4. A pack of cards is laid on a table ; each projects in the direc- tion of the length of the pack beyond the one below it ; if each projects as far as possible, prove that the distances between the 286 CENTROIDS [CENTEES OF GRAVITY]. [175. extremities of the successive cards will form a harmonic progression. (Walton, p. 183.) 5. A rectangular column is formed by placing a number of smooth cubical blocks one above another, the base of the column resting on a horizontal plane ; all the blocks above the lowest are then twisted in the same direction about an edge of the column, first the highest, then the two highest, and so on, in each case as far as is consistent with equilibrium. Prove that the sum of the sines of the inclinations of a diagonal of the base of any block to the like diagonals of the bases of all the blocks above it is equal to the sum of the cosines. (Walton, ibid.) CHAPTER XII. THE PRINCIPLE OP VIRTUAL WORK APPLIED TO ANY SYSTEM OF BODIES. 176.] Forces Applied to a Particle. It has been shown in Art. 136, p. 214, that the resultant of any number of forces applied to a particle may be represented by the side required to close the polygon of the forces. And whether the polygon OP 1 P 2 ... P n be plane or gauche, it is clear (as in p. 67) that the sum of the projections of the sides, taken in order, along any line OA, is equal to zero. Let the projections of the sides be denoted by Q 19 Q 2 , ... Q n . Then Q : + Q 2 + ... -f Q n = 0. Multiplying this by OA, an arbitrary length along the line OA, we have Q l .OA+Q 2 .OA+. ..+&. 04 = 0. vbC * But if/>! is the projection of OA along 0P 15 we have (see p. 67) ~j> Q l .OA=OP l . Pl . If, then, the sides 0P 13 P 1 P 2) ... be denoted by P 15 P 2 , ... we have P 1 .^ 1 -fP 2 . A + ... +P n . A =0; and if the sides represent forces, each term in this equation is the virtual work of the corresponding force for the displacement OA. Since the resultant, R, of n 1 of the forces is P n3 we have B.r = P 1 . Pl + P 2 .p 2 ...; and if the displacement is small, this equation is written (as in P-70 Rbr = P 1 bp l + P^ j p 2 + .... (1) In particular, if X, Y, Z denote the rectangular components of R, we have Rbr = Xbx+Y&y + Ztz. (2) 177.] Extension to any Number of Connected Particles. If two particles, m : and m 2 , are connected by a rigid inextensible rod, and are in equilibrium under the action of forces, P 15 Q l} ... 288 THE PRINCIPLE OF VIRTUAL WORK. [177. applied to m l and P 2 , Q 2 , .., applied to m 2 , it is evident (as in p. 1 1 8) that the force arising from the connexion acts in the line joining % to m 2 . If, then, this force be denoted by T, and the distance between the particles by r, we have for the equilibrium ^r denoting the change in r arising from an arbitrary small displacement of %. The equation of equilibrium of m 2 is and if in the new positions of % and m 2 the distance between them remains unaltered, 8 1 r + 8 2 r = 0. Hence, by adding these equa- tions, we obtain the equation P 1 tp l + Q l t>q l + ... +P 2 ^ 2 +Q 2 ^ 2 + ... = 0, (1) which is free from the internal force T. This is exactly the same as the investigation already given for coplanar forces in Chap. VI. The extension to any number of particles, that is, to any body, proceeds just as in that chapter, and the enunciation of the principle of virtual work there given applies in general without the limitation that the forces are coplanar. If in the case of the two particles % and %, considered above, their new positions are such that the distance between them is altered by 8r, the equation of virtual work will be A^i+Qi^i... + PM + Q 2 &?2 + ...+nr=0; (2) and, generally, if the virtual displacement is such that the internal forces do virtual work, these forces will enter into the equation of virtual work in exactly the same manner as the applied forces. The theorem of virtual work may, therefore, be thus enunciated : When a material system is in equilibrium under the action of any external and internal forces ', the sum of the virtual works of the external and internal forces is equal to zero for any small virtual displacement whatsoever. Instead of saying that the total virtual work is zero> we should in strictness say that it is an indefinitely small quantity of the second order, the greatest of the displacements being considered as a small quantity of the first order. This has been already explained in p. 1 23. The proof of the converse proposition namely, that when the virtual work vanishes for all imagined displacements, the system 1 79-] KINEMATICAL THEOREM I. 289 is in equilibrium has been already given in p. 122 for coplanar forces ; and as the proof obviously holds for non-coplanar forces, it is unnecessary to reproduce it here. 178.] Displacements along Smooth Surfaces. If any body or system of connected bodies be in contact with smooth curves or surfaces, and the system be imagined to receive any small displacement along these curves or surfaces, it is clear (as in p. 71) that, since the point of application of each of the geo- metrical forces (reactions of the curves or surfaces) moves in a plane at right angles to the corresponding force, these forces will contribute nothing to the equation of virtual work for such a displacement. If any of the bodies of the system are connected by strings or rods whose lengths are unaltered in the virtual displacement chosen, the tensions of these strings or rods will not enter into the equation of virtual work. But, as already explained in pp. 80 and 1 20, we may choose virtual displacements of the system which violate the imposed conditions at the expense of bringing into our equation the corresponding forces. 179.] Kinematical Theorem I. When all the points of a rigid body move parallel to a plane, the motion may be produced by a pure rotation round an axis perpendicular to this plane. DBF. A motion of a body round an axis whereby each point in the body describes an arc of a circle having its centre on the axis and its plane perpendicular to it is called a pure rotation. The position of the body will evidently be known if the positions of any two points in a plane parallel to the plane of motion are known. Let A and B be any two points in such a plane, and suppose that after the displacement of the body they occupy the positions A' and ]? (fig. 210). At the middle points of AA' and BB? erect two perpendiculars^ which meet in I. Then in the triangles AIB and A'Iff, AI = A'I, BI = B'I, and AB = A'ff\ therefore the triangle A' IB? is nothing more than AIB turned round the point / through an angle AT A' or BIB'. Hence the line AB can be brought into its new position by a pure rotation about /, and the same is true of every point rigidly connected with A and B in the plane AIB. If through / an axis be drawn perpendicular to the plane of motion, it is evident that the body can be brought into its new u 290 THE PRINCIPLE OF VIRTUAL WORK. [l8l. position by a pure rotation about this axis through an angle = AIA', however complicated the paths along which A and B have travelled to A' and B'. When the motion of the body is small, this axis is called the Instantaneous Axis ; and it is obviously constructed by drawing two planes normal to the directions of motion of any two points in the body. The intersection of these planes is the instantaneous axis. When the body is a plane figure, the point /is called the Instantaneous Centre; and the consideration of this point is of very extensive use in Kinematics, Statics, and Geometry. To construct the instantaneous centre, at any two points erect perpendiculars to the directions of motion of these points, and their intersection is the required point. Fig. 210. 180.] Kinematical Theorem II. The motion of a rigid body round a fixed point is at every instant a pure rotation round an axis. One point, 0, in the body being fixed, the position of the body will be known if the positions of any two points, A and B, not in directum with are known. Bound let a sphere, forming part of the body or rigidly connected with it, be described with arbitrary radius, and let A and B .(fig. 210) be any two points on the sphere. After the motion of the body let A' and B' be the positions of A and B. Imagine the lines AB, A'B', AA', and BB" in this figure to be arcs of great circles on the sphere instead of right lines. Then, at the middle points of A A' and BI? draw two great circles perpendicular to AA' and BB', respectively, and let them meet in J. In exactly the same way as in the last theorem, we have the spherical triangles AIB and A' Iff equal ; that is, the latter triangle is the former turned round the axis 01 through an angle AIA' or BIB'. Hence the whole body is brought by rotation through this angle round the axis 01 from the old to the new position. 181.] Kinematical Theorem III. If a body has a motion of translation represented in magnitude and direction by a right line OA, and at the same time a motion of translation repre- 182.] KINEMATICAL THEOREM IV. 291 sented in magnitude and direction by a right line OB, the resulting motion of translation is represented in magnitude and direction by the diagonal, OC, of the parallelogram determined by OA and OB. This proposition has been already illustrated in p. 6. It follows immediately that any motion of translation can be resolved by the papallelopiped law into three motions along the axes of x, y, and z, after the manner of forces. 182.]^ Kinematical Theorem IV. If a body receives a motion of rotation round an axis OA, the rotation being repre- sented in magnitude by OA, and at the same time a motion of rotation (of the same sign as the first) round an axis OB, the rotation being represented in magnitude by OB, the resulting motion is one of rotation round the diagonal, OC, of the paral- lelogram determined by OA and OB, and is represented in magnitude by this diagonal. [The signs of rotations are determined by the rule given in Art. 137, Chapter X. We shall, for defmiteness, suppose that when a watch is held with its face perpendicular to AO, so that OA passes up through the glass, the rotation about OA takes place in a sense opposite to that of the hands ; and similarly for OB.} Let P be any point on OC, p the perpendicular from P on OA, q the perpendicular from P on OB, and Jc.OA and Jc.OB the angular motions round OA and OB, respectively. Then in virtue of the rotation round OA, P moves upwards from the plane of the paper through a space equal to Tcp . OA ; and in virtue of the rotation round OB, P moves downwards from the plane of the paper through a space equal to kq . OB. Therefore the whole motion of P upwards is equal to O A .'03 ,'/ &^~ & k(p.OA-q.OB}. But this is obviously zero ; therefore P is at rest, and so is every point on OC. The motion is, then, a rotation round OC. Let 12 be the angular rotation of the body round OC. Then the point A moves upwards from the plane of the paper through a space equal to 12 . OA sin A OC, since OA sin AOC = the per- pendicular from A on OC. But A in turning round OB moves through a space equal to k . OB . OA x sin AOB. Hence 12 . OA sin AOC = k.OB.OA sin AOB, TJ 3 292 THE PRINCIPLE OP VIRTUAL WORK. [183. sm Therefore the resulting angular velocity is represented by OC, if the component rotations are represented by OA and OB. This proposition is known as the 'parallelogram of angular velocities.' It follows at once that an angular motion about any axis, OL, may be decomposed into three angular motions about three axes, Ox, Oy, and Oz. If these latter are rectangular, an angular motion o> about OL is equivalent to angular motions, o> cos a, CD cos /3, and o> cos y, of the same sign, round the axes of no, y, and z, the direction angles of OL being a, ft, y. 183.] General Displacement of a Rigid Body. The position of every point in a rigid body is known when the positions of any three points in it are known, provided that these points are not in one right line. The general displacement of a rigid body is, therefore, the same as that of a system of three points forming a triangle. Let A, B, C be the positions of three points in the body before the displacement, and A', If, C' the positions occupied by these points after the displacement. Then the triangle ABC may be brought into the position ABC' by moving A directly to A while B and C move parallel to AA' through spaces equal to AA', and then turning the triangle about A' until B and C coincide with I? and C'. But (Art. 180) this latter motion is one of rotation round some axis through A ' . Hence the general displacement of a rigid body consists of a motion of translation which is the same for all its points, and a motion of rotation round an axis through an angle which is the same for all its points, To find the changes produced in the co-ordinates, x, y, z, of any point in the body by a general displacement, we may con- sider the motions of translation and of rotation separately. Although we shall be concerned only with small displacements, it is well to investigate the changes produced in the co-ordinates of a point by a rotation through any angle, 6, round an axis whose position is given. Let the direction angles of the axis, OL (fig. 31 1), be a, /3, y; let P be the J)oint (x, y, z) which, after the body has rotated through an angle 183.] GENERAL DISPLACEMENT OF A KIGID BODY. 293 6 round OL, occupies the position Q; let PL ( = p] be the per- pendicular from P on OL, and Q r a perpendicular from Q on LP. Now the/a? of Q is the projection of on the axis of a? ; there- (^- fore the change in x is the projection of PQ along Ox, or the sum of the projections of Pr and rQ. But Pr = p (l cos 0), and u'-u Again, if the direction angles of ^ are A, p, v, since Qr is at right angles to OL and f~L } the direction cosines of -Qr are cos cos v cos y cos ju, &c. Hence, if the x of Q is a?', a x x =. p sin (cos /3 cos v cos y cos /x) 2jt? cos A sin 2 (1) LT 2 But jj cos A is the projection of -PL- along the axis of x, or the projection of OP the projection o^OL, and since O.Z/ = a? cos a +y cos /3 + # cos y, p cos A = x (x cos a+^ cos /3 + cos y) cos a ; similarly p cos /x = y(oc cos a-J-y cos fi+z cos y) cos /3, pcos v = 2 (#cos a-j-y cos j8 + ^ cos y) cosy. Substituting these values in (1), we have a x' x = sin (z cos /3 y cos y) + 2 sin 2 - [(^? cos a +y cos /3 ^ -f^ cosy) cos a #], (2) and similar values for the changes in y and z. If the angular rotation 6 is very small, we have bay = (z cos /3 y cos y) 0, g^ = (# cos yz cos a) g# = (y cos a a? cos and if the components of the rotation bd along the axes be denoted by 50 1} S0 2 , 60 3 , these equations give x ( (3) / Of course these equations can be obtained very simply by con- sidering the separate changes in the co-ordinates produced by successive rotations 80 15 b0 2 , bO B round the axes of #, y, z> re- spectively. (See Routes Rigid Dynamics.) If the components of the motion of translation common to all points in the body be ba, bb, be, the complete changes in the co-ordinates for a small displacement will be 294 THE PRINCIPLE OF VIRTUAL WORK. [184. (4) 184.] Deduction of the Six Equations of Equilibrium. Replacing the virtual work of each force in equation (1) of Art. 177 by the virtual work of its three components, the general equation of virtual work becomes 2 (J804- Yby + Zbz) = 0, (1) and substituting in this equation the values of 8 x, 8^, and bz given by (4), we have ba. 2JT+ bb. 27+8*. 2^+8^.2(^-7^) + 80 2 .2(Xz-^) + 80 3 .2(7# Xy) 0. (2) Now, the displacement being quite arbitrary, its components 80, bb, be, 80j, 80 2 , 80 3 , are completely independent. Hence in (2) we may consider all of them zero except one, and the equation then gives the coefficient of this one equal to zero. Thus (2) involves the six equations 2Z=0, 27=0, 2^=0, 2 (Zy Tz) = 0, 2 (Xz-Zx) = 0, 2 (YvXy) = 0, which are the equations of equilibrium before obtained (see p. 232). In addition to the following Examples, the student will do well to solve some of those in p. 179 by the Principle of Work. EXAMPLES. 1. Four rigid bars, freely jointed together at their extremities, form a quadrilateral, A BCD ; the opposite vertices are connected by strings, AC and BD, in a state of tension ; compare the tensions of these strings. Let the bar AB be considered as fixed, and let the quadrilateral undergo any slight deformation. Then the bars AD and BC will turn round the points A and J3, that is, the points D and C will describe small paths, Dd and Cc, perpendicular to AD and BC. Hence (Theorem I) the point, /, of inter- section of AD and BC is the en- Fig. 212. stantaneous centre for the bar CD, and the angles DId t and CIc are U?J 184.] EXAMPLES. 295 equal. Denote their common value by SO. Then Dd ID . bO, and Cc = ic.se. Now, since in the displacement of the system none of the geo- metrical conditions namely, the constancy of the lengths of the bars are violated, the stresses of the bars will not enter into the equa- tion of virtual work. Hence if T and T' denote the tensions of the strings AC and BD, this equation will be (see p. 78), T.bAC + T'.bBD = 0. (I) But bAC = projection of Cc on AC = Oc . sin A CB = 1C . sin A CB . 80; and similarly bBD = -ID . sin BDA . 80. Hence (1) becomes T.IC.smACB = T / .ID.smBDA. (2) 1C AC sin CAD Substituting in (2) we obtain - OA . OC ~ OJB.OD' ^ ^ ^ ;- JJ % ; ?* Another solution of this problem (quoted from Euler) will be found ,; 77/*c. o3'< in Walton's Mechanical Problems, p. 101. = T-.'J&^.O^.C u/ 2. Four rigid bars, freely jointed at their extremities, form a '77/4C, ~J~, J quadrilateral, ABCD ; the bars AB and AD are connected by a string, 0/Toc *" 0% a a in a state of tension, a being a given point in AB, and a a given point in AD ; in the same way, BA and BC are connected by a string b(3; CB and CD are connected by a string cy ; and DC and DA by a string db; find the relation between the tensions of these strings. If the lengths of the strings a a, 6/3, cy and db are denoted by x, y, z, and w, and the tensions in them by X, Y, Z, W, the equation of virtual work for a slight deformation will be Xbx+ Yby + Zbz+ Wbw = 0. (1) Now si? therefore x b x = 2 ' - BD . bBD. AB.AD Substituting this value of bx, and similar values of by, bx, bw, in (1), we have X Aa.Aa Z Cc.Cy But from the last Example, we have bBD bAC~ 3 have /* *' sS <> ^ >A X-- C BD.OA. OC ^- V. AC.OB.OD ' \ \ c^ & ^ ^ f 0- V 296 THE PRINCIPLE OF VIRTUAL WORK. [184. .X Aa.Aa Z Cc.Cy^ BD* hence, finally, (_ . ^ + -_ . g^-^g) ^-QD T Bb.Bfi W Dd.Db AC 2 ~> ' BA.BC*^' DC.DA* OA.OC' For a different solution, see "Walton, ibid. v 3. Six equal heavy beams are freely jointed at their extremities ; one is fixed on a horizontal plane, and the system lies in a vertical plane ; the middle points of the two upper non-horizontal beams are connected by a rope in a state of tension. Show that the tension ofthiaropeis W being the weight of each beam, and the inclination of the non- horizontal beams to the horizon. Let x be the length of the rope, y the height of the centre of gravity of the system, 2 a the length of each beam, and T the tension of the rope. Then the virtual work of the tension is Tb x (see p. 78), and the virtual work of the weight of the system is 6Wby. Hence But x = 2 a (I + cos 0), and y = 2 a sin 0, and the deformation imagined is one in which the upper horizontal beam moves vertically through a small space. Hence the values of y and x will be of the same forms as before, and ^ = _^ ^ ^ by==2a cos Substituting these values of bx and by, we have 4. A body receives a small general displacement parallel to one plane ; find the co-ordinates of the instantaneous centre. If the components of the motion of translation parallel to the axes of x and y are b a and bb, and the rotation is 6\o, the equations (4) of Art. 183 give for the displacement of any point whose co-ordinates are x. y. s * * g# = oa 2/oco, by = bb-\-fcb(). Now, the displacement of the instantaneous centre is zero j hence, if (x } y) be its co-ordinates, we have o* 7 o^ bb ba x T"' y = ^~' OO) 0(0 A particular case may be noticed. If any body in contact with a surface receives any small displacement parallel to one plane, the body still remaining in contact with the surface, the instantaneous centre lies on the normal to the surface of contact. In the rolling of one figure on another the point of contact is the instantaneous centre. 184.] EXAMPLES. 297 5. A uniform beam, AS (fig. 133, p. 149), rests as a tangent at a point P against a smooth curve in a vertical plane, one extremity, A, resting against a smooth vertical plane ; find the position of equi- librium, and the nature of the curve so that the beam may rest in all positions. Let the weight of the beam through 6r, and the normal reactions at A and P meet in the point ; take the vertical line AD as axis of y ; and let 2 a = the length of the beam. Then, if x is the abscissa ryt of P, we have AO = . n ? and also AO = a sin 0. Hence, equating sm 2 6 these values, oo = asin 3 0. (1) Now, from the equation of the given curve, 6 is known in terms of * in the form =/(*?). (2) From (1) and (2) the value of #, and therefore the position of equi- librium, can be found. For example, if the curve be a circle of radius r whose centre is at a distance c from the vertical plane, we find a sin 3 + r cos 6 c = 0. If f = 0, we get the result in Ex. 7, p. 149. If (1) holds in all positions in which the beam is placed, every position is one of equilibrium. Now, since tan = ? (1) gives dy = ^/a?oc* . x~*dx, and since this equation holds in all positions, we may integrate it. Hence y + k = (a% x% ) f , or k being an arbitrary constant. We may, without loss of generality, assume k = 0, and the curve will be 222 x^ + y*=. a*. The equation of virtual work shows that in this case the centre of gravity of the beam is at a constant height. For if y denote the ordinate of G, this equation is way = o, and since this holds in all positions, we have, by integration, y = constant. v 6. Four rigid bars freely jointed at their extremities form a quadri- lateral ABGD (fig. 213) ; the middle points of the opposite pairs of bars are connected by strings, mm' and nn', in a state of tension. Compare the tensions of these strings. Let I and If be the lengths of the strings mmf and nn', and let the tensions in them be T and T' } respectively. 298 THE PRINCIPLE OF VIRTUAL WORK. [184. Then, assuming the quadrilateral to receive any small deformation, the equation of work will be Now, it may be left to the student as an exercise to prove that that is, r 2 Z 2 is constant however the quadrilateral may be deformed. Hence j8 Z _r8r=0; (2) B and from (1) and (2) we have a remarkable result, since it shows that one of the tensions must be negative ; i. e., if the bars AS and CD are pulled together, equilibrium will be impossible unless the bars AD and BC are pulled asunder. It is well to notice an apparent exception to the result (3). The student will easily prove that if the sides AB and DC are parallel, equilibrium will be maintained by the single string mm' in any state of tension, i. e., T' 0, a result which contradicts (3). The difficulty is easily removed, however, by reverting to (1), which in the case under consideration is identically satisfied. For, since AB and CD are parallel, the line mm' passes through/, the instantaneous centre, and therefore for a slight deformation the point m' moves perpendicularly to Im', that is, to mm'. Hence bl = 0, and equation (1) is satisfied by having at once T'= and 8Z = 0. The combination of (1) and (2) is therefore irrelevant. 7. A number of bars are freely jointed together at their extremities and form a polygon ; each bar is acted on perpendicularly by a force proportional to its length; all the forces emanate from one point and all act inwards or all outwards ; prove by virtual work that for equilibrium the polygon must be inscribable in a circle. Let the polygon be ADCBEF ... (fig. 213), of which the vertices E,F... are not represented in the figure. [AB is not one of the -bars.] Choose a virtual displacement in which all the bars except the three AD, DC, CB remain fixed, and let the extremities A and B be fixed in the displacement. Then / is the instantaneous centre for DC. Let be the point from which the forces emanate ; let m, n,j) be the feet of perpendiculars from on AD, DC, CB, respectively; let Q be the foot of the perpendicular from / on DC ; let IQ meet mO in L and jpO in M\ and let the forces in Om, On, Op be Jc.AD, k.DC,k.CB. If AD turns round A through the small angle 6"$, the displacement of D is AD . b ; and if DC turns round / through 6co, the displace- ment of D is ID . 6 co. Hence 184.] EXAMPLES. 299 AD.bcji = ID.ba>. Similarly EC . b 6 = 1C . b o>, if b6 is the angle through which EC turns round B. Now the equation of virtual work is k.AD.Am.b, we have cos a) = 2ab 1 12. If the forces in Example 7 are each transferred to the middle point of the bar on which it acts, prove by virtual work that the polygon must be inscribable for equilibrium. 185.] Lagrangian Meaning of the Virtual Moment of a Force. We see that in the general equation (2) of virtual work, each of the displacements, 80, &c., is multiplied by a function of force which tends to produce this displacement. Thus 8 X is multi- plied by the whole moment of the forces round the axis of x, and the tendency of this moment is to produce a rotation round the axis; ba is multiplied by the whole component of the forces along the axis of x, and the tendency of this component is to produce a motion of translation in this direction. In the same way, in equation (2) of Art. 177, each force is multiplied by a variation which it tends to produce. Thus the tendency of the force P l is to drag its point of application in its own direction. If PI is the distance, OA : , of the point, A 1 , of application of the force from a fixed point, 0, in the line of action of the force, the tendency of P : is to alter the distance p 1) and accordingly the term P l bp l appears in the equation of virtual work. Similarly, the tendency of the internal force T is to alter the distance, r, between the points % and m z , and accordingly the 1 86.] EQUATIONS OF CONDITION. 301 term Tbr enters also into the equation. Each of these terms is, in fact, the elementary work which the corresponding force tends to do, and which it would do if the system were displaced or deformed ; and hence all such terms must appear in a com- plete equation of virtual work. Hence Lagrange defines the virtual moment, or virtual work, of a force as the product of the force and the variation of the function which it tends to alter (Mecanique Analytique, 5, p. 29 ; 9, p. 33 ; 6, p. 73 ; 26, p. 126, Bertrand's edition), and in every case he obtains the general equation of equilibrium of a system by adding together all such products, whether they belong to the given external forces, the geometrical forces (reactions of smooth surfaces, or forces of connexion), or to the internal forces (mutual attractions or repulsions) of the system. This method of the solution of statical problems (which is obviously only the method of virtual work) is one of great power and generality, and its nature will be rendered more clear in the sequel. 186.] Equations of Condition may be Replaced by Forces. Suppose a system of n particles whose co-ordinates are connected by k equations of condition, ^=0, 4 = 0,... 4=0, (1) each of these equations being of the form that is, involving the co-ordinates of all the points in general. Then the equation of virtual work for the position of equilibrium of the system is which, when written at full length, is (2) Now if the virtual displacements of the particles were all independent, this equation would involve the vanishing of the coefficient of each displacement (see Art. 184); but the displace- ments of the particles must be such as still to satisfy the equa- tions (1). Hence the quantities 8#, &c., are connected by the k equations 302 THE PRINCIPLE OF VIRTUAL WORK. [186. dJj-\ dJj-t dJL/-t + (3) Solving these k equations for any k of the displacements suppose 8# 15 8 # 2 5 ... bsn k and substituting their values in (2), we obtain an equation connecting the remaining 3n k displace- ments of the form A * Now, the remaining quantities, 8#j. +1 , &c.^ are completely independent, and therefore (see Art. 184) every coefficient in this equation must = 0. Thus, we obtain 3n k equations involving the forces, that is, statical equations of condition. Combining these statical equations with the equations of con- nexion (1), we have finally 3n equations for the 3n co-ordinates of the particles. The elimination of the displacements from the equations can, however, be exhibited in a more symmetrical and useful form. Multiply the equations (3) by A 15 A 2 , ... X^. in order, these multipliers being undetermined quantities j then add the equa- tions together, and finally equate to zero the coefficient of every displacement in the resulting equation. Thus we shall have the following 3 n equations : 1 86.] EQUATIONS OF CONDITION. JL -t -f- A-i -= -|~ An ~~: "f- . . . "T" AT.~^ == U^ -I- * A ft m & // /yt ** ft HP 303 + ... (5) If from these equations we eliminate the k undetermined multipliers, we shall have 2n k statical equations of condition, as before. Now this method of elimination has the advantage of dis- covering the geometrical forces, or forces arising from the con- nexions, of the system. For, suppose that we suppress the condition L^ = ; then the system will begin to move ; but it may be kept at rest by applying a special force to each particle. Let the components of the force applied to m l be X-f, T/, Z-f, those of the force applied to m^ X 2 ', Y 2) Z%, and so on for all the others. The equations of equilibrium of % will then be similar equations holding for the other particles. Subtracting each of these from the corresponding equation in (5), we have Hence^ Y ' Y ' 7 ' * * and 304 THE PRINCIPLE OF VIRTUAL WORK. [186. If, now, all the co-ordinates involved in the equation L^ = are considered constant except x l9 y lt and z lt this equation will denote a surface on which the particle m 1 is constrained to lie, and dz l ^ each divided by A - + (~) + will be the direction cosines of the normal to this surface at the point (#,_ , y l , ^). It is evident, therefore, that the force required to keep the particle m at rest, when the condition LI is suppressed, is a force acting normally to this surface, its magnitude being ~ ~ ~ : In the same way the force required to keep m 2 at rest acts normally to the surface denoted by L^ = when a? 2 , y 2) z% are considered as the only variable co-ordinates in the equation, and the magnitude of this force is If the condition I/ 2 = were suppressed, it follows in like manner that forces should be applied to the particles %, &c., in directions normal to the surfaces represented by the equation I/ 2 = when the sole variables in it are the co-ordinates of %, &c., in succession. It is easy to see that sdL* ^ dL, dL, . \ A! (-J+ 8a? + A gy + A ^ ) l ^d%i dy^ ' dz *' is equal to ^ ( cos a . 8^ + cos ^3 . 5^ -f cos y . 8^), where F l is the force of connexion acting on m in virtue of the condition L^ = 0, and a, (3, y the direction angles of the normal to the surface denoted by L = when the co-ordinates of m l are regarded as the only variables in it. Now, the multiplier of F 1 in this expression is evidently the projection of the displacement of % along the normal to this surface. If this projection be denoted by n, n being the 1 86.] EXAMPLES. 305 length of the normal at the position of m l measured from some fixed point on the normal, we have A^^ = F^n, in which the variation of I/ has reference solely to the par- ticle m-y. Now, as the force F^ acts along the normal, and tends directly to alter its length, or to produce the displacement bn, we may, in conformity with Lagrangian language, regard the term X^L-^ as the virtual work of a force tending to vary the function L^ This is true without regard to the nature of the function L . It may, then, be a function not pnly of co-ordinates, but of differential coefficients of co-ordinates. It may, for example, express the imposed condition of inextensibility in the case of a string, and then it will take the form ds = constant ; or it may express the same condition in the case of a membrane, or, finally, the incompressibility of a fluid, and then it will be dxdydz = constant. Except in the case of continuous systems (such as springs, membranes, and strings), this method is not a simplification of the ordinary statical methods. Nevertheless, for the sake of showing its application in practice, we add a few examples solved by means of it, deferring its more useful application for the present. EXAMPLES. 1 . A number of heavy particles are attached at given intervals to a weightless string the extremities of which are fixed ; investigate the circumstances of equilibrium (Funicular Polygon). Let (a, b) be the co-ordinates of one of the fixed extremities, (o^, y^), (#2, y 2 ), ... the co-ordinates of the particles taken in order from this extremity, Z 01 , Z 12 , ... the lengths of the portions of the string between these points, and W lt W z , ... the weights of the particles. Then the equations of connexion of the system are (a -0 Hence the Lagrangian equation of virtual work is i) ... = 0. 306 THE PRINCIPLE OF VIRTUAL WORK. [186. Equating to zero the coefficients of the several displacements, A i( - l )-X a (aJ l - ! ,) = 0, A 2 (x l ^ 2 ) A 3 (a? 2 # 3 ) = 0, ^2-^2 (2/1-2/2) + ^3(2/2-2/3) = 0, The first set of these equations evidently give M^-^i) = ^2(^1-^2) = A 3 (tf 2 -# 3 ) = ... = T, suppose, and by substituting in the remaining set, 1 W^ a x l OC-LXZ T 2/1-2/2 _ 2/2-2/3 j W r i *Oj ~""" t/o ^o ~~" 3 But -- - is the tangent of the inclination of the portion Z 01 of the CL #7j string to the horizon. Hence we have as in p. 32. Also the tension of the string joining (a, b) to (x l1 y^ is j^- acting from the first point towards the second, and so on for the ^01 other tensions. 2. Deduce by the method of Lagrange the conditions of equilibrium of a system of three particles forming a rigid triangle, each particle being acted on by given forces. Let (a?!, y 19 %) be the co-ordinates of one particle, and (Jfj, T lt Z^ the components of the force acting on it, with similar notation for the other two particles. Then, if 12 , Z 23 , I 3l denote the sides of the triangle, the equations of connexion are -*) 1 = V, Hence the Lagrangian equation of equilibrium is -^ 2 ) ... =0, ^ the undetermined multipliers being A 12 , A 23 , and A 31 . Equating to zero the coefficients of the displacements, we have X 1 + A 12 (# 1 -ag-A 31 (* 3 -*i) = 0, (1) Y i + ^12 (2/i-2/ 2 ) ~ A 3i fe-2/i) = 0, (2) 187.] EXAMPLES. 307 ^i + Aia (*i - a) ~ A si ( V- i) = 0, (3) with similar equations for the other particles. By addition, we have at once X = or 2JT=0 ^ + ^2 + ^3 = 0, or 2^=0, which are the ordinary equations of translation. Again, multiplying (1) by y l and (2) by x 19 and subtracting, r i os l --T 1 y 1 -A ia foy a -.y 1 a! 2 )-A 81 (x^-y^) = 0, and by taking the similar equations for the other particles, and adding, we get 2 (Yx Xy) = 0. Similarly, 2 (Xst - Zx} = 0, and S(y-7*) = 0. These last three are the equations of moments, and they constitute with the first three six equations of equilibrium. Now these are all the conditions that can be obtained among the forces and co-ordinates. For if n particles be connected by Jc equations of condition, there are (Art. 186) 3n k final equations. But here n 3, k = 3, therefore 3n k = 6. It is to be observed that the equations of equilibrium of any rigid body must be the same in number as those for three particles forming a rigid triangle, because if three points of a rigid body are determined in position, the position of the body is determined. 3. Show that the equations of equilibrium of a system subject to given conditions may be expressed as the vanishing of the differential coefficients of a single function of the co-ordinates of the system. Suppose that or 2 (Xdx + Ydy + Zdz\ =dV where V is a function of the co-ordinates i, 2/i> *p 2 2/ 2 > *2 Then > taking where L^ = 0, Z 2 = 0,... are the equations of condition, we shall have dV dL dL d\ d\ But since the co-ordinates make L : = L 2 = , . : = 0, dV _ dL dL and comparing with equations (5), we see that the equations of equi- librium are dU dU dU dU -= = 0, - r ~ = 0, ... -r = 0, -T = 0, &C. dx : dx 2 dy l dy^ 187.] Distinctive Feature of the Lagrangian Method. If the first method of eliminating the displacements described in the last article is adopted, we arrive at an equation such as (4) of that Article, from which the conditions of equilibrium are x a 308 THE PRINCIPLE OP VIRTUAL WORK. [187. obtained by equating to zero the coefficients of the displace- ments. But in proceeding thus, we fail to obtain the values of the internal and geometrical forces of the system. Now these forces are, as we have seen, intimately related to the undeter- mined multipliers ; and as these latter are found from the Lagrangian equations, it follows that The method of Lagrange gives not only the conditions of equi- librium^ but also the internal forces of the system. A single very elementary example will suffice to render this clear. Two heavy particles of weights W l and W 2 are connected by a rigid rod, and each particle rests on a smooth inclined plane. The incli- nations of the planes are ^ and i 2 and their intersection is horizontal ; find the position of equilibrium and the internal and geometrical forces. Let the line of intersection of the planes be taken as axis of z, let the axis of y be vertical and that of x horizontal. Also let (x l y^ zj, (#2 2/2 z z) be the co-ordinates of the particles, and I the length of the rod connecting them. Then the equations of connexion are y 1 x 1 tsaii l = 0, 2 = 0, Hence the Lagrangian equation of equilibrium is , A 2 , and T being the undetermined multipliers. Equating to zero the coefficients of the separate displacements, j tan i^ r (x^ x^ = 0, From the last equation we have z^ z 2 = 0, which shows that both particles must lie in a vertical plane perpendicular to the line of inter- section of the inclined planes. If 6 be the inclination of the line joining the particles to the horizon, the other equations give (^ i+ JF 2 )tan0 = _ - W l cos cos ij 1 ~ cos(^ d) W 2 cos 6 cos i 2 2 " 189.] STABILITY AND INSTABILITY OF EQUILIBRIUM. 309 The student will easily perceive from Art. 186 that rl is the tension of the rod, and A x sec tj and A 2 sec i z the reactions of the smooth planes. Thus we have the same values of the inclination of the rod and of the internal forces as we should have obtained by the ordinary statical methods. Suppose now that the equation of virtual work is employed according to the first method ; that is, let us write = 0, and eliminate the displacements without employing undetermined multipliers. Then we obtain simply the equations 2j-2 2 = 0, ( JPi + W^ tan 0=W 1 cot i 2 TF 2 cot i 1} which define the position of equilibrium, without giving the values of the unknown forces of the system. V 188.] Potential of a System of Forces. Let there be any number of particles, m 1 , m 2) ... acted on by forces X 19 Y lt Z lt X 2J Y 2 , Z^ ..., and let the co-ordinates of the particles be 0*1 y\ ' z i)> 0*2 > y* Z 2 )> Then, if V be such a function of the co-ordinates that *L Y ^L V dV 7 ~~ 15 ' 1} we have or, as it may be written for shortness, V = ^f(Xdx + Ydy + Zdz\ 2 denoting a summation of the integral for all the particles of the system. The integral may be considered either as indefinite, or as performed from any fixed position which the system can geometrically occupy to the position which it occupies at the moment under consideration. Of course it may happen that the forces are such that 2 (Xdx + Ydy + Zdz) is not the differential of any function of the forces and co-ordinates ; when it is the differential of some function, the system belongs to what Thompson and Tait call Conservative Systems (Nat. Phil.). The function V which belongs to a conservative system is called the Potential of the given forces. 189.] Stability and Instability of Equilibrium. When a body in equilibrium under the action of given forces is slightly 310 THE PRINCIPLE OF VIRTUAL WORK. [190. disturbed from its position, it will not, in general, be in equili- brium in the new position. Now the effect of the forces in the new position of the body may be either to drive it back to its original position, or to deviate it still further from this position. In the former case the equilibrium is said to be stable, and in the latter unstable. For example, take a heavy rod, AB, moveable round a smooth hinge at one end, A. If the rod is placed in a vertical position it will evidently be in equilibrium ; but if the end B is vertically above A, a slight disturbance will cause the rod to fall from this position ; while if B is vertically below A, after a slight disturbance the rod will revert to its original position. 190.] Maximum and Minimum Potential. When a body or a system of bodies assumes such a position that the potential of the forces acting on it is a maximum, the body or system is in a position of stable equilibrium. When, on the contrary, the position of the system is such that the potential has a minimum value, the equilibrium is unstable. Here the terms maximum and minimum are to be understood as they are defined in the Differential Calculus. The complete proof of this principle is kinetical, and it will be found at great length in the Mecamque Analytique (6th section of the Dynamique^ p. 320). In a very useful particular case, however, a statical proof may be given. Suppose a system, subject to certain geometrical conditions, to be in equilibrium, and suppose, moreover, that, subject to these conditions, the position of the system is defined by a single variable. In general (Art. 186) the Aquations of equilibrium are ^ = 0, 2 = 0, ...^=0, and 2 (Xbso + Yby -f Zb z) = 0. Assuming the forces to have a potential, F, the last equation is 5F=0. (a) Now if all the co-ordinates, # 15 y l9 z l9 ... , in conformity with the geometrical conditions, L 0, . . . , are expressible in terms of a single variable, q, V is simply a function of q, and the statical equation can be written ' ^ = o. (ft) Hence, in the position of equilibrium -= = 0, and there- 190.] MAXIMUM AND MINIMUM POTENTIAL. 311 fore V is, in general, either a maximum or a minimum, since d z V y-g- will not, in general, vanish. Now it has already been explained (Arts. 185 and 186), that in the equation of equilibrium the coefficient which multiplies each variation is proportional to a force which tends directly to produce this variation ; therefore from (/3) we see that 7 is proportional to a force which tends to produce the displacement dV denoted by bq, or, in other words, -= is proportional to a force do which tends to increase q ; and (/3) shows that in the position of equilibrium this force must vanish. Suppose now that, the geometrical equations of condition being still satisfied, the system receives a small displacement for dV which q becomes q + bq. Then if = is denoted by f (q), the dV % value of -7 in the new position will beffq + bq) j that is, the do force called into play by the displacement is dV or -= -- h dq dV But, by hypothesis, = 0, therefore the force called into play is If this force has the same sign as bq, the force called into play increases the displacement, and the equilibrium is unstable ; whereas if the sign of the force is opposite to that of the displacement, the force destroys the displacement, and the equilibrium is stable. In the former case -r-^- is positive and V a minimum, and in the latter -=-^ is negative and V a maximum. Whether the position of the system depends on a single variable or on several variables, equation (a) is satisfied in every position of equilibrium ; but the vanishing of the first differential of a function of several variables is not a sufficient condition for a maximum or minimum value of the function. Hence we 312 THE PRINCIPLE OF VIRTUAL WORK. [191. cannot assert that every position of equilibrium of such a system is one in which V is either a maximum or minimum. On the contrary, when the position depends on a single variable*, Fis, in general, either a maximum or a minimum, and the equilibrium is, in general, either absolutely stable or absolutely unstable. A position of equilibrium is said to be absolutely stable when, after all possible small displacements, the system reverts to its position of equilibrium ; and absolutely unstable when, after all possible small displacements, it deviates still further from that position. Since maxima and minima values of a function succeed each other alternately, it is clear that the same is true of the posi- tions of stable and unstable equilibrium of a system. r 191.] Maximum or Minimum Height of the Centre of Gravity. When gravity is the only force acting on a system of bodies, the potential is simply ^ -W.i, W denoting the weight of the system, and z the height of its centre of gravity above any fixed horizontal plane. For if w 1 be the weight of any one body of the system and % the height of its centre of gravity above a fixed horizontal plane, the virtual work of w 1 for a small increment of % will be (Art. 66, p. 79) -wj.8^. Hence f 5 V = w-^z^w^z^ But TT.z = w 1 .z l + w 2 .z 2 + . . . ; therefore 6 V = W. bz, and 7=-W.z. Now the maximum value of V will occur when z is least ; hence when the centre of gravity of any system of bodies is in the lowest position that it can occupy consistently with the geometrical conditions of the system, that system is in a position of stable equili- brium; and when its centre of gravity is in the highest position, the system is in a position of unstable equilibrium. Unless the position of the system depends on a single variable, we cannot assert conversely that a position of equilibrium is one in which the height of the centre of gravity is either a maximum or a minimum. * The system in this case is called by French writers un systbme & liaisons completes. t This assumes that none of the geometrical forces required for a position of equilibrium are infinite ; for the term \8L cannot be assumed to vanish, even though 8Z = 0, if \ is infinite. 