Baixe statics and dynamics - s.l. loney e outras Manuais, Projetos, Pesquisas em PDF para Matemática, somente na Docsity! 9<9 /nJ.+ IJ i \ ^ -̂ ^ fM I X *-o i ii \ r*/^ % L - Q N 1! i U 3 BY THE SAME AUTHOR THE ELEMENTS OF COORDINATE GEOMETRY. Crown 8vo. Complete 12s. Part I. Cartesian Coordinates. Twenty -fifth Impression. Is. Part II. Trilinear Coordinates, etc. Third Impression. 6s. Key to Part I. 10s. net. Key to Part II. 7s. M. net. THE STRAIGHT LINE AND CIRCLE. Chapters I-IX of Part I of The Elements of Coordinate Geometry. Crown 8vo. 4s. AN ARITHMETIC FOR SCHOOLS. Seventeenth Impression. Globe 8vo. With or without Answers, 5*. Or in two Parts, with Answers, 3s. each. The Examples alone, 3s. Qd. The Answers alone, Qd. A NEW EDITION OF DR TODHUNTER'S ALGEBRA FOR BEGINNERS. Twelfth Impression. Globe 8vo. is. without Answers. 5s. with Answers. Answers only, Is. 3d. Key, 10s. 6d. net. A NEW EDITION OF DR TODHUNTER'S EUCLID. Fourth Impression. Globe 8vo. 5s. Also Book I, Is. Bd. 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Fifth Edition (revised and enlarged), July, 1901). Reprinted 1907, 1908, 1911, 1914, 1918, 1920 (twice) 1923, 1925, 1927, 1932 PRINTED IN GREAT BRITAIN PREFACE TO PART I. TN the following work I have aimed at writing a -*- working text-book on Statics for the use of Junior Students. Throughout the book will be found a large number of examples ; most of them, with the exception of many of those at the end of the Chapter on Friction and the Miscellaneous Examples at the end of the volume, are of an easy type. I have tried to make the book complete as far as it goes; it is suggested, however, that the student should, on the first reading of the subject, omit every- thing marked with an asterisk. I must express my obligations to my friend Mr H. C. Eobson, M.A., Fellow and Lecturer of Sidney Sussex College, Cambridge, for his kindness in reading through the proof-sheets, and for many suggestions that he has made to me. Any corrections of errors, or hints for improvement will be thankfully received. S. L. LONEY. Barnes, S.W. December, 1890. PEEFACE TO THE TENTH EDITION. THE book has been somewhat altered, and I hope improved, for this edition, and the type entirely re-set. Graphic solutions have been introduced much earlier, and more use has been made of graphic methods throughout the book. More experimental work has also been introduced. The chapter on Work has been placed earlier, and much greater stress has been laid upon the Principle of Work. Sundry somewhat long analytical proofs have been relegated to the last chapter, and here I have not scrupled to introduce alternative proofs involving the use of the Differential Calculus. For ten of the new figures in this book I am much indebted to the kindness and courtesy of Dr R. T. Glazebrook, who allowed me to use the blocks prepared for his Statics. Most of these figures have the additional merit of having been drawn from actual apparatus in use at the Cavendish Laboratory at Cambridge. S. L. LONEY. KoYAl. HoLLOWAY CoLliEGE Eng^efield Gkeen, Subbey. July 23rd, 1906. STATICS. CHAPTER I. INTRODUCTION. 1. A Body is a portion of matter limited in every direction. 2. Force is anything which changes, or tends to change, the state of rest, or uniform motion, of a body. 3. Rest. A body is said to be at rest when it does not change its position with respect to surrounding objects. 4. Statics is the science which treats of the action of forces on bodies, the forces being so arranged that the bodies are at rest. The science which treats of the action of force on bodies in motion is called Dynamics. In the more modern system of nomenclature which is gradually gaining general acceptance, the science which treats of the action of force on bodies is called Dynamics, and it has two subdivisions, Statics and Kinetics, treating of the action of forces on bodies which are at rest and in motion respectively. 5. A Particle is a portion of matter which is in- definitely small in size, or which, for the purpose of our investigations, is so small that the distances between its different parts may be neglected. L. S. 1 2 STATICS A body may be regarded as an indefinitely large number of indefinitely small portions, or as a conglomeration of particles. 6. A Rigid Body is a body whose parts always preserve an invariable position with respect to one another. This conception, like that of a particle, is idealistic. In nature no body is perfectly rigid. Every body yields, perhaps only very slightly, if force be applied to it. If a rod, made of wood, have one end firmly fixed and the other end be pulled, the wood stretches slightly if the rod be made of iron the deformation is very much less. To simplify our enquiry we shall assume that all the bodies with which we have to deal are perfectly rigid. 7. Equal Forces. Two forces are said to be equal when, if they act on a particle in opposite directions, the particle remains at rest. 8. Mass. The mass of a body is the quantity of matter in the body. The unit of mass used in England is a pound and is defined to be the mass of a certain piece of platinum kept in the Exchequer Office. Hence the mass of a body is two, three, four... lbs., when it contains two, three, four... times as much matter as the standard lump of platinum. In France, and other foreign countries, the theoretical unit of mass used is a gramme, which is equal to about 15*432 grains. The practical unit is a kilogramme (1000 grammes), which is equal to about 2*2046 lbs. 9. Weight. The idea of weight is one with which everyone is familiar. We all know that a certain amount of exertion is required to prevent any body from falling to the ground. The earth attracts every body to itself with INTRODUCTION 3 a force which, as we shall see in Dynamics, is proportional to the mass of the body. The force with which the earth attracts any body to itself is called the weight of the body. 10. Measurement of Force. "We shall choose, as our unit of force in Statics, the weight of one pound. The unit of force is therefore equal to the force which would just support a mass of one pound when hanging freely. We shall find in Dynamics that the weight of one pound is not quite the same at different points of the earth's surface. In Statics, however, we shall not have to compare forces at different points of the earth's surface, so that this variation in the weight of a pound is of no practical importance ; we shall therefore neglect this variation and assume the weight of a pound to be constant. 11. In practice the expression " weight of one pound " is, in Statics, often shortened into "one pound." The student will therefore understand that "a force of 10 lbs." means "a force equal to the weight of 10 lbs." 12. Forces represented by straight lines. A force will be completely known when we know (i) its magnitude, (ii) its direction, and (iii) its point of application, i.e. the point of the body at which the force acts. Hence we can conveniently represent a force by a straight line drawn through its point of application; for a straight line has both magnitude and direction. Thus suppose a straight line OA represents a force, equal to 10 lbs. weight, acting at a point 0. A force of 5 lbs. weight acting in the same direction would be repre- sented by 0, where B bisects the distance OA, whilst a 12 6 STATICS This statement will be found to be included in Newton's Third Law of Motion [Part II., Art. 73]. Examples. If a ladder lean against a wall the force exerted by the end of the ladder upon the wall is equal and opposite to that exerted by the wall upon the end of the ladder. If a cube of wood is placed upon a table the force which it exerts upon the table is equal and opposite to the force which the table exerts on it. 17. Equilibrium. When two or more forces act upon a body and are so arranged that the body remains at rest, the forces are said to be in equilibrium. 18. Introduction, or removal, of equal and opposite forces. We shall assume that if at any point of a rigid body we apply two equal and opposite forces, they will have no effect on the equilibrium of the body; similarly, that if at any point of a body two equal and opposite forces are acting they may be removed. 19. Principle of the Transmissibility of Force. If a force act at any point of a rigid body, it may be considered to act at any other point in its line of action provided that this latter point be rigidly connected with the body. Let a force F act at a point A of a body in a direction AX. Take any point B in AX and at B introduce two INTRODUCTION- 7 equal and opposite forces, each equal to F, acting in the directions BA and BX ; these will have no effect on the equilibrium of the body. The forces F acting at A in the direction AB, and F at B in the direction BA are equal and opposite; we shall assume that they neutralise one another and hence that they may be removed. We have thus left the force F at B acting in the direction BX and its effect is the same as that of the original force F at A. The internal forces in the above body would be different according as the force F is supposed applied at A or B ; of the internal forces, however, we do not treat in the present book. 20. Smooth bodies. If we place a piece of smooth polished wood, having a plane face, upon a table whose top is made as smooth as possible we shall find that, if we attempt to move the block along the surface of the table, some resistance is experienced. There is always some force, however small, between the wood and the surface of the table. If the bodies were perfectly smooth there would be no force, parallel to the surface of the table, between the block and the table; the only force between them would be perpendicular to the table. Def. When two bodies, which are in contact, are perfectly smooth the force, or reaction, between them is perpendicular to their common surface at the point of contact. CHAPTER II. COMPOSITION AND RESOLUTION OF FORCES. 