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H i I 1 t DON Entered according to Act of the Parliament of Canada, in the year one <}housand eight hundred and seventy six, by Adam Miller & Co., in the Oflace of the Minister of Agriculture. I the year ki Miller Elementary Statics BY J. HAMBLIN SMITH, M.A., GONVILLE AND CAIUS COLLEGK, AND LATE LECTURER AT ST. PETER'S COLLEGE, CAMBRIDGE. ^WITH ^^I»FE]SriDI25: BY THOS. KIRKLAND, M.A., SCIENCE MASTER NORMAL SCHOOL, TORONTO. fiT ■N CANADIAN COPYRIGHT (Authorized by theCowicil of Public Instrutlion for Ontario. \ "'■'-^-''Ofv.msl TOHOXTO : ADAM M1|/iJ^:R &' CO., 6^/9?^/ I Z lb To the Canadian Edition has been added an appendix by Thomas Kirkland, M.A., Science-Master, Normal School, Toronto. This appendix contains (1) a few general hints for solving statical problems; (2) a collection of typical problems worked out, shewing tlie application of statical principles, and especially the application of elementary trigonometry to the solution of questions in statics ; (8) a collection of Examination Papers, principally selected from the questions set at the examinations in the University of Toronto, during the past twenty years. The whole, it is hoped, will be found valuable to students preparing for University Examinations, and to teachers preparing for first-class certificates. I I PREFACE TO THE THIRD EDITION. ippendix Normal jiuts for typical statical raentary statics ; incipally as ill the y years, students teachers This treatise was originally designed to explain the part of Statics required in the Previous Examina- tion and the Second Examination for Ordinary Degi-ees in the University of Cambridge. It is now published in such a form, that, while serving its primary purpose, it may meet the requirements of Students in Schools, especially those Avho are pre- paring for the Local Examinations. It may also be regarded as an introduction to the works on Mechanics, which will appear in due course in Rivinqton's Mathematical Series. The Examples have been selectea from Papers set in University Examinations. The propositions requiring a knowledge of Tri- gonometry are marked with Romaii numerals. For some explanations of the Elementary Defi- nitions I am indebted to the late Dr. Whewell's work on ** The Philosophy of the Inductive Sciences;'' and a special acknowledgment is duo from me to Dr. Parkinson, for permission to make free use of his treatise on Meclianics. In this, as in my other publications, I have been assisted in no slight degree by Mr E. J. Gross, of Gonville and Caius College, who has taken the most lively interest in revising and correcting all that 1 have written. J. HAMBLIN SMITH. Cambbidqk, 1871. I If Defini lii! 9-^ On Co The r On th On te On P^ On t] I CONTENTS. CHAPTER I. PAQS Definitions ^ CHAPTER II. On Component and Resultant Forces .... 9 CHAPTER III. The Parallelogram op Forces 15 CHAPTER IV. On the Triangle and Polygon of Forces ... 27 CHAPTER V. On the Resolution of Forces 3* CHAPTER VI. On Parallel Forces ^2 CHAPTER VII. On the Equilibrium of a Body moveable bound A Fixed Point . *® Vlll CONTENTS. CHAPTER Vin. On the Centbb of Gbavitt • PAOS . 64 CHAPTER IX. Of Moments 71 CHAPTER X. Of Mechanical Instruments . . , . , . 8,3 Answers to Examples 125 5. FAOB 64 ELEMENTARY STATICS. 71 83 126 CHAPTER 1. DEFINITIONS. 1. Matter is that, which can be perceived by the senses of eight and touch. A Body is any portion of matter. A RiGW Body is one, in wliich the different portions are held together in invariable positions with respect to each other. A Particle or Material Point is a portion of matter, mdefinitely small in all its dimensions : so that its length, breadth, and thickness are less than any assignable linejir magnitude. 2. Rest. When a body or pai-ticle constantly occupies the same position, it is said to be at rest. Motion. When the position of a body or particle is being changed continuously, it is said to be in motion. 3. Force. Any cause, which changes or tends to change the state of rest or motion of a body or particle, is called force. 4. Statics is the scienco, which treats of the conditions, tinder which forces, acting on matter, produce rest. 5. Line op Action. The line of action of a force is the tine, in which a particle would begin to move in consequence of the action of the force. 6. The Forces, with which we are chiefly concerned in this treatise, may be roughly divided into three classes : (1) Pressures. (2) Tensions. (3) Attractions. Of the first and second kinds of force we have illustra- tions in many actions of our daily life. Whether we push, s.s. 1 DEFINITIONS. pull, or lift a body, we bring into action a force acting by pressure or by tension. Imagine a gimlet to be firmly fixed in a block of wood. If we push the gimlet, we apply to the block a force acting by pressure. If we pull the gimlet, we apply to the block a force acting by tension. Honoe we obtain the following definitions : Pressure. If one body be forced against another, each body is subjected to a force acting at the point of contact; such force is called pressure. Tension. When a body is pulled by means of a string or rod, the force exerted along the string or rod is called tension. If we consider a string as a line of consecutive particles, when a force is applied at each end of the string, each particle of the string is pulled in opposite directions by the forces, which the consecutive particles on either side of it are com- pelled to exercise upon it. These forces are called tensions, and are the same at every particle of the string. Suppose an engine, attached to a truck by a coupling-chain, to be just on the point of moving the truck. Each link of the chain is then acted upon by two equal and opposite forces, which act by means of the other links on either side of any particular link. The force, with which the part of the chain on one side of any particular link resists the force exerted along Uie chain on the other side of the link, is called the tension of the chain. 7. Attraction is a force less easily conceived than pres- sure or tension, because it arises from the action of one body on anotlier at a distance from it. Such is the influence of the magnet on the needle : such is the influence by which the Earth attracts to itself all bodies about it: and such is the influence by which the Sun and planets attract eaeh other. 8. All bodies fall, if unsupported, or tend to fall, if sup- ported, towards the surface of the earth. The direction, in which a particle would fall freely at any place, is called the vertical li7ie at that place. A plane perpendicular to this vertical line is said to bo htyrizontal. DEFINITIONS. acting by irmly fixed pply to the gimlet, we )ther, each [)f contact: a string or cd tension. ) particles, ch particle the forces, t are com- d tensions^ ling-chain, ch link of site forces, ide of any e chain on rted along tension of than pres- one body e: such is all bodies I Hun and ill, if sup- ( ely at any aid to be , \ M If a ball of lead be suspended at one end of a string, and we hold the other end of the string, we must exert a certain force to sustain the ball, equal to the force, with which the Earth attracts the ball. This latter force is called the Weight of the ball. Hence we obtain the following definition: Weight or Gravity is the name given to the force with which the earth attracts a body. The tendency of bodies to the Earth results from their attraction or Gravitation to the Earth. This tendency is only a particular instance of the attraction^ which is exerted by every body upon those about it; and this attraction of one body to another arises from the attraction of every pa' tide of matter to every other, which is called Universal Gf'av 'tation, 9. Density. The Density of a substance is the degree of closeness, with which the particles composing the substance are packed together. 10. Volume is the amount of space occupied by a body. The Volume, Bulk, or Solid Content of a body is measured by the number of times a certain cubical unit must be repeated, to fill up the space occupied by the body. Thus, when we say that the volume of a body is 8 cubic inches, wo mean that a cubical unit, which we call a cubic inch, nmst be repeated 8 times, to fill up the space occupied by one body. 11. Equilibrium. If several forces acting on a particle, or on a body, are so related, that no motion of the particle or the body takes place, the forces are said to be in equilibrium. TiQo forces, which, acting in opposite directions, keep eack other in equilibrium, are necessarily and manifestly equ.'^l. If we see two boys pulling at two ends of a rope, so that neither of them in the smallest degree prevails over the other, we have a case in which two forces are in equilibrium. If three hooks be fixed in a log of wood, two at one end and one at the other, and if the efforts of two boys pulling at ropes, attached to the two hooks at one end, be just counteracted by the effort of a man pulling at a rope, attached to the hook at the other end, we have a case in which three forces are in equilibrium: and tliis illustration may bo extended io four ^ Jive or more forces. 1—9 DEFINITIONS. ■■\ \ ! Again, if a number of rings be inserted round a block of wood, if a rope be attached to each ring, and a boy set to pull at each rope, it is easy to conceive such a disposition of the forces exerted by the boys, that no motion of the block may take place. Here then we have a case, in which a number of forces, not acting in parallel directions, are in equilibrium. 12. When two men pull at a rope in the same direction, wo know that the force, which they exert, is equal to the sum of the forces, which they would separately Rxert. When two stones are put in a basket suspended by a string, their weights are added and the sum is supported by the string. Thus we see that forces, acting together in the same direction, may bo added together to obtain their combined effect. Since two oi)i)osite forces, which balance each other, are equal, each force is measured by that which it balances; and since forces are capable of addition, a force of any magnitude is measured by adding together a proper number of such equal forces. Thus a heavy body, which, appended to an elastic spring, will draw it through one inch, may be taken as the unit oj weight. Then, if we remove this body, and find a second heavy body, which will also draw the spring through one inch, this second body is also a unit of weight. In like manner we might go on to a third and a fourth equal body; and adding together the two, or the three, or the four heavy bodies, we have a force twice, or three times, or four times the unit oi weight. And with sucii a ^;ollection of heavy bodies, or weights, we can readily measure all other forces ; for, since forces that keep a body at rest nrast bo equal in their opposite effect, wo con- clude that any statical force is measured hy the weight which it will support, 13. To in aa sure forces wo fix upon some definite force for our standard, or unit, and then any othor force is measured by the number of times it contains this unit, and this number is culled the measure of the force. It is usual to take as the unit of force that force which will sustain, when acting vertically upwinds, a weight of one pound avoirdupois. The measures of forces, which will sustain lib., Slbs., albs., /*lbs., will then bo 1, 2, 3, P respectively. 14. Two forces are conrmeusurahle when a f(>rco can bo i DEFINITIONS. block of 3t to pul) n of the ock may timber of rium. lirection, 1 to the . When ng, their e string, lirection, t. Lher, are ces; and agnitiide ich equal ) spring, unit qt id heavy iich, this ^0 might together have a weight. we can keep a we con- t which brce for urcd by mbcr is lidi will 3 pound tin lib., can bo i taken as the standard of measurement, such that it is contained in each an exact number of times. 15. Method of estimating foires. The three elements specifying a force, all of which must be known in order to estimate the effect of the force, are (1) The point of application of the force. (2) The direction in which the force acts. (3) The magnitude of the force. 16. Method of represent in ff forces. Forces may be represented by sti-aight lines : for (1) A stniight line can be drawn from any point, and thus it will represent a force with respect to the point of application. (2) A straight lino can be drawn in any direction, and tJius it will represent the direction of a force. (.3) A straight line can be drawn of such a length, as to contain .as many units of length as the given force contains units of force, and thus it will represent the magnitude of a force. Thus, suppose we are speaking of a force of 5 lbs., acting at the middle point of a horizontal rod, and inclined at an aiglo of 45° to the horizon. Let BC represent the rod, A the middle point of the ro making an angle of ^.V* with AC. Mark off a portion of the line AD, suppose AP, containing 6 units of length, that is, as many units of length as there arc units of force in the given force. Then wo may say that AP represents the given force in every particular : DEFINITIONS. I; i (1) In point of application, at A the middle point of the rod. (2) In direction, as being inclined at an angle of 45** to the horizon. (3) In magnitude, by the number of units in its length. Next, suppose that we have to represent two forces of 5 lbs. and 7 lbs., applied to a point in directions at right angles to each other. Taking any line we please to represent the unit of length, we draw two lines AB^ AG 2X right angles to each other, ■LJ-I I I I I the one containing our unit of length 5 times, and the other containing it 7 times, and the lines AB^ AG will properly represent the two i rces acting at the point A. 17. On the Trammissihility of Force. It is plain that two equal and opposite forces, P, Q, applied .-. 4 ? — ? -s I 1 at the extremities of a straight rigid rod AB^ and acting in direction of the rod, will be in equilibrium. This result will be true whatever be the lengtli of the rod : and hence we infer that P will balance Q, at whatever point of the rod Q be applied ; in other words, tlio effect of Q is the same, at whatever point of the rod, whetlier at By C7, or any other point, it may bo applied, the direction remaining the same. Suppose a piece of wood of any shape, say like a horse- shoe, to be laid on a smooth table, and to bo acted on by a force, A^ B, G, D being any four points in the lino of action of tho force, and in the wood. Then iho only force tending to move iho body is tho aforesaid force, and it i ' DEFINITIONS. le point of e of 45*' to its length. forces of •ight angles i of length, each other. i the other ill properly is found that the body will move in precisely the same manner, whether the force be applied at ^, 5, C, or D, These considerations lead us to the following principle, called the principle of the trammissihility of force. The effect op a force on a rigid body, to which it is applied, will be the same, if we suppose the porcl to be applied at any point in the line of action, providkd the point be rigidly connected with the body. The following is another example of this principle. If ABC be a chain fastened at ^ to a hook fixed in a beam, and a weight W be suspended from (7, then the pressure on the hook is ^ + weight of chain AG. If now we remove the part of the chain below B^ and suspend W from J3, the pressure on the hook is W + weight of chain AB, , Q, applied ^ I acting in of the rod : cr pouit of i Q \% the , (7, or any aining the e a horso- , \ Oj3 Thus the pressure on the hook caused by JV acting at exceeds the pressure caused by IV acting at B by the weight of the chain BG, and W, so far as its own effect is concerned^ produces the same pressure on A, whether it be applied at CorB. DEFINITIONS. Note. It may here be remarked, that if a weight W be w attached to one end of a fine string, and the string be made to pass over a small grooved wheel, fixed in position and moveable about its centre, it is proved by experiment that the force P, which has to bo applied at the other end of the string to keep W at rest, is the same, whatever may be the angle between the two parts of the string WA and PA. The same is true with regard to the tension of a string that passes over a smooth peg. Thus, if a picture frame be suspended symmetrically by a cord passing over a nail <4, the tension of the portions of the cord AB and AC \& the same. eight W be ng be made losition and riment that r end of the may be the iPA. of a string *e frame be r a nail A, AC is the ti CHAPTER II. ON COMPONENT AND RESULTANT FORCES. 18. Suppose to be a material particle, and ABa. straight line of unlimited length passing through O. a ■^ o M Let a force P act on in the direction AB^ and Q BA. Set off in OB the line OM^ containing as many units of length as P contains imits of force ; and in OA set off ON^ containing as many units of length as Q contains units of force : then OM and ON will represent P and Q in the point of ap- plication, in direction and in magnitude. The learner must be careful to speak of a line not as a forcey but as representing a force. Thus we say here that OM represents the force P. The order, in which the letters and M stand, serves to indicate the direction, in which the force P acts: OM representing a force that acts from O towards i)f, whereas MO v/ould represent a force of equal magnitude act- ing from M towards O. 19. Suppose to be a material particle, and A Bo. straight line of unlimited length passing through O. O Q. If two forces, P and Q, act simultaneously on in the same direction OjB, they are equivalent to a single force P-¥Q acting along OB, lo o,v coMPo^yEhn^ and resultant forces. 20. If the forces P and Q act in (ypposlte directions, Ol\ a o ~->T antl OA respectively, they are equivalent to a single force equal i?i magnitude to the difference between P and Q, and acting in the direction of the greater of the tw^o forces. And, generally, a system of forces, acting in the same line A 13 on a particle O, is equivalent to a single force, equal in magnitude to the excess of the sum of the forces which act in one direction, over the sum of those which act in the opposite direction, and acting in the direction of the greater sum. 21. If P and Q be two forces acting simultaneously in the lines AB, AG, not in the same straight line^on a free particle A, the particle will begin to move in some definite direction eqi lin wit m res" for liui ani AD within the angle BAC^ which is less than two right angles. The particle will move continuously along AD furtlier and further the longer P and Q continue to act in their original directions. iS'ow suppose P and Q to bo removed and a single force R to act on the particle in the direction AD^ so as to cause the l)article to move over the same space in any given interval of time, as it would have moved over in that interval had it been acted on by P and Q. The force R is called the Resultant of the forces P and Q, and we obtain the following definition : Def. a force, whose effect is equivalent to the combined effects of two or more other forces, is called their Resultant^ and, with reference to this resultant, each of the other forces is called a Co7npoi\ent, ORCES. ections, Ol\ ON COMPONENT AND RESULTANT FORCES 1 1 single force and Q, and two forces. > same line ;o, equal in vhich act in !ie opposite • sum. ously in the rco particle e direction two right D furtlier t in their le force R cause the nterval of 'al had it illed the ) following combined Resultant y r forces is Note. In Arts. 1!) and 20 we treated of a 8inj,do force equivalent to a systeui of tV)rces acting ia the smnc straight line, to which force the term Resultant may, in accordance with the definition just given, be applied. 22. If P and Q be equal forces, the particle A will move in a straight lino bisecting iho angle BAV; for there is no reason why it should move more towards the direction of one force, than towards the direction of the other. Hence the line of action of the resultant of two equal forces bisects the angle between their lines of action. 