192.] CONTINUOUS EQUILIBRIUM. 313 If any bodies of the system rest on rough curves or surfaces, the equation of virtual work will involve the reactions of these curves or surfaces for displacements along them. Hence we have no longer the equation JT.bz=:0, and the principle of maximum or minimum height of the centre of gravity does not hold. Even when the position depends on one variable, it may happen that in a position of equilibrium the height of the centre of gravity is neither a maximum nor a minimum. Take, for example, the case of a heavy particle placed at a point of in- flexion on a smooth curve in a vertical plane, the tangent at the point being horizontal. The particle is evidently in equilibrium, since for a small displacement Pbz is zero, P being the weight and z the height of the particle. But z is neither a maximum nor a minimum, and the equilibrium, accordingly, is stable for a small displacement along the upper part of the curve, and un- stable for a displacement along the lower part. When the connexions of the system are complete (see note, p. 312) the centre of gravity describes, in all positions of the system compatible with the given conditions, a curve which is sometimes very easily found. In the position of equilibrium the centre of gravity will be the point of contact of a horizontal tangent to this curve, and in this manner we can most readily perceive the nature of the equilibrium of the body. When the connexions of the system are not complete, it may happen that its centre of gravity is constrained, in all displace- ments compatible with the connexions, to describe a fixed surface. In this case the position of equilibrium will be one in which the tangent plane to this surface at the centre of gravity is horizontal ; and if the surface lies entirely below the tangent plane in the neighbourhood of the point of contact, the equilibrium will be unstable, as in the case of a curve ; if the surface lies above the tangent plane, the equilibrium will be stable ; and if the tangent plane intersects the surface in a real curve in the neighbourhood of contact, the equilibrium will be stable for some displacements and unstable for others. 192.] Continuous Equilibrium. If in all positions of the system, compatible with the geometrical conditions, the statical equation aF=0 314 THE PRINCIPLE OF VIRTUAL WORK. [192. is satisfied, every position is one of equilibrium. Writing- down this equation in all positions, and adding the equations thus obtained is evidently the same thing as integrating it. Hence if all positions of the system are positions of equilibrium, the applied forces must satisfy the equation V = constant. In the particular case of a heavy system under the action of gravity alone, V is W. z ; therefore if a system be continuously in equilibrium under the action of gravity, the centre of gravity of the system for all displacements compatible with the condi- tions moves in a fixed horizontal plane, or, in other words, maintains a constant height. EXAMPLES. 1. A heavy beam, AB (fig. 127, p. 145) rests on two smooth in- clined planes ; find the nature of its equilibrium. It is very easy to prove that if the right line AB moves between two fixed right lines, OA and OB, the given point G on AB describes an ellipse whose equation with reference to OA and OB as axes of x and y is __ The centre of this ellipse is the point 0. In the position of equi- librium G is the point of contact of a horizontal tangent to this ellipse. Now two such tangents can be drawn, one above the inter- section of the inclined planes and the other below it. There are, therefore, two positions of equilibrium; that with which we were concerned in the example of p. 145 is obviously the position in which G is at a maximum height, and it is, therefore, unstable; the other requires the planes to be prolonged below their line of intersection, and as it also requires the reactions of the planes to assume impossible directions, it is physically impossible. It would, however, be possible if the planes were replaced by smooth fixed rods to which the extremities of the beam are attached by rings. The second position of equilibrium would then be stable. The impossibility in a certain case of any position of equilibrium, except one of continuous contact with either plane, which has been signalized in p. 146, is now easily explained. It occurs when the point of contact of the horizontal tangent to the ellipse locus of G falls underneath the plane (a) or the plane (/3), so that it is not a possible position of G. The problem may be solved by a purely analytical method. If z is the height of the centre of gravity of the beam, it will be easily found that in the position of equilibrium 192.] EXAMPLES. 315 sin a sin /3 cos dd* {(a + 6) 2 + (a cot a - b cot /3) 2 } 2. Two given points of a body rest each in contact with two smooth inclined planes; show that the equilibrium of the body is unstable. We know that if two vertices of a given triangle move along two fixed right lines, the locus of the third vertex is an ellipse whose centre is the intersection of the given lines. Hence if we consider a given triangle in the body to be formed by the centre of gravity and the two points which are in contact with the planes, we see that the locus of the centre of gravity is an ellipse whose centre is at the intersection of the inclined planes. Now in the position of equilibrium the centre of gravity is the point of contact of a horizontal tangent to this ellipse. Hence the only possible position of equilibrium is one in which the height of the centre of gravity is a maximum ; therefore the equilibrium is unstable ; and if, as explained in the last Example, the point of contact of the tangent falls underneath either plane, the only po- sition of equilibrium of the body is one of continuous contact with one of the planes. The student will find several particular examples of this problem in Walton's Mechanical Problems (pp. 164, &c.), where the solutions are ana- lytical. 3. A heavy body has two plane surfaces, OP and CQ (fig. 214) which rest against two smooth fixed pegs, P and Q, the line PQ making any angle with the horizon; show that the positions of equilibrium are determined by drawing horizontal tangents to a Limaon. The centre of gravity and the pegs must lie in one vertical plane, which is that of the figure. Since P and Q are fixed points and the angle at C between the plane faces is constant, the circle described round the triangle PCQ is fixed in space. Again, let G be the centre of gravity of the body. Then since CG and CP are lines fixed in the body, the angle GCP is given ; and if CG meet the circle in 0, the point is fixed in space ; also the distance CG is given. Hence in all positions of the body i.e., in all positions of C on the circle the centre of gravity is found by drawing the line OC from to the circumference of the circle, and taking a constant length, CG, on this line. The curve deduced in this way from a circle is a Limaon, which is, therefore, the locus of the centre of gravity. A particular example has been already discussed in p. 1 50. 4. A heavy plane body of any shape is suspended from a smooth peg, fixed in a vertical wall, by means of a string of given length, the Fig. 214. 316 THE PRINCIPLE OF VIRTUAL WORK. [192. extremities of which are attached to two fixed points in the body. Determine the nature of the equilibrium. This problem, so far as the positions of equilibrium are concerned, has been already discussed (Ex. 11, p. 153). We propose here to show that there are two positions of stable and one position of unstable equilibrium. In the figure of the Example referred to, the point of contact of GP S with the evolute is between G and P 3 ; the point of contact of GP^ is between G and P x ; and the point of contact of GP 2 is on PJJT produced. Now it is easy to see that GP B is a line of maximum length drawn from G to the ellipse. For, let Q be a point on the ellipse close to P 3 , and let QC be the normal at Q. Then C is the centre of curvature, and therefore the point of contact of GP 3 and the evolute. 2I , Hence CP 3 = CQ, therefore GP 3 = GO + CQ, which is > GQ, therefore GP 3 > GQ, and GP S is, therefore, a maximum. In the same way GP l is a maximum and GP 2 a minimum distance of G from the ellipse. Hence, in the positions of equilibrium, GP : and GP 3 are maximum distances of the centre of gravity from the peg. The positions in which these lines are vertical are, therefore, positions of stable equi- librium. And since GP 2 is a minimum depth of G, the position in which GP 2 is vertical is one of unstable equilibrium. 5. To find the nature of the equilibrium of the beam in Example 5, p. 297. Take any position of the beam (in which, of course, the lines GW, AR, and PS (p. 149) do not meet in a point). Then, if y is the ordinate of P, the point of contact of the beam and the curve, referred to a fixed horizontal axis, the ordinate of G will be y + (GA-PA)cose, or y + acosO xcotO. Denoting this by y, we have dy dy dy dy Hence sin 2 Q -= - = a sin 3 + a?. dv Differentiating this, and remembering that in the position of equi- librium ~ = 0, we have au (1) dv du 192.] EXAMPLES. 317 Again, since cot 6 = -^- j we have ax * cosec*0 But if p is the radius of curvature of the curve at P, * cosec*0 = 5-. ax ax* rl A 1 Therefore = -. - } and (1) gives dx p sin 6 sin -=-^ = p 3 a sin 6 cos Q ao = p-3PO. Hence, since sin is necessarily positive, ~ > will be positive, and y therefore a minimum if p > BPO. The equilibrium will therefore be stable or unstable according as p>or <3P0. To arrive at this result, it would have been sufficient to demon- strate it for a circle, which is very easily done. The curve in the neighbourhood of P may be replaced by the circle of curvature at this point. 6. Prove geometrically that the equilibrium of the beam in Example 2, p. 147, is stable. 7. Two uniform heavy rods freely jointed together at a common extremity rest on a smooth parabola whose axis is vertical and vertex upwards; find the position of equilibrium. Ans. Let the weights of the rods be P and Q, their lengths 2 a and 26, and let them make angles and 0, respectively, with the vertical in the position of equilibrium; then these angles are deter- mined from the equations Pa sin 3 + (P + Q) m cot < = 0, Qb sin 3 $ + (P+ Q) m cot = 0, 4m being the latus rectum of the parabola. [Taking the tangent at the vertex as axis of y, the abscissa of the point of intersection of two tangents, y = tx and y = t'x r > t t is 7 Hence (P + Q)x = Pa cos + Qb cos(f) + (P + Q)m cot cos(/>. Then x is to be a max. or min.] 8. A uniform heavy rod, AB, moveable about a smooth hinge fixed at A t has its extremity B connected with a string which, passing over a smooth pulley at a point C vertically over A, sustains a given weight 318 THE PRINCIPLE OF VIRTUAL WORK. [193. which rests on a smooth inclined plane passing through C. Find the positions of equilibrium, and the nature of each. Ans. Let W and 2 a be the weight and length of the rod ; P the weight on the plane whose inclination to the horizon is i; 2c the distance AC, and the inclination of the rod to the vertical. Then, if (c a) W< 2 PC sine, there will be three positions of equilibrium defined by the equations 2ac W* ' The first and last positions are stable and the intermediate one is unstable. If (c a) W>2Pc sin i, there is no intermediate position, and the first and last positions are unstable and stable, respectively. 9. One end of a beam rests against a smooth vertical plane, and the other on a smooth curve in a vertical plane ; find the nature of the curve so that the beam may rest in all positions. Ans. An ellipse whose axis major is the horizontal line described by the centre of gravity of the beam, the axis minor lying in the vertical plane. 10. A uniform heavy rod rests inside a smooth fixed sphere whose diameter is equal to the length of the rod. In all positions of the rod its centre of gravity is fixed ; hence the rod should rest in all positions ; but, except in the vertical position, it is impossible that the acting forces can give equilibrium. Explain this. (See note, p. 313.) 1 1 . A uniform rod rests in all positions with its extremities on two smooth curves in a vertical plane; given the equation of one, find that of the other. Ans. Let the axis of y be vertical, 2 a the length of the rod, h the constant height of the centre of the rod, and x $ (y) the equation of one curve ; then the equation of the other will be x = (> 2k- 12. Find the general equation of a smooth curve (in a vertical plane) on which if the ends of a uniform rod are placed, the rod will rest in all positions. Ans. If the line described by the centre of gravity is axis of x, the equation is of the form [$ (y 2 ) + x] 2 + y z a?, where 2 a = length of rod, and (y 2 ) is a function which does not change sign with y. 193.] Expansion of the Ab- scissa and Ordinate of a Curve in Powers of the Arc. Let A and B (fig. 216) be any points on a curve, and let Am and An be the tangent and normal at A. Also let \l/ be the angle between the normals at A and B, and let I94] EQUILIBRIUM OF A HEAVY BODY. 319 Am ( = x) and Bm ( = y] be the co-ordinates of B with reference to the tangent and normal at A as axes. Then, by Maclaurin's Theorem we have $ denoting the arc AB, and \jf Q) (~r-) > the values of *// and v -a where p is the radius of curva- ture. Hence p 1 . 2 ^s 1.2.3 ds 2 the suffix being omitted, it being understood that p is the radius of curvature at A. Again, we have also rui -_!:, an d ^ = I^. But (y) = 1, and (-J-) = ; therefore (-y^) = 0, (yf ) = - and the successive differential coefficients are calculated with ease. We thus obtain 194.] Equilibrium of a Heavy Body resting on a Fixed Rough Surface. Let AD (fig. 217) be a fixed rough surface on which a heavy body, AC, rests, under the action of gravity, at a single given point A ; and let this body receive a slight dis- placement of rolling on the fixed surface. We propose to investigate the nature of the equilibrium. The figure represents a section of the bodies made by the vertical plane through their common normal, AO, in which the rolling takes place. We suppose the normal AO to be vertical. 320 THE PRINCIPLE OF VIRTUAL WORK. [194. Then, since in the position of equilibrium the body AC is acted on by only two forces namely, its own weight and the total resistance of the fixed surface its centre of gravity, G, must be vertically over the point of contact. Let the point A of the rolling body come to A', and G to G', the new point of contact being B, and the new common normal 0(/. Draw the vertical line BF, meeting AC/ in V. Then, if A'7 is >A'G', the weight of the body acting through G' will produce a rotation round B which will send the body back to its original position; while, if AV is A'G'. (I) Let p and p' be the radii of curvature of the curves AD and AC at A, and let ^ and $' be the angles AOB and A'&B. Then drawing Bn perpendicular to A0\ we have AY- A'n + n7= An + Bn coiA'FB-, but / A'TB = \lr + i/r'j therefore the condition for stability is An + Bn cot (^ + f) > A' G', or, denoting A'G'(or AG) by k, Bn > (k - An) tan (^ + $'). (2) Now, carrying approximations as far as s 3 , it will be found from equation (1) of last Article that s being the common length of the arcs AB and AB. Substituting this, and the values of Bn and An from last Article, in (2), the condition for stability is Ip4-] EQUILIBRIUM OP A HEAVY BODY. 321 A + 2 t Neglecting all powers of s, the first condition for stability is , A 1\ P P f / If 7^ > 2__ the equilibrium will be unstable. P + P A special case occurs when k = - ? , and this is commonly called the ' neutral ' case, or the equilibrium is said to be neutral. We shall, however, call this the critical case. To find the real nature of the equilibrium in this case, we revert to the general condition (3), and neglect all powers of s beyond the first. The condition for stability now is Hence when k = , , the equilibrium will be stable or un- stable according as j \ d d is negative or positive. The bodies are, however, frequently in contact at vertices, or points of maximum or minimum curvature, and then 7 1 ,1 d- d~, and - 322 THE PRINCIPLE OF VIRTUAL WORK. [194. are both zero. Hence the condition (5) fails to determine the nature of equilibrium. Reverting to the condition (3), the terms as far as s 2 destroying each other on both sides, we see that equilibrium will be stable if or, substituting - -, for 7^, if . ~d? * ds'* ' />V 2 and in the contrary case the equilibrium will be unstable. If the lower surface is concave, instead of convex, to the upper, the conditions are obtained by changing the sign of p. Thus, the equilibrium will be stable or unstable, according as P P and in the critical case, the equilibrium will be stable or un- stable, according as d\ d- p _p ~ds' ds is negative or positive ; and in case of contact at vertices the condition (6) is to be similarly modified. If the body rest on a plane surface, p = oo, and the differential coefficients of are all zero. Hence the limiting value of h for P stability is p ; but if h = p', the equilibrium will be stable or unstable according as -p is positive or negative ; and if the point of contact is a vertex, equilibrium will be stable or un- stable, according as \ ds'* is negative or positive*. * Different methods of arriving at the conditions for stability have been published in the Quarterly Journal of Pure and Applied Mathematics by Professor Curtis (vol. ix, p. 41), and Mr. Routh (vol. xi, p. 102). The kinetical method of 1 94.] EXAMPLES. 323 EXAMPLES. 1. If a cone of the same substance and of equal base with a hemi- sphere be fixed to the latter, so that their bases coincide, find the greatest height of the cone in order that the equilibrium may be stable, when the hemisphere rests symmetrically on a horizontal plane. (Walton's Mechanical Problems, p. 185.) Ans. The height of the cone must be < r A/3, r being the radius of the hemisphere. 2. Prove that any body with a plane base, resting on a fixed rough spherical surface, will, when the height of its centre of gravity has the critical value, be in unstable equilibrium. 3. A heavy body whose section in the plane of displacement is a catenary, resting on a rough horizontal plane, has its centre of gravity at the critical height ; prove that the equilibrium is really stable. *4 (The condition (6) reduces in this case to ~ < for stability.) cts 4. A heavy body in the shape of a paraboloid of revolution, placed on a rough horizontal plane, has its centre of gravity at the critical height ; determine this height, and find the real nature of the equi- librium. Ans. The critical height = the radius of curvature of the gene- rating parabola at the vertex, and the equilibrium is really stable. 5. In the critical case, if both of the conditions (5) and (6) fail, prove that the equilibrium will be stable or unstable, according as d P is negative or positive, the surfaces being convex towards each other. 6. A uniform heavy bar, AB, moveable in a vertical plane round a fixed smooth axis passing through A has a string attached to the end B; this string passes over a fixed pulley C vertically over A. Find the positions of equilibrium, and determine whether they are stable or unstable. treatment adopted by the latter is very exhaustive. The method in the text was employed independently by Professor Wolstenholme and the author. It may be well to caution the student against the error of replacing the sections, AD and AC, of the surfaces in contact by their osculating circles at A. For, if we do this, the condition (5) necessarily disappears, and the application of (6) is not allowable, since, to the third power of the arc, the value of A'n is not the same for the circle of curvature as for the curve AC, as at once appears from the expression for A'n given by equation (3) of last Article. The nature of the equilibrium, therefore, as determined from the osculating circles is erroneous. Y 2' 324 THE PRINCIPLE OF VIRTUAL WORK. [195. Ans. Let W= weight of bar, 2 a its length, P = suspended weight, A = h, L CAB. Then the positions of equilibrium are given by the equations = 0, cos = ! + (^-|r 2 )^ and = *. The first will be stable if - > 5 and then the second (when it exists) will necessarily be unstable and the third stable. If the second does not exist, the third will be opposite in nature to the first. [To find the condition for stability in this problem, take any position of the bar and calculate the moment of force tending to turn it round A. If M = this moment, and $ = Z ACB, M=Phsm(j)- Wa sin 6. (I) Also h sin $ = 2 a sin (0 + ). (2) Now M = in a position of equilibrium ; and if -= is positive, a dQ slight increase of 6 will call into play a moment tending to restore equilibrium. In the position = 0, we have from (2) dc) 2a ~dO~ h-2a' Therefore, &c. Of course this might have been solved by Art. 191.] 7. If the equilibrium in the first position is critical, find its real nature. Ans. It is really unstable. In the position = 0, it will be found from (2) that -=- = 0, _ " ' _ 1 ~ "J dO s (h~2a) 3 8. Determine whether the equilibrium of the beam in example 12, p. 138, is stable or unstable. Ans. Unstable. [Either by taking the restoring moment about 0, or by the maximum or minimum value of the potential; the potential = P (a + b) cos BWa sin 0.] 195.] Definition of Work. If a force actually displaces its point of application in such a manner that the displacement has an orthogonal projection along the direction of the force, this force is said to perform work ; and if the force is constant during- the displacement, the product of the force and the projection of 195-] DEFINITION OF WORK. 325 the displacement along its direction is called the work done by the force. Thus, suppose that during the passage of a material particle along a curve ABD (fig. 218) it is continually acted on by a force, P, constant in both magnitude and direction. Then if dp denote the pro- jection of any elementary arc of the curve along the direction of P, the work done by P in this displacement is P. dp-, and the work done in this passage from A to B is f ' Pdp, or P x (the projection of AB along the direction of P\ since P is con- A ..,. p fe ' Fig. 218. stant during the motion. Suppose the point D to be such that AD is perpendicular to the direction of P. Then the whole work done by P on the particle during the motion from A to D is zero, whatever be the shape of the path pursued between A and D. _/ e/ J- J- When the forces acting on a particle are variable with the position occupied by it, we have to consider the elementary work done for a small displacement of the particle ; and to find the whole work done by the forces during the passage of the particle from one position to another, this elementary work must be integrated between the extreme positions considered. In the most useful application of the principle of work the forces acting on a given system are functions of the co-ordinates of their points of application, and do not depend on the velocities of these points ; and it is solely with forces of this description that we shall be concerned. It must be pointed out, however, that in considering the work which such forces are capable of doing on a particle or system of particles while this system is displaced from one position to another, all conceptions of time are here left out of consideration. The work which a given system of applied forces performs on a given material system during the passage of this system by any route from the position (A) to the position (B) in no way involves the time or the manner in which the passage is effected. The different particles of the system may have in one case moved more or less swiftly than in another from the first to the second position, and yet the work done by the forces (which are func- tions of co-ordinates only) is the same in both cases. 326 THE PRINCIPLE OF VIRTUAL WORK. [195. The theorems which have already been given for virtual work apply evidently to work actually done. Thus, as in p. 287, we see that for a small actual displace- ment of a particle occupying the position (#, y, z) the work done by the force acting on at is Xdx+Ydy + Zdz, the components of the force along the axes being X, Y, Z, and the components of the displacement being dx, dy, dz. The work per- formed on the particle in moving from one position to another is then f(Xdx + Tdy + Zdz), the force acting on the particle being a function of the co- ordinates of the particle, and the integration being performed from the values of the co-ordinates in the first position to their values in the second. If there are several particles in the system, each acted on by given forces, the work performed on the system will be the integration being performed for each particle from its first to its second position, and 2 denoting (as in Art. 188), a sum- mation of this integral for all the particles of the system. The case of most usual occurrence is that in which the forces belong to a conservative system (see Art. 188), or, in other words, when 2 ^ dx + Ydy + Z dz) is the perfect differential of a function, T 7 , of the co-ordinates of the particles acted on. In this case if dW denote the elementary work done on the system for a small displacement, we have and the work done in the passage of the system from one position to another is given by the equation r= r-r,, TO denoting the value of V in the first position. Since V is a function of co-ordinates only, the value of V F depends merely on the original and final positions of the material system, and not at all on the route by which the system has moved from the one to the other. I97-] ENERGY. POTENTIAL WORK. 327 196.] Unit of Work. Since, by definition, work is the product of a force and a line, the unit of work will be the product of a unit of force and a unit of length. If the unit of force is a kilogramme, and the unit of length a metre, the unit of work will be done when a force of one kilogramme drags its point of applica- tion through one metre along its line of action. Thus, if a body whose weight is a kilogramme is lifted vertically through a metre by a force which just overcomes its weight, this applied force does a unit of work, which is called a kilogramme-metre. In the same case the weight of the body does a negative unit of work (see Art. 54). [/ 197.] Energy. Potential Work. Energy means capacity for doing work. This capacity is possessed by a body in motion. For the velocity of the body might be made use of for causing the body to ascend vertically against the attraction of the earth, i.e., to do work against resistance. The exact measure of the amount of work which a particle weighing w grammes moving with a velocity of v centimetres per second can do against Q resistance before its motion is completely destroyed is t/ gramme-centimetres, where g is the velocity in centimetres per second acquired in one second by a body falling vertically in vacuo. Work is always done against some resistance. The work which is done by a force in moving a particle from one position to another is done against the inertia of the particle, or its resistance to acceleration. Thus work is the equivalent of energy, and energy is reconvertible into work at the rate indicated by the expression 2- If a system consists of any number of particles moving ** wv 2 in any directions, its total energy is 2 - - > the summation in- 2 y eluding all the particles, and the different directions of their velocities being of no account. This is an anticipation of elementary kinetics, and is here used only for the purpose of pointing out to the student what work done on a system is converted into. The Potential Work of a system of forces in any given con- figuration of their points of application is the amount of work which they are capable of doing in moving their points of application from any chosen confgnration to the given one. 328 THE PRINCIPLE OF VIRTUAL WORK. [198. This chosen configuration is quite arbitrary. Thus, the quantity of work which the forces applied to a system are capable of doing- during the passage of the system from one position to another is (Art. 196) V V " 'O' Fand P being the potentials, or certain functions of the co- ordinates, belonging to the two positions considered ; and the two positions may be taken as denned by the values of the func- tion V belonging to them respectively. The zero position of the system (that corresponding to F" ) is generally chosen in such a way that in any other position, practically considered, V shall be > F (Thompson and Tait, Nat. Phil.] ; or, in other words, the zero position of the system is such that the work done by the acting forces in moving it to any other position considered in our investigation shall be positive. 198.] Inclusion of Internal Forces. When any of the bodies of a system, acted on by given forces, are connected by elastic rods or strings, or when they mutually attract or repel each other, as has been already explained (Art. 97), these forces may or may not be brought into the equation of virtual work, ac- cording to the nature of the virtual displacement chosen. In finding the figure of equilibrium of such a system we have hitherto supposed it known, and determined the requisite con- ditions accordingly. We may, however, include in the potential work of the forces not only the potential work of the external (or applied) forces, but also that of the internal forces. Thus the total potential work of the system of forces will be the sum of the works of the applied and internal forces; and equation (a) of Art. 190 shows that in the position of equilibrium the variation of the total potential work of the forces of the system is zero. This principle will serve to determine the figure of equilibrium of the system without presupposing it. 199.] Criterion of Stability and Instability. Since in a position of absolutely stable equilibrium V is a real maximum, and in a position of absolutely unstable equilibrium V is a real minimum (Art. 190), it follows that in the former case the applied and internal forces would do negative work on the I99-] EXAMPLES. 329 system if its position were slightly altered ; and in the latter they would do positive work. A configuration of absolutely stable equilibrium is, then, such that the applied and internal forces cannot do positive work in any small displacement of the system ; and a configuration of absolutely unstable equilibrium is one in which every change of position involves the doing of positive work. And in general (see Art. 190) in a position of equilibrium these forces will do positive work for some displacements and negative for others. EXAMPLES. 1. Find the work done in drawing up a Venetian blind. Ans. Let n be the number of bars, a the interval between n + 1 them, and W the weight of the blind ; then the work is jp^ 2. A and B are two fixed points which are connected by any curve, APB; at each point, P, of this curve there acts a force, F, directed towards a fixed point, 0, the force being a function of the distance OP. If is the angle between OP and the tangent to the curve at P, and ds an element of the curve at P, prove that f F cos Ods taken from A to B is independent of the curve. 3. Prove that the work done in dragging a heavy body up a rough inclined plane, without acceleration, by a force parallel to the plane, is equal to the work done in dragging the body along the base of the plane (supposed equally rough), together with the work done in lifting it vertically through the height of the plane. 4. A heavy body is dragged, without acceleration, up a rough in- clined plane by a force whose line of action always passes through a fixed point ; prove that the work done in dragging the body through a given height, h, is 8 -\- W U> Wh (1 + fj. cot i) /ut Wp cos i (/x + tan i) log 5 where t is the inclination of plane, p the perpendicular from the fixed point on the plane, s the initial, and s' the final distance of the body from the foot of this perpendicular. MISCELLANEOUS EXAMPLES. 1. Two equal heavy spheres rest inside a hollow right cone, and against each other \ the cone (which has no base) rests on a horizontal plane, the vertex being uppermost ; only one sphere rests in contact with the ground. Find the least weight of the cone consistent with equilibrium. 330 THE PRINCIPLE OF VIRTUAL WORK. [199. Ans. Let a vertical plane through the centres of the spheres cut the cone in a triangle ABC, in which C is the vertex of the cone ; let LCAE LCEA = /3 ; let be the angle between the line joining the centres of the spheres and the side BC '; let r and c be the radii of the spheres and of the base of the cone, respectively, and W the weight of each sphere ; then the least weight of the cone is t. p Qt g_ co o ( c v r cos 2. A heavy triangular lamina, ABC, of uniform thickness and density, is suspended successively from the vertices A and B ; show that if any side in the second position is at right angles to its first position (The bisectors of the sides CA and CB drawn from B and A must be at right angles to each other.) 3. A heavy rod hangs from a fixed smooth pulley by means of a string attached to its extremities ; find the tension of the string. (See Example 30, p. 157.) Ans. If W is the weight of the rod, the tension 2c with the notation of the example referred to. 4. A heavy rectangular block is laid on the less steep of two smooth inclined planes which slope in the same direction and intersect in a horizontal line, an edge of the block coinciding with the line of inter- section of the planes. To the middle point of the upper edge is attached a cord which passes over a smooth pulley and sustains a weight ; determine the condition of equilibrium, and supposing that in any case equilibrium is about to be broken, find how this will happen. 5. A uniform board in the shape of an isosceles triangle rests on two smooth planes equally inclined to the horizon, the base of the triangle being horizontal, and the vertex upwards ; the board is cut into two equal portions by a plane passing through its vertex ; find the inclination of the planes if equilibrium continues to exist. Ans. If h is the length of the perpendicular from the vertex on the base, and c the length of the base, 6. A solid right cone rests with its base in contact with two smooth planes equally inclined to the horizon, the base being hori- zontal and the vertex upwards ; find the inclination of the planes such that if the cone is cut into two equal portions by a plane through the vertex, the equilibrium of the pieces will not be troubled. Ans. If h is the height and r the radius of the base of the cone, (,-I)r. CHAPTER XIII. EQUILIBRIUM OF FLEXIBLE STRINGS. 200.] Perfectly Flexible String. A string- is said to be perfectly flexible when at every point in its length it can be bent round all lines passing through the point perpendicularly to the tangent line without the expenditure of work. From this definition it follows that the internal force, or mutual action between the particles at each side of any normal section of such a string, has no component in the plane of the section ; this force must, therefore, be entirely normal to the section ; or, in other words, the internal force in a perfectly flexible string is at every point directed along the tangent line to the string. This internal force we have called the tension of the string, and, like all internal forces in a system, it is a mutual action between parts of the system. This has been sufficiently ex- plained already (p. 25). In the sequel we shall use the term flexible string as equivalent to perfectly flexible string. 201.] Imperfectly Flexible String. No effort is required to bend a perfectly flexible string at any point; but if we attempt to bend an imperfectly flexible string, or a wire, we encounter a certain amount of resistance according to the degree of inflexibility or rigidity of the string or wire. If we consider the nature of the mutual forces existing between the particles on each side of a normal section of such a body, we shall find that these forces are not necessarily reducible to a single re- sultant at all. In the general case of a wire bent and twisted by the action of any external forces, these internal actions on the particles at one side of a section may, of course, be reduced to a single resultant force and a single couple; and the re- sultant force may be applied at any point in the section, the couple varying according to the point chosen. All this is 332 EQUILIBRIUM OF FLEXIBLE STRINGS. [202. evident from the general reduction of a system of forces in Chapter X. 202.] Three Methods of Investigation. There are three methods by which the equilibrium of a string or wire may be treated namely, 1. We may isolate an infinitesimal element of the body, supplying to it at each extremity the action exercised by the neighbouring portions which are imagined to be removed (see p. 161). 2. We may apply the general condition that the variation of the whole potential work of the system of forces, internal as well as external, is zero (see p. 328). 3. We may consider the equilibrium of any finite portion of the body, treating it, when the figure of equilibrium has been assumed (see p. 13), as a rigid body. (See Thomson and Tait, Nat. Phil.} We begin by considering the equilibrium of a perfectly flexible string which suffers no elongation under the action of the forces which will keep it in equilibrium. Such a body is called a flexible inextensible string, and it is scarcely necessary to add that it exists only in the abstractions of Rational Statics. SECTION I. Flexible Inextensible Strings. 203.] Tangential and Normal Resolutions. Let A (fig. 219) represent a flexible inexten- sible string in equilibrium under the action of any system of forces applied continuously throughout the string. Then the force acting on a unit mass of matter placed at any point of the string will, in the Fdm general case, be expressed as a -,. function of the co-ordinates of this Jb ig. 210. point and their differential co- efficients with respect to the arc. Thus, if the co-ordinates of P 203.] TANGENTIAL AND NOKMAL RESOLUTIONS. 333 are x, y, z, the force acting on a unit mass placed at P will be dx On an element containing dm units of mass the force will be j f dx x (x,y,z,-j s >...)dm. We shall denote by F the coefficient of dm in this expression. Suppose, then, that we consider an element PQ of the string, whose length is ds, apart from the rest of the string ; let the mean density of the element be /, and let a be the area of its mean section; then the mass of PQ is Jcards, and the external force acting on it is kvFds. Now, the element PQ is kept in equilibrium by three forces namely, the tension (T) at P, the tension (T+dT) at Q, and the external force (kvFds), which acts at the middle point of PQ. These three forces must be coplanar and meet in a point. Now, the two tensions act along two consecutive tangents to the string, and as the plane of two consecutive tangents to any curve in space is the osculating plane, we see that The resultant applied force at any point of a flexible string acts in the osculating plane of the string at the point. If the string is stretched over any smooth surface by means of two forces applied at its extremities, the only applied force which is continuously distributed throughout the string is the reaction of the surface ; and as this reaction is everywhere normal to the surface, we see that A string which is stretched along any smooth surface, and acted on by no external forces, except the reaction of the surface and two terminal tensions, has its osculating plane at every point normal to the surface. The string in this case assumes the form of a shortest line, or geodesic, on the surface. Let Pt be the tangent and Pn the normal at P ; let dO be the angle between the tangents at P and Q ; and let < be the angle between Fdm and Pt. Then, resolving along Pt the forces acting on the element, we have (T+dT)cosd9 + & T >fcds is the pro- jection of ds on the direction of F. Denoting this projection b y &> T = C-fkvFdf. (3) But fkvFdf is evidently the potential of the applied forces if they are a conservative system*. Hence, if V and F denote the potentials at two points in the string at which the tension are I 7 and T , we have T=T Q (7- F ), (4) or the difference of the tensions at any tivo points is equal to the difference of the potentials a result which we shall find to be true also in the case in which the string rests on a smooth surface. 204.] Cartesian Equations of Equilibrium. Let the force F acting on the unit mass at any point P whose co-ordinates are #, y> z be resolved into three components, X, T t Z, parallel to three fixed rectangular axes. Then the components acting on the element PQ are JccrXds, kvYds, kvZds. Also the components of the tension acting on the extremity P are dot dy dz ~~ -^ 7 ' ~~ -* ~~r~ ' *~~ 7~ > ds ds ds the components of this tension are affected with negative signs, since, when the element PQ is considered apart, the tension at * A simple case in which the external forces are not a conservative system will be presently given. (See Art. 208.) 204-] CARTESIAN EQUATIONS OF EQUILIBRIUM. 335 P will be directed towards the left-hand side of fig. 219, where the origin of co-ordinates is supposed to be. These components of the tension will at any point be functions of the length of the arc measured from some fixed point, A, of the string up to the point considered. Thus, if AP = -j- j and -=- > respectively, adding, and remembering that 336 EQUILIBRIUM OF FLEXIBLE STRINGS. [204. dx dy dz d 2 z _ ~~^ ' we obtain dT 'ds dx ydy dz \ __ da ds ds ' or T = CfJc* (Kdx + 7^ + Z and the equations become The first equation shows that the horizontal component of the tension is the same at all points of the string (see p. 32). Denoting this component by r, we have Hence, from (2) =Ti ... ds dx 338 EQUILIBRIUM OF FLEXIBLE STRINGS. [306. or ^ , 9 ds fo (3) It is to be observed that ha- is the mass per unit length of the string at the point #, y. This last equation, therefore, determines the mass per unit of length at any point when the form of the curve in which the string hangs is given; and, conversely, it determines the curve in which any string will hang when the laws of variation of its section and density are given. Ify-be denoted by^, and the independent variable changed from os to y, equation (3) becomes dp r di/ ~9 206.] The Common Catenary. When the mass of a unit length of the string is everywhere constant, the form of the string is that of a curve called the Catenary. The name Cate- nary is sometimes employed to denote the form of a string in general, whatever be the law of variation of its density. In the present case k we obtain and by integration c,- --\ + e c ) where c" is an arbitrary constant. Now, taking the origin, 0, at a distance equal to c below A, we have y = c when x = 0. This gives c" = 0, and the equation of the catenary referred to axes chosen as above is The point of intersection of these particular axes we shall in the sequel call the origin of the catenary. We shall next find the length of the arc, AP, measured from A to any point, P, on the curve. If ds is the element of arc, ds= Vdx 2 4- dy* V* _ 2 1 + 1 ( e c _ e ^ . dxy from (l), L JL = \(e c + e c )dx\ no constant being added because s = when x = 0. z a (2) 340 EQUILIBRIUM OF FLEXIBLE STRINGS. [206. From (1) and (2) we have (3) and from (3) dy_ ~ y ds ' Let PM and PT be the ordinate and tangent at P, and let fall a perpendicular MT on PT. Then dy i \ y cos ~ $ d s 9 hence s = PT-, (5) and since/ = PT 2 +MT*, we have from (3) and (5) c = MT. (6) Hence, given the catenary to construct its origin and hori- zontal axis On the tangent at any point, P, measure off a length, PT, equal to the arc AP ; at T erect a perpendicular TM to the tangent meeting the ordinate of P in M; then the horizontal line through M is the axis of the curve. In making a proper figure this rule will be found of great use. The involute of the catenary -which starts from the lowest point is the Tractory. To get a point on this involute we measure on the tangent, PT, at any point, P, a length equal to the arc AP. From (5) we see, therefore, that T is a point on the involute ; and since PT is a normal to the involute, its tangent at T must be TM. But from (6) TM is constant ; hence the involute is a curve such that the length of the tangent between its point of contact and a fixed right line, Ox, is constant. The involute is, therefore, a tractory (see p. 195). The tension at any point of the catenary is equal to the weight of a portion of the string whose length is equal to the ordinate of Consider the equilibrium of the portion AP of the string apart from the rest. This portion is kept in equilibrium by three forces namely, the tension at P in the direction TP, the horizontal tension at A in the direction QA, and its weight acting through its centre of gravity, G. Hence the vertical through G must pass through Q. Resolving vertically, we have 20;.] THE CATENARY OF UNIFORM STRENGTH. 