21. Suppose a flat piece of wood is resting on a smooth table and that it is pulled by means of three strings attached to three of its corners, the forces exerted by the strings being horizontal ; if the tensions of the strings be so adjusted that the wood remains at rest it follows that the three forces are in equilibrium. Hence two of the forces must together exert a force equal and opposite to the third. This force, equal and opposite to the third, is called the resultant of the first two. 22. Resultant. Def. If two or moreforces P, Q, S. . . act upon a rigid body and if a single force, R, can be found whose effect upon tlie body is tlie same as that of tlie forces P, Q, S... this single force R is called tlie resultant of tlie oilier forces and the forces P} Q, S... are called tlie com- ponents of R. It follows from the definition that if a force be applied to the body equal and opposite to the force R, then the forces acting on the body will balance and the body be in equilibrium ; conversely, if the forces acting on a body balance then either of them is equal and opposite to the resultant of the others. COMPOSITIONAND RESOLUTION OF FORCES 1 1 Taking some convenient scale (say three inches, or less, per one lb.) mark off OA, OB, and OB to represent P, Q, and R lbs. Complete the parallelogram OACB. Then OC will be found to be equal in length, and opposite in direction, to OB. But, since P, Q, and R balance, therefore R must be equal and opposite to the resultant of P and Q. Therefore the resultant of P and Q is represented by 0C } i.e. by the diagonal of the parallelogram whose sides represent P and Q. This will be found to be true whatever be the relative magnitudes of P, Q, and R, provided only that one of them is not greater than the sum of the other two. In the figure P, Q, and R are taken respectively to be 4, 3, and 5 lbs. In this case, since 5 2=42+ 32 , the angle AOB is a right angle. "When the experiment is performed, it will probably be found that the point may be moved into one of several positions close to one another. The reason for this is that we cannot wholly get rid of the friction on the pivots of the pulleys. The effect of this friction will be minimised, in this and similar statical experiments, if the pulleys are of fairly large diameter; aluminium pulleys are suitable be- cause they can be made of comparatively large size and yet be of small weight. Apparatus of the solid type shewn in the above figure is not necessary for a rough experiment. The pulleys F and G may have holes bored through them through which bradawls can be put ; these bradawls may then be pushed into a vertical blackboard. The pulleys and weights of the foregoing experiment may be replaced by three Salter's Spring Balances. Each of these balances shews, by a pointer which travels up and down a graduated face, what force is applied to the hook at its end. 12 STA TICS Three light strings are knotted at O and attached to the ends of the spring balances. The three balances are then drawn out to shew- any convenient tensions, and laid on a horizontal table and fixed to it by hooks or nails as shewn. The readings of the balances then give the tensions P, Q, and B of the three strings. Just as in the 0^ V N C preceding experiment we draw lines OA, OB, and OC to represent P, Q, and B on any scale that is convenient, and then verify that OC is equal in magnitude and exactly opposite in direction to OD, the diagonal of the parallelogram of which OA and OB are adjacent sides. 26. To find the direction and magnitude of the re- sultant of two forces, we have to find the direction and magnitude of the diagonal of a parallelogram of which the two sides represent the forces. Ex. 1. Find the resultant of forces equal to 12 and 5 lbs. weight respectively acting at right angles. B ^C COMPOSITION AND RESOLUTION OF FORGES 13 Let OA and OB represent the forces so that OA is 12 units of length and OB is 5 units of length; complete the rectangle OACB. Then OC2=0^'2 +^C2=122+ 52 =169. .-. 0C=13. 5 Also tanCO^=^ = ] Hence the resultant is a force equal to 13 lbs. weight making with the first force an angle whose tangent is y%, i.e. about 22 37'. Ex. 2. Find the resultant of forces equal to the -weights of 5 and 3 lbs. respectively acting at an angle of 60. Let OA and OB represent the forces, so that OA is 5 units and OB 3 units of length ; also let the angle AOB be 60. Complete the parallelogram OACB and draw CD perpendicular to OA. Then OC represents the required resultant. Now AD=ACeosCAD=3cos60=%; OD: Also DC=4Csin60=3' and OC= JoiP+DC*= \/^F+?= V49=7, tnCOD= = *g=.SM. Hence the resultant is a force equal to 7 lbs. weight in a direction making with OD an angle whose tangent is *3997. On reference to a table of natural tangents this angle is easily seen to be about 21 47'. 27. The resultant, R, of two forces P and Q acting at an angle a may be easily obtained by Trigonometry. For let OA and OB represent the forces P and Q acting at an angle a. Complete the parallelogram OACB and draw CD perpendicular to OA, produced if necessary. Let R denote the magnitude of the resultant. 16 STATICS Exs. I 11 Find the angle between two equal forces P when their p resultant is (1) equal to P, (2) equal to . 12. At what angle do forces, equal to (A + B) and (A -B), act so that the resultant may be Ja^+B2? 13. Two given forces act on a particle; find in what direction a third force of given magnitude must act so that the resultant of the three may be as great as possible. 14. By drawing alone solve the following : (i). IfP=10; Q = 15; a= 37; find P. (ii). IfP= 9; g= 7; o=133; find P. (Hi). IfP= 7; Q= 5; P=10; find a. (iv). IfP=7-3; P=8-7; a= 65; find Q. 28. Two forces, given in magnitude and direction, have only one resultant ; for only one parallelogram can be con- structed having two lines OA and OB (Fig. Art. 27) as adjacent sides. 29. A force may be resolved into two components in ail infinite number of ways ; for an infinite number of parallelograms can be constructed having OC as a diagonal and each of these parallelograms would give a pair of such components. 30. The most important case of the resolution of forces occurs when we resolve a force into two components at right angles to one another. Suppose we wish to resolve a force F, represented by OC, into two components, one of which is in the direction OA and the other is perpendicular to OA. Draw CM perpendicular to OA and complete the paral- lelogram OMCN. The forces represented by OM and ON have as their resultant the force OC, so that OM and ON are the required components. COMPOSITION AND RESOLUTION OF FORCES 17 Let the angle AOC be a. C o Then OM = OC eos a = F cos a, and 0N=MO = OC sin a = Fain a. [If the point M lie in OA produced backwards, as in the second figure, the component of F in the direction OA = - OM= - OC cos COM= - OC cos (180 - o) = OC cos a=Fcos a. Also the component perpendicular to OA =ON=MC= OC sin COM=Fsin a.] Hence, in each case, the required components are ^cosa and i^sina. Thus a force equal to 10 lbs. weight acting at an angle of 60 with the horizontal is equivalent to 10 cos 60 (= 10 x = 5 lbs. weight) in a horizontal direction, and 10 sin 60 (= 10 x ^- = 5 x 1-732= 8-66 lbs. weight) in a vertical direction. 31. Def. The Resolved Part of a given force in a given direction is the component in the given direction which, with a component in a direction perpendicular to the given direction, is equivalent to the given force. Thus in the previous article the resolved part of the force F in the direction OA is F cos a. Hence The Resolved Part of a given force in a given direction is obtained by multiplying the given force by the cosine of the angle between the given force and the given direction. 32. A force cannot produce any effect in a direction perpendicular to its own line of action. For (Fig. Art. 30) there is no reason why the force ON should have any tendency to make a particle at move in the direction OA \* a 2 18 STATICS rather than to make it move in the direction AO produced ; hence the force ON cannot have any tendency to make the particle move in either the direction OA or AO produced. For example, if a railway carriage be standing at rest on a railway line it cannot be made to move along the rails by any force which is acting horizontally and in a direction perpendicular to the rails. 33. The resolved part of a given force in a given direction represents the whole effect of the force in the given direction. For (Fig. Art. 30) the force OC is completely represented hy the forces ON and OM. But the force ON has no effect in the direction OA. Hence the whole effect of the force F in the direction OA is represented by OM, i.e. by the resolved part of the force in the direction OA. 34. A force may be resolved into two components in any two assigned directions. Let the components of a force F, represented by OC, in the directions OA and OB be required and let the angles A OC and COB be a and fi respectively. Draw CM parallel to OB to meet OA in M and complete the parallelogram OMCN. Then OM and ON are the required components. Since MC and ON are parallel, we have OCM=p; also OMC= 180 - CMA = 180 - (a + /3) . Since the sides of the triangle OMC are proportional to the sines of the opposite angles, we have . 