23. Now suppose a particle A to be kept at rest by three forces P, Q, S. Then S may be regarded as neutralizing the joint effect of P and Q. Hence, if R bo the resultant of P and Q, R and S must be LNjual and opposite forces, for since R produces the same itlcct as that which is produced by the combined action of P and Q, R and S are also capable of keeping the particle A at rest, and this they cannot do, unless they be equal in magnitude and opposite in direction. 24. Illustrations of Component and Resultant Forces. Since a clear conception of the meaning of the terms Component and ResulUmt is necessary for a right under- standing of Statical principles, we shall give in this Article two rough illustrations, wliich may serve to expbia the denni- tiou given iu Art. 21. 12 OAT COMPONENT AND RESULTANT FORCES. -4 is a block of stone to be drawn along a level road RD, 4 Tt B n J} Suppose two horses to be attached to the stone at O, in such a manner, that each exerts a force along the line OHB, one at H and the other at B. Now suppose the horses oo removed, and a traction- engine to be applied to the block at 0, so as to move it in the direction OHB with the same drawing-power as that of the two horses. The horses will represent the Component Forces. The traction-engine will represent the Resultant. CAj MN are the straight and parallel edges of the banks of a river. a- B Wr M^ -dV -5 is a barge in the middle of the stream. Suppose two horses of equal power to be pulling at ropes attached to the same point of the barge, the ropes being inclined at equal angles to the line B T which passes along the middle of the stream, parallel to the banks. Then the barge will move along the line BT. Now suppose the horses to be removed, and a steam-tug to be attached to the barge, so as to move it in the direction i?T, just in the same manner as it moved when the horses were pulling it. The horses will represent the Component Forces. The steam-tug will represent the Resultant. VRCE^'. road RD, ►ne at O, in I line OHB, a tracfcion- move it in ' as that of s. t. ges of the ON COMPONENT AND RESULTANT FORCES. 13 25. From the two illustrations of Component and Resul- tant Forces, which have been given, we may derive examples of forces in Equilibrium. For, first, suppose that while the horses are pulling at the stone the traction-engine is applied to the opposite side of the stone, so as to pull the stone with the same drawing-power as that of the horses. Then the force exerted by the engine will counteract the forces exerted by the horses, and the three forces will be in equilibrium. Then also it is plain that when three forces are in equilibrium, one of them is equal and opposite to the resul- tant of the other two. Precisely the same results will follow, if in the second illustration we suppose the steam-tug to be applied to the opposite end of the barge. 26. Let P, Q be forces acting at a point A, and P', Q be other forces acting at a point B rigidly connected with A. If P', ^ produce the same effect as that which ia g at ropes pes being sses along steam-tug direction he horses i produced by P, Q, then the resultant of P, Q lies in the straight line joining A and B. For suppose the resultant of P, (/ to be in the direction BT. Then the force counter- acting the effect of P', Q will lie in the line BS opposite 14 ON COMPONENT AND RESULTANT FORCES. to BT. Now this force is to counteract tiio effect of P and Q, and therefore it must be in the same line vith AR, the direction of the resultant of P, Q. (Art. 23.) Hence the re- Bultant of P, Q passes through B, 27. Before proceeding further, we will state three axioms which arc the groundwork of much that follows : Axiom I. A force may be supposed to act at any point in the lino of its direction. Art. 17. Axiom II. Forces may have equivalent forces substituted for them. Axiom III. When two or more forces are in equilibrium, their resultant is zero. '>RCES. of P and h AR, the ice the re- ree axioms ; any point substituted quilibrium. CHAPTER III. THE PARALLELOGRAM OF FORCES. 28. We proceed to establish an important theorem, which enables us to determine the resultant of any two forces acting at a point. The theorem is called the Parallelogram of Forces, and may be thus enunciated. If two forces, acting at a point, he represented in magni- tude and direction hf/ two straight lines draicn from that point, and if a parallelogram he constructed, having these two lines for adjacent sides, then that diagonal of the paral- lelogram, which j)asses through the point if application of the forces, inill represent their residtant in magnitude and direction. That is, if the two forces P, Q be represented by AB, AC, and the parallelogram A B DC ho completed, their resultant 7? will be represented by the diagonal A D. In other words, if AB and AC contain as many units of length as P and Q contain unit? of force, the resultant, /?, of P and Q will act in the line AD, and will contain as many units of force as ^Z> contains units of length. i6 THE PARALLELOGRAM OF FORCES. 29. The truth of this theorem is illustrated by the follow- ing experiment. Two small puUies M and N are attached to a vertical wall, and a string is passed over them, having weights of 3 lbs. and 4 lbs. attached to its ends. A weight of 6 lbs. is then sus- pended from a point A of the part of string between the pullies : this mil draw the string down so as to form an angle MANf and the apparatus will settle itself in a state of rest as represented in the diagram. The tensions of the strings AM^ AN^ are equivalent to 5 lbs. and 4 lbs., and their resultant is equal and opposite to the weight 5 lbs. \ rei rei Now mark oflf a distance ABy along the string AM^ con- taining three units of length, and a distance AG^ in the direction of the string AN^ containing 4 units of length, and complete the parallelogram ABDG. Then it will be found that the diagonal AD m ^ vertical lino, and therefore in the direction of the lino in which the weight of 5 lbs. acts, and \i AD be measured, it will bo found to contain 6 units of length, and therefore it represonts the weight of 5 lbs. in magnitude. This shows that AD represents in direction and magnitude the resultant of the forces represented in direction and magnitude by ABy AC, f THE PARALLELOGRAM OF FORCED. 17 he follow- tical wall, 3 lbs. and then sus- ;ween the 1 an angle of rest as 30. U AB and AC^ the sides of a rhombus ABDC represent in magnitude and direction the equal forces P and Q, which act at the point J, then nmst the diagonal AD represent in directior Ry the resultant of P and Q. For V AB = AGy and AD is common, and BD=CDj :. L BAD= L CAD, (Eucl. i. 8.) .*. AD bisects the angle between the directions of P and Q ivalent to pposite to Jf, con- in the th, and I found in the cts, and mits of > lbs. in ion and irection i But the direction of R bisects the angle between thf directions of P and Q. (Art. 22.) .'. AD represents R in direction. Now, by the principle of the trans- missibility of force, R^ acting along a rigid rod AD, will produce the same effect when applied at Z>, as that which it produces when applied at A. Hence for P and Q acting at A we may sub- stitute i?, acting in the direction AD, atZ>. Again, R acting at D may be replaced by two forces P and Qy acting in directions parallel to AB and AO respec- tively, that is, acting in the directions CD and BD, thus : Hence if ABDC be a rhom- bus, whoso sides are rigidly con- nected, two equal forces P and Q will have the same effect on a particle A, whether they bo appHed along AB and AC, or along CD and BD. 8. s. I ,1 ■' 1 I I ^( i8 77/A' PARALLELOGRAM OF FORCES, 31. Wo now proceed to the Mathematical Proof of the Parallelogram of Forces, whiyh is divided into three Parts. Part I. To prove that the resultant acts in the direction of the diagonal, when the component forces are commen- surable. First, to shew that the proposition is true for forces P and P. When the component forces are eq^ial their resultant bisects the angle between the directions of the forces, and therefore acts along the diagonal. Thus the proposition is true for P and P. Next, to shew that the proposition is true for forces /' and 2P. Let P, Q, R be three equal forces. ^B Lot P act at A in the direction A By lot Q and R art at A in the direction AGE. Take AB and AC to represent P and Q in magnitude, and since R may be supposed to act at uny point in the lino AGE^ which is rigidly connected with A (Ax. I.), let R act at (7, and take CE to represent R in magnitude. Complete the parallelograms j?(7, DEy and draw the diagonals ADy GF. The resultant of P and Q acts along AD, Let P and Q be replaced by this resultant (Ax. ii.) and lot it act at D. Then for this resultant acting at D we may substitute the two forces P and Q, acting in the linos CDf DFy which are respectively parallel to AB, AC. Now suDDoso P to act at C and to act at F. (Ax. i.^ ( s. THE PARAIJJil.OGRAM OF FORCES. »9 'roof of the BO Parts. ho direction ro commen- orccs P and ir resultnnt forces, {ind oposition is or forces /' JB id R art at opreaont P Bd to act at lectcd with osont R in draw tho acts along t (Ax. II.) g at 7> wo n tho linen n (Ax. O i Then P and /?, acting at C, have a resultant acting along GF: let them be replaced by this resultant, and let it act SiiF. For this resultant wo may substitute tlie forces /* and //, acting at i^in the lines FFumX DF. (Ax. ii.) Thus wo have shewn that tho forces P and Q \^ R which are applied at A^ may bo supposed to act at F without altering their combined effect ; .•. F is a point in tho direction of tho resultant of P and Q -\ A' (Art 25), .*. AFia tho direction of tho resultant of /* and Q + R, thatis, ofPand2A By a similar process we can shew that tho proposition ie true for P and 3/*, using tho annexed diagram. Similarly, it may bo shewn to bo true for /* and 4P, for P and 5P, and so for P and m/', m being any whole number. Now since the proposition is true for mP and I\ it may be dhewn to be tine for mP and 2P, by using the annexed figure. So also it may bo shewn to bo true for mP and 3P, for mP and 4P, and so for 7nP and nPj n being any who'e number. Now any two commcnsu- rable forces nuiy, by a.-signing a proper value to P, bo ex- pressed by mP and nP, lleuce Part 1. is proved. 2—2 ViP, :^ V li it) THE PARALLELOGRAM OE FORCES. 32. Part II. To prove that the resultant acts in tho irection of the diagonal, if the forces arc incommensurahle. i)^ Let A By ^C represent two incommensurable forces. Complete the parallolof^ram ABDCj and if AD be not the direction of the resultant, let it be some other line, as A V. Let ^C be divided into an integral number of equal parts, each less than DV^ which is always possible, and mark oft' from CD portions equal to these, the last division E clearly falling between D and V. Complete the parallelogram GF by drawing ^JF^ parallel to AC. Then AC, ^F represent commemmrdhfe forces, and the resultant of tiio forces represented by AC^ AFwWX be in the direction AE, and wo may suppose this resultant to be sub- stituted for them. The resultant then of the forces represented by u\C and AB \% equivalent to the resultant of two forces, one acting in the direction AE^ the other represented by /'Vi, and which may therefore bo supposed to act at A in the direction AB\ ami this resultant must lie within the angle BAE. But, by hypothesis, it acts in the direction A V, without the same angle, which is absurd. In like manner it may bo shewn that no direction but A D c»n be that of the resultant of the forces reprosontod by AB^ AC. f acts in tlia ensurahle. 10 THE PARALLELOGRAM OF FORCES. 21 Thus the theorem has been proved, so far as the direction qfthe residtant is concerned. 33. Part III. To prove that the diagonal repments the re.^ultant in magnitude. rccs. bo not the as A V. )qual parts, d mark ofl* 1 -^clearly parallel to s, and the 1 be in tho to be sub- »y AC and 3 acting in ind Avhich ction AD\ ithout tlio n but A 1) >d by AD, Let All, ^C represent the component forces in magnitude ami direction. Complete the parallelogram ABDC: join AD, In DA produced take y/JE7of such a length, as to represent the magnitude of tho resultant of the forces represented bj AB^AC. Complete the parallelogram AEFC: join AF. Now AB, ACy AE represent three forces which are in equilibrium. Therefore AB represents a force equal aiuf opposite to tlie resultant of the forces represented by A(\ AE (Art. 23). But the resultant of the forces rei)resented by ylT, A E !!■ n in the direction of A F. Therefore AB is in the same straight line with AF. Therefore AFCD is a parallelogram ; and :.ADFC\ but FC^AE\ :.ADAE; .'. AD represents in magiutmle the resultant of tho forcr.« represented by AB, AC. ii l> 12 I I I THE PARALLELOGRAM OF FORCES. 34. When we determine, by means of the Parallelogram of Forces, the single force, whicli is equivalent in its eflfect to the joint eflfect of two other forces, we are said to compound those forces. We shall now give some simple examples on the composi- tion of forces, so far as the method can be illustrated by easy Geometrical processes. The following Theorems are of fre- quent use : 1. The diagonals of a Square bisect the angles. 2. The diagonals of a Rhombus bisect the angles. 3. The diagonals of a Parallelogram bisect each other. 4. The perpendicular, dropped from the vertex of an equilateral triangle on the base, bisects the base, and also the vertical angle. Consequently, if the angles, yl, J?, (7, of a triangle be 90°, 60° and 30'^ respectively, then will the side BC be double of the side BA. 35. If two forces act at the same point, in directions at right angles to each other, to find tJie magnitude and direc- tion of their resultant. Let AG, AB represent two forces ^ P, Q, acting at riglit angles to each other at the point A. Complete the rectangular parallelo- ^' gram ABDG. Then the diagonal AD will represent R, the resultant of P, Q. Now, since the angle DCA is a right angle, AD' = AC^ + CD'; .'.AD' = AC' + AB'; .'.R'^P'+Qi; \ .'.R = JP'-i-Q'; and thus wo obtain the inagnitude of the resultant. The direction of the resultant is known, if we know tht size of the angle DAC. V ■ ec i THE PARALLELOGRAM OF FORCES. »3 lelogram of ts eflfect to compound le composi- ted by easy are of fre- es. [i other. "tex of an nd also the glo be 90", ) double of Irections at and direc- D know tht For certain simple relations between the sides of the tri- angle ADG^ we can determine the angle DAG by geometry. Thus, '\i AB and AC represent equal forces, AG and GD are equal, and DA G is half a right angle. Examples worked out. 1. Two forces of 12 lbs. and 16 lbs. act on a particle in lirectlons at right angles to each other : Jind the magnitude f their resultant. Let AB and ^(7 represent the component forces. Complete the rectangle ABDG. Then AD represents the resultant. Let the measure oi ADhQ x. Then x^ = {Uf-\-{V2Y; .-. 0?" = 256 + 144; .-. 072 = 400; .-. a? = 20. Hence the magnitude of the resultant is 20 lbs. 2. Two forces o/8lbs. each act at an angle of Q(f on varticle: fnd the magnitude and direction of the resultant. Let ABf -4 C represent the component forces. Complete the rhombus ABDG, and draw DE at right angles U) AG pro- duced. Then •.• z DGE ^ lBAG^ 60^ ^ and lCED=-W, :. I GDE-= 30'; and .'.GD-2GE. (Art, 34) Let X be the measure of AD. Then v AD^^AG^ + GD^ + 'lAG.GE', {Va\c\. ii. 12^ .-. 472 = 82 + 8^ + 2x8x4; .-. a?' = 64 f (;4 + (;4; .*. a?2.r.G4x3; Hence the inagnitude of the resultant is 8 \/3 Ibi. a 4» N ^u I': 24 EXAMPLES WORKED OUF. The direction of the resultant can also be determined. For since the diagonal of the rhombus ABDC bisects the angle BAG, wo know that the resultant makes an angle of 30° with each of the components. Note. By the use of Trigonometry we are enabled to work examples of this kind, by means of a general fommla, as we now proceed to shew. xxxvi. If two forces act at the same point, and the angU between their lines of direction is given, to find expressions for the magnitude and direction of their resultant. Let AC, AB represent two forces P, Q, acting upon the point A, and let a be the angle between their lines of di- rection. Complete the parallelogram ABDC, and produce AC to xV. ThentLeaiigleDC7iV = a. Join AD. Then AD will represent R, the resultant of P, Q. Now we know by Trigonometry (Art. 179) that AD^ = AC^ + CD"^ -2 AC. CD. COB ACD; also (Trig. Art. 101), cos ACp=- cos DCN — -cos a; .-. AD^ = AC* + AB' + 2AC.AB.cosai :. Ii*=P^+Q^ + 2PQ.cosa; and thus we obtain an expression for the magnitude of the resultant. The direction of the resultant is known, if we know the lize of the angle DAC. 4 4 EXAMPLES,—!. 25 riiiined. For ts the angle 3 of 30" with ) enabled to 1 formula, as nd the angh expressions U. ig upon the lines of di- •oduce AC Q. CD', '4e of the know the 4 4 4 i I L(itLDAC=0; thenLADC==a-9. sin 6 sin (a— ^) Tl.en -. (Trig. Art. 178.) Q P .: P . sin e=Q. sin a cos 0- Q . cos a , sin ^ ; .*. sin 6. (P -f Q . cos a) = Q . sin a cos d ; Q . sin a .-. tan = -75-7-7^ P-f Q.cos a Examples. — I. 1. Three forces of 3 lbs., 4 lbs. and 5 lbs., respectively, act on a body, their directions being all in the same straight line: find a fourth force which will balance them. 2. Find the resultants of the forces 5lbs., 11 lbs., 13lbs., according to the different possible arrangements of them;— all three acting in the same straight line. 3. P and Q are two forces applied to a particle in direc- tions at right angles to one another; P is 90 lbs., Q is 120 lbs.; find the magnitude of the resultant. 4. Forces of 3Glbs. and 4Slbs. act on a particle in dircc tions at right angles to one another: find the magnitude of the resultant. 5. Three forces, whose magnitudes are 6, 8 and 10 lbs. respectively, acting upon a point, keep it at rest: prove tiia the directions of two of the forces are at right angles to each other. * 6. Place three forces, which are in the ratio of 3, 4 and ,">. so that they may keep a particle at rest. 7. If two forces, acting at right angles to each other, be in the ratio of 1 : >/3, and their resultant be 10 lbs., find the forces. 8. Forces of 3 lbs. and Slbs. act on a point at an angle of 30". Find the magnitude of the resultant. 9. Forces of 10 lbs. and 7 lbs. act on a point at an angle of 60**. Find the magnitude of the resultant. 10. Forces of 9lbs. and 11 lbs. act on a point at an angle of 135^. Find the magnitude of the resultant. i» y ! 36 EXAMPLES,—!, 11. Twc» foiTcs of libs, and r>lbs, aro inclined to one another at an angle of 45", determine the magnitude of their resultant. 12. Two equal forces act upon a point. If the angle between their directions be 60", find the resultant. 13. Two forces act at a point. Shew that the forces are equal, if, when the direction of one of the forces is reversed, the direction of their resultant is at right angles to the direc- tion of their resultant before the change. 14. A weight of 10 lbs. is suspended by a string AB from the fixed point A . A force F acts horizontally at IS on tlie string. What must bo the magnitude of Fj in order that the ttiiglo ABFnmy bo 120^^ 4 i •lib '1 I ^y clined to ono itude of thoir If the angle it. tho forces aro >8 is reversed, s to tho direr- ring AB from ' at B on tlie rder tliat the CHAITKR IV. ON TMK TKfANCiLK AND POLYC.ON OF FOKCKS. # 37. The Trianole of Forces. 1/ three forces, acliny at a point, can he represented in magnitude and direction hy the sides of a triangle, taken in order, they will be in equilibrium, i Let Ali, JiC, CA, tho sides of tho triangle ABO, taken in order, represent in niagn itude and direction three forces, /*, Q, /^, acting at tho point O. Ccanplete the parallelogram A BCD. i Then, since AD m equal and parallel to BC, tho force represented in magnitude and direction by BC will also bo represented in magnitude and direction by AD. Therefore the forces P, Q will bo represented in magnitude and direction by ABf AD. Now yl (7 represents tho resultant of two forces represented by A B, AD. Hence AC represents in magnitude and direction tho combined effect of P and Q. Therefore AC, CA represent in magnitude and directiop the combined effect of P, Q, li. But forces represented by AC, CA will clearly be in equi librium. Therefore P, Q, R will be in equilibrium. !