341 TcosTPM = = mgy, from (5). (7) COB. It follows from this that if a uniform inextensible string hangs freely over any two smooth pegs, the vertical portions which hang over the pegs must each terminate on the horizontal axis of the catenary. In the catenary the length of the radius of curvature at any point is equal to the length of the normal between that point and the horizontal axis. By equation (2) of Art. 203, we have ~ = mg sin TPM, P which by means of (7) gives p = Tfrpivr'* ^ u ^ ^is is evidently si n j-j^jjU. the length of the normal between P and the axis of x. It will be readily seen that the differential equation of the 70 catenary can be written in the form c 2 ^ = y, and that the area clots OAPM = twice the area of the triangle PTM. It is well to observe that if a weight is suspended from a given point of a catenary, the continuity of the curve ceases at that point, and the portions of the string at opposite sides of the point must be treated as branches of two distinct cate- naries. 207.] The Catenary of Uniform Strength, If the area of the normal section of the string at any point is made propor- tional to the tension at that point, the tendency to break will be the same at all points, and the curve is therefore called the Catenary of Uniform Strength. To find its equation, we have a- = \T, A. being a constant ; and since T = r -7- > we have dx ds . Then 1 p o> T W h ds W J> x Also, o- = = cot --- -j- = cot - sec - p p a ax 2p a a S _ But it is easy to prove that sec - = \ (e a + e a ) W - --\ 0) Hence a = ( e p +e M.cot > p ^ p which is the expression for the area of a section at a distance s along the chain from the middle point. The student will verify the homogeneity of this equation. 208.] The Parabola of Suspension Bridges. Suppose a string to be attached to two fixed points, and let each element of its length be acted on by a force in a constant direction, the magnitude of the force being proportional to the projection of the element on a line perpendicular to the direction of the force. Then it can be shown geometrically that the figure of the string is Fig. 221. that of a parabola. Let Oy (fig. 221) be the direction opposite to that of the force on each element; Ox a tangent to the curve, perpendicular to this direction; P and Q any two points on the string, the tangents at them being PI 7 and QT; PM and QN perpendiculars on Ox. Consider the separate equilibrium of the portion PQ. The forces acting on it are the tensions in the directions TP and TQ, and the resultant of the parallel forces on the elements of PQ. This resultant must pass through T, and it also passes through the middle point of MN> since its constituent forces are all proportional to the elements of the line MN. Hence drawing 344 EQUILIBRIUM OF FLEXIBLE STRINGS. [209. TV parallel to Oy, and meeting PQ in V, the point V must bisect the right line PQ. The curve of equilibrium of the string is therefore such that a right line drawn from the point of intersection of any two tangents parallel to a fixed direction bisects the chord joining their points of contact. This well known property identifies the curve with a parabola. If we make use of the equations of equilibrium in Art. 205, doc we shall have X = 0, J" = jut -7- > //, being the constant. There is cl/s no difficulty in arriving at the result just found. It is to be observed that the acting forces in this case are not a conservative system. Hence the function T (see Art. 203) does not exist. The connexion of this parabola with Suspension Bridges has been already explained in Chap. II. 209.] String Acted on by a Central Force. When the lines of action of the forces applied to the various elements of the string pass all through the same point, the force acting on the string is said to be central, and this point is called the centre of force. It is easy to prove that in this case the string must lie in a plane passing through the centre of force. For (Art. 203) the osculating plane at every point contains the centre of force ; and since two consecutive osculating planes have a tangent line to the string common, these two planes, having in addition a point (the centre of force) common, must be identical. Hence the osculating plane is the same at all points ; or the string must lie wholly in one plane. To find the form assumed by a string acted on ~by a given central force. Let (fig. 222) be the centre of force (supposed repulsive), PQ an element of the string whose equilibrium is considered apart, r the radius vector OP, the angle POA between OP and a fixed initial line, s the length of the arc AP, and p the perpendicular from on the tangent at P. Then, for the equilibrium of the element PQ, taking moments about 0, we have 309.] THE PARABOLA OF SUSPENSION BRIDGES. 345 moment of tension at P = moment of tension at Q ; or Tp = Tp + d(Tp}, .'. Tp = h, (1) where k is a constant *. Denote the tensions at P and Q by I 7 and T + dT respectively. Resolve the forces acting on PQ along the tangent at P, denote kcr by m, and let the central force be mFds. Then this force passes through the point of intersection of tangents at P and Q, and the cosine of the angle between its direction and the tangent at P is - J- e, where e is indefinitely small. In the ds equation of resolution the component of mFds is so that may be neglected, and we have dT= -mFdr. (2) Equations (1) and (2) determine the form of the curve. If the central force is attractive, the sign of F must be changed in (2), and the curve of equilibrium will be convex towards 0. It is usual in problems concerning central forces to denote r by-- Making this substitution, and eliminating T from the above equations, we have 7?* = But (Williamson's Differential Calculus, Chapter XII), 1 f du^ f =U+ (d^ ' Hence, denoting by < (&), and f (n) du by fa (u\ an arbitrary constant being implied in fa (u), we have from (3) 5) i +"-it*iW} i - w It is often more convenient to retain a differential equation of * Of course this proof holds whether the portion PQ is an element of length or a portion of any length, however great. 346 EQUILIBRIUM OF FLEXIBLE STRINGS. [209. the second order for w*. Differentiating (4) we have, dividing du out by -j^ i and remembering that fa' '(u) = < (#), 1 , / \ / \ /e\ -it = -77- d)j ( u) . a) ( wj. ^oj (7" # Now, since the integration of (4) gives u in terms of 0, and introduces an arbitrary constant in addition to that already involved in fa (u\ we see that the solution of the problem involves only two arbitrary constants. But (5) will require two integrations to express u, and each integration will intro- duce an arbitrary constant. Hence it appears that in this way we get three arbitrary constants, instead of two. These three 7 2 are, however, easily connected, since the values of u and (-j^\ given by the complete integral of (5) must satisfy (4) for all values of u. As an example, let it be required to discuss the form of a string of uniform section and density when the central repulsive force varies inversely as the square of the distance. In this case m is constant, and F = fjfu 2 , // being a constant which obviously denotes the magni- tude of the force on a unit mass of matter placed at the unit distance from the centre of force. Hence we have, putting m^jf = jut, T = C being a constant. If T Q denote the tension at a point A of the string whose distance from the centre is ? we have, evidently, = p. (u + c), suppose. Hence, '($.*#'*(+#, (6) which gives, by differentiation, M u First suppose that -&<'li an d denote 1^- by X 2 . Then this equation becomes fl* u 1 A 2 * This method of treating the equilibrium of a string acted on by a central force is taken from a paper by Professor Townsend in the Quarterly Journal of Pure and Applied Mathematics, 1874. 209.] STRING ACTED ON BY A CENTRAL FORCE. 347 the integral of which is u= c + A cos A (0 a), A and a being the constants of integration. Substituting this value (8) of u in (6), we have A = - c, and therefore The value of a is found by putting u = a and = the angle belonging to the point A. When = a, = 0, and there is an apse. If the initial line be taken through the apse, and T Q and a belong to this point, we have m T c = - -a = ( -- 1) a, and (8) assumes the simple form /A JJ. u=-?+cos\6), (9) 1 + I which differs from the focal polar equation of a conic in having the angle multiplied by a number, A, less than unity. u 2 V? If > 1, we must put j^ 1 = A 2 , and putting pa T Q = pc, equation (6) becomes du* 2 _ f_ , 2 W H " A 2 " ^~~ c ' ' which gives it = ~~ c + Ae + Be ~ XQ , (10) the constants A and jB being connected by the equation AB = Equation (10) can obviously be written or When d = a, there is an apse, and if the initial line be taken through the apse, we have, in the same manner as before, If T = 1, both (9) and (11) give w = a, a constant; and the figure of equilibrium is a circle. 348 EQUILIBRIUM OF FLEXIBLE STRINGS. [310. For the remarkable analogy between the curve of equilibrium of a flexible string and the orbit of a particle under a given force, see Professor Townsend's paper, and Thomson and Tait's Nat. Phil. 210.] Problem. To find the angle between the apsides in a string which, under the action of a central force, assumes a form nearly circular. DEP. An apse is a. point on a curve at which the radius vector is at right angles to the tangent. Since the form of the string is nearly circular, u will differ from a constant value, a, by a small variable quantity, x. Let, then, u = a + x. In this case fa (u) = fa (a) + xcj) (a), neglecting higher powers of x ; and (f> (u) = (f> (a) + x(j/(a). For shortness, denote fa (a), (/> (a), and '(a) by fa, , and <' respect- ively. Then (5) of last Art. becomes ^++* = -^{' +>} (i) But if the string were exactly circular, x and -y- 2 would always = ; therefore a = ^-~ > or Hence (1) becomes The constant a may be chosen as the reciprocal of the radius of any circle which nearly coincides with the figure of the string ; but simplicity is gained by taking it equal to the reciprocal of the radius of that circle in which the tension at each point is equal to the mean tension in the string. Now in a circle of radius - the tension (see (2), Art. 203) is <; and (2) of last Art. gives T in the curve equal to fa (u), and there- fore the mean tension = fa. Hence a = 0. (1) Resolving normally, --&(rFsm(l>-R= 0. (2) P These are the most useful resolutions in the case of a string resting on a curve. Equations of resolution along arbitrary axes may, of course, be obtained by introducing the components of R into the general equations of Art. 204. From (1) we obtain T = Cf&o-Fcos $ds, C being a constant. But Fcostyds is obviously the virtual work of the force F. Hence if the acting forces are conservative, and V is their potential at P, we have, as in Art. 203, * This investigation is taken from the paper by Professor Townsend previously referred to. f The student will observe that in considering the equilibrium of an element of length ds we represent the reaction of a curve on it by lids, and the applied force by JcffFds, while we represent the tension by T, and not by Tds. The reason of this is that the tension depends merely on the cross section of the ele- ment and not on its length, while the magnitude of the reaction depends evidently on the length of the element in contact with the curve. 350 EQUILIBKIUM OF FLEXIBLE STRINGS. [213. 212.] String on Bough Plane Curve. If the curve in the preceding Article is rough, and the string in limiting equi- librium, slipping being about to take place in the direction QP, we have merely to include among the forces acting on the element PQ a tangential force pRds, the coefficient of friction being \L and the normal reaction It ds, as before. Equations (l) and (2) of last Article now become dT = 0, T -- ^o-jPsin dR = 0. P 213.] String on a Smooth Surface. When a string acted on by two terminal forces only is stretched over a smooth surface, we have seen that it assumes the form of a geodesic on the surface, and that the tension is constant throughout its length. The general Cartesian equations of equilibrium are readily obtained by adding to the components of the given applied forces the components of the reaction of the surface. Let R be the magnitude of the normal reaction per unit of length of the string. Then, the direction angles of the normal to the surface at the element ds being A, \L, v, the components of the reaction on this element parallel to the axes are R cos \ds, R cos ^ds, R cos vds ; and (1), (2), (3) of Art. 204 If we multiply these by cos X, cos jm, and cos v respectively, and add, we have z N cos A + ^-f cos p, + -^ cos v) + k ? i and > re- ds ds ds spectively, and add, we obtain where S = X-j- + Y-j-+Z-j- = the component of the applied force along the tangent to the string. The integral of this equation, when the applied forces are conservative, gives, as in Art. 211, In the particular case in which a uniform inextensible string rests on a smooth surface under the influence of gravity, this equation gives T- T mg (yy Q ), mg being the weight of a unit length of the string, and the axis of y a vertical line. From this it follows that at all points of the string which are in the same horizontal plane the tension of the string is the same ; hence the free extremities lie in the same horizontal plane. The curve of equilibrium of the string on the surface is obtained by eliminating T and E from equations (l). If the equation of the surface is u = 0, the result of eliminating dx du 5 ds dx dT A-P- -^- and K is -, - ds dy dz du = 0, in which the value of T must be substituted from (3). The general results arrived at in Art. 203 can be easily verified here. 352 EQUILIBRIUM OF FLEXIBLE STRINGS. [314. 214.] String on a Rough Surface. If a string, acted on by no forces, is stretched over a rough surface it need not, as in the case of a smooth surface, assume the form of a geodesic or shortest line. One simple case in which it will be a geodesic is that in which it is about to slip on the surface at every point in the direction of the tangent to the string at this point. Consider the equilibrium of an element, PQ, of the string, whose length is ds> and suppose that it is about to slip in the direction QP. The element is acted upon by three forces namely, a tension T, at P, a tension T+dT, at Q, and the total resistance of the rough surface, which must pass through the intersection IT, Rds Fig. 223. of the tangents at P and Q. It is evident that we may consider this total resistance as acting at P, ultimately, since it is of the form R^ds, R l being a finite quantity, and if it be assumed to act at " any point between P and Q, its components in any directions will differ from those of the total resistance supposed to act at P by in- finitesimals of the order of (ds) 2 . Resolve the total resistance at P into a normal force, Rds, and a force in the tangent plane, pRds, fj, being the coefficient of friction between the string and the surface. Now the component pRds must act along the tangent at P, since (see p. 57) slipping is about to take place along this tangent. Hence the three forces T, T+dT, and y.Rds being all in the osculating plane of the curve at P, the remaining force, Rds, must also lie in this plane; that is, the osculating plane at every point of the curve contains the normal to the surface. Hence the string assumes the form of a geodesic. Denoting the angle between the tangents at P and Q by dO, we have, by resolving along the tangent at P, dT+fj.Rds=0. (1) Again, resolving along the normal at P, TdO-Rds = 0. (2) From (1) and (2) we have 2I4-] STRING ON A ROUGH SURFACE. 353 rJT ^ + ^0 = 0, /. T=Ce~>* 9 C being the constant of integration, and the sum of the angles of contingence, or angles between successive tangents to the string from any chosen point, A, to the point, P. Let T^ be the tension at A. Then T = T when = ; therefore T=T eri*. (3) Hence, as the angle through which the string turns increases in arithmetical, the tension diminishes in geometrical, progression. The general investigation of the equilibrium of a string on a rough surface under the action of given forces is a problem of much difficulty, and in the sequel we shall confine our attention to the case in which the string assumes the form of a plane curve on the surface. When the string lies in one plane, 0, the sum of the angles of contingence is simply the angle between the tangents at A and P. Suppose that (the weight of the string being neglected) two weights, P and Q, are suspended from the extremities of a string which passes over a fixed rough cylinder whose axis is hori- zontal, the string lying in a plane perpendicular to this axis ; it is required to find the relation between P and Q when the equilibrium is limiting. Let A (fig. 223) be the point at which the portion of the string next P leaves the cylinder, and B the point at which the portion next Q leaves it. Then from (3) by putting T = P and = TT, we have Q=Pe~^, (4) when P is about to overcome Q. If P is on the point of ascending, the sign of //, in this equation is to be changed. If the string makes a complete revolution and a half round the cylinder, the value of 6 corresponding to Q is 37i, and we have in this case Q =Pe"^ ir . The factor e~f* e diminishes very rapidly as the angle increases, and thus we see how it is that a small force applied at one extremity of a rope coiled several times round a fixed rough cylinder can overcome a large force applied at the other extremity a practical example of which occurs when the small motion of a ship in harbour is stopped by a small force applied at the extremity of a rope coiled round a p fixed post. For example, if /* = i, * air = 4 . 8, and Q = A a 354 EQUILIBRIUM OF FLEXIBLE STRINGS. 215.] Work done against Friction for a given Arc of Slipping. If the string slips through a space bs in the direction BA, the work done against the friction, pRds, acting on any element is bs . ^Rds, and the work done against the friction acting all over the string is bs .f^Rds. But from (1) of last Article, fyRds = -fdT = T -T lt Hence the work done against the friction is 1 being the sum of the angles of contingence between A and B, or the angle between the tangents at these points if the curve of the string is a plane curve. EXAMPLES. 1. A uniform chain of length I hangs over two fixed points, which are in a horizontal line ; from its middle point is suspended by one end another chain of equal thickness and length I'. Supposing each of the two tangents of the former chain at its middle point to make an angle with the vertical, to find the distance between the two fixed points, and to show that can never exceed a certain value. ("Walton's Mechanical Problems, p. 123.) Let the fixed points be P and Q (fig. 224), EQCPM the string hanging over them, CD the string of length I' suspended from 6', the middle point of the first string, and 2d the distance PQ. Then (Art. 206) the arcs PC and QC belong to the distinct cate- naries. Suppose the semi-catenary to which PC belongs to be com- pleted, and let A be its lowest point. Then if the portion A C were supplied to the string CPM, and the point A fixed, the string CD and the portion CQR might both be removed, and we should have the string APM hanging in equilibrium. Hence (Con., Art. 206) PM terminates on the horizontal axis of this catenary. The same remarks apply to the portion CQR, and since the two portions CPM and CQR are exactly similar, it follows RM is the *\Q p y horizontal axis of the catenary AP. We shall next prove that Let T be the common tension of the portions CP and CQ at C. Then re- solving vertically for the equilibrium of the point C, But Tmg. CN (Art. 206), N being the point in which CD meets the axis. Hence 2 CN cos = l'\ but it is evident from figure 220 that CN cos 6 = AC; therefore A C = J I'. 215.] EXAMPLES. 355 Again, e being the parameter of the catenary, we have c = AG tan ; therefore c _. j f tan 0m ( y Also, denoting $.# by x, being the origin of the catenary, we have x x or - = -tane?' e 2 4 ?eow a* t , .'. 2 cot = e v 4 i' Squaring both sides of this equation, adding 4 to each side, and taking the square root, we have which, by addition to the last equation, gives easily I' x = - tan B log cot - (2) J 4 c x+d x+d Again, AP = -(e e c ), , PM" C f ~ \ 2 therefore by addition we have, since CP + PM = \l, l + l f _ 2 Substituting in this equation the values of c and x given by (1) and (2) ; and taking logarithms, we have r i if j. i /i >' (3) which is the required distance between P and Q. Since d cannot be negative, the expression whose logarithm is taken in (3) must be > 1 . Hence (I + T) tan \Q>1' tan 6 ; and substi- tuting for tan0 in terms of tan \0, we find the limiting value of Q given by the equation Q I I' 2. A uniform chain hangs over two smooth pegs in the same hori- zontal line, and at a given distance apart; find the length of the chain when the pressure on each peg is a minimum. Let P and Q be the pegs, 2 a the distance between them, 21 the length of the chain, the angle which the tangent to the chain at P makes with the vertical, PM the portion which hangs over the peg P, and C the lowest point of the chain. A a 2, 356 EQUILIBRIUM OF FLEXIBLE STRINGS. Then CP+PM = ee 7 (by adding the values of CP and PM), or , ':'. | = cei . - . (1) an equation which determines Z in terms of c. Again, CP = c cot 0, and PM = c cosec 6, therefore by addition ft a tan - = "" (2) Now, the pressure on the peg P is the resultant of two equal tensions, one along PM and the other along the tangent to the chain at P. Hence, if R denote the pressure, and T the tension at P, $ 2* fL _fL Substituting for T the value \rngc (e c +e c )> and for cos- its value obtained from (2), we have 2a 1 7 IT) Now, c must be determined so that R is least ; hence -3 = 0, and we obtain easily c for the determination of c in terms of a ; Z is then known from (1). 3. A uniform inextensible string, acted on by gravity and by two terminal tensions, rests in contact with a smooth curve in a vertical plane j find the form of this curve so that the pressure which it exerts on the string may at every point be inversely proportional to the radius of curvature. Let vertical and horizontal lines in the plane of the curve be taken as axes of y and x, respectively, and let the concavity of the curve be upwards. Then R being the pressure on a unit of length at any point, and T the tension at this point, we have, by resolving along the tangent, dT=mgdy, mg being the weight of a unit of length of the string. Hence T=T. + mg(y-y Q \ (1) T and y Q belonging to one end of the string. Again, resolving normally, (dO being the angle between two consecutive tangents), or -, = A (2) 215.] EXAMPLES. 357 k Let R = -ik being a constant. Then from (1) and (2) p ds y\ dx = Ts> < 3 > denoting the numerator of the left-hand side of the previous equation by >9 (y~^\ f r simplicity. To integrate (3), put dx 1 ' (1+p 2 ) , dy = = 5 and p = 5 - =$-- where = - ds */l+p* }; for, identifying the two expressions, we have Hence we have where e^ a = 6/x. This is, of course, the equation of a common catenary whose para- meter is > and whose origin is the point (A, a). 4. A uniform inextensible string, acted on by two terminal tensions, and any system of conservative forces in one plane, rests in .contact with a smooth curve in this plane; if at every point the 358 EQUILIBRIUM OF FLEXIBLE STRINGS. [215. pressure against the curve is inversely proportional to the radius of curvature, then, without any change in the forces, the tension at one extremity can be so varied that the constraining curve may be re- moved, and the string will rest in free equilibrium. For, if V denote the potential of the applied forces at any point, we have (Art. 211) T = T - ( F F ), (1) Again, if N denote the normal component of the applied forces at any point measured towards the convex side of the curve, and JS the pressure per unit of length at this point, (2) k Suppose that E = Then, from (1) and (2) we have P Let us now change the terminal tension T into T Q Jc, and in- vestigate the pressure of the curve at the point considered above. Denoting the new pressure by R', and the new tension by T f t there being no change in any of the applied forces, we have T' T 1? ( V V \ from which R = * (y - ^ - N ; P but the right-hand side of this equation is zero by (3). Hence there is no pressure at any point, and the curve is one of free equilibrium. It is obvious that the last example is a particular case of this. 5, Find the law of variation of the mass per unit of length at each point of a string acted on by gravity in order that it may hang in the form of a semicircle whose diameter is horizontal. Let AB(=2a) be the horizontal diameter, the centre of the semicircle, P any point on the curve, and the LAOP = 0. Then taking horizontal and vertical lines through as axes of x and y, respectively, we have . n dy dO 1 dx y x = 00000, y sm0, -/- = cot0, - = -- , = sm0 = - dx dx y ds a dfy_ 1 de _ a* -~"-'~' Also, denoting KCT in equation (3) of Art. 205 by w, we have T a 9 y which proves that the mass per unit length at any point varies inversely as the square of the depth of the point below the horizontal EXAMPLES. 359 6. A heavy chain of variable density, suspended from two fixed points, hangs in the form of a curve whose intrinsic equation is s =f(0), the lowest point being origin ; prove that the density at any point will vary inversely as cos 2 6 ./' (0). (Wolstenholme's Book of Mathematical Problems.} We have here \MU - \MtU .. i tl/0 // / /1\ r 1 = tan 0, -=- = cos 0, and - - = f (0). a# as c/0 d 2 ^ 1 c?0 1 c?0 c?s 1 Jience ~~ dx* cos 2 dx cos 2 <& ^x cos 3 6f (0) ' and equation (3) of Art. 205 gives m = 7. A string is kept in equilibrium in the form of a closed curve by the action of a repulsive force tending from a fixed point, and the density at each point is proportional to the tension ; prove that the repulsive force at any point is inversely proportional to the chord of curvature through the centre of force. (Wolstenholme, ibid.) The equations are (Art. 209), Tp = h, (1) dT = -mFdr, (2) Now, m == kv, and by hypothesis k or T, and a- is constant ; there- fore we have m = pT, ^ being a constant. Hence from (2) .. (3) But from (1), dT = -- -dp, therefore = -- -> and we have from (3) _ 1 dp_ 2 W -t 7 * _ J p dr y where y is the chord of curvature passing through the pole (see Williamson's Diff. Cat, p. 293, third ed.). As a particular case, we may notice that the vertical chord of curvature at any point of the catenary of uniform strength (under gravity) is constant, as the student can easily prove otherwise. 8. A heavy inextensible string rests, in limiting equilibrium, on a rough curve in a vertical plane ; find the tension at any point. Let fig. 223 represent the string lying on the curve ; let a hori- zontal line above the curve AB be the axis of x, and let the axis of y be drawn vertically downwards. Then, if 6 be the angle made by the tangent at any point, P, with the axis of x, mg the weight of a unit length of the string at P, and x, y the co-ordinates of P, we get by a tangential resolution (slipping being on the point of taking place from P to Q), and by a normal resolution TdQ Rds + mgdx = 0. 360 EQUILIBRIUM OF FLEXIBLE STRINGS. [215. Eliminating R, we obtain AT . dx d. = mg (fj, cos sin 0) p, (l) where p is the radius of curvature at P. This is a linear differential equation of the first order, the solution of which is (Boole's Differential Equations, p. 39), T = e* e {C+fmgp (/* cos 0-sin 0)e~* e d0}, (2) C being a constant. When the curve of constraint is given, p is known in terms of 0, and the integration may then be performed. For example, let the string rest on a circle of radius a, one ex- tremity being at the highest point, and free from tension. It will be easily found that f(\JL cos 0-sin 0) e-^dB = -- (2jx sin + (1 -ft 2 ) cos 0}, therefore T= <7^ + L{ 2/ / S in0 + (I-fx 2 )cos 0}. 1 jn 2 At the highest point = and T= ; therefore C = mga 1+ a * Hence T= If the length of the string is that of a quadrant, we have T = edi 2ft when =- j and then /u is determined from the equation 9. A, B, C are three unequally rough pegs in a vertical plane ; P is the greatest weight that can be supported by a weight W when both are connected by a string (whose weight is neglected) passing over A, B, and C; Q is the greatest weight that W can support when the string passes over A and B ; and B is the greatest that W can support when the string passes over B and C. Find the coefficients of friction for the pegs. Let the inclinations of AB and EG to the vertical (measured in the same sense) be a and /3, respectively; /z, JJL', //' the coefficients of friction of A, B, G. Then, if the string passes over all the pulleys, and W hangs from A, it follows from equation (3) of Art. 214, that the tension, T, in the portion A B is We* a ; and, by the same equation, the tension, T, in EG is TeM-4 ; and, finally, P = TW), 362 EQUILIBRIUM OF FLEXIBLE STRINGS. [215. T = Pe(- ), Wa sin B 2 Tc sin 0. 15. Prove that the area of the normal section at any point in the catenary of uniform strength is proportional to the radius of curvature. 16. Find the law of variation of the mass per unit of length in order that a string may hang, under the action of gravity, in a parabola. Ans. The mass at any point is proportional to the horizontal projection of the unit length at the point. (Compare Art. 208.) 17. If a string hangs, under the action of gravity, in the form of an ellipse whose axis major is horizontal, prove that the mass per unit of r b 3 length at any point is - , , 2 > y being the distance of the point from the axis major, and If the length of the semi-conjugate diameter cor- responding to the point. 18. One extremity of a uniform string is attached to a fixed point, and the string rests partly on a smooth inclined plane ; prove that the horizontal axis of the catenary determined by the portion which is not in contact with the plane is the horizontal line drawn through the extremity which rests on the plane. 19. If, in the last example, i is the inclination of the plane, a the inclination of the tangent at the fixed extremity, and I the whole length of the string, prove that the length of the portion on the plane is Zcosa (Walton, p. 1 1 9.) cos * cos ( a - *) ' 20. Given two smooth pegs in a horizontal line, find the least length of a uniform heavy string which will rest over them. Ans. If 2 a is the distance between the pegs, and e the Napierian base, the least length is ae. 21. A uniform inextensible string assumes the form of a circle under the influenca of a repulsive force emanating from a point on its circumference ; find the law of force. Ans. It varies inversely as the cube of the distance, 22. A uniform inextensible string is in equilibrium under the action of a central repulsive force ; prove that at each point of the string this force oc where p is the perpendicular from the centre of force on the tangent, and y the chord of curvature passing through the centre of force. 23. If the curve of equilibrium is an ellipse whose focus is the centre of force, the force at any point oc -T? where >' is the semi- conjugate diameter corresponding to the point, and r the focal distance of the point; 215.] EXAMPLES. 363 24. If the string assume the form of an ellipse under the influence of a repulsive force emanating from the centre, find the law of force. Ans. The force is directly proportional to the distance, and inversely proportional to the conjugate diameter. 25. If an inextensible string can assume the same plane figure of equilibrium under the separate action of any number of forces, it can assume this figure under their combined action. (To prove this, suppose the string under the combined action of the forces to be constrained to a smooth curve of the given figure, and it will follow that the pressure at every point of this curve varies inversely as the radius of curvature. The theorem follows, then, from example 4.) 26. A uniform inextensible string rests against the inner side of a smooth elliptic wire, and is repelled from the foci and the centre by the following forces : ~r, and -777 emanating from the foci, and -77- from the centre, the distances of a point on the string from the foci being r and /, respectively, its distance from the centre being of, and the semi-conjugate diameter corresponding to the point being &'. Find the pressure on the wire at any point. Ans. If T Q is the tension of the string at the extremity of the minor axis, R = pressure per unit length = - (The student will easily see from examples 4 and 25, that if the curve of constraint of a string is a possible curve of free equilibrium under the action of the given forces, the pressure will, at every point, be C > where G is a constant. The result, in this example might, there- fore, be at once obtained by this principle. By direct calculation, however, the result is obtained with little trouble. The equations of equilibrium are ro and the first gives, by integration, --f*"6' = const.). r The student will do well to apply the principle explained here to he kinetical examples in Walton, pp. 295, and 259 second edition. 364 EQUILIBRIUM OF FLEXIBLE STRINGS. [316. SECTION II. Flexible Extensible Strings. 216.] Experimental Law of Extension. The strings which we now proceed to consider are extensible, i. e. such as have their lengths increased when they are in a state of tension. For such strings we shall still assume the property of complete flexibility as defined in Art. 200. The law of extension which we proceed to enunciate applies not only to flexible strings but also to straight bars of iron, steel, &c. Let 1 denote the length of any string or straight bar of uniform section when it is not subject to the action of any external force. This is called the natural length of the string or bar. Let a be the area of the normal section, F the magnitude of the force applied at one extremity in the direction AB, of the string or bar. Then supposing the extremity A to be fixed, the force F will produce an extension, BC, of the body. Denote this extension by x. Then ex- perience proves thatybr small values of the ratio j- in the case of solid bars there is for the same bar a constant ratio between this fraction and the quantity ; and there is the same constancy of ratio in the case of F strings, but for some of these latter bodies the value ofj lg ' 225 ' may be very much greater than for bars. We have, then, F - x , . =jfc 7" J E being a constant quantity which is called the modulus of elasticity of the matter of which the string or bar is formed. Since j- is a number, it follows that E is a force per unit of section. This force is also known as Young' *s modulus, and it is evidently a measure of the longitudinal rigidity of the substance. If the law expressed by equation (l) be supposed to hold for an extension x equal to 1 , and if the force applied to the body 2 1 6.] EXPERIMENTAL LAW OF EXTENSION. 365 p to produce this extension be called P, we have E = ; and if or is a section of unit area, U = P. The modulus of elasticity of any substance might then be defined as that force which, if applied at the extremity of a bar of the material of unit section, would double its length this force being fictitious in the case of bars or strings for which (1) holds only within extremely narrow limits. For bars of iron and steel this equation is true only within narrow limits called the limits of elasticity while for flexible strings of such substances as India-rubber its range is much wider. If the limiting amount of extension has not been surpassed, the body will, after a time varying with the sub- stance, retui :i to its original state when the stretching force F is removed. The law expressed by equation (l) is also true within narrow limits in the case of a straight bar which is compressed without bending. An idea of the magnitude of the modulus of elasticity of a solid body may be formed from the fact that in the case of iron, the unit of force being a kilogramme and the unit of area a square centimetre, E is about 2,000,000. For what are commonly called elastic strings, E is of course very much smaller than for bars of iron or steel. In the case of an elastic string it is usual to put equation (1) into another form. If I is the length which the string assumes under a tension T, we have x = I / , and or, as it is usually written, + -, (2) the quantity A being called the modulus of elasticity of the string. This quantity is obviously the force which must be applied to the string to double its length. The law expressed by (l) or (2) is known as Hooke's Law, from the name of its discoverer, and is sometimes expressed in the 366 EQUILIBRIUM OF FLEXIBLE STRINGS. [217. form the tension of any elastic string is proportional to its exten- sion beyond its natural length. 217.] Work done in slowly extending a String or Bar. If at each instant during the extension of a string or bar the stretching force applied at the extremity is exactly equal to that which would keep the body in its state of deformation at this instant, there is continuous equilibrium between the (gradually increasing) applied force and the elastic force of the body, and therefore the total amount of work done by the applied force is equal to the work done against the internal force. [The more advanced student will see that this would not be true if the extension were suddenly produced, so that oscillations would take place in the body.] Now if x is the extension of the body at any instant, the cor- fio- responding force is -j- x, and the work done against this force in . E(T a further extension dx is -j- xdx. Let a be the final extension ; 4) then the total work done is / Jc\ or the extension being, of course, confined within the limits of elasticity. Now the applied force which is required to keep the body in its final state of extension is, by (l) of last Article, j Hence if the force applied in the final state be denoted ^o by P, the whole amount of work done is or half the work which would be done by the final force of extension in moving its point of application through a space equal to the final extension. 218.] Equations of Equilibrium of an Extensible String. Suppose the string to have assumed its figure of equilibrium under the action of given forces. At any point in the string let ds be the stretched length of an element whose length before the action of the forces was ds Q ; and at this point let m be the mass of a unit length of density equal to that at the point, the mass per unit length at the same point in the natural state of the string being m Q . EQUILIBRIUM OF AN EXTENSIBLE STRING. 367 Then since the quantity of matter in the element is unaltered by stretching, mds = m Q ds . (l) Also by Hooke's law, + ) dr (2) But, the string having assumed its form of equilibrium, we have, as in the inextensible string, (a) Also ds = -/ (5) where s is the length of the arc of the original string measured from some fixed point up to the element ds Q . Now the general problem of extensible strings may be stated as follows : an extensible string, the law of variation of whose density in its natural state is given, is, under given circumstances, submitted to the action of given forces ; find the form which it will assume. To solve this problem it is necessary to find two equations between a?, y, z, the co-ordinates of any point in the stretched string ; and as the equations just given contain, in addition to these co-ordinates, the quantities m, m QJ s, s , and T, these latter must be eliminated. But from the seven equations above, these five quantities may theoretically be eliminated, by differentiation or otherwise, and there will result two independent equations, which are the equations necessary for the determination of the curve of equilibrium. The problem in its general form is one of great difficulty, and one which it would be practically impossible to solve. We shall, therefore, in the sequel confine our attention to the case in which 368 EQUILIBRIUM OF FLEXIBLE STRINGS. [218. the string in its natural state is such that m , the mass per unit length, is constant at all points, and to the case in which the acting forces are constant. Let us first consider m constant. By multiplying the equations (3) by ^-> -y-> and -j- re- spectively, and adding, we have + s) = 05 (6) ds ds j - 2 and from (1) and (2) we have m = - 2. . Hence (6) becomes T (l + ) dT+m (Xfa+ Ydy + Zdz) = 0. (7) Hence by integration, \ yr 2 /* - ( 1 + - ) + m I (Xdx + I dy 4 Zdz) const. 2 ^ A. ' J Denote the integral in this equation by 7, the potential of the acting forces, and let the constant of integration be A. Then we have o,b y{2 ), ^=^0, from which the relation between s and s is found, and hence the extension of the string. Equation (8) is the analogue of (3) of Art. 230. If V is the potential at a point of the string at which the tension is J", this equation gives ' +^) = r - r - ( 10 ) The equations of the curve of equilibrium are obtained by sub- stituting the value of T given by (8) in any two of the equations 2l8.] EQUILIBRIUM OF AN EXTENSIBLE STRING. 369 which are deduced from the equations (3) by substituting for m in terms of m Q . Secondly, suppose that the applied forces, X, J", Z, are constant. Then the first of equations (3) gives T dx _ ^_jy W( ^ 05 ( H ) A being the constant of integration. The remaining two of these equations give T^r = B-Yfm Q ds Q , T ~ = C-Zfm Q ds . (12) as ds Hence, by squaring and adding, /772 ( i ~\T f,,,^ 3 a \2 i / 7? V /'/vn il ' \2 i //"* V r/wt /Jo ^2 /I Q\ j. :=: ( // ^i. / winds i -7- ( D ^ J. / Wlc\ CvSn I T I v/ ^ V "2n O/ V / This equation gives ^ the tension at any point in the stretched string, in terms of the length of the arc of the un- stretched string corresponding to this point ; or, in other words, r =<#,(*). (14) Hence, from (2) we have which gives the relation between the stretched and unstretched lengths of any arc. The equations of the curve are obtained from (11) and (12) by substituting for ds in terms of ds Q . Thus we have Integrating these equations and eliminating s between them in pairs, we obtain the two equations of the curve. As an example, let it be proposed to investigate the form of an elastic string suspended from two fixed points and acted on by gravity, the string being uniform in its natural state. Taking axes as in Art. 206, we have Bb 370 EQUILIBKIUM OF FLEXIBLE STRINGS. [218. UiS wo rf &y Hence T = r = m gc, suppose; and T -^ = B + m Q gs . But if s be measured from the lowest point, ^ = and s = at the same time. Hence 5 = 0, and we have dx from which T = Hence, putting X = w <7a, we have CS , . v (15) The relation between x and y is obtained by eliminating s from these equations. An approximate relation between them may be obtained when the string is only slightly extensible, i. e. when A. (or a) is very great. In this case (16) gives V = fe/'- 2 )l- + (17) to the second order of the small quantity Now, writing (15) and (16) in the forms we know that 77 = ~(e c +e c ). s 2 c _o _ fo Hence 2/-^-=o( eC * e a + e c -e a ) EXAMPLES. 371 fo _o j by expanding e a and e as far as and denoting by u and v the c - -- c - _- quantities (e + e c ) and - (e c e c ) Substituting in this equation the value of s given by (17) in which it is evident that the term of the second order may be rejected if we wish to obtain y to this order only in terms of x we obtain an equation of the form p Q y = u+ ^ + *' < 18 > in which P and Q are both functions of x and y. Now assume v = w -J f- - } where A. and a are functions of x a a* alone, and substitute this value of y in every term of (18). This will give us, with a little trouble, 1 1 , Hence, finally y = u - -^ + ^ , to the second order of the small quantity 219.] Extensible String on Smooth Surface. It is clear that the equations (l) of Art. 213 are applicable to an exten- sible string-, as are also the results arrived at in that Article without integration. The result arrived at by integration, which expresses the tension in terms of the potential, is to be replaced by equation (10) of Art. 218; and from this equation it follows that if an extensible string, uniform in its natural state, rest on any smooth surface under the action of gravity, the free extremities are in the same horizontal plane. EXAMPLES. 1. An elastic string, uniform in its natural state, is suspended from one extremity, which is fixed, and has a given weight attached to the other ; find the extension of the string, taking its own weight into account. Let W be the weight of the string, P the suspended weight, A the modulus of elasticity, and m the mass of a unit length of the un- stretched string. Then the equation of equilibrium is If Z is the natural length of the string, m^glg = W j therefore this equation gives by integration B b 2 372 EQUILIBRIUM OF FLEXIBLE STRINGS. [219. W T+ s = const, ^o When 8 9 = 0, T is evidently W + P ; therefore T Again, since ds (1 +-T-) ds^, we have A. TF no constant being added because s = when s = 0. If s = Z , and is the whole length of the stretched string, we have ' 2. A heavy uniform elastic ring is placed round a smooth vertical cone ; find how far it will descend. Let W be the weight of the ring, 2ira its natural length, A its modulus of elasticity, y the distance of the plane of the ring from the vertex of the cone in the position of equilibrium, and I the stretched length in this position. Then if the ring be shoved down through an indefinitely small vertical distance, by, the equation of work is T being the tension of the ring. If a is the semi-vertical angle of the cone, I = 27ty tan a ; hence bl = 2ir tan a . by, and But, by Hooke's Law, T --) A W cot a) i 7TA 3. An elastic string, uniform in its original state, is placed on any smooth curve and acted on by given forces ; find its extension. The tension at any point is determined by the equation T or A (1 + ) 2 + 2m f(Xdx + 7% + &) = const. (1) Let m f(Xdx+Ydy + Zdz) be denoted by V. Now, take any point, 0, in the string as the point from which s and s are measured, and let A be the value of 7 at a free extremity of the string. If one 219.] EXAMPLES. 