0M -J1. 0C BinOGM " iinMOC ~ sinOMC COMPOSITIONAND RESOLUTION OF FORGES 21 Hence the resultant of two forces, acting at a point and represented by the sides AB and BC of a triangle, is represented by the third side AC. 37. In the Triangle of Forces the student must carefully note that the forces must be parallel to the sides of a triangle taken in order, i.e. taken the same way round. For example, if the first force act in the direction AB, the second must act in the direction BC, and the third in the direction GA ; if the second force were in the direction CB, instead of BC, the forces would not be in equilibrium. The three forces must also act at a point ; if the lines of action of the forces were BC, CA, and AB they would not be in equilbrium ; for the forces AB and BC would have a resultant, acting at B, equal and parallel to AG, The system of forces would then reduce to two equal and parallel forces acting in opposite directions, and, as we shall see in a later chapter, such a pair of forces could not be in equilibrium. 38. The converse of the Triangle of Forces is also true, viz. that If three forces acting at a point be in equi- librium they can be represented in magnitude and direction by the sides of any triangle which is draum so as to have its sides respectively parallel to the directions of the forces. Let the three forces P, Q, and R, acting at a point t be in equilibrium. Measure off lengths OL and OAI along the directions of P and Q to represent these forces respec- tively. Complete the parallelogram OLNM and join ON. Since the three forces P, Q, and R are in equilibrium, each must be equal and opposite to the resultant of the 22 STATICS other two. Hence R must be equal and opposite to the resultant of P and Q, and must therefore be represented by NO. Also LN is equal and parallel to OM. Hence the three forces P, Q, and R are parallel and proportional to the sides OL, LN, and NO of the triangle OLN Any other triangle, whose sides are parallel to those of the triangle OLN, will have its sides proportional to those of OLN and therefore proportional to the forces. Again, any triangle, whose sides are respectively per- pendicular to those of the triangle OLN, will have its sides proportional to the sides of OLN and therefore proportional to the forces. 39. The proposition of the last article gives an easy graphical method of determining the relative directions of three forces which are in equilibrium and whose magni- tudes are known. We have to construct a triangle whose sides are proportional to the forces, and this, by Euc. I. 22, can always be done unless two of the forces added together are less than the third. 40. Lami's Theorem. If three forces acting on a 2>article keep it in equilibrium, each is proportional to the sine of the angle between the other two. Taking Fig., Art 38, let the forces P, Q, and R be in equilibrium. As before, measure off lengths OL and OM to represent the forces P and Q, and complete the parallelo- gram OLNM. Then NO represents R. Since the sides of the triangle OLN are proportional to the sines of the opposite angles, we have OL LN_ NO sinLNO ~ sinLON ~ smOLN' COMPOSITION AND RESOLUTION OF FORCES 23 But sin LNO = sinNOM= sin (180 - QOR) = sin QOR, sin LON= sin (180 - ZOff) sin ROP, and sin 0LN= sin (180 - PCX?) = sin i>0#. 6> OM NO sin #0# _ sin ROP ~ sin P0#' P _#_ H sin 2?OP Also Hence sin QOR sin POQ 41. Polygon of Forces. 7/* any number offorces, acting on a particle, be represented, in magnitude and direction, by tlie sides of a polygon, taken in order, the forces shall be in equilibrium. P B Let the sides Ah, BC, CD, DE, EF and FA of the polygon ABCDEF represent the forces acting on a particle 0. Join AC, AD and AE. By the corollary to Art. 36, the resultant of forces AB and BC is represented by AC. Similarly the resultant of forces AC and CD is repre- sented by AD ; the resultant of forces AD and DE by AE ', and the resultant of forces AE and EF by AF. Hence the resultant of all the forces is equal to the resultant of AF and FA, i.e. the resultant vanishes. 26 STATICS Exs. Ill 9. P= 251bs. wt., Q=201bs. wt. and = 35; find R and a. 10. P=50 kilog., Q=60 kilog. and P = 70 kilog. ; find a and 0. 11. P=30, P = 40 and a= 130; find Q and 0. 12. P= 60, a= 75 and 6= 40 ; find Q and P. 13. P=60, P= 40 and 0=50; find Q and a. 14. P= 80, a= 55 and R= 100 ; find Q and 0. 15. A boat is being towed by means of a rope which makes an angle of 20 with the boat's length; assuming that the resultant reaction R of the water on the boat is inclined at 40 to the boat's length and that the tension of the rope is equal to 5 cwt., find, by drawing, the resultant force on the boat, supposing it to be in the direction of the boat's length. EXAMPLES. IV. 1. Two forces act at an angle of 120. The greater is represented by 80 and the resultant is at right angles to the less. Find the latter. 2. If one of two forces be double the other and the resultant be equal to the greater force, find the angle between the forces. 3. Two forces acting on a particle are at right angles and are balanced by a third force making an angle of 150 with one of them . The greater of the two forces being 3 lbs. weight, what must be the values of the other two ? 4. The resultant of two forces acting at an angle equal to -gds of a right angle is perpendicular to the smaller component. The greater being equal to 30 lbs. weight, find the other component and the resultant. 5. The magnitudes of two forces are as 3 : 5, and the direction of the resultant is at right angles to that of the smaller force ; compare the magnitudes of the larger force and of the resultant. 6. The sum of two forces is 18, and the resultant, whose direction is perpendicular to the lesser of the two forces, is 12; find the magni- tude of the forces. 7. If two forces P and Q act at such an angle that P=P, shew that, if P be doubled, the new resultant is at right angles to Q. 8. The resultant of two forces P and Q is equal to *JSQ and makes an angle of 30 with the direction of P ; shew that P is either equal to, or is double of, Q. 9. Two forces equal to 2P and P respectively act on a particle ; if the first be doubled and the second increased by 12 lbs. weight the direction of the resultant is unaltered; find the value of P. COMPOSITION AND RESOLUTION OF FORCES 27 10. The resultant of two forces P and Q acting at an angle is equal to (2t + 1) >Jp~ + Q2 ; when they act at an angle 90 -0, the resultant is (2?-1)n/P*+Q 2 ; prove that m + l faoNT covert , 11. The resultant of forces P and Q is R; if Q be doubled R is doubled, whilst, if Q be reversed, R is again doubled ; shew that . P : Q : R :: SI2 : ^3 : J2. 12. If the resultant, R, of two forces P and Q, inclined to one another at any given angle, make an angle with the direction of P, shew that the resultant of forces (P+R) and Q, acting at the same a given angle, will make an angle - with the direction of (P+R). 13. Three given forces acting at a point are in equilibrium. If one of them be turned about its point of application through a given angle, find by a simple construction the resultant of the three, and, if the inclination of the force continue to alter, shew that the inclination of the resultant alters by half the amount. 14. Decompose a force, whose magnitude and line of action are given, into two equal forces passing through two given points, giving a geometrical construction, (1) when the two points are on the same side of the force, (2) when they are on opposite sides. 15. Two given forces act at two given points of a body; if they are turned round those points in the same direction through any two equal angles, shew that their resultant will always pass through a fixed point. 16. A, B, and C are three fixed points, and P is a point such that the resultant of forces PA and PB always passes through C ; shew that the locus of P is a straight line. 17. A given force acting at a given point in a given direction is resolved into two components. If for all directions of the components one remains of invariable magnitude, shew that the extremity of the line representing the other lies on a definite circle. 18. Shew that the system of forces represented by the lines joining any point to the angular points of a triangle is equivalent to the system represented by straight lines drawn from the same point to the middle points of the sides of the triangle. 19. Find a point within a quadrilateral such that, if it be acted on by forces represented by the lines joining it to the angular points of the quadrilateral, it will be in equilibrium. 