• 23 OiV THE TKIAXCLE OF FORCES, NoTK. The convorso of UHh propOHition is true, that iw, if tlirco foivcM, acting at a point bo in o([nilibn(ini, tlioy can bo roprosontod in njnijnitndo and direction by the KJdcH of a triangle, tiiken in onle.-. 'I'iiis is a particular oaso of a more general theoreni, wliicli we now proceed to prove. 3S. If three forces, (ictin(/ at a point, Ite in eijnilibrium, and any triamjle be constnicted, hariinj its aides jxiralld tit the directions e proportional to the forces. Let /', (?, A* be three forces, whicli, acting at tlie point A, are in equilibrium. is; II' Tict A /?, ylC represent P and Q. Then DA, the diagonal of the parallelogram ACDB, will represent N. Now construct.a triangle MNO, whoso sides are parallel to the aides of the triangle ADD. Then ADD, MNO are similar triangles. Iloiice and and MN :NOr=AD iDD = P :Q NO:OM==DD:DA = Q :D OM :MN=DA'.AD = R :P. (Eucl. VI. 4.) Cor. If two forces P aiid Q act at a point, and have a resultant R', and if a triangle MNO be constructed, whoso sides are parallel to P, Q, R\ then must MN • NO OM r= P -.Q.Rf, ':s. ON THE TRIANGLE OE EOKCES, 29 *«i 1 true, that irt, lilibriuiu, tltcy ticular cane of I to prove. // rquffihn'tnn, 'iangle shnll hr it tlio point A, im A CDB, will a arc parallel to (Eucl. VI. 4.) For let R bo Mh3 force, which would bo in (••juilibrium with /* and Q : then the direction of li \h opp(»Hite to that of R', and therefore tho nideH of the triangle AfNO .re parallel to /', (^, li^ which arc in (Mpiilibriuni : ^ .-. MN : NO : OM P Q . It, II lid tho magnitude of II' ~-- niagniludu of It, .-. MN : NO : OM P :Q H'- xxxix. //* three Jhrccs, actuKj at a, point, he in rtim- , ,i>ri am, Mch force in projtorl.ionol. in the, xiim ([f the aiigk II /(lined beticccn the directions of (he other tico. loint, and have 3e constructed, ust Q Let Py Qf R bo the throe forces : a, ^, y tho angles between tho lines of direction of tho orcos. Construct a triangle MNO, whoso sidcjs MOy ON, NM are parallel, and thcrcforo proportional, to P, Q, II. Produce tho sides to Z>, E, F. Then tho exterior angles ONF, NME, MOD are equal to a, i3, y respectively. Now P.Q'.R=MO .ON.NM = sin ONM : sin NMO : sin MON, (Trig. Art. 1 78), = sin ONF : sin NME : sin MOD, (Trig. Art. 101), = 8in a : sinjS : einy o ■f ir 30 ON THE POL YGON OF FORCES. 40. Tub Polygon op FoRCKa If any number qf forces^ acting at a pointy can he re- presented in magnitude and direction by the sidct of a polygon taken in order, they will be in equilibrium. lict any number of forces P, Q, Ry S, T, acting at the point Of bo represented in magnitude and direction by tlio aides of the polygon ABODE, taken in order, thus, AB, BC, CD, DE, EA. Join AC, AD. Now AB, ^C represent P, Q in magnitude and direction , .'. AC represents the joint ofTect of P, Q .'. A C, CD represent the joint effect oi P,Q, R .'. AD represents .•.-^Z),/)J5^ represent the joint eflfect oi P, Q, R, S :.AE represents .-. AE, EA represent the joint effect of P, Q, R, S, T. Now forces rcpres* .itod by AE, EA will clearly bo in c as diagonal. Hence we can obtain an infinite number of pairs of com- ponents, having as their resultant the force represented hyAD. 42. If we have a single force given, represented by yl Z> and we are required to find a pair of components, one of which shall act in a given direction Ax^ wo can describe an infinite number of parallelograms, such as ABDG^ having AB as diagonal and one side AB lying along Ace, Hence we can obtain an infinite number of pairs of com- ponents, of which one acts in the given direction. i ON THE RESOLUTION OF FORCES. 35 :es oted by AD, •ams, such as 43. But there is only one parallelogram having yli> as diagonal, one side AB lying along Ax^ and another side AG making with AB a right angle at A. Hence we can obtain only one pair of comimnenta of a given force, such that one force acts in a given direction, and the other perpendicular to that direction. These are called the resoleed parts of the given force in their own respective direc- tions, and we give the following detinition : Dep. The resolved part of a j^nven force in any dircctioiv is the force acting in this direction, which will, with another force acting in the perpendicular direction, have the given force for their resultant. pairs of corn- represented mted by AD 1 44. To find the resohcd part of the force represented hg AD in the direction Ax. Drop DB perpendicular to Ax. Complete the parallelogram CABD, having CA/^ a right iOlglo. lients, one of describe an 7, having AB bairs of com- I i Then the forces represented by AC, AB act at right angles to each other, and have the force represented by A/) for their resultant, and the force represente in R^a line of action drop perpendiculars />/^, JjC, on Ax and Ay. Then, if AD represent R in magnitude, AB represents the resolved part of R in direction AjTt AG Ay. Now AB=AD.coaB\ :, resolved part of iZ in direction Ax=R.Qoa6, Hence wo obtain the following nilo : To find the resolved part (\f a force in any given directiil> I f Casb II. WhoD the forces act towards opposite parts. Let A and B be any two points in the lines of action of the two forces P, Q, which act in the parallel directions AP^ BQ, and suppose Q to be greater than P. At A apply any force aS* in the direction BAS, and at B apply a force AS"=^in the direction ABS'. Then aS' and P, acting at A, have a resultant R'; and S' and Q acting at B have a resultant B". Let the directions of iZ' and R'^ meet in O. Draw SOS' parallel to AB. ^-iSF ^if S At resolve the force R into two components, S, acting along OS^ and P, acting along ROG^ parallel to AP and BQ. Also resolve R" into two components, aS" acting along OS'^ and Q acting along C0j?2. Then S and aS", being equal and opposite, will counteract each other, and may be removed, and if R bo the resultant of P and Q, there will remain a force acting along CORy such that R^^Q-P. {'>) EXAMPLES.— IV. 45 osite parts. 3f action of the tions AP, BQ, 3AS, and at B 'hen aS' and P, Q acting at B -). Draw SOS' ^^ Also, '.• the sides of a AGO are parallel to P, Ry S, and the sides of A BCO are parallel to Q, R'\ S\ ,K , no n ^P OC ,S' BC .'.(Art. 38. Cor.), 0= -77,, and yz = P S' OG BG P S "" Q~ AG"^ OG'^ "Q OG' BG AG' Note. We have supposed Q to be greater than P We will now show that, this being the case, the diagram ia J correctly drawn, that is, the directions of R' and R" will ^ meet in a point on the side of Q remote from P. If two forces act at a point, the more one force is increased, the smaller will be the angle between it and the resultant. Now, aS" = aS*, and Q is greater than P, and .'. R", the resultant of Q and S\ makes with Q ;in angle less than that, which R', the resultant of P and S, makes with P. Hence z 0-^C is less than lOBG, and .'. z • OAG, OB A are together less than two right z ■ : .*. the directions of R" and R' meet, when produced, on the upper side of AB. But the part of the direction of JR' on this side of AB lies also on the side of Q remote from P. ^1 .', must be on the side of Q remote from P, >nonts, S, acting ,0 AP and BQ. cting along 0S\ will counteract tho resultant of long GORy such Examples. — IV. Suppose A and B to be two points in a rigid body, and P and Q to bo two parallel forces, acting in tho same direction through A and Z?, their resultant R passing through the point Cin the line AB, then solve the following questions: (1) If P = 20lb8., <2 = 30ll)s., and y) P^ 10 lbs., AG \S incho8 ^\B=^ii inches, find Q, .'!i B I N 11 46 OJV PARALLEL FORCES. 51. Centre op Parallel Forces. The proposition, which we have just proved, enables us to find the resultant of any number of parallel forces, acting at different points of a rigid body. For let P, Q, R, be any number of parallel forces, acting &t Aj By G, points in a rigid body. Then P and Q are equivalent to a single force P + Q, acting at a point E in the straight line AB, such that P '.Q^BE'.AE. Hence P, Q and R are equivalent to the parallel forces P+ Q and Ry acting at the points E and G. Then P + Q and R are equivalent to a single force P-i-Q + R acting at a point F in the straight line EG, such that P+Q:R==GF:EF, and thus any number of parallel forces may bo reduced to a single resultant. Now the position of the point, at which this resultant acts, does not depend on the direction, in which the component forces act, but only on their relative magnitude and their points of application. Hence if these comi)onent forces bo turned about their points of application in any manner, Htill remaining parallel, the point, at which their resultant acts, will still bo the same. For this reason that point is called thg centre qf the parallel forces. ON PARALLEL FORCES. 47 Bnables us to oes, acting at arallel forces, 62. If three forces^ acting upon a rigid Indy, balance each otfier, the lines, in which tfiey act, must eith r he parallel or pass through a point. Fig. I. Fig. II. force P + (?, I that }arallel forces single force ine EC, such reduced to a esultant acts, 10 component do and their ent forces bo r manner, still tant acts, will is called th$ Let P, Qy R be the forces. First, suppose P and ■} to be parallel (as hi fig. i.). # Then they will have some resultant, acting ui a direction panillel to each of them. But this force, since it counteracts R, must be in direction exactly opposite to the direction in which R acts. Consequently the line, in which R acts, must be parallel to the directions of P and Q. Next, suppose the lines of direction of P and Q to meet in a point (as in fig. ii.). Then the resultant of P and Q will pass through the point 0. |K But this force, since it counteracts R, must be in direction ^exactly opposiio to tlie direction, in which R acts. Consequently the lino, in which R acts, must pass through the point 0. flfV m m II Q m I I CHAPTER VII. ON THE EQUILIBRIUM OF A BODY MOVEABLE ROUND A FIXED POINT. 53. If P and Q be two forces acting in one plane at points A and J5 of a rigid body, and be a fixed point of the body, about which the body might trni freely, then, if under the action of P and Q the body be at .est, the resultant of P and Q must pass through 0. For it is plain that any si7ig/i' force, acting on the body and not passing through O, will cause the body to turn about 0. Fig. I. For the sake of simplicity let us suppose that the body in question is a rigid rod, without weight, moveable about a fixed point 0. Then, if P and Q be parallel, their resultant, /?, will bo parallel to both (as in fig. i.). * And if P and Q be not parallel, and G be the point, in which their linos of direction meet, COR will be the lino of direction of the resultant of R (as in fig. ii.). TI] erted Bultan this r( Th iP and 0. ON EQUILIBRIUM ROUND A FIXED POINT, 4V OVEABLE one plane at I point of the ihen, if under esultant of P lat any single hO,will ciiuso [. 54. A rigid rod, capable of turning round a fixed point in the rod, is called a Lever. The point, about whicli it can turn, is called the Fulcrum^ and the parts, into which the rod is divided by the fulcrum, are called the arms of the lever. When the arms are in a straight line, the rod is called a ttrai^ht lever : in all other cases it is called a bent lover. 55. If two forces f acting at right angles on a straight lever, prodmc equilibrium, the forces are inversely as their distances from the fulcrum. M If X ^p B" a Ithat the body in >lo about a fixed lltant, /2, will bo be the point, in 111 bo the lino of if 3r -r- ■A Let P and Q bo the forces acting at the points M, N, and balancing each other round the fulcrum G. Then the lever is kept at rest by P, Q, and the force ex- erted by the resistance of the fulcrum C. TIenco the ro- •ultant of P and Q must be equal and opposite to the force of this resistance, and must therefore pass through 0. Then, since Cis the point, through which the resultant of P and Q passes, it follows from Art, 50, that P'.Q^CN'.CM. B.B. i so ON EQUILIBRIUM ROUND A FIXED POINT. ■' Also, if R be the pressure on the fulcrum, in fig. 1, R = P + Q, in fig. 2, R = Q-P. 66. If two parallel forces^ acting on a lever^ produce equilibrium, they are inversely as the perpendiculars dratcn from the fulcrum to the directions in which the forces act. il' Let P and Q be two parallel forces balancing each otner on the lever AB round the fulcrum C. Then the lever is kept at rest by P, Q, and the force ex- erted by the resistance of the fulcrum G. Hence the resultant of P and Q must be equal and opposite to the force of this resistance, and must therefore pass through G. Draw MCN at right angles to the directions of P and Q. Then, since M and N are points in the lines of action of P and Qy and G the point in MN^ through which the resultant of P and Q passes, P .Q = GN .GM, 67. If two forces which are not parallel, acting on a lever, produce equilibrium, they are inversely as the perpendicularly drawn from the fulcrum to the directions in which the forces act. Let P and Q be two forces, not parallel, balancing euci) other on the lever AGB round the fuLinim C7. din pre au( POINT, ON EQUILIBRIUM ROUND A FIXED POINT. $1 culars drawn I forces act. cing each otnor nd the force ex- ucc the resultant the force of this J. )ns of P att^ Q- los of action of V ich the resultant Produce tlic lines of direction of P and Q to meet in 0. Then the lever is kept at rest by P, Q, and tlic force oxcrted by the resistance of the fulcrum C. Hence the re- sultant of Pand Q must be equal and opposite to the force of til is resistance, and must therefore pass through C. acting on a lerei\ prrpendicularif which thf 'le ns %n el, balancing eadi a Let R bo the resistance of the fulcrum ; then the direction of R must pass through 0. Art. 52. Di-aw CD parallel to OP and CE paraiie] to OQ, and CM and CTVat right angles to OP atul OQ respectively. Then the sides of the triangle CDO, being parallel to the directions of the three forces /*, Q, R, may bo taken to re- present P, Qy R in magnitude. Art. 38. Then P Q CD OD CD CE • Now the triangles GME, CND are similar, since the right angles CMEy CND are equal, and angle CEM= angle DOE = anglo CDN. (Eucl. i. 29.) CD CN (JE^ CW llenco P Q CN CM 4—2 5« EXAMPLES.— K Examples. — V. 1. Weights of 5 lbs. and 7ll)s., hung at ilio extremities of a straight lever 6 feet long, balance each other; find the length of the arms. 2. A weight of 5 lbs., hung from one extremity of a straight lever, balances a weight of 15 lbs. hung from the other extremity : find the ratio of the arms. 3. When the pressure on the fulcrum is 99 lbs., and the arms of the lever are as 4 : 7, determine the weights. 4. The weight at the extremity of one arm of a straight lever is 12 lbs., the length of the arm is one foot, and the pressure on th" fulcrum is 16lbs. ; what is the length of the arm at which the other weight acts ? 5. If the weights on a lever are as 5 : 7, and the length of the lever is 36 inches, find the position of the fulcrum. 6. If two weights of 2 lbs. and 5 lbs. balance on a lever, one of whose arms is 5 inciies longer than the other, compare the arms. 7. Two weightsi, P, Q^ balance on a straight lever : if they be interchanged, detennine the weight, which must be added to or subtracted from either to produce equi- librium. 8. Weights, proportional to 8 and 3, balance each other on a straight weightless Lever whose length is 2 feet 9 inches. Find the position of the Fulcrum: and state the ambiguity that would exist in the problem, if the equilibrium were pro- duced by parallel forces instead of weights. 9. The weights on the extremities of a lever 8 feet long are as 1 to 3. Find the position of the fulcrum. 10. The length of a horizontal lever is 12 feet, and the balancing weights at the ends are 3 lbs. and Gibs, respectively. How far ought the fulcrum to bo moved for equilibrium, if each weight be placed 2 f(\t from the ends of the lever? 11. A le.ver is 4G inches in ; ngth : the fulcrum is 8 inches from one end : what weight applied at this end will balance a weight of 22 lbs. at the other end? EXAMPLES,— V. 53 remitics oi •: find the emity of a 11 the other bs., and the its. f a straight )ot, and the Migth of the the length of n'uni. ►n a lever, one compare the 12. At one end of a lover a fulcrum is placed, at the Twiddle point of the lever hangs a weight of lolbs., and this is supported by an upward force of 10 lbs. acting 6 inches from the other end. Find the length of the lever. 13. A lever with a fulcrum at one end is \\ feet in length. A weight of 14 lbs. is suspended from the other end. At what distance from the fulcrum will an upward force of 'if) lbs. preserve eipiilibrium / 14. A lever AB witk a fulcrum at D is divided in the points C and D into three equal parts. From C and 1) weights of 12 lbs. are suspended. What force acting upwards at A will just support them ? 15. AOB is a straight lever, and forces P and Q^ acting at A and B, are in equilibrium about 0, the fulcrum, which is 3 feet from A and 5 feet from B: solve the following questions: (1) IfP-4lbs., Z./>MP = 90°, ^(2Z?O-60^find^. (2) 1 f P - 3 lbs., L BA P = ()0°, L QBO = CO'', find Q. (3) If P .-: 2 lbs., L BA P = 45'', L QBO -= SO", fiud Q. ht lever : if which must Drodiice equi- co each other feet 9 inches, ho ambiguity ium were pro- cr 8 foot long )9 feet, and the _. respeetivcly. equilibrium, if ho lever 1 jrum is 8 inches I will balance a i;! CHAPTER VIII. ON THE CENTRE OF GRAVITY. I 58. The directions of the forces, which the Earth exerts on the different particles composing a body, are not, strictly speaking, parallel. But since the dimensions of any body, which we shall have to consider, arc very small compared ^ith its dis- tance from the centre of the Earth, we may consider these directions to he parallel. The resultant of this system of parallel forces is the weight of the body ; and the point in the body, at which this resultant acts, is called the Centre of Gravity of the body. Thus the Centre of Gravity is the centre of Parallel Forces, which act in a vertical direction. "We may suppose the whole weight of the body to be col- lected at the Centre of Gravity, and if it be in rigid connection with all the points in a body or a system of bodies, then the body or system would be in equilibrium in all positions, if the Centre of Gravity were supported. Having thus explained the reasoning, on which we proceed to investigate the position of the Centre of Gravity of a bodyi we may give the foUowiuj,- definition : " The point at which the weight of a body or system may always be supposed to act is called tlie Centre of Gravity of the body or system." 69. Every system of heavy particles has one and only on€ Centre of Gravity. Let A, By (7,...be any number of heavy particles, /*, Q, i?,...the weights of ^, B^C... and nu gr« an( Q' sys| an( pos tinil ON THE CENTRE OF GRA VITY. 55 Suppose, first, tliat A and B are counected by a rigid rod without weight. Earth exerts •0 not, strictly ly body, which sd ^ith its dis- consider these s is the weight 1 this resultant Parallel Forces, )ody to be col- igid connection .odies, then the )Ositions, if the lich we proceed ivity of a bodyi or system may re of Gravity of me and only one tides, y P 'B Now P and Q, being parallel forces acting in the same direction, are equivalent to a single resultant, the magnitude of which is P + Qt and which acts through a point E in the line AB, such that P : Q = BE : AE. E is then the centre of gravity of A and B, and the effect of P and Q will be the same as if A and B were collected into one particle, of weight P + Q, and placed at E. Now suppose P + Q to act at £': then we may find the centre of gravity oi P + Q acting at E and /i acting at C, as before, by taking a point F in the line EC, such that P + Q : R=CF :FE, and we may suppose P, Q, R all collected at F: acid so we may proceed for any number of particles. Therefore every system of particles has a centre of gravity. Also a system of particles can have hut one centre of gravity. For, if possible, let a system have two such points G and G\ and let the system be turned about till the lino joining O and O' is horizontal. Then we shall have the resultant of the system of parallel forces acting in a vertical lino through (?, and also in another vertical line through G' \ which is im- possible, since it cannot act iti two different lines at the same time. ■I'A 56 ON THE CENTRE OF GRAVTVY. 