373 extremity is fixed, it will be well to measure s and s from it. Putting T ds -T=Q, V=A, and also 1+ = , (1) gives (*)' = i + _r). ( 2 ) Suppose the curve of constraint to be given by the three equations *=/iW y=fM *=/ 8 W- Then (2) gives fo_ __ =^-F)~ *" or, by integration, ^(s, A) = S Q + $(O, A ), (3) s and 5 being both measured from 0. Let Z and ? ft be the stretched and original lengths of the portion between and the free extremity considered. Then we have 4>(1,A) = l + (o 9 A). (4) But A is evidently a function of the co-ordinates of the extremity, and these co-ordinates are, by supposition, f^l), / 2 (Z), f 3 (l)', hence A is a known function of I, and by substituting its value in (4) we deduce the value of I. 4. One extremity of an elastic string, originally uniform, is fixed at the highest point of a smooth cycloid in a vertical plane, the string lying along the convex side of the curve ; find the extension produced by gravity. If the tangent at the highest point is taken as axis of as, and if - - is denoted by c, we find easily, for any curve of constraint, g as ds Q Vc+hy VG h being the ordinate of the free extremity. In the cycloid s 2 = Say. Substituting this value of y in the equation, and integrating, we have If I be the length from the fixed to the free extremity, and 1 the natural length of the string, VA v Also P = Bah. These equations combined give 1 = 2 where 4 m = latus rectum of generating parabola. 4TrmA. 8. An elastic string, uniform in its original state, rests on a rough inclined plane with its upper extremity fixed ; prove that its extension will lie between the limits p g j n ( t ' + 6 ) 2c cos e where i = inclination of plane, e = angle of friction, I = natural length of string, and c = length of a portion of the string in its natural state whose weight is the modulus of elasticity. ("Wolsten- holme's Math. Prob.) 9. A weight P just supports another weight Q by means of a fine elastic string passing over a rough circular cylinder whose axis is horizontal ; A is the modulus of elasticity, and a the radius of the cylinder ; prove that the extension of the part of the string in contact with the cylinder is r\ x - log ^r ' (Wolstenholme, ibid.) fJ, 221.] COMMUTATIVE PROPEKTY OF d AND S. 375 10. Two uniform ladders, freely jointed at a common extremity, rest in a vertical plane with their other extremities on a rough horizontal plane, these extremities being connected by an elastic rope ; find the greatest angle between them consistent with equi- librium. Ans. If a is the length of each ladder, 2 a sin a the natural length of the rope, 2 6 the greatest angle between the ladders, and A the modulus of elasticity of the rope, A (sin 6 sin a) = TFsin a (p + \ tan 6). 11. A heavy uniform elastic ring is placed horizontally round a rough right cone whose axis is vertical and vertex upwards, the stretched ring being uniform; find its extreme positions of equi- librium. W Ans. y = a [I + cot (a+ e)} , with notation of Ex. 2. 2 TT A i SECTION III. The Method of Virtual (or Potential) Work 220.] Distinction between the Symbols d and 8. In the sequel we shall use the symbol d to denote the increment which any function receives when we pass from a given point P in a body, which occupies a given position, to any indefinitely near point Q in the body^ the position of the body being invariable ; while by the symbol 8 we shall denote the increment which the function receives as we pass from the point P when the body occupies a given position to the same point P in the body when it is displaced, or imagined to be displaced, from this position to any one indefinitely close to it. This use of the symbol 8 has been already exemplified in the Chapters on Virtual Work. 221.] Commutative Property of d and 8. If V denote any function of the co-ordinates of a point P in a body, we propose to show that This will be rendered plain by a very simple illustration. Let P and Q (fig. 226) be two very close points in a body occu- pying a given position, and let P' and Q' be the positions of these points when the body receives any Fig. 226. slight displacement. Let Ox be the axis of #, and let the co-ordinates of P and Q be Or and 376 EQUILIBRIUM OF FLEXIBLE STRINGS. [22T. On, those of P* and Q' being O/ and On', measured along Ox. Then if x is the co-ordinate of P, dx = rn, and bx = rr. Also 8 (dos) = value of ^ in the new position value of dx in old position = rn rn ; and d(bx) = value of 8 a? for Q value of bx for P = #' rr'. But obviously /#' rn = ww' rr'\ therefore b(dx) = ^(8#). From this it follows that if Fis any function of x, 8 (^F) = d(b7). For, by the elementary princi- ples of the Differential Calculus 8 (uv) = ubv +vbu. Now, d7= d ^dx, dx and 8r=-j- dx The two expressions are^ therefore, identical ; and the same proof may be applied to show their identity when V is any function of the co-ordinates. Again, since by the Differential Calculus it follows that b/7dx = /8 ( Vdx\ Suppose that any integration in which the element of arc PQ (fig. 226) is taken as the constant infinitesimal, ds, is performed over a curve, and let the integral be fVds. Then the change in the value of this integral when it is found for the same curve in a displaced position is bfFds. Now the infinitesimal in the new position of the curve is P'Q', which is equal to PQ; therefore 8 (ds) = 0, and If Yds = /8 ( Yds) = / (8 7) ds, that is, the change in the value of the integral of a function = the integral of the change in the function, both integrations being performed over the same curve, the arc of which is taken as independent variable. The same remains true if the integration fVds is performed over a surface or through a solid, and ds denotes the element of 222.] METHOD OF WOEK APPLIED TO A STRING. 377 superficial area or of volume. Again, since by Differential Calculus, b- = - 9 > it follows that if ds is constant, V V 2 dx bdiK dbffi ~J ^~ 7 ~~ 7 J as ds ds dx dx 7t> $ d-r- 79 . , a* as ds d* osc and = g __ = __ ds* ds ds and generally 8 -= = -=- V * /Vo** //o>* EXAMPLE. Every element of a solid body is multiplied by the product of its two co-ordinates x and y, and the sum of all such products is taken. If the body receives a small displacement of rotation round the axis of z, find the variation of this sum. The element of mass at any point x, y, z being dm, the sum in question isfxydm. Now bfxydm = fb (xy) . dm = f (xby + ybx) dm. But bx = ybd, by = x8B, if the angular rotation of the body is 80. Hence the variation = b6f(x 2 y' i )dm. 222.] Method of Work Applied to a String. First suppose the string to be perfectly inextensible. Now if the particles of a system are dm lt dm 2 , ..., and if they have to fulfil conditions denoted by L^ 0, L 2 = 0, . . . , the equation of Art. 186 becomes In the present case the particles are portions ds lt ds 2 ... of a string at points (x-^y^ %), (#2,^2^2)5 an< ^ eacn nas ^ satisfy the condition of having its length unaltered in any displacement of the system. Hence the geometrical equations are d*! = const., &c.; and equation (l) becomes or f(Xbx+ Yby + Zz)dm+fXds = 0, (2) the number of particles being indefinitely great. Now, as in Art. 186, we express all the variations in terms of the variations of the co-ordinates #, ^, z. For this purpose, put .'. dsbds = 378 EQUILIBRIUM OF FLEXIBLE STRINGS. or s= ds ds * ds Hence (2) becomes .., /* . dx , / dx \ f dx . \ /\ ^ /. Now / A d8# = (\--8a?) (A 8#) / Stf-y-fA J ds ^ ds /! V ^ y o J ^ v by integration by parts, the term(A-r-8#A being the value of d# . . * X -^-8# at one of the limits of integration, i.e. at one extremity ds -I of the string; and (\ 8 a?) being its value at the other ., ^ ds ' n extremity. Performing similar integrations for the other terms, (3) becomes (dx . dy ^ dz . \ ,dx . dy . (x, y, z), a function of the co-ordinates of a point, the equation V C, (2) where C is any constant, will denote a surface at every point of which the potential of the forces has a constant value. More- over (l) shows that at all points on this surface T has the constant value K C. Although it may happen that there is no portion of the string on the surface denoted by (2), still we shall say that the tension has a constant value on this surface, since T has an analytical value given by (l) ; and, in the same sense, we shall speak of the tension at any point whatever in space, although no part of the string exists at this point. By attributing different values to C in (2), we get a series of 224.] PROPERTY OP MINIMUM. 381 surfaces called Equipotential Surfaces. These surfaces are called by French writers Surfaces de Niveau, or Level Surfaces, from the part which they play in hydrostatics. Some of the principal properties of these remarkable surfaces will be given in a subsequent Chapter. 224.] Property of Minimum. If a uniform inextensible string, in equilibrium under the action of a given conservative system of forces, joins two fixed points, A and B, the variation of the integral fTds will be zero when we pass from the curve of the string to any in- definitely close curve which passes through A and J3. Let us calculate the variation of this integral, fads = fbT.ds + Tb ds) Now, from (l) of last Art., bT = -bF= Hence by integration by parts (as in Art. 222), we have to)-T.*+ ds ' "\ds ds 4 ds Now the right-hand side of this equation is zero, since, the extreme points of the curve being fixed, the coefficients of T Q and 7\ both vanish, and the coefficients of 8 a?, 8y, bz under the sign of integration vanish by the general equations of Art. 204, the mass of a unit length of the string being here taken as unity. Hence the proposition. This theorem leads to a remarkable property of the common catenary. Of all curves of the same length joining two given points in a vertical plane, the common catenary is that whose centre of gravity is lowest. For if y be the depth of the centre of gravity of this curve, whose length is L, we have 382 EQUILIBRIUM OF FLEXIBLE STRINGS. But (Art. 206), T = mgy ; therefore y - j ; therefore, by the theorem of this Article, we have That y is in this case a minimum in the true sense of the word does not, of course, appear from this ; the proof that it is so depends on the criterion for maxima and minima furnished by the Calculus of Variations, for which see Jellett's Calculus of Variations^ p. 80. It is there proved, that when the variation of any integral of the form \ 1 Udx vanishes (the limits being t Jxo fixed) the value will be, in general, an algebraic maximum or d z U minimum according as -= ^ is continually or continually + between the limits of integration, -~ being denoted byjo n , and [/being any function of #, y,^ 1} j 2 , ,., p n . In the present case U = y ds = y \/l -\-p^dx, a change of the independent vari- able from s to x being necessary since it is the limits of x that are assigned. The application of the criterion is then obvious. CHAPTER XIV. SIMPLE MACHINES. 225.] Functions of a Machine. A machine may be defined either from a statical or from a kinematical point of view. Regarded statically, it is any instrument ~by means of which we may change the direction^ magnitude, and point of application of a given force ; and regarded kinematically, it is any instrument ty means of which we may change the direction and velocity of a given motion. In Statics it is usual to consider the points or machines to which forces equilibrating each other are applied as absolutely motionless ; nevertheless, it appears from our definition of force (Art 1), that a system of forces acting at a point will be in equi- librium when the point has a uniform motion in a right line. If a particle describes any curve whatever with uniform velocity, a little reflection will show that at no point of its path can there be any force in the direction of the tangent or, in other words, the force acting on it must everywhere be normal to the path. It follows (see Art. 195), that there is no work done by this force in the passage of its point of application from any one position to any other. Extending this a little, we shall so far anticipate the results of Kinetics as to assume that when the parts of any machine are each in a state of uniform motion, the forces applied to the machine are in equilibrium among themselves. By the extension of the equilibrium of forces to this case, we comprise both the statical and kinematical definitions of a machine in the following : a machine is any assemblage of different pieces whose displacements, resulting from their mode of connection, depend on each other ly geometrical laws, and whose object is to transform into mechanical work the result of the action of given applied forces. (See Resal, Mecanique Generale, vol. iii, P-3-) 384 SIMPLE MACHINES. [226. It has been already pointed out that in applying the equation of virtual work to a system of connected bodies, advantage is gained by choosing such displacements as do not violate any of the geometrical connections of the system. This principle we shall use largely in the discussion of machines, and the dis- placements which we shall choose will be those which the different parts of a machine actually undergo when it is em- ployed in doing work. Thus, instead of equations of virtual work, we shall have equations of actual work ; and in future we shall speak of the principle referred to as the Principle of Work. Since in the motion of a machine the work done by a force applied to any part of it depends on the magnitude and direction of the displacement of the point of application of this force, we see at once the importance of the discussion of the motions pro- duced in the several parts of a machine by a definite motion given to some one part. This discussion, which is a problem of pure geometry, constitutes the Kinematics of Machinery, for which the student may consult ResaFs Mecanique Generate, Willis's Principles of Mechanism, or the treatise of B-euleaux. 226.] Moving Forces and Resistances. Every machine is designed for the purpose of overcoming certain forces which are called resistances; and the forces which are applied to the machine to produce this effect are called moving forces. The distinction between these forces is easily drawn by the Prin- ciple of Work. For, when the machine is in motion, every moving force displaces its point of application in its own direc- tion, while the point of application of a resistance is displaced in a direction opposite to that of the resistance. A moving force is, therefore, one whose elementary work is positive, and a re- sistance one whose elementary work is negative. A moving force applied to a machine is often (but impro- perly) called a power. The resistances against which a ma- chine works are divided into two classes, viz. useful resistances and wasteful resistances. The former constitute those which the machine is specially designed to overcome, while the over- coming of the latter is foreign to its purpose. For example, if a pulley is employed for the purpose of lifting a weight by means of a rope, a part of the effort employed is spent in over- coming the friction between the pulley and its spindle, and another part is spent in overcoming the rigidity of the rope. 227-] EFFICIENCY OF A MACHINE. 385 Friction and rigidity in this case are the wasteful resistances, and the weight of the body lifted is the useful resistance. The distinction between the resistances overcome gives also the distinction between useful work and (so-called) lost work. Useful work is that which is performed in overcoming useful resistance, while lost work is that which is spent in overcoming wasteful resistances. 227.] Efficiency of a Machine. The ratio of the useful work yielded by a machine to the whole amount of work performed by it is called its efficiency. Let W be the work done by the moving forces, W u the useful and Wi the lost work, when the machine is moving uniformly. Then W^W^+W^ and if rj denote the efficiency of the machine, r Since some of the work expended in moving the machine must be expended in overcoming wasteful resistances, the efficiency is always less than unity, and the object of all im- provements in the machine is to bring its efficiency as near unity as possible. The counter-efficiency is the reciprocal of the efficiency. If the useful work to be performed is given, the amount of work to be expended on the machine is obtained by multiplying the former by the counter-efficiency. Let P be the moving force applied at any point of a machine to perform a given amount, W UJ of useful work ; let W t be the work lost, and let s be the space through which P drives its point of application in its own direction. Then we have Ps= W U +WL Let P be the force which would perform the same amount of useful work if the wasteful resistances were removed. Then P,s = JT U . W P But 77 = -- = ; hence the efficiency is the ratio of the force which would drive the machine against a given useful resistance, if the wasteful resistances were removed, to the force which is actually required to do so. In many cases this definition is useful in practice. c c 386 SIMPLE MACHINES. [228. As regards the wasteful resistances in machines, the most noticeable are friction, the rigidity (or rather imperfect flexi- bility) of ropes, and the vibrations which are produced in the various pieces. Of these the first is that with which alone we shall be concerned. The student who desires information on the experimental laws of the rigidity of ropes may consult Coxe's translation of Weisbach's Mechanics of Engineering and of the Construction of Machines, vol. i, p. 363 (New York, 1872). 228.] Simple Machines. By simple machines are meant the Lever, the Inclined Plane, the Pulley, the Wheel and Axle, the Screw, and the Wedge. Of these, the Lever, the Inclined Plane, and the Pulley may be considered as distinct in principle, while the others are only combinations of pairs of these three. 229.] The Lever. A lever is a solid bar, straight or curved, which is constrained to turn round a fixed axis. This fixed axis is called the fulcrum of the lever. It is usual to define three kinds of levers. If the fulcrum is between the moving force and the resistance the lever is said to be of the first kind; if the resistance acts between the moving force and the fulcrum /Q ^X ( as i n a wheelbarrow, an oar, or a Fi &- 22 7- pair of nutcrackers), the lever is of the second kind ; and if the moving force acts between the fulcrum and the resistance (as in the construction of the limbs of animals), the lever is of the third kind. In the last kind the moving force is always greater than the resistance to be overcome, and levers of the third kind are there- fore seldom employed. To find the efficiency of a lever, the wasteful resistance being friction Let the moving force, P, be applied at the point A (fig. 227) in the direction OA perpendicular to the axis, and the useful resistance at B in the direction OB, also perpendicular to the axis ; let EDF be a section of the axis on which the lever turns, made by the plane of P and Q, the contact between the beam and its axis, although it may be very close, being still such that they can be considered as touching along a single line when the machine works. In this case (see Art. 114) the reaction of the axis consists of a single force touching the circle of radius r sin A concentric with EDF, X being the angle of friction for the lever and its axis; and since this reaction must also pass 330.] THE INCLINED PLANE. 387 through 0, its direction is obtained by drawing from this point a tangent to the circle. Let p and q be the perpendiculars from C, the centre of the axis on OA and OB respectively, and let &c., and adding, we have 2 n ~ l y+p = constant. Now the equation of work or and W 2 n l 390 SIMPLE MACHINES. [233. 233.] The Wheel and Axle. This consists of a horizontal cylinder, I, (fig. 231) moveable round two journals (or small cylinders projecting from the centres of its faces), one of which is represented in section at e; a wheel, + Q 2 , o) being the angle between the directions of P and Q, exactly as in Art. 229 ; and the efficiency is the same as that investigated in the Article on the lever. Economy of power is attained in the wheel and axle by diminishing b, the radius of the axle ; but in this way the strength of the machine is diminished. To avoid this disadvantage a Differential Wheel and Axle is sometimes employed. In this instrument the axle consists of two cylinders of radii b and b' (fig. 232), and the rope, wound round the former in a sense opposite to that of watch-hand rotation (suppose), leaves it (at the point b in fig. 231), and, after passing under a moveable pulley to which the weight to be raised is at- tached, is wound in the opposite sense round Fig. 232. the remaining portion (that of radius b') of the axle. The power P is applied, as before, tan- gentically to the wheel. For the equilibrium (or uniform motion) of the machine, the tensions of the rope in bm and Vn are each equal to \ Q ; and taking moments round the centre of 234-] THE SCREW. 391 the journal, c, for the equilibrium (or uniform motion) of the rigid system consisting of the wheel and axle alone, we have Thus, by making the difference # V small, the requisite moving force can be made as small as we please ; but since the amount of work to be done is constant, this economy of power is accompanied by a loss in the time of performing the work. For it is easily seen that if the wheel turns through an angle 80, the point of application of P will describe a space a$0, and the weight will be raised through a space i (-&') 80, which latter will be very small if 6 If is very small. 234.] The Screw. The screw consists of a right circular cylinder on the convex circumference of which there is a uniform project- ing thread, GH (fig. 234), of a helical form. The helix is a curve traced on the circumference of a cylinder in the following manner. Take a sheet of paper on which are drawn two in- definite right lines, AB and AC, and let the paper be wound round the cylinder in such a way that the line AB coincides with the circumference of the base ; then the other line, AC, will appear on the cylinder in the shape of a spiral curve which is called the helix. (Fig. 233 represents a projection of the helix on a plane through the axis of the cylinder.) A screw with a rectangular thread (which is that represented in fig. 234) is obtained by making a small rectangular area, abed, move so that one side, a b, always coincides with a generating line of the cylinder, the middle point of ab describing the helix, and the plane of the rectangle always passing through the axis of the cylinder. If a small triangle is used instead of the rectangle, we should have a screw with a triangular thread. Let p and q be two points on the indefinite line AC, and draw pn perpendicular to AB and gn parallel to it. Then pq becomes a portion of the arc of the helix, and qn a portion of a section of the cylinder perpendicular to its axis,jtm remaining a straight line coinciding with a generator of the cylinder. Fig. 233 392 SIMPLE MACHINES. [234. Hence the relation holding between the sides of the triangle pqn before the paper was wound round the cylinder will hold also after the winding. But if the angle between AB and AC is i, we have evidently pn = qn . tan i, pq = qn . sec i. The thread GH works in a block on the inner surface of which is cut a groove which is the exact counterpart of the thread. The block in which the groove is cut is often called the nut. It is clear, then, that if the screw moves in the nut until the point jo of the thread occupies the position q, the axis must move in its own direction through a space pn> and the angular rotation of the screw about its axis is > r being the radius of the cylinder. Hence, if the angle through which the screw turns is denoted by o>, we have pn = or tan i, pq = cor sec i. If o> = 2-jr, or if the screw make a complete revolution, any point on the surface of the screw describes a space 27rrtan& parallel to the axis. This is obviously the distance between two portions of the thread measured on a generator, and is called the pitch of the screw. We shall consider the screw as driving a resistance Q applied in the direction of the axis, and the moving force, P, as applied in a plane perpendicular to the axis, at the extremity of an arm whose length measured from the centre of the axis is a. Suppose that the screw rotates through an angle w. Then the work done by P is P#co, and the work done against Q is Qr<0 tanz. If no work is lost against wasteful resistance, we must have Pa Qr tan i. If there is friction between the thread and the groove, let R be the normal pressure at any point p of the thread (acting towards the under side of pq in the figure), and ^R the friction at this point. Then, in a small angular motion, 8co, of the screw the work done against the friction is pR.pq (taking pq as an ele- mentary portion of the thread), or pErSa sec i. Hence P5a> = Qrbu tan fc + fir 8 co seci 335-] PEONY'S DIFFERENTIAL SCREW, 393 denoting the sum of the normal reactions at all points of the thread. But, for the equilibrium of the cylinder, resolving along its axis, we have Q = 2 (R cos ipR sin i), or Q = (cos ju sini) 27?. (a) Hence, substituting this value of 272 in the previous equation, A being the angle of friction. This result could have been obtained without the principle of work by combining by (a) the equation of moments round the axis of the screw. By taking moments round the axis, we have Pa = 2 (R sin i + fj,R cos i) r, or Pa = r (sin i + JJL cos i) 2 E. (fi) Dividing (/3) by (a) we obtain the relation between P and Q. The efficiency of the screw is evidently tan i tan (i -f- A) which will be a maximum when i = --- 235.] Prony's Differential Screw. If k denote the pitch of a screw, the relation between P and Q when friction is neglected is therefore economy of force in C3S overcoming a given resistance is gained by making h very small. But it is impossible to do this in Fig. 235. practice, and to attain the result desired a differential method is resorted to. Let the screw work in two blocks, A and B (fig. 235), the first of which is fixed and the second moveable along a fixed groove, n. Let li be the pitch of the thread which works in the block A, and Ji r the pitch of that which works in the block B. Then one complete revolution of the screw impresses two opposite motions on the block B one equal to h in the direction in which the screw advances, and the other equal to h' in the opposite direction. If, then, the 394 SIMPLE MACHINES. [236. resistance, Q, is driven by this block, we have by the principle of work 2Pna = Q(h-h')> and the requisite moving force will be diminished by dimi- nishing h h'. 236.] The Wedge. The wedge is a triangular prism, usually isosceles, which is used (as represented in the figure) for the purpose of separating two bodies, A and JB, or parts of the same body which are kept together by some considerable force, molecular or other. The figure represents a section of the wedge made through the line of action of the moving force, P, perpendicular to the axis of the wedge. Suppose that the line of action of P passes through the vertex of the wedge, and that slipping is about to take place; then the total re- sistances of the surfaces A and B against the wedge will make the angle, A, of friction with the normals at the points, m and n, where they act; but these points are indeterminate themselves. To find the efficiency of tlie wedge. Let the wedge be driven through a vertical space equal to dp, and let 2 a be its vertical angle. Then the useful work performed is the separation of A and B in directions normal to the faces of the wedge in contact with them ; in other words, the useful work is that done by the normal components of the total resistances, R. Now the point m moves vertically down through a space dp, and the projection of this displacement along the normal at m is evidently Fig. 236. Hence the work done by the normal components is 2 R cos A sin a dp, and the whole work expended is Pdjp. Hence 2R cos A sin a r? = _ But by resolving vertically for the equilibrium of the wedge, we v P = 2R sin (a + A) sin a cos A tan a r? = sin (a + A) p + tan a 237.] THE BALANCE. 395 Having given the theory of the simplest machines, we proceed to discuss a few of their most useful forms. 237.] The Balance. The common balance is a lever of the first kind with two equal arms, from the extremity of each of which is suspended a scale pan, the fulcrum being vertically above the centre of gravity of the beam when the latter is horizontal. Let (fig. 237) be the fulcrum, AB the line joining the points of attachment of the scale pans Fig. 237. to the beam, G the centre of gravity of the beam, and let AB be at right angles to OC, the line joining the fulcrum to the centre of gravity of the beam. Then, if AC=CB = a, OC = h, OG = Jc, W= weight of the beam, and = the inclination of AB to the horizon when two weights, P and Q, are placed in the pans, we have for the position of equilibrium (by moments about 0), Now, the most important requisites for a good balance are Sensibility and Stability. The first requires that the beam should be sensibly deflected from the horizontal position by the smallest difference between the weights P and Q ; hence the sensibility may be measured by the angle of deflection from the horizontal position caused by a given difference, P Q. The stability of the balance is measured by the rapidity of the oscillation of the beam when it is slightly disturbed, and will be greater the smaller the time of oscillation. Hence the in- vestigation of the stability of the balance is a kinetical problem. For sensibility, tan Q must be as great as possible for a given value of PQ. Hence (l) a must be large, (2) k must be small, (3) JFmust be small, and (4) k must be small, i.e. the distance of the fulcrum from the centre of gravity of the beam must be small. The last condition is obtained in balances in which great sensibility is desired by making OC an axis along which a heavy nut moves with a screw motion ; by moving the nut towards 0, the centre of gravity of the machine can be made to approach the fulcrum. The time of a small oscillation can be shown (see Thomson and Tait, p. 423) to be proportional to the square root of 396 SIMPLE MACHINES. [238. where K is the radius of gyration of the beam about 0. For stability this must be small ; it is evident that, with the exception of the third condition above, the conditions for sta- bility are the very reverse of those for sensibility. 238.] RobervaFs Balance. Roberval's Balance is an excel- lent illustration of the principle of work. Two equal bars, AB and CZ), (fig. 238) revolve round axes through their middle points, H and E, which are fixed in a vertical support, HN\ these bars are connected by smooth joints to two equal bars, AC and BD, and to these latter bars are rigidly attached N two plates or scale pans, P and Q, the Fi 8 points of attachment being any what- ever, and one or both of the plates may lie towards the vertical support, or away from it (as in fig. 238). Suppose P and Q to be the magnitudes of two weights placed in the pans P and Q, respectively. Then if for any displacement of the bars round the points H and E, the pans describe vertical spaces p and q, respectively, we shall have for equilibrium Pp-Qq = 0. Now, the bars AC and BD, being always parallel to the fixed line HE, will be always vertical, and the vertical space through which one moves up is obviously equal to that through which the other moves down. Hence p = q, and we have for equi- librium p .... Q whatever be the lengths of the pans {provided their weights are neglected] , whatever be their points of attachment to BD and AC, and whatever the points in the pans at which P and Q are placed. If the weights of the pans are taken into account, the same results follow if they are of equal weight. If the pan P were replaced by the pan P / ) and the weight P placed at P f t the other pan, Q, remaining unchanged, and the weights of the pans being either equal or neglected, equilibrium would still subsist a result which seems at first sight very strange. 239.] BALANCE OF QUINTENZ. 397 If the lengths AH and HB, CE and ED are not equal, it is easy to prove that - = 77-7 , and the condition of equilibrium is M: 239.] Balance of Quintenz. This is a compound balance formed of a combination of several levers, and is used for weighing very heavy loads. This machine also furnishes an admirable example of the principle of work. AB (fig. 239) is a lever moveable about its fixed extremity, A ; MN is another lever moveable about a fulcrum, F, fixed at its N. - middle point ; CD is a moveable ^ platform, which receives the load Q, whose weight is to be found ; this platform is con- B H~~A nected with the lever MN by a Fig. 239. rigid vertical bar, DI, articulated at D and 7; and the platform further rests against the lever, AB, by an edge of contact at a fixed point, ff, on the latter ; finally, the two levers are connected by a rigid vertical bar, BM, articulated to both. The weight, P, employed to measure Q is attached to the upper lever at N. Let the system receive any slight displace- ment, then the lever, AB, will turn round A through an angle 5 0, suppose, and the lever MN will turn round F through an angle 8$. "We shall arrange the dimensions of the machine in such a manner that the platform, CD, may remain horizontal in the displacement. The vertical descent of the point H is evidently AH . 6 9, and this is also the vertical descent of the point in the platform above H. The vertical descent of the point D is the same as that of /, and this latter is obviously FI.bQ- hence if the platform remains horizontal, Again, the vertical descent of M is the same as that of B ; or , FM.IQ = AB.be. Hence from these equations we have MF _BA FI " AH ' 398 SIMPLE MACHINES. [240. which is the condition for the horizon tali ty of the platform. 7? A Denote -r= by n. The equation of work is obviously P x descent of N = Q x descent of D, or the result is the same as if Q were suspended from the point I of the upper lever. Loads placed on the platform may all be weighed by means of a constant weight, P, by merely moving the point of suspension of this latter along the arm NF; thus, if P is suspended from the point K between N and F 9 we shall have * 240.] Toothed Wheels. Motion may be transferred from one point to another and work done by means of a combination of toothed wheels, each one of which drives the next one in the series. The discussion of this kind of machinery possesses great geometrical elegance ; but the space at our disposal renders it impossible to do more than give a slight sketch of the simplest case that in which the axes of the wheels are all parallel. For the investigation of the proper forms of teeth, the student is referred to Willis's Principles of Mechanism, Collignon's Statique, and Resal's Mecanlque Generate. Fig. 240 represents a toothed wheel, A l9 moveable round a horizontal axis, ab -, the moving force, P, is applied by means of a Bi handle, cd> which, when turned, causes the axis ab to rotate in its bearings at a and t> and to turn the 2 wheel A ; this wheel causes another, AJ 19 in contact with it, to rotate Fig. 240. round a horizontal axis which also moves in fixed bearings at its ex- tremities; on this latter axis is fixed another wheeU 2 , whose rotation in like manner turns B 2 on its axis, which in the figure is the axis of a cylinder to which the resistance, Q, is attached. Suppose that there are n wheels, A lt A 2 , ... A n , whose radii 340.] BALANCE OF QUINTENZ. 399 are 1} a 2 , ... a n) and n wheels, E 19 13 2 , ... B n wliose radii are ^u ^25 ^n an ^ k* ^ c == J !9 J an d ^ e ra dius of the cylinder (or wheel) to which Q is attached = q. Then if coj is the angle through which the radius be revolves, the moving force being always applied tangentially to the circle described by its point of application, the work expended is and if &> M is the angle through which, in the same time, the cylinder rotates, the weight Q will be raised through a space g<*> n , and the work done against the resistance is Qg** Supposing then that no work is lost either by the friction of the axes in their bearings or by the friction of the teeth against each other, we must have Pp^ = Q00),, (1) when the machine is moving uniformly. To determine the kinematical relation between coj and o> n , let the angle through which B^ turns be o) 2 . Then since the spaces described by the points of A 1 and B^ which are in contact are the same, a o) t = b o> 2 . Also if co 3 is the angle through which B 2 turns, we have 2 o) 2 = b 2 a> 9 * Proceeding in this way, we have by multiplying the corresponding sides of these equations together a^^... a n . ^= b^...b n . tt . (2) Hence from (l) and (2), Q PW*-ln P ' For the calculation of the work lost by the friction of the teeth among themselves see Collignon's Statique^ p. 468. CHAPTER XV. ATTRACTIONS. THEORY OF THE POTENTIAL. SECTION I. Solid Distributions of Matter in General. 241.] Universal Law of Attraction. Every particle of matter m the universe attracts every other particle with a force whose direction is that of the line joining the two particles, and whose magnitude is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This law of universal attraction is a generalization from experience, verified in its consequences as to the motions of all bodies in the universe which come within the reach of our observation. That two particles of matter universally exercise upon each other an attractive action denned as above we observe by experiment ; and this action is called Gravitation. Over and above this particular force, they may exert other forces, at- tractive or repulsive, upon each other, depending on particular states, transitory or permanent, in which they may exist in presence of each other. Among forces of the latter class are magnetic and electric attractions and repulsions, and the mole- cular forces of natural solids. At present we are concerned with the bare fact that such an action as that of gravitation is exercised between two particles, without attempting to account either for its cause or for its precise mode of operation that is, without any speculation as to whether it is really an action at a distance, or an affection of some medium intervening between them. 243-] POTENTIAL DUE TO AN ATTRACTING SOLID. 401 We shall, it is true, investigate in certain cases the conse- quences which would result if the particles of matter exercised on each other a force whose magnitude did not follow the natural law of the inverse square of distance ; but these cases must be regarded as mere f examples of an analytical method, and not as the expressions of any observed natural phenomena. 242.] Action between Two Particles. Let there be two particles whose masses are dm and dm', and let r be the distance between them. Then the attraction of gravitation between themis dm. dm' where //, is a constant quantity depending on the unit of force adopted. Suppose that we take as a unit force that exerted by two elementary units of mass placed at a unit distance apart ; then the above expression must be unity when dm, dm', and r are units. Denote the unit of mass by \m\, the unit of distance by \cf\, and the unit of force by [/] ; then the force,/, between the particles at the distance r is given by the equation It would be tedious to introduce the unit factor ^=^L into W 2 our equations, and we shall for the future omit it, remembering, at the same time, that it is implied in our results, and that the value of every force /^ -x^ ^-^ p subsequently given must be multiplied by this unit factor. 243.] Potential due to an Attracting Solid. Let P be any point at which a fig. 241. unit mass is placed ; M any point in the solid at which the element of mass is dm ; and r the distance PM. Then the force between the particles at P and M is -^ , and the virtual work of this force is ^-dr, since the force tends to diminish r, and since dr signifies an increment of r. Hence (Art. 188), if Fis the potential at P, Dd 402 ATTRACTIONS. THEORY OF THE POTENTIAL. [244. the sign 2 denoting a summation of the integral for all elements of mass of the solid. This evidently gives but as the solid consists of an infinite 1 number of elements, the summation here is an integration, and we have finally 7 = /am / ' If in Art. 195 T is the value at infinity of the gravitation potential of a given mass, T is of course zero, and from that article we have W -=V Hence we may define the gravitation potential of a given mass at any point to be the quantity of work required to move a unit mass of matter from that point to an infinite distance. If the law of attraction is other than that of nature, let it be a function of the distance denoted by ' (r). Then the force between P and Mis <$>'(r) . dm, and the virtual work of this force being <' (r) . dm . dr (supposing the force attractive), V = I,dmf(t>' (r)dr = ^^(r).dm = /< (r) dm, <(r) being the integral of $'(r)dr. For example, if the at- traction is proportional to the # th power of the distance, = -=TT/' For a repulsive force given by the law $> (r) the virtual work is <'(r) . dm . dr, and the sign of 7 is simply changed. In general, then, to get the potential of any system of forces, write down the expression for their elementary virtual work, and integrate it (see Mecanique Celeste). 244.] Calculation of the Potential in Special Cases. The law of attraction considered is that of nature. (1) Let the attracting solid consist of two particles of masses m l and m 2 placed at two points, N and S (fig. 35, p. 39). Then if the distances NP and SP are denoted by ^ and r 2 , F^- 1 *- 2 , (1) i r 2 and if the action of m l is repulsive, F = _^ + 5s. (2) 244-] CALCULATION OF THE POTENTIAL IN SPECIAL CASES. 403 (2) Let the attracting solid be a bar of uniform density and small uniform section. From P (fig. 242) let fall PO perpendicular to the bar, AB\ let P s denote the distance of any point M of //j\ the bar from ; let 6 be the angle PMO, / / \ \ and let p and k be density and section of /' / the bar. Then the element of mass at M / I \ . r* = kp PM But s PO cot 6, /. ds = PO cosec 2 0d6, and PM = PO cosec ; F=-& r P Jir-B sinO' the angles PAB and P.5J. being denoted by A and J5. Hence This may be put into another form. If PA = r, PB = /, and AB = 2c, we have from trigonometry, (4) p. being the semi-axis major of the ellipse described through P, with A and B for foci. (3) Let the attracting solid be a spherical shell of uniform density and small uniform thickness. First suppose the point P, at which the value of the potential is required, to be outside the shell. Let T and p be the thickness and density of the shell, its centre, and M any point on it. Then if /. M OP = 0, OM = a, and is the angle made by the plane MOP, with a fixed plane through OP, the element of mass at M is pro? sin 6d6di and if PM = r, r= Fig. 243. by performing the integration in < at once. D d 2, 404 ATTRACTIONS. THEORY OF THE POTENTIAL. [244. Now, if OP = c, we have r z = a 2 2ac cos0 + c 2 , /. rdr = ac sin QdO, 27Tpra r PB , 47rpra 2 mass of shell .... and V = - / dr = - = -- (5) c J PA c c Secondly, let P be inside the sphere at P f . Then we have exactly as before 2irpr /""* , fi , 7= - / drkitpTa. (6) Since P'B-P'A = a + c-(a-c) = 2c. It is to be carefully noted that in this case V has the same value at all points inside the shell. (4) Let the attracting solid be a sphere of uniform density. First, suppose P to be outside the sphere. Let the sphere be broken up into an indefinitely great number of spherical shells, and since the potential due to each of these is given by (5), we have for the sphere (whose radius is a), v _ 4 TT/O a 3 _ mass of the sphere , . 3c c Secondly, let P be inside the sphere. Then the potential of the sphere concentric with the given one, and passing through P, is 4 Trp c 3 4 TTO c 2 > or ; and the potential of the portion included between O C O these spheres must be found by dividing it into shells. Let r and dr be the radius and thickness of one of these shells ; then the potential due to it at P is by (6), and the integral of this from r c to r = a is 2irp (a 2 c 2 ). Adding this to the first portion of V, we have 7= ZTTpa? l^r/oc 2 . (8) o (5) Let the attracting solid be that inclosed between two concentric spherical surfaces of given radii, a and a', the density being uniform. First, let P be completely outside the mass, and suppose a > a. Then the potential is obviously the given mass divided by c ; or 7 =. (9) O C Secondly, let P be inside the space between the bounding surfaces, i. e. inside the mass. Then evidently 346.] CONTINUITY OF THE POTENTIAL. 