28 STATICS Exs. IV 20. Four forces act along and are proportional to the sides of the quadrilateral ABCD; three act in the directions AB, BC, and CD and the fourth acts from A to D; find the magnitude and direction of their resultant, and determine the point in which it meets CD. 21. The sides BC and DA of a quadrilateral ABCD are bisected in F and H respectively ; shew that if two forces parallel and equal to AB and DC act on a particle, then the resultant is parallel to HF and equal to 2 . HF. 22. The sides AB, BC, CD, and DA of a quadrilateral ABCD are bisected at E, F, G, andH respectively. Shew that the resultant of the forces acting at a point which are represented in magnitude and direction by EG ana HF is represented in magnitude and direction by^C. 23. From a point, P, within a circle whose centre is fixed, straight lines PA 1 , PA 2 , PA 3 , and PAi are drawn to meet the circum- ference, all being equally inclined to the radius through P; shew that, if these lines represent forces radiating from P, their result- ant is independent of the magnitude of the radius of the circle. COMPOSITION AND RESOLUTION OF FORCES 31 These two equations give F and 6, i.e., the magnitude and direction of the required resultant. Ex. 1. A particle is acted upon by three forces, in one plane, equal to 2, 2^2, and 1 lbs. weight respectively ; the first is horizontal, the second acts at 45 to the horizon, and the third is vertical ; find their resultant. Here X=2 + 2N/2cos45+0= 2 + 2V2 . -7o= 4, 2^/2.-^ + 1= 3.7=0 + 2^2 sin 45 + 1 Hence Poos 0=4; JFsin0=3; f. 1^=^42+32=5, and tan 0= . The resultant is therefore a force equal to 5 lbs. weight acting at an angle with the horizontal whose tangent is , i.e. 36 52'. Ex. 2. A particle is acted upon by forces represented by P, 2P, 3*J3P, and 4P; the angles between the first and second, tlie second and third, and the third and fourth are 60, 90, and 150 respectively. Shew that tlie resultant is a force P in a direction iticlined at an angle of 120 to that of the first force. In this example it will be a simplification if we take the fixed line OX to coincide with the direction of the first force P; let XOX' and YOY' be the two fixed lines at right angles. The second, third, and fourth forces are respectively in the first, second, and fourth quadrants, and we have clearly P.OX=60; COX'= 30; and DOX=60. 32 STATICS The first force has no component along OY. The second force is equivalent to components 2Pcos60 and 2Psin60 along OX and OY respectively. The third force is equivalent to forces 3^3? cos 30 and 3^/3Psin30 along OX' and OY respectively, i.e. to forces - 3J3P cos 30 and 3JSP sin 30 along OX and OY. So the fourth force is equivalent to 4Pcos60 and 4Psin60 along OX and OY', i.e. to 4Pcos 60 and -4Psin60 along OX and OY. Hence X=P+ 2P cos 60 - 8 /3P ooa 30 + 4P cos 60 =P+P^+2P= _2- and 7= + 2P sin 60 + 3J3P sin 30 - 4P sin 60 = P^3+ 3 -^P-4P.^ = ^P. Henoe, if F be the resultant at an angle 6 with OX, we have Y and tan === - /3= tan 120, JL so that the resultant is a force P at an angle of 120 with the first force. 45. Graphical Construction. The resultant of a system of forces acting at a point may also be obtained by means of the Polygon of Forces. For, (Fig. Art. 41,) forces acting at a point and represented in magnitude and direction by the sides of the polygon ABCDEF are in equilibrium. Hence the resultant of foroes represented by AB, BC, CD, DE, and EF must be equal and opposite to the remaining force FA, %e., the resultant must be repre- sented by AF. It follows that the resultant of forces P, Q, R, S, and T acting on a particle may be obtained thus ; take a point A and draw AB parallel and proportional to P, and in succession BC, CD, DE, and EF parallel and proportional respectively to Q, R, S, and T; the required resultant will be represented in magnitude and direction by the line AF. COMPOSITION AND RESOLUTION OF FORCES 33 The same construction would clearly apply for any number of forces. Ex. Fourforces equal to 2, 2h , 1 and 3 kilogrammes wt. act along straightlines OP, OQ, ORandOS,"suchtliat L POQ= 40, L QOR= 10Q, and l ROS= 125; find their resultant in magnitude and direction. :i25 O 2 P A ---... '3 3 Draw AB parallel to OP and equal to 2 inches ; through B draw BC parallel to OQ and equal to 2-5 inches, and then CD parallel to OR and equal to 1 inch, and finally DE parallel to OS and equal to 3 inches. On measurement we have AE equal to 2-99 inches and L BAE equal to a little over 14. Hence the resultant is 2*99 kilogrammes wt. acting at 14 to OP. EXAMPLES. V. [Questions 2, 3, 4, 5, and 8 are suitable for graphic solutions. "\ 1. Forces of 1, 2, and ^/3 lbs. weight act at a point A in directions AP, AQ, and AR, the angle PAQ being 60 and PAR a right angle; find their resultant. 2. A particle is acted on by forces of 5 and 3 lbs. weight which are at right angles and by a force of 4 lbs. weight bisecting the angle between them ; find the force that will keep it at rest. 3. Three equal forces, P, diverge from a point, the middle one being inclined at an angle of 60 to each of the others. Find the resultant of the three. L. S. 3 3G STATICS 48. Ex. 1. A body of 65 lbs. weight is suspended by two strings of lengths 5 and 12 feet attached to two points in the same horizontal line wlwee distance apart is 13 feet; find the tensions of the strings. Let AC and BC be the two strings, so that AG=5it., BC=12ft., and ^B=13ft. A D Since 132= 122+ 52 , the angle ACB is a right angle. Let the direction CE of the weight be produced to meet AB in D ; also let the angle CBA be 0, so that lACD=90- lBCD= iCBD= 0. Let T x and T2 be the tensions of the strings. By Lami's theorem we have T x T 2 65 sin ECB ~ sin EGA ~ sin ACB ' y, ra _ 65 "' sin BCD ~ sin 6 ~ sin 906 ' .-. Tj= 65 cos 6, and T = 65 sin 0. * BC 12 . . . AGBut cos 6 = r = , and sin 6 = -7^= .-. T^eO, and T2=25 lbs. wt. 13' Otherwise thus ; The triangle ACB has its sides respectively per- pendicular to the directions of the forces Tlt T2 , and 65 ; BC~CA~AB' HP AC ;. r1 = 65^ = 60, and T2 = 65^s = 25.AB i AB Graphically; produce BC to meet a vertical line through A in 0. Then A CO is a triangle having its sides parallel to the three forces Tlt T2 , and W. Hence it is the triangle of forces, and " AG~ CO~ OA' COMPOSITION AND RESOLUTION OF FORCES 37 Ex. 2. A string ABCD, attached to tioo fixed points A and D, has two equal weights, W, knotted to it at B and G and rests with the portions AB and CD inclined at angles of 30 and 60 respectively to the vertical. Find the tensions of the portions of the string and the inclination of BG to the vertical. Let the tensions in the strings be Tj , T2 , and T3 respectively and let BG be inclined at an angle to the vertical. [N.B. The string BG pulls B towards G and pulls G towards B, the tension being the same throughout its length.] Since B is in equilibrium the vertical components and the hori- zontal components of the forces acting on it must both vanish (Art. 46). A IS0 N I Ti\B IT,,v Vrc Tw | w Hence Tx cos 30 - T2 cos =W (1), and ^ sin 30 -ra sin 0=0 (2). Similarly, since G is in equilibrium, r 8 cos6O o+:ra cos0=jr (3), and T8 sin6O-r3 sin0=O (4). From (1) and (2), substituting for Tlt we have W= T2 [cot 30 sin - cos 0] = r2 [v/3sin0 -cos 0] (5). So from (3) and (4), substituting for T3 , we have W= T2 [cot60sin + cos 0]= T3 T sin + cos 0~| (6) ; therefore from (5) and (6), ^3 sin - cos = -^ sin + cos ;V3 .-. 2 8^0=2^/3 008 0; .-. tan O^sjS, and hence 0=60. Substituting this value in (5), we have 38 STATICS Hence from (2), we have sin 60 ^ 2Ih7805-MWd ' and from (4) T3= T2^- = T2= W. Hence the inclination of BG to the vertical is 60, and the tensions of the portions AB, BG, and CD are TT^/3, W, and W respectively. EXAMPLES. VI. 1. Two men carry a weight W between them by means of two ropes fixed to the weight ; one rope is inclined at 45 to the vertical and the other at 30 ; find the tension of each rope. 2. A body, of mass 2 lbs., is fastened to a fixed point by means of a string of length 25 inches ; it is acted on by a horizontal force F and rests at a distance of 20 inches from the vertical line through the fixed point ; find the value of F and the tension of the string. 3. A body, of mass 130 lbs., is suspended from a horizontal beam by strings, whose lengths are respectively 1 ft. 4 ins. and 5 ft. 3 ins., the strings being fastened to the beam at two points 5 ft. 5 ins. apart. What are the tensions of the strings ? 4. A body, of mass 70 lbs., is suspended by strings, whose lengths are 6 and 8 feet respectively, from two points in a horizontal fine whose distance apart is 10 feet; find the tensions of the strings. 5. A mass of 60 lbs. is suspended by two strings of lengths 9 and 12 feet respectively, the other ends of the strings being attached to two points in a horizontal line at a distance of 15 feet apart ; find the tensions of the strings. 6. A string suspended from a ceiling supports three bodies, each of mass 4 lbs., one at its lowest point and each of the others at equal distances from its extremities; find the tensions of the parts into whioh the string is divided. 