60. To find the centre of gravity qf a material straight line. Let ABhQ the given straight line. a s ): lb We may regard AB ^% a line of equal particles uniformly arranged. We may then divide the line into a series of pairs of equal particles, each pair being equidistant from (7, the middle point of A B. Let P and Q be such a pair of particles. Then C will be the centre of gravity of P and Q. Similarly each pair of the particles, of which AB is com- posed, will have C for its centre of gravity. Therefore C will be the centre of gravity of the whole line AB. 61. To find the centre of gravity of a parallelogram. H Xi) iJSi^ tl f V M J) o Let A BCD be a parallelogram, regarded as a unifornt lamina, or thin sheet, of matter. Draw EF parallel to AL and DC, bisecting AD, BG in the points B, F; and HK parallel to AD and BC, bisecting AB, DC in the points H, K. We may suppose the parallelogram to be made up of a series of lines of particles, parallel to one of its sides, as BC. Then it is plain that EF bisects each of these lines, and hence the centre of gravity of each of the lines composing the parallelogram is in EF. Hence the centre of gravity of the whole parallelogram lies in EF. Une^ the oenti «thel M ON THE CENTRE OF GRA VITY, 57 ^rial straight B icles uniformly jries of pairs of nC, the middle ndQ. ich AB ia com- ity of the whole led as a uniform ' parallel to ^^ E^ F\ and HK 1 the points H, K. be made up of a its sides, as BC. )f these lines, and les composing the lole parallelogram Similarly it may be shewn to lie in HK. Therefore O, the point of intersection of EFy HK, is th« centre of gravity of the parallelogram. 62. To find the centre of gravity of a plane triangle. Let ABC be a plane triangular lamina of matter. We may suppose this triangle to be made up of a series t i lines of particles, nmning parallel to one of the sides, as BC. Let he be one of these lines. Bisect BC in F^ and join AF, cutting he in / Now Af . fh = AF : FB (by similar triangles Afh, AFB) ^ =AF:FC (since FB=^FC) = Af : fc (by similar triangles AFC^ Afc) ; :./hr=/c. Similarly it may be shewn that A F will bisect each of the lines parallel to BC; and hence the centre of gravity of each of the Hues composing the triangle is in AF, and therefore the centre of gravity of the triangle is in AF. Now bisect AB in E and join CE. Then the centre of gravity of the triangle will be in CE. Therefore the point 0, in wliich AF and CE cut each other, will bo the centre of gravity of the triangle. ,'i#' ffT ' I I'll iili i5i| 58 O// THE CENTRE OF GRAVITY. 63. To shew that if a line he drawn from any angle to the middle point of the opposite side^ the centre of gravity of the triangle lies in this line at a distance from the angular point equal to two-thirds of the line. Draw BD and AFio the middle points oi AC and BG. Then O, the intersection of AF^ BDy is the centre of grafity of the triangle. We have now to shew that BO-IOD. Join I'D. Then, since FD bisects BC and AC^ it must be parallel id AB. (Eucl. VI. 2.) I m 4 iM And since ABC^ DFC&re similar triangles, TI ft! A ! AB : DF=AG : DC = 2 : 1. TJi of gni Again, since AOB^ FOD are similar triangles, taking BO : OD^AB : DF = 2:1; .-. BO = 'iOD. And hence BO is two-thirds of FV' UOi 51 Y. I any angk <'^ J of gravity oj n the angular OAT THE CENTRE OF GRAVITY, 59 64. The centre of gravity of a triangle coincides inpotir tion with the centre of gravity of three equal particles placed at the angular points. Let three particles, cacli of weight P., be placed ad Take D the middle point of AC. >f AG and BG. is the centre of must be parallel [les, Then D will be the centre of gravity of the particles acting at A and C, and wo may suppose both to act at Z). angles, Then we have 2P acting at />, and P at B, and the centre of giiivity of thcHc weignts will be found by joining BD and taking in it a point 0, such. that 60 : 0Drr2P : p 2 : 1. Lf» is the centre of gravity of the triangle. "TTT' 60 ON THE CENTRE OF GRA VITY. 65. If a body he suspended from a point, about which it can swing freely, it will rest with its centre of gravity in th vertical line, which passes through the point of suspension. 6 falli centt Let ABC be the body, G its centre of gravity, S the poii: in tlio body, by which it is suspended by a string fastened t the fixed point D. Then there aro two forces acting on tlie body : Su; gravit; Buppos There in the opposi< acting OW, (1) the weight of the body acting in the vertical llii be pro( (2) tlie tension of the string DS. When the body is at rest, tl»eso two forces must act in tli same straight line ; .*. DS is a vertical line ; /. O is in the vertical line passing through a.V and D. G6. The position of the centre of gravity of a hody im he sometimps determined by ejcpcriment in the foUoich manner: J3ut tbe base and the contact string d The point of intersection of the two lines in the body is t ni. centre of gravity of the body. of wh' -l Let the body be suspended from any point in its surfai and let the line, which is vertical and passes through the pui of suspension, be marked. Then let the body be suspended from another point in 1 surface, and let the line, which is vertical and passes throiii the point of suspension, be marked. TV. about which a f gravity in Iht if suspension. ON THE CENTRE OF CRA VITY. 6i 67. A body, placed on a horizontal plane, will stand or Jail over^ according as the vertical line drawn through the centre qf gravity of the body falls tcithin or icilhout t/ie base. Fig. I. Sup|K)80 the vertical lino GO, passing through the centre of gravity G to fall within the base, as in fig. i. Then we may suppose the weight of the body ( JV) to be concentrated at G. There will then bo a vertical pressure of IV downwards acting in the line GC, which will bo counteracted by an equal and opposite pressure of the plane, on which the body is placed, acting upwards in the direction (76r, and so equilibrium will L ^Ijq rertical I'm be produced, and the body will stand. Fig. II. ravity, S the poiu string fastened t body : OS must act in tli ugh *.V and D. Wjity of a body »»<' in the folloich . . ^^^^.j..^^ But suppose, as in fig. ii., that the line GC falls v/ithout )Oint mi ' ^ base: ihon there is no pressure equal and opposite to W^ s tinoug I ^^^ ^j^^ l^^^jy ^,.jj 1^^ turned round /?, the nearest iK)int of contact in the base to the vertical lino GC, and will fall. noUicr point m i jj jj jj^ ^^^^ ^,^^^^ j^ j^^^^.^^ ^^^^^^^^ ^^^^ ^^^^^^^ bounded by a ind i)absca throii: ^^^ ^^^^^,^ tightly round the parts of the body in contact with the horizontal plane. s in the body is t Thus the base on which a chair stands is the quadrilateral of which the feet are the four corners. .:M m li •ii! 62 ON THE CENTRE OF GRAVITY. 68. The following illustratijn may enable the learner to grasp more completely the fact enunciated in the preceding Article. Cut a piece of cardboard of uniform thickness in the form of a square. In the corners A, D, C insert three pins of equal length and place them in a vertical position with their points M, Ny O resting on a horizontal table. The cen- tre of gravity of the card- board is G the middle point of AC, and this point is vertically above H the mid- dle point of MO. If now we put a small weight on the triangle ABC, tho centre of gravity of the cardboard and weight will be vertically above some point within the triangle MON, and the system will stand. But if we put tho small weight on the triangle AGDy tho centre of gravity of the cardboard and weight will be verti- cally above some point outside the triangle MON^ and tho system will fall. V I 69. On stable and unstable equilibrium. (1) If a body under tho action of any force bo in a position of equilibrium, and a very smdi displacement be given to the body, if ii then tend to return to the original position of equi- librium, thai position is called ono of stable equilibrium. (2) If the body tend to move further from its original position, that posiMon is called one of unstable equilibrium. (3) If it remain in the new position which the dlsplacr ment has given it, the position is said to be neutral. ^ V. ON THE CENTRE OF GRAVITY. 03 tho learner to the preceding )ss in the form ianglo ABC, tho will be vertically [, and the system •iangle AGD, tho ;ht will be verti- MOl^, and tho Examples. (1) A weight suspended by a strinc: is an instance of stable equilibrium. (2) A stick balanced on the finger is an instance of uit- stable equilibrium. (3) A sphere resting on a horizontal table is an instance of neutral equilibrium. 70. Having gicen the centre of gr a city of a body and also the centre of gravity of a part of the body, to find the centre of gravity of the remainder. DO be in a position it be given to the position of equi- juilibrium. from its original le equilibrium. ich the displuco leutral. Let G and // be the centres of gravity of tho two parts of the body, w and x the weights of tho parts respectively. Then, if bo the centre of gravity of the whole body, GO: OH=x\w\ .'. ioxGO = xxOHi wxGO .-. 0//= X m.^ • Hence if G and bo given we can find 77, by producing XmO 10 far that the part produced -^ — . WirWiWMnM-^Ml. I M 64 ON THE CENTRE OF GRAVITY. 71. To find the centre of gravity qf a number 0/ particles lying in a straigM line. \0 ', E, F be the middle points of the sides of the pro- posed triangle -4 5C7. I Then the centre of gravity of the perimeter ABC w'lW be m tlie same position as the centre of gravity of three particles placed at Z>, E, F, whose weights are proportional to AB, BCy CA respectively. Draw EM bisecting the angle DEF, and DN bisecting EDF. Now DM : MF - DE : EF, by Euclid, vi. 3, ^AC \AB, by sim'. triangles ABC, EFD Hence M is the centre of gravity of the two sides AB, AC\ and therefore tlio centre of gravity of the whole perimeter lies \\\EM. Similarly it lies in DN. Therefore O the point of intersection of EM, DN is the centre of gravity required, and this, by Euclid, iv. 4, is the centre of the circle inscribed in the triangle DEF» 1.1. 1^ ! I t 66 EXAMPLES.— VI, Examples— VI. 1. A uniform rod, of 3 feet in length and 8 Iba. weight, has a i)Ound weight placed at one end ; find the centre of gravity of the whole system. 2. A ball of weight 2 lbs. lies in the middle between two balls, a foot apart, of weights 4 lbs. and 1 lb. ; find the centre of gravity of the three. 3. Four equal particles are placed in a straight line, the distance between tlio first and second being one inch, between the second and third two inches, and between the third and fourth ; i'e«^ .i.;heg; find their centre of gi'avity. 4. 1' centre of gravity of weights whose measures are 16, 8, 4, 2, 1, , ^ced at distances of 1 ft. from each other in a straight lino. 5. Shew that if the centre of gravity of three heavy particles, placed at the angular points of a triangle, coincide with the centre of gravity of the triangle, the particles must be of equal weight. 6. Shew that if a number of triangles be described upon the same base and between the same parallels, their centres of gravity lie on a straight line. 7. If the angular points of one triangle lie at the middle points of the sides of another, .show that the triangles will have the same centre of gravity. 8. If two triangles are upon the same base, shew that the line joining their centres of gravity is parallel to the line joining their vertices. 9. Two equal particles are placed on two opposite sides of a parallelogram ; shew that their centre of gravity will remain in the same position, if they move along the sides so as to be always equidistant from opposite angles. 10. Having given the positions of three particles Ay B^Cy and the position of the centres of gravity of A, B and A, C, find the position of the centre of gravity of By G. 11. An equilateral triangle is inscribed in a circle; shew that its centre of gravity will be the centre of the circle. EXAMri.ES.-VI. 61 12. If the centr ity of )S. weight, has tro of gravity between two i the centre of aight lino, the 1 inch, between the third and hose measures 1 each other in >f three heavy angle, coincide particles mufit iescribed upon their centres of 3 at the middle ingles will have shew that the el to the line pposite sides of rity will remain les so as to be rticles Ay B^C, 4, B and A, C, n J. a circle ; shew ho circle. triangle inscribed in a coincide with the centre of the circle, shew that the le is equilateral. 13. Why does a man, when ho is carrying a weight with one arm, extend the other ? 14. A uniform flat rod whose length is 14 inches and weight 3 lbs. rests on a horizontal table ; if a weight of 4 lbs. be placed on one end of the rod, find the greatest distance, which the other end may be made to juojeet beyond tlio table, without the rod falling off. 15. Find the centre of gravity of four equal particles A^ B^ (7, Z>, when the straiglit line AB bisects the straight hneC/). 16. A triangle suspended from one of its angles has its base horizontal ; shew that the triangle is isosceles. ^i 17. A square and a triangle have the same *)a~ ; find the altitude of the triangle, if the centre of gravi ' of tae two lie in that base. 18. A right-angled triangle is suspended '.eely against a wall from its right angle ; and again fror one of the other angles. If in these two cases the positions of the hy})oteniiso be at right angles to each other, compare the length of the sides of the triangle. 19. If ABC be an equilateral triangle, and weights 3 lbs., 4 lbs., ,'5 lbs. bo placed at its corners, find their centre ol gravity. 20. Find the centre of gi'avity of three weights at the angular points of an equilateral triangle, two of the weights being each double of the third. 21. If an equilateral parallelogram bo suspended from an angular point, one diagonal will bo iiorizontal. 22. If a parallelogram be divided into four triangles by its diagonals, and the centres of gravity of the four triangles be joined, the joining lines will form another parallelogriun. 23. If a triangle ABC, right-ani;led at C, and having the side opposite A double that opposite B, bo suspended iuccessively by A and C from a peg in a vertical wall, find the Itfigle between the two positions of /6'. 6 2 •' ' ' 1 1 1 1 1 i 1 : * 68 EXAMPLES.— VI. 24. A right-angled triangle, whoso acuto angles are to each other as 1 : 6, is suspended from the right angle; determine the inclination of the hypotenuse to the vertical. 25. An isosceles triangle is suspended successively from the angular points of its base ; shew that the two positions of the base will be at right angles, if the base of the triangle be two-thirds of its altitude. 26. AB h a straight line, G a point in it such that AC = 2GB. Weights of 1 lb., 2 lbs., 3 lbs. are placed at ^, i? and C respectively. Find their centre of gravity. 27. ABGD is a straight line divided into three equal l)art8 at the points B and G, and equal heavy particles are placed at the points A^ B, D. Find their centre of gi'avity. 28. ABGD is a parallelogram having the angle ABC - 60**, and the l)ase BG six inches in length, what length must the side AB not exceed, if tlio figure is to stand upon BG^ 29. O is the centre of a circle, AOB a diameter; G the middle point of the arc AB\ D, E points on the arcs AG, GB respectively, at distances from AOB each equal to half the radius AO. I*]qual heavy particles are placed at 0, A^ By G, Z>, E. Find their centre of gravity. 30. If a body, whose centre of gravity is 6r, be divided into two parts, and if Gi, G^ be the centres of gravity of the two parts, shew that Gi GG^ is a straight line. 31. AB the base of a square ABGD is divided in ^and the triangle GBE removed; slicw that the remainder will stand or fall according as BE bears to EA a less or greater ratio than sf^ : 1. 32. Find the centre of gravity of a trapezium, of such form, that tlic line joining ono pair of its opposite angles divides it into two equal triangles. 33. From a given square shew how to cut a triangle haviiig ono side of the scpiare for its base, so that the centre of gravity of tiie remaining portion may bo at the vertex of the triangle. 34. If a trir.iigle have a side upon which it will not stand, that side is the shortest, and is not half the length of the longest side. ^ EX AM PL ES.— Vr. 69 iigles are to right angle; ) vertical. (ssively from ) positions of 3 triangle be t such that Lced at A^ B three equal particles arc of gi'avity. angle ABC length must iponi?(7? neter; G the arcs AG, GB 1 to half the t 0, A, B, G, ', be divided gravity of the led in ^and inainder will !ss or greater ium, of such )08ite angles t a triangle at the centre vertex of the it will not he length of 35. If a quarter of a triangle be cut off by a line drawn parallel to one of its sides, find the centre of gravity of the remainder 36. In a right-angled triangle, of which the greatest side is 6 inches long, what will be the distance of its centre of gravity from the right angle ? 37. A BCD is a square surmounted by an isosceles triangle on one side BC, which forms the base of the triangle. Determine the greatest height of this triangle, so that the figure (which is cut out of one piece of board) can be placed on its side DC without tumbling over. 38. Find the position of the centre of gravity of a figure formed by an equilateral triangle and a square, the base of the triangle coinciding with one of the sides of the square. 39. Find the centre of gravity of the quadrilateral formed by drawing a straight line parallel to the base of an isosceles triangle and bisecting the sides. 40. A square is divided into 9 smaller squares. At the centre of each of the latter a weight is placed of the magnitude indicated in the margin. Find their centre of gi'avity. Suppose the weight 9 to be removed; where will be the centre of gravity of the rest ? 41. The diagonal of a square AC EG is divided in A' so that GK-2KG. Through K are drawn two lines parallel to the sides of the square, meeting the sides AG in B, CE in Z>, EG in F, GA in //. At all the angles of the figure are placed weights, 1 lb. at A, 2 lbs. at B, 3 lbs. at G, 4 lbs. at /), 6 lbs. at E, (J lbs. at F, 7 lbs. at G, 8 lbs. at //, 9 lbs. at K. Find their centre of gravity. 42. Construct a figure as in Ex. 41, except that no weights are to be placed at the angles: remove the portion IlKFG: suppose matter distributed uniforirly over the remainder : find the centre of gravity. 43. A heavy bar 14 feet long is bent into a right angle, the legs of which are 8 feet and 6 feet long respectively; prove that the distance of the centre of gravity of the bar so bent from the point in it, which was its centre of gravity when it 4 3 8 9 5 2 7 1 6 was straight, is liH feet. i I !' 70 EXAMPLES.~VI. 44. A wire is bent into the form of a hexagon with one side omitted. Find its centre of jjravity. 45. A uniform triangular board of given weight rests in a horizontal position on three props; find the portion of the weight sustained by each. 46. A heavy circular disc is suspended from its centre by a string and rests horizontally. Weights of 3, 3 and 6 lbs. are suspended by strings, attached to the middle point of the disc and passing over its edges. Shew how to arrange them, so i/hat the disc may remain horizontal. 47. Assuming the position of the centre of gravity of a triangle, fimi that of any quacb-ilateral cut off by a straight line parallel to one side. 48. If a parallelogram suspended from an angular point have one diagonal horizontal, prove that the four sides aro equal. 49. Find the centre of gravity of weights, ?r, 2'm?, 3«* placed at the angular points of a triangle, and determine the ratios, in which the lines, drawn from the angular points through the centre of gravity to the opposite sides, are divided at that point. 