405 Thirdly, let P be inside the surface of radius of. Then 245.] Continuity of the Potential. The gravitation potential of any attracting solid mass varies in a continuous manner from point to point in space, whether the points chosen be inside any portion of the mass or outside it. For if r be the distance of any element of mass, dm, of the attracting body from P, the point at which the potential is /j Let P be taken as origin, and let the T position of the element Am be denned by the radius vector, r, and two angles, 6 and , as in p. 280, and let p be the density of the element. Then Am = pr 2 sin 6 drdO dcf), and 7 - fffpr sin 6 drdO d$. This form of 7 shows that even if r is zero, i. e. if P is inside the mass, the value of the potential is finite, no infinite term being introduced by the indefinitely close proximity of P to some of the elements of mass. Hence the potential varies continuously throughout space, and diminishes from the vicinity of the attracting mass towards the space very remote from it in all directions. 246.] Continuity of the First Differential Coefficients of the Potential. At each point in space the potential of a given mass has a definite value. Let the co-ordinates of P, a particular point considered, be x, y, z. Then if #', y', / be the co-ordinates of the attracting element dm, we have ^ = (*-O 2 + (y-/)3 + (*-/) 2 . (1) And since ,K [dm dV r. *U fx-x' 7 = / > we have -= = / dm 7 = / 5 dm. J r dx J dx J r 3 Hence dV Cx-x' dV ./>-/, dV [z-z _ .. = / tr-dm, -j-=^^-^-dm, -r-= I ^-dm. (2) dx J r 3 dy J r 3 dz J r 3 The continuity of these expressions can be shown by putting xx' = r sin 6 cos $, yy'=. r sin sin <, z /= r cos 0, dm = pr 2 sin Odrdddfa where and $ are the same as in last Article. Then dV = psm 2 0cosdrd0d(t); 406 ATTRACTIONS. THEORY OF THE POTENTIAL. [247. and thus, even when the point P is inside the mass no infinite term is introduced into any of the differential co-efficients of V. Each of these differential coefficients varies, therefore, in a con- tinuous manner throughout space, whether the points at which their values are calculated are inside the mass or outside it. It must be carefully observed that this result has been proved true only when the attracting element of the mass is one of finite volume. It will be subsequently shown that if the attracting element is superficial^ i. e. if its volume is zero, the continuity of some of the first differential coefficients of V ceases. 247.] Discontinuity of its Second Differential Coefficients. d*(-\ Q . _ [dm , d*V r V, ., ,. Since V I - > we have -7-5-= / 7-?- w&\ ^he co-ordinates J r eur J do? of the point, P, at which the potential is T 7 ", being #, y^ z. Now from (l) of last Art. we find and since 3 x-x C<2 f dr** 1 d*r) , = J fa (^ ) - ^ ^ dm, Similarly =-, (2) If in these expressions we substitute for xx',yy> zz', and dm, as in last Article, we have ~ = f( 3 sin 2 cos 2 -\)^s\nO CvX J T hence, when r = 0, i. e. when P is inside the attracting mass, the expression under the integral sign becomes infinite, and the value of -y-g ceases to be continuous from points inside to points outside the mass. Fig. 244 represents the values of V, -r-jand -> when the attract- dx dx z H 8.] COMPONENTS OF ATTRACTION. 407 c ing solid is that contained between two concentric spherical surfaces whose radii are Oaf and Oa, and the point P occupies positions along a fixed diameter, Ox, varying from to infinity. The distance of P from is here denoted by x, which is therefore the same as c in case (5) of Art. 244. The values of V are given by the ordinates (distances from Ox) of the continuous curve ABCD, of which the portion AB is a right line correspond- ing to the constant potential within the inner surface. Fig. 244. dV The values of are given by the doc ordinates of the continuous curve Oafbc, of which Oaf corresponds to the constant zero value within the inner surface. The values of are given by the ordinates of the discontinuous doo curve Oa'nmpq. From case (5), Art. 244, when P is completely outside the mass we have ^ a / , and when P is inside the shell between dc 2 3c 3 the two surfaces dc 2 ' 3 c 3 By putting c = a in the first of these values we have the value, ap, of -ry when P comes to the outer surface from the outside ; and putting c = a in the second, we have the (negative) value, am, of -=-j when P conies to this surface from the inside*. 248.] Components of Attraction. The attraction between a unit mass at P and the element dm at M (fig. 241) is -^- in the line PM\ and since PM makes with the axis of x an angle whose cosne s x x (the co-ordinates of P and M being x, y, z and of, y ', /, respectively), the component of this attraction parallel xx' to the axis # is -- r dm. Hence if X be the attraction of the whole mass parallel to the axis of #, Cx-x' J X dm. Fig. 244 is from Thomson and Tait's Nat. Phil. 408 ATTRACTIONS. THEORY OF THE POTENTIAL. [249. Similarly, if Y and Z denote the components of attraction parallel to the axes of y and z, r.- /'-#*,, *=-/?- o t A = 7 -5- ~w ' or ~J~ ' ~r~ ' ~^~~ ' eac " divided by R R R dx d dz J Let the value of 7 at the point P be denoted by C ; then, 7 being a function of x, y^ z, the equation 250.] CHANGE OF ATTRACTION THROUGH A SHELL. 409 denotes a surface passing through P, and at each point of this surface the potential has the constant value C. Now the direc- tion cosines of the normal to this surface are exactly the same as those obtained above for R. Hence At each point in space the resultant attraction on a particle is normal to the surf ace of constant potential passing through the point. Let APS (fig. 245) be the surface of uniform potential described through P for a given attracting mass ; let PQ be an element of the normal to this surface at P ; and let CQD be the surface of Fig. 245. uniform potential described through Q. Then if Fis the potential at P, and V that at Q, the resultant attraction at P in the direction PQ is the limit of the ratio V'-V or if the element of normal is denoted by dn, the resultant at- traction is ( IY dn in the direction in which the increments dV and dn are taken along the normal at P. As has been already mentioned (Art. 223), surfaces of uniform potential are also called Level Surfaces, or Surfaces de Niveau, the appropriateness of this name depending on the fact that no work is done against the acting forces in displacing a particle in any manner whatever on such a surface. 250.] Change of Attraction in passing through an Attract- ing Shell of Small Thickness. It will be proved (see Example 4), that the attraction of a circular plate of uniform density (p) and thickness (/) on a unit mass placed on the perpendicular to the plate through its centre is 2-TT^p (l cos a), where a is the semivertical angle of the cone whose vertex is the attracted particle and whose base is the plate. Hence, if the particle is very close to the plate, the attraction will be since a is sensibly a right angle ; and this result is independent of the radius of the plate. 410 ATTRACTIONS. THEORY OF THE POTENTIAL. [251. Let P and Q (fig. 246), be two points on the normal at op- posite sides of an attracting surface of small thickness, and consider the separate attrac- -^ tions of a small circular plate in the vicinity Fi 6 of P and the remainder of the surface. The attraction of the latter portion will be sensibly the same at P as at Q ; and by what precedes, the attraction of the plate at P will be a force Zirkp in the sense PQ, and at Q an equal force in the opposite sense. Hence it is evident that the whole attraction at P is the resultant of the whole attraction at Q and a force equal to lirkp along the normal from P towards Q, where k and p are the thickness and density of the shell at P ; so that if the attraction at Q is zero, the attraction at P is 1-nJcp. 251.] Lines and Tubes of Force. If the element PQ (fig. 245) be indefinitely prolonged in such a manner as to be at all its points normal to the level surfaces which it meets, it becomes what Faraday called a Line of Force, which may therefore be defined either as a curve intersecting perpendicularly all the level surfaces, or as a curve at every point of which the resultant force is directed along the tangent to it. If a superficial element of the level surface at P is taken, and lines of force are described along the contour of this element, these lines form a tubular surface which is called a Tube of Force. As a simple example let us consider the level surfaces of a uniform bar (Art. 244). Since F= Tcp log , if F is constant, ju is constant, or the axis /x c major of the ellipse whose foci are the extremities of the bar is constant. Hence the level surface at P is an ellipsoid of revolution round AB ; and since the curve drawn through P cutting at right angles a series of confocal ellipses in the plane of the figure is a hyperbola whose foci are A and B, the lines of force are hyperbolas confocal with the ellipses. Moreover, since the resultant attraction at P is normal to the ellipse through P, we see that the, attraction of the bar AB on a particle at P bisects the angle APE. Again (case (l), Art. 244), the level surfaces of a magnet whose magnetism is supposed to be concentrated in equal and opposite quantities at its poles are given by the equation = const. r i r* They are obviously surfaces of revolution round the magnet bar, 252.] SURFACE-INTEGRAL OP NORMAL ATTRACTION. 411 generated by the plane curve whose equation is the above. One of these surfaces is a plane bisecting NS perpendicularly, and the lines of force are the magnetic curves, one of which is represented in p. 39. 252.] Surface-integral of Normal Attraction. Let any closed surface be described so as to contain an element, dm, of attracting matter completely inside it, at a point ; and let the attraction of this element on a unit mass imagined to be placed at P, any point on the surface, be resolved, either constantly inwards or constantly outwards, along the normal to the surface at P and then multiplied by dS, an element of the surface at P. The integral of this taken all over the closed surface is called the surface-integral of normal attraction. To find its value, let OP = r, and let be the angle between OP and the normal at P measured towards the interior of the closed surface. Then the normal attraction on a unit mass at P is 9- cos 0*. Hence the surface-integral of normal attraction measured outwards from the surface is 'cos d r , If a sphere of unit radius is described round 0, and if lines drawn from to the contour of dS intercept a portion 'of the surface of this sphere equal to da>, it is well known that cosOdS = r 2 da. Hence the surface-integral becomes the integration being performed over the whole sphere since is completely surrounded by the closed surface. But /Y/o> = surface of sphere of unit radius = 4 TT ; therefore, the surface-integral is which is independent of the position of dm. Hence if n denote the magnitude of the normal attraction of dm on a unit mass imagined at P, fndS = 47T dm; and if the surface inclose any quantity of attracting matter whose mass is M i , we have, denoting by N the normal compo- nent of its attraction on a unit mass imagined at P, (1) * Or rather this multiplied by the unit factor, as explained in Art. 242. 412 ATTRACTIONS. THEORY OF THE POTENTIAL. [253. If there is repulsion between the mass inside and the unit mass on the surface, we b&vefNds = -}-^TrM i , ^being still measured outwards. We shall now suppose that the element dm is completely outside the closed surface. From draw a right line OPQ meeting the surface in P and Q ; let OP = r 19 OQ = r 2 , t QPa = 19 Z.PQm = 2 , Pn and Qm being the p. 2 ? normals to the surface at P and Q. Then, if ^ is the element of surface at P, ^o> the element of surface intercepted on a sphere of unit radius described round as centre by lines from to the contour of dS 19 and dS 2 , the corresponding element of surface at Q, we have CO B0 1 d8 1 = tfda, cos 2 ^$ 2 = r 2 2 da> ; , cos 9 _ ft cos 0, _ rt hence r^i r * iSSj = Oj but the expression on the r 2 r i left-hand side of this equation, when multiplied by dm, is the sum of the normal attractions at P and Q, each measured in- wards and multiplied by the corresponding superficial element. Hence if any closed surface is described in such a manner as to include none of the attracting matter fNdS = 0, (2) the integration being performed over the whole of the closed surface. In the figure we have represented a line from as meeting the surface in two points only ; but since the surface is closed, a right line must meet it in an even number of points, and it is evident that the elements of the integral considered destroy each other at the points where the line meets the surface, as in the above figure. 253.] Surface-integral for a Tube of Force. Let PA QB represent any portion of a tube of force, P and Q being elements of two level surfaces intercepted by the tube. Then the atttaction on a unit mass at P is Fig. 248. normal to the section P, and the attraction on a unit mass at Q is normal to the section Q, while at every point, A or , on every portion of the lateral surface of the tube the attraction is wholly tangential to the surface. Let F be the force at P, IF that at Q, and o> and a/ the areas of the sections P and Q. Then, supposing that the tube 253-] SURFACE-INTEGRAL OF NORMAL ATTRACTION. 413 contains none of the attracting matter, equation (2) of last Article gives JFYa -FV= 0, (1) since the only portions of the closed surface PAQJ3 which con- tribute elements to the surface-integral of normal attraction are the sections P and Q. Hence, at all points in empty space on a given line of force the resultant attraction on an imagined unit mass is inversely prop or- tional to the normal sections of the tube of force at these points. This simple theorem gives the law of attraction very readily in certain cases. For example, let the attracting body be a sphere whose density is the same at the same distance from its centre. Then the lines of force are obviously right lines drawn from its centre ; the tubes are therefore cones whose vertices are the centre, and since the normal sections of these cones are directly as the squares of their distances from the centre, the attraction of the sphere at any external point is inversely proportional to the square of its distance from the centre. Again, let the attracting body be an infinite cylinder whose density is the same at the same distance from its axis. Here the lines of force are right lines emanating from the axis per- pendicularly, the tubes become wedges, and the areas of their normal sections are directly proportional to their distances from the axis ; hence the attraction of an infinite cylinder at an external point is inversely proportional to its distance from the axis. Finally, for an infinite attracting plate, the tubes are cylinders and the attraction is constant at all points in empty space. These elegant applications of equation (1) are given by Thomson and Tait (Nat. Phil.). If the tube of force contain within it a quantity of the attracting matter whose mass is dq, we have by (1) of last Article JFo> - JV = 4 TT dq. (2) This equation can in like manner be employed to find the re- sultant force inside a sphere, a cylinder, or a plate. In the case of a sphere of uniform density, let the tube be contained between the spheres of radii r and r+dr. Then dq = p&dr, p being the density at the attracted point, and (2) becomes d(F<*) = lirpudr, or 414 ATTRACTIONS. THEORY OF THE POTENTIAL. [254. since to is proportional to f 2 . Integrating this last equation, Fr* = -irpr* + C. 4 Now Fis evidently zero at the centre, therefore C = 0, and . For a point inside an infinite cylinder at a distance r from the axis we have, since o> is ultimately a rectangle of breadth pro- portional to r, d(Fr) In general, if the tube is terminated by two level surfaces whose distance measured along the lines of force forming the tube is dsj we have dq = pw ds, and (2) gives for the determina- tion 254.] Equations of Laplace and Poisson. We propose to prove that if V is the potential of an attracting solid mass on a unit mass at the point P (#, y, z) we shall have d 2 7 d 2 7 d 2 7 according as there is not or is some of the mass at P. The first is Laplace's equation and the second Poisson's. At P draw a small parallelepiped whose edges, measured from P, are dx, dy, dz, and find the surface-integral of normal at- traction over the surface of this parallelepiped. The normal force, N, measured outwards from the surface on the face dy dz dV passing through P is y- dy dz ; the normal force on the opposite face is ( - f- -^ dx) dy dz ; and the sum of these two ftiy \f MX is y-g- & x ty fa- Similarly, the sums contributed, fNdS, by the d 2 F d 2 7 faces dz dx and dx dy are -=-^- dx dy dz and -=-^ dx dy dz. Hence if there is no mass inside the little parallelepiped we have ftty ^Y d 2 7 -j-p + -j-2 + -j-% =0. If there is mass, equal to pdxdy dz, we have by equation (l), Art. 252, d 2 F 254-] EXAMPLES. 415 In this very simple way we may also find the differential equation in polar co-ordinates satisfied by V. Take as the surface over which the normal attraction is integrated the polar element of volume msqt (fig. 208, p. 280). Let the polar co- ordinates of the point s be r, 6, $ ; let the normal force on the face msql>e denoted by R and the area of this face by ^. Then this face will contribute the term Es l to the surface-integral; and the opposite face (through t) will contribute Rs l -\ ^ *' dr ; fj I 7? \ therefore these faces give conjointly ~ dr. Let the normal dr forces on the faces ms t and tsq be T and S, and the areas of these faces s 2 and s 3 ; then the first and its opposing face will furnish the term ^ 2 ' dO, and the second and its opposing face , , 2 ln j Hence dr+ de + -d* = 0, or = - according as there is not or is mass inside the element of volume. -XT T? AT T IdV 1 dV Now jft = > r = - -=-r S = : - - ; dr r dO r sm d(p s l = r* sin 6 dO d$, s 2 = r sin dr d<$> 3 s 2 = r dd dr ; and the differential coefficients are, of course, all partial. Hence, if m i is put = p r 2 sin 6 dr dO d<[>, p being the density at c', we have . A d , 9 dFx d , . dT. 1 d z 7 sm^--(r 2 - r ) + (sm0 ) + -P- - =0, dr\ dr' d6\ dd' sm ci 2 or = d 2 d 2 d 2 It is usual to denote the operation -j-^ + -^ + -~- by the , T __ d% 2 dy* dz 2 J symbol v. EXAMPLES. 1. To find the attraction of a magnet on a magnetic particle whose distance from the centre of the magnet is very great compared with the length of the magnet. Let NS (fig. 35, p. 39) be the magnet, and suppose equal and opposite quantities of magnetism, m and m, to be concentrated at 416 ATTRACTIONS. THEORY OF THE POTENTIAL. [254. its poles N and S, respectively. Then, assuming a quantity \i at any point P t whose distances from N and S are r l and r 2 , respectively, Let be the centre of the magnet, OP = r,ON = a, an Then r* = r 2 - 2 ar sin A + a 2 , OP = r,ON = a, and PON = - -A. and rejecting ( -) > we have rj * r r . . Hence r = - s sin A. If N is the attraction in the direction PO, we have N = -- > or or . sin A. r dv If ^ is the attraction on P perpendicular to PO, we have T = - } raA since the element of a circular arc at P whose centre is is rd A ; 2mjua .'. ^ = f cos A. r 3 If the direction of the resultant attraction at P makes an angle - i with OP. we have 2 tan ^ = 2 tan A, the well-known equation which expresses the magnetic dip (i) in terms of the magnetic latitude (A), on Biot's hypothesis of terrestrial magnetism. 2. To find the attraction of a uniform bar on a particle. First, by the method of potentials. It has been shown (case (2), Art, 244) that ..,, . (i) and also that the resultant attraction acts in the bisector of the angle APS (fig. 242). Let ds be an element of this bisector at P. Then the attraction is (TV -j~ ; and from (1) dV 2Jcpc dp ds * p?-c* ds' dT Now if 2 = /.APB, and AP = r, we have ~ = cos <. Also 2/x = r + r', and since along the bisector, or, in other words, along a hyperbola through P confocal with the ellipse whose axis major is 254-] EXAMPLES. 417 2 ju, r / is constant, we have dr = dr, therefore -3- = - cos < as ti6' Hence = = 5- cos . c?s ft 2 tf* r Again in the triangle APS, cos $ == A/ ^ ^- 5 by an elementary formula in trigonometry. Therefore -7- = , ; and if y is the perpendicular from P on AB, we have 2cy r/sin2^>; therefore finally, which is the resultant attraction. Secondly, by direct calculation. The attraction of the element kpds Jcpds kpd\l/ at M on a unit mass at P (fig. 242) is ^^ > or (if Z J/PO = \^) > where PO = y. Eesolve this along and perpendicular to PO, and denote the components of the resultant attraction in these directions by Y and X, respectively. Then Y = / cos \l/d\j/ = (sin a + sin j3), y J-fi y X= smtid\l/ = (cos/3 cosa), y J-JB y where a = Z^P<9, /3 = Z5PO. From these values we have where which shows that the direction of R hisects the angle APB. 3. If a circular arc of uniform thickness and density (equal to those of the bar) is described with P as centre, touching the bar at and terminated by the lines PA and PB, the attraction of this arc at P is the same in magnitude and direction as the attraction of the bar at P. For, draw PN to a point N on AB very near M, and let PM and PN meet the circular arc in ra and n respectively. Then the attraction of the element MN on P is kp ~pM*' From ^ let fal1 ^ perpendicular to PN. Then MQ _MQ.PM _ hence the attraction of the element E e 418 ATTRACTIONS. THEORY OP THE POTENTIAL. [254. MN of the bar is equal to that of the element mn of the circular arc. Therefore, &c. 4. To find the attraction of a circular plate of uniform density and small thickness on a unit mass placed anywhere on an axis through the centre of the plate perpendicular to its plane. Let z be the distance of the attracted particle, P, from the centre, 0, of the plate. Divide the plate into an infinitely great number of circular rings whose common centre is ; let r be inner radius of one of these rings, dr its breadth, a the radius of the plate, k its thickness, and p its density. Then the potential of the ring at P is 2irkprdr _ since each particle of the ring is at a distance Vz*+r z from P. Hence F= 2-nkp and the attraction = -- = Zirkptl Let a be the semi-vertical angle of the cone whose base is the plate and vertex P. Then z = a cot a, and the attraction is 2 7T&/>(1 cos a). The same result follows easily by direct calculation. For if is the angle made with PO by lines drawn from P to the circumference of the ring of radius r, the attraction of this ring resolved along PO is 2-nkprdr T- 2 cos ; but r-=z tan0, therefore this expression=2T7&psm Odd, the integral of which from = to 6 = a is Zirkp (1 cos a). 5. To find the attraction of the frustum of a right cone on a particle placed at the vertex of the complete cone. Let the frustum be divided into an indefinitely great number of circular plates, each of the thickness, dz. Then since a is the same for all the plates, and k in the above case is now dz, rh whole attraction = 2irp (1 cos a) / dz, J h' where h and h' are the distances from P of the faces of the frustum. Hence the attraction is 27^-^(1 -cos a), and the attraction depends merely on the thickness, hh', of the frustum and not its proximity to the attracted particle. This remarkable proposition is true also in the case of an oblique cone standing on any plane base whatever, the attracted particle being at its vertex. To prove this we have merely to show that if two plates of the same thickness, each parallel to the base, be taken anywhere in the cone, these plates exert equal attractions at the vertex. 254-] EXAMPLES. 419 Through the vertex, P, draw an infinite number of rays forming a very slender cone intersecting the two plates in two small similar elements of surface, dS and dS', at the points M and M', suppose. Then the attraction of dS on P is --^ > k and p being the thickness Z kpdS' and density of the plate ; and the attraction of dS f on P is . These attractions are in the same line, PMM' \ and since the contours of the elements dS and dS' are similar curves, -77^ = nil/r ,<> ; therefore ab rM the attractions of these elements on P are equal. Similarly for all other corresponding elements of the plates; therefore the plates attract P equally. The attraction of any frustum on P depends, then, only on the number of plates of given small thickness in the frustum, i. e. on the thickness of the frustum. 6. To find the attraction of a spherical shell of uniform density and small thickness on an external particle. The potential has been proved (Art. 244) to be - > or . o F= - Also the attraction measured in the direction in which c dV c increases (Art. 248) is > therefore the attraction towards the dV d A centre is -j- > or *vpr or 4 that is, the attraction of a sphere on an internal particle varies as the distance of the particle from the centre. 8. To find the attraction of a circular plate of uniform thickness and density on a particle in its plane, the law of attraction being that of the inverse cube of the distance. E e 420 ATTRACTIONS. THEORY OF THE POTENTIAL. [254. From P, the attracted point, draw two very close radii vectores intercepting a narrow strip of the plate between them. Let be the centre of the plate, let Q be the angle OP A made by one of the radii vectores, and let 6 + dO be the angle made by the other, with OP. Let Q be a point on PA, and PQ = r. Then the mass of the ele- ment at Q included between circles of radii r and r + dr described with P as Fig. a 49 . centre is kprdrdO, Jc and p being the thickness and density of the plate. The attraction of this element on P resolved along PO is kpdrdd J V~~ C S ^ hence the resultant attraction is 'drde COS0, the integration in r being performed from r = PA to r = PB, and that in 6 from = sin" 1 - to = sin" 1 - > where a is the radius of c c the plate and c = OP, the extreme values of corresponding to the two tangents that can be drawn from P to the circle. Now denoting PA by r^ and PB by r 2 , and integrating first with respect to r a , the attraction is The values of ^ and r 2 are given by the equation r 2 2crcos0 + c 2 a 2 = 0, _!_ J_ 2\/a 2 -c 2 sin 2 ^" = c 2 -a 2 Hence the attraction is where is put for c sin 6. In this case we might have found the attraction from the Potential. The latter is easily found by dividing the plate into rings with as centre. If r is the radius of one of these rings we have V = = to = , and - doubling the result we have V = nkp I 2 ^ j in which r runs from Otoa. Hence F=^log Q> 254-] EXAMPLES. 421 But F may also be easily found from the attraction, thus, _ do = c(c 2 -a 2 )' Tr irkp . c 2 V = -~- log -^ - ^ -f const. J c & Now, since F= - f , it is clear that at infinity F= 0, or F= when c = oo . This gives the const. = 0, 9. If V n and F^ denote the potentials of an attracting mass when the law of attraction is the n^ and (n 2)& power of the distance, respectively, prove that T7 F n -~ d* d 2 d 2 where V + + > the co-ordinates of the attracted particle being x, y, z. We have F B = -- ^r. w+ 1 therefore, as in Art. 246, and Adding to this the similar values of -=- and -r-j > we have This equation enables us, generally, to find the potential for the (n 2) th power of the distance when that for the n th is known; but it fails in two most important cases, namely, when n = 1 and when If the attracting mass is a plate, r 2 = (xyff + tyy'y*, and the result is easily proved to be In the last example we find the potential of a circular plate for the inverse third power ; hence we have at once the potentials, and there- fore the attractions for the inverse fifth, seventh, &c., powers of the distance. 10. Calculate the attraction of a uniform spherical shell of small thickness on an external particle when the attraction varies as the n^ power of the distance. With the notation and figure of case (3), of Art. 244, we have * This equation, as well as that in Ex. 19, is Mr. Jellett's. 422 ATTRACTIONS. THEOEY OF THE POTENTIAL. [254. n+1 9-rrnrn C* as- p I (c 2 -2ccos0 + w+1 Jo +! This integral is of the form / x 2 c?a?. Hence If we wish to find the attraction of a full sphere of radius r, we observe that T is da, and we integrate this expression from a to to a = r. In each case the attraction towards the centre is -- = dc 11. Let there be two distributions of mass denoted by M and M l '; if at any point in M, where the element of mass is dM, the potential due to M' is denoted by V, and if at any point in M', where the element of mass is dM ', the potential due to M is denoted by F, we shall have /V'dM = fVdM', the integration (or summation) on the left-hand side being performed throughout the mass M, and that on the right-hand side through- out M'. For the element V-dM of the first integral means that all the elements of M' are to be multiplied each by the element dM at a fixed point in M, and each of these products is to be divided by the distance between dM and the corresponding element chosen in M' ; nn&fVdM means, therefore, that the elements of the mass M are to be combined in pairs in all possible ways with the elements of M' and each product divided by the mutual distance of the two elements in it. But this is manifestly also the meaning of f VdM'. Of course this is also true if the elements are multiplied by any function of their mutual distance, and it is also true whatever elements may be denoted by dM and dM' they, one or both, may be elements of space, for example. Hence the mean potential over a spherical surface due to matter entirely outside the sphere is equal to the potential of this matter at the centre of the sphere. (Gauss, Papers on Forces varying inversely as the square of the distance, Taylor's Scientific Memoirs, vol. iii. part x.) For let mass of uniform density p and small uniform thickness r, be supposed to be distributed on the sphere ; let dS be an element of its surface, V the potential at this element of the attracting mass, and a the radius of the sphere. Then since the potential of a shell at an external point whose distance from the centre is r it follows that if dm is an element of the attracting matter 254.] EXAMPLES. 423 p T f VdS = 47r / ora 2 /? = 47rpra 2 F , if F is the potential at the centre of the sphere. Hence fVdS _ "- which proves the proposition, since f\ r ds divided by the whole surface of the sphere is the mean value of the potential over its surface. 12. Given the whole mass of a solid, find its shape so that its attraction on a particle placed at a given point may be a maximum. It is clear that the surface of the solid must pass through the attracted point, and a little consideration shows that the component of the attraction of any element of the solid along the direction of the resultant attraction must be constant. Hence the surface of the solid is one of revolution round the line of action of the resultant, and the equation of the generating curve is COS0 1 9 = z- = constant, r 2 a 2 the attracted point being pole, and the line of action of the resultant the initial line. Hence if R be the whole attraction, and / the unit force (Art. 242), I Jo ==!**> y ) will also satisfy it (where r 2 = or J + 2/ 2 + 2 2 ). If is the origin, P the point (#, y, z), Q a point on OP produced such that #$ = 777,5 the co-ordinates of Q are 3-5 ^ -=- Let C/jT V f* 7* OQ = p, let (f, TJ, Q be the co-ordinates of 0, and let Then satisfies the equation But p 2 - = a 2 - ; therefore this equation becomes aa ar A d . . A dU. 1 ramd , + (sing -+ - = 0. x The first term being the same as sin -r- (r z ) , this equation is, by Art. 254, the equivalent of _ dx* d*cW = 15. A homogeneous fluid, self-attracting according to the law of nature, completely fills the space between two spherical non-concentric 254-] EXAMPLES. 425 surfaces one of which entirely surrounds the other ; find the resultant attraction at any point of the fluid, and also the level surfaces. Let be the centre of the larger and 0' the centre of the smaller sphere ; P any point in the fluid ; 00' = c ; radius of smaller sphere = b ; OP r,0'P = r f ', p = density of fluid. To calculate the resultant force at P, imagine that the place of the smaller sphere is occupied with fluid ; then the larger is completely full, and there is a force ^Ttpr in the line PO towards 0. Now let the effect of the fluid which we have introduced be annulled by com- bining with the above force the force exercised at P by a repulsive fluid of same density filling the smaller sphere. This latter force would be 7^ on the scale adopted; and this would act in the line O'P from 0'. The resultant of these forces is the resultant force at P. If V is the potential at P, r', [Art. 243] 3 o T 2 47TP& 3 .-. F = -- npr* -- ^- + const. This value is otherwise evident, since the potential at a point due to any attracting bodies is the sum of their separate potentials at the point. If a is the radius of the larger sphere (see p. 404), 2 4-irpb* V = - The level surfaces are given by the equation 26 3 r 2 H T = const. 16. "Whatever maybe the law of attraction, prove that the attraction of the smaller of two concentric spheres at a point situated on the surface of the larger is to the attraction of the larger at a point situated on the surface of the smaller as the square of the radius of the smaller is to the square of the radius of the larger. Draw any radius meeting the surface of the smaller sphere in q and the surface of the larger in Q. Let /'(r) represent the law of attrac- tion ; let a = radius of larger, b = radius of smaller, and be their centre. Then the attraction at Q of an element of the smaller at the point xvz is JV / \ U - f(r) along QO, which is taken as axis of x. Performing the integration with respect to x, considering y and z constant, the attraction at Q of a thin prismatic bar (parallel to OQ) of the smaller sphere is 426 ATTRACTIONS. THEORY OP THE POTENTIAL. [254. where r^ and r 2 are the distances from Q of the ends, A and B, of this bar. Draw OA and OB meeting the surface of the larger sphere in A' and B' y respectively ; then A'B' will be the axis of a prismatic bar of the larger sphere whose attraction at q is (y' and s being co-ordinates of A' or B*) since qA' = QA = r lt and qB'=QB = r 2 . But ^ = 4 = - 5 therefore -^ = ^; therefore y z a dydz a? the attractions of these corresponding bars bear to each other a constant ratio, and hence taking all such bars for the two spheres, the proposition is proved. 1 7. From the last proposition prove that the only law of attraction for which a homogeneous spherical shell of uniform thickness will exercise no attraction at any internal point is the law of nature. For if such a shell exercises no internal attraction, it follows that the matter contained between the surfaces of the two spheres exercises no attraction at q, however great OQ may be. Hence however great OQ may be, the attraction of the greater at q is constant ; therefore the attraction of the less sphere at Q oc -$ Q. E. D. 18. To express the amount of work done by the mutual attractive forces of the particles of a self-attracting solid when the body changes from one figure to another. Let m lt m 2 , ... be any elements of the solid, and r 12 , r 13 , &c. their mutual distances. Then in the alteration of the distances between m^ and the other elements the work done on m l is , .Win T??/,, x or Wi( f(_2 + _J +...), Qr where Fj is the potential of the whole mass at the position at m lt Hence the work done on m l is m 1 (F/ 7 F/), where F/' denotes the potential at m l in the final figure of the solid, and F/ the potential in the initial figure. Similarly the work done on m 2 is m 2 (F 2 7/ F/) ; and since the term * 2 2 dr 12 is common to the expressions for the work on m 1 and the work on m 2 it is clear that whole work is expressed by l/ ran this integral extending over the whole solid in its first and second figures, and the first result being subtracted from the second. This expression is the potential work of the internal forces of the attracting solid. 1 9. If R n and 7? w _ 2 denote the resultant attractions of a given solid at a given point when the law of attraction is that of the n th power, 254-] EXAMPLES. 427 and that of the (n 2) th power, of the distance, respectively, prove that -( 20. Find the attraction of a circular plate of uniform thickness and density on an external particle of unit mass in its plane, the law of attraction being that of the inverse distance. A ns. The mass of the plate divided by the distance of the particle from its centre. 21. Prove that if a material lamina attract according to the law of the inverse distance and if N is its attraction on a unit mass at any point of a closed curve, measured outwards along the normal, we shall haVe fNds = 0, Or = -27im i) according as there is no mass or mass m i inside the closed curve, and hence that VF = 0, or = 27r/o. 22. Prove that the values of V F calculated for external points and for internal points do not agree for points on the surface of a solid sphere. 23. Prove that neither Laplace's nor Poisson's equation holds for points on the bounding surface of an attracting solid. 24. Find the attraction of a uniform hemispherical shell of small thickness on a unit particle placed at a distance, x, from its centre on the diameter perpendicular to the plane of the rim of the shell. Ans. If r is the radius of the shell, p its density, and T its thickness, the attraction is - % (1 -- . )* the unit of force x V r^ + x 2 being that between two units of mass at a unit distance apart. 25. If a number of uniform bars of the same section and density form any closed polygon with no re-entrant angle, prove that they produce the same potential (for the law of the inverse square) at any point inside the polygon as a polygon of bars formed by joining the feet of the perpendiculars from the given point on the sides of the given polygon. Extend this proposition to any curve. (See Case (2), Art. 244.) 26. If a self-attracting sphere of uniform density and radius a changes to one of uniform density and radius a', find the amount of work done by its mutual attractive forces. Ans. The unit of work being that done when two particles, each of unit mass and placed at a unit distance apart, are drawn to an infinite distance apart, the work done will be !*.-> where M is the mass of the sphere. 27. Two equal uniform bars of given sections and densities are placed parallel to each other and at right angles to the lines joining 428 ATTRACTIONS. THEORY OF THE POTENTIAL. [254. their extremities ; find the amount of work done against their mutual attraction in drawing them a given distance asunder. Ans. If y is the distance between the bars in any position, I the length of each, m and m are their masses, and the unit of work is the same as in last example, the work done in changing the distance from y^ to y 2 will be the difference of the values of the expression y when y t and y a are successively put for y. 28. The gravitation potential of an attracting mass cannot have a maximum or minimum value in empty space. [Let it have a maximum value at A. Then round A, and in- definitely near it, can be described a closed surface, at every point of which V is less than it is at A. Therefore if dn is an elementary length along the normal (measured inwards) to this surface, dV /** is positive all over the surface ; but N = -7 ; hence equation (2), Art. 252 is contradicted.] 29. A particle in equilibrium under the attraction of any system of masses (for the law of nature) is in unstable equilibrium. (This follows from last example. See Art. 199. See also Clerk Maxwell's Electricity and Magnetism, vol. i, p. 139. The Theorem is known as Earnshaw's.) 30. If a level surface contath none of the attracting mass, the potential is constant throughout its interior, and equal (of course) to that on the surface. (Gauss, in Taylor's Scientific Memoirs.) For if not, it must have either a maximum or minimum value at some point within. This very simple proof is given by Thompson and Tait, Nat. Phil. 31. If all the attracting mass lies on or within a level surface on which the potential is zero, then in all space outside this surface the potential is constantly zero. (Gauss. ) [If possible, let the potential at any external point, P, be A, which is > 0. Then, since lines drawn from P to the given level surface meet it in points of zero potential, it is possible to find a series of points on these lines at which the potential has a constant value, < A and = B, suppose. Also since the potential is zero at all points at infinity, it is evidently possible to describe round P a closed level surface on which the potential = B, and which includes none of the mass. This surface is subject to the result of last example, which contradicts our supposition. Therefore A cannot be > ; and by changing the sign of every mass in the system, the supposition that A is negative may be rejected ; therefore A = in all external space.] 32. If all the attracting mass lies on or within a level surface, then in all space outside this surface the potential is less than on the surface, and has the same sign. 255-] SHELL BOUNDED BY SIMILAR SURFACES. 429 33. If in any portion of empty space of finite volume the potential has a constant value, it will have this value throughout all space, which can be reached without passing through any of the mass. (Gauss). SECTION II. The Attraction of Ellipsoids. 255.] Shell bounded by Similar Surfaces. Let vr'p' and rqp be two concentric, similar, and similarly situated surfaces whose normal distance from each other is at all points very small. Sup- pose the space between these surfaces to be filled by attracting mat- ter of uniform density, and let be an at- tracted particle in the interior of the shell. F S- 2 5- With as vertex let any slender cone be described, intercepting on the shell two frustums whose thicknesses measured along the generator pr of the cone are pp' and rr'. Then, since by the property of similar, similarly situated, and concentric surfaces, the intercepts pp' and and rr' are equal, we see by example 3 of last Article that the attractions of these frustums on are equal and opposite. Hence the corresponding frustums of all such cones exert equal and opposite attractions on ; and the resultant attraction of the shell on any internal particle is therefore zero. Hence, generally, if the law of attraction is that of nature, every shell of uniform density and small thickness, bounded ly similar ', similarly situated, and concentric surfaces produces a con- stant potential at all points in its interior, and exerts, therefore, at these points no attraction. The same is true for a solid of uniform density and any thick- ness bounded by two similar, similarly situated, and concentric surfaces, since the thicknesses of the frustums intercepted bet ween its bounding surfaces will still be equal. 430 ATTRACTIONS. THEOEY OF THE POTENTIAL. [256. 256.] Corresponding Points on Confocal Ellipsoids. Let rqp and PQ (fig. 250) be two confocal ellipsoids, let the axes of the first be a', /3', y', and those of the second a, /3, y, let the co- ordinates of a point p on the first be of, y, /, and those of a point P on the second x, y, z. Then, if x _ x' y _ y' z _ / a = Z' J = J' X~7' the points P and p are called corresponding points on the ellip- soids. Also, let Q and q be two other corresponding points. Then it is very easy to prove that the distance Pq is equal to the distance Qp. (Salmon's Geometry of Three Dimensions -, Art. 181.) 257.] External Potential of an Ellipsoidal Shell. Let it be required to find the potential at an external point, P, of a shell bounded by the similar, similarly situated, and concentric ellip- soids v/jp' and rqp. Through the point P describe an ellipsoid, PQ, confocal with rqp, and describe also an ellipsoid, mm, con- focal with vr'p' and similar to PQ. This latter surface is completely determinate, since its axes must be pa, fj.fi, py, and since ^ (a 2 -/3 2 ) must be equal to ju' 2 (a' 2 -' 2 ), where /xV, //', /m'y' are the (given) axes of the ellipsoid vr'p' ; or /u = ju', since a 2 /3 2 = a' 2 -/3' 2 . Now, let ', r/', f' be the co-ordinates of any point q on the inner shell, and f , 77, f those of the corresponding point, Q, on the outer. Then if p is the density of each shell, the element of mass at p is pd'dr[d', and the potential produced by this element at P is But since ^7 = ^, &c., we have a a ap y therefore the potential of the element is apy Pq and the potential produced at p by an element of mass at Q is Qp And since Qp = Pq (Art. 256), potential at P due to element of mass at q potential at p due to element of mass at Q " a(3y _ mass of shell rqp ~ mass of shell PQ 258.] ATTRACTION OF AN ELLIPSOID. 431 Taking all the elements of the inner shell, and all the cor- responding elements of the outer, and thus exhausting both shells, we see that the potential of the inner shell at Q mass of inner shell the potential of the outer shell at p mass of outer shell Now since these shells are bounded each by similar surfaces, the potential of the outer shell is constant at all internal points, and (in virtue of the continuity of the potential) this potential is the same as the potential of the outer shell at P. Hence the potential of an ellipsoidal shell bounded by similar surfaces is constant at all points on the surface of any ellipsoid con focal with the surface of the shell that is, the level surfaces of an ellipsoidal shell are confocal ellipsoids, and its attraction at any point is therefore normal to the confocal ellipsoid through the point. Let Fand V be the potentials of the shells PQ and rqp at p ' then r'_ a '^' v. '~^/ y> and if #, y, z be the co-ordinates of P, we have dV ^a'PY dV , dx afiy dx hence the components of the attractions of the two shells in the same direction are to each other in the ratio of the masses of the shells. For this reason the calculation of the attraction of an ellipsoidal shell at an external point is reduced to that of a shell on its surface. 