7. Two equal masses, of weight W, are attached to the extremities of a thin string which passes over 3 tacks in a wall arranged in the form of an isosceles triangle, with the base horizontal and with a vertical angle of 120; find the pressure on each tack. 8. A stream is 96 feet wide and a boat is dragged down the middle of the stream by two men on opposite banks, each of whom pulls with a force equal to 100 lbs. wt. ; if the ropes be attached to the same point of the boat and each be of length 60 feet, find the resultant force on the boat. COMPOSITION AND RESOLUTION OF FORCES 4 1 Let P be the force of the wind on the sail ; resolve it in directions along and perpendicular to the sail. The component (KA = ) Pcos (a - 6) along the sail has no effect. The component (LA = ) P sin (a - 6) per- pendicular to the sail may again be resolved into two, viz. (NA = ) Psin(a-0)cos0 perpendicular to AB and (MA = ) P sin (a - 6) sin 6 along AB. The former component produces motion sideways, i.e. in a direction perpendicular to the length of the ship. This is called lee-way and is considerably lessened by the shape of the keel which is so designed as to give the greatest possible resistance to this motion. The latter component, P sin (a - 6) sin 6, alongAB is never zero unless the sail is set in either the direction of the keel or of the wind, or unless a is zero in which case the wind is directly opposite to the direction of the ship. Thus there is always a force to make the ship move forward ; but the rudder has to be continually applied to counteract the tendency of the wind to turn the boat about. This force = jP[cos (a - 26) - cos a] and it is therefore greatest when cos (a -26) is greatest, i.e. when a -20=0, i.e. when = ~, i.e. when the direction of the sail bisects the angle between the keel and the apparent direction of the wind.] 49. Examples of graphical solution. Many problems which would be difficult or, at any rate, very laborious, to solve by analytical methods are comparatively easy to solve graphically. 42 STATICS These questions are of common occurrence in en- gineering and other practical work. There is generally little else involved besides the use of the Triangle of Forces and Polygon of Forces. The instruments chiefly used are : Compasses, Rulers, Scales and Diagonal Scales, and Protractors for measuring angles. The results obtained are of course not mathematically accurate; but, if the student be careful, and skilful in the use of his instruments, the answer ought to be trust- worthy, in general, to the first place of decimals. In the following worked out examples the figures are reduced from the original drawings ; the student is recom- mended to re-draw them for himself on the scale mentioned in each example. SO. Ex. 1. ACDB is a string whose ends are attached to two points, A and B, which are in a horizontal line and are seven feet GRAPHIC SOLUTIONS 43 apart. The lengths of AC, CD, and DB are 3|, 3, and 4 feet respec- tively, and at C is attached a one-pound weight. An unknoivn weight is attached to D of such a magnitude that, in the position of equilibrium, CDB is a right angle. Find the magnitude of this weight and the tensions of the strings. Let T x , T2 , and T3 be the required tensions and let x lbs. be the weight at D. Take a vertical line OL, one inch in length, to represent the weight, one pound, at C. Through O draw OM parallel to AC, and through L draw LM parallel to CD. By the triangle of forces OM represents 2\ , and LM represents !Ta . Produce OL vertically downwards and through M draw MN parallel to BD. Then, since LM represents T2 , it follows that T3 is represented by MN, and x by LN. By actual measurement, we have <W= 3-05 ins., LM= 2-49 ins., MN=5-1 ins., and NL= 5-63 ins. Hence the weight atD is 5-63 lbs. and the tensions are respectively 3-05, 2-49, and 5-1 lbs. wt. Ex. 2. A and B are two points in a horizontal line at a distance of \& feet apart; AO and OB are two strings of lengths 6 and 12 feet 46 STATICS Exs. VII 7. The jib of a crane is 20 feet long, the tie 16 feet, and the post 10 feet. A load of 10 cwts. is hung at the end of a chain which passes over a pulley at the end of the jib and then along the tie. Find the thrust in the jib and the pull in the tie. 8. In the figure of Ex. 3, Art. 50, the tie DG is horizontal and the chain coincides with it; if W=500 lbs., AG=11 feet, an<* X>C=5 feet, find the actions along DG and AC. 9. In the figure of Ex. 3, Art. 50, the angle CDB=45, and the angle ACD=15; the chain EG coincides with DC; if W be one ton, find the forces exerted by the parts AG, CD. 10. In the figure of Ex. 3, Art. 50, DA = 15 feet, DG=20 feet and AC=30 feet, and a weight of one ton is suspended from C, find the thrusts or tensions produced in AC, CD, and DA when the chain ooincides with (1) the jib CA, (2) the tie CD. 11. In the figure of Ex. 3, Art. 50, the jib AC is 25 feet long, the tie' CD is 18 feet, AD=12 feet and AE= Q feet; find the tensions or thrusts in AC and CD, when a weight of 2 tons is suspended from the end of the chain. 12. ABCD is a frame-work of four weightless rods, loosely jointed together, AB and AD being each of length 4 feet and BG and CD of length 2 feet. The hinge G is connected with A by means of a fine string of length 5 feet. Weights of 100 lbs. each are attached to B and D and the whole is suspended from A. Shew that the tension in AC is 52 lbs. weight. 13. In the preceding question, instead of the string AC a weight- less rod BD of length 3 feet is used to stiffen the frame ; a weight of 100 lbs. is attached to G and nothing at B and D. Shew that the thrust in the rod BD is about 77 lbs. weight. 14. In question 12 there are no weights attached to B and D and the whole framework is placed on a smooth horizontal table; the hinges B and D are pressed toward one another by two forces each equal to the weight of 25 lbs. in the straight line BD. Shew that the tension of the string is about 31*6 lbs. weight. 15. ABCD is a rhombus formed by four weightless rods loosely jointed together, and the figure is stiffened by a weightless rod, of one half the length of eaoh of the four rods, joined to the middle points of AB and AD. If this frame be suspended from A and a weight of 100 lbs. be attached to it at C, shew that the thrust of the cross rod is about 115-5 lbs. weight. CHAPTER IV. PAKALLEL FOKCES. 51. In Chapters n. and in. we have shewn how to find the resultant of forces which meet in a point. In the present chapter we shall consider the composition of parallel forces. In the ordinary statical problems of every-day life parallel forces are of constant occurrence. Two parallel forces are said to be like when they act in the same direction ; when they act in opposite parallel directions they are said to be unlike. 52. Tofind the resultant of two 'parallel forces acting upon a rigid body. Case I. Let theforces be like. Let P and Q be the forces acting at points A and B of the body, and let them be represented by the lines AL and BM. Join AB and at A and B apply two equal and opposite forces, each equal to S, and acting in the directions BA and AB respectively. Let these forces be represented by AD and BE. These two forces balance one another and have no effect upon the equilibrium of the body. Complete the parallelograms ALFD and BMGE ; let the diagonals FA and GB be produced to meet in 0. Draw OG parallel to AL or BM to meet AB in G. 48 STATICS The forces P and S at A have a resultant Px , represented by AF. Let its point of application be removed to 0. So the forces Q and S at B have a resultant Qx repre- sented by BG. Let its point of application be transferred to 0. The force P2 at may be resolved into two forces, S parallel to AD, and P in the direction OG. So the force Qx at may be resolved into two forces, S parallel to BE, and Q in the direction OG. Also these two forces S acting at are in equilibrium. Hence the original forces P and Q are equivalent to a force (P + Q) acting along OG, i.e. acting at G parallel to the original directions of P and Q. To determine the position of the point G. The triangle OCA is, by construction, similar to the triangle ALF ; OGALP " CA~LF~S' so that P.CA^S.OC (1). PARALLEL FORCES 51 Hence ^rn ~T *- e - @ divides the line AB externally in the inverse ratio of the forces. To sum up ; If two parallel forces, P and Q, act at points A and B of a rigid body, (i) their resultant is a force whose line of action is parallel to the lines of action of the component forces; also, when the component forces are like, its direction is the same as that of the two forces, and, when the forces are unlike, its direction is the same as that of the greater component. (ii) the point of application is a point G in AB such that P.AG^Q.BC. (iii) the magnitude of the resultant is the sum of the two component forces when the forces are like, and the difference of the two component forces when they are unlike. 