50. An isosceles triangular lamina weighs 6 lbs. "What weight, placed at the vertex, will make the triangle balance about the middle point of the perpendicular from the vertex on the base ? 51. ABCis a right-angled triangle without w«.'ight, A the right angle: place such weights at J3 and C, that, when the triangle is suspended from A, it may rest with the side BC horizontal. 52. A cubical block of stone of uniform density rests on a rough horizontal plane. It is tilted up, by having a wedge tlriven under it with its edge parallel to an edge of the cube, Shew that the block will not be overtunied, unless the angle of the wedge be greater than 45". 53. Find the centre of gravity of three squares described on the sides of an isosceles right-angled triangle. 'M xagon with one eight rests in a portion of the )m its centre by 3 and 6 lbs. are loint of the dis, we mean a distance which contains the unit of distance / times. 'I ■ V ir Ni -\ 72 OF MOMENTS. 75. Now suppose A, By C, D... to he any number of rods o/eqiml lengthy fixed in a circular wheel moveable about its centre. -B. JL Y p a Then t 2P force P to turn tlu IS evident that the power ol wheel will be the same, when applied perpendicularly at the extremities of any one of the rods. Also, if equal forces P and P be applied perpendicularly at the extremities of the rods A and B^ so as to turn the wheel from right to V*/> it is plain that the rotatory i)ower of these forces will be just counteracted by a force 2P, applied perpendicularly at the extremity of the rod Z>, acting in a contrary direction, that is, tending to turn the wheel from Icjt to right. Hence we infer that a force 2P, acting at the extremity of I), will have ttvico the rotatory ctfect of a force P, applied in the same direction at the same point. And tlius we conclude generallif that, if two forces JF\ and F^ be applied at the sitnio point of a rod in the same direction, moment of /\ : moment of Z', = /\ : i^g, that is, the moment of any force acting at a constant distance varies as that force. 76. If Fhe the measure qfaj\>rc6 acting on a hothj^ and D the measure of the 2)erpenii{cular distance q/' t/ie linn m pi'i'i tliat 1 same I out 0. Also, the resultant of the parallel forces F and P must pass through O, .'.P.OB^F.OCy (Art. 50) and .*. P varies as F. OG, since 0/J is constant. Hut moment of /* varies as /', (Art 75); .*. moment of F varies as F. 00^ that is, moment of F varies as F . D. Now, if we take, as our unit of moment, the moment of « unit of fence about a point at a unit of tlistanco from the direction of the force, moment of /•"' : unit of mwinent = F . D . \\ .*. moment of F'm F.D times tho unit of moment; .'. the measure of the moment of F=F. D. Cor. Hence the moment of a force about a point in ita own lino of action is zero. } n i 74 OF MOMENTS. Tl. We have shewn that forces can be represented geometrically by straight lines, and we can now shew that moments can be represented geometrically by areas. Let us suppose that a force JP, represented in magnitude and direction by the line AB^ acts at -4, a point in a straight rod AO, moveable about 0. ill •It" I Draw OM at right angles to the lino of action of P, ami join OB, » Then, since the force P contains P units of force, the line AB must contain P units of length. (Art. 16.) .*. the measure ^A AB'va P. Let the measure of OM bo Z>. P D Then the measure of the area of the triangle AOB is — -~ ; .*. measure of twk6 the area of triangle AOB is P .D. Also, measure of moment of P about k P.D. (Art. 7(5) | Hence twice the area of the triangle AOB will represent the nioniont of P about O. We conclude, then, that the moment of a force about a point may he represented geometrically by ticice the area ({f Vie triangle, whoso vertex is the point, and irhose base is a line representing the force in magnitude and direction. )0 represented now shew that vreas. I in magnitude Lt in a straight OF MOMENTS. 75 78. Let us now take the case of two forces P and Q, represented by the linos AB, AC, acting at A on the rod AO in such a way, that the perpendicular drawn from the fixed point to P's line of action falls on one skle oi AO, and the perpendicular drawn from to Q'& line of action fulls on the other side oi AO. ction of P, and ' force, the lino P 7) H'\^P,D, \D. (Art. 7(1.) > will represent force about a ice the area <{( Those base is a lirection. I m « Tlio monient of P about O will then be represented by twice the triangle AOU, and the moment of Q about O will bo represented by twice the triangle AOC. Now the force P tends to cause the point A to move along the cir(!nlar arc AD, and the force Q tends to cause the point J to niove along the circular arc .//^. Thus the forces tend to turji the rod AO in contrary dirccti(mH, and tluH difference we can express by the terms positive and nojatire. These terms are only roliitive, and may be applied, at discretion, to express causes or etfects that are directly opposed to each other, but for convenience sake we make tlie following statement. The moment of a fierce may he considered ncffatire or positiee, according as th force tends to turn the hndj/ in the same direction as tfiat, in ichich hands dive, Art. 78^ algebraic sum of moments of P and Q round O -Q.Oh-P.Oa ^Q.{Oc-l)c)-P.{Or..ac) Q . Oc-Q,bc-P.Oc f P.m ^Q.Oo- P.OCy from(l) = (Q-P).0(; ^momrnt of H nmnd O^ i l.l I ( 'I J' ||i 78 OF MOMENTS. 80. y/t^ algelrraic mm, of the moments of two force9y meeting in a point and acting in one plane^ about any point in the plane, i^ equal to the moment of their restdt- ant about that point. Case I. When the point is within the angle between the forces. Let AB, ^(7 represent the two forces /*, Q. Complete the parallelogram ABDC. Draw MON parallel to AB and CD. Now, as the figure is drawn, the moments of P and R about arc posUive, and the moment of Q about in negative J m^l wo have to show that 9 triangle AOB -2 triangle AGO ^2 triangle AOD. Now parallelogram i?iV = parallelogram /?C - parullulogram Mi'\ .'.. 2 triar.glo A OB = 2 trianglo ADC- 2 triangle D0( ' = 2 (trianglo ADC- triangle DOC) =r- 2 (trianglo yl 06^ + triangle AOD) = 2 trianglo AOC+2 trianglo AOD\ .*. 2 triangle AOB-2 triangle AOC 2 trianglo AOD, frhicli proves iho proposition. the OF MOMENTS. 79 two forcei^ about any heir result- Ic between )f P and R about \^ AOD. ogram .^/<'i jglo VOi ' DOC) lo AOD) UqAOD, 1 1 Case II. Wh'^n the point is without tlie angle between the forces. t Draw ONM parallel to CD and AD. As the figure is drawn, the moments of P, Q, R, about O are all positive, and we have to shew that 2 triangle AOD + 2 triangle ^0C=: 2 triangle yiOI>. Now 2 trifuigle ^ 02) --= 2 (quadrilateral ylO(7Z> - triangle OCD) = 2 (triangle AOC + triangle ^C7> - triangle OCi>) = 2 triangle ^06'+ 2 triangle ACD-2 triangle OCZ) = 2 triangle AOC + parallelogram CVi - parallelogram CM = 2 triangle AOC + parallelogram DN = 2 triangle ^OC + 2 triangle AOD, which proves tiie propOHition. Ohs. ^Froni this and the preceding article we see, that tho algebraic sum of the moments of any two forces, acting in one pliuiu, about any point in that plane, in (M|ual to tho moment of their resultant about that point. 1! 8o OF MOMENTS, 81. The algebraic sum of the moments of two forces, acting in one plane, and meeting in a pointy about any point in the line qf action of the resultant is zero. ! ') Let A By -4 C represent the two forces, P, Q. Complete the parallelogram ABDC, First, to shew that the sum of the moments of P and Q about the point D is zero. moment of P about Z> = 2 triangle ADD, moment of Q about /> = 2 triangle ADC^ and the triangles ADB^ ADC are equal ; .'. moment of P about D = moment of Q about D, and the moments of P and Q are respectively positive and negative ; .*. the algebraic sum of the moments of P and Q about D is zero. Next let d bo any point in the lino of action of R. Now triangle AdB : triangle ADB ~ Ad : AD, •nd triangle AdC : triangle ADC - Ad : AD\ -ik OF MOMENTS. 8l ,*. triangle AdB : triangle J />/? ^ triangle AdC : triangle ADC; and ' , since the triangles AD/i and ADC are equal, H triangle AdB - triangle AdC; 1. e. moment of P about d = moment of Q about d ; and the moments of P raid Q are respectively positive and negative, .'. algebraic sum of moments of P and Q about d is zero. 82. To show that the moments of two parallel forces about any point in the lino of action of their resultant are cijual in magnitude and opposite in direction, wo make use of the figures in Article 79, and taking moments of I* and Q round any point c in the line of action of the resultant 7?, wo observe (1) Since P.ar QJ>r, the moments of P and Q are equal in magnitude (2) Since /* and Q act in contrary directions with respect to their tondem-y to turn tirh in Case 1. and ahc in Cmo n., regarded as rods moveable round the point c, the moments of P and Q arc opposite in direction. , 0.8. 82 OF MOMENTS, Note. We can readily extend the propositions proved in the preceding articles to any number of forces in one plane. For since the sum of the moments of two forces is equal to the moment of their resultant, we may substitute the resultant for the two forces; we may now combine this resultant with a third, and so on for any number of forces. Hence we obtain the following conclusion : The moment of the resultant of any number offerees in one planCf taken with respect to any point in that plane, is equal to the algebraic sum of the moments of the several forces with respect to the same point i ' "^ ions proved in in one plane. is equal to the e resultant for resultant with '»• of forces in that plane, is of the teveral ■if CHAPTER X. OF MECHANICAL INSTRUMENTS. 83. A MECHANICAL Instrument, or Machine, is a con- trivance for making a force, which is applied at one point, available at some other point. The Simplest Machines are rods used in pushing, and ropes used in pulling, but what are called The Simple Machines, or Mechanical Powers, aro 1. The Lever. 2. The Wheel and Axle. 3. The Pulley. 4. The Inclined Piano. 5. The Screw. 6. The Wedge. 84. THE LEVER. We have already defined a Lever as a rigid rod, capable of turning round a fixed point in the rod, called the Fulcrum. Li the simplest cases, in which the Lever is employed as a Machine, three forces are brought into play : 1. The Power applied to raise a weight or to over- come an obstacle. 2. The Weight or obstacle to be overcome. 3. The Reaction of tlic fixed point or fulcrum, 6—2 J0 IMAGE EVALUATION TEST TARGET (MT-3) 1.0 14^128 {|2.5 1^ *m i2 2 S ■:£ ||||2^ I.I U 1.25 1.4 1.6 — ^ 7] ys // "3 .%^^^ vV^ 7 Photographic Sciences Corporation as WIST MAIN STRUT WHSTIR.N.Y. MSSO (716) •7a-4S03 '^ / ^ >? ^ M i/. 1^ « 84 OF MECHANICAL INSTRUMENTS, 85. Levers may be divided into throe classes, according to the relative position of the points, where the power and weight are applied, with respect to the fulcrum. In levers of the first class, the power and weight are applied on opjyosite s'des of the fulcrum G, and act in the same direction. r> "A" i Examples. A poker between the bars of a gnite, raising the coals. A spade. A pair of scissors is a double lever of this class, the rivet being the fulcrum. In levers of the second class, the power and weight are applied on the same side of the fulcrum, and act in opposite directions, the power being applied at a greater distance from the fulcrum than the weight is. i i: a ExAMPLEa. A whcelbaiTow : the fulcrum being the point where the wheel presses on the ground. The chipping knife, in which the fulcrum is at the end attached to a bench, and the weight is the resistance of the substance to be cut, acting upwards, the power being applied by the hand dowU' wards. This instrument is used for cutting turnips, sugar and tobacco. Another instance is an oar, the blade of the oar in the water being tlie fulcrum. A pair of nut-crackers is a double lovor of this cluaiB. J api dir we rest this weij per[ dire fore mon of Ci J inoii dire the the i, according power and weight are in the same ^ite, raising dlo lover of weight arc in opposite stance from the point ling knife, a bench, to be cut, ind down- lipa, sugar of the oar crackers is OF MECHANICAL INSTRUMENTS, 85 In levers of the third dans, tlio power and weight are applied on the same side of the fulcrum, and act in opposite directions, the power being nearer to the fulcrum than the weight is. B A Examples. A man lifting a long ladder with one end resting on the ground. A pair of tongs is a double lever of this class. 80. Conditions of Equilibrium of a Lever. Let Fi and Fi be tlie measures of t ro forces acting on u weightless lever, and let Z>j and D^ be the measures of the perpendiculars drawn from the fulcrum to meet the Mnes of direction of Fi and F^. Al H Then, since the lever is kept in e! F^ X Dj. If more than two forces act on a lever, the sum of the moments of the forces, which tend to turn the lever in one direction, about the fulcrum, must be equal to the sum of the moments of the forces, which tend to turn the lever in the contrary direction, al>out the fulcrum. II 86 OF MECHANICAL INSTRUMENTS. u On Mechanical a''vantage and disadvantage. 87. The force applied to a macliiiio to set it in motion is (as we said in Art. 84) called the Power (P), and the resistance to be overcome is called the Weight ( IV). In the propositions which wo are now discussing we determine the value of the Power which would suffice to balance the Weight, and any increase in this value of P will enable us to work the machine. The efficiency or working power of a machine is measured W by the fraction -p , which is often called The Modulus of the machine. When W is greater than P, the machine is said to work at a mechatiical advantage^ and when W is less than P, at a mechanical disadvan tage. To illustrate this from the cases of the straight and weightless Lever, acted on by two forces, in directions perpendicular to the Lever, wo refer to the diagrams of A rt. 85, and observe (1) In a lever of i\\Q first class, P will be less than (r, if AC he greater than BC. P will be greater than JF, if AC be less than BC. Hence, in this case, mechanical advantage is the result, if P be further from the fulcrum than IV^ mechanical disadvantage is the result, if P be nenror to the fulcrum than W. (2) In a lev(n' of the second class P is always less than W. Hence in this case mechanical advjuitage is always gained. (3) In a lever of the third class P is always greater than }V. Hence in this case mechanical disadvantage is always the result. EXAMPLES. — VII. 87 ige. 1 motion is resistance jussing wo suffice to b of P will I measured ulus of the to work at m P, at a •aiglit and directions iagrams of wBG. nBC. 88. If two weights he in equilibrium on a straight heavy lever in any position, which is not vertical^ they will he m equilibrium in every position of the lever. Let P and Q be the weights suspended from the points Ay B of the lever, whose fulcrum is (7, centre of gravity G^ and weight W, Then, since P, Q, TV are parallel forces, acting always at the same points in the lever, their resultant will always paw through the same point of the lever. Now, since in one position the lever balances about (7, the resultant of P, Q, IV, passes through C, and therefore in all positions of the lever the resultant will pass through G, and will therefore be counteracted by the resistance of the fixed point. Hence the lever will balance in every position. 3r from the \roY to the always less fo is always [lys greater vantage is Examples. — VII. 1. Pressures of 4 lbs. and Gibs, act, at points A and B, perpendicularly to a straight lever AB, 6 feet long. Find the position of D the fulcrum and the pressure on it, when the pressures act in the same direction. 2. In a lever of the first class the arms are as 7 : 9. The pressure on the fulcrum is Ji^lbs. Find the weights. 3. A uniform rod, If) feet long, balances on a i)oint 4 feet from one end, when a weight of JJlbs. is suspended from that end. Find the weight of the rod. lit 88 EXAMPLES.— VII. lil 4. ABG is a straight lever; the length o( AB - 1 inches, that of BC — 3 inches ; weights of Gibs, and 10 lbs. hang at A and B, and an upward pressure of (Jibs, acts at C. Find the position of Ji fulcrum, about which the lever, so acted upon, would balance, and determine the pressure on it. 5. A heavy rod balances itself on a point, one-third of its length from one end : ii the rod be carried bv two men, one at each end, what part of the weight will be supported by each ? 6. In a lever of the first chiss, if the distance of the fulcrum from one of the forces be a third of the whole length of the lever: shew that, whon the direction of either of the forces is 'oversed, the fulcrum must then bo placed at three times its former distance from the same force. 7. A straight uniform lever, whoso weight is 50 lbs. and length 6 ft., roous in equilibrium on a fulcrum, when a weight of 10 lbs. is suspended from one extremity. Find tho position of the fulcrum, and the pressure upon it. 8. It is observed that a beam A //, whose length is 12 feet, will balance at a point 2 feet from the end A, but when a weight of 100 lbs. is hung from the end B, it balances at a point 2 feet from that end. Find the weight of the beam. 9. If a force of 2i lis., acting vertically iipwards at one extremity of a heavy rod, of length 4 feet, can just support 2^ lbs., suspended at a distance of lA feet from the other extremity, about which the rod can turn, find the weight of tho rod and the pressure on the hinge. 10. A rod, G feet long, is laid horizontnlh across two pegs, which are four feet ajnirt, so that each ciid projects one foot beyond the pegs: weights of 3 lbs. and 5 His. lu'inghung at tho ends, required the pressures on the pegs. Also, if an upward pressure be applied to the rod, at a point 2 feet from tho luore heavily laden end, in consequence ,>f which the pegs now bear each the same pressure, determine the amount of pressure applied. 11. Suppose weights of 2, 7, 5 and .'Jibs, to be suspended from a straight lod, at successive distances of 2, 2 and 13 inches from each other. Where nmst a fulcrum be placed, for the rod to balance i r.XAMri.E^.— VII. 89 : 7 inches. li.ing at A Find the ited upon, hird of ita ) men, one ported by ice of the olc length ler of the d at three lbs. and weight of K)Hition of is 12 feet, t when a dances at c beam. lis at one t support the other weight of jross two >io(jts one 'inghung rod, at a 1 sequence loterniine uspended 2 and 13 laced, for • l« 12. llow would the mechanical advantage of an oar bo modified by lengthening that part of it which is within tlio rowlock? 13. A heavy uniform bar, 10 feet long and of given weight W, is laid over two proi)8 in the same horizontal line, so that 1 foot of its lengtii projects over one of the prop'i. What must bo the distance between tlio props, that the I)ressure on one may be double that on the other \ 14. If the weights on a lever be 3 lbs. and 7 lbs., and the arms 8 inches and 9 inches respectively, at what point must a force of 1 lb. bo applied perpendicularly to the lever, in order to keep them at rest ? 15. Two weights, each equal to 8 lbs., hanging on a straight lever at points 12 inches and 18 inches from the fulcrum, and on tlie same side of it, are bahmced by a single vertical force, acting at a point IG inches from tlie fulcrum. Find the magnitude of the force, and shew whether it acts at mechanical advantage or disadvantage. 10. Two men carry a uniform l)eam, G feet in length and weighing l(j stone, upon tlieir shoulders, and at two feet from one end a weight of 4 stone is placed ; what weiglit does each sustain, supposing the ends of the beam to rest on their shoulders ? 17. Two men su[>port a uniform heavy beam on their slioulders, which are at distances a and b from the ends; if the pressure on one man be r times that on the other, find the length of the beam. 18. Two uniform cylinders, whose lengths are 3 feet and 5 feet, and weights 15 lbs. and i) lbs. respectively, are fastened together, so as to have thoir axes in the same straight line. Find the point about which they wouhl bahince. IJ). A uniform beam, whose weight is 130 lbs. juid length 12 feet, has weights of 15 lbs. and 25 lbs. hung on its ends. ^\\\\ the point on which the system will balance. 