258.] Attraction of an Ellipsoid at an External Point. Let ABD (fig. 250) be a solid homogeneous ellipsoid, and let it be required to find its attraction on a unit mass placed at P. Break the ellipsoid up into an infinite number of thin shells bounded by ellipsoids similar to each other and to the surface ABD ; let one of these shells be that between the surfaces v/p' and rqp. Denote this shell by (s) ; and describe the ellipsoids PQ and msn, similar to each other and confocal with the surfaces of (s), as in the preceding Articles. Denote this shell by ( therefore P =* . ^. Bat -^ = ^ Q{ pq k + dk ., Ps dJc , _ pdk = r > therefore -^- = ~ 5 an ^ Pn = - Substituting this value in (1), We find the attraction of (o-) parallel to the axis of so to be Multiplying this by the ratio of the mass of (s) to that of ( C' are vari- ables, but if inside, constants. The whole force parallel to the axis of x on a unit particle is obviously X -(Z'cos a, + T cos 2 + Z'cos a s ), with similar expressions for the components along the axes of y and z. If the attracted particle is inside the cavity, the level surface F f 2 436 ATTRACTIONS. THEORY OF THE POTENTIAL. [258. passing through it is easily found. For, the virtual work of the attraction of the whole ellipsoid is Xdx + Ydy + Zdz, or \d (Ax* + By 2 and that of the attraction of the small ellipsoid is Z'daf+Tdtf+Z'ds/, or \d(A'vf* + Hence the level surfaces inside the cavity are given by the equation Aa? + By'* + Cz' i -A'x*-B'y'*-C'z z = const. They are therefore quadrics. We could in the same way find the effect due to an ellipsoidal mass which contains in its interior another ellipsoidal mass (or nucleus) of density different from that of the remainder. If p and // are the densities of the two portions (p' > p), imagine the whole to consist of a homogeneous mass of density p, and add the effect due to the nucleus, supposed of density p'p. 4. Prove that an oblate spheroid of uniform density cannot have its own surface for one of its level surfaces. [The condition that its own surface should be a level surface is tan^e = - 5 } which cannot be satisfied by any value of e, except zero.] 3 + e ' 5. Prove that a prolate spheroid of uniform density cannot have its own surface for a level surface. [By putting e = k*/ 1 in the last result, the required condition becomes !+& 3 which gives by expansion +... = 3, or ., O . O t> . i which is, of course, quite impossible.] 6. Prove that in the spheroid considered in example 2 the re- sultant attraction at any point on the surface is proportional to the length of the normal between that point and the axis of revolution. 7. Express gravity on the surface of such a spheroid in terms of the latitude. [The latitude of a point on the surface is the angle made with the plane of the equator by the normal at the point. If E denotes the value of gravity at the equator, G the value in latitude A, and e the eccentricity of the generating ellipse, so that if e is small, the increase of gravity at any point above the equatoreal value is proportional to sin 2 (latitude).] 8. The components of attraction of a homogeneous ellipsoid at an internal point (x, y, z) being Ax, By, Gz (as in p. 434), prove that A+B + C = 477/0, where p is the density at the point. 259-] QUANTITY OF ELECTRICITY. 437 SECTION III. Superficial Distributions. 259.] Quantity of Electricity. The student is supposed to be familiar with the elementary phenomena of electric attractions and repulsions that is, with certain forces which bodies are observed to exhibit when they have been rubbed with resin, catskin, and some other substances. The mode in which these forces are brought into existence is called the process of electri- fication, and the bodies which exhibit them are said to be elec- trified. In the older theories of electricity such bodies were assumed to have been charged by electrification with a certain quantity of fluid, the process consisting either of directly com- municating the fluid to the bodies, or of altering its arrange- ment within them (if they naturally possessed it themselves) in such a manner as to render possible the play of electrical forces. Whether this fluid theory is true or false, there appear to be at present some strong objections to its adoption. Following the views of Clerk Maxwell, we shall regard electrification merely as a state of a body, without speculating more closely as to the nature of this state. Suppose that two small electrified bodies of equal surface, acting in exactly similar circumstances on the same third elec- trified body, produce exactly equal forces of attraction or of repulsion on this body; then we say that the two bodies con- sidered have the same quantity of electricity; and if one of them attracts, while the other repels the third body, we should say that they have equal quantities of electricity with opposite signs. The phenomenal effect of electrification being, then, a measurable quantity, this state of electrification itself becomes also a measurable quantity if it is, as we assume it to be, fully represented by this effect. We have thus attained the notion of equal quantities of electricity, as equivalent to equal states of electrification. A unit quantity of electricity, in electrostatics, will therefore be that quantity which, when acting on an equal quantity placed at a unit distance from it, repels the latter with a unit force the unit force called the dyne, or any other convenient one. 438 ATTRACTIONS. THEORY OF THE POTENTIAL. [260. 260.] Electric Potential. An electrified body exerts force on other electrified bodies in its neighbourhood ; and, just as in distributions of matter, the potential at any point in space due to an electrified body is the amount of work which must be done against the repulsive force of the body to bring a unit of elec- tricity from an infinite distance to the point considered. (This supposes that the proximity of the unit does not modify the state of electrification of the body a supposition which will be allowable when the unit is small.) An electrified body is found by experiment to be capable of impressing its state of electrification, with more or less success, on bodies which are put in contact with it, according to the nature of these bodies ; and bodies which very readily allow this transference of state are called good conductors, or simply conductors^ while those which best resist it are called non- conductors, or dielectrics. We may speak of 2^ flow of electricity > instead of a transference of state, if we are careful to avoid including in the expression any hypothesis of the material nature of electricity. Where this flow can take place it will take place. It follows, therefore, that when a conductor is in complete electrical equi- librium there must be a uniform electric potential throughout its substance. And from this it follows by Poisson's equation that when a conductor is in electrical equilibrium, the electricity resides entirely at the surface. For since V is constant throughout the mass of the conductor, ^V ; and therefore p = 0. A distri- bution throughout the mass could exist only in a non-conductor. Also the external surface of the conductor itself must be a level surface of the electricity; for if not, a flow would take place from a point of high potential on it to one of low potential. In other words, no electrical force can be exerted anywhere in the conductor except at its surface of contact with a resisting (dielectric) medium. This is usually stated thus in an elec- trified conductor the electricity resides wholly at the surface. It is to be noted that in all cases the potential of the Earth is assumed as zero, and that the potential of any body in com- munication with the Earth by means of a conducting wire is therefore zero. 261.] Free Charge. Induced Charge. The quantity of electricity on a conductor is called its charge. A charge may be 263.] DENSITY AT EACH POINT. 439 communicated to a body in two ways either by actual contact with another charged body (such as a piece of glass which has been rubbed with catskin), or by the influence , at a distance, of such a charged body. In the first case the electricity on the conductor will be everywhere of one kind viz., the kind which is on the body touched ; and in the second the charge will consist of quantities of positive and negative electricity in two portions of the surface of the conductor. 262.] Surface Density. The electric density at any point of a surface is the limiting ratio of the quantity of electricity within a sphere whose centre is the point to the area of the surface contained within the sphere when the radius of the sphere is diminished indefinitely (Clerk Maxwell). We have already spoken of solid distributions of matter over a surface, meaning that the thickness of the material stratum is everywhere very small. If p denotes the density of the matter, and k its thickness, the quantity on a small unit surface is kp. If now we imagine p to be increased and k to be diminished indefinitely, we shall have a truly superficial distribution, in which the product fcp becomes the surface density here considered. Although this mode of conception may assist us in understanding a true superficial distribution, it is not necessary to imagine that electrical distribution is really produced in this way. We shall denote the surface density by (l) which determines the law of density at each point. Here N (being supposed repulsive) is of course j > where dn is the element of normal measured outwards. 264.] Torce Exerted by an Electrified Conductor on its own Electricity, At each point on the surface there is a certain force produced on a unit quantity of the conductor's own electricity, which, it must be very carefully observed, is not 7 1 dV equal to ^ dn For suppose ABD (fig. 250) to represent the surface of an elec- trified conductor, and take any very small element of its area at the point B. The repulsion of the remainder of the surface on a unit of electricity at B is the same as its repulsion on a unit just inside or outside B. This latter is 27T0-, as at once appears by the method of Art. 250. Also the action of the element at B on itself is zero. Hence the* resultant repulsion of the charge on a unit quantity at B is 2 TT cr, and it acts in the normal at B ; and if dS is the area of the small element at B, the quantity on it is adS' } on which the repulsion, dp, is 2n(T 2 dS. Hence dp = 2TT(T 2 dS, Or ^ = 27T(T 2 , ao which is the repulsion of the electricity on itself per unit of surface at a point where the density is a: This quantity, 2ircr 2 , is what Sir W. Thomson calls the electric diminution of air pressure on the surface (Papers on Electrostatics and Magnetism, p. 254), for the following reason : each element of surface of an electrified soap-bubble being repelled by the force 2iro- 2 per unit of surface, the bubble expands, just as it would do if the air pressure diminished, and, when discharged it contracts. Hence the electric diminution of air pressure at any point of a conductor is jya 27T(T 2 , or > 8 77 N being the electrical repulsion on a unit in the air just outside the point. EXAMPLES. 441 265.] Theorem. The sections of a tube of force made by the surfaces of two conductors, which mutually act on each other, contain equal quantities of opposite electricities. Let A and B (fig. 251) be two portions of the conductors whose adjacent surfaces, P and Q, are electrified, and let the tube of force contain the elements P and Q. Then in the substance of each conductor, no matter how thin it may be, F is a constant (Art. 260). Let the tube be prolonged to any distance in the conductors, and let it be closed at its extremities. Apply the surface integral of normal force to this tube, and let the areas cut off at P and Q be dS and dS* t and the surface densities at these points cr and c/. Now N (Art. 253) is zero all over the surface of the tube ; hence which proves the theorem. If the surfaces are very close together we may take dS = dS', and then we shall have cr = cr'. EXAMPLES. I. To find the quantity of electricity on each of two very close parallel plates, the potential of each plate being given. Let F and V be their potentials, and h the distance between them (or the thickness of the dielectric). Then the surface density at any point on either is where dn is the element of normal dV measured from the point towards the other plate. But -= is sensibly equal to = ; hence cr = ; and if be the surface of the h 47T/Z. plate and Q the quantity of electricity on it, If one plate communicates with the ground, its potential is zero, VS and Q will then be = > where V is the potential of the other. This case is approximately that of the Leyden Jar. 2. To find the work done by the electric forces in the discharge of a Leyden Jar. If one armature is connected with the ground (whose potential is zero) and the other with a source of electricity whose potential is F, VS the charge will be - Now on the armature whose potential is F, 442 ATTRACTIONS. THEORY OF THE POTENTIAL. [265. the potential work of the forces, or energy of the electrification, is (p. 426) y-rdq, or \Vfdq, or \VQ, where Q is the whole charge on the armature and dq the elementary charge at any point. Substituting for V its value in terms of Q, we have the energy equal to This is the work done in changing the potential from V to zero ; or, in other words, the work done in discharging the jar is pro- portional to the square of its charge. 3. To find the surface density at any point of an ellipsoidal conductor. If we regard surface density as the limit of volume-density (Art. 262) when the thickness is indefinitely diminished, it follows at once by Art. 258 that, as the normal distance between two very close concentric, similar, and similarly placed ellipsoids is proportional to the length of the central perpendicular on the tangent plane, = / U-j-dS- /( + -. +_)<$,> t a \ J J dn J \dx dx dy dy dz dz' where the integral on the left-hand side and the second integral on the right-hand side are taken throughout the whole of the space inside any closed surface, the element of volume of this space being denoted by da ; and the first integral on the right- hand side extends over the whole superficies of the given closed surface, the elementary length of whose normal measured out- wards is dn. For d< = dxdydz, and frr fff J UvYd =JJj 444 ATTRACTIONS. THEOEY OF THE POTENTIAL. [266. Integrate U ' -^ dxdydz with respect to #, considering y and z as constant, the extreme values of x belonging to the two (or any even number of) points, p l and p 2 , in which the surface is intersected by the line along which y and z are constant. The contribution to the integral on the left-hand side made by a long and slender parallelepiped parallel to the axis of SB will Fig. 252. f i then be dY dV - dUdY , , where the suffixes denote the values of the quantities in brackets at the points p 2 and p v Now if dS 2 is the element of surface cut off by the parallele- piped at p 2 , and if A 2 is the angle (represented by dotted line) made with the axis of a? by the normal measured outwards at j? 2 , cos A 2 . dS 2 = dydz. Similarly, if dS 1 and X 1 (measured in the same sense as A 2 ) denote these quantities at p^_ , cos A x . dSi = dydz. Hence (l) becomes 7 -WVK/ i\.w"~f i | i v> i \sw VM//^ I ^~Cvytv& j Z Z Cvtfc 1 and hence '^S^ = /"^ cos ^-^/'//??*4r^ ( 2 ) A denoting the angle made by the normal at any point with the axis of x. In the same way, if // and v are the angles made by the normal with the axes of y and z, we have r rr d z r _ r AV rrrdUdr . , . U^du = / U-j-cospdS- 1 l-j r dxdydz } (3) J dy 2 J dy JJJ dy dy Adding (2), (3), and (4) together, we obtain the equation (a). COR. Writing down the value of fFVUdv, and subtracting the result from (a), we obtain 268.] CASE OF GKEEN'S EQUATION. 445 a remarkable and very useful equation in which the volume integral on the left-hand side is changed into the surface integral on the right. 267.] Case of Green's Equation. In (a) of last Article let V be the potential of an attracting mass and let U = V. Then we have, since V7 7 ~ = 4-77/5 at every point inside the volume where there is mass (and V V at points where there is no mass), = fr M- J an where 72 2 is put for (-7-) + (-7 ) + (-7) , the square of the v dx ' \oy'' ^ dz ' resultant force on a unit mass at the element cl CD of volume. Hence f R*d = \ V ^~ dS+4v f 7 P du, (A) J J Cln J an equation which will be found useful. 268.] THEOREM* If on a conductor , removed from the influence of all electricity except its own, the total quantity of electricity is zero, the only possible distribution of the electricity is one in which the density is zero at all points i.e. the conductor must be in its natural state. For if the conductor be not in its natural state, it will have a potential V different from zero ; and it will be possible to describe round it and completely enclosing it a surface on which the potential has a constant value #, less than V suppose. (For on lines drawn in all possible directions from any point on the conductor we can find points at which the potential = a.) Denote this surface by A, and apply the equation (A) of last Article to the surface of A and the volume enclosed by it. This equation becomes since wherever inside the surface A there is mass, the potential is constant the mass existing only on the surface of the conductor. = charge on the conductor = ; and f^-ds=- (ms J dn J (N being the electrical repulsion on a unit of + electricity) = 4:r x charge on conductor = 0. Hence fl&d<& = 0, i.e. E = at every point inside the surface A ; but since at each 446 ATTRACTIONS. THEORY OF THE POTENTIAL. [469. p point on the conductor the density = > at each point the density is zero, i. e. the conductor is in its natural state. In the same way it can be proved that in a system of insulated conductors placed in given positions, if the total charge on each of them is zero, the only possible distribution is one in which each conductor is in its natural state. For if possible let there be a distribution in which the poten- tials on the conductors are, in order of descending magnitudes, V L9 T 29 Fg, ... . Then it is evidently possible to describe round the conductor whose potential is /^ a closed surface which will not meet any of the other conductors and on which the potential has a constant value, a, < F lt and > F 2 . Applying equation (A) to this surface and its enclosed volume, we have s.fl f^-dS+ivF! fpdu. J dn J Now fpdu> = 0, by hypothesis ; therefore, as before, E 0, and the first conductor is in its natural state. Proceed to the second, &c. 269.] THEOREM. A charge of given amount can be distributed in only one way on a given conductor which is removed from the influence of all other electricity. For, if it be possible, let the same quantity be distributed in two different ways, and let o- and or - > at the centre. Hence if F be the potential on the sphere, To find the potential produced in the shell, consider the potentials produced by the sphere and by the shell itself separately. The charge on the sphere produces (p. 404) potential -j- at all points 271.] EXAMPLES. 449 distant b from the centre ; and 1 the potential on the shell by U, Q' distant b from the centre ; and the charge Q' produces -JT- Denoting Hence V-U = 0()> or Q = --(V-U). If the shell is in metallic connection with the earth, U 0, and the capacity of the system will be - -- which increases as b a diminishes. By this arrangement the charge accumulated may be very great. Such a compound instrument is called a condenser. The capacity of an insulated sphere removed from all conductors is equal to its radius. 3. To find the capacity of a very long thin cylinder or wire. Except near the ends, the density of the electrification will be sensibly constant, and since the potential everywhere inside is constant, we have only to find its value at the middle point of the axis. Take a section of the wire normal to its axis at a distance x from the middle point, and another section at a distance x + dx. Then the quantity of electricity on the surface between these sections is 2TTprdx, where r = radius of wire and p = density of electrification. The potential of this at the middle point is = Hence if I is the length of the wire, rL = 4:7tpr 2 /o This = 4 npr log j if r is very small compared with I. Now Q=27tprl, ^ c _ Hence if r is exceedingly small in comparison with I, G will be small, and if the wire is used to connect two electrified conductors, the charge on the wire may be neglected. 4. Find the capacity of a condenser consisting of a very long cylinder of radius r surrounded by another of radius R, the two being separated by a given dielectric. Ans. K- s where K is the specific inductive capacity of the dielectric. [This is the case of a cable.] 271.] Case of Green's Equation. Let V be the potential of a system of masses, M, M' (fig. 253), and let U = - > where r Gg Fig. 253. 450 ATTRACTIONS. THEORY OF THE POTENTIAL. [271. denotes the distance of any point from a fixed point, 0. Suppose, moreover, for simplicity, that S is a level surface of the system of masses and that it in- cludes M' in its interior, while M is outside it. First, let be outside the space included by S. Then it is easy to prove that V - = ; and since in all parts of the space internal to S we have V7 = 0, except in those parts occupied by M', in which VF = 47rp, where p is the density of M' at each point, the equation (/3) of Art. 266 gives (y) dn denoting the element of normal at each point of the surface 8 measured outwards. , M _ r dn Now dS= 3- cos OdS, where is the angle made by the normal at any point on S with the line joining this point to ; and exactly as in Art. 252, for an external point, the integral of this expression taken over the surface vanishes. Hence (y) gives for an external point Secondly, let be a point in the interior of S. In this case the distance, r, of a point in the volume from becomes zero, and we cannot assert that V = ; but this difficulty is avoided by surrounding with an infinitely small spherical surface, and taking as the volume through which the integration is per- formed that contained between the given surface 8 and the surface of this sphere. In this way ceases to be a point within the volume considered, and consequently V is always = 0. We shall, however, have to perform the surface integra- 27 1 -] EXAMPLES. 451 tion on the right-hand side of (y) over the surface of this small sphere (on which we may consider the potential of the system as constant) as well as over S. Equation (y) now becomes Fpdu fldV, {\dV' , w r 7CI ,\,w _47r / = - dS + I --dS'Vl d8F'J T7^0j J r J r dn J r' dri J dn J dn' where V is the potential of the whole system, M and M' y at 0, and dS f an element of surface of the small sphere, whose radius is /. Now, dS' = r' 2 ds, where ds is the element of surface of a sphere of unit radius cut off by a cone whose base is d& and vertex ; or, in other words, d& is of the form / 2 sin (Art. 171). Hence, since / is indefinitely small, / -/-j-fdS' =Q ; i J T (tin (r' and V ] -j-rdS' 4*7', since dn is evidently d/. We have, then> " dn Coda 1 or _/>^, i n^r _ r> J r 4irJ r dn At the point denote the potential of the external mass, M, by 7 and that of the inside mass M', by J^. Then obviously / J T is 7^ ; and this equation becomes, since V = J^+7, _ r dn while (6) becomes \ f\dV * ; ^- EXAMPLES. 1. Any mass contained within one of its level surfaces may be distributed, according to a simple law, over this surface as a thin shell so as to produce the same effect as the given mass at all points outside the level surface. Let the mass M f alone exist. Suppose that matter is distributed over 8 so that kp, the product of the density and thickness, or the surface density, if the distribution is truly superficial, at any point is equal to I dV 4-7T dn 452 ATTRACTIONS. THEORY OF THE POTENTIAL. [271. Then the potential of this distribution at an external point, 0, is F -dS. i_ r dv_ 471V r dn But, by (f ) of this Article, this is equal to F;, the potential of M' at 0. Again, the whole quantity of matter on the surface is 1 but, remembering that dn is here measured outwards, this is equal to M' (Art. 252). 2. In the same case the superficial distribution which replaces M ' produces constant potential at all points inside the surface. This at once follows from (e), since V e = 0, there being no external mass. The constant internal potential is, therefore, the same as that on the surface. 3. Instead of the system, M, M', of which one portion, M' , is internal, and the other external, to a given level surface, S, of the system, may be substituted a distribution on the surface itself, with these results : (1) The effect at all points outside S is the same as that of M 1 '. (2) The effect at all points inside S is equal and opposite to that of M. Let kp, or for an infinitely thin distribution, the surface density, o-, be (Art. 263). Then these results follow at once from (e) and (f), p. 451. In this case also the mass of the distribution 4. Let M and M' be two quantities, q and g', of opposite electricities concentrated at two points, / and /', and S their zero potential surface. Then, since the level surfaces are given by the equation ^ - , = const., r r tne surface of zero potential is a sphere whose centre and radius are thus found : divide the line //' internally at A so that -^-r = -^- j 10 q I A q and produce //' to G so that - = -^ ; then G is the centre and CA the radius of the sphere. The distribution on this sphere which will produce the effect of I' at all external points and of / at all internal points is got by taking the surface density, o-, at any point P on the surface equal to - times the resultant of the forces T' acting from / to P, and q -TTfi-i ' acting from P to I'. TT)T' Now this resultant, N, = y|^- //p ,? since it acts in PC ; and sin J. Jr(^ 271.] sin/P/' II'.CA EXAMPLES. 453 -JL.^^i. This is the re- sultant measured inwards ; hence by Art. 263, _ IP*' iv.cr* so that the density varies inversely as the cube of the distance from 7. It is easily proved that CI . CT = CA 2 , which shows that / and /' are inverse points with regard to the sphere. 5. When a con- ductor envelops elec- trical charges and also has electrical charges outside it, show that the internal charges together with the in- duced charge on the inner surface of the conductor form a sys- tem in equilibrium by itself, producing no action at any external point ; and also that Fig. 254. the external charges together with the induced charge on the outer surface of the con- ductor form a system in equilibrium by itself, producing no action at any internal point. Let A and B (fig. 254) be the outer and inner surfaces of a con- ductor; let S be any closed surface drawn in the body of the conductor ; and let the finely-dotted lines at the outside of the outer and the inside of the inner surface represent the induced charges existing on these surfaces. The internal mass here consists of the given internal charges and the induced charge on the inner surface. Employ the equation (f), p. 451. Now all through the body of the conductor the resultant force = 0, therefore all over the surface S we have = 0, therefore (f) gives no, that is, the given internal charges and the induced charge on the inner surface give a constant zero potential at all external points, and there- fore a zero force at all such points. Hence at external points the action of this internal system is null. Of course this system produces zero potential at all points in the body of the conductor also ; for the surface S can be taken as close as we please to the inner surface of the conductor, and all points in the body of the conductor will be external to S. Employ now equation (e). It refers to points internal to S, and it gives V e = V, i. e. ? at all internal points the external charges together with th 3 in- 454 ATTRACTIONS. THEORY OF THE POTENTIAL. ground. duced charge on the outer surface of the conductor produce a constant potential which is equal to that in the body of the conductor, and by what we have just proved this latter is entirely due to the external masses. The conductor is therefore an Electrical Screen which protects charges (or other smaller charged conductors) inside it from the dis- turbing action of external electricity. This explains why delicate electrical instruments are protected from external disturbance by screens of wire gauze connected with the [If the conductor is connected with the ground, the constant potential inside due to the external elec- tricity will be zero.] 6. Calculate the surface-tension of an electrified soap-bubble. When a membrane is acted on by 255. forces of any kind, there will be along every line traced on the membrane a tendency of the two portions separated by this line to tear away from each other ; in other words, one of these portions exercises on the other a set of internal forces along the line of separation. In the neighbourhood of any point P of the membrane (fig. 255) consider a very small rectangular portion, ABCD, of the membrane isolated from the remainder. Then there will be forces exerted on its sides at their middle points, m, m', n, n, by the removed portion. These forces will, if the rectangle ABCD is chosen, at random, be oblique to its sides ; but we shall afterwards see that it is always possible to choose the rectangle at P so that these forces are at right angles to the sides on which they act. Suppose this done. The amount of force exerted on AB is, of course, proportional to the length AB] so that if tfj is the amount exerted on AB per unit of length, the force at m in the sense rnf m is ^x AB. Similarly, if t 2 is the force per unit of length on AD, the force on AD is t 2 x AD. The quantities ^ and t^ are called the surface-tensions at P perpendicular to AB and AD. For the equilibrium of the rectangle resolve forces along the normal to its plane at P. Then, exactly as in Art. 203, if r^ and r 2 are the radii of curvature of the curves mm,' and nn', and N the amount of external normal force exerted at P per unit area, we have or J.v =- +- (ra 1 ) = Qrrni cr /2 . if cr' is the density of the second electrification, m (n 1) n (m-l) 8. In the interior of a hollow conductor are placed given electrical charges, there being no external charges. Show that the induced charge on the outside of the outer surface consists of only one kind of electricity. dV [-y- has the same sign at all points on the surface. See ex. 32, p. 428.] 9. A spherical soap-bubble is electrified in such a manner that it is just in equilibrium when the pressures of the external and internal air are equal. Calculate the surface-tension in terms of the potential. (Mr. Orchard, Educational Times.) 72 Ans. t = 456 ATTRACTIONS. THEORY OF THE POTENTIAL. [272. 10. Find the law according to which a given uniform attracting bar may be distributed over any one of its level surfaces so as to produce the same attraction at all points outside the surface. 11. An electrified point I is placed in front of a given spherical conductor which is connected with the ground ; find the density of the charge induced on the sphere at any point. [In example 4 it is evident that if c[ could be replaced by the superficial distribution on the sphere, this distribution could be re- placed by an electrified point at I' whose charge is equal to that on the sphere ; and the law of density is that given in example 4.] 272.] Electric Images. The theory which has been illus- trated in these examples, and which is founded on equations (e) and (f) of last Article, is Sir William Thomson's theory of Electric Images. If an electrified point, /, is held outside any conductor connected with the earth by a wire (so as to have zero potential) there will be induced on the conductor a certain charge of opposite electricity, the effect of which at all points outside the conductor is the same as that of an imagined electrified point, I', inside the conductor ; and the effect of which at all points inside is equal and opposite to that of /. If there is a continuous series of external points forming an electrified body, M, there will be a continuous series of imagined internal points forming an oppositely electrified body, M f ; the latter is called the electric image of the former body in the con- ductor, and the distribution on the surface exerts at all external points the same effect as would be produced if this distribution were actually replaced by the image M'. Mr. W. D. Niven has treated the subject of Electric Images on the basis of the secondary solution of Laplace's equation (ex. 14, p. 424). See the Proceedings of the London Mathematical Society, Dec. 1876. CHAPTER XVI. ANALYSIS OF STRAINS AND STRESSES. 273.] Definitions of Strain and Stress. When a natural solid (such as iron, wood, &c.), or any material medium, is not acted upon by any external forces,, its particles assume certain determinate distances from each other, and the body is then said to be in its natural state* But when forces act on it either at its surface or throughout its mass, or when any disturbance is pro- pagated through its interior, these natural distances between its particles suffer alteration, and the body is said to be in a state of strain. Thus a fluid exerting pressure, a medium propagating sound, and the luminiferous ether when it is propagating light are instances of a body in a state of strain. The change of the natural distances between the particles is always attended by the production of internal forces, or, as they are called, internal stresses, or simply stresses ; and these stresses will depend, as we shall see, both on the nature of the body and on the nature of the strain in any case. SECTION I. Analysis of Small Strains. 274.] Displacements of a Rigid Body. It has been already pointed out (Chap. X) that the general motion of a rigid body consists of a motion of translation which is the same for all its particles, together with a rotation round an axis through an angle which is the same for all its particles. These displacements 458 ANALYSIS OF STRAINS AND STRESSES. [275. do not alter the distance between any two particles of the body, and they are therefore unaccompanied by the development of stress in its interior. Stress results only from the alteration of distances between pairs of particles, and hence in treating of strains and stresses all displacements, whether of translation or of rotation, which are impressed, with common magnitude, upon all particles of the body, may be discarded; and again any such common displacement may be freely introduced if it is found convenient for the discussion. 275.] Changes in Relative Co-ordinates. Let a system of rectangular axes, Ox, Oy, Oz, (fig. 256) be fixed in space ; through any point, P, in the natural solid under consideration let Px, Py, Pz be drawn parallel to the fixed axes. Let the particle at P be displaced to P', and suppose that the co- ordinates (#, y, z) of P referred to the axes through are increased by small quantities, u, v, w, Fig. 256. respectively. The co-ordinates of P / are therefore x + u,y + v 9 z + w. Now these displacements u, v, w depend on the position of the point P, i. e., they are functions of its co-ordinates depending on the law according to which the strain is produced. We have then, when the kind of strain is specified, some such equations as =/2 ( y> 4 w = where f\>f %>/?, are symbols of functionality. Let Q be a particle very near P, and let its co-ordinates with reference to the axes drawn through P be (f, 77, f). Then the displacements of Q parallel to the axes are obviously that is, by Taylor's Theorem, 275-] CHANGES IN EELATIVE CO-ORDINATES. 459 ..du du .du ,.dv dv .dv u + (-r + ^T- + f~r ' ^ + CT~ + ^T~ + f ^~ d# dy dz dx dy dz ..dw dw .dw w + ~r + *1 -7- + C~r* dx dy dz Suppose Q to come to Qf by displacement. Then in consider- ing the nature of the strain in the neighbourhood of P, we may, by last Article, impress on every particle of the body a motion of translation represented in magnitude and sense by P'P, so that P' will be brought back to P without in any way interfering with the strain of the solid. By drawing Q'Q" equal and parallel to P'P, the particle which was originally at Q may now be considered to be at Q"; and a similar process is to be repeated for all other particles. The part of the strain, therefore, due to the alteration of the distance between P and Q will depend on the co-ordinates of Q" with reference to Pa?, Py, Pz. These co- ordinates are, of course, the excesses of those of Q' over those of P'} and therefore the relative co-ordinates of Q" are .. , du\ du du dv , dv\ .dv .. dw dw . / div\ ^+^ +c ( 1 + ^) ; in other words, the changes, Af, Ar;, A in f, 77, are .. jlu du .du ..dv dv dv Af =fy- +^7- +C^~; Ar ? = f +77^- +(^~; dx dy dz dx dy dz i/ y .dw dw .dw , ^ A *-s +1 VE+*ar (a) COR. 1 . All particles near P which in the natural state lie in one plane will after strain also lie in one plane. For if the co- ordinates of Q" are denoted by ', rf, f, we have ., .. / du\ du ,.du ... f =^ 1 + &) + "^ + ^"'=-' f = -' which equations, being linear, give , rj, linearly in terms of ', ij', f . Remembering- that > -=- > ... are all small, these dx , $ du equations give f = f ' + small quantities of the order of -^ &c. ; y clx so that in any terms multiplied by j- ... ' may be put for , 6ft?? ?;' for rj, and ' for f 460 ANALYSIS OF STRAINS AND STRESSES. [276. Hence we have, to the order of accuracy adopted, f ,-,(, du\ r du du dz Therefore if all the points (, 77, () lies in the plane all the points (', 77', C 7 ) will also lie in a plane. That is, every plane curve is strained into a plane curve in a different plane. COR. 2. All particles near P which in the natural state lie in one right line will after strain also lie in one right line. For if we have =Q and A'-\ -B'n + &+& = 0, we shall have ', 77', f, also satisfying two linear equations. COR. 3. Two parallel right lines in the natural state are changed into two parallel right lines (ivith a different direction) in the strained state. For one of the two lines being given by the equations the other will be given by two equations in which the terms D and J/ alone are altered. But by substituting for f, 77, f their values in terms of f , rj ', f, the values of D and D' do not in- fluence the direction cosines of the line into which any one is converted by strain. 276 ] Elongation in any Direction. DEF. Supposing P and Q to be, as before, two particles in the natural state of the body, the elongation in the direction PQ is defined as the ratio of the change produced by strain in the distance between these same particles to the original distance between them. Hence the elongation in the direction PQ is > or > if p denotes PQ, and Ap the change in p. Now 2 = 276.] ELONGATION IN ANY DIRECTION. 461 or if we substitute for Af, A 17, and Af their values from last Article, Let the cosines of the angles made by PQ with P#, Py, P^ be I, m, , respectively, let = a . = o, = c, dx ' dy dz du dv dv dw dw du and denote the elongation by e ; then the last equation gives e = al 2 + /ft 2 -I- en 2 + 2 lms s + 2 m^5j + 2 w& a . ( 1 ) The elongation in any direction may be graphically represented as follows : Construct at P the quadric surface whose equation referred to the spatially fixed axes Px, Py, Pz is ^ + ^ + ^ + 2*3^ + 2^^+2*2^=^ (2) where Tc is any constant linear magnitude. If r is the length of the line PQ intercepted by this surface, we have r 2 (al 2 + lm 2 + en 2 + 2 Ims 3 + 2 mns l + 2 nls 2 ) = k* ; e = ^-> (3) or the elongation in any direction varies Inversely as the square of the radius vector of the Elongation Quadric in this direction, if we agree to call the above surface the Elongation Quadric. It is possible, however, that equation (2) may fail to represent the elongation in all directions. For there may be contraction (negative elongation) in some directions, and then (2) will repre- sent a hyperbolic surface, the radii of which will give as in (3) the elongations, while the contractions must be given by con- structing the surface -^ 2 , (4) 462 ANALYSIS OF STRAINS AND STRESSES. sc Fig. 257. which is the hyperboloid conjugate to that which gives the elongations. Unless, then, all lines are contracted or all lines elongated, there will really be two quadrics required, one to represent elon- gations and the other to represent contrac- tions. For example, consider the simple case in which the strain is made by drawing out all lines perpendicular to the plane yz in the same proportion, and contracting all lines perpendicular to the plane xz in the same proportion ; so that u ax> v = by, w = 0. Then the elongation is given by the equation e = al 2 bm 2 . Now this expression is negative when bm 2 > al\ and if we con- struct a surface whose equation is a 2 brf = 0, i. e., two planes through the axis of z, this surface will form the boundary between lines which are elongated and lines which are contracted. The elongations are given by the radii of the surface a 2 br) 2 = k 2 , a hyperbolic cylinder, the section of which by the plane xy is represented in fig. 257 by the curve (DAC, 1/A'Cf) ; and the contractions by the conjugate surface brfa 2 = & 2 , which is represented by (DBC\ I/B'C) ; the planes of no elongation or contraction being the asymptotic planes, DD', CC', of these surfaces. All lines through P along which the elongation is the same lie on a cone whose equation is easily found from (l). For, putting e (I 2 + m 2 -f- n 2 ) for e, we have (a e) I 2 + (b e) m 2 + (c e) n 2 + 2 S 3 Im + 2 S 1 mn + 2 S 2 nl = ; and if f, 77, f are the co-ordinates of any point on the line (I, m, n), we have I : m : n = f : 77 : f ; therefore this equation gives (a-O^ + C*-*)? 2 + (*-) ^ + 2*3^ + 2^1,^+2*2^=0, which, if e is constant, denotes a cone whose vertex is P. This is called the cone of equal elongation. If is taken = 0, we get a cone of no elongation, and it is evidently (when real) the ELONGATION IN ANY DIEECTION. 463 asymptotic cone both of the Elongation Quadric and of the Compression Quadric. COR. I . The elongations in the directions of the axes of #, y^ z .. 7 , du dv dw are, respectively > a, 0, c, or -=- ? -y- , - . dx dy dz COR. 2. The elongation is the same along all parallel lines in the neighbourhood of P. For if R is any point very near P, the value of e along a direction (I, m, n) at R is got by using the values of a, b, c, s^ ? 2 , s 3 at E in equation (I). But these values at R differ from the values at P by infinitesimals of the second order. Therefore, &c. COR. 3. Any small parallelogram or parallelopiped in the natural state in the neighbourhood of P is changed into another parallelo- gram or parallelopiped l)y the strain. For (Cor. 3_, Art. 275) any two parallel lines are strained into two parallel lines, and (Cor. 2, Art. 276) they are equally elon- gated. Therefore, &c. COR. 4. A small circle very near P in any plane is strained into an ellipse in a different plane. For, let AQB (fig. 258) be a circle in the natural state; let OA and OB be any two rectangular radii, Q any point on the circle, and QM and QN perpendiculars on OA and OB. Let the lines OA and OB become oa and ob (in a different plane) by the strain, and let Q become q. The circle will become a u curve in the plane of oa and ob / N \^( by Cor. 1, Art. 275. Also if f | \\A qm and qn are drawn parallel V J to ob and oa, the lines QM and QN will become qm and Fig. 258. qn ; for M must become some point on oa (Cor. 2, Art. 275), and OB and QM must become parallel lines (Cor. 3, Art. 275). Again, if c is the elongation along OA, oa = (1 +e) OA and om = (l + e) OM ; OH _Om L , '* 'OA ~ = Oa ' ON on similarly = ' 464 ANALYSIS OF STRAINS AND STRESSES. [276. OM* , ON 2 __ + __ = 1} , p omZ , on * therefore $ + "Ta = *> 00 2 0$ 2 which shows that the curve on which q lies is an ellipse having the lines oa and ob for conjugate semi- diameters. Hence every pair of rectangular radii of a circle is strained into a pair of semi-conjugate diameters of an ellipse ; and since among these latter there is one rectangular pair (the axes of the ellipse), it follows that some two rectangular diameters of the circle are strained into two rectangular lines. Hence in every plane near P can always be found two rectangular lines which are strained into two rectangular lines. COR. 5. Any two small cqplanar artas in the natural state are strained into two coplanar areas having the same ratio to each other as the unstrained areas. For let CAB and C'A'B' be any two elementary rectangles in the same plane near P such that AB is parallel to AB' and AC parallel to A'C'. Then by Cor. 3 these will be strained into two parallelograms, cab and ca'b', such that ab is parallel to a'U and ac to ac'. area cab ac x ab Hence 7-777 = -7-7 TF/ area c ab ac x ab Let be the elongation in the direction AB and e' that in the direction AC-, then ab = (l+c) AB, a'b' = (1 + e) A'ff ; ac = (1 + e') AC, a'c = (1+ e') AC' ; area, cad ACxAB area, CAB therefore area (fa' If ~ AC' x AB' area C'AB' Now, whatever be the two areas, they can each be broken up into an infinitely great number of small parallel rectangular strips, and the ratios of the strained areas of these strips being the same as those of the unstrained, the whole strained areas are to each other as the unstrained ones. COR. 6. Every small sphere in the natural state is strained into a small ellipsoid. This is evident from Cor. 4, since the sphere, being a surface every section of which is a circle, must alter into 277-] LINES OF NO ROTATION. 