53. Case offailure of the preceding construction. In the second figure of the last article, if the forces P and Q be equal, the triangles FDA and GEB are equal in all respects, and hence the angles DAF and EBG will be equal. In this case the lines AF and GB will be parallel and will not meet in any such point as ; hence the construction fails. Hence there is no single force which is equivalent to two equal unlike parallel forces. We shall return to the consideration of this case in Chapter vi. 42 52 STATICS 54. If we have a number of like parallel forces acting on a rigid body we can find their resultant by successive applications of Art. 53. We must find the resultant of the first and second, and then the resultant of this resultant and the third, and so on. The magnitude of the final resultant is the sum of the forces. If the parallel forces be not all like, the magnitude of the resultant will be found to be the algebraic sum of the forces each with its proper sign prefixed. Later on (see Art. 114) will be found formulae for calculating the centre of a system of parallel forces, i.e. the point at which the resultant of the system acts. 55. Resultant of two parallel forces. Experimental verification. Take a uniform rectangular bar of wood about 3 feet long, whose cross-section is a square of side an inch or rather more. A face of this bar should be graduated, say in inches or half inches, as in the figure. Let the ends A and B be supported by spring balances which are attached firmly to a support. For this purpose a Salter's circular balance is the more convenient form as it drops much less than the ordinary form when it is PARALLEL FORGES 53 stretched. On the bar AB let there be a movable loop C carrying a hook from which weights can be suspended; this loop can be moved into any position along the bar. Before putting on any weights, and when C is at the middle point of AB, let the readings of the balances D and E be taken. The bar being uniform, these readings should be the same and equal to R (say). Now hang known weights, amounting in all to W, on to C, and move C into any position Cx on the bar. Observe the new readings of the balances D and E, and let them be P and Q respectively. Then P-R(=P1) and Q-R(=QX) are the additional readings due to the weight W, and therefore P1 and Q1 are the forces at D and E which balance the force W at Cv It will be found that the sum of Px and Qx is equal to W (1). Again measure carefully the distances AGX and BCX . It will be found that P 1 xAC1 = Ql.C1 (2). In other words the resultant of forces Px and Qx at A and B is equal to Px + Qx acting at C1} where But this is the result given by the theoretical investigation of Art. 52 (Case I). Perform the experiment again by shifting the position of Clt keeping W the same; the values of Px and Qx will alter, but their sum will still be W, and the new value of P1 .AC1 will be found to be equal to the new value of Qt.BCx . Similarly the theorem of Art. 52 will be found to be true for any position of Cx and any value of W. 56 STATICS Exs. 5. Find two like parallel forces acting at a distance of 2 feet apart, which are equivalent to a given force of 20 lbs. wt., the line of action of one being at a distance of 6 inches from the given force. 6. Find two unlike parallel forces acting at a distance of 18 inches apart which are equivalent to a force of 30 lbs. wt., the greater of the two forces being at a distance of 8 inches from the given force. 7. Two parallel forces, P and Q, act at given points of a body ; P2 if Q be changed to -~- , shew that the line of action of the resultant is the same as it would be if the forces were simply interchanged. 8. Two men carry a heavy cask of weight 1^ cwt., which hangs from a light pole, of length 6 feet, each end of which rests on a shoulder of one of the men. The point from which the cask is hung is one foot nearer to one man than to the other. What is the pressure on each shoulder? 9. Two men, one stronger than the other, have to remove a block of stone weighing 270 lbs. by means of a light plank whose length is 6 feet ; the stronger man is able to carry 180 lbs. ; how must the block be placed so as to allow him that share of the weight ? 10. A uniform rod, 12 feet long and weighing 17 lbs., can turn freely about a point in it and the rod is in equilibrium when a weight of 7 lbs. is hung at one end ; how far from the end is the point about which it can turn? N.B. The weight of a uniform rod may be taken to act at its middle point. 11. A straight uniform rod is 3 feet long; when a load of 5 lbs. is placed at one end it balances about a point 3 inches from that end ; find the weight of the rod. 12. A uniform bar, of weight 3 lbs. and length 4 feet, passes over a prop and is supported in a horizontal position by a force equal to 1 lb. wt. acting vertically upwards at the other end; find the distance of the prop from the centre of the beam. 13. A heavy uniform rod, 4 feet long, rests horizontally on two pegs which are one foot apart ; a weight of 10 lbs. suspended from one end, or a weight of 4 lbs. suspended from the other end, will just tilt the rod up ; find the weight of the rod and the distances of the pegs from the centre of the rod. 14. A uniform iron rod, 2^- feet long and of weight 8 lbs., is placed on two rails fixed at two points, A and B, in a vertical wall. AB is horizontal and 5 inches long ; find the distances at which the ends of the rod extend beyond the rails if the difference of the thrusts on the rails be 6 lbs. wt. VIII PARALLEL FORCES 57 15. A uniform beam, 4 feet long, is supported in a horizontal position by two props, which are 3 feet apart, so that the beam pro- jects one foot beyond one of the props; shew that the force on one prop is double that on the other. 16. A straight weightless rod, 2 feet in length, rests in a horizon- tal position between two pegs placed at a distance of 3 inches apart, one peg being at one end of the rod, and a weight of 5 lbs. is suspended from the other end ; find the pressure on the pegs. 17. One end of a heavy uniform rod, of weight W, rests on a smooth horizontal plane, and a string tied to the other end of the rod is fastened to a fixed point above the plane; find the tension of the string. 18. A man carries a bundle at the end of a stick which is placed over his shoulder ; if the distance between his hand and his shoulder be changed how does the pressure on his shoulder change ? 19. A man carries a weight of 50 lbs. at the end of a stick, 3 feet long, resting on his shoulder. He regulates the stick so that the length between his shoulder and his hands is (i) 12, (ii) 18 and (iii) 24 inches ; how great are the forces exerted by his hand and the pressures on his shoulder in each case? 20. Three parallel forces act on a horizontal bar. Each is equal to 1 lb. wt., the right-hand one acting vertically upward and the other two vertically down at distances of 2 ft. and 3 ft. respectively from the first; find the magnitude and position of their resultant. 21. A portmanteau, of length 3 feet and height 2 feet and whose centre of gravity is at its centre of figure, is carried upstairs by two men who hold it by the front and back edges of its lower face. If this be inclined at an angle of 30 to the horizontal, and the weight of the portmanteau be 1 cwt., find how much of the weight each supports. CHAPTER V. MOMENTS. 57. Def. Tlie moment of a force about a given point is the product of the force and the perpendicular drawn from the given point upon the line of action of the force. Thus the moment of a force F about a given point is F x ON, where ON" is the perpendicular drawn from upon the line of action of F. It will be noted that the moment of a force F about a given point never vanishes, unless either the force vanishes or the force passes through the point about which the moment is taken. 58. Geometrical representation of a moment. Suppose the force F to be represented in magnitude, direction, and line of action by the line AB. Let be any MOMENTS 61 Take the bar used in Art. 55 and suspend it at C so that it rests in a horizontal position ; if the bar be uniform G will be its middle point ; if it be not uniform, then C will be its centre of gravity [Chapter ix]. The beam must be so suspended that it turns easily and freely about C. When the forces are parallel. From any two points A, B of the bar suspend carriers on which place weights until the beam again balances in a horizontal position. Let P be the total weight, including that of the carrier, at A, and Q the total weight similarly at B. Measure carefully the distances AC and BC. Then it will be found that the products P.AC and Q.BC are equal. The theorem can be verified to be true for more than two forces by placing several such carriers on the bar and putting weights upon them of such an amount that equi- librium is secured. In every such case it will be found that the sum of the moments of the weights on one side of C is equal to the sum of the moments of those on the other side. When tlie forces are not parallel. Arrange the bar as before but let light strings be attached at A and B which after passing over light pulleys support carriers at their 62 STATICS other ends. Let these carriers have weights put upon them until the beam balances in a horizontal position. Let P and Q be the total weights on the carriers including the weights of the carriers themselves ; these will be the tensions of the strings at A and B. Measure the perpendicular distances, p and q, from C upon OA and OB respectively. Then it will be found that P.p = Q.q. 61. Positive and negative moments. In Art. 57 the force F would, if it were the only force acting on the lamina, make it turn in a direction opposite to that in which the hands of a watch move, when the watch is laid on the table with its face upwards. The force Fx would, if it were the only force acting on the lamina, make it turn in the same direction as that in which the hands of the watch move. The moment of F about 0, i.e. in a direction ^), is said to be positive, and the moment of F1 about 0, i.e. in a direction ^), is said to be negative. MOMENTS 63 Algelveio sum of moments. "The algebraic sum of tho moments of a. set of forces ubout a given point is the sum of the moments of the forces, each moment having its proper sign profixcd to it. Ex. ABCD is a square; along the sides AB, CB, DO, and DA forces 8 aê equal respectively to 6, 5, &and D T 8 12 ls. wi. Yind the algebraio sum i uf their moments about the centre, O, ufthe square, if the side of the square de 4 fecê. The forces along DA and 4B tend fo tum the square about O in the positive direction, «hilst the forces along the sides DC and CB tond to turn dt in the negativo direction. The perpendienlar distaneo o O A 6 8 from each force is ? feet. Tlonea the moments of tha foroes are respectively +0x9, —5x2, —8x9, and 4+12x2. Their algobraio sum is therefore 2[6-5-8412] or 10 units of moment, i.e. 10 times the moment of a force equal to 1 lb. wt. acting at tha distance of 1 foo from O. 62. Theorem. “he algcbraio sum of the moments of amy tivo forces about amy point im their plane is equal to the moment of their resultant about the same point. Case I. Let the forces mect in q point, Let P and (Q acting at the point 4 bo the two forces und O the point about which tho moments are taken. Draw 00 parallel to the direction of P to mect the line of action of Q in the point C. ú Let AC! represent () in magnitude and on the same sealo lot AB represent 2; complete the parsllelogram ABDO, end jom OA and OB. N Then 4D represents the resultunt, 2, of P and Q. 66 STATICS Since the sides of the triangle ACM are respectively parallel to the sides of the triangle DBL, and since AC is equal to BD, :. AM=LD, ;. aOAM= aOLD. First, let fall without the angle CAD, as in the first figure. M Then 2aOAB+2aOAC =2aOAL + 2aOAM =2aOAL+2aOLD =2aOAD. Hence the sum of the moments of P and Q is equal to that of R. Secondly, let fall within the angle CAD, as in the second figure. The algebraio sum of the moments of P and Q about =2aOAB-2aOAC =2aOAL-2aOAM =:2aOAL-2aOLD =2aOAD =moment of II about 0. 64. If the point about which the moments are taken lie on the resultant, the moment of the resultant about the point vanishes. In this case the algebraic sum of the moments of the component forces about the given point MOMENTS 67 vanishes, i.e. The moments of two forces about any 'point on the line of action of their resultant are equal and of opposite sign. The student will easily be able to prove this theorem independently from a figure ; for, in Art. 62, the point will be found to coincide with the point D and we have only to shew that the triangles AGO and ABO are now equal, and this is obviously true. 65. Generalised theorem of moments. Ifany number offorces in one plane acting on a rigid body have a resultant, the algebraic sum of their moments about any point in their plane is equal to the moment of their resultant. For let the forces be P, Q, R, S,... and let be the point about which the moments are taken. Let P 1 be the resultant of P and Q, P2 be the resultant of Px and R, Ps be the resultant of P2 and S, and so on till the final resultant is obtained. Then the moment of P1 about = sum of the moments of P and Q (Art. 62); Also the moment of P2 about sum of the moments of P x and R = sum of the moments of P, Q, and R. So the moment of P3 about = sum of the moments of Pa and S = sum of the moments of P, Q, R, and S, and so on until all the forces have been taken. Hence the moment of the final resultant = algebraic sum of the moments of the component forces. Cor. It follows, similarly as in Art. 64, that the alge- braic sum of the moments of any number of forces about a 52 68 STATICS point on the line of action of their resultant is zero ; so, conversely, if the algebraic sum of the moments of any number of forces about any point in their plane vanishes, then, either their resultant is zero (in which case the forces are in equilibrium), or the resultant passes through the point about which the moments are taken. 66. The theorem of the previous article enables us to find points on the line of action of the resultant of a system of forces. For we have only to find a point about which the algebraic sum of the moments of the system of forces vanishes, and then the resultant must pass through that point. This principle is exemplified in Examples 2 and 3 of the following article. If we have a system of parallel forces the resultant is known both in magnitude and direction when one such point is known. 67. Ex. 1. A rod, 5 feet long, supported by two vertical strings attached to its ends, has weights of 4, 6, 8, and 10 lbs. Ming from the rod at distances of 1, 2, 3, and 4 feetfrom one end. If the weight of the rod be 2 lbs., what are the tensions of the strings 1 LetAF be the rod, B, C, D, and E the points at which the weights R< A C G D id v* 8 + 10 are hung ; let G be the middle point ; we shall assume that the weight of the rod acts here. Let R and S be the tensions of the strings. Since the resultant of the forces is zero, its moment about A must be zero. Hence, by Art. 65, the algebraic sum of the moments about A must vanish. V V MOMENTS 71 EXAMPLES. IX. 1. The side of a square ABCD is 4 feet; along the lines CB, BA, DA, andDB, respectively act forces equal to 4, 3, 2, and 5 lbs. weight; find, to the nearest decimal of a foot-pound, the algebraic sum of the moments of the forces about C. 2. The side of a regular hexagon ABCDEF is 2 feet ; along the sides AB, CB, DC, DE, EF, and FA act forces respectively equal to 1, 2, 3, 4, 5, and 6 lbs. wt. ; find the algebraic sum of the moments of the forces about A. 3. A pole of 20 feet length is placed with its end on a horizontal plane and is pulled by a string, attached to its upper end and inclined at 30 to the horizon, whose tension is equal to 30 lbs. wt. ; find the horizontal force which applied at a point 4 feet above the ground will keep the pole in a vertical position. 4. A uniform iron rod is of length 6 feet and mass 9 lbs., and from its extremities are suspended masses of 6 and 12 lbs. respec- tively; from what point must the rod be suspended so that it may remain in a horizontal position ? 5. A uniform beam is of length 12 feet and weight 50 lbs., and from its ends are suspended bodies of weights 20 and 30 lbs. respec- tively; at what point must the beam be supported so that it may remain in equilibrium? 6. Masses of 1 lb., 2 lbs., 3 lbs., and 4 lbs. are suspended from a ^tiniform rod, of length 5 ft., at distances of 1 ft., 2 ft., 3 ft., and 4 ft. respectively from one end. If the mass of the rod be 4 lbs., find the position of the point about which it will balance. 7. A uniform rod, 4 ft. in length and weighing 2 lbs., turns freely about a point distant one foot from one end and from that end a weight of 10 lbs. is suspended. What weight must be placed at the other end to produce equilibrium ? 8. A heavy uniform beam, 10 feet long, whose mass is 10 lbs. , is supported at a point 4 feet from one end ; at this end a mass of 6 lbs. /is placed ; find the mass which, placed at the other end, would give equilibrium. ,9. The horizontal roadway of a bridge is 30 feet long, weighs 6 tons, and rests on similar supports at its ends. What is the thrust borne by each support when a carriage, of weight 2 tons, is (1) half- way across, (2) two-thirds of the way across? 10. A light rod, AB, 20 inches long, rests on two pegs whose distance apart is 10 inches. How must it be placed so that the reactions of the pegs may be equal when weights of 2W and 3W respectively are suspended from A and B ? 72 STATICS Exs. 11. A light rod, of length 3 feet, has equal weights attached to it, one at 9 inches from one end and the other at 15 inches from the other end ; if it be supported by two vertical strings attached to its ends and if the strings cannot support a tension greater than the weight of 1 cwt., what is the greatest magnitude of the equal weights? 12. A heavy uniform beam, whose mass is 40 lbs., is suspended in a horizontal position by two vertical strings each of which can sustain a tension of 35 lbs. weight. How far from the centre of the beam must a body, of mass 20 lbs., be placed so that one of the strings may just break? 13. A uniform bar, AB, 10 feet long and of mass 50 lbs., rests on the ground. If a mass of 100 lbs. be laid on it at a point, distant 3 feet from B, find what vertical force applied to the end A will just begin to lift that end. 14. A rod, 16 inches long, rests on two pegs, 9 inches apart, with its centre midway between them. The greatest masses that can be suspended in succession from the two ends without disturbing the equilibrium are 4 lbs. and 5 lbs. respectively. Find the weight of the rod and the position of the point at which its weight acts. 15. A straight rod, 2 feet long, is movable about a hinge at one end and is kept in a horizontal position by a thin vertical string attached to the rod at a distance of 8 inches from the hinge and fastened to a fixed point above the rod ; if the string can just support a mass of 9 ozs. without breaking, find the greatest mass that can be suspended from the other end of the rod, neglecting the weight of the rod. 16. A tricycle, weighing 5 stone 4 lbs., has a small wheel sym- metrically placed 3 feet behind two large wheels which are 3 feet apart ; if the centre of gravity of the machine be at a horizontal distance of 9 inches behind the front wheels and that of the rider, whose weight is 9 stone, be 3 inches behind the front wheels, find the thrusts on the ground of the different wheels. 17. A tricycle, of weight 6 stone, has a small wheel symmetri- cally placed 3 ft. 6 ins. in front of the line joining the two large wheels which are 3 feet apart ; if the centre of gravity of the machine be distant horizontally 1 foot in front of the hind wheels and that of the rider, whose weight is 11 stone, be 6 inches in front of the hind wheels, find how the weight is distributed on the different wheels. 18. A dog-cart, loaded with 4 cwt., exerts a force on the horse's back equal to 10 lbs. wt. ; find the position of the centre of gravity of the load if the distance between the pad and the axle be 6 feet. 19. Forces of 3, 4, 5, and 6 lbs. wt. respectively act along the sides of a square ABCD taken in order; find the magnitude, direction, and line of action of their resultant. IX MOMENTS 73 ( 20. ABCD is a square; along AB, GB, AD, and DC equal forces, P, act ; find their resultant. 21. ABCD is a square the length of whose side is one foot; along AB, BC, DC, and AD act forces proportional to 1, 2, 4, and 3 respec- tively ; shew that the resultant is parallel to a diagonal of the square and find where it cuts the sides of the square. 22. ABCD is a rectangle of which adjacent sides AB and BC are lal to 3 and 4 feet respectively; along AB, BC, and CD forces of 30, 40, and 50 lbs. wt. act; find the resultant. 23. Three forces P, 2P, and 3P act along the sides AB, BC, and CA of a given equilateral triangle ABC; find the magnitude and direction of their resultant, and find also the point in which its line of action meets the side BC. 24. ABC is an isosceles triangle whose angle A is 120 and forces of magnitude 1, 1, and JB lbs. wt. act along AB, AC, and BC', shew that the resultant bisects BC and is parallel to one of the other sides of the triangle. 25. Forces proportional to AB, BC, and 2CA act along the sides of a triangle ABC taken in order; shew that the resultant is repre- sented in magnitude and direction by CA and that its line of action meets BC at a point X where CX is equal to BC. 26. ABC is a triangle and D, E, and F are the middle points of the sides; forces represented by AD, \BE, and iCF act on a particle at the point where AD and BE meet; shew that the resultant is represented in magnitude and direction by \AC and that its line of action divides BC in the ratio 2:1. 27. Three forces act along the sides of a triangle ; shew that, if the sum of two of the forces be equal in magnitude but opposite in sense to the third force, then the resultant of the three forces passes through the centre of the inscribed circle of the triangle. 28. The wire passing round a telegraph pole is horizontal and the two portions attached to the pole are inclined at an angle of 60 to one another. The pole is supported by a wire attached to the middle point of the pole and inclined at 60 to the horizon ; shew that the tension of this wire is 4^/3 times that of the telegraph wire. 29. At what height from the base of a pillar must the end of a rope of given length be fixed so that a man standing on the ground and pulling at its other end wfth a given force may have the greatest tendency to make the pillar overturn ? 76 STA TICS 69. Theorem. The algebraic sum of the moments of the two forces forming a couple about any point in their plane is constant, and equal to tlte moment of tJie couple. Let the couple consist of two forces, each equal to P, and let be any point in their plane. Draw OAB perpendicular to the lines of action of the forces to meet them in A and B respectively. The algebraic sum of the moments of the forces about = P.OB-P.OA=P(OB-OA) = P.AB = the moment of the couple, and is therefore the same whatever be the point about which the moments are taken. 70. Theorem. Two couples, acting in one plane upon a rigid body, whose moments are equal and opposite, balance one another. Let one couple consist of two forces (P, P), acting at the ends of an arm p, and let the other couple consist of two forces (Q, Q), acting at the ends of an arm q. Case I. Let one of the forces P meet one of the forces Q in a point 0, and let the other two forces meet in '. From 0' draw perpendiculars, O'M and O'N, upon the forces which do not pass through 0', so that the lengths of these perpendiculars are p and q respectively. COUPLES 77 Since the moments of the couples are equal in magni- tude, we have P.p^Q.q, i.e., P. 0'M=Q.O fK Hence, (Art. 64), 0' is on the line of action of the resultant of P and Q acting at 0, so that 00' is the direction of this resultant. Similarly, the resultant of P and Q at 0' is in the direction O'O. Also these resultants are equal in magnitude ; for the forces at are respectively equal to, and act at the same angle as, the forces at 0'. Hence these two resultants destroy one another, and therefore the four forces composing the two couples are in equilibrium. Case II. Let the forces composing the couples be all parallel, and let any straight line perpendicular to their directions meet them in the points A, B, G, and D, as in the figure, so that, since the moments are equal, we have P.AB = Q.GD (i). Let L be the point of application of the resultant of Q at G and P at B, so that P.BL^Q.GL (Art. 52) (ii). Q Q 78 STATICS By subtracting (ii) from (i), we have P.AL = Q.LD, so that L is the point of application of the resultant of P at A, and Q at D. But the magnitude of each of these resultants is {P + Q), and they have opposite directions ; hence they are in equilibrium. Therefore the four forces composing the two couples balance. 71. Since two couples in the same plane, of equal but opposite moment, balance, it follows, by reversing the directions of the forces composing one of the couples, that Any two couples of equal moment in the same plane are equivalent. It follows also that two like couples of equal moment are equivalent to a couple of double the moment. 72. Theorem. Any number of couples in the same plane acting on a rigid body are equivalent to a single couple, whose moment is equal to the algebraic sum of the moments of the couples. For let the couples consist of forces (P, P) whose arm is p, (Q, Q) whose arm is q, (B, P) whose arm is r, etc. Replace the couple (Q, Q) by a couple whose components have the same lines of action as the forces (P, P). The magnitude of each of the forces of this latter couple will be X, where X.p = Q . q, (Art. 71), so that X = Q^. P So let the couple (R, P) be replaced by a couple \ P P/ whose forces act in the same lines as the forces (P, P). COUPLES 81 These two resultants are equal and opposite, and there- fore balance. Hence we have left the two forces (P, P) at A l and Bx acting in the directions A 1C1 and BXDU i.e., parallel to the directions of the forces of the original couple. Also the plane through A 1C1 and B1Dl is parallel to the plane through AC and BD. Hence the theorem is proved. Cor. From this proposition and Art. 71 we conclude that A couple may be replaced by any other couple acting in a parallel plane, provided that the moments of the two couples are the same. 74. Theorem. A singleforce and a couple acting in the same plane upon a rigid body cannotproduce equilibrium, but are equivalent to the single force acting in a direction parallel to its original direction. Let the couple consist of two forces, each equal to P, their lines of action being OB and XC respectively. Let the single force be Q. Case I. If Q be not parallel to the forces of the couple, let it be produced to meet one of them in 0. Then P and Q, acting at 0, are equivalent to some force R, acting in some direction OL which lies between OA and OB. Let OL be produced (backwards if necessary) to meet the other force of the couple in U and let the point of application of J? be transferred to X . Draw OyAi parallel to OA. Then the force R may be resolved into two forces Q and P, the former acting in the direction 0^, and the latter in the direction opposite to XC. L. s. 6 82 STATICS This latter force P is balanced by the second force P of the couple acting in the direction XC. Hence we have left as the resultant of the system a force Q acting in the direction 1A 1 parallel to its original direction OA. Case II. Let the force Q be parallel to one of the forces of the couple. A Q Let X meet the force Q in 0%. The parallel forces P at and Q at 2 are, by Art. 52, equivalent to a force (P + Q) acting at some point S in a direction parallel to OB. The unlike parallel forces (P + Q) at 3 and P at X are, similarly, equivalent to a force Q acting at some point A in a direction parallel to 3D. Hence the resultant of the system is equal to the single force Q acting in a direction parallel to its original direc- tion. COUPLES 83 75. If threeforces, acting upon a rigid body, be repre- sented in magnitude, direction, and line of action by the sides of a triangle taken in order, they are equivalent to a couple whose moment is represented by twice tlte area of the triangle. Let ABC be the triangle and P, Q, and R the forces, so that P, Q, and R are represented by the sides BC, CA, and AB of the triangle. Through B draw LBM parallel to the side AC, and in- troduce two equal and opposite forces, equal to Q, at B, acting in the directions BL and BM respectively. By the triangle of forces (Art. 36) the forces P, R, and Q acting in the straight line BL, are in equilibrium. m Hence we are left with the two forces, each equal to Q, M acting in the directions CA and BM respectively. These form a couple whose moment is Q x BN, where BN is drawn perpendicular to CA. Also Q x BN= CA x BN twice the area of the triangle ABC. Cor. In a similar manner it may be shewn that if a system of forces acting on one plane on a rigid body be represented in magnitude, direction, and line of action by the sides of the polygon, they are equivalent to a couple whose moment is represented by twice the area of the polygon. 62