20. A uniform rod rests with its extremities on two props. Find at what point a weight, c'um be one foot from that end, to which the weight of 3 lbs. is attached. 26. A bent lever consists of two straight rods, of the same uniform material and thickness; the length of one rod is three times that of the other; what weight must be attached to the middle point of the shorter one, that the rods may be equally inclined to the horizon i 27. A uniform rod AB i» kept at rest by a prop placed 2 inches from the end A and a force of 5f lbs. acting vertically upwards at B. The fulcrum is shifted 1 inch further from ^1, and, when a weight of 1 lb, is hung on to the middle point of the rod, it is found that a force of 5 lbs. acting upwards at B is Butficient for equilibrium. Find the length nnd weightof the rod. EXAMPLES,— VI L 9« t long, is 10 ladder »y each of ts length p, 10 feet beam is )s. at the 20 lbs. at he beam. nd length stances 2, point on hich (not weight P that end, er end be find the d weighs the other eh can be im, if the weight of ' I the same 1 is three Bd to the e equally 3p placed vertically • from Ay 3 point of lIs at B is f the rod. 28. A uniform rod, to one end of which a weight of 5 lbs, is attached, will balance at a point 2 ft. from that end. When an additional weight of G lbs. is attached, it will balance at a point 1 ft. from that end. Find the length and weight of the rod. 29. AB is a lever 1,5 ft. long, moveable round a fulcrmn at a distance of 4 ft. from B, A weight of 12 lbs. being suspended from A, and a weight of 8 lbs. from J5, what fore© must act upwards at the middle point of AB to preserve equilibrium ? 30. If weights of 3, 5 and 7 lbs. be suspended, at distances of 2, 4 and 8 feet from the fulcrum, on one arm of a straight lever, and weights of 5, 7 and 9 lbs. be suspended, at distances of 3, 6 and 7 feet from the fulcrum, on the other arm, where must a weight of 1 lb. be placed, so as to keep the lever at rest? 31. The arms of a bent lever are 3 and 5 feet. A force of 6 lbs. acting per[)endicularly at the end of the shorter arm, is balanced by a force, acting at an angle of 30° to the longer arm at its end. Find this force. 32. In a straight lever of the second class, a force of 10 lbs. inclined at an angle of G0° to the lever, acts at one end, and balances a weight of 20 lbs. hanging 4 feet from the fulcnim. Find the length of the lover. 33. Two straight rods without weight, each four feet long, are loaded with weights lib., 3 lbs., 5 lbs., 7 lbs., and 9 lbs., placed in order a foot apart. Shew how to place one of the rods upon the other, so that both may balance about a fulcrum at the middle point of the lower : and explain liow this may be done in two ways. 34. Two forces of 2 lbs. and 4 lbs. act at the same point of a straight lever, on opposite sides of it, the less one being perpendicular to the lever, and keep it at rest. Determine the direction of the greater force, and the pressure upon the fulcnmi. 35. A bent lever consists of two unifoiin heavy beams whose lengths are as I : 2. Find the weight, which must bo attached to the extremity of the shorter, in order tAmt 4tu> arms may make cfpial angles with the horizon. 'i 99- OF MECHANICAL INSTRUMENTS. 36. A beirt lever ABC, having B for its fiilcruni, and the angle ABC between its arms a right angle, has equal weights suspended from A and C, its extremities. Sliow tlwit, in the position of cquilibriuni, either arm is equally incliucd to a vertical line, and to the line -(4(7. 37. The arms of a lever are inclined to each other ; sliew that the lever will be in equilibrium with equal weights suspended from its extremities, if the point, midway between the extremities, be vertically above the fulcrum. 38. ABC is a weightless triangle having a right angle BACy^ndi AB=^2.AG\ if the triangle be suspended from A, and two weights P, Q, hanging at 7?, C, keep it at rest, with the sides AB, AC equally inclined tp the vertical, find the ratio of P to Q. • 39. The arms of a bent lever, of length 18 inches and 12 inches, are inclined to the horizon at angles of 45** and 30® respectively. If the greater weight be IG lbs., find the less. 40. A BCD is a quadrilateral figure, having the angles at B and Z> equal. Forces acting m AB and AD^ tend to turn the figure about C, which is supported. Shew that, when there is equilibrium, the forces are proportional to the opposite sides. THE COMMON OR ROMAN STEELYARD. jD C /? i -/ • T o Y w 89. This balance consists of a stroight lever AB sus- pended by the point C, and capable of turning round this point. At the point A in the shorter ann is attached a hook, from whioh is suspended the substance, whose weight W is required. or or I, and the il weights tluit, iu iucd to a lor; sliew I weights ^ between flit angle from A, rest, with , find the iches and i** and 30» le less. angles at d to turn lat, when ) opposite D. O -J w AB sus- und this a hook, 3ight W OF MECHANICAL INSTRUMENTS. 93 A ring 2>, carrying a weight P of constant magnitude, can slide along the graduated arm CB^ till P and W balance each other about C, when the" lever is horizontal. The graduation, at which P rests, when this is the case, indicates the weight of the substance. In graduating the arm BC^ account must be taken of tlie -vcight of the lever : let Q bo the weight of the lever and G its centre of gravity, D the point from which P is suspended when it balances WB.t A : then taking moments about (7, we have P.CD-rQ.CG=JV.CA ....(a). If on the arm CA we take a point (9, such that P.CO = Q.CG, the equation (a) becomes ^ P.CD + P,CO=lF,CA, or P{CD + CO)=JF.CA, or P.OD=W.CA; .-. OD:CA:=}V: P. Now wo may graduate OB by taking distances, measured from 0, successively equal to CA, 20 A, 30 A, and marking them 1, 2, 3...... and so on. When P rests at the first of those graduations, ' OD = CA and :. TV = P. When P rests at the second of these graduations, CD = 20 A and .-. W = 2P, and so on. Thus, when the weight of P is known, tlic weight of W is known also. THE COMMON BALANCE. 90. This balance consists of a lever AB, called the beam, suspended from a fulcrum 0, about which it can turn freely : the point (7 is a little above the centre of gravity G of the beam, and from the extremities A, B of the arms GA, GB, wliich ought to be similar and equal, are suspended two 94 OF MECHANICAL INiiTRUMENTS. scale-pans, in one of which is placed the substance, whose weight W \9, required, and weights of known magnitude are placed in the other, till their sum P just balances W\ this being the case, when the beam is exactly horizontal in n i 1 position of rest. In this case, if the arms bo perfectly equal and similar, and the scale-pans also of equal weight, P will be exactly equal to W. If these weights differ by ever so little, the horizontality of the beam will be disturbed, and, after oscillating for a time, it will rest in a position inclined to the horizon, and the greater this inclination is for a given difference of P and W<, the greater is the sensibility of the balance. 91. The Requisites for a good balance. (1) The balance ought to be true; i.c, the beam should be horizontal when loaded with equal weights in the scales at A and B. This will be the case, if the scales bo of equal weight, and if the lino drawn through G at right angles to AB divide the beam into two equal and similar arms. (2) The balance ought to be sensible; i.e. the angle, which the beam makes with the horizon ought to be easib perceptible, when the weights P and W differ by a very small quantity. (3) The balance ought to be stable; i.e. if the equili- brium be a little disturbed cither way, there ought to be a decided and rapid tendency to return to the original position of rest, so as to ensure speed in the performance of a weighing. m I to mo^ li ! ice, whose nitudo arc 8 W; this mtal in a OF MKCHANrCAL INSTRUMENTS, 95 The comparative importance of these quahties in a balance will depend upon the service, for which it is intended. For weighing heavy goods, sfahility is of more importance. For weighings requiring great accuracy, as in practical chemistry, sensibility is the quality desired. A simple way of testing the accuracy of a balance is by interchanging P and W in the scales. The balance ought to retain the same position, when this is done. / THE DANISH STEELYARD. f :ectly equal b, P will be er so little, and, after lued to the 3r a given Uity of the cam should le scales at of equal angles to ms. the angle, bo easib by a very the equili- ^ht to be a lal position a weighing. 'K V IP ir 92. This instrument consists of a bar AB, terminating in a ball B, which serves as the power, and the substance to be weighed is suspended from the end A ; the fulcrum C, which is generally a loop at the end of a string, being moved along ABy till P and IV balance. To graduate the instrument. Let P be the weight of the steelyard, acting at the centre of gravity : and let C be the position of the fulcrum, when P and IV balance. Then there is equilibrium, when IV'.P==0C:AC. Now if W= Py 0C= AC, and .-. AO^^, if^-2P, OC ^ 2 AC, and .\ AC = if W - 3 /', 0C=3 A C, and :. AC = 2 AO 3 ' AO 4 And thus the graduations may be determined. 96 EXAMPLES.- VI IL ExAMPLEa — VIII. 1. A steelyard is formed out of a unifona bar, 38 inches long and weighing 12 lbs., the fulcrum being placed 5 inches from one end. If the moveable weight be 1^ lbs., find the greatest weight, which can be determined by means of the instrument. 2. If the moveable weight, for which a common steel- yard is constructed,* be lib., and a tradesman substitute a weight of 2 lbs., using the same graduations, shew that he defrauds his customers, if the centre of gravity of the steel- yard be in the longer arm, and himself, if it be in the shorvcr arm. 3 If the common steelyard consist of a uniform rod, whose weight is of the moveable weight, and the fulcrum be ; p 4 of the length of the rod from one end; shew that the greatest weight that can be weighed is '^ — times the moveable weight. 4. Where must bo the centre of gravity of the common steelyard so that any moveable weight may be used with it ? 6. If the common steelyard be correctly constructed for a moveable weight P, shew that it may be made a correctly constructed instrument for a moveable weight nP^ by suspending at the centre of gravity of the steelyard a weight equal to w - 1 times the weight of the steelyard. 6. The arms of a balance are in the ratio of 19 : 20. The pan, in which the weights are placed, is suspended from the longer arm. What is the real weight of a body, which apparently weighs 38 lbs. '\ 7. A body, the weight of which is 1 lb., appears to weigh 14 ounces, when it is placed in one scale of a false balaiu^e. What will bo its apparent weight, when placed in the other scale 1 8. One pound is weighed at each end of a false balance, and the sum of the apparent weights is 2^ lbs., what is the ratio of the lengths of the arms ? at ', 38 inches id 5 inches J., find the (Jins of the OF MECHANICAL INSTRUMENTS. 97 9. If a balance bo false, having its arms unequal and in the ratio of ir> : 16, find how much per lb. a customer really pays for tea, which is sold to him from the longer arm at 35. 9c?. per lb. 10. If a Danish steelyard weigh nibs., shew how to graduate it by ounces. mon stccl- bstitute a iv that ho the steel- bo in the iforni rod, Icrumbe 4 10 greatest ! moveable common with it? onstructed 9 made a eight nP, teelyard a ird. : 20. The from tho iy, which s to weigh balwi^o. tho other balance, lat is the THE WHEEL AND AXLB. 93. This machine consists of a cylinder HIl\ called the axle, and a wheel AD, the two having a common axis, terminating in pivots C and (7, about which the machine can turn; tho pivots resting in fixed sockets at C, C A rope, to one end of which the weight W is attached, passes round tho axlo, and has its other end fixed to the axle. Another ropo passes round the wheel, being attached at one end to tho circumference of the wheel and at tho otlier end tho power P is applied. The ropes pass round tho wheel and axle in opposite directions, and thus tend to turn tiie maciiino in oppdsito directions. The windlass and capstan are examples of tho practical use of this mechanical instrument. 6. B. f 98 OF MECHANICAL INSTRUMENTS. 94. To find ths condition of equilibrium on the Wheel and Axle. i! ! ■|i I * -" 1. Suppose the Wlieel and Axle to be cut by a vertical plane at the point of thoir junction, and that this figure represents the section. We may then suppose P and IF to act in this plane, and that their lines of action touch the circles at M and N, From the common centre draw OM and ON. Then OM and OiV, being drawn from the centre of tlio circles to the points of contact, will be perpendicular to PMmd \VN. The axis of the machine being at rest, wo may consider the machine as a lever, moveable roun \ as a fulcrum. Then there will be equilibr'r.nn, when P: W=^ON'.OM, i. e. when P : W= radius of axle : radius of wheel. Note. If the thickness of the ropes cannot be neglected, we must suppose P and W to act along the middle of the ropes, and in this case ON- radius of axle f radius of ropo, OM^ radius of wheel + radius of rope. \e Wheel EXAMPLES.— IX. 99 cal phiDC epresents lane, ami e of tlio icular to sider tlio )f wheel. 3glectc(l, of the Examples. — IX. 1. If the radius of the wheel be 3 feet, the weight 18 lbs., and the power 3 lbs,, what must bo the radius of the axle '{ 2. If the radius of the wheel be G feet, the radius of the axle 2 feet, the weight 3 lbs., what must be the power, in order to produce equilibrium ? 3. If the radius of the axle be 3 feet, and the radius of the wheel 9 feet, what power will be necessary, in order to keep a weight of 12 lbs. in equilibrium ? 4. Explain how the "capstan" possesses mechanical advantage, and if the radius of the axle be 2 feet, .and 6 men push, each with a force of 1 cwt., on spokes ;"> feet long, find the weight they will just bo able to supiMjrt, 5. In what way must the power act so that the pressure on the axle may be the least possible \ 6. If the string, to which the weight is attached, be coiled in the usual manner round the axle, but the string, by which the power is applied, be nailed to a point in the rim of the wheel, find the position of equilibrium, the power .*nd weight being equal. 7. The radius of the wheel being three times that of the axle, and the string on the wheel being only strong enough to support a tension equivalent to 36 lbs., find the greatest weight which can bo lifted. 8. If the axis of the axlo do not coincide with the centre of the wheel, shew that the sum of the weights, which a given power will support in the two positions, in which tho diameter of the wheel containing the centre of the axle is horizontal, is double tho weight, which it would iupport, if the machine were accurately made. 9. Compare the greatest and least weights which can be supported by a force acting on a \s\w\A with a square axle. 7-« lOO OF MECHANICAL INSTRUMENTS. THE PULLEY. 96. The pulley is a small circular disc or wheel, having a uniform groove cut on its outer edge, and it can turn freely about an axis, which passes through its centre. This axis rests in sockets within the hlocky to which the pulley is attached. When the block is fixed, the pulley is said to be fixed ; in other cases it is moveable. A cord passes over the pulley along the groove, and at its extremities the power and wciglit are applied. I' u \ \\ The pulley is very useful for changing the direction of a string; and, assuming that the tension of a string is not altered by passing over a small pulley, the tension at all points ^)f the string between the points of application of P aud W will bo the same. When the pulley is fixed, no mechanical advantage is gained by its use, beyond tiiat of gi'enter convenience in applying the force. OF MECHANICAL INSTRUMENTS. lOI 96. To jind the conditions of equilibrium on a single moveable pulley. (1) When the strings are parallel. Let the string PA DC have one extremity fixed at C, and, after pass- ing under the moveable pulley, sup- / pose it to be passed over a fixed p pulley D and pulled Ky a force P. The weight W is suspended by a string whose direction is in a line with O the centre of the moveable pulley. ^ Then the tension of each of the strings DA^ CB is P. Hence the pulley is acted on by tliree parallel forces P, P iuid W\ and therefore, when there is equilibrium, W = 2 P. Obs, If weight of pulley {w) be taken into account, IP ^ W+w. (ii) When the strings are not parallel. Let the string quit the pulley at A and B. Then, since the tension along AP \% equal to that along BG^ their resultant will bisect the angle between them, Art. 22, and this resultant niust be equal and opposite to the weight W suspend- ed f 'oni the axis of the pulley, and acting in a vertical direction. Hence AP^ BC must be equally inclined to the vertical. Let 6 bo this inclination. Then the resultant of the two tensions, which we may regard as acting at A and B^ is 2P. cos By and this must bo equal to W] /. 2P.cos^ -: W is the condition of cquilibrinni. I I02 OF MECHANICAL INSTRUMENTS. 97. To find tfie condition of equilibrium for a system of pidleySy in which each pidley hangs by a separate string, the strings being parallel. K X <' I il.l In this system a string, acted on by the power P, passes under the pulley A, and is fastened to the block at M. A string, attached to A, passes under the pulley B, and is fastened to the block at N, and so on for any number of pulleys. The weight JV is suspended from the lowest pulley. Then, since JV is supported by the tension of the strings RC, JBCy W 2' tension of BC W '^ 4 ' AgaiD, BC^s tension = tension oi AB- and tension of PA = 2 tension oi AB W 8 • But tension of PA = P, W ' p= -L Thus, when there are three pulleys, P = --.3, and, similarly, when there are n pulleys, P = ~ , .-. P: [r=l :2\ W 2' system of tringf the OF MECHA^flCAL INSTRUMENTS. 103 Note. If the pulleys have weight, an additional force p will be required to assist P. Calling the weights of the pulleys, commencing with the highest^ Wu ic^, w^ w, "J w^ «r. w, ^=-2 + 4+1"+ *2"' - 1 the terms on the right-hand side of the equation being W obtained by taking the formula P= -^ , and making W = Wi, W2' ' .'M'n, successively, and n - 1, 2. . .«, successively. Note. This is usually called the First St/stem of Pulleys. Py passes pulley B, y number By. .0 strings V W 8 • 98. To find the condition of cqiiilihriam for a system oj pulleys^ where there are two blocks and the same string passes round the pulleys. In this system of pulleys the same - string passes round each of the pulleys as in the figure, and the parts of the string between successive pulleys are taken to be parallel. The tension of the string is the same throughout, and is equal to P, Hence, if n be the number of strings at the lower block, nP will be the resultant of the upward tensions of the strings upon the lower block. This resultant must bo equal to \V, when there is equilibrium. that is. nP=W is the condition required, which may be expressed thus, P : W=^ 1 :n. Note. This is usually called the Second System of Pulleys. 104 OF MECHANICAL INSTRUMENTS. 99. To find the condition of equilibrium in a system oj puUeySf in which all the strings are attached to the weight. The figure represents the system. The weight W is supported by the tension of the strings EA, SB, TO, attached to a bar AD. Now tension of RA = P, tension of JSB = tension of MO = tension of RA + tension of OP ^ 2P, tension of TC = tension of NM = tension of SB -f tension oi MO = 2*.P, and so on. Therefore TV = P + 2P + 2KP+ , and if there be n pulleys, W = P + 2P + 2^P + + 2'»-».P =:P(1 4- 2 + 22+ 4.2'-^) /2'' — 1\ = P . ( J, by G eonictrical Prog ression, = P.