465 a surface every section of which is an ellipse. Nevertheless for clearness we may re- peat the proof of that Cor. Let OA, OB, OG be any three rect- angular semi - dia- meters of the sphere, Q any point on the sphere, QR a line parallel to OC ter- Fig. 259, minated by the plane OAB, and EM, RN parallels to OB and OA. Let the lines OA, OB, OC be strained into oa, ob, oc, and Q to q. Then, by Cor. 3, Art. 276, QR, RM, and RN will be strained into qr, rm, and rn which are parallels to oc, ob, and oa terminated by the planes oab, oac, and obc. Also by Cor. 2, om OM oa om similarly, But therefore on_ ON qr __ QR ob == US' ~oc"~OC' OM 2 ON 2 QR* _ ~fi7v *J om" on" at* , I l_ JL oa' ob 2 oc- which shows that the surface on which q lies is an ellipsoid having- oa, ob, oc for a system of conjugate semi-diameters. Hence every rectangular set of radii of a sphere in the natural state is strained into a system of conjugate semi-diameters of the ellipsoid into which the sphere is changed ; and it follows that there is one rectangular set which is strained into a rectangular set and altered in directions, the latter being the axes of the ellipsoid into which the sphere is strained. Con. 7. Any two small volumes in the natural state are strained into two small volumes which bear the same ratio to each other as the unstrained volumes. The proof of this proceeds exactly as in Cor. 5. 277.] Lines of no Rotation. Let us enquire whether, with the given strain, it is possible to find a particle Q, in the natural state of the body, such that its displaced position, Q' 7 , shall be Hh 466 ANALYSIS OF STRAINS AND STRESSES. [278. on the line PQ. If this is so, all particles (near P) on the line PQ will retain the same direction with respect to P; i.e., the line PQ will not suffer rotation by the strain. ^x du\ du du The direction cosines ofPQ"are p ^ > ... , and those of PQ are r, ... . Hence if these are the same, / du\ du pdu \ dx' dy dz PQ" PQ ' with two similar equations. Now PQ"= (1 + ) PQ; hence /du \ , du du ( e )c~l *H C = QJ with two similar equations ; or if I, m, n be the direction cosines of PQ, a line of no rotation, , /du \ du du dv dv ,dw dw ,dw \ - \-m-j-+n(- e) == 0. , dx dy \dz ' (i) By eliminating ^, m, n from these equations, we obtain the cubic equation for e, du du du Jz dv dx dy dx dz dw = 0, (2) which gives necessarily one real value of e and may give three real values. Hence in the small general strain of an elastic solid there is at every point at least one line of no rotation. 278.] Change of Inclination of Two Lines. In the unstrained state let there be two points, Q a and Q 2 , very near P, and let be the angle between the lines PQ l and PQ 2 . We propose to find the angle between the lines into which these are strained. Let (f i ^i ti) and ( 2 rj 2 f 2 ) be the co-ordinates of Q and Q 2 wifh 278.] CHANGE OF INCLINATION OF TWO LINES. 467 reference to Pa, Py, Pz (fig. 256) ; and supposing that the strained positions of Q 1 and Q 2 are Qi" and Q 2 ", whose co-ordinates are V faO* we have > bv Art - 275 > du . du .. dv .. dv dw dw Hence neglecting squares and products of a, b, . . , , -= , . . . 3 we ay have &'/ + 77/r// + d'C/= f i f 2 + ft i? a + fi ^ If <^'is the angle between PQ/'and P&'.PQ," so that if 6 X and 2 are the elongations in the directions PQ l and PQ 2 , and (I 1 m l n 1 ), (I 2 m 2 n 2 ) the direction cosines of the lines and PQ 2 , the above equation gives i) (l + f 2 ) COS ^ = COS H" 2 ( a h or dividing out by (l + 6j) (1 + e 2 ), cos ^ x = cos ' in (l). For if (/> = - and also '= - > 2 2 the directions of the lines PQ 1 and PQ 2 are connected by the equation -f s l m + 1 or ($ 2 /i + *i% + <%)#2 = 0, which shows that PQ X and PQ 2 are conjugate diameters of the quadric fl fa + ^H-^ + 2 8 ft + 2 1 i,f+2* 2 rf = ^, ^ being any constant. COE. 3. 7^ e 8* COR. If the axes of co-ordinates at P are taken in the direc- tions of the axes of the strain ellipsoid, the quantities s l9 s 2 , and s 3 are all zero, as is evident from (2) of last Art., and the equation of this ellipsoid will be (t-'i)H(i-*)ii'+(i-Of i -if* = o. (a) 282.] Pure Strain. A strain is said to be pure when the lines at P which become the axes of the strain ellipsoid are unaltered in their directions by the strain. 283.] Conditions for a Pure Strain. Since a, b, c, ... are infinitesimals of the first order, it follows from the value of e given by equation (1), p. 461, that the elongation along the direction PQ" (fig. 256, p. 458) may be taken as equal to the elongation along the direction PQ ; so that if e is the elongation in the direction of any radius vector of the strain ellipsoid, we have / 1 . x p = r(l + ), where p is the length of this radius vector and /the radius of the sphere which becomes by strain the strain ellipsoid. 283.] CONDITIONS FOR A PURE STRAIN. 471 Hence if the axes of this ellipsoid are a, ft, y, we have (itf) I s 3 . ms 2 . n = \t, -\ s 3 . l+(\ ft) m s l . n = \m, > s 2 .l s-L.m + C^ c)n \n ; ) Now if I, m, n are the direction cosines of any axis, it is well known (see Salmon's Geometry of Three Dimensions, or Frost's Solid Geometry) that ... ,. ,^ ^ the three values (A 1} A 2 , A 3 ) of A obtained from these equations being such that the equation of the ellipsoid referred to its own axes would be ~ 2 r 2 2 r 2 Hence \ 1 = =%e 1 ; A 2 =i-d? 2 ; A 3 = | e s> ^ e equations (l) become, for the direction of any axis, = 0, (2) a<)l+s 3 m + s 2 n = 0,-\ 3 l+(b )m + s 1 n = 0, > 2 l+s l m + (c e) n = Q.) Now if there are three unrotated lines, they are given by equations (l), Art. 277 ; and if the same lines are determined by (2), we must have du dv du dw dv dw dj = fa = '~ S3 ' Tz^Tx"'*^ Tz^Ty^ 8 ^ and the conditions for pure strain are that the displacements u y v, w satisfy the equations du dv du dw _ dv dw _ , . dy dx ~ ' dz dx " ' dz dy These are the well-known conditions that the expression in which u, v, w are functions of x, y, z, should be the perfect differential of a single function, < (#, y, z). When this function exists, i. e., when the strain is pure, it is called the Displacement Potential of the strain. Hence the components, Af, Ar/, A of the strain (given in Art. 275) become when the strain is pure 472 ANALYSIS OF STRAINS AND STRESSES. [284. i. e., the coefficient of rj in A is the same as the coefficient of f in Ar;, &c.; and this is the distinguishing character of a pure strain. A pure strain is also called an irrotational strain. The values of the principal elongations of a strain are the roots of the cubic equation ac, s 3 , s 2 = 0, ( 284.] Theorem. Every strain can le resolved into a pure strain and a rotation. By a rotation here is meant such a displacement as a rigid body undergoes in turning round an axis, and we propose to show that the general small strain at any point P of a body, may be produced by two operations, viz., first holding fixed in directions the principal axes of the strain and straining the body to a certain extent, and then rotating it as a rigid body about a certain axis. It has been shown (p. 293) that if a rigid body receives small angular displacements 8^, 80 2 , b0 3 round three fixed rectangular axes, the displacements of the co-ordinates, 17, of any point in it are /* & /j 5/3 %. A / s Q x/a t $. A /i\ C C/o 77 C/q , c I/ o C (/-I , 77 (j-\ c t/o. ( 1 I (Such a displacement has, of course, no displacement potential; for if these displacements are denoted by u y v, w, we have du dv , . , -j j2 equal to 280 3 and not equal to zero.) Now the component, Af, of the displacement of Q along the axis Px is (Art. 275), and this A 1 /f ^ ^ N . , ,du . dw^ fc . , t du dw* Hence, with the same values of * 1? * 2 , s 3 as before, we have 285.] SIGNIFICATIONS OP 8 l9 S 2 , S 3 . 473 A comparison with (l) shows that the portions in brackets in these expressions denote rotations, as of a rigid body, about the axes through the small angles which are equivalent to a rotation through v^^) 2 4- (S0 2 ) 2 + (^s) 2 about one line (p. 292) ; while the portions of Af, A??, Af out- side the brackets denote a pure strain by Art. 283. If the axes of reference, Px, Py, Pz, are chosen in the direc- tions of the principal axes, the pure portion of the strain will be expressed by A ._ ^ A7? _ ^ r?j A ^ _ ^ ^ i.e., the pure strain is produced simply by multiplying the co- ordinates of every particle by the numbers l-f 1 + ^s- A simple elongation of a body in a direction perpendicular to any plane means the drawing out from the plane of every particle through a distance proportional to the perpendicular from the particle on the plane, so that those particles farthest from the plane in the natural state are most drawn away, but all in the same proportion to their original distances from it. By this Article we see that every small strain at a point P can be produced by three successive simple elongations followed by a rotation, as of a rigid body, about an axis through P. 285.] Significations of s l , s 2 , s 3 . Let the axes Px and Py become by strain Px" and Py", fig. 260. (Of course it is supposed, as in Art. 275, that P is brought back to its ori- ginal position after the strain.) All particles in the plane of Px and Py originally are in the (different) plane of Px" Fi g . 2 6 . and Py" after the strain ; and if A is a particle on the axis of y and AB a line parallel to 474 ANALYSIS OF STRAINS AND STRESSES. [286. Px, the line of particles AB will become (Cor. 3, Art. 275) a line of particles A"B" parallel to Px". Let fall a perpendicular, A"p } from A" on Pa/'. Then the particle (A") which was at A has advanced in front of P parallel to the line Px" through the distance Pp. Now Pp PA" cos a?"P/' = 2PA". S B (Cor. 1, Art. 278); andP^"= (l +b)PA ; therefore Pp = 2(l+d)s 3 .PA; or, neglecting the product bs 3 , Pp _ T> A 3* rA. Hence the quantity 2 s 3 is the rate (per unit of distance be- tween the two lines) at which particles on any line AB parallel to Px have slid beyond the corresponding particles on Px. Evi- dently it is also the rate at which sliding has taken place between particles on Py and lines parallel to Py. Or again, imagine a little parallelepiped at P having its edges along the lines Pa?, Py, Pz. Then 2s 3 is the rate at which the face parallel to that in the plane xz has slid in front of the latter ; or the rate at which the face parallel to the plane yz has slid in front of the face in the plane yz. Similarly for the values of ^ and s 2 . DBF. When a plane is held fixed in a body and all planes in the body parallel to it are slid in the same direction and sense parallel to the fixed plane, each through a distance proportional to its distance from the fixed plane, the strain so produced is called a shearing strain. Those planes which are nearest to the fixed plane are least dis- placed, and those which are farthest from it are most displaced. The ratio of the distance through which any plane has slid to its distance from the fixed plane is called the amount of the shear. Hence the quantities 2s , 2s 2 , 2s 3 are the small shears of the axes of {y,z), (z } x), (%>&) respectively, at the point P. From fig. 260 it is clear that the change in the cosine of the angle between any two lines at right angles in the natural state is the shear in their plane of lines parallel to either. 286.] Shearing Strain. The two fundamental kinds of strain of what are called isotropic bodies (i. e., bodies whose constitution is the same at all points and in all directions round every point) are Cubical Dilatation and Shearing Strain. We propose, therefore, to consider this latter more particularly here. 286.] SHEARING STRAIN. 475 Confining our attention to a shear, 2s 3 , of the two rectangular lines Ox and Oy, the elongation quadric would be 2* 3 ^ = 2 , the axes of co-ordinates being the lines Ox and Oy. But this equation denotes a hyperbola in the plane xy referred to its asymptotes ; and if we alter the axes of co-ordinates to the axes of the curve, the equation referred to them will be A comparison with the general equation of the elongation quadric shows that this equation denotes an elongation s 3 (half the shear) of the body along one axis of the curve accompanied by an elongation s 3 (i.e., an equal compression) of the sub- stance along the other axis. Hence the shearing strain of a body can be produced by a simple elongation (equal to hat/ the shear) along one line and a simple compression of equal amount along a perpendicular line. We have been considering small displacements; but let us now consider an elongation of any amount along a line Ox, and an equal compression along a perpendicular Q. Oy (fig. 261). Sup- pose that all lines im the body parallel to Ox are increased in the ratio a: 1, and that all lines parallel to Oy are diminished in the ratio 1 : a ; and consider displace- ments in the plane xy. There will, of B"' Fig. 261. course, be similar displacements in all planes parallel to xy. The displacement of the point may be impressed in reversed direction on all points, so that may be considered as at rest. Draw OA, of any length, making the angle AOx = tan~ 1 a. From A let fall An perpendicular to Oy. Then An becomes elongated by the strain parallel to Ox into a . An ; but a.An nO', therefore by this strain A is drawn out to a, Aa being parallel to Ox, and a a point on the bisector, Oa, of the 476 ANALYSIS OF STRAINS AND STRESSES. angle xOy. From a draw am perpendicular to Ox. Then, by the strain parallel to Oy, am becomes shortened into Now if we draw OA' making with Ox an angle equal to AOy, this line will meet am in a point, A', such that A'm = Hence after the two strains A will come to A' ; and we see that OA' is equal in length to OA, and that they are both equally inclined to the bisector of the angle xOy. In the same way if OB be drawn making LBOx' ' tan- 1 **, the length of OB will be unaltered, the point B will come to J?, and the lines OB and OB f are equally inclined to the bisector of the angle x'Oy. Also OA is perpendicular to OB'. Hence, since parallel lines are all altered in the same ratio, all lines parallel to OA are unaltered in length, and all lines parallel to OB are unaltered in length. Imagine a plane through OA perpendicular to the plane of the paper, and let any curve whatever be traced out in this plane. The curve will remain perfectly undistorted after the strain. For all lines perpendicular to the plane of the paper obviously remain so and are unaltered in length, and all lines parallel to the plane of the paper remain parallel to this plane, while of these latter those which are parallel to OA remain unaltered in length. Hence ordinates and abscissae of the above-named curve parallel to OA and to a normal to the plane of the paper remain perpendicular to each other and unaltered in length. The curve, therefore, as regards magnitude and shape remains exactly as it was ; its plane only is altered (to the plane through OA per- pendicular to the paper). It follows, of course, that all lines, whatever be their directions, in the plane through OA perpendicular to the paper remain un- altered in length. Similarly all lines in the plane through OB and the normal to the paper remain unaltered in magnitude ; and all figures in this plane also remain undistorted. The planes through the normal to the paper and the lines OA and OB are called the planes of no distortion. Suppose that we impress on the body a common motion of rotation about the normal to the paper at so as to bring OA' into coincidence with OA. This motion will, of course, be un- 286.] SHEARING STRAIN. 477 accompanied by any strain (Art. 274). Then OB' will come to 0-5"," and BB" is perpendicular to OB and parallel to OA, as is very easily seen. Draw BQ parallel to OA. Then since the length of BQ remains unaltered, Q will come to Q", a point such that B"Q"=BQ. Hence all particles in the line Q are slid parallel to AO through a space BB". Now if p is the length of the perpendicular from B on OA, BB" _ _l as is easily found. P a Consequently in this strain if the undistorted plane OA is held fixed, every plane ', BQ, parallel to it is slid parallel to it through a space proportional to the perpendicular distance between J3Q and OA ; and this is the usual way of representing a shearing strain. Of course the strain may otherwise be produced (neglecting the effect of mere rotation common to all points) by holding fast the other undistorted plane, OB, and sliding all planes parallel to it. The plane (ocy) perpendicular to the two planes of no distortion is called the plane of the shear ; and the lines ( One and Oy) which bisect, in the plane of the shear, the angles between the planes of no distortion are called the axes of the shear. Since a sphere described about as centre becomes an ellip- soid, and since there are two sections of an ellipsoid which are circles, the planes of these sections must be OA' and OB', the strained positions of the planes of no distortion. The quantity, a j which is the fractional sliding per unit of distance between the parallel planes is called the amount of the shear. If the strain is small, a = 1 + s, where s is a small quan- tity; and - = lSj nearly, so that the amount of the shear = 14-* (1 s) = 2s, which agrees with the analytical result at the beginning of this Article. The expression for the displacement in a shearing strain can be simplified by taking the fixed plane as that of xy and the axis of x in the direction of the sliding. Then u = 2 sy, v = 0, w = ; 478 ANALYSIS OP STRAINS AND STRESSES. [387. so that a shear is a homogeneous strain, but not a pure one (Art. 283). 287.] Traction and Torsion. Suppose a cylindrical bar of an isotropic body to have its base held fixed while the bar is pulled in the direction of its length. Then each particle of the bar will be displaced in a direction parallel to the axis through a distance proportional to the natural distance of the particle from the fixed base ; and in addition, the particle will be displaced towards the axis through a distance proportional to its natural distance from the axis. That is, at each point there will be uniform elongation and uniform contraction. Hence if the axis of the bar is taken as that of z t and the axes of x "and y are in the plane of the fixed base, u = ax, v = ay, w = cz will express the displacements of any point, the quantities a and c being constant throughout the bar. This is the case of Traction. Suppose that, the base being still held fixed, the free extremity is twisted round through any angle (measured by the angle through which any diameter of the section revolves) ; then every other normal section of the bar will turn through an angle pro- portional to the distance, z, of this section from the fixed base. If I =. length of bar, a = angle through which its free end is twisted, every point in the section considered will be twisted through an angle a j Hence the displacements of a point as, y in this section are (the twisting taking place from axis of x towards axis of y) azy azx u = j- , v = , w = 0. I C This strain is called Torsion. 288.] Lines of Flow and Vortex Lines. Just as a Line of Force has been defined (p. 410) as a curve at every point of which the resultant force of attraction of a system is directed along the tangent, so a Line of Flow is defined to be a curve at every point of which the resultant displacement of the particle existing there is directed along the tangent. Again, we have seen that the whole strain at any point can be produced by a pure strain together with a rotation round an axis through the point. A curve such that at every point of it 289.] EQUIPOTENTIAL SURFACES. 479 the rotation corresponding to that point takes place round the tangent is called a Vortex Line. In analogy with a Tube of Force, we have a Tube of Flow. If through points constituting the contour of any area we draw Lines of Flow, these lines form a surface called a Tube of Flow. Similarly if through the points constituting the contour of any area we draw Vortex Lines, these lines will make a surface which may be called a Vortex Tube. When the normal section of the Vortex Tube is everywhere very small, it is called a Vortex Fila- ment. Such a filament, AB, is represented in figure 262. Fi - 262. 289.] Equipotential Surfaces. When the strain at every point is irrotational, the quantity udx + vdy + wdz is a perfect differential of a function (x, y, z). Describe in the body a series of surfaces the equation of any one of which is (a>,y,z)=C. (1) Then by giving C a series of different values we shall have a series of surfaces, exactly analogous to the equipotential surfaces of an attracting mass (Chap. XV) ; and these equipotential surfaces of strain will be related to the lines of flow exactly as the equipotential surfaces of attraction are to the lines of force ; that is, at every point the line of flow is perpendicular to the equipotential surface. For the direction cosines of the normal to the surface (l) at any point (#, y, z) are proportional to d(j> d$ d(f) . ~ ' ~~ ' ~~ ' 1 * e ' ) u > V ' W ' U) Vy w " ein> *" e com " ponents of the displacement, are of course proportional to the direction cosines of the line of flow. Therefore, &c. The potential function of any small strain being <, we see that -j- is the displacement parallel to the axis of x ; and since the axis of x may be in any direction, the displacement in any direction is the rate of variation, per unit of length, of potential in this direction. It follows that the resultant displacement (which is perpen- dicular to the surface = C ) is -j- > where n denotes length 480 ANALYSIS OF STRAINS AND STRESSES. [290. measured along the normal to the surface, and the displacement is measured in the same sense as n. Let two very close equipotential surfaces, < = C 19 = C 2 , be described. Denote these by fa and fa. Then at all points on fa the resultant displacement is inversely proportional to the normal distance at this point between the surfaces fa and fa. For if at any point on the surface fa the normal distance between it and is A n. the displacement is - or An An But for all points considered fa fa = C 2 C = a constant ; therefore the displacement varies inversely as A n. 290.] Circulation. Suppose any curve, AB, to be traced out in the body, and let the displacement of each particle, P, on the curve between A and B be resolved along the tangent to the curve at P (the resolution taking place between A and B in a sense opposite to that of watch-hand rotation); then the sum obtained between A and B by multiplying this resolved part of displacement by the element, ds> of the curve at P and adding all such products, is called the circulation between A and B. Hence, by definition, the circulation from B to A is equal and opposite to the circulation from A to B. The components of the displacement parallel to the axes being, as before, it>, v, w, and the direction cosines of the tangent ,, . . dx dy dz .. to the curve at P being -^ , ~> -=-t the circulation is & ds ds ds dx dy dz\ , +v ie ... are constant for all points on the curve) du dv dv du r . ufdq-\-vfd"n-}- /^~f~ /^?^ J ?~^~ J d*] ~^~ ~j~JT]d*a) dx dy dx dy of which all the integrals except the last two vanish, since the curve is closed. Now fdrj = area of curve = A] and fi)d = A, since the two integrations are carried round at the same time from x to y. Hence the circulation = A (-= --- =-) \ j = 2A . dd 3 , (p. 473) d6 B being the rotation round axis of z at P, i. e., perpendicular to the plane of the curve. Suppose that any surface, plane 01 curved, bounded by any curve, A BCD, (fig. 264) is traced out in the body and that at each point on this surface we take the component of rotation round the normal to the surface, multiply this component by the ele- ment of superficial area at the point, and take the sum of all such products. This sum is called the surface-integral of normal rotation. The normal must be supposed to be drawn away from the same side of the surface i i Fig. 264. 482 ANALYSIS OP STRAINS AND STRESSES. [290. at every point, and the rotation is supposed to take place oppo- site to that of the hands of a watch held so that the normal passes up through its face. It is very easy to prove that this surface-integral of rotation is equal to one half the circulation round the edge> ABCD, of the surface. For, let the surface be broken up into an indefinitely great number of little plane areas. Then the sum of the circu- lations round these areas is twice the surface integral of rotation (by what has just been proved). But the circulations in the common portions of every two contiguous areas are directly opposed, and therefore mutually destructive, as is seen by drawing any two such little areas, a and #, apart ; hence the circulation exists only along lines which do not form common parts of contiguous areas, i.e., along the edge which bounds the surface. If the surface has no bounding edge, i.e., if it is a closed surface, the surface-integral of rotation over it is zero. If the surface, without being closed, is such that at every point of it the rotation takes place about a tangent line to the surface, the circulation round its bounding edge is zero. Such a surface is that of a vortex filament (fig. 262, p. 479) or that represented in fig. 265, which consists of a vortex tube whose ends are any two irregular curves whatever. The sum of the circulations round the terminal sections D and E of this tube, estimated in the cyclical order indicated round the Fig. 265. contour in fig. 264, is zero, i.e., the circulations^ estimated as represented by the arrows in fig. 262, round any two sections whatever of a vortex tube are equal ; or, in other words, the circulation round any section, normal or oblique, plane or tortuous, of a vortex tube is constant. EXAMPLES. 1. Prove analytically that the shear of any two rectangular lines intersecting at any point is equal to the difference between the elongations along the internal and external bisectors of the angle between them. Let the axes of co-ordinates be the principal axes of the strain at the point. Then the value of s given in equation (2), Art. 278, becomes s = ej,l' + e z mm' + e s nn', the direction-cosines of the lines being (I, m, n) and (/', m', n'), and 290.] EXAMPLES. 483 the shear 2s. Now the direction-cosines of one bisector are I I', mm', n n, each divided by the square root of the sum of the squares of these quantities, i. e., by >/2, since the lines are rectan- gular j and the direction cosines of the other bisector are I -f l' t m + m', n + n, each divided by \/2. Let e and e' be the elongations along these bisectors. Then, by Art. 276, therefore c' = 2 (^ IV -f- e 2 mmf + e s nn'), or e' e = 2s, which proves the proposition. 2. Find the pair of rectangular lines in a given plane for which the shear is greatest. In any plane the elongation is greatest along one axis of the conic in which this plane cuts the Elongation Quadric, and least along the other. Therefore the difference of elongation along two rectangular lines is greatest for this pair; and therefore, by last example, the shear of the two rectangular lines of whose angle these axes are the external and internal bisectors is greatest. Hence the shear in a given plane is greatest for two lines making angles of 45 with the axes of the conic in which the given plane cuts the Elongation Quadric. The magnitude of the shear for any two rectangular lines in the plane is easily found and represented by a curve. Let the axes of x and y be taken in the given plane and coincident with the axes of the section of the Elongation Quadric in the plane. Then s s must = for these axes. Also let one of two lines along which we wish to find the shear make an angle 6 with the axis of x. Then in the expression for s (Art. 278) we have 7 X = cos 0, m l = sin 0, 1 2 = sin 6, m z = cos 0, n = n 2 = ; therefore s = I (b a) sin 20, or 2 s = (b a) sin 20 = shear, which of course shows that the shear is a maximum along lines bisecting the angles between the axes of the section. The curve whose polar equation is r = (b a) sin 2 6 consists of four loops, one in each quadrant, and its radius -vector gives the shear for any directions, denoted by and - + 6. 2 It follows that the two rectangular lines whose shear is absolutely the greatest at a point in the body are those in the plane of the greatest and least axes of the Elongation Quadric (or of the Strain Ellipsoid) and making angles of 45 with them, and that their shear is e s e v if we assume e ly e z , e 3 to be in ascending order of magnitude. 3. Prove that a simple elongation in any direction is equivalent to a uniform cubical dilatation together with two shears, each having the given direction for one axis, the other axes being at right angles to it and to each other. li % 484 ANALYSIS OF STRAINS AND STRESSES. [290. Consider a cube whose three edges at the point are Ox, Oy, Oz, and suppose the given simple elongation, e, to take place along Ox. We may consider this as Je + * + J* along Ox, and we may suppose an elongation Je along Oy together with an elongation Je (or a contraction) in the sense of yO ; and similarly Je and ^e in Oz. Now ^ e along Ox, Oy, and Oz (and of course along all lines parallel to these) constitutes (p. 468) a cubical dilatation e ; while -Jc along Ox and ^c along Oy constitute (Art. 286) a shear, whose amount is | (Art. 286). Therefore, &c. 4. Eesolve a simple elongation e in a given direction into its components with reference to three rectangular axes. Am. If the direction-cosines of the direction of elongation with reference to the three axes are I, m, n, the elongations and shears to which e is equivalent are el 2 , em 2 , en z , 2 elm, 2mn, 2enl. For, if f, 17, f be the co-ordinates of any point before strain, the length of the perpendicular from this point on the plane through the origin perpendicular to the direction (I, m, n) is Zf+mrj + wf ; and the point (f, 77, f) is drawn out along this perpendicular through a distance e(l + mri + n). The projection of this distance along the axis of x is tl(l-\-mr}4-nC) : hence the strained co-ordinates (f , if, O, are rf = Comparing these values of ', if, with those given at p. 459, we see that el 2 = a, cm 2 = b, en 2 = c, eZm = s 3 , emn = s 1 , enl = s. 2 , which are the required components of the elongation with reference to the axes. 5. Find the condition that, in the general small strain, there should be two planes of no elongation. Ans. a, s 3 , s <2 s 3 , b, Sj = 0. Hence one of the principal elongations ? 2 , ^ G must be zero (see p. 472). 6. Given two small strains, (a, b, c, 2s 1} 2s 2 , 2s s ), (a, b', c' , 2s/, 2s/, 2s 3 '), find the resulting elongation quadric and strain ellipsoid. Ans. In the previous equations of these surfaces put a + a for a, i.e., the rate at which the force is o distributed per unit of area. Thus the atmospheric pressure on any area at the surface of the earth is roughly i5lbs. on every square inch, and if the unit of force is a pound weight and the unit of length an inch, the intensity of atmospheric pressure is represented by the number 15. If force acts over an area in such a way that there is not the same amount exerted on the same area everywhere, the distri- bution is not uniform ; and in this case we can speak only of the intensity of force at each particular point. If about any point we describe a very small area, ds, on which we may assume the distribution of force to be constant, and if dF is the amount of force on it, the intensity of force at the point selected is An instance of this occurs when the area pressed is any non- horizontal area in a heavy liquid. The intensity of pressure at points in the upper part of the area is less than the intensity at points in the lower part. 292.] Stress at a Point. At any point, P, of the body consider a small plane surface of area ds and any position. This may be regarded as separating the part (A) of the body at one side of it from the part (H) at the other side. Then the particles in this element plane, when the body is strained in any manner, are subject to certain forces proceeding from the particles at the side (A) and resulting from the elongation or contraction of the natural distances. The resultant of these forces is called the stress on the side (A) of the element plane. The particles in the element plane are also subject to forces proceeding from particles at the side (B) of the plane ; and the 293-] EQUILIBRIUM OF AN ELEMENT. 487 resultant of these latter is, of course, a stress equal and opposite to the first-mentioned stress. The resultant stress (on either side of the element plane) divided by the area, ds, is the intensity of stress on the plane ; and the resultant stress may be either normal to the plane, oblique to it, or in it. If at the same point P we consider a small plane surface of the same area as before, but of different position, the resultant stress on it will, generally speaking, be different both in magni- tude and in direction from the previous stress. In the case of a perfect fluid body the magnitude of the stress is constant and its direction is normal to the element plane, whatever be the position of the latter at the point P. Hence in the case of a strained body the term ' stress at a point ' has no definite meaning until we specify the element plane on which the stress acts. 293.] Equilibrium of an Element. At any point, (fig. 13, p. 19) whose co-ordinates with reference to three fixed rectangular axes are (x, y, z) let a very small rectangular parallelepiped of the substance be separated in imagination from the rest of the body by means of element planes perpendicular to the fixed axes; and through draw the lines Ox, Oy, Oz parallel to the fixed axes. We may then, if we actually produce on the faces of this element the stresses which are produced on them by the neighbouring portions of the body, consider the equilibrium of the element apart from the remainder *. Let the stress per unit of area on the face BOCF have for com- ponents along Ox, Oy, Oz the values P x , P y , P z , respectively; let the corresponding components for the plane AOCH be Qx> Q v , Qz> and let those for the face AOBD be R x , R y , R z . The stress on each face may be supposed to be applied at the middle point of the face, and each component is supposed to be measured in the positive sense of the corresponding axis. Let OA doc, OB = dy, OC = dz. Now these component * In considering the equilibrium of an element of a fluid body it is customary to say that we consider it as solidified and acted on by the stresses (pressures) which the fluid exerts on its surface. This solidification is however wholly unnecessary and misleading if, indeed, it is not actually wrong. The element while regarded as forming part of the body is not solidified, but is kept in its condition by the very forces which, by supposition, are produced on it by other means. If these forces were by themselves sufficient in the one case, they must be so in the other .without the aid of solidification. 488 ANALYSIS OF STRAINS AND STRESSES. [293. stresses are all functions of the position of 3 i. e., each of them is some function of (#, y, z). And the co-ordinates of A are (0 + *), the P x for the face DAHtf \$f(x+dx, y, z), i. e., it is P x + -j- dx, neglecting (dx) 2 ' &c. This component is, of course, directed in the sense AO, since the stresses produced on the faces BOCF and DAH(/ by the portions of the body removed are opposed. Hence the components of intensity of stress on DAHC/ are Similarly for the components of intensity of stress on the faces DBFtf and HCF(/. To get the whole amount of stress in any direction on any face, the intensity in this direction must, of course, be multiplied by the area of the face. Let us calculate the whole amount of stress parallel to Ox exerted on the parallelepiped. The face BOCF will contribute P x xdydz, while the opposite face, HALO*, will contribute (P x -\ =-^- dx) ** z ; and the sum of these is - - x dxdydz. The face ax AOCH will give a stress Q x x dzdx parallel to Ox, and the oppo- site face will give (Q x + -j^dy}dzdx; and the sum of these do y is j^dxdydz; similarly, the faces AOBD and HCFO' will 77? give j^ dxdydz. Hence the whole stress force acting on the element in the direction Ox is * ax ay az Some external force (gravity, or other) may also act on each element of the body. Such a force will always be proportional to the quantity of matter in the element. Suppose p to be density of the body at ; then, approximately, the quantity of matter in the parallelepiped is p dxdydz. Let the components of the external force which is felt at along the axes of x, y, z be X, Y, Z, per unit of mass. Then the component of the external force along Ox exerted on the element is pXdxdydz. Equating to zero the sum of the components along Ox of all forces exerted on the element, we have 293-] EQUILIBRIUM OF AN ELEMENT. 489 dP dQ dR m _ "~"~''- PA - dq z dR. the last two equations being 1 obtained by resolving forces along the axes of y and z. In a strained medium in which the stress on every plane is normal the equations of equilibrium are dP dQ dR d^ = pX > fy =pr > ^ = pZ ' since the tangential components P y , P z , Q x , ... are zero; and if, in addition, as in a perfect fluid, the intensity of stress is the same on all element planes at a given point, P Q = R, and these equations become the well-known hydrostatical equations dP dP dP -= = pJL -y- = pi, -=- = p/J. dx dy dz For any kind of body we obtain another valuable set of equations by expressing the equilibrium of the moments of the forces acting on the parallelepiped. For example, take moments about the line joining the middle points of the opposite faces -BOC^and DAHV. The external force* acting on the parallele- piped may be considered to act at its middle point; it will therefore contribute nothing to the moments about the axis chosen. Neither will the stresses on the faces BOGF and DAIIC/, since these stresses act at the middle points of the faces. Of the stresses on the faces AOCH and DBFO' the * It is important for the student to distinguish two species of external force acting on any body. There may be external forces which act only at particular points on its surface as, for example, when a beam rests against the ground and against a wall, the reactions of the ground and wall and there may be external forces which affect every element inside the body as, in the same case, the attraction of the earth which produces a force (the weight) on each element of the beam. The latter are called continuous forces. Thus a strained body may be affected by both the above beam, if slightly flexible, will be bent. The forces (per unit of mass), X, Y, Z, in equations (1) belong exclusively to the second kind. Forces of the first kind do not enter into these equations ; they are like the terminal tensions of a string, and are required for determining the values of constants which occur in the integrals of the differential equations (1) of equili- brium. 490 ANALYSIS OF STRAINS AND STRESSES. [293. components ^xdxdz and (Q z + -j^dy]dxdz, which are parallel to Oz will alone contribute moments. The moment of the first is Q z x dxdz x > or \ Q z dxdydz, and the moment (in the same sense) of the second is (Q z + -~dy)dxdzx -j-> or ay & \Q z dxdy dz, neglecting the term dx(dyfdz. The sum of these moments is Q z dx dy dz. Again, of the stresses on the faces AOBD and HCFO' the components, R y x dxdy and (R y -f ~~ dz)dxdy, will alone con- tribute ; and the sum of their moments is R y dxdydz, which is obviously in the sense opposite to that of the previous moment. Hence equating the sum of these moments to zero, Similarly, P y = Q x , I ( 2 ) which are obtained by taking moments about the lines joining the middle points of the faces (AOBD, HCFtf] and (AOCH, DBFO'), respectively. The stress (per unit of area) on the face BOCF can be resolved into two, viz., one normal to the face and the other in the face. The first is P x , and the second (which is the shearing force on the face) is P^ + P*. Equations (2) obviously assert that if we take any two element planes at right angles to each other at any point of the body, the component along the normal to the second of the stress per unit area on the first is equal to the com- ponent along the normal to the first of the stress per unit area on the second. We shall now see that this very important result is true for two element planes inclined at any angle to each other. To save a multiplicity of symbols, use N^ for P x , N 2 for Q y , JV 3 for R n , T 3 for P y and Q X9 T 2 for P z and R x , ^ for Q z and E,\ N standing for normal and T for tangential intensity of stress. Consider now the equilibrium of a tetrahedral element of the body included between the plane ABC (fig. 13, p. 19), and the planes BOG, AOC, AOB. Let the components along Ox, Oy, Oz of the stress per unit area on the triangular face ABC be P, Q, R\ and let the direction-cosines of the perpendicular on this plane 293.] EQUILIBRIUM OF AN ELEMENT. 491 be /, m, n. Resolve along Ox the forces acting- on the tetrahedral element. The face BOG will contribute N^ x BOG (where BOG means the area of the face) ; the face AOC will contribute T 3 xAOC; the face AOB will contribute T 2 xAOB; the face ABC will contribute P x ABC ; and the external force pXx%0A.OB.OC. Hence P x ABC = JVi x BOC+ T 3 x AOC+ T 2 xAOB +%pX*OA.OB.OC. Divide out by ABC. BOC AOC AOB OA.OB.OC . Therefore P = IJ\\ + mT 3 + nT 2 + %pXxl. OA. But by taking all dimensions of the element very small, the term ^pX. OA proceeding from the external force, ultimately vanishes, and we have accurately Similarly, Q = lT 3 + mN 2 + nT lt (3) by resolving along % and Oz. These very important equations give us the intensity of stress in magnitude and direction on any assigned element plane when the stresses on three rectangular element planes are known ; they are, in fact, the composition and resolution of stress. Any one of these equations (3) suffices for the proof of the im- portant general theorem of projection already referred to. For P is the projection, along the, normal to the element plane BOC, of the intensity of stress on the element plane ABC, and is the projection, along the normal to the latter plane, of the intensity of stress on the former. This theorem is true therefore for any two element planes at a point. Remark. The components of stress on an element plane at the bounding surface of the body are to be equated to the com- ponents of the external force applied to the surface at the element. COR. It follows immediately from this theorem of the projec- tions of two stresses that if there is at a point in the body any plane on which the stress is zero, the lines of action of the stresses on all other planes at this point lie in this plane of zero stress. 492 ANALYSIS OF STRAINS AND STRESSES. [293. A^r, When the stress on an element plane, -GT, exercised by the part, A, of the body on one side of it consists of a force whose com- ponent normal to tsr is directed from this plane towards the part A, the stress on ta is called tension ; and when the normal com- ponent is directed from A to OT, it is called pressure. All fluid stress is pressure. In general at every point inside a strained body there will be some planes on which the stress is pressure, and others on which the stress is tension. It may assist the student to understand the nature of the action of stress on an element plane if we draw a figure representing the equilibrium of these stresses on an element of the body. Thus if we take the elementary parallelepiped OCf (fig. 13, p. 19) to be a cube, and also take (as we may) the stress on any face as acting at its middle point, the forces in the plane of xy may be represented as in fig. 266, which is that of a section of the cube through its centre and parallel to the plane of xy. If there were no stresses on planes parallel to %y> this figure would com- pletely represent the equilibrium of the cubical element. (Since the faces have been all taken as equal in area, the intensities of stresses are proportional to the stresses acting on them.) It is evident, of course, that when the stresses on any three planes at a point (rectangular or not) are known, the stress on every plane at this point can be found both in magnitude and in line of action. For we may con- sider the equilibrium of the tetrahedral element contained by the assumed plane and the three given ones, and the re- quired force will be equal and opposite to the resultant of three given forces. Let it, for example, be given that the stress at any point P is a shearing stress in each of two rect- angular planes, there being no stress on planes perpendicular to both of them. Suppose that all planes in the neighbourhood of P which are perpendicular to the plane of the paper and parallel Fig. 267. 293-] EQUILIBRIUM OF AN ELEMENT. 493 to CD (fig-. 267) are subject to a shearing stress, and that all planes parallel to AD and perpendicular to the paper are also subject to shearing 1 stress, and that planes parallel to the paper are not subject to stress. The intensities of these shearing stresses are obviously equal (either by what precedes, or by con- sidering the equilibrium of a small prism whose base is the square ABCD and whose edges are perpendicular to the paper. The equality of moments round an axis through P perpendicular to the figure gives the equality of the intensities of these shears). Let their common intensity be S, and suppose them represented by the arrows. Draw the plane AC, and consider the equilibrium of the portion ABC of the body (or rather of a little right prism whose base is ACB). It is kept in equilibrium by the forces 8 acting in the lines DC and DA and by the stress on the face AC. This last must (since it may be supposed to act at the middle point of AC) act in the line PD from P to D. If h is the height of the prism, the areas of its faces are h x AC, h x CD, h x DA ; so that the forces acting in DC and DA are Sxhx DC and Sx%x DA ; and their resultant, F, which is equal and opposite to the stress on AC, is given by the equation h 2 x DC 2 + S*xtf F i. e., the intensity of stress on the face AC is equal to the intensity of the shearing stress on each of the other two faces ; moreover, the stress on AC is normal to AC. This stress is the action of the portion of the body at the right-hand side of AC on the particles in the plane AC, and since it acts in the sense PD, it is a pressure. Hence if the portion of the body at the right-hand side of AC, or of any plane parallel to it and near it be removed, a pressure of intensity S must be applied to the plane in the sense PD. The action of the part of the body at the left-hand side of AC, or of any parallel to and near it, consists, of course, of a pressure in the opposite sense ; so that if we draw two element planes HI and JK parallel to AC and consider the portions of the body at the right of the first and at the left of the second as removed, two pressures (indicated by the arrows pointing to B and D) must be applied to the portion of the body contained be- tween these planes. 494 ANALYSIS OF STRAINS AND STRESSES. [294. Similarly, by drawing BD and considering the equilibrium of the prism standing on the base BCD, we see that the action of the portion of the body at the lower side ofBD on the particles in this face consists of a normal stress of intensity 8 directed in the sense CP, i. e., towards the parts considered as removed ; in other words, this stress is a tension. Consequently if we isolate in imagination a small prism of the body standing on the square HIJK, we regard it as acted on by two pressures on its faces HI and JKj and by two tensions on its faces // and KH. The state of stress of the body at P may just as well be produced by applying normal stress (pressure), of the same intensity as the shearing stress, to all planes parallel to AC and near it, and normal stress (tension), of same intensity, to all planes parallel to BD and near it ; in other words, we may sub- stitute this state of stress for the shearing stress. e/ e/ *7 Hence a shearing stress on two rectangular planes at any point produces equal normal stresses of opposite signs (pressure and tension) and of intensities equal to * that of the shearing stress on the two planes which Used the angles between them. This result follows, of course, from equations (3) by taking the lines from P perpendicular to CD and BC as axes of x and y, and putting N = 0, N 2 = 0, N 3 = 0, 7\ = 0, T 2 = 0, T 3 = S, I = m = =, n=0. From these equations also we deduce the V2 magnitude and line of action of the stress on any plane near P. The student will do well, however, to deduce from the figure the stress on any plane through (or near) P perpendicular to the figure. 294.] Problem. Given the condition of stress of a body at any point in it with reference to one set of rectangular planes, to find the condition of stress at the same point with reference to any other set of rectangular planes. Let the given stresses at a point 0, on three rectangular planes of aoy t yz> zx, be N 19 N 2 , N 3 , T 1} T 2 , T 3 , as in last Article. Then the components along the axes of #, y, z of the stress per unit area on an element plane at the point the direction-cosines of whose normal are I, m } n are given by equations (3) of last * Compare with the corresponding result in the case of shearing strain. The shearing strain may be replaced by two simple elongations, the magnitude of each being Jtalf that of the shear. (See p. 447.) 295-] CONE OF SHEARING STEESS. 495 Article. The resolved part, T, of this stress along any line whose direction-cosines are A, /u, v is \P-\- pQ + vR; i.e., T l\N 1 + miJiN 2 + nifN 3 + (l^-^m\)T 5 + (mv + n^')T 1 + (n\ + lv}T 2 . (1) If the line along- which the stress is resolved is the normal to the element plane itself, the component, N, is IP + mQ+nlt; i.e., N=l 2 N 1 + m 2 N. 2 + n 2 Ns+2lmT 3 +2mnT 1 + 2nlT 2 . (2) Let it be required to find the intensities of stress on three other rectangular element planes at whose normals are Ox', Oy ', Oz'', and let the direction-cosines of these normals with respect to Ox, Oy, Oz be (I, m, n), (/', m', n'\ (I", m", n"), re- spectively. Denote the components of the intensity of stress on the plane // by N{ along Ox', T 3 ' along Oy, and T 2 along #/; the components of the intensity of stress on the plane z'x by T 3 ' along Oaf, N 2 along O/, and T along Oz'; and those of the intensity of stress on the plane x'y' by T 2 along Oof, T{ along 0/, and N B ' along Oaf. Then N{ is given by (2) ; N 2 ' is obtained by using (I', w! ', n') for (I, m, n) in (2) ; N 9 ' by using (I", m" ', n") for (J, m, n) in (2) ; T; by using (r, < n') for (A, M , r) in (l) ; T{ by using (r, m , n"} for (A, /LI, v) in (l) ; and T-[ by using (^, ^ 7 , n') for (/, w, ), and (I", m", n") for (A, M , v ) in (l). It will be seen from this that in transforming from one set of rectangular axes Ox, Oy, Oz to another, the quantities N^,N 2i JV 3 , T 3 , T l9 T 2 transform like x 2 , y z , z 2 , xy, yz, zx. The system of stress, thus calculated, on the new planes may be substituted for the original system of stress the two systems are, in other words, perfectly equivalent, and either will produce the other. 295.] Cone of Shearing Stress. The expression (2) for the normal component of intensity of stress on a plane may for all values of I, m, n (i. e., for all element planes at the point con- sidered) retain a positive value. In this case the normal com- ponent of stress is a tension on all planes. Or the expression may be negative for all planes, and then the normal stress will be pressure all round. Or, finally, it may be positive for some directions and negative for others. It will then be zero for some directions ; i. e., there will be planes on which the stress is entirely tangential. The directions of the normals to these planes are given by the equation 496 ANALYSIS OF STRAINS AND STRESSES. [296. n z + 2T 3 lm-\-2T 1 mn+2T 2 nl= 0, and therefore the normals trace out the cone fl 1 a? + N< t ^ + N 3 z 2 -t-2T 3 SE/y+2T 1 yz+2T 2 za! = 0, (l) the planes themselves tracing out the cone whose generators are perpendicular to the generators of this cone. This latter cone, when it exists, is called the Cone of Shearing Stress. 296.] Principal Planes of a Stress. The angle between the direction of stress and the plane on which it acts depends on the plane chosen. Let us try whether, with any given stress, it is possible to find a plane on which the stress is normal. If F is the resultant stress on a plane the direction-cosines of whose normal are (I, m, n), and if F acts in the normal, P = IF, Q = mF, R = nFj and equations (3) of Art. 293 become = IF, \ = mF, (1) and these give, by elimination of the direction-cosines, the cubic Xi-Z T 3 , T, T 3 , N,-F, 21 N t -P = 0, toil 1 or - -T 3 ")F-(N 1 N 2 N 3 -N 1 T^-N, 1 T/-N 3 T^+2T 1 T 2 T 3 ) = 0. This equation, as is well known, gives three real values of F, and equations (l) will give the direction-cosines of the planes subject to these normal stresses. The coefficients of this equation have, as is also well known, the same values, no matter what three rectangular planes are taken as those of reference. All theorems, therefore, concerning stress may be simplified by supposing that we have selected as planes of reference the three on which the stresses are normal. These are called the principal planes of the stress at the point considered. Let the stresses on them (per unit area, of course) be denoted by A, 12, C. The equations (l) which determine the planes and magnitudes of the principal stresses show that these planes are the principal planes of the quadric N^n* + N 2 f + N^ -f 2 T.,xy+ 2 T^z + 2 T 2 zx =/, (2) f being any constant force magnitude. 296.] PRINCIPAL PLANES OF A STRESS. 497 The equation of the tangent plane to this quadric at the point #', /, / is Let a normal be drawn to any element plane at the point, 0, considered, and let r be the length of this normal from to the surface of this quadric. Then by putting Ir, mr, nr for af, #', z', the tangent plane at the extremity of this normal is (by the values of P 3 Q, R in p. 491) * (3) The direction cosines of the perpendicular from on this POT? plane are -= > - > -^ > where F is the resultant stress (per unit area) on the element plane ; and these show that the resultant stress acts in this perpendicular. Again, if p is the length of the perpendicular from on the plane (3), we have F=-?-, (4) pr the value of the resultant stress. If the axes of the quadric (2) are taken as those of co- ordinates, we have JVi =A,N t = B, N 3 = C,T 1 =T 2 =T 3 =0; and the quadric has for equation The cone traced out by the normals to the planes of shearing stress is obviously the asymptotic cone of the quadric (2) ; and if this cone is real its reciprocal cone (the cone of shearing stress) will separate the planes on which the stress is pressure from those on which it is tension. When the cone is imaginary, all planes at the point will be subject to stress of one kind either pressure or tension. When the cone is real, the quadric (2) must be accompanied by another whose equation is obtained by merely changing f to /", as has been explained in the analogous case of strain (p. 461). Another graphic mode of connecting the stress on a plane with the position of the plane is this. Let the principal planes be taken as the co-ordinate planes ; then the components of the Kk 498 ANALYSIS OF STRAINS AND STRESSES. [297. intensity of stress on any plane (l } m, n) are by equations (3), Q = mS 9 (5) R = nC P 2 O 2 H 2 Hence 2 + ~ + -^ = 1. Of coarse P, Q, .# are the co- A Jj C ordinates of the extremity of the line representing the intensity of stress on the plane (I, m, n). Hence the extremities of lines representing in magnitude and direction the intensities of stresses on all planes at lie oh the ellipsoid x 2 v 2 z 2 ^ + j? + C*= 1 > ( 6 ) whose semi-axes are in magnitudes and directions the principal intensities of stress at 0. If a tangent plane be drawn to this ellipsoid parallel to the plane whose stress is considered, the length of the perpendicular from the centre on the tangent plane represents the magnitude of the intensity of stress, as is obvious by squaring and adding the sides of equations (5). The ellipsoid (6) may for shortness be called the Stress Ellipsoid. In proving general properties of stress simplicity is, of course, gained by taking the principal axes of the stresses as those of reference. Thus, with these axes, the cone of shearing stress is and that traced out by the normals to planes of shearing stress is Ax* -f By* -f Cz 2 = 0; so that for the reality of these cones (i.e. for the existence of planes subject wholly to shearing stress) the principal stresses must consist either of one tension and two pressures, or two tensions and one pressure. With any system of axes the equation of the cone of shearing stress is = 0. x y z 297.] Work done in Strain. We propose to investigate the work done in the strain of any small volume of the body. 297-] WORK DONE IN STRAIN. 499 About the point P (fig. 256, p. 458) let any small closed surface be drawn in the natural state of the body. Let ds be any element of this surface, and let the direction cosines of the normal to this element, measured outwards, be I, m, n. Then the components of intensity of stress (resulting* from strain) on the element plane ds being P, Q, R, and the final displacements of the mean point of the element being (see Art. 275) Af, Ar/, A the work done in the displacement of the element will be (see Art. 217, p. 366) Hence the work done in the strain of the volume contained in the whole surface is Substituting for P its value (p. 491), the term Pds becomes But if do- lt do- 2 , do-% are the projections of ds on the planes of yz t zx, and xy, respectively, Ids = dcr l9 mds = da 2 , nds = do-- ; so that the work done becomes The intensities of stress N l9 N 2 , ... may be considered as con- stant over the surface and taken outside the integral signs. Also substituting for Af, A 17, A f their values (Art. 275), we have u du ,.du\ ., du C du Now, the surface being closed, f^dcr^ = da> = volume enclosed by surface ; and fr)d(T 1 = ffd^ = 0, since, the normal being always drawn outwards, the elementary projections da^ on one side of the plane yz must be given a sign opposite to the sign of those on the other side. In this way we have also filda- 2 =f(d(T Q = d<*', fd(T 2 = fd n"\ Then for the system of planes on which the stresses are given we have N^= NJ = N 3 '= 0, and also T^= T z '= 0, since there is no stress on AOBD. Therefore if P', Q', R' are the components along OA, OB, OC of the intensity of stress on a plane whose direction-cosines with respect to these lines are A, p, v, we have Hence the components along OA, OB, 00 of the intensity stress on the plane yz are p , = ^ Q , = ^ R , = Q . and Ni is the sum of the components of these along the axis of a? ; therefore N, = IP' + I'tf + l"R' = 2 II' S. Also T 3 = mP f + m'Q + m"R' = (lm' + I'm) S, T 2 = nP' + n'Q' + ri'R? = (In' + I'n) S ; and hence the components of the given shearing stress are 2 ITS, Zmm'S, 2nn'S, (lm' + rm)8, (ln'+l'ri)S, (mn' + m'n) S. (Compare with the resolution of a shearing strain, p. 484.) 2. Two normal stresses on two rectangular planes are combined with two shearing stresses on the same planes j find the principal planes and intensities of the resultant stress. Let fig. 266, p. 492, represent the normal stresses N 1 and N 2 acting on planes at right angles to each other and to the plane of the paper, and combined with shearing stresses, T 3 , or S, in these planes. (Of course the figure represents the equilibrium of an element of the body.) Since there is no stress on any plane parallel to the plane of the paper, the stress on every plane lies in the plane of the paper (p. 491). Then T 3 = S, N^ = T^ = T 2 = ; and the principal planes are obviously perpendicular to the plane of the paper. Let the normal to 297-] EXAMPLES. 501 any plane perpendicular to the paper make an angle 6 with the direc- tion of N! . Then the components of stress on this plane are For a principal plane P = F . cos 0, Q = F . sin 0, where F is a principal stress. Hence m0 = 0, S.cos0 + (N 2 F) . sin 6 = 0. From these equations we find the two principal intensities of stress and the directions of the principal planes are given by the equation tan 20= 2 _5 . 3. If the stress on a plane is wholly a shearing stress, prove that its line of action is the line of contact of the plane with the cone of shearing stress, and find its magnitude. Since P = IA, Q = mB, R = nC, a point whose co-ordinates are x 2 v 2 z 2 P, ft R will lie on the cone T + -%- + -77 = 0, if A P + Bm 2 + Cn 2 = 0; ^1 Jj / that is, the extremity of the line representing the intensity of stress will lie on the cone of shearing stress if the stress is wholly shearing. Therefore, &c. Since the magnitudes of all stress intensities are repre- 3? y 2 z 2 sented by the radii vectores of the ellipsoid + -|- -f- = 1, the A. Jj C intensities of shearing stress will be represented by the radii vectores of this ellipsoid measured along the edges of the cone of shearing stress. 4. If at any point in a body the principal stresses consist of two tensions of intensities A and B (A>B) and a pressure of intensity C, prove that the maximum intensity of shearing stress is */AC, and find the plane on which it is exerted. 5. If at any point in a body the principal stresses consist of a tension of intensity A and two pressures of intensities B and C (B > (7), prove that the maximum intensity of shearing stress is VAB, and find the plane on which it- is exerted. Ans. l.= 6. Find the conditions that a stress whose components with respect 502 ANALYSIS OF STRAINS AND STRESSES. [298. to any three rectangular axes are given should produce shearing stress on two planes only, and these rectangular. NTT 1V 1 a 3> ^2 TN * 3 > " 2 Ans. + , ^ 2 , = 0, and N t -{-N 2 + N 3 = 0; the first TV T 19 N expresses that the product of the three principal stresses = 0, and the second that their sum = ; so that one principal stress must be zero and the other two a tension and a pressure of equal intensities., 7. Given the components of the stress with reference to the principal axes of the stress, find the components of the same stress with reference to any set of rectangular axes. ,= A IT + B.m'm" + Cn'n", T,= A II" + Emm" + Gnu", SECTION III. Expression of Stress in terms of Strain. 298.] Coefficients of Elasticity. The strain at any point depends, in the first instance, on the nine quantities du du du dv dv dv dw dw dw dx dy dz dx dy dz dx dy ' dz Now the strain being small, we may evidently assume that if these components of strain are all increased in the same ratio, the stress components which correspond to them will all be increased in the same ratio. Hence each of the six stress com- ponents, NI, N 2 ,N^ T-^ T 2 , T 3) is a linear function of the nine strain components ; so that we have, for example, , r du du du dv dv dv dw *M = *i -5 h C 9 1 H ^q -^ h dy dx dy dx 298.] COEFFICIENTS OF ELASTICITY. 503 which is a shear ; and similarly we have two other pairs, which are also shears. Hence, as the strain really involves only six components, #, #, c, 2^, 2<5 2 , 2s 3 , each stress component is a linear function of only six quantities ; and there are therefore only thirty- six distinct coefficients. There is a further reduction of this number to twenty-one in all cases of strain, irrespective of the nature of the strained body a reduction which is thus made by Green (see the Mathematical Papers of the late George Green, pp. 249, &c.). The work done in bringing a body from any one state of strain to any other must be simply a function of the quantities which define the magnitudes of the two strains ; i. e., it cannot depend on the order or nature of the series of states of strain, through which the body may pass from the first state to the second j in other words, the stresses must be a conservative system (see p 309). For, if this were not the case, we might bring the body from a state (A) to a state (B) through a certain series of states by the expenditure of a certain amount, 'W t of work, and then (by constraint, implying no expenditure of work) make it return from (B) to (A) through another series of states, and in this series we might receive from the stresses an amount, W+ W, of work done against external resistances. Each cycle of changes would therefore create an amount of work, and perpetual motion would be possible. The reasoning would be conclusive were it not for the fact (well pointed out and explained by Thomson and Tait, Nat. Phil.) that compression (as a rule) generates heat and extension (as a rule) causes a loss of heat; and this alteration of temperature at every moment affects the elasticity of the body, and therefore the stresses. Hence even when the body is at two different times in the same state of strain, the stresses may not be the same in these states ; and the above reasoning for the existence of a potential of stress falls to the ground. If, however, the states of strain are produced slowly, so that the temperature may be sensibly constant, the stresses will always be the same in the same state of strain ; and the work done in strain will be simply a function of the strain. By p. 499, the work done in the very small strain (da, ... d* lt ...) of an element of volume da is 504 ANALYSIS OF STRAINS AND STRESSES. [299. and if (a, b, c, s l} s 2 , s a ) is the potential of the strain per unit of volume, this work mus-t be - db Hence Ai=2^, N,= 2^ , ... T 1 = ^ , .. da db ds 1 Since N lt . . . are linear functions of #, ... , is obviously a homo-| geneous quadratic function of the six components of strain, and it has therefore twenty-one distinct coefficients, which are those entering into the values of the components of stress. For the particular case of Isotropic Bodies (p. 474) these coefficients reduce to two, as has been differently shown by Green, Lame, and Kankine. Green's method consists essentially in so determining the constants in $ that it shall be sym- metrical all round each of three axes as it must be for iso- tropic, as distinguished from crystalline, bodies. 299.] Method of Cauchy. This simple method consists in assuming that at every point in a strained isotropic body the principal axes of the strain coincide with the principal axes of the stress. Here then we have i=* 2 =* 8 =0, ^ = ^=^=0. Also we can assume A = (\ + 2 t j)e l + \e 2 -\-\e 3 , where \ and ju, are constants; for e 2 and e% must evidently have the same coefficient in the value of A, since the body is elastically symmetric with regard to the axes of y and z (and, of course, with regard to all axes) and the plane on which N^ acts is also symmetrically placed with respect to them. Thus A = A0 + 2/Z*!, \ .=A0-f2 M * 2 , V (1) C = A0+2/xtf 3 , ) where 6 = 1 + 2 +6? 3 = the cubical dilatation, and e lt e 2 , e 3 are the principal elongations. It is required to express the components, N 11 N 2 ,N 5 ^ T v T 2 ,T. At of the stress at the point considered in the body with reference to three rectangular axes at the point arid the corresponding components of the strain. Let (I, m, n\ &c., be the direction- cosines of the new axes with reference to the principal axes of strain and stress. Then by multiplying both sides of equations ^00.] METHOD OF THOMSON AND TAIT. 505 (l) by I 2 , m 2 , n 2 , respectively, and adding, we have by example 7, p. 502, and example 9, p. 485, Similarly JVi = \0 + 2 pa. (2) And by multiplying the sides of equations (I) by I'l", and adding, we have by the same examples Similarly, T : = (3) 300.] Method of Thomson and Tait. If a spherical portion of an isotropic body be subject to pressure of uniform intensity all over its surface, it must in yielding retain its spherical form, i.e. it experiences no distortion* And if a cube of it be subject to shearing stresses in the planes of its faces, it must, for a small strain, undergo distortion (into the shape of a slightly oblique parallelepiped) without alteration of volume, and the amount of this distortion (defined as in Art. 286) must be the same no matter to what side of any face the shearing stress is parallel. Consequently the elastic quality of a completely isotropic body depends on two, and only two, constants which are the same throughout its mass viz., its resistance to dilatation (or compression) and its resistance to distortion. Resistance to Dilatation. To find this constant, let a uniform tension (or pressure) of intensity N be applied all over the surface of any portion of the body and let it produce a small dilatation (or compression) of this portion, the amount of this dilatation being (defined as in Cor. 4, Art. 278) ; then the re- sistance to change of volume is N_ T' This resistance (since is a number) is a force per unit of area. Resistance to Distortion. To find this constant, let a shearing stress of intensity 8 be applied to any pair of parallel planes* and let the amount of the shear (defined as in Art. 286), be denoted by 2 s ; then the resistance, to distortion is ^ 2s' This resistance (since s is a number) is a force per unit of area. 506 ANALYSIS OF STRAINS AND STRESSES. [300. Denote these two coefficients respectively by k and //. The values of the shearing- stresses, T 19 T 2 , T 3 , in terms of the shears (given in equations (3) of last Art.) follow at once. To find the stresses called into play by a simple elongation, a, along the axis of x, resolve this elongation exactly as in example 3, p. 483, into a cubical dilatation a together with two shears. Now, by our above definition, the dilatation will cause a normal intensity of stress equal to ka on each face of a cubical element whose edges coincide with Ox, Oy } and Oz at the point 0. Consider the elongation ^a along Ox and the accompanying contraction %a \H along Oz. These give shears each equal !/ |\* to \a on the planes OCHD inclined at X j "X angles of 45 to Ox and Oz; and these *^ - ..... shears will, by the above definition, give pj 2 g 8 rise to shearing stresses each of intensity ju# on these planes. Again, by p. 493, these shearing stresses will give rise to normal stresses each of in- tensity \\j.a on planes parallel to 0/Tand CD ; and it is obvious that the normal stress on the plane OH (or rather the plane through OH perpendicular to the paper) produced by the portion of the body to the right of OH will be tension, i. e., it will be in the sense Ox ; while on the plane CD (or Ox) the normal stress produced by the portion of the body at the upper side of the figure will be pressure, i. e., it will be in the sense zO. Similarly by considering* the other shear (that which consists of elongation J a along Ox and contraction \ a along Oy) we have a further normal tension equal to fx# on the plane perpendicular to Ox-, and normal pressure f /u# on the plane perpendicular to Oy. Hence the elongation a gives normal stresses on the planes perpendicular to Ox, Oy, Oz, respectively. Similarly the elongation & (which is along Oy) gives normal stresses (A-| M ), (* + * M ), (*-!)* in the same directions ; and the remaining elongation, c 3 gives (k-Sfic, (k-lp)c, Hence we have 300.] EXAMPLES. 507 J\i = \K-ii- and N 2 = (k f/x)0-h2ju#, V (A) where 6 = a + & + c = the cubical dilatation. EXAMPLES. 1. To express Young's modulus in terms of the resistances to dilatation and distortion. Let a bar of the body be subject to traction, as in Art. 287. Then we have N s = (k f jut) (c 2a) + 2/xc ; N t = N 2 =. (k fju) (c 2 a) 2fJ,a. But the intensity of the elongating stress is -/V 3 , and the elongation (per unit of length) is c ; therefore if E = Young's modulus, \r JSss-J. C Also since there is no force on the sides of the bar ;, and - 3 = (k c When a bar is elongated, it thus appears that there is lateral con- traction (a) in all directions perpendicular to the axis of the bar, and the ratio of this to the elongation (c) is 2. One end of a bar of isotropic material is held fixed, and the bar hangs vertically ; find its elongation caused by its weight. Let AB be the bar in its natural state, P a point in AB at a distance z from A ; let A'B represent the elongated bar, and let P' be the displaced position of P. Then the intensity of stress on a normal section at P'= E -=-, where E is Young's modulus. But if co is the area of the section at weight of length PB W l-z P , the intensity of stress = - - ' - = -- =- > where 0) 60 / W and I are the weight and length of the bar. r dw W l-z Hence J ' -=- = -- r- dz o) I where C is a constant. Now the value of w for the fixed end is zero, 508 ANALYSIS OF STRAINS AND STRESSES. [300. therefore 0=0; and the value of w for the free end, B t is the amount of elongation. Hence, putting z = I, Wl amount of elongation = ^ &Ja<0 It is immaterial whether o> means the section of the bar A'B' or the section of A B, since these areas differ by a small quantity of the first order. 3. To find stresses produced at any point in a circular cylinder which undergoes torsion round its axis. With the notation of p. 478, we have by Art. 299. The torsion may be produced either by fixing one end of the cylinder and applying a couple to the other end, or by applying two equal and opposite couples to the ends, each of which is free. By considering the equilibrium of a portion of the cylinder between one end and a section made at any point, 0, (fig. 269) on the axis perpen- dicularly to the axis, we see that the stress system exerted over this section by the remaining portion of the cylinder must be a couple equal in amount to the applied couple, (F, F). Let the fixed axes of x and y at be Ox and Oy, and let P be a point in the section whose co-ordinates are x and y. Then the above values of the intensities of stress show that on the element area dS at P the two components of stress on the lower side of dS are ^- ydS in the direction Ox, and ~ xdS in the direction yO. The I sum of their moments about Oz is p> 2 g "- (x* -f- 2/ 2 ) dS in a sense opposite to that of the applied couple. Hence if the mo- ment of this couple is denoted by T, --y where r =OP, and the integration is extended over the whole area of the section at 0. Now fr^dS is the moment of inertia, 7, of the section about Oz. Therefore Let j 5 which is the rate of twist per unit length of the cylinder be I denoted by r, and we have ^ T f = T. () 300.] EXAMPLES. 509 Fig. 270. If the cylinder is solid (having no hollow part), / = The result in equation (a) is known as Coulomb's Law. 4. To show that Coulomb's Law cannot apply to a non-circular cylinder when it is acted on only by twisting couples at its ex- tremities. In order that the law of torsion strain expressed by the equations u = ryz, v = TXZ, w = may hold we shall show that force must be applied over the bounding surface of the cylinder parallel to its axis. Let fig. 270 represent a section of the cylin- der perpendicular to its axis, the axis passing through ; let P be a point on the bounding surface, PT the tangent to the section, and OQ a perpendicular to PT. Let OQ be taken as axis of x, the axis of z being the axis of the cylinder ; and let us calculate the stress on an element plane which touches the bounding surface at P. We have for this plane ? = 1, w = 0, n = 0; and equations (3) p. 491 give (by last example) P = 0, Q = 0, R = ~/*ry = -fir . PQ ; i.e., the stress on this plane is proportional to PQ, and there must be an applied force to balance this stress, since there is none of the material of the cylinder at the right-hand side of the plane. (See Remark, p. 491.) 5. Let there be a straight solid bar or beam subject to a slight bending strain such that the fibres (mean fibres) which lie in a certain plane, although bent, are not elon- gated, and that the elongation (posi- tive or negative) along every other fibre is proportional to its (positive or negative) distance from this plane, the bending of all fibres taking place parallel to a single plane which cuts the normal section of the bar perpendicularly. It is required to find for any normal section the sum of the moments, round the line in which it inter- sects the plane of the mean fibres, of the stresses which are exerted at the section by the strained fibres. Suppose that after the bending any one section, AHB (fig. 271), is brought by a motion as of a rigid body (Art. 274) back to its old position, and let a neighbouring section then occupy the position A'H'B'. Let HH' t cc' be two of the mean fibres which reach across from one of the sections to the other. Then the original distance Fig. 271. 510 ANALYSIS OF STRAINS AND STRESSES. [ 3 00. between the sections is HH' or cc. Let this be denoted by ds. If PP f is any other fibre reaching across, Pn and P'n the perpendiculars from P and P' on the right lines cH and c'H', the elongation along PP' (i.e. - j ) is proportional to Pn. Let the planes of the sections A HB and A'H'B' intersect in a line OL, let p denote the length of the radius of (cc 7 or nn'), and Pn = y. curvature (cO) Then evidently PP' p + y of the bent mean fibres nn p PP'-nn _ y nn' p which is the elongation along PP'. For fibres at the lower side of cZT, there is contraction, or negative elongation, and for these y is reckoned as negative. Now, by Hooke's Law, if we consider a small prism whose sides are the fibres emanating from points on a very small area, dcr, at the point P, the longitudinal stress of this prism is (p. 364) The moment of this force about cH is da; therefore the sum of Tji these moments all over the section AHB is --fy*d the angle made by R with the vertical ; a the angle which the tangent to the beam at A makes with the horizon ; h and \'W 2 7 2 - 300.] EXAMPLES. 511 k the distances, Ax and Bx, of the extremities from the line of intersection of the ground and wall. Let P be any point in the beam, at which we shall calculate the Bending Moment, i.e., the sum of the moments of all the forces acting on the beam between P and A ; let the horizontal and vertical lines through A be taken as axes of x and y ; let Q be any point between P and A ; let the co-ordinates of P and Q be (x, y) and (a/, y'), respectively; let the original length of the beam be /, and its weight W. W Then the weight of an element of length, ds', at Q is ds ', and the V moment of this force tending to produce curvature at P round a line (such as cH in fig. 271) perpendicular to the plane of the figure is Also the moment of R about this axis is Hence if p is the radius of curvature of the mean fibre at P, we = R (x cos (j> y sin $) j-f(xaf) ds', (1) P the integration being performed from A to P. If P is taken very close to A, the Bending Moment on the right side of (1) is zero, therefore p at A = oo , i.e., A is a point of inflexion ; and B is also a point of inflexion for a similar reason. Assume y = x iana + a^ + a^ + a^ ... t (2) where 3 , a 4 , a B , ... are all very small quantities; there being no &y term in x 2 since y-^ = 0(p=oo)at^l. From (2), we find = sec a + sin a (3a 3 # 2 + 4a 4 x 3 + 5# 5 # 4 + . . . ) u/x j 2 Now i = -- X - and if we ne lect P roducts of a sj 4 ~ we shall have - - cos 3 a(6a 3 #+ 12a 4 Also sin a . f*(x-x') -^- f -dx \x* sec a + Ja 3 sin a . Jo " Making these substitutions in (1), and equating to zero the 512 ANALYSIS OF STRAINS AND STRESSES. [300. coefficient of every power of x, we have _ R sin $ (cot (j) tan a) a * ~ fi PJ 7 Pn 3 n ' E1 cos 3 a JF while a 5 , a 6 , ... are of the order ^ and may be neglected. C? 2 W Also at the extremity B, -=-^ must be zero ; therefore and the equation of the mean fibre is W y = x tan a + x sec a. By putting k and A for y and #, this equation gives tan a = Wh 3 sec 4 a. Putting sec a = in the same term, we get = - where V is used for -//k 2 + A*. Substituting this value of a in the equation of the mean fibre, we have k irr 4 3 C-2^ 3 + C 4 ), which is the equation of the mean fibre, to the first power of = It will be easily found that -4iV, the abscissa of the centre of gravity of the beam, is A 7. A rigid bar is supported nearly horizontally on three given vertical props which are slightly elastic ; to determine the pressures on these props. Suppose that the props are fixed .__ * c . in the ground at D, E, and F (fig. 273), and that their extremities were originallly a, 6, c, which are in a horizontal line; but that when the shrinking has taken place, their extremities, A, B, (7, lie in a Fig. 273. let the pressures on them at A, the centre of gravity of the bar and W its weight. line slightly inclined to the horizon. j^ ^eir original lengths be p, q, r, so that Aa = 6>, Bb = bq, Cc =dr; and be P, Q, and R ; let G be 300.] EXAMPLES. 513 Then we have B-R.GC=Q, (1) the second being obtained by moments about G, Now if the areas of the normal sections of the props are a, /3, y, we have (Art. 216) p V *P Q F *V * F ** = & 3 = Mi > = ti > (2) a p p q y r supposing that Young's modulus is the same for all. Again, we must express the fact that ABC is a right line. Drawing through C a parallel to abc, we have bp-br _ AC b^-br~BC' . /. BC . bp-AC. bq + AB. br = 0, (3) or , by(2 ), t^^^^jr-a (4) The three equations (1) and (4) determine P, Q, R. 8. A heavy rigid slab is supported nearly horizontally on four given vertical props ; to determine the pressures on these props. Let the extremities, A, B, C, D, of the props when the shrinking has taken place be represented in fig. 212, p. 294; let the original lengths of the props be p, q, r, s ; let the perpendiculars from A and C on the diagonal ED be p' and /; let those from B and D on A C be (f and /; let the perpendiculars from G, the centre of gravity of the slab, on A C arid BD be x and y ; let P, Q, R, S be the pressures on the props, whose sections are a, /3, y, 8, respectively; and let JF= weight of slab. Then we have obviously the statical equations P+Q + R + S= W, Pp'-RS-Wx=Q, Qq'-Ss'+Wy = Q, (1) \G is supposed to lie within the area AOD] the two latter being equations of moments round BD and AC. We must now express the fact that A, B, C, D lie in one plane. To do this we shall calculate the vertical descent, b , of the point from the descents of A and C and also from those of B and D. Just as in last example, we have bp-br __A_ p' + r' f _ r'bp +p'br a^a7 " ~oc " ~7~ +/ (2) P _ Q , _ 7? o-o /o\ " ' ' V+ ^" o- -T i */ Similarly gf q +s xl , r'&p+p'br s'bq + q'bs therefore . ~ = / 7 -- p' + r' /+/ Also, as before, = E , &c., therefore (2) becomes a p The four equations (1) and (3) determine the pressures. Ll 514 ANALYSIS OF STRAINS AND STRESSES. [300. 9. When the external forces have a potential (for the law of inverse square), prove that the cubical dilatation satisfies the equation and that each component (u) of displacement satisfies the equation V 2 w = 0, d* d* d* where V = T- + -71 + ^~z * dx* dy z dz* These results follow easily. For if X, 7, Z in equations (1), p. 489, are -^-j -=-> -3-5 and if VF= 0, we obtain V0 = by dec dy dz differentiating the first of these with respect to x, the second with respect to y, the third with respect to z, and adding, using the values of tf lt T 3 , &c., given in equations (2) and (3) of p. 505. 10. If 7= is the equation of the surface of any solid subject to strain, but having no superficial stress prove that at all points on the surface, ^ 0, where U, = -y- > U 2 = -=- U 3 = -,- ; and that the stresses on all ax ay dz planes passing any point on the surface lie in the tangent plane at this point. 11. Investigate an expression, in terms of stress alone, for the work done in the small strain of a body. It has been shown (p. 500) that the work done in the strain of an element, dca, of volume is %(Ae 1 + J3e z + Ce 3 )da>. Now in equations (A) p. 507, using the principal stresses A, B, C for N lt N z , N 3 , and du dv dw the principal strains e lt e 2 , e 3 for -=- -r-> -=- we have A+B+C ' = 3kO ; and multiplying them by A, B, C and adding, we have Therefore if A + B + C = S, and AB + BC+ CA = 2, where E is Young's modulus. Hence the whole work of deformation the integration being extended throughout the whole body. 300.] EXAMPLES. 515 If we do not employ the principal stresses and strains, but those having reference to a given set of axes, the same expression gives the work, and S will stand for and 2 for N This expression for the work of deformation is Clapeyroris Theorem. (See Lame's Lemons sur L' jZlasticite, p. 83.) 12. Find the work done in the uniform compression of a body. Ans. If P is the intensity of external pressure exerted all over the surface, V the original volume and V' the final volume, the work is \P(V-V). [In a uniform compression u = ax t v = ay, w=- az' ) and of V V course 6 = The principal stresses are equal at all points, and each = JP.] 13. Prove that, although the volume of a solid body may not have changed during a small strain, there is work done in its deformation, and find an expression for this work. Ans. The work = ~ f(N* + N* + N* + 2 F* + 2TJ + 2 T*) da>, or - / (^. 2 + J5 2 + (7 2 ) c?co if we express it in terms of the principal stresses at each point ; and this cannot possibly vanish unless all the components of internal stress vanish. (Lame", p. 85.) [Assuming no change of volume at any element, 0=0, therefore Ni + N 9 + N a .= 0. In a fluid the stresses are all of the same kind (pressures) therefore the work = if = 0.] 14. If throughout a body there is only one principal stress (A), which is constant, prove that the work of deformation is VA* TT where V is its volume. (Lame*, p. 83.) 15. A weight is placed on an ordinary rectangular table which rests on the ground; calculate the pressures on the four legs, supposing that the legs may be treated as rigid in comparison with the ground. Ans. If the adjacent sides at any corner A are 6 and a, and if x and y are the distances from these sides, respectively, of the point of application of the resultant of the sustained weight and the weight of the table, the pressure on the leg through A is where W = sum of sustained weight and weight of table. 516 ANALYSIS OP STRAINS AND STRESSES. [300. 16. Prove that a circular cylinder can be subject to the strain u = ryz, v = TZX, w = cxy, (its axis being axis of z) provided that surface stress parallel to the axis is supplied. 17. Determine the components of strain as quadratic functions of the co-ordinates so that at all points we shall have and show that such strain will require the application of external surface stress. [Assume u px + qy + rz + \ (ax* + If -f cz* + 2fyz + 2gzx + 2 Tixy\ with similar values of v and w ; then let the equations be satisfied at all points, i. e., equate to zero the coefficient of each variable.] 18. Construct a diagram of the work done in slowly extending a cylindrical bar. [On the axis of x measure off from the origin a length, OA, equal to Z 05 the natural length of the bar; at A draw a line making with the axis J?fT of x an angle whose tangent is the numerical value of (see p. 364). ^o The ordinate, PM, of this line at any point, P, will represent the force which produces a length in OM in the bar; and the area of the triangle APM represents the work of extension. The result in p. 366 is graphically evident.] 19. A slightly elastic beam rests horizontally at any number of points against fixed vertical props, and is loaded uniformly between each successive pair of props. Prove that if M lt M v M 3 denote the bending moments at three successive points, A lt A 2 , A 3 , of support, we shall have 8 where a = A^A^ b = A Z A S , w = load per unit length throughout A^A Z , w' = load per unit length in A 2 A 3 . [The Bending Moment at any point = El -=-J > since, -j- being dx^ dx everywhere small, we may neglect its square. This is known as TJie Equation of Three Moments^ [The four following examples were communicated to the Author by the Rev. Professor Townsend.] 20. A horizontal beam, supported at both ends, being loaded with any number of isolated weights, if the bending moments be equal at any pair of contiguous weights, P and Q, they are equal throughout the entire interval PQ. 21. A uniform load, PQ, is moved along a horizontal beam sup- ported at both ends, A and B ; prove that at a given point, 0, in the beam the bending moment will be greatest when PQ occupies such a ... ,, . OP OA position that ^ = . 300.] EXAMPLES. 517 22. A uniform beam is tangentially faced at both extremities A and B, D is its point of greatest deflection, C is the foot of the perpendicular from D on AB ; X is any point in the line AB ; a perpendicular to AB at X meets the bent beam in T and the circular arc through A, D, B in Z. Prove that _XZ* ~-~CD' 23. A uniform beam is supported by four equidistant props, two of which are terminal ; prove that the two points of inflexion of its middle segment lie on the horizontal line of the props. MISCELLANEOUS EXAMPLES. 1 . Let the magnetic curves of a magnet be described, and suppose electric currents to run in wires coinciding with the curves. Prove that if C is the strength of the current in any wire and k the constant sum of cosines (see p. 39) corresponding to it, the force which it will exert on either pole of the magnet is proportional to [Hence the curve which cuts the magnet perpendicularly exerts the maximum force.] 2. If the walls of a room and an insulated electrified body inside it are at the same potential, prove that no electrical effects (attractions or repulsions) will be observed in the room. 3. A uniform beam, AB, is supported horizontally at two points, C and D, in its length, C being adjacent to A and D to B. Prove that if two circles be described with C and D for centres and CA and DB for radii, respectively, the two points of inflexion of the beam are the two limiting points of the coaxal system determined by the circles. (Rev. Professor Townsend). 4. A force, R, given in magnitude, line of action, and sense, is resolved into two components, P, Q, which are subject solely to the condition of passing each through a given point ; find a relation (in- volving only given quantities) between P, Q, and R. 5. Two equal bars, OA and 00, are freely jointed at the fixed point ; four equal bars forming a lozenge, A BCD, are freely jointed at A, B, C, and D, and the system (called a Peaucelliers Cell) is held in equilibrium by two forces applied at B and D. If the force at D is of constant magnitude in all positions of the cell, as it suffers defor- mation about 0, prove that the force at B will be one varying in- versely as the square of the distance OB. (Mr. G. H. Darwin, Pro- ceedings of the London Math. Soc., April 8, 1875. See the same paper for Mr. Darwin's most ingenious mechanical description of the M m 518 ANALYSIS OF STRAINS AND STRESSES. [300. Equipotential Lines of any number of magnetic poles by means of Peaucellier's Cells). 6. A given system of forces is to be reduced to two inclined at the angle a ; prove that the shortest distance between their lines of action cannot be less than -=r-cot- (Wolstenholme's Book of Math. Prob., JK i p. 387, second ed.) UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. ASTRONOMY LIERARY r AU2 NOV 7-4869 LD 21-100m-ll,'49(B7146sl6)476 M177047 . THE UNIVERSITY OF CALIFORNIA LIBRARY