(2''- 1). If the pulleys have weight, th(jy all, except the uppermost, assist P ; and if wo call the assintancc, which they afford, p, and designate the weights of the pulleys beginning with the I system of J weight. the strings 'P. l2".P, and EXAMPLES.—X. 105 agression, ppermost, ifford, /?, I with tho ''>ti'est by Wi, w-z, Ws, ... ;> = (2"-^-l)?ri + (2''-=-l)?c,4-...+ {2=-n.-^r,.24-(2-l).ir„_i, the terms on t!ie right-liund member of tho equation being obtained by taking the formula JV = P (2" - 1), and making P equal Wi, ic^. «r„-2, «^n_i, successively, the number of pulleys being n — 1, n - 2 2, 1, successively. Note. This is usually called tho Third System of Pulleys. Examples. — X. 1. In a single moveable pulley, if the strings be not parallel, and P— IF, what must be the angle between the strings ? 2. Shew that no mechamcal advantage is gained by the single moveable pulley, unless the weight of the pulley be less than the power. 3. If the angle between the strings of the single moveable pulley be two-thirds of a right angle, what must be the ratio of the Power to the Weight, in order to produce equilibrium ? ■^ First System. 4. In a system of three pulleys a weight of 8 lbs. is attached to the lowest pulley. Neglecting the weights of the pulleys, find the power necessary to produce equilibrium. 6. In a system of pulleys in which each pulley hangs by a separate string there are 3 pulleys of equal weights. The weight attached to tho lowest is 32 lbs., and the power is 11 lbs. Find the weight of each pulley. 6. In a s,^^8tem of 3 pulleys a weiglit of Gibs, is attached to the lowest pulley. Supposing tho weight of each pulley to bo 3 lbs., fijid the force required to sustain equilibrium. 7 If there be 4 pulleys, whose weights, commencing with the iiighest, arc 1, 2, 4, and 8 lbs. respectively, and W bo 1()(H.)).^., find P. M io6 EXAMPLES.— X. » 1 Wkm 8. If there be three pulleys, and the weight of each bo 1 lb., find the force capable of supporting a weight of 9 lbs. 9. If there be three pulleys, the weight of each being W, but no weight attached to the lowest, shew that theic will be equilibrium, when P : ]^ : : 7 : 8. 10. If there be three pulleys of equal weight, what must be the weight of each, in order that a weight of 56 lbs. attached to the lowest may be supported by a power "qual to 7 lbs. 14 oz.? 11. What must be the weight of each of three pulleys that P may equal W^ the pulleys being all of equal weight ? 12. If all the pulleys, except the lowest, be considerc(i weightless, and the weight of the lowest and the power be each y lbs., and the weight attached be w lbs., shew that w is some odd multiple of jt?. 13. If P = 2 lbs. in a system of 4 pulleys, each hanging by a separate string, and the weight of each pulley together witli the string beneath it be 1 lb., shew that W— 17 lbs. 14. If on the same system of pulleys, in which each pulley hangs by a separate string, P can support Wy and P' causes a pressure W on the beam, to which the strings are all W W attached, shew that p- — -p, = 1. 15. If the beam, to which the strings are attached, be fixed at one point only, about which it is capable of revolving, and there arc only two moveable pulleys, find the position of the point, in order that the beam may remain horizontal. r T)ass( weig toth \i I S tl "'\ Secoxd System. 16. In the system of pulleys, in which the same string passes round all the pulleys and the parts of it between the pulleys are vertical, if there be three pulleys at the lower block, and this block weigh 8 lbs., find the power which will support 1 cwt. dght of each sight of 9 lbs. ach being W, V that there it, what must 5 lbs. attached qual to 7 lbs. three pulleys ual weight ? )e considered the power be lew that ic is ch hanging by together witli )S. :h each pulley md P' causes ings are all [attached, be of revolving, position of Kzontal. EXAMPLES.— X. 107 jamo string )etween the It the lower which vdll 17. In the system of pulleys, in which the same string passes round all the pulleys, if P = ^lbs., the lower block weigh 8 lbs., and contain 3 pulleys, and the string be fastened to the lower block, shew that W-^ lbs. 18. A man supports a weight, equal to half his own weight, by a system of pulleys, in which the same string passes round all the pulleys, the upper block being attached to the ceiling: if there be 7 strings at the lower block, find his pressure on the floor, on which he stands. 19. What weight will be supported, if there be 3 pulleys in the lower block, the string being fastened to the upper block, and the weight of the lower block being equal to 3 times the power? 20. Supposing that a power of 3 lbs. will just support a weight of 10 lbs. suspended from the lower block, the number of strings being 4, what is the weight of the lower block? 21. If the weight of the lower block and the power be each jolbs., and the weight attached to the lower block be tclbs., shew that w is some odd or even multiple of jt?, according as the end of the string is fastened to the upper or lower block. Third System. 22. A power P and a weight W are in equilibrium on a system of pulleys, in which all the strings are parallel and attached to a uniform bar, from which the weight is suspended, the weights of the pulleys being neglected. If the number of pulleys be three, and the strings be equidistant, from what point of the bar ought the weight to be suspended, that thf bar may rest in a horizontal position ? 23. In a system of 6 pulleys of equal weight, where each pulley is attached to a string, which is attached also to the weight, find the ratio, which the weight of each pulley must bear to the weight supported, in order that there may be equilibrium, without any power being applied. v|' I io8 OF MECHANICAL IiXSTRUMENTS. THE INCLINED PLANE. 100. By an inclined plane, as a mechanical instrument, is meant a plane inclined to the horizon. The figure represents a section of the inclined plane, made by a vertical plane perpendicular to the inclined plane. lOL Incline Let P actin i; I ■ ij k ^-5 is called the length of the plane. EG, which is taken to bo perpendicular to ACy is called the height of the plane. AGys, called the base of the plane. The angle BAG is called the inclination of the plane. When a body is in contact with a stnooth plane, there is ii mutual action between the body and the plane, acting at right angles to the plane. The force thus brought to bear on the body is called the reaction of the plane, and the reason for this reaction being equal to the pressure of the body on tho plane is to be explained thus :— Reaction is always contrary and equal to action: or, the mutual actions of two bodies upon each other are always equal, and directed towards opposite parts. Whatever draws or presses another, is as much drawn and pressed by that other. If any one press a stone with his finger, his finger is also pressed by the stone. If a horse draw a stone tied to a rope, tin horse will be equally drawn back towards tho stone. Theb P. 1 Prodi] From Then, pa7'allel order are Again t and .-. th( Hence that is P Simila ^TS. 1 instrument, is ed plane, made d plane. ACf is called ;he plane. ane, there is ii ane, acting at ^ht to bear on the reason for le body on tho iction; or, the er are always latever draws jssed by that r, his finger is ;one tied to a towards tho OF MECHANICAL INSTRUMENTS. 109 i 101. To find the condition of equilibrium on a smooth Inclined Plancy when the Poicer acts parallel to the plane. Let a body, i>, whose weight is IVt be pulled by a force P acting parallel to tho plane, and let the body be at rest. The body is acted upon by three forces : P, the power, acting parallel to AB, Wy the weight, acting parallel to BC^ i.e. vertically, i?, the reaction of the plane, acting at right angles to AB. ^ Produce RD to meet AC'm E. From E draw EF parallel to BC. Then, since the three sides of the triangle DFE are parallel to P, W^ R respectively, the sides taken in proper order are also proportional to P, W, R, by Art. 'AS ; .. P: W=DF.FE. Again, EFD, ABC are similar triangles, for the right angle FDE = the right angle BCA^ and t>ie angle EFD = the angle ABC, and .*. the remaining angles FED, BAG avo equal. Hence DF :FE^ CB : BA; and.-. P : W=GB:BA, that is P : Wr= height of plane : length of plane. Similarly we can prove that P : R = CB : CA, and W:Rr=BA: CA, no OF MECHANICAL INSTRUMENTS, 111 i' !' 102. To find the condition of equilibrium on a smooth fndined Plane, when the Power acts horizontally. Let the body Z> be kept at rest by three forces : P, the power, acting horizontally, Wf the weight of D, acting vertically downwards, Rj the reaction of the plane, acting at right angles to AB. Produce RD to meet AG in M, Now, angles MDO, ODA make up a right angle, and angles DAO, ODA are together equal to a rig^t angle; .'. angle MDO = angle DAO; that is, angle MDO = angle BAG, Also, angle MOD -^ angle BGA ; .*. MDO, BAG arc similar triangles. Then, since the sides of the triangle MOD are parallel, and therefore proportional, to P, JV, R, P : JV-MO \D0 =r. BG : AG. Also P : R =BG : AB, and ^ VF: R ^AG '.AB. cii of oqi Let smooth Mh^kes r Let Thei n 1 Then 'P Iwo (1 (2 008 "S. OF MECHANICAL INSTRUMENTS, III on a smooth lly. ces: vnwards, right angles igle, Iri^t angle; are parallel, ciii. By the aid of Trigonometry, we can find the condition uf equilibrium on a smooth inclined plane in a more general Let a body O, whose weight is W^ bo supported on a smooth inclined plane by a force P, the direction of which makes an angle 6 with the plane. Let a be the inclination of the plane. Then the forces acting on tiio body O are JVy the weight of the body, acting vertically downwards, /?, the reaction of the plane, acting at right angles to the plane, P, the given force. , " Then, since there is equilibrium, we have by Art. xxxix. P : IV : R r- sin RO W : sin ROP : sin PO IV, r-. sin (180" - a) : sin (})0'' - 6) . sin (90° + a + ^), = sin tt : cos ^ : cos (a + 6). (Trig. Art. 101, 99, 1 02.) Two i)articular cases are to bo especially noticed : (1) When P acisparallel to the phme, d-0, cos 6=^ I, and .'. P : W : R- sin a : I : cos a. (2) When P acts hnrizontaUtf, B = - a^ cos 6 - cos a, and cos (a + 6) -- 1 , (Trig. Art. 104, 67), and .'. P : JV : R = sin a : cosa : I. 112 EXAMPLES.— XI. Examples.— XL ofl INCLINED PLANE. )■' 1. If W^ bo 3 tons, find P, acting parallel to the plane, when the height of the plane is to its base as 5 : 12. 2. Find the pressure on the plane, when the height of the plane is to its base as 3 : 4, and the weight supported is lOlbs., the power being pa.^allel to the plane. 3. If, when P acts along the plane, R : P = S : 4, express R and P in terms of W. 4. Find the horizontal force necessary to support a body whose weight is 12 lbs. upon a plane, whose base is to its length as 4 : 5. 5. If the pressure on the plane be 2 lbs., and the power acting horizontally lib., what is the weight? and what the inclination of the plane 1 « 6. A force of 15 lbs., acting horizontally, supports a weight of 20 lbs. on an inclined plane: what force, acting along the plane, will support the same weight 1 7. A railway train, weighing 50 tons, is drawn up an incline of 1 in 30 by a rope : what is the least strain on the rope if 10 lbs. per ton be allowed for resistance to motion on level rails i 8. A force of 15 lbs., acting horizontally, supports a weight of 5^3 lbs. on an inclined plane; find the inclination of the plane to the horizon. 9. The weight supported upon an inclined ])liine is 2\/2lb8., and the plane is inclined at half a right angle to the horizon ; find the power, which, acting along the plane, will support tho weight* EXAMPLES.— XI. "3 3 the plane, 2. eight of the mpported is : 4, express port a body use ia to its d the power d what the rts a weight s along the iwn up an ain on the motion on rts a weight Ition of the is 2\/2lbs., he horizon ; lupport the 10. A weight of 66 lbs. rests upon a smooth plane inclined at 45'' to the horizon. What is the smallest horizontal force required to move it up the plane ? 11. What force, acting horizontally, will sustain a weight of 12 lbs. on a plane inclined to the horizon at an angle of 60' ? 12. What force, acting horizontally, will sustain a weight of 10 lbs. on a plane inclined to the horizon at an angle equal to half of one of the angles of an equilateral triangle ? 13. If the force, which will support a weight, when acting parallel to the plane, be half that, whick will do so, acting horizontally, find the inclination of the plane. 14. Equal weights are attached to the ends of a string; one of which rests on a plane inclined at 45*^ to the horizon, and the other hangs vertically over the summit of the plane, and rests on the ground beneath. Find the pressure of the latter on the ground. 16. What power, acting parallel to a smooth plane inclined at an angle of 30*^, is necessary to sustain a weight of 4 lbs. on the plane? 16. Show that, if P, instead of acting parallel to the plane, were to make the same angle with the vertical as the pressure of the plane on the body, tiio pressure on the plane would be equal to P. 17. Which will support the greater weight, a power acting horizontally, or the same power acting parallel to the plane ? 18. A weight of 20 lbs. is supported by a string fastened to a point in an inclined plane, and the string is only just strong enough to support a weight of 10 lbs.; the inclination of the plane to the horizon being gradually increased, find when the string will break. 19. Two unequal weights W and W\ connected by a string, are placed upon two smooth inclined pianos, the string passing over the intersection of the pianos. Find the ratio between the weights, when there is equilibrium. 20. The least power which will sustain a body on a certain incUucd piano in sufficiout, whou acting horizoutoUy, to support 8,(1. 8 114 EXAMPLES.— XI. a body weighing half as much again as the other on a plane, whoso inclination is half that of the former : find the angle of inclination. 21. The powers required to keep a given weight at rest on an inclhied plane, are 10 lbs. and 20 lbs., in the cases of the power acting along the plane and horizontally; find the in- clination of the plane. 22. A weight of 20 lbs. is supported on an inclined plane by a power of 12 lbs., acting parallel to the plane. Shew that, if it were required to support the same weight on the same plane by a force applied horizontally, the force must be increased in the ratio of 4 to 5, whilst the pressure on the plane >vill be increased in the duplicate of the same ratio. 23. If the power, acting along an inclined plane, the pressure on the plane, and the weight be as 1 ! ^3 : 2, find the inclination of the plane to the horizon. 24. When a certain inclined plane ABG^ whose length is A G, is placed upon AB as a base, a power of 3 lbs. can support on it a weight of 6 lbs. ; what weight could the same power sup- port, if the plane were placed on BG as base, so that AB '\» then the height of the plane? 25. If a plane bo inclined to the horizontal at an angle a, and a force W. tan a act up\/ard8, parallel to the plane, on a weight W which is in the pi me, what force must act down the plane to keep the weight in equilibrium ? 26. Give a geometrical construction for determining the direction, in which the power must act, when it is equal to the weight, and shew that, if Ri bo the pressure on the plane in this case, and R the pressure, when the power acts parallel to theplaneji?i=2/i?. ii t )n a plane, le angle of OF MECHANICAL INSTRUMENTS. "5 ^ht at rest ises of the ind the in- lined plane Shew that, 11 the same e must be are on the } ratio. plane, the : 2, find the ise length is can support power sup- that AB'xA ,t an angle |e plane, on act down piining the (qual to the fe plane in I parallel to THE SCREW. 104 The screw is a spiral thread, running along the surface of a circular cylinder, which may be imagined to bo generated thus: Let AG be a rectangle, whoso base AB is exactly equal to the circumference of the cylinder; make the rectangles BD^ GF^ EH... equal in every respect, and draw the straight lines AG^ DE, FG...; then, if the rectangle BH be applied to the surface of the cylinder, so that the base AB coincides with the base of the cylinder, the broken lines AG, DE, FG.., will form a continuous line on the surface of the cylinder, the point G coinciding with Z>, E with F, and so on. If we now suppose this Una to become a protuberant thread, perpendicular to the plane of the rect- ^ angle, we obtain a screw, in which the distance between any pomt of one thread and the one next below it, measured parallel to the axis of the cylinder, is everywhere the same and equal to BG. The angle GAB, which the thread at any point makes with the base of the cylinder, is called the pitch of the screw. The screw formed on the solid cylinder, as above, works in a holloio cylinder of equal radius, in which a spiral groove is cut exactly equal and similar to the thread on the solid cylinder, and in whicli groove the thread of the solid screw can work freely. A solid and hollow screw, related as above, are called companion screws; and, when in action, one of them is fixed and the other is turned by means of a lever, fixed into the cylinder at riglit angles to its axis. Hy turning tbe lever a weight is laised, or a pressuro produced, at 8— *i ii6 OF MECHANICAL INSTRUMENTS. if I the end of the screw, which pressure acts in direction of the axis of the screw. When the hollow screw is small, it is some- times called a nut. The annexed figure, representing the ap- pearance of a solid screw, will assist the reader in understanding that a screw is nothing more than an inclined plane, constructed on the sur- face of a cylinder. P, c?. To find the condition of equilibrium on the Screw. I I ( Tho forces acting on the screw will bo OF MECHANICAL INSTRUMENTS. 117 lircction of \ the Screw. P, the power, acting horizontally at riglit angles to the rod MN {which is at right angles to the axis of the Screw, and whose length is a), and producing a vertical pressure upwards in direction of the axis ; Wj a weight placed a^ the end of the screw and acting vertically downwards* and a series of forces R', R". . . arising from the pressure of the hollow screw on each point of the solid screw, wit' which it is in contact. The forces iZ', 72"... act at right angles to the thread, and will therefore make an angle a, equal to the pitch of the screw, with the lines drawn vertically from the points of contact Resolving R\ R"... vertically and horizontally, we shall have R' . cos a, R". cos a,... acting vertically upwards, and R. sin a, R". sin a,... acting horizontally, and tending to turn the screw in a direction contrary to that, in which P tends to turn it. Also, each of these horizontal forces acts at an arm r, equal to the radius of the cylinder. Hence, taking moments, (7i^'4- 72'' 4- . . .) . sin a.r-P .a.. .(1), and (72' 4- 72'' -»-...). cos a- W (2). Dividing (1) by (2) P. a tan a . r or -^- W ' r . tan^ a This condition of equilibrium may be expressed in another form, thus, since 27r . »• . tan a = distance between two threads, measured parallel to the axis, and 27r . rt = circumference of the circle described by M, P 2ir . r . tan a distance betw een two threads ~ circumference of circle whose radius is M^t' ii8 OF MECHANICAL INSTRUMENTS. ii;' THE WEDGE. 106. The Wedgo is a solid triangular prism. Its two ends are equal and similar triangles. Its three sides are rectangular parallelo- grams. ^J5 is called its edge: GDEF its Jiead. It is used for separating bodies, or parts of the same body, which adhere strongly to each other. The edge of the Wedge is introduced into a small cleft, and it is then driven forward by blows of a hammer applied at its head. The mode of working this machine is quite different in principle from the method used in the other machines, which have been described. They are worked by the regular and steady application of a power, acting uniformly at that point of the machine, to which it is applied, and gradually producing motion : but in this machine the power is applied by sudden impulses. Hence any i)ivestigation for finding the relation between the power and weight in this machine must involve considera- tions, which caniiot be explained by the principles of Elemen- tary Statics. Hatchets, chisels, nails, carpenters' planes, swords, are modifications of the wedge. be inti to of am< ON FRICTION. 107. In our investigation of the Conditions of Equilibrium for the Mechanical Instruments we assumed that the surfaces under consideration were perfectly smooth. But, since in practice no surface is perfectly smooth, it is necessary, in applying the laws of equilibrium to particular problems, to take into consideration the resistance to motion, which is brought into play by an attempt to move a rough body over a rough surface. This resistance is called Friction. the same small cleft, iier applied different in lines, which application le machine, ig motion : impulses. n between considera- f Elemen- ^vords, are juilibrium le surfaces i, since in [•essary, in )blems, to which is [body over OF MECHANICAL INSTRUMENTS. 119 108. Friction is a force, of which the practical use may be seen from the following instances : (1) At ever)' step we take in walking we bring friction into play. If we attempt to walk on a surface approaching to perfect smoothness, as a polished oaken floor, or a sheet of ice, our feet have a tendency to shp, because but a slight amount of friction can be brought into action. (2) If a wedge be driven by a blow into a block of wood, but for the friction between the wedge and the block, the wedge would fly back. 109. On attempting to displace a body at rest on a rough horizontal plane, we experience three kinds of re- sistance : (1) On trying to UJt the body off the plane, we experience an opposition as it were in the body itself, arising from the attraction of the Earth. (2) On attempting to press the body against tho plane, on which it rests, we find that the plane resists our effort. (3) On trying to pus?i the body along the plane, we experience a resistance, varying according to the nature of tlie surfaces in contact. This resistance is called Friction, and its laws are as follows : (a) Its direction is opposite to that of attempted motion. 0) Its magnitude is just sufiicient to prevent motion, but no more than a certain amount can be called into play. If more than this amount bo required to prevent motion, motion will ensue. This greatest possible amount is termed Limiting Friction, and when this is jush called into action, the body is said to be in a state bordering on motion. 110. The statical laws of Limiting Friction are I. Wlien the substances in contact remain the same, the Limiting Friction varies n.s the Pressure between the bodies. I20 OF MECHANICAL INSTRUMENTS. II. The amount of Limiting Friction is independent of the area in contact. If any oily matter be introduced between the sub- stances, a smaller amount only of friction is capable of being called into play, that is, the Limiting Friction is then less. All the Laws of Friction have been obtained by experiment. 111. If -K be the pressure between two bodies in contact, and F the amount of Limiting Friction, then by Law I, F varies as /?, or F=iiR, where fi ii a constant quantity, to be determined by experiment. 112. When wo say that [i. is constant, we mean that it is independent of 72, independent of the extent of surface in contact, and dependent only upon the nature of the surfaces. /i is called The Coefficient of Friction. cxiii. ]fa body he on the point of motion, when placed 07i an inclined plane^ inclined at an angle a to the horizon, then the coefficient of friction between the body and the inclined plane = tan a. Let ABC be a rough plane, on which a body O when placed is in a state just bordering on motion. Then the body is kept at rest by three forces, W, the weight of the body, R, the resistance of the plane, Fy tho force of friction acting along the plane. MTSCI-: LLA A'EOUS EXERCISES. T21 idcpendent 1 the sub- >le of being ;n less, sxperiment. in contact, wl, xperiment. can that it of surface lie surfaces. I placed 071 e horizon, y and the when Hence, by Art. 101, Now, by Art. Ill, R AC " R A& _BC "^~AC' Hence, if a be the inclination of the plane to the horizon, /x — tan a. Miscellaneous Exercises. 1. If two equal forces, acting on a point, are balanced by two other equal forces acting at the same point, shew that the latter two are equally inclined to the former two. Are all four forces necessarily equal % 2. The resultant of two forces P and Q, acting on a particle, is the same, when their directions are inclined at an angle By as when they are inclined at an angle . — ^ to each other : shew that tan tf = ^2 - 1. 3. If the actual lines of action of three forces, which keep a point at rest, be represented by hues drawii from the point of apphcation, the greatest of the three will always be opposite to the least of the angles made by tho lines, and the least force will be opposite the greatest angle. 4. A pole AB is supported with one end B on the ground, and is prevented from falling by means of two ropes, of equal length, attached to the end A\ shew that the powers exerted by the men, who hold tho ropes, are inversely as their own distances from B. 5. Equal weights are supported on two inclined planes of equal length, but of diflferent heights; and the power required on the steeper plane is equal to tho resistance exerted by the other: also, this resistance is three times tho resistance of the steeper plane. Compare the two powers employed, and their relation to W. '1 f il I '•:i 122 MISCELLANEOUS EXERCISES. \- I I I 6. A point is taken within or outside a triangle, and linea are drawn from it to the angular points of the triangle. Prove that the resultant of the forces represented by these lines is represented by 3 times the lino joining the point and the centre of gravity of the triangle. 7. A line is drawn parallel to the base of an isosceles tri- angle, bisecting the perpendicular from the vertex on the base and meeting the sides. Shew that the distance of the centre of gravity of the quadrilateral figure thus formed from the vertex is seven-ninths of the perpendicular. 8. A weight of 25 lbs. hangs at rest, attached to the ends of two strings, the lengths of which are 3 and 4 feet, and the other ends of the strings are fastened at two points in a liorizontal line, distant 5 feet from each other ; find the tension of each string. 9. A uniform heavy rod is su^'^nortcd by two strings attached to its two ends and to a fixed point, so that the strings and the rod form an equilateral triangle. Compare tlie tension of the strings with the weight of the rod. 10. The resultant of two forces P and Q makes an angle of 30** with the direction of P, and is equal to ^3 . (? ; shew that P may be equal either to Q or to 2Q. 11. If a pole, one end of which is on the ground, bo supported by means of two ropes fastened to the other end and pulled in a vertical plane on opposite sides of the pole ; shew that when the persons who hold the ropes are equidistant from the foot of the pole, each must pull with a force pro- portional to the length of rope he is holding. 12. Three forces, acting in the same plane, keep a point at rest, the angles between the directions of the forces being 135", 120°, and 105^ Compare their magnitude. . 13. A boat, moored by the bank of a rapid river, is also attached by a rope to a stake higher up the stream on the opposite bank. Shew that, if the boat be loosed from its moorings, it will swing round to the opposite bank. 14. If three forces, acting at a point, be represented in magnitude and direction by three straight lines drawn from th:| fo! pr( shel tudi an( MISCELLANEOUS EXERCISES. "3 ;, and line» gle. Prove Bse lines is it and the sosceles tri- tex on the mce of tlie lus formed r. to the ends et, and the points in a the tension iwo strings 10 that the onipare the 33 an angle 3 . (? ; shew round, be other end the pole; equidistant force pro- a point at roes being [^er, is also Im on the from its ^sented in lawn from i that point, then, if the forces are in equilibrium, the triangles fonned by joining the extremities of the lines are equal. 15. If a point be acted upon by three forces, parallel and proportional to the three sides AB^ BG^ DCoi a quadrilateral, shew that the resultant of the forces is represented in magni- tude and direction by EGF^ E being the middle point of AD^ and CF being equal to EC. 16. Three pegs, A, B, Cy are stuck in a wall in the angles of an equilateral triangle, A being the highest and BC hori- zontal. A string whose length equals four times a side of the triangle is hung over them, and its two ends attached to a weight IV. Find the pressure on each peg. 17. A spiiere, of weight JV, is hung from a point in a wall by a string of length I attached to its centre. If T is the tension of the string, R the pressure on the wall, r the radius of the sphere, shew that II T: ir=r:l: {P-r-f. 18. Two inclined planes of equal bases, 6, and altitudes, h, are placed facing each other on a smooth table ; a sphere of weight W is placed between them, and they are prevented from sliding apart by a string. Shew that the tension of the string r- tyW. J . 19. A line of telegraph posts is erected along a road, and a wire is carried along the tops in the usual manner. At a certain point, where a post stands, the road makes a turn tlu'ough 60". To keep the post in its vertical position one extremity of a wire rope is attached to the middle point of the post, and the other to a point in the ground, midway between the two directions of tho road. If the rope make an angle of 30** with the post, and the telegraph wire be supposed hori- zontal, shew that the tension of the rope is 4 times the tension of tho wire. 20. Prove the following construction for finding the centre of gravity of a quadrilatenil. E is the point of intersection of the diagonals, F and G are points in the diagonals, dividing them into segments equal to those, into which they are divided at E. Tho centre of gravity of the quadrilateral coincidei with the centre of gravity of the triangle EFG. 124 MISCELLANEOUS EXERCISES, i i''^: A 21. Three smooth pegs are fastened in a vertical plane, so as to form an isosceles triangle with its base horizontal, vertex downwards, and vertical angle equal to 120°. A fine string, with a weight W attached to each extremity, is passed under the lower peg and over the other two. Find the pressure on each peg. Find also the vertical pressure on each peg, and shew how the whole weight is supported. 22. Six forces, of 1, 2, 3, 4, 5 and 6 lbs., act from the centre, O, of a regular hexagon towards the angular points, A^ B^G^ Z), E^ F, respectively. Shew that the direction of their resultant is 0E\ and its magnitude equals 6 lbs. 23. A heavy rod, AB, is moveable about a fulcrum at C, ind is kept in a liorizontal position by a string which has its ends attached to A and B, and passes over two fixed pullies, A E, which are respectively vertically above A and B. Shew tliat wherever G is, the tension of the string equals half the weight of the rod. 24. If triangles be described upon a given base AB^ antl the sides bo taken to represent forces, two of which tend from A, and the third towards B^ shew that the resultant of these forces is constant in direction and magnitude. 25. A uniform beam rests against a smooth peg ^, and has its lower extremity connected, by means of a string o) given length, with a point B in tlio same horizontal line as A. Determine the length of the beam, that it may rest at an angle of 45° to the horizon. 2G. The two arms of a lever are equal, and they form two sides of a square, the fulcrum being at their intersection. Two equal forces act at their extremities along the remaining sideB<; (ind the pressure on the fulcrum. %\ plane, so ital, 7ertcx fine string, ssed under iressure on h peg, and ANSWERS TO EXAMPLES. from the lar points, ireetion of )S. crum at C, ich has its :ed pullies, B. Shew half the ABy and tend from it of these >cg A , and i string o) line as A. rest at an form two bion. Two ling sides>: I. (i). 25,) 1. 12 lbs., or 6 lbs., or 4 lbs., or 2 lbs. 2. 29 lbs., or 19 lbs., or 7 lbs., or 3 Iba. 3. 150 lbs. 4. 60 lbs. 7. 5 lbs., and 5 V3 lbs. 8. \/344-15V3lb8. ^219 lbs. \/202^ir72lbs. V4r4-"2(V21bs. ^3 times either of the ec[ual force*. I, T 9. 10. 11. 12. 14. 10 3 V3lbs. fP II. (p. 31.) 2. The forces are equal. 4. If F bo one of the efjual forces, third force- 6- VllJlbs. 7. (V6-l)lbi. 8. Direction, North. Magnitude, P. 10. P:(2:22=l;l; V2. 12. 1:2:V3. 13. V3.P. J 4. It makes an angle of CO' with the smaller. 15. s/2 : 1. la 1 : V3. >J3.F, 126 ANSWERS TO EXAMPLES. Mm m m ir.i ' 17. (1) Resultant makes an angle of 30" with greater. ('!) Resultant makes an angle of 67 i" with force that acta Eaat. 22. 120°. 23. The forces must be opposite, two and two. 2.3. R is equal and opposite li'. 26. X lies in EA produced so that AX = AE. 111. (^p. 40.; 1. 4N/3lb8. and8x/3lbfl. (2) 10 lbs. 8. 1 : x^ 3. - 2. (1) ^- lbs., 6. 24 lbs. 7 11 9. -andj^. 11. The radius through Q's position makes with the vertical p an angle whose tangent is ^ . 13. Resultant = 2 \/(47 + 18x73), and makes with OA an angle whose tangent = — — — . 10. that AO 13. gravi 14. of^i 18. v/ trianiJ the tance 20. wei side. the ai 27. 28. IV. (p. 45.) 1. 3J^ inches. 2. 4 inches. 4. 2ft. bin. 5. 15 lbs. 3. 4ft. 10 in. m y. (p. 52.) 1. 34 ft. and 24 ft. 2. 3:1. 3. 63 lbs. and 36 lbs. 4. 3 ft. 5. 21 inches from the smaller weight. 0. 5 : 2. 1. P -Q, 8. 9 inches from the larger weight. If the forces were parallel they might act in opposite directions, and tlion the fulcrum would not bo between tlie points of application. 9. 6 ft. from the smaller weight. 10. 8 inches nearer to the smaller weight. 11. 1044 lbs. 12. 2 ft. 13. 20. /^ inches. 14. 12 lbs. 15. (l)-^''^lbs. 6 (2)1 Jibs. (3)-^^ lbs. VI. (p. 65.) 1. 16 inches froni end to which the 1 lb. is attaclicd. i. 3^ inches from thu 4 lbs. 3. 3^ inches from the 42. ANSWERS TO EXAMPLES. 137 WvoX acta must be )Osito R'. 7 , n vertical OA an :ft. 10 in. lbs. d. lorn tho fourth wciffht. 26 4. - ft. from the weight 16. 01 10. Let D bo the centre of gravity of A and jB, and £ that of A and C. Join CD, BE, and let them intersect in O. AO produced will cut BG in the point required. 13. To keep the vertical lino of tho common centre of ^'ravity of himself and the weight within his base. 14. 11 inches. 15. Midway between tho middle points of AB and CD. 17. v'3 times the side of the square. l?s. s^2 : 1. 19. If be tho centre of gravity of the trianG;le, and the 4 lbs. be at B, and the 5 lbs. at C, the centre of gravity required is one-fourth of the ilis- tance of from the point of tri section of BG nearest C. 20. Draw from the angular point, at which the smallest weight acts, a line to the middle point of the opposite side. The centre of gravity is four-fifths of this line from the angular point. 23. 45®+^. 24. 30». 26. At C. 27. If G bo tho centre of gravity, AG :yiD = 4 :9. 28. 12 inches. 29. If G bo tho centre of gravity, OG G divides OG, so that OG = „ . 32. Tho middle point o of tho lino joining tho centres of gravity of tho equal triangles. 33. If 2a be tho side of tho square, the vortex of the triangle is at a distance 2= 1:3: v'lO. 8. 15 lbs. and 20 Ibe. 12. v'4:s/6:V3+l. JV /3 9. Tension of each string = — ^- 16. Pressure onA=W, Pressure on each of B and C= V3. 21. Pressure on each upper ]^eg=W^3. Pressure on lower peg=^. Vertical pressure on each upper peg=?^; on lower peg= W. Thus the whole weight supported, i.e., 2W, is so distributed as to produce a downward pressure on each of the upper pegs= — Wy and an upward pressure on the lower peg=rr. 24. Resultant =2ilJ5. 25. Length of beam = -y-, where a = length of string, 26. Pressures »J2 , P, where P represents each force. ft u ■%:. ' ' f • ^ • wer acting jlination is ths of the 23. 30". .and 20 lbs. re on lower jd, i.e., 2fF, •e on each of )n the lower 15. Length APPENDIX TO CANADIAN EDITION. In solving statical problems the following rules will be found useful : 1. Draw a figure of the system as accurately as pos- sible, representing all the forces by straight lines, and arrows showing the directions in which they act. 2. Resolve all the forces acting on the system in two directions at right angles to each other, and equate each result to zero. Take moments about any point and equate the result to zero. Choose the directions in which you resolve the forces, and the point about which you take moments, so as to make your equations as simple as pos- sible. 8. Count the unknown quantities in the statical equa- tions, and then add as many more equations deduced from the geometry of the figure as will make the whole number of equations equal to the whole number of unknown quantities involved. 4. The general method of solving statical problems here described may be frequently abbreviated by particular artifices, some of which may be learned from a careful study of the following problems. If only three forces act on a body, bear in mind that their direction musu always pass through the same point. This alone will often afford sufficient data for the solution of the problem. 132 APPENDIX. %i n EXAMPLES WORKED OUT. 1. A uniform beam rests upon two perfectly smooth inclined planes ; to find its position and its pressure upon the two planes. Let DM, DN be the inclined planes, a, ^, their respective inclinations; ^5 the beam, G its middle point, FT its weight, 2a its length, jR, R the reactions of the two planes, 9 the an- gle which AB makes with the horizon. The beam is kept at rest by three forces, R, R' and W; resolving these horizontally and vertically, and taking the moments about A , we obtain, R sina— i2 sin /3 R QOS(X-\'R cos/3==Pr and R' 2a cos ABE= Wa cos BAE or R' cos(/3-d)=^ Jf'cosQ. Eliminating R from the first and second, we get ij/ W sin a it =^ 8in(a + p) ^^ W sin /3 sin(rt + ^) Substituting the value of R in the third equation, and simplifying, we get 2 sin a sin fS ' By symmetry APPENDIX. 133 r smooth ure upon and W; aking the et ion, and We might find R and R ' more easily by the triangle of forces. Produce PA, PB, till they meet the horizontal line through D, then it is easily seen that the angle APQ—ix^ and BPG=fi. The point P is kept at rest by R^ R' and W, and by the triangle of forces, each is proportional to the sine of the angle between the other two ; hence R _ R' _ ^V sin fi sin a sin(a+/i) from which we obtain R and R' as before. To find we have, \ng\Q PAG-^^O—GAD =90— (a +0) PG "~ _sin(90— a+(/) AG~~ sin r a fixed when V ther rise r , IS BUS- attached the cord V. 1. State the experimental law on which the science of Stoics is based. If a rod be pressed against a wall in the direction of its length, are you to infer from the above law that the actual pressure thus caused at any point in the wall, in the direction of the rod produced, is equal to that at the point of contact ot the rod and wall. Explain clearly the meaning of the law. 2. Two forces act along opposite sides of a quadri- lateral in a circle, towards the same parts, and are respec- tively proportional to these sides ; prove that the resultant will pass through the intersection of the diagonal. 3. Find the centre of gravity of a sytem of particles in one plane. Three weights are placed in the angles of a right-angled triangle, and are proportional to the squares of the oppo site sides ; required the distance of the centre of gravity from the right angle. 4. Find the ratio of P to TF in the single moveable pulley when the strings are not parallel. If a weight W be supported by a weight of P hanging over a fixed pulley, the strings being parallel, show that in whatever position they hang the position of the centre of gravity is the same. 6. State the laws of statical friction, and give a practi- cal method of determining the co-efficient of friction for two given substances. If the roughness of a plane, which is inclined to the horizon at a known angle, be such that a body will just rest on it, find the least force along the plane requisite to drag the body up. H4 J VVh: