IMAGE EVALUATION TEST TARGET (MT-3) J/, / .1^^^ /<^^..^' & C/j ^ 1.0 I.I 1.25 S b£ 12.0 M 2.2 U II! 1.6 V] <;/,.-/» Baker, B.A., Math. TJtor UniveraUycSS^X^o^^tiy CANADIAN €!Oj»YRiaHT EDITIONS. HAMBLIN Si^lTH'S STATICS, With Appendix by Thot. Kirkland, 3f.A., Seienct Master Normal School, Toronto. Price 00 cents. Hamblin Smith's Ilydrostatics 75 cents. *' There are few books in elemontary mathematical science that I can more confidently recommend. The arrangement of the book work' la admirable, its treatment clear, and the subject made as elementary ar it can be, consistent with a scientific treat- ment of ift. In such a study, the working of problems will excite the same pleasure and impart to the subject the same interest that the conducting of experiments does in chemistry. The ex- amples at the end of each chapter are well selected- simple at the begmning, progressive, and ending with problems of 8ufllcier.t difficulty. The appendix lends an additional interest to the Cana- dian ediUon. Alfred Bakkr, B.a., "Math. Tutor, Univ. College, Toronto." Bi & ^0\, mmmoml ^n-iw ELEMENTARY STATICS. CY TIIF. SAME AUTHOR, ELEMENTARY HYDROSTATICS, DESIGNED FOR CANDIDATES FOR FIRST AND SECOND CLASS CERTIFICATES, AND FOR THE INTERMEDIATE EXAMINATION. IN rREr.VRATION, Elementary Statics, Principally designed for the use of Candidates for First and Second Class Certificates, and for the Intermediate Examination. WITH NUMEROUS EXAMPLES AND EXERCISES, BY THOMAS KIRKLAND, M.A., Science Master, Normal School, loroma TOROIKTTO! ADAM MILLER & CO. 1^77. Eiuered according to the Act o» the Parliament of Canada, in the year one thousand eight hundred and seventy seven, by Adam Miller & Co., in the Office of the Minister of Agriculture. PREFACE. r I iHE following pages contain the substance of the -*- lectures on Statics, delivered to the students of the Second Division in the Normal School, Toronto, during the past five years. They are now published in compli- ance with resolutions passed at several Teachers* Asso- ciations, and at the request of many excellent teachers in High and Public SchooNi. Throughout the book, I have constantly kept in view the requirements of Candidates for Second Class Certifi- cates and for the Intermediate Examination, who have not, in general, obtained the knowledge of mathematics, which most elementary books on Statics presuppose. In the perusal of the following pages, whi h will be found to contain all that is usually comprised in Ele- mentary Treatises on Statics, the only thing required of the student is a competent knowledge of the First Book of Euclid and simple equations in Algebra. Each chapter is divided into sections with the same fundamental idea running through the section. The student will not, therefore, be perplexed by trying to learn too much at once, but will be obliged to fix his attention on one subject at a time. PREFACE. He will thus the more easily master the difficulties it pre- sents. Each section contains one or more examples fully worked out ; and amongst the answers, at the end of the book, very full hints are given for the solution o all the more difficult problems. These features, it is hoped, may be of service to those who study the book without the aid of a teacher. To each section is ap- pended a collection of questions and problems designed to test the student's knowledge of the subject-matter of the section, to awaken activity of thought, and to exercise his invention in the solution of problems by the applica- tion of the principles contained in the section. Great pains have been taken in constructing and selecting the questions for the different exercises. The object has been not to select intricate and puzzling questions, bu such as will serve to determine from the form of solution, whether the student has grasped the fundamental princi- ples of the particular subject, and is capable of applying them. As a guide to beginners the more important pro- positions have been printed in black-letter type, the less important in italics. Throughout the work I have endeavored to explain as clearly as I could the leading ideas of ElemCiitary Sta- tics and to get rid of all difficulties that are not inherent in the subject itself But it may be well to remind the student that after all that is possible has been done in the way of exposition and illustration, the subject will still present difficulties to beginners — difficulties which can only be overcome by the labor of close thinking. PREFACE, y\\ Students unfamiliar with geometrical deductions are recommended, at the first reading, to omit the following: Exercise III., Chapter II., and Sections II. and III., Chapter III. ; Sections I. (except the Introduction), III., and Exercise IV., Chapter VI. ; Section III., Chapter VII. After reading the chapter on the Mechanical Powers, the student will be better able to master the omitted sec- tions. By this division into sections, it is hoped that the book has been adapted to the requirements of beginners as well as to the wants of advanced students, I have to tender my thanks to several friends for sug- gestions and assistance which have been of the greatest service to me, and particularly to Professor Young for suggesting several important improvements in the work, and for the excellent collection of Examination Papers in Chapter XII., which add much to the value of the book. Any remarks on the work, and especially indications of errors, omissions, or difficulties will be thankfully received. It will give me much pleasure if these pages shall in any way contribute to advance the taste for a science that is at once useful and interesting. Thomas Kirkland, Normal School, April 1877. CONTENTS. CHAPTER I. Definitions and Preliminary Notions. "'"", CHAPTER II. Parallelogram of Forces. Sect. II.— Forces acting at angles of fn° ,«" /<> ,* V *.' ^ supplements. ^ co , 30 , 45°, and their CHAPTER III. The Triangle and Polygon of Forces. Sect. I.— Proof of Triangle cf Forcp<; ^nri t? t- SECT II.-Exte„.-o„ of the Pn^dp,rof .he'r T'": " 25 I CHAPTER IV. The Resolution of Forces '" wliilT^Slil-tin^a given force into two others '''force;7cuf '^Tn^^^^^ wh;n';;;'umbe;of '' 36 CHAPTER V. Parallel Forces. ^ CONTENTS. Lc CHAPTER VI. Moments of Forces, PAGE. Sect. I. — Principis of Moments, Equilibrium of a body capable of turning round a fixed point. Equilibrium of a body acted on by any number of forces in one plane. Hints for the solution of problems 48 Sect. II.-- The application of the Principle of Moments to th» Lever 57 Sect. III. — Equilibrium of a body acted on by any number of Forces 63 CHAPTER VII. The Centre of Gravity. Sect. I. — Centre of Gravity of a uniform straight rod, and of bodies lying in the same straight line 69 Sect. II. — Properties of the Centre of Gravity 74 Sect. III. — Centre of Gravity of plane areas, &c , 78 CHAPTER VIII. Mechanical Powers. Sect. I. — The Lever, Balances &c 89 Sect. IL— The Wheel and Axle 98 Sect. III.— The Pulley 100 Sect. IV. — The Inclined Plane 112 Sect. V.— The Wedge 1 19 Sect. VI. — The Screw 121 CHAPTER IX. Virtual Velocities as Applied to Machines. CHAPTER X. Friction. CHAPTER XI. Tne Application of Similar Triangles to the Solu- tion of Statical Problems , 140 CHAPTER XIL Examination Papers set to Candidates for First and second • Class Certificases, froni 1871 to 1876 inclusive 146 CONTENTS, APPENDIX. Proof of the Parallelogram of Forces. Results, Hints, Ac, for the Exercises. MHM ELEMENTARY STATICS. CHAPTER L Definitions and Preliminary Notions. 1 Mechanics. The science of mechanics treats of the effects oi force ; we must then begin by explaining what we mean by the term/^;r^. 2 Def. of Force. Any cause which moves or tends to move, abodjror which changes or tends to change its motion/is called force. 3 In order to conceive the existence of a force, we must conceive that there is something upon which it can act and which may be called matter. A limited portion of matter is called a body. When the body is so small that for the purpose of any discussion the relative posi- tion of its parts need not be considered, it is called a material particle, 4- Def. A material particle may, therefore, be de- fined as a portion of matter occupying an indefinitely small space. It is spoken of for shortness as z. particle, 5- Rigid Body. A rigid body is one in which the relative position -.f the particles cannot be altered by the action of any finite force. 6. Equilibrium. VVhen several forces act simultane- ously upon a body, their individual tendencies to pro- duce motion may so counteract each other that no motion will ensue ; these forces are then said to be in equilibrium. 7- Statics. Statics is the science that investigates the relations existing among forces in maintaining rest or preventing change of motion ; it is the Science of Equili- brium. * 8. Magnitude of a Force. The first thing to be done in order to bring force under mathematical treat- ment, is to ^d some means by which the magnitudes of ■napivM ELEMENTAR V STA TICS. different forces may be immerically represented. In order to do this we must fix, by common consent, upon some standard force which, producing a known effect, may be taken as a unit of force. Nothing can be more convenient for this purpose than a given weight. Take, for instance, a weight of iB). If a force, when acting vertically upwards, will just sustain this weight, the force may properly be described as a force of ifi). If it will just sustain two such weights fastened together, it may be described as a force of 2fi)s. ; and so on. And generally, if a force is just capable of sustaining a weight of P ft>s., it is called a force of P lbs., or more briefly a force P ; and a force just capable of sustaining a weight of Q ft)s. is called a force Q. Whenever, therefore, a force is denoted by a letter of the alphabet, as P,Q,R, &c., it will be understood that these letters represent nuiri- bers, and that these numbers are the number of pounds which the forces are just capable of sustaining. 9. Force measured in terms of weight for con- venience. The student will understand that it is simply for convenience, and not of necessity, tliat refer- ence is made to weight as a measure of force. Instead of taking as the unit of force that force which will just sustain a weight of ifi)., a force might be taken that would just bend a given spring, but this would not be so convenient a method. 10. Representation of Forces by Lines. The three elements specifying a force, or the three elements which must be linown, before a clear notion of the force under consideration can be formed, are ; I. Its point of application, 3. Its direction. 3. Its magnitude. Hence we may conveniently represent forces by straight lines. For we may draw a straight line from the point of application of the force in the direction of the force, and containing as many units of length as the given force contains units of weight An example will make this plainer : k DEFhVITIONS AND PRELIMINARY NO'IlONS. 3 Ex. Suppose a force of sibs., inclined at an angle of 30 to the horizon, to be acting at the centre of a hori- zontal rod. f Let BAG be the rod, A the centre ; draw at A a straight hne making an angle of 30^ with the horizon, lake a portion of this line AP containing five units of length that is, as many units of length as there are units of weight in the given force. Then we say that AP repre- sents the given force in every necessary particular. In magnitude, by the number of units in its length. In direction, by its inclination to the horizon viz., 30". In point of application, viz. at A the centre of the rod. It is evident that any force may in this manner be fully represented by a straiglit line ; and th.s mode of representing forces by lines has the following advan- tages : — 1. The three elements specifying a force are all pre- sented to the eye at once. 2. It simplifies the enunciation of many theorems in mechanics. 3. It enables us to infer mechanical propositions from corresponding propositions in geometry. II Distinction between the force AP and the force PA. It may be inferred from the preceding article that a force AP applied at A has not the s-.n-io effect as a force PA applied at the same point ; fui a 4 . ELEMENTARY STATICS, ^orce AP would tend to move the body at A in the line AP towards P, but a force PA would tend to move the body at A in an opposite direction. The two forces are equal in magnitude and act along the same straight line, but tend to produce motions in opposite directions. The " force AP " is generally used to indicate the force represented in magnitude and direction by the line AP, acting in a direction from A towards P. 12. Ways in which force may be exerted. A force may be exerted in various ways, but only the following three will come under our notice. 1. Pressures. We may push a body by the hand or by a rod \ a force so exerted is called a pressure. 2. Tensions. We may pull a body by means of a string or by means of a rod ; the force exerted by means of the string, or by means of the rod used like a string, is called a tension, 3- Attractions. The earth exerts a force tending to draw bodies towards its centre ; a magnet exerts a force on certain bodies brought under ks influence ; such forces are called attractions, 13. Gravity. All bodies if left to themselves would fall towards the centre of the earth, and the force which produces this tendency is called the force of gravity. This force is generally of different magnitudes for diflferent bodies, varying in some degree with their size and substance, but it is found to be constantly of the same magnitude for the same body at the same place on the earth's surface ; but acts with slightly different force at different parts of the earth's surface. Its direction is always perpendicular to the surface of still water, and is usually designated as the vertical line. A plane perpen- dicular to this line is said to be horizontal. If a body be prevented from falling by the inter- position of some object such as a hand or a table, the body exerts a pressure on the hand or table. Hence wc obtain the follovvinii definition. DEFINITIONS AND PRELIMINARY NOTIONS. 14. Weight. T/ie precise amount of pressure 7vhtch any body at rest exerts in the direction of the earth is called the weight of that body. 15. The Transmissibility of Force. If a weight be attached by a cord to a spring balance, and the weight of the cord be neglected, the effect will be the same at whatever point in the cord the weight is attached. The same would be the case if the cord were passed over a perfectly smooth peg before being attached to the balance. Similarly, a force may be applied to a body by means of a rigid rod, and if the rod be supported independently, the result will be the same. The general principle here illustrated, and which we derive from experiment and observation, may be stated as follows : — of Force. the effect whatever provided, be rigidly Principle of the Transmissibility When a force acts on a rigid body of the force will be unchanged at point of its direction it be applied, this point be a point of the body, or connected with the body. 16. Object of Statics. It is the principal object of Statics to investigate what must be the relations between a given set of forces, as regards their magnitude and direction, in order that they may be in equilibrium. To this end it is necessary to investigate under what circumstances and in what manner it is possible to replace a given set of forces by another of a simpler or more convenient kind without affecting the state of equilibrium. 17. Bodies in nature not perfectly rigid. In the following investigations we will suppose strings per- fectly flexible, surfaces perfectly smooth, and bodies perfectly rigid. Now, there is no such thing in nature as a rigid body. Some are more nearly rigid than others. Iron is more rigid than wood, and wood than water. It may at first sight appear useless to make in- vestigations upon an hypothesis which never corresponds to reality. The method of proceeding is, however, 6 ELEMENTA RY STA TICS. to base the calculations upon the hypothesis of perfect rigidity, &c., and then to make such an allowance in the result, for want of ^his flexibility, smoothness, and rigidity, as experiment shows tc he necessary for the particular substance with which we happen to be dealing. In this way our final results, though not mathematically exact, are yet in most cases sufticiently near the truth for all practical purposes. QUESTIONS ON CHAPTER I. - I. Define Statics. 2. Define force. Mention any forces you know of, and show how your definition applies to them. Is a book lying on a table a force ? 3. How is force measured in statics ? Show how your measure of force applies to a force acting in any direc- tion whatever. 4. What would you understand by a force of 10, or a force of P in the absence of other information respecting such a force ? 5. Define the direction of a force. 6. Show clearly how a straight line may be made to represent a force in magnitude and direction. 7. Show the propriety of representing statical forces by straight lines, and mention any advantages which belong to this mode of representation. 8. If a force which can just sustain a weight of 5lbs. be represented by a straight line whose length is i ft. 3 in., what force will be represented by a straight line 2 ft. long? 9. When a weight hangs by a string, what is meant by the statement that the tension of the string is the same throughout ? Is the tension the same if the string has weight ? 10. What is meant by weight ? In what direction does it act? In how far is the weight of a given substance invariable? Does ilb. at the equato*- weigh the same as i lb. at the pole ? 11. State the " Principle of Transmissibility of Force,'* and give illustrations. CHAPTER II. to by THE PARALLELOGRAM OF FORCES. SECTION I. Experimental Proof ; and Examples of Forces acting at right ambles to each other. i8. Resultant and Components. Two or more forces acting on a particle in any direction may be represented by a single force which will be equivalent to the whole system. For suppose such a system did not keep the panicle at rest, it is evident that the particle would begin to move in a certain direction. A single force of the necessary magnitude and acting in a direction exactly opposite to that in which the particle would begin to move, would keep it at rest. This force would counter- balance the original set of forces, and a force equal and opposite to it would produce the same effect as the first set of forces, and is therefore called their resultant. 19. Def. of Resultant. — Hence the single force which represents the combined effects of two or more forces, is termed their resultant, 20. Def. of Components. Those forces which form a system equivalent to a single force are called its com- ponents 21. Composition of Forces. The method of find- ing the resu'tant of two or more forces is called the Composition of Forces. 22. To find the resultant of two or more forces acting on a particle along the same straight line. If a force of P fi)s. acts upon a particle in a certain direction, and another force of Q Rs. acts upon the 8 ELEMENTARY STA TICS, ' 'M: II particle in the same direction, it is evident that the two forces together will produce the same effect as a single force of (P4-Q)ft>s. acting in the direcnon of the two forces. Or, if K be the numerical value of the equiva- lent force in lbs., P + Q = R. If P act in one direction and Q in the opposite direction, and P be the greater force, we shall have P — Q = R. 23. The Parallelogram of Forces. If, however, the forces P and Q, instead of acting in the same line, act upon the particle in two different directions, then their combined effect is the same as that of a single force, whose magnitude and direction are found by the follow- ing proposition which is called the Parallelogram oj Forces : — If two forces acting on a particle be repre- sented in magnitude and direction by straight lines drawn from the particle, and a parallelo- gram be constructed having these straight lines as adjacent sides, then the resultant of the two forces is represented in magnitude and direction by that diagonal of the parallelogram which passes through the particle. 24. Illustration of the Parallelogram of Forces. The following illustration will help the student to under- stand this enunciation : — Let P and Q be two forces acting on the particle A in the directions AP, AQ, respectively. " ** 3a> Take AP, AQ, each as many units of length as there are units in the forces P and Q respec- tively. m i I PARALr.ELOGRAAf OF FORCES. f For example, if P =^- 5 lbs., and Q = 3 lt)s., and one inch be chosen as the unit of length, take AP = 5 inches, AQ = 3 inches, then the lines AP, AQ represent the forces P and Q in magnitude, direction, and point of application. Complete the parallelogram APRQ, and draw tlie diagonal AR. Then the proposition states that the diagonal AR will represent the resultant of P and Q in magnitude and direction, />., the resultant of P and Q contains as many pounds as AR contains inches. P and Q might, therefore, be removed, and the single force, represented by the line AR substituted in their stead, 25. Experimental proof of the ParaUelogram of Forces. Let A and B be two pulleys fixed in a vertical board. Let P and Q be two weights at- tached to fine cords passing over the pulleys and united at the point O. At this point let another weight be attached by a fine cord. Let the system be now abandoned to its own action and it will settle into a fixed posi- tion represented in the figure. The proof of the Parallelogram of Forces usually given in works on Statics is unsuitable for those who are not familiar with mathe- matical reasoning. It is deduced from the partieular case of two equal forces acting upon a particle ; but as the case of two equal forces must itself be deduced from experience, there appears to be little gained by this method of proof either in point of clearness or of certainty. We have therefore inserted an experimental proof of the proposition itself, giving the usual proof in the appendix. 10 ELEMENTAR V STA TICS. Forces equal to the weights P and Q now act at the point O, in the directions OA, OB, and the effect of these IS counteracted by the weight R, acting vertically down- wards. Therefore, the weight R is equal and opposite to the resultant of P and Q. If, now, the line Op be measured off in the direction OA containing as many inches as there are ounces in the weight P, and the line O^ be measured off in the direction OB containing as many inches as there are ounces in the weight Q, and the parallelogram Oprq be completed; then it will be found that the vertical line OR produced backwards passes through the point r, and that Or contains as many inches as there are ounces in the weight R. Hence, if two forces O/, Oq, act upon a particle at O, their combined effect is equal to the effect of a single force represented in magnitude and direction by the diagonal of the par- allelogram Oprq. Care must be taken in constructing the parallelogram of forces, that the components of the forces act /row the angle of the paiallelogram from which the diagonal is drawn. 26. Remarks on the Experimental proof of the Parallelogram of Forces. The principle of the parallelogram of forces may be regarded as the foundation of the science of Statics. By the means indicated in the last article it can be proved, so far as such a proposition can be proved experimentally. The more accurately the experiments are made, and the more they are varied in their circumstances, so much the more certain will they render the truth of the proposition. But the best evidence we have of its truth is of an indirect character. Direct experiment, as we have seen, affords a very strong presumption in favor of the truth of the proposition. We then assume it to be true, and construct on this basis the whole science of statics. Then we compare many of the results obtained with observation and experiment, and we find the agreement so close in every case that we may fairly infer that the principle 01. which the theory rests is true. f I 1 I! • I if I PARALLELOGRAM OF FORCES. n force; ^Ih ^^ h^?''"^ ^^* ^" ^"^ P^^ne. AH the orces with which we are concerned are supposed throughout this treatise to act in one plane, aTd'tS b evL "°^rn'"' ^^^'^ '^^ P^^»^ «f ^he p^per. Fo brevity we shall generally leave this to be 'understood m the investigations. Of course all the operations are supposed to be carried on in the same plane. EXAMPLES. I. Two forces of 4lbs. and 3lbs. respectively act on a particle at right angles to each other; find the magnitude and direction of the resultant Let AB, AC, represent the forces acting on the particle at the point A, AB = 3, and AC = 4- Complete the rectangle ABDC and join AD. Then AD repre- sents the resultant in magnitude and direction, AD2 = AC2 + = AC2 + s= 42 J. 5- 2. Two forces which are to each other as t to 4 j^cf on a particle at right angles to each other and produce a* resultant of islbs. ; find the forces. ^ c,-n^!^j^^r ^^ ^'^'^ ^^-^ represent the forces. Then nnTnf /'l'?K ^'\^' .3 to 4, one is three times some unit of which the other is four times respec^velV^'' ""^'' '^' ^"'"'^ ^'^^ *^^" ^' 3^ ^"^ ^ AD2 = AC2 + CD2 = AC2 + AB2 = (4^)2 + or 15 = 25x2 , A ' CD2 (Euc. I. 47) AB2 (Euc. ,. 34) 3^ AD = (3^)' ; II 13 ELEMENTAR Y S7A TICS. And therefore the forces are gibs, and i2lbs. Or thus : When AB = 3, and AC = 4, then AD But in this case AD = 15 or 5 x 3. Hence AB = 3 x 3 or 9. And AC = 4 X 3 or 1 2 as before. EXERCISE I. ==5- ^1 ■'J T. Define the resultant of any number of forces. ^ How would you find the resultant of two or more forces ^ which act on a particle and in the same straight line? 2. Two forces of yRs. and 8Ibs. act upon a particle in ^ the same direction ; what will be the magnitude and direction of their resultant ? 3. What is the resultant of two forces measured by 3RS. S and 4lbs., respectively, acting in opposite directions ? 4. Three foices of 4lbs., 2&s., and 30Z., respectively, ^ act uoon a particle in the same direction ; and in the opposite direction, forces of 8oz., and slbs., act. What ^other single force will keep the particle at rest ? 5. Can the resultant of two forces in any case exceed S the sum of the forces ? Under what circumstances is it least ? Can it be zero ? 6. Can the resultant in any case be equal to one of S the components ? If so, what are the conditions ? 7. State the Parallelogram of Forces. Explain the . meaning of the terms employed in your statement ; apply ^ \t to show that if four forces acting on a particle be re- presented by the sides of a rectangle taken in order, they will be in equilibrium. 8. Two forces of gibs, and i2!bs. act upon a particle S in directions at right angles to each other ; find the mag- nitude of their resultant. 9. If two forces of 5RS. and i2lbs. act at the same point <, at right angles to each other, what single force will pro- duce the same eflfect ? 10. Two forces, one of which is three times the other, 5 act along the adjacent sides of a square ; find the resul- tant i 1 PARALLELOGRAM OF FORCES. 13 It i ■--'- "-S'e having , If AC or CB = 5, then AB = 5 •j ' IfAB= I, ACorCB = -L; and if AB = c, « 5 EXAMPLEa I. To find the resultant of two forces of i2lbs and 81bs. respectively, acting on a particle at an anofeof 6o^ 16 ELEMENTARY ST^. TJCS. I Let AG = 8, AB = 12, and the angle BAG = 60° ; it is required to find the resultant AD. Produce AB, and from D draw EfE perpendicular to AB produced. Then BD = AG = 8, and -^ ^ DBE = CAB = 60° (Euc. 1.29), and therefore BDE = 30''. Since BD = 8, then BE = 4, and DE = 4y'3 (Art. 28) ; and AD2 = AE* + DE2 = (12 + 4)2 + (4i/3)« = 304 =16 X 19; .*. AD = 4\/i9. 3. Two forces of 10 lbs. and 42 lbs., act on a particle at an angle of 120° ; find their resultant. Let AB = 10, AG = 42, and the angle BAG = 120°. Since the angles BAG, AGD are together equal to 180°, and BAG = i2o^ then ACD = 60", (Euc. 1.29), and therefore CDE = 30". Also, since CD = AB = ic I>E=5v'3 (Art. 28.); and AD2= AE^ + ED* = (42 - Sf + (5 i/3)« = 1444; AD = 38. \ EXERGISE IL 1. Two forces of gRs. and 56ft>s. act upon a particle at an angle of 60'' : find their resultant. 2. The directions of two forces represented by 3^)5. and 5iba. respectively, include an angle of 60** : find the magnitude of their resultant. 3. Two forces, whor.e magnitudes are represented by 5 the numbers 2 and 3v'3, act upon a particle at an angle of3c'' S \Qr : find their resultant. PARALLELOGRAM OF FORCES, 17 \ '3 tide LUJ at )S. 4. Two forces of 45lbs. and 325lbs. act upon a particle ^ at an angle of 120'' : find their resultant 5. Two rafters making an angle of 60**, support a c, chandelier weighing 90&S. ; what will be the pressure along each rafter? 6. Two forces of 2lbs. and 3lbs. respectively, act at an S angle of 45® : find their resultant. 7. Forces of lylbs. and 24\/3lbs. act on a particle S at an angle of 135° : find the magnitude of their resultant. 8. Show that the resultant of the forces 7 and 14 act- c^ ing at an angle of 120", is the same as the resultant of the forces 7 and 7 acting at an angle of 60°. 9. Three posts are placed in the ground so as to form an equilateral triangle, and an elastic string is stretched ^ round them, the tension of which is 6tt>s. : find the pres- sure on each post. ^«^ ^^ A,*<.p^ «a, ^c^ =» ^ ^ii^ ^ 10. Three pegs A, B, C, are stucK in a wall in the f angles of an equilateral triangle, A being the highest and BC being horizontal ; a string, the length of which is equal to four times a side of the triangle, is hung over them, and its two ends attached to a weight W : find the pressure on each peg. SECTION III. Questions requiring for their solution a knowledge of easy dediutions from the propositions of Etulid^ Book i, EXERCISE III. i 1. The resultant is always nearer to the greater force. 2. The greater the angle between two forces the less ^ is their resultant. 3. If the angle between two equal forces acting on a i particle be 120**, what is their resultant ? 4. A string passing round a smooth peg is pulled at each end by a force equal to the strain upon the peg ; tnd the angle between the two parts of the string. 5 5. Show that the resultant of two forces, which act on a |8 ELEMENTARY STATICS, particle in directions not in the same straight line, must be less than the sum of the two forces, and greater than tbeir difference. 6. Two forces act on a particle at right angles to each other, and the resultant is double of the less ; show that ^ the angle which the resultant makes with the less is double the angle which it makes with the greater. 7. Let ABC be a triangle, and D the middle point of BC ; if the tlwee forces represented in magnitude and ^ direction by AB, AC, DA, act upon the point A, find the magnitude and direction of the resultant. 8 ABCD is a square, a force of ilb. acts along the side AB from A to B : a force of iBb acts along the side AD from A to D ; and a force of 2lbs. acts along the 5 side CB from C to B ; determine the magnitude and position of the resultant of the three forces. 9 AB and AC are adjacent sides of a parallelogram, and AD a diagonal ; AB is bisected in E : show that the resultant of the forces represented by AD and AC is '*• double the resultant of the forces represented by AE and AC. 10 Two strings, each bearing a weight of 6tt>s. pass over two pulleys A and B, lying in the same horizontal c line, and 20 inches apart ; the ends of the cords are joined at C to a third weight : what must that weight be that C may be exactly 10 inches below the line AB ? 11 If two forces acting at a point be represented by the two diagonals of a parallelogram, their resultant will ^ be represented by a line equal to twice one of the sides of the parallelogram. 12. If two forces be represented by the lines joining the bisections of two^ sides of a triangle with the oppo- ^ site vertices, and both forces act either towards or from ^ these vertices, then will the line joining the bisection of the third side and the opposite vertex represent the resultant. 13 ABC is an isosceles triangle, A and B the equal Angles, CD a perpendicular from the vertex on the base ; every line that is parallel to this ling, is said to be in the direction of the force. \ tlVMUlT Ht II ■■Qlllll TRIANGLE AND POLYGON OF FORCES. ai EXERCISE I. J I. Enunciate and prove the triangle of forces. Can a particle be kept at rest by three forces whose ^ magnitudes are P, Q, and P + Q respectively ? 2. Show that the following statement cannot always f be true, " If three forces acting on a body are parallel "^ to the sides of a triangle they will keep it at rest." ^ State the correct form of the theorem which thii *-* resembles. 3. Is it possible for three forces represented by the J mimbers i, 4, 7, to a keep a particle at rest? 4. Show that when three forces are in equilibrium, no ^ one of them is greater than the sum of the other two. 5. Three forces whose magnitudes are 6, 8, and lolbs., respectively, act upon. a particle and keep it at rest; 5 prove that the directions of two of the forces are at right angles to each other. 6. How can three sides of a triangle, not passing 5 through a point, represent three forces which act at a point ? 7. If a particle be kept at rest by three forces of 55>s., Q 61bs., and 7lbs., respectively, draw lines that will cor- rectly represent their directions. 8. Show how to keep a particle at rest by means of «: three forces, each equal to Pibs. 9. Two forces whose magnitudes are ^3 x P, and P, respectively, act at a point in directions at right angles ^ to each other ; find the magnitude and direction of the force which will balance them. TO. If two forces, acting at right angles to each other, t, have a resultant, which is double the smaller of the two forces, find its direction. II. If two forces be inclined to each other at an angle ^ of 135*^, find the ratio between them, when the resultant "^ is equal to the smaller force. in ELEMENTARY ST A TICS. ) \ » SECTION II. Extension of tJic Principle of the Triangle of For ses^ and its application to find the tension of strings. Hx. 3. A weight W is sustained by two cords of given lengths CA, CB, fastened to two points A and B, lying in the same horizontal line ; to find the tensions in the cords, when they are at right angles. Let AC = rJf BC = b, then AB = V^(a»+^«)(Euc. % ^ -7? 1.47)- Let P, Q, be the tensions in BC and AC respectively. The point C is kept at rest by three forces, VV act- ing vertically downwards, and P, Q, the tensions of the strings. In the vertical line through C take CD, equal to AB, and through D draw 1)K, parallel to CB, meeting AC produced in E. Then in the triangles CDE, ABC, we have the side CD in the one equal to the side AB in the other, the right aiigla CED, equal to the right angle ACB, and the angle DCE, equal to the angle ABC, each being the complement of the angle BCG, therefore, the triangle CDE is equal in every respect to the triangle ABC (Euc. 1.26), and its sides are respectively parallel to the directions of the three forces W, P, Q, and therefore proportional to their magnitudes. Hence, by the Triangle of Forces, I- W DE CD AC a Similarly, AB "" V{a*+d*)' •.P = W(2 V{a^+b*) Q = \\b 1^ ■' V{a^+b^) TRIANGLE AND POLYGON OF FORCES, n or 32. Extension of the Principle of the Triangle of Forces From the preceding article we see that , P jA£ _Q _ BC^ _P AC W •" AB' W "~ AB' *"*^ Q " ^C* P side perpendicular to direction of P W side perpendicular to direction of W and similarly for the others. The above is only a particular case of a more general proposition. If the sides of a triangle be parallel and proportional to three forces which keep a particle at rest, it is evident that the triangle may be turned so that its sides shall be at right angles to their former position and still remain proportional to the forces. We may, there- fore extend the enunciation of the triangle of forces given in art. 29, as follows : — If <"hree forces acting at a point be in equili- briur -, and if any triangle be drawn, the sides of which are respectively parallel or perpendicu- lar to their directions, the forces will be to one another as the sides of the triangle ; and con- versely, if the three forces are to one another as the sides of the triangle, they will be in equili- brium. Ex. 4. If a string ACDB be 21 inches long ; C and D two points in it, such that AC = 6, CD = 7 ; and if the extremities A and B be fastened to two points in the same horizontal line at a distance of 14 inches from each other ; what must be the ratio of two weights, which, hung at C and D, will keep CD horizontal ? A E Let ACDB ; AC CD: DB and AB If AE 21, 6, 7. 8, 14, 24 ELEMENTAR Y STA TICS, then AC* — AE» == CE« = DF» = DB« — BF«i or6»— «» = 8«— \ 14- (a; + 7) !►» ... a; = 1} , and BF = 7 — a; =sJ. • Let T be the tension of the string CD ; the point C is kept at rest by three forces P, T, and the tension of AC, and these three forces are parallel to the sides of the triangle ACE, and are therefore proportional to them ; 2_ __ CE T " AE* T;^ _ BF _ BF Q ~ DF "~ CE* . . P BF __ 5i^ II Multiplying. Q- = Xe "" iT " T Hence, Similarly, exercise II. 1. A weight of loolbs. is sustained by two cords whose lengths are 21 and 28 inches respectively, fastened to two points lying in the same horizohtal line ; required the tensions in the cords when they are at right angles to each other. 2. A weight of lofts. is suspended by two strings attached at their upper ends to two points in the same horizontal line; their lengths being 39 and 52 inches respectively, and the angle between them being 90°, find the tensions in each. 3. A weight of 24tt>s. is suspended by two flexible strings, one of which is horizontal and the other is in- clined at an angle of 30° to the vertical direction ; what is the tension in each string ? 4. The lower end B of a rigid rod without weight, loft. long is hinged to an upright post, and its other end A is fastened by a string 8ft. long to a point C vertically above B, so that ACB is a right angle. If a weight of 2ocwt. be suspended from A, find the tensic i of the string. 5. A cord having equal weights attached to its extre- mities passes over pulleys A and B in the same hori- i TRrANGLE AND POL YGO.V OF FORCES. as zontal line 70 inches apart> and through a smooth ring C, from which a weight of ioott)s. is attached ; what must be the magnitude of th'^ equal weights, that C may rest exactly 1 2 inches below the line AB ? 6. A thread 12 feet long is fastened at points A and B in the same horizontal line", 8ft. apart. At C and D points 4ft. and 5ft. respectively, from A and B, weights are attached ; what must be the ratio of the weights that CD may be horizontal? 7. A and B are points 14 inches apart in a vertical wall, to which are attached the extremeties of a cord ACDB. AtC a weight of iilbs. is attached, and at D another weight W. AC = 6 inches, CD = 7 inches, and DB = 8 inches ; what must be the weight of W in order that CD may be horizontal ? SECTION III. The reaction of hinges and smooth surfaces ; and three forces keeping a body at rest, 33. Reaction of S irfaces and Hinges. It nearly always happens that amongst the forces which keep a body at rest is the reaction of one or more surfaces. Suppose a weight of loolbs. to rest on a sinooth table ; the weight must be supported by the table which must, therefore, exert iipwards a force of looD&s. in a direction opposite to the direction of the weight. If we consider the case particularly, we shall see that this is an instance of what may be called a distributive force ; for the under surface of the weight will be in contact with the table at many points, and at each po'nt there will be a reaction. But it will be shown hereafter that in case of a force distri- buted over a surface, it is possible to find a single point and a single line such, that a certain force acting at that point in that line would produce the same eflfect as is really produced. This resultant reaction is called the reaction of the surface. The reaction of a smooth surface on a body in contact with it, is always exerted in a direc- 26 ELEMENTARY STATICS. tion perpendicular to the surface. If two rods be hinged together the reaction at the hinge is generally unknown both as regards magnitude and direction ; but if a body is kept at rest by the reaction of a hinge and two other forces which are not parallel, the reaction of the hinge always passes through the* intersection of the lines of action of the other two forces. 34. Three forces acting on a rigid body. When- er'er three forces which are not parallel act on a body and keep it at rest, their directions pass through the same point. Two, at least, of the forces meet in one point, and have a single resultant ; this resultant must balance the third force. But two forces which balance must act in the .same straij^ht line and in opposite directions ; therefore the third force must pass through the intersection of the other two forces. Ex. 5. A beam AB has one end attached to a hinge A, and the other end attached to a cord BC one end of which is tied to a peg. The weight of the beam is 5olbs. and may be supposed to act at its middle point. The beam and cord make angles of 60*^ on opposite sides of the vertical. Find the tension of the cord. i Let AB be the beam, F its middle point, and BG the vertical Lne. Let T be the tension of the cord BC. Through F draw a vertical line to represent the direc- tion of the weight, meeting CB produced, in D. The reaction of the hinge must pass through the point D, (art. 34). 4 TRIAiVGLE AND POLYGON OF FORCES. 27 Join AD, and through F draw EF parallel to BD. The angle BFD = angle FBG = 60^, (Euc. 1.29). And FDB = CBG, = 60* II .*. triangle BFD is equilateral and .-. DF = BF == AF, .-. the angle FAD = angle FDA = \ angle BFD =30**; .-. angle ADB = 90°. Let DF = 2, then EF = 1, and ED = ^3, (art. 28.) The three sides of the triangle DEF are parallel to the three forces which keep the beam AB at rest, and are therefore proportional to them ; hence 50 ■* FD "~ *' .'. T = 25!bs. Similarly the reaction of the hinge is found to be aSv^3. I k EXERCISE III. 1. A sphere weighing 200*63. rests between two planes inclined to the horizon at angles of 30^ and 60*^ respectively ; find the pressures on the planes. 2. A carriage Wheel whose weight is VV and radius r, rests upon a level road ; show that the force F necessary to draw the wheel over an obstacle of height h is F = W V^(2 rh—h^) r—h 3. A rigid rod, the weight of which is f olbs., acting at its middle point, moves at one end about a hinge, and is supported at the other end by a piece of string attached to a point vertically over the hinge, and at a distance from it equal to the length of the rod ; find the tension in the string when the rod rests in a horizontal position. 4. A rod AB without weight, can turn freely about a fixed point or hinge at one end B ; it is held in a hori- zontal position by a force of 5oIbs. which acts vertically downwards through its middle point, and by a force P which acts at the end A, in such a manner that the angle BAP equals 3o^ ; find P. s8 ELEMENTARY STA TICS. \'\ i I' |i II 5. A rod AB whose weight is W acting at its centre, and which can turn freely about a hinge at A, rests with its end against a smooth vertical wall, making an angle of 45° with it ; find the direction and magnitude of the reaction at A. 6. A uniform beam AB, whose weight is W, and may be supposed to act at its middle point, rests with one end A against a smooth vertical wall ; the other end B is supported by a string fastened to a point C in the wall. If the length of the beam be 3ft., and the length of the string 5ft. ; find CA, and the tension of the strii g. 35. The Polygon of Forces. If any number of forces acting on a particle be represented in magnitude and direction by the sides of a polygon^ taken in order ^ they will be in equilibrium. In the polygon ABCDE, from the Parallelogram of Forces, we know that the forces AB, BC are equivalent to a force represented by AC, that the forces AC, CD are equivalent to AD, and that AD, DE are equivalent to AE. Therefore the forces represented by AB, BC, CD, DE, EA, are equivalent to AE, EA and will therefore balance each other. Hence the proposition is true. The following converse of this proportion is also true, viz., if any number of forces acting on a particle be in equi- librium^ they can be represented in magnitude and direction by the sides of a polygon taken in order. 36. Resultant of any number of Forces acting at a point. If any number of forces be in equilibrium, a force equal and opposite to any one will be the result- ant of the remaining forces. Hence, any side of a poly- gon taken in reverse order, will represent the magnitude and direction of the resultant of any number of forces acting upon a point, when these forces are represented in magnitude and direction by the remaining sides of the polygon taken in order. t i i _ -i JJKILtlMISL.™ I HI, BS J^SHPHHR^^Pwiji! TRIANGLE AND POLYGON OF FORCES, 29 Thus if AB, BC, CD, and DE in the last figure repre- sent in magnitude and direction, four forces acting upon a particle, the remaining side AE (not EA) will represent the magnitude and direction of the resultant. 37. Remarks on the Polygon of Forces. The remarks of art. 31 apply equally to the Polygon of Forces. If the forces were actually acting along the sides of the Polygon and represented by them in magnitude, they would clearly have a turning tendency, and could not produce equilibrium. ■i. ! CHAPTER IV. THE RESOLUTION OF FORCES. SECTION I. Method of resolving a given force into two others which act in given directions. 38. Direct and Inverse problems. Direct pro- blems are those in which the resultant of forces is to be found ; inverse problems are those in which the com- ponents of a force are to be found. The former class is fixed and determinate ; the latter is quite indefinite without limitations to be stated for each problem. A syst-em of forces can produce only one effect ; but an in- finite number of systems can be obtained which will pro- duce the same effect as one force. The problem, therefore, of finding components must be, in some way or other, limited. This may be done by giving the lines along which the components are to act. Finding components is called Resolution of Forces. 39. A force R acts on a particle A, it is required to resolve it into two others P and Q, which shall act in given directions. Let AR be the direction of the force R. Take aD representing R in magnitude. Through A draw AP, RESOLUTION' OF FORCES. 3» AQ parallel to the given directions of P and Q re- spectively. Draw DB, DC parallel to AQ, AP respec- tively. Then AB, AC will evidently represent the required forces in magnitude and direction. The following is a form of this proposition which is constantly occurring in statical problems. 40. To find the effect of a Force in a given Direction. Let ABP, in the preceeding figure, be drawn from the point of application parallel to the required direc- ';ion. Then we want to find what tendency the force R has to j)ull the particle along the line ABP. From D draw DB perpendicular to ABP, and com- plete the paralellogram x\BDC. Then the forces represented by AB, AC, are equivalent to the force R. But it is found by experiment that no force can produce any effect in a direction at right angles to itself. The force AC, therefore, being perpendicular to AB can produce no effect in the direction of AB. Therefore, the force AB represents the whole effect of R in a direction parallel to ABP. In all future investigations, if AB represents a lorce in magnitude and direction, and it is required to find the effect of that force along the line AM, draw BN perpen- dicular to AM, and AN will represent the portion of the force required. 41. If the hyppthenuse of a right-angled triangle represents a force, the sides, taken in order represent its components. In any right- angled triangle ABN, if the liypothenuse AB, represents 3a ELEMENTARY STA TICS. a forde in magnitude and direction, the sides AN, NB will represent its components. These components are called the resolved parts of AB in their respective directions. EXAMPLES. 1. A man pulls a weight along a road by means of a rope, with a force of 2ott)s. The rope makes an angle of 30° with the road ; find the force he would need to apply parallel to the road to pull the weight. In the fig. Art. 40, let AB represent the force of 2ott)s. and let AM represent the direction of the road. From B draw BN perpendicular to AM. Then AN represents the required force. Since the angle BAN is 30", and ANB is 90°, ABN will be 60° ; and since AB represents a force of 2oR)s. AN will represent a force of 10 -4/3 D)s. (art. 28), the force required. 2. Three forces act at a point, and include angles of 90° and 45°, respectively. The first two forces are each equal to 2P, and the resultant of them ail is P-^/io; find the third force. Let Q be third force. ( Resolve Q vertically and hori- zontally, and we have for each com- p o n e n t Then the sum of the vertical forces = 2P + — ; 1/2 and algebraic sum of horizontal forces=2P . i/a !ip*iipmippppiipp"m RESOLUTION OF FORCES. 33 And the sum of the squares of these forces = square of resultant, that is ;) = (P ^10)' ; (2P + ^) + (2P .*. Q« = 2P», or Q = P ^2, the third force. 3. Show how it is possible for a sailing vessel to make way m a direction at right angles to that of the windL For simplicity let the sail be one of those attached to the yards of a ship, so that it extends on both sides of O. It is evident that the sail must not be placed along the line AB, for then the only effect of the wind would be to blow the vesse' sideways; nor could the sail be placed with its edge to the wind , that is, along the line OW, for then the wind would merely glide along the sail without producing any effect. Let, then, the sail be placed between the two positions, as in the direction PQ. Let the line WO represent in magnitude and direction the force of the wind on the sail. Through O draw OR at right angles to PQ, and from W let fall the perpen- diculars WX, WR, on PQ, OR respectively. Then the force WO can be resolved into the forces XO and RO. The force XO, in the direction of the plane of the sail, produces no effect in advancing the vessel, and may therefore be left out of consideration. The other com- ponent RO is perpendicular to the sail, but not in the 34 ELEMENTARY STATION. -\ \ direction in which the ship is sailing. Resolve RO into LO and MO. The effect of LO is to propel the vessel in a direction perpendicular to that in whi( h she is sailing. This force is partly counteracted by the keel and the form in which the vessel is built, but more espe- cially by the rudder which turns the head of the vessel towards the wind, and makes her sail sufficiently to the windward to counteract the effect of LO in driving her leeward. There remains MO which acts directly to push the vessel in the required direction. Hence it is seen how the wind, aided by the resistance of the water, is able to make the vessel move in a direction perpendicu- lar to that in which the wind blows. EXERCISE I. 1. Explain the meaning of the words Composition and Resolution of Forces, and show how forces may be com- pounded and resolved. A particle is acted on by a force whose magnitude is unknown, but whose direction makes an angle of 60° with the horizon. The horizontal com- ponent of the force is known to be 1.35 ; find the total force and also its vertical component. 2. Show how to resolve a given force into two com- ponents, one of which has a given magnitude and acts parallel to a given straight line. As a special case, resolve a force of magnitude 12 acting horizontally from left to right, into two components one of which is a force of magnitude 25 acting vertically upwards. 3. Show why the traces of a horse ought, in general, to be parallel to the road along which he is pulling. 4. Show by a diagram the forces which keep a kite in equilibrium. 5. Show how it is possible for a sailing vessel to make way in a direction making half a right angle with that of the wind. Why cannot a round tub be steered at as great an angle to the direction of the wind as a long-boat ? 6. When a horse is employed to tow a barge along a canal, the tow-rope is usually of considerable length ; RESOLUTION OF FORCES, 35 give a definite reason for using a long rope instead of a short one. Show whether the same considerations hold good in relation to the length of the rope when a steam- tug is used instead of a horse. 7. A horse walking by the side of a canal is to draw a boat along the canal by means of a horizontal rope attached to the boat near the bow. Point out the posi- tion in which the rudder must be placed in order that the boat may move parallel to the bank ; and show by a diagram all the forces acting upon the boat when in mo- tion. 8. Find the horizontal and vertical pressures, when a force of loolbs. acts in a direction making an angle of 60° with the vertical line. 9. It is required to substitute for a given vertical force, two forces, one horizontal, the other inclined at an angle of 45*^ to the vertical ; determine the magnitude of these two forces. 10. Find the horizontal and vertical pressures, when a force of Solbs. acts in a direction making an angle of 30** with the vertical line. — 11. Find the resultant of any three forces, the least of which is lolbs., which are represented by, and act along OA, OB, OC, two sides and the diagonal of an oblong whose area is 60 square inches, and shorter side 5 inches. 12. Three forces of 20, 20, and 15^)5, respectively, act upon a particle ; the angle between the first and second is 120^, and the angle between the second and third is 30^ ; find the magnitude of the resultant. ^ 13. Three forces 99, 100, loilbs. respectively act upon a particle in directions making an angle of 120** with each other, successively \ find the magnitude of the resultant^ and the angle it makes with the force 100. 14. A man and a boy pull a heavy weight by ropes inclined to the horizon at angles of 60*^ and 30^ with forces of 8ot>s. and ioofi)s., respectively. The angle between the two vertical planes of the cords is 30** ; find the single horizontal force that would produce the same effect. i i 34 ELEMENTA ?y STA TICS. SECTION ir. Conditions of equilibrium when any number of forces act at a point. 42. To find the conditions of equilibrium when any num- ber of forces act at a pointy their directions being in one plane, c o A Ji E Let O be the particle on which the forces act. Through O draw any two lines AOB, and COE, at right angles to each other. Resolve all the forces along AOB, and COE, respect- ively. Let Xj = sum of resolved pts. of fcs in the direct ^1j, Xg = II 11 ti OA, Y^ = It 11 H OC, 1^8 = W H ft OE. Then in order that particle may be a rest, we must have separately, Xj = Xg, Hence the conditions necessary and sufficient for equilibrium are as follows* : — If any number of forces acting on a particle and maintaining equilibrium be resolved along any two lines at right angles to each other, the sums of the resolved parts in opposite directions along each of these lines, must separately be equal^ • By the words "necessary and sufficient for equlibrium " is meant that, on the one hand, if the forces are in equilibrium the above equations will be satisfied, and on the other hand, if the above equations are satisfied the forces will be in equilibrium. RESOLUTION OF FORCES. Since Xi — X2 = o, and ^ Y, — Y2 = o, then, if we consider forces acting in one direction positive and those acting in the opposite direction fu^ative^ the conditions of equilibrium may be stated as follows : — The aif^ehraic sums of the fonts resolved into two per^ pendicular directions shall separately vanish. EXAMPLES. I. A weight of lolbs. is supported by two strings each of which is 3 ft. long, the ends being attached to two points in a horizontal line 3 ft. apart; fuid the tension of each suing. A n B Let A and B ^e the points of support Let AC and BC the strings. Let W be the weight of the body. Draw CD perpendicular to AB. ABC is an equila- teral triangle, and therefore, = fo Ihs, the angle CAD = angle CBD = 60", and angle ACD = angle BCD = 30''. The tension in AC = tension in BC = t. Resolve each of the tensions along CD. To do this, suppose CB = 2, then CD = ^3 ; if, therefore, the force acting along CB were 2, its resolved part along CD would be -/3 (art. 28) ; but the force acting along CB is t, its resolved part along CD is therefore }ty/;^. Since there is equilibrium, the sum of the resolved pans must be equal to the weight, (art. 42) ; hence i8 ELEMEATAR V STA TICS. w Ys 2t-^ = > 2 10 lO .*. t = — lbs. V3 43. Resolution along one line often sufficient. The student will observe that in the solution of pro- blems in which the forces are in equilibrium it is frequently sufficient to resolve the forces along one line only. Any line whatever may be chosen and thtn the sums of the resolved parts, in opposite di^ctions, along this line must be equal to each other. The line must be chosen so as to make the resulting equation as simple as possible. In the preceding article we inferred from the symmetry of the figure that the tension of each string was the same, and therefore the vertical resolution was sufficient. Had the angles ACD, BCD been un- equal, it would have been necessary to have resolved iferticallyy and horizontally^ and from the two equations thus obtained, the two tensions could have been deter- mined. 2 A weight W is sustained on an inclined plane by a certain force ; the inclination of the force to the inclined plane is 30°, and the inclination of the plane to the hori::on is 30^^ : find the force and the pressure on the plane. Let ABC be the inclined plane. R = the reaction of the plane. P = the sustaining force. (( « RESOLUTlOhi Ol< FORCES. 39 Then, resolving the forces along and perpendicularly to the inclined plane, we have P~ = iW (i), and R + iP = W 1/3 (2). We have from (i), P — Substitute in (2), and R = vi The second figure shows tlie two groups into which by resokuion we have decomposed the three forces P,W R ; the equations (i) and (2) representing the con- diti us lor equ'libruiin ot the two new groups, each regarded as independent of the other. EXERCISE II. 1. Wliat power acVig p:;:.'lel to a smooth plane inclined to the horizon at an angle of 30°, will sustain a weight of 4tt)s. on t 'le plane ? 2. What lurce ac'ig horizontally, will sustain a weight ot 1 2ft)s. on a plane inclined to the horizon at an angle ot 60°? 3. A cord is attached to two fixed points A and B in the same horizontal line, and bears a ring, weighing ioR)s., at C, so thit ACB is a right angle; find the tension ot the cord. 4. Two strings at right angles to each other support a weight, and one string makes an angle ot 30"* with the vertical line. Compare the tensions of the strings. 5 A weight ot 24tbs. is suspended by two flexible stnngs, one ot which is horizontal and the other inclined at an angle 01 30° to tiie vertical direction ; find the tjnsion of each string. 6. Two pictures of equal weights are suspended symmetrically by cords passing over smooth pegs; the two portions of the cord m one case making an angle of 60°, and in the other an angle of 120'' with each other. Compare the tensions ot the strings. ' It; I ! .Hi 40 ELEMENTAR Y STA TICS. 7. A force of 40&S. acting on a particle between two forces of 2oll>s. and 2o-/3lbs. respectively, makes an angle of 60° with the former and 30° with the latter ; find the magnitude and direction of the force which will keep the particle at rest. 8. On AB an inclined plane whose base is AC and which has the angle BAG equal to § of a right angle, a heavy body is kept at rest by two equal forces, the one acting in the direction of AB, and the other in the direc- tion of AC and towards the plane. Prove that the reaction of the plane on the body is equal to the weight of the body. 9. A weight W is supported on an inclined plane, the inclination of which to the horizon is 60", by three forces each equal to P, one acting along the plane, ano- ther horizontally, and a third in a direction inclined to the plane at an angle equal to the inclination of the plane ; required the value of P, and the pressure on the plane. 10. Two strings fastened to pegs A and B in a vertical wall, of which A is the higher, are joined at the point C to a weight W. the strings are of such lengths that BC is horizontal, and the angle BCA is 135°; find the tensions in the strings. n. Two planes of equal altitude are inclined at angles of 60° and 45° to the horizon ; what weight resting on the latter will balance 2olbs. on the former, the weights being connected by means of a string passing over the common vertex ? ^i It [lL CHAPTER V. PARALLEL FORCES. 44. Definition of Parallel Forces. Parallel forces are those acting at difierent point? of a body and in directions parallel to one another. 45. To find the resultant of two paraUei Jorccs acting on a rigid body in l/ic same direction. T C T 1. The magnitude of the resultant. Let P and Q represent two parallel iorces acting on a rigid body, at points A and B and in the same direction. Join AB ; and apply at A and B two forces, T, T, acting in opposite directions in the line AB ; this will not affect the equilibrium 01 the system. Let the forces T and P at A have a resultant R, and the foi-ces T and Q at B have a resultant S. Let the forces R and S meet in C. They can now be here i i ^ 1 1 •i I 4* ELEMENTARY ST A TICS. resolved into their original components P, T, and Q, T. The forces T, T, acting at C will balance and may be removed. There will remain P + Q acting at C, in a direction CD parallel to AP or BQ and this is the resultant of P and Q acting at A and B repectively. Let R represent the resultant of P and Q, then R = P + Q. Hence, the resultant of two parallel forces P and Q^ acting in the same direction^ is equal to their sum and acts parallel to their diredions in a straight line ivhich cuti AB in D ; so that it may be supposed to act at D. 2. The position of the point D. The sides of the triangle AC I) are parallel to the three forces P, T, R, which keep the point A at rest, and are, therefore, pro- portional to them j hence by Art. 30 we have ^ _ CD T ~" AD Similarly Q CD Multiplying (i) and (2) we have Q ~ AD' that is, the point D divides AB into segements which are inversely as the forces P and Q respectively. Let AB = a, and AD = x, Z. - 5P p. AD = Q.BD; P^ = Q (a — a?) ; Since or that is therefore a; = „ ^ . P + Q 46. To find the resultant of two parallel forces acting on a rigid body and in contrary directions.* * Contrary signifies parallel and in dissimilar directions ; opposite signifies, contrary and in one line. PARALLEL FORCES. 43 Let P and Q be two parallel forces acting in contrary directions, and let Q be greater than P. By the same method as in the preceding article it may be shown that R = Q — P, and that it acts parallel to the directions of P and Q in a straight line which cuts AB produced at a point D, such that Q" ~ AD ^ that is, D divides AB produced through B into segments which are inversely as the forces P and Q respectively.* Let AB = «, and AD = x, Z. - ?P "Q " AD' P.AD = Q.BD; Vx = Q (^ — «) ; Q_a Since or that is Therefore X = Q-P' *When a line is cut at any point the intercepts oetween the point of section and its extremities are called its segments. When the point of section lies between the extremities of the line it is said to he cut internaUy ; but when it is not the line itself but its produc- tion that is cut, and therefore the point of section lies beyond one of its extremities, it is said to be cut cxtenuillv. '44 ELEMEi\TARy STATICS, \ I II 47« Resultant of two parallel forces. If we extend to parallel lorces the same method of indicating opposition of direction by difference of sign, which was used in the case of forces acting in the same line (art. 42), v/e can include the results of the two preceding articles in the following statement : — The resultant of two parallel forces is equal to their algebraic sum, and is in the parallel line which divides any line drawn across their lines of action into segments inversely as their magnitudes. 48. Resultant of any number of parallel forces. We may find the resultant of any number of ])arallel forces by repeated application of the process of the last two articles. First find the resultant of two of the forces ; and then find the resultant of this and the third force ; and so on. 49. Couples. In Article 46 it was shown that if P and Q are two parallel forces acting at A and B, and in contrary directions, then if Q is greater than P, their result- ant R will be a parallel force acting in the same direction as P and Q through a point D, given by the equation, AD = Q-P' Now, if we suppose P to be gradually increased, but a and Q to remak unaltered, the magnitude of R, (or Q — P), will continually diminish and AD will continually increase and when P becomes equal to Q, the point D is removed to an infinite distance, and R = o ; yet the forces are not in equilibrium, since they are not directly opposed. Hence two equal and contrary forces neither balance nor have a single resultant. It is clear that they have a ten- dency to turn the body to which they are applied. Such a pair of forces constitute what is called a couple. This case must be excluded from the statement in Art. 47. 50 Centre of Parallel Forces. The expressions tor X or AD in Articles 45 and 46 are independent of ihe angle which the direction of the lorces makes with the line PARALLEL FORCES, 45 joining their points of application. The position of the point D is not, therefore, altered by changing the direc- tion of the forces, if the points of application remain the same. Generally, if we conceive any system of Parallel Forces, and suppose that each force acts at a particular point, then if we suppose the lines along which the forces act to be turned around the points through any equal angles so that they still continue parallel, it will be found that there is a certain fixed point through which their re- sultant will always pass, whatever be the magnitude of the angles ; this fixed point in the line of action of the re- sultant is called the centre of that system of parallel forces. If the parallel forces are the weights of the parts of a heavy body, or of the members of a system of heavy bodies, the centre of these parallel forces is the centre ol gravity of the body or system of bodies. EXAMPLES. I. Two persons A and B, carry a weight of loofts. on a pole between them. The weight being placed two feet from A and three feet from B, find what portion of it they respectively support. Let C be the point of the rod at which the weight W is hung, and let R, R' be the forces which A and B respectively exert at the ends of the pole in a vertical di- rection ; by supposition, these with W acting downwards at C preserve equilibrium in the system ; therefore W is equal and opposite to the resultant of R and R' ; or m ^ H' IV' and W = R + R'; R ^ BC ^ 3 R' AC 2 Hence we obtain R = — W = 6olbs.; and R' = ~ W = 4olb8. S ■^■' •i!! 46 ELEMENTARY STATICS, 2. The resultant of two parallel forces acting in contrary directions is i2lbs., and acts at a distance of 5in. from the greater and 7 in. from the lesser force, find the forces. Let P = the greater force, K = Z'— Q = 12, P = 12 + Q, 5 P = 7 Q, 5(12 +Q) = 7Q; Q = 3olbs. ; P = 42lbs. then or And or therefore and EXERCISE I. 1. Two men of the same height, carry on their shoul- ders a pole 6ft. long, and a weight of i2ilbs. is skmg on it, 3oin. from one of the men ; what portion of the weight does each man support ? 2. Two men carry a weight of i52lbs. between them on a pole, resting on a shoulder of each ; the weight is three times as far from one as from the other ; find how much weight each supports, the weight of the pole being disregarded. 3. Two men are carrying a block of irdti, weighing I76lbs. suspended from a uniform pole 14ft. long; each man's shoulder is ift. 6in. from his end of the pole. At what point of the pole must the heavy weight be sus- pended, in order that one of the men may bear ^ of the weight borne by the other? 4. The resultant of two parallel forces actmg in con- trary directions is 6lbs., acting Sin. from the greater force, which is lolbs. ; find the distance between them. 5. A weight of i2lbs. is placed on a square board 5in. from one side and loin. from the opposite side, along the line that bisects its width. The board is suspended at its corners by four vertical strings ; find the tensions in each string. 6. A flat board i2in. square, whose weight is i2lbs., acting at its centre, is suspended in a horizontal position by strings attached to its four corners A, B, C, D, and a weight equal to the weight of the board is laid upon it at a point 3in. distant from the side AB, and4in. firom AD ; find the tensions in the four strings. PARALLEL FORCES. 47 7. A weigtit of i5olbs. rests on a triangular table at a point in the line joining one of the angular points with the middle of the opposite side, and at a distance frem the angular point equal to tvv'ce its distance from the side. The table is supported on three props at its angular points ; find the pressure on each prop. 8. A beam AB, loft. long, rests horizontally upon two vertical props, A and B, and another beam CD, 2ort in length, rests upon two vertical props, C and D ; and a third beam 30ft. in length, lies across the two former beams in such a way that the one extremity E is 3ft from the prop A, and the other extremity F, is 5 feet from the prop D ; a weight of 6olbs. is attached to the third beam at a distance of loft from F; find the pres- sure on A, B, C, D respectively. 9. If a weight rests in the middle of a square rough table, will the pressure on each leg be altered if one pair of legs are longer than the opposite pair ? 10. Two forces of lolbs. each, act on a body along parallel lines, and in contrary directions ; why should it be impossible to balance these forc(!S by any one force ? 11. The weight of a window-sash 3t't. wide is 5lbs., acting at the middle point of the sash ; each of the weights attached to the cord is 2lbs. ; if one of the cords be broken, find at what distance from the middle of the sash the hand must be placed to raise it with the least effort. CHAPTER VT. MOMENTS OF FORCES. w \ SECTION I. Principle of moments. Equilibrium of a body capable of turning round a fixed point Equilibrium of a body acted on by any number of forces in one plane. Hints for the solution of the problems. 51. Introduction. The moment of a force about a given point is its tendency to produce rotation about the point. Suppose a rod OD capable of turning about the fixed point O, to be acted on by a force P ; it is g^ shown by experiment that the tendency of the force to turn the rod about O depends on the magnitude of the force and on its distance from the point O. We might, for example, double this tendency either by [_ doubling the force, or by keeping the force^H the same and causing it to act at twice the distance from O. Hence the tendency of the force to turn the rod about O is measured by the product of the force into the perpendicular OD. When one point in a body is fixed, in order that the body may be at rest it is evident that the moment in one direction about that point must be equal to the moment in the opposite direction. But when a body is at rest under the acti#n of forces it is evident that we may imagine any point in it to be fixed, for the fixing of a point which is already at rest without introducing any new forces at the point would not ^ MOMENTS OF FORCES, 49 disturb the equilibrium. Hence, the tendency to turn in one direction about any point must be equal to that in the opposite direction, or if we consider the moments in one direction /(7j-///z/s. ? MOMENTS OF FORCES, <1 3. On a horizontal rod, 45in. long, whose extremities ^ are supported, where must a weight be placed so that the pressure on the supports may be as 5 to 4 ? 4. Two men of the same height bear a weight sus- pended from a pole, which rests on their shoulders. VVhere must the weight be placed so that one man may support twice as much as the other? 5. If the forces at the end of the arms of a horizontal ^ lever be Slbs. and ylbs. and the arms Sin. and Qin. respectively ; find at what point a force of ife. must be applied i>erpendicularly to the lever to keep it at rest. 6. A lever with a fulcrum at one end is 3ft. in length. A weight of 28lbs. is susi)ended fiom the other end. If the weight of the lever is 2Bbs. and acts at its middle point, at what distance from the fulcrum will an up- wards force of 5oft)s. preserve equilibrium ? 7. The length of a horizontal lever is 12ft. and the balancing weights at its ends are 3S)s. and 6&s. resi)ect- 5 ively ; if each weight be moved 2ft. from the end of the lever, find how far the fulcrum must be moved for equili- brium. 8. The whole length of each oar of a boat is loft., and from the hand to the rowlock the distance is 2ft. „ 6in. ; each of 8 men sitting in the boat pulls his oar with a force of 5olbs. Supposing the blades of the oar not to move through the water, find the resultant force pro- pelling the boat. 9. A straight rod, movable about one end, makes an angle of 30" with the vertical. A weight of yfes. hangs at ;; the other end, what force acting perpendicularly to the rod at its middle point will preserve equilibrium ? 10. A straight lever is inclined at an angle of 60^ to the horizon and a weight of 36oIbs. hung freely'at the distance of 2 in. from the fulcrum is supported by a 5 power acting at an angle of 60*^ with the lever, at the dis- tance of 2 ft. on the other side of the fulcrum ; find the power. 1 1. ACB is a bent lever ; the arms CA, CB, are c; straight and equal, and inclined to one another at an angle of 135**. When CA is horizontal a weight of P j«^. T 6a ELEMENTARY STATICS, at A just sustains a weight of W at B ; and when CB is 5 horizontal the weight VV at B requires a weight Q at A to balance it ; find the ratio of Q to P. 12. A bent lever has equal arms making an angle of 120**; find the ratio of the weights at the ends of the S arms when the lever is in equilibrium with one arm hori zontal. 13. ACB is a bent lever with its fulcnim at C, the ^ angle ACB is a right angle, the arms AC and BC are ^ 10ft. and 7ft. long, and AC is in a vertical position j a horizontal force of 21 lbs. acting at A, is balanced by a vertical force P, acting at B ; find the magnitude of P and the pressure on the fulcrum, and show by a diagram the line along which the latter acts. M MOMENTS OF FORCES. j, SECTION in.. Ejumrium of a tody acU.l on by a„y „un,bcr of Foras its lower end upon VsLon.hK""'''"" !"""'' ''^■'^'^ '^"^ dl/'poirt^ "'^ ''^'^'" "'' "'' '"^'^"■' ^"'"S "' "» '"''1- ^^' » ~' "'^ '"'^'^^ *''''^'' P'^^ei's sliding, respect^'ej' ^ """ '"■^''"'■'' °° "'^ P'^"'' ""<^ ''"P^ 23 = the length of the ladder. From A draw AG perpendicular to direction of R'. Then, taking moments about A, we have R'.AG = W. AK. (i). IMMMi ^^l^^i 1^ .' ■i ri ■ III III ELEMEXTAK Y ST A TICS, The angles ABG, ADK, cnch = 6o*i therefore AG = — ^3 ; and AK = — ^3. Substituting the above values of AG and AK in (i), we have R'aV3 = W. — \r3. . • . R' = — . a Next, resolve R' verticallv and horizontally, and we W W have — and — y/^ respectively. Since there is equilibrium, we must have, Art. 6l| W , F=~-v^3, 4 W and R + — = W; 4 . • . R = } W. We can obtain the same results by taking moments only. Produce the directions of R and R' till they meet in E, and produce R' backwards till it meets the direction of F in O. Then, ABE is an equilateral triangle, and therefore, AE = AB = 2a. Taking moments about E, we have F. AE = W. AK, or -F. 2 « = W. — 1/3 ; W . • . F = — V3, as before. 4 Again, since AEO i= 6o^ AOE = 30**, and since AE, the side opposite 30" = 2a.y the hypothenuse OE == 4^; and therefore, AO = 2a V^. MOMENTS OF FORCES, And KO = AO — AK 2 Taking moments about O, we have R. AO = W. KO, or R.2aV^ W -^ v'a ; 3W ♦ • . R = ^— , as before. 4 EXERCISE III. : A r"5f'>*ii: beam AB, 17 feet long, whose weight 18 lao Ids. acting at its centre, rests wilh one end against a smooth wall, and the other end on a smooth floor, this end being tied by a string 8 feet long, to a peg at the bottom of the wall ; find the tension of the string. 2. A beam AB rests with one end A against a smooth vertical wall, and the other end B on a smooth horizon- tal plane ; it is prevented from sliding by a cord tied to one end of the beam and to a peg at the bottom of the wall ; the length of the beam is 10 ft. 6 in., and the length of the string 9 feet. Suppose the weight of the beam to be ii2tt)s. and to act vertically through its middle point, find the forces acting on the beam. 3. Solve the first four questions under Exercise III, page 27, by the princi|)le of moments. 4. A ladder, the weight of which is QoTbs., acting at a point one-third of its length from the foot, is made to rest against a smooth vertical wall, and inclined to it at an angle of 30°, by a force applied horizontally to the foot ; find the force. 5. A uniform beam 6 feet in length, rests with one end against a smooth vertical wall, the other end resting on a smooth horizontal plane, and is prevented from sliding by a horizontal force applied to that end, equal A M ELEMENTARY ST A TICS, to the weight of the beam, and by a weight equal to two- thircs the weight of the beam, suspended from a certain point on the bea'm. Find the distance of this point from the lower end of the beam, the beam being inclined at an angle of 45° to the horizon. 6. A roof ACB consists of beams which form an isosceles triangle, of which the base AB is horizontal. Given the weight of each beam looJbs., and 45° the angle at which it is inclined to the horizc.., find the force necessary to counttjrbalance the horizontal thrust at A. 7. A ladder the weight of which may be regarded as a force acting at a point one-third the length from the foot, rests with one end agiinst a peg in a smooth hori- zontal plane, and the other end on a wall. The point of contact with the wall divides the ladder into parts which are as i : 4 ; having given that the ladder weighs i2oIbs., and makes an angle of 45° with the horizontal plane, find the pressure on the peg, and the reaction of the v/all. Ex. 5- AB is a string, whose length is 20 feet, and BC a pole, whose length is 13 feet, and v/eight igRjs. acting at its cen- tre; together *')cy support a we'.gl'.t . of i6ofi)s;findthe ^ ^ tension of the string, if AC -- \i ft. Let CD = X, and BD = ?y ; AB* == fAC + x)« + BC« = i« + BC8 = AC8 + 2 AC. av then Ati* = f^AU + X)* + y and BC* = ir* + 2/^ Hence AB« or 20* — i3* = II* + 22 a;; therefore a; = 5. And 132 = 52 + y2 . . •. y == 12. The I Gibs at F, the centre of BC, may be resolved MOMENTS OF FORCES, 67 into 5fts. at C and stt)s at B ; the latter may be aaded to the i6oIbs. Draw CE perpendicular to AB. then 20 X CE = 2 area triangle ABC, and II X 12 = 2 area of same triangle j hence 20 x CE =11 x 12; . • . CE = 6f Let T be the tension of the string AB. Taking moments about C, we have T X 6| = i65 X 5; . *. T = I25tt>S. EXERCISE IV. This exercise requires a knowledge of easy deductions from the propositions of Euclid Book I. 1. ABC is an isosceles triangle, and D any point in the base BC. If equal forces act along the sides AB, CA, prove that the sum of the moments about D is in- dependent of the position of D. 2. Three equal forces act in order along the sides of an equilateral triangle ; show that the sum of the moments about any point within the triangle is invariable. If the point be outside the triangle, show that the algebraic sum of the moments is invariable. 3. Show by the Princi;)le of Moments that if three forces act in consecutive directions round a triangle ABC, and be represented by its sides, they are not in equilibrium. Show also that if two forces act in conse- cutive directions along two sides of a triangle, and be re- presented by them, no force acting along the third side and represented by it, can produce equilibrium, 4. One end of a uniform beam is placed on the ground against a fixed obstacle, and to the other is attached a string, which runs in a horizontal direction to a fixed point vertically above the obstacle, and, passing freely over it sustains a weight VV at its extremity, the beam being thus held at rest at an inclination of 45° to the horizon : prove that, if the string were attached to the !^P 68 ELEMEN lARY STA TICS. centre instead of the end of the beam, and passed over the same fixed point, a weight at the end of the string, equal to WVz, would keep the beam at rest in the same position. 5. At what point of a tree must a rope of given length a be fixed, so that a man pullmg at the other end may exert the greatest force in upsetting it? I't .% CHAPTER VII. CENTRE OF GI WITY. SECTION I. Definition ; Centre of gravity of a wiiform straight rod, ana of bo(iiy.i lying in the same straight line. 65. Introduction. The attraction of the earth, w'rich causes a body to have weight, acts on every particle of the body ; if a stone, for example, bs broken into small fragments, the stmi of the weights of the particles will be equal to that of the whole body. If one of these particles be attached by a fine thread to a fixed point, the thread will talce the direction of the vertical through the point, and if several of them be suspended from points near together, the threads will be parallel. When, therefore, the particles are united so as to form the body, we may regard their weights as a system of parallel forces. Regarding the earth as a sphere, it is true that the vertical lines would all converge to its centre, and there- fore, strictly speaking, the direction of the forces which the earth exerts on the difi'erent particles composing a body are not parallel. But since the dimensions of any body we have to consider are ver;/ small compared with the radius of the earth, we may consider these directions to be appreciably parallel. For a similar reason we hiay suppose that the weight of any body is the same in r latever position it may be placed. The resultant of aiis system of parallel forces is the weight of the body. f '^' < m. I 70 ELEMENTAR Y STA TICS. It has been shown (Art 50), that the resultant of a sys- trin of parallel forces acting on a rigid body passes through a fixed point, the position of which is indepen- dent of the direction of the forces ; the point at which this resultant acts is called the centre of gravity of the body. We have, therefore, the following definition : — 66. Definition of Centre of Gravity, The centre of gravity of a body is the centre of par- allel forces due to the weights of the respective parts of the body. 67. If Centre of Gravity be fixed the body will rest in every position. If the resultant of the forces acting on a body be equal to the weight of the body, and act vertically upwards through the centre of gravity, it is evident from the d«-finition that the body will be at rest. This statement is sometimes taken as the definition of the Centre of Gravity, and expressed thus : — The Centre of Gravity oj a system of hea^y particles is a point such t/iat^ if it be supf-orted and the parts ri^idiy connected with it^ the system, will rest in any position, 68. Centre of Gravity of a pniform straip:ht rod. The centre of gravity of a unif< rm straight rod is at its middle point. For we may suppose the rod to be made up of an indefinitely large number of equal p.irticles. Take two of these which are equidistant from the middle point of the rod ; their centre of gravity is at the middle point. And since this is true for every such pair of particles, the centre of gravity of tiie whole rod is at the middle point of the rod. Hence, the weight of a uniform straight rod may always be supposed to be col- lected at its middle point cea: ^r^: of gra^^ity. EXAMPLF<^. 71 ^ I. On a uniform straight le ,er weighing sibs., and ^{\* in length, weights of i, 2, 3, 4lbs., respectively, are hung at distances i, 2, 3, 4ft., respectively, from one end, find the centre of gravity of the system. A B D E G Let AF be the lever, and let G be its centre *' ^^ravliy. The whole weight of the lever may be supposed^ to act at G. Let the given weights act at the points B, C, D, E, respectively. Let a; = the distance of the C.G. of the whole system from A. The resultant of all the forces which act on the lever IS the sum of the weights, together with the weight of the lever, and acts at the centre of gravity of the system. Take moments about A ; and since the moment'of the resultant about any point is equal to the sum of the mo- ments of the forces about the same point, we have 152;= 1x1+2x2 + 5x21 + 3x3 + 4x4 = 44; .'. X =r 2|ft. The C.G. is, therefore, 2ft. loin. from A. 2. A uniform beam AB, 20ft. long, is suspended from a nail by a string which is fastened to the beam at 9. distance of 2ft. from its centre ; a weight of 2oft)s. is attached to the other end to keep the beam horizontal. What is the weight of the beam ? A a w^ T O B Let G be the C.G. of the beam ; then W, the weight of the beam, may be supposed to act at G. 'pip I 7» eleAaEn tar y sta tics. !!i Ill Let T be the tension of the string, and O the point in the beam to which it is attached. The beam is kept at rest by three forces W, T, and aolbs. ; T is evidently equal to the resultant of the other two. Take moments about O ; and since the moments about any point in the direction of the resultan* are equal and opposite (Art. 56), we have W X OG = 20 X OB, or W X 2 = 20 X 8 ; .-. W = 8olbs. EXERCISE I. 1. Find the centre of gravity of i2lbs. and 2oIbs. res- pectively, the line joining their centres of gravity being 2ft. Sin. 2. Three weights of ilb., 2lbs. and 3lbs., are placed a foot apart, along a straight line ; find their centre of gravity. 3. Two weights of 61bs. and i2lbs. are suspended at the ends of a uniform horizontal rod, whose weight is iSIbs., and length 2ft. ; find the centre of gravity. 4. Weights of 2lbs., 4lbs., 6Ibs. and Bibs., are yjlacedso that their centres of gravity are in a straight line, and six inches apart ; find the distance of their common centre of gravity from that of the larger weight. 5. Three weights of 4lbs., 61bs. and 81bs., respectively, are placed at intervals of 9in. along a weightless rod ; find the distance of the centre of gravity from the middle of the rod. 6. A bar of uniform thickness and density, and 4 feet in length, has a weight of lolbs. attached to one end ; it balances about a point 9 inches from that end. What is the weight of the bar? 7. A bar of uniform thickness and density, and weigh- ing 5lbs., has a weight of lolbs. attached to one end, and a weight of i2lbs. is suspended from the other ; it balances about a point 4 inches from the middle. What is the length of the bar ? :n^' h .1 ':' CENTRE OF GRAVITY. 73 8. A uniform stick 6 ft. long lies on a table, with one end projecting beyond the edge of the table to the ex- tent of two feet ; the greatest weight that can be siis- S pended from the end of the projecting portion without destroying the equilibrium is ilb.; find the weight of the stick. 9. Four weights of 3lbs., slbs., 4lbs., and ylbs., respec- tively, are at equal intervr of 8 inches on a lever with- ^ out weight, two feet in length ; find where the fulcrum must be in order that they may balance. 10. A ladder 20 ft. long weighs 6olbs- ; its centre of gravity is 8 ft. from its thicker end ; it is carried by two men, one of whom supports the heavier end on his S shoulder ; where must the other stand that the weiglit may be equally divided? 11. A heavy tapering rod, having a weight of 2oIbs. attached to its smaller end, balances about a fulcrum placed at a distance of loft. from the end ; the weight V of the rod is 2ooK)s. ; find the point about which it will balance when the attached weight is removed. 12. A uniform bar ot iron loft. long, projects 6ft. over the edge of a wharf, there bemg a weight placed upon the other end ; and it is found that when this is ^ diminished to 3 cwt. the bar is just on the point of falling over ; find its weight. 13. A cylindrical vessel weighing 4lbs., and the inter- nal depth of which is 6 in. will just hold 25)S. of water. If the centre of gravity ot the vessel when empty is ^ 3-39in. from the top, determine the position of the centre of gravity of the vessel and its contents when full of water. 14. Prove that if the centre of gravity be found for c one position of a body, it will be the centre of gravity when the body is turned into any other position. 15. What hypotheses not realized in nature, are made >S in the definition ol the ceatr« of gravity ? mt mmm :l\\ '$ ■I i I ill III I! 74 ELEMENTARY S'jA TICS. SECTION 11. Properties of the Centre of Gravity. 69. Every body has a centre of gravity. For, if it be divided into an indefinite number of small particles, these particles are the points of application of a system of parallel forces equal to their weights ; the centre of gravity of two of these may be found by the method already given ; join this point with a third par- ticle and find the centre of gravity of the sum of the first two particles and the third, and so on to the last particle. The point thus found will be the centre of gravity. 70. No body can have more than one centre of gravity. For, if it be possible, let a body have two centres of gravity, and place it so that the line joining them shall be horizontal ; then since the resultant of the weights of the several particles of the body passes through each of these centres, its direction must be the right line joining them, that is to say, a horizontal line, which is absurd. Therefore the body cannot have two centres of gravity. 7 1. When a body is suspended from a point about which it can turn freely^ it ivill rest ivith its centre of gravity in the vertical line passing through the point of suspension. For the body is acted on by Iwo forces, viz., its own weight in a vertical direction through the centre of grav- ity, and the force arising from the fixed point. The body will not rest unless these two forces are equal and opposite. Therefore, the centre of gravity must be in the vertical line which passes through the point of sus- pension. 72. Experimental method of finding centre of gravity. The preceding article suggests an experi- mental method of finding the centre of gravity of a body. Let the body be suspended from two points suc- cessively, and let the vertical line through the poiftt of suspension in each position be marked upon it. Each of these lines must pass through the centre of gr^ /ity, therefore, their intersection is the point required. CEXTKE OP GRAVITY, 75 73 A body placed on a horizontal plane will stand or fall according as the vertical through Its centre of gravity falls within or without the b^se. o Let G be the cenir^ of gravity of a body resting on a horizontal plane. The weight of the body, W, is a ver- tical force acting through G. If the body fall over, it must turn about a Hne touching the base. In (i) this rotation will raise G, and therefore, cannot be produced by gravity alone ; but in (2) this rotation will cause G to descend, and the tendency of gravity is to bring G lower if possible, or to induce the motion of falling. Hence in (i) the body will stand, and in (2) it will fall over. If G be vertical above A or B the body will stand, but the slightest disturbance will upset it. If the plane be inclined, the vertical through G must still fall wiihm the base, or equilibrium will be impossible. 74. DeC of base- By the term base in the preceding article, is meant the area enclosed by a string drawn tightly round the points of support of the body ; thus, if the body stand on three legs, by joining these a triangle will be formed and this will be the space within which the vertical line from the centre of gravity must fall. 75. Remarks. A man must stand in a vertical position, in order that the centre of gravity may fall within the limits of his feet, which form the base on which he stands. This base may be enlarged by separ- ating the feet, and the man's steadiness is correspond- ingly increased. If a person raise one foot from the . tills j ■P" I II f il![! 7^ ELEMEA TARY STATICS, ground, then his base is reduced to the sole of his foot iind his steadiness is diminished. Men, and indeed all animals accjuire the habit of instinctively shifting their position, so as to satisfy this condition of equilibrium ; thus, if a man walking upon a narrow plank feels hin>«elf in danger of falling ujjon one side, he throws out the opposite arm ; a man carrying a bundle upon his back leans forward; in walking down a hill, we lean backwards; in rising from a chair we must either lean forward lo bring the centre of gravity over the feet, or else we must put the feet backwards under the chair to produce the same effect. 76. Stable Equilibrium. A body is said to rest in Stable Equilibrium, when, upon being disturbed in a very slight degree from its position of equilibrium, it will, upon the disturbing cause being withdrawn, return to its first position. Unstable Equilibrium. If however, the body tend to move furthtr from its original position, that position is called one of unstable equilibrium. Neutral Equilibrium. It it remain in the new position, which the displacement has given it, the posi- tion is said to be neutral. A weight suspended by a string from a fixed point is an example of stable equilibrium, for if it be slightly pulled out of its position, it will tend to return to its original position. A pencil balanced on the end of the finger is in un- stable equilibrium, for if it be in the least degree dis- turbed, it will fall away from the finger. A sphere resting on a horizontal table, will remain in its new position if slightly disturbed, and is therefore in neutral equilibrium. EXERCISE II. 1. Show that a body has one centre of gravity and S only one ? 2. Why does a person carrying a heavy weight in his .5, hand lean towards the opposite side ? CENTRE OF GRAVITY. 77 S S S I1U,I J lew osi- 6 t is .S uly its ;S un- v.iia* S 1 in ; in % VI 6 i3 %. Explain under what circumstances a body placed on a horizontal plane will remain at rest. 4. Explain how the centre of gravity of a body may be experimentally determined, by suspending it from a point. 5. When a body is placed on a horizontal plane, it will stand or fall, according as the vertical line, drawn Irom the centre of gravity, falls within or without the base. Explain fully what you mean by the term base in this proposition, and give familiar examples. 6. How is it that an inclined tower, such as that of Pisa, does not fall, although its top hangs about twelve feet over the base ? 7. Supposing such a tower to be built in the form of an oblique cylinder, so that the slant side is to the height as 5 : 4 ; what is the greatest height to which it may be built in theory, if the radius of the base be 30 feet ? 8. Find the height of a cylinder which can just rest on an inclinea plane, the angle of which is 60" ; the diameter of the cylinder being 6 inches. 9. Define the terms stable, tuisiable^ and ?ieutral equili- brium. Give familiar examples. 10. If a body be in stable equilibrium, how is the centre of gravity aftected by a small displacement of the body ? 11. A body cannot be in stable equilibrium upon a horizontal plane, if it rests on less than three points of sui)port. 12. Why is a solid cylinder resting on a horizontal plane more difficult to upset, than a hollow cylinder of the same dimensions ? 13. Why is it necessary that a table should have thiee legs at least ? 14. Why cannot a pin practically be made to stand upon its point ? 15. Why is it more dangerous to place luggage on the top of a coach, than in the body ? 16. Two carriages have the samt base, but the centre of gravity of the one is higher than the centre of gravity of the other ; which of the two is more easily upset ? Illustrate by a diagram. IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I " 112 M 22 2.0 1.8 1.25 1.4 1,6 -• 6" ► ' ^ V*" '^^ Sdences Corporation # <^^^ .•V <^ :\ \ ^9> V ^^^. 'o 6^ ■^ '% 23 WEST MAIN STREET WEBSTER, NY. 14S80 (716) 872-4503 ^r m ■1 i 1 i 78 ELEMENTARY STATICS, 17. Two books similar in every respect, each 10 inches long, lie one exactly upon the other on a table, over the S edge of ^hich they project 3 inches. How much farther may the upper book be pushed out before they fall ovej. 18. A body is in shape a sphere, but loaded in suchn manner that its centre of gravity is not at its geometrical 5 centre ; when it is placed on a horizontal plane, what are its positions of stable, and unstable equilibrium ? 19. Explain how a long rod is more easily balanced j> oif its end than a short one. 20. An equilateral triangle is placed upon an inclined plane, its lowest angle being fixed ; find how high the «S plane may be elevated before the triangle rolls, 21. Aright-angled triangular board, hangs at rest from ^ the right-angle, and the hypoihenuse is inclined at 60^ ^ to the plumb line ; compare the lengths of the sides. SECTION III. Centre of Gravity of plane areas, &c 77. To find the centre of gnivify of a triangle. A Let ABC be the triangle which is supposed to be crt small and uniform thickness. We may conceive this triangle to be made up of a number of rods all parallel to BC. Let be be one of the CENTRE OF GRA VI TV, 79 inches Dver the 1 farther ill ovej. n such a metrical le, what Lim? alanced inclined ligh the £st from I at 60' sides. 3 be (»1 jp of .1 5 of tlic seriaj of rods ; the centre of gravity of be will be d its middle point (Art. 68); and /will be a point on the me AD drawn from A to the bisection of BC And tha centre of gravity of every other one of the series of rods must also lie on this Hne AD. Therefore the centre theTne ad! "^^'^^ '"'"^^' '""^•^ ^'^ soi.c.h.r. in Now we might have supposed the triangle to have beeri made up m the same way of a series of rods a U parallel to the side AC, and then we should have obta ned the result that the centre of gravity of the whole triangle must be somewhere m the line BE, drawn from B to the bisection of the side AC , Since, therefore, the centre of gravity of the triangle IS m the line AD, and also in thi line BE, it must be the pomt G where these two lines intersect. 73. Distance of C.G. from middle point of base. W e may find the value of FG, or i^G, as follows : 41™' ^^' ^""^ ^^' ^^' ^" ^' ^' ^' ^> -d ^. Then since AC is bisected in F SimilX^'T'^c'E^fr * ^^- f "^^^^ABC (Euc. I. 38). Therefore „ BFC == the triangle CEb! Md therefore EF is parallel to BC (Euc. I. 39). 8o ELEMENTARY STATICS. nm In a similar manner it may be shown that FD is par- allel to AB. Therefore BDFE is a parallelogram. Hence EF is equal and parallel to BD (Euc. I. 34). Similarly MN ,1 11 BD. therefore MN n n ¥.¥. Hence the angle EFG = the angle GMN (Euc. I. 29); and "1. EGF = .. MGN (Euc. I. 15). And the side EF has been shown to be equal to the MN. Therefore FG = GM = MB, because BG is bisected in M, = JBF. Similarly it may be shown that EG = i CE. Hence, to find the C. G. of a triangle ABC : from an angle draw a line to the bisection of the opposite side ; the cent re. of gravity lies on that line, and at a point onc-ihird its length from the side which it bisects. Hence also, the right lines drawn from the angles of a triangle to bisect the opposite sides all pass through the same point, for they all pass through the centre of gravity. 79. Remarks on C. G. of plane areas. When we speak ol the centre of gravity of a triangle, it would be more correct to speak of the centre of gravity of a portion of matter bounded by two plane triangles, the surfaces of which are i)arallel and very near to each other, or of the centre of gravity of a very thin lamina on a triangular base. No confusion can arise from speaking of the centre of gravity of a plane figure, if the student bears in mind that his results are applicaole to indefi- nitely thin plates or lamina ; or if thickness be considered he must regard the centre of gravity as lying inside the body and at equal distances from liie two bounding sur- CENTRE OF GRA VII Y, 81 80. Heavy pirtUIes are placed at the angles of a triangle, their 7veighU being proportional lo the sides opposite to them- to find their centre oj gravity. ' mc ma Let a, b c be the sides of the triangle, and ma mb mc the weights at the opposite angles. ' ' ar.v?flr:ffh *' ^^T^% ^^^ distances of the centre of f ^d 7f f , *^".^^^Sh^« ^o.'" the sides a, b, ., respectively ; and let/ be the perpendicular upon the side a from the opposite angle. ^ Take moments about the side a, and we have {ma + mb + /«<:) a? = map ; therefore fl; = u litre A is the area of the triangle. Similarly y ___ a + b-^c 2 A a + ^+tf ' 2A and M = tf + /J + f * 2A These rcsuhs show that the centre of gravity of the heTri^e^' Th^"'">' ^'''T ''""^ ^^^ thiee ^sides of lie triangle. This point is the intersection ol the Imes bisecting any two angles of the triangle. I!: 88 ELEMENTAR Y STA TICS, ^1. To find the centre of f!,ravtty of the periphery of triangle formed by a piece of uniform wire> A R V r Bisect the sides of the triangle ABC in D, E and F. The weight of each side may be supposed to be collect- ed at its middle point, it is also proportional to the side, and therefore proportional to the opposite side of the triangle DEF ; hence the problem becomes the same as the one just solved. ^%, To find the centre of gravity of a parallelogram, D Let ABCD be any parallelogram ; draw the diagonals AC, BD, intersecting in G. These diagonals mutually bisect each other. The centre of gravity H of the triangle DAB is in the line AG at a distance = J GA from G (Art. 78) : and the centre of gravity H' of the triang-e BCD is at a distance = J GC froniG ; also the triangles are both equal, and consequently their weights are so, therefore the resultant of the two weights is half- |!-:;i: CENTRE OF GRAVITY, n rryof andF. collect- he side, of the same as im. way between H and H'; or, in other words, the centre oi" gravity of the whole parallelogram is at G. 83. Centre of Gravity of Symmetrical Figures If we can by inspection determine a point around which the material of a body is symmetrically s^'Uiated, that point will be the centre of gravity of the body. It has been shown (Art. 68), that the centre cf a uniform straight rod is its centre of gravity. This result might have been arrived at by the general consideration that around the centre of the rod, the material, being symme- trically situated, will be equally drawn towards t^e earth's centre ; and therefore the centre of the rod must be the centre of the system of forces produced by the attraction of the different parts of the rod towards the earth. In the same v/ay the centre of gravity of a sphere is evidently its geometrical centre, for any change in the position of the sphere can produce no change in the disposition of the material about its centre. Hence if the centre were fixed, the sphere would have no ten- dency to turn about that point, from one position into any other. The centre of gravity of the parallelogram, in the last article, might have been determined in this way ; and in the same manner the centre of gravity of many geometrical figures may be found. EXAMPLES. igonals [utually lof the iGA of the (so the weights half- 3. At the corners of a square, taken in order, are placed weights i, 3, 5, 7 ; find their centre of gravity. Let a, be the length of a side of the square. Let X = the distance of the centre of gravity from the side (i, 7) ; and let y = the distance of the centre of gravity from the side (5, 7.) The sum of the weights may be supposed to act at their cen- tre of gravity. 3 a o 84 ELEMENTARY ^ TA IICS. Take moments about the side (i, 7), and we have 16 a; = 3 ^ ^^ 5 d!; a, therefore x = — - 2 Take moments about the side {5, 7), and 1 6// = 3 rt X a\ therefore y •= \a. Hence, the distance of the centre of gravity from (i, 7), i«j ^a, and from (5, 7), is \a. 4 Find the centre of gravity of a uniform circular disc out of which another circular disc has been cut, the di- ameter of the latter being *he radius of the former. Let R be the radius of the circle ABC, II r N « AGO. From the symmetry of the figure we see that the centre of gravity of the disc will lie on the diameter AC, which passes through ^y the centres of both circles. Let X = the distance of the centre of gravity of the disc from O. The Weights of the circles are proportional to their areas and may be supposed to act at their centres of gravity which are their geometrical centres. Take moments about O, and we have therefore o = r* + (R^ — r'^)x = r* + (4r* —r*);c, since R= ar; . ^' — ^ 3 6 The negative sign shows that x lies to the left of O. Hence the centre of gravity of the disc lies between O and C and at a distance from O eq ual to ^th the radius, *I? r be the radius of a circle, it« area =» %r*t where *— 3. I : T 9 CENTRE OF GRAVITY, EXERCISE III. v5 N$ 1. At two of the angular points of an equilateral tmngle are suspended weights of one pound, and at the third angular point is suspended a weight of two pounds ; find the centre of gravity of the system. 2. Find the centre of gravity of three equally heavy particles placed at the three angles of a triangle, and show that their centre of gravity coincides with the centre of gravity of the triangle. 3. Find the centre of gravity of a right-angled isosceles triangle, and the squares described on the two equal sides. 4. A. bent bar of uniform thickness and density, forms an equilateral triangle ; show that the centre of gravity of the bar corresponds with the cenire of gravity of the triangle. 5. Four heavy particles, whose weights are 4, 6, 5 and 3lbs. respectively, are placed in the corners of a square plate whose sides are 26 inches, and weight 81bs. ; re- quired the distance ot the centre of gravity from the centre of the plate. 6. From a circular plate, whose radius is 2^, a circular piece, whose radius is a is cut away ; the distance between the two centres is \a ; required the distance of the centre of gravity of the remainder from the centre of the plate. 7. From a circular disc of radius R a circular disc of radius r, is removed, the distance between their centres l)eing d\ determine the centre of gravity of the re- mainder. 8. A round table whose weight is W, stands on three legs placed on the circumference at equal distances ; what is the greatest weight which may be placed upon any part of the table without upsetting it ? 9. A circular table, weighing i681l)s. is supported on four legs in the circumference which form a square ; find the least weight which being 4t the circumference will overturn the table. T 86 ELEMENTARY STATICS. TO. A triangular slab of uniform thickness is supported r at its three angular points ; sho'v that the pressures on the supports are equal to one another. 1 1. An isosceles triangle is placed with its base, which is a ft. in length, upon a plane whose inclination is 30**, and c is prevented from slidmg by a small obstacle placed at the lowest point of its base ; what is the greatest height which the triangle can have without toppling over ? 12. Three weights are placed in the angles of a right- angled triangle, and are proportional to the squares on the ^ opposite sides \ required the distance of the centre of gr wity from the right angle. 13. Find the centre of gra,vity of three uniform beams fo'ming a right-angled isosceles triangle, the greatest ^ si( e of the triangle being 20 ft. 14. An isosceles triangle rests on a square, and the he»f;ht of the triangle is equal to a side of the square : > fird the centre of gravity of the figure thus foimed. . 15. If a piece of uniform wire bent into the for of a trangle ABC, be suspended freely from t'le angular r point A, prove that it will only rest with the side BC horizontal when the angle ABC is equal to the angle ACB. 16. A number of cent pieces are cemented together so that each just laps over the one below it by the ninth part of its diameter ; find how many may be thus piled without falling, when the lowest stands on a horizontal plane. 17. A short circular cylinder of wood has a hemis- pherical end. When placed with its curved end on a smooth table it rests in any position in which it is placed. Determine the position of its centre of gravity. 18. P/Xplain how a plane area is said to have a centre of gravity. Find tlie centre of gravity of two material plane isosceles triangles on the same base, their veitices being on opposite sides of the base at dist-nices hj, hg, from it. 19. The centre of gravity of a body being given, and also that of a portion of it, show how to find the centre of gravity of the remaindei. ot a CENTRE OF GRAVITY, «7 20. State and prove the proposition which suggests a practical method of finding the centre of gravity of any plane area. 21. A quadrilateral is such that when it is suspended from any angular point the diagonal through the point of suspension is vertical ; show that it must be a parallelo- gram. f I CHAPTER VIIT. SIMPLE MACHINES OR MECHANICAL POWERS. 84. Def. of Machine. Any contrivance which enables us to change the point of application, direction, or intensity of a force may be called a machine. 85. Object of machines. The object of all machines, considered in a statical point of view, is to enable a certain force, as P, which is called the Power^ to be in equilibrium with a second force, as W, which is termed the Weight. These distinctive names are given to the two forces because we are most familiar with the use of machines, when employed for the purpose of raising, or moving heavy bodies, by the application of a small force or power. 86. Mechanical Advantage. The mechanical advantage of a machine may be defined as the ratio of the weight to the power when there is equilibrium ; thus if a weight of 5 pounds sustains a weiglit of 80 pounds, the mechanical advantage is 80 -f 5 or 16. This value, W -— , is also the measure of the efficiency or working power of a machine. When W is greater than P the machine is said to work at a mechanical advantage^ when W is less than P at a tnmhanical disadvantage. THE LEVER, «9 87. Mechanical Powers. The following six machines are, for convenience, regarded as simple machines or Mechanical Powers. I. The Lever. a. The Wheel and Axle. 3. The Pulley. 4. The Inclined Plane. 5. The Wedge. 6. The Screw. Every machine, however complicated, mny be shown to be composed of combinations of these simple machinec. THE LEVER. I. 88. Kinds of Levers. We have already considered the principle of the lever as a general mechanical {)rin- ciple, and we have shown that two forces will baljnc e about a fulcrum when the moments about it are equal ; but the lever may also be regarded as one of the Mechanical Powers, and so considering it we distinguish three kinds of levers according to the position of the fulcrum with respect to the power and the weight. A lever of the first kind has the fulcrum between the power and the weight. In this case any amount of mechanical advantage may be gained by making the arm upon which the power rests sufficiently long. A crovv- bar used to lift great weights, a poker, a pair of scissors, are examples. In the poker, the coals are the weighty the bar of the grate the fulcrum, and the force exerted by the hand, the power. A lever of the second kind has the weight between the fulcrum and the power. The oar of a boat is an example, in which the water forms the fulcrum, the resistance of the boat, applied at the rowlock, is the weight, and the power is applied by the hand of the rower. A lever of the third kind has the point of application of the power between the fulcrum and the weight. The most Lateresting example is the human arm when applied w <': «! ill 90 ELEMENTARY STATICS. to lift a 'S'eiglit by turning about the elbow ^ here the elbow-joint is the fulcrum, and the power is applied by means of the muscles at a point between the joint and the weight. A man lifting a long ladder with one end resting on the ground is also a familiar example. 89. Remarks. These distinctions are, however, of but little practical importance. The turning effect of a force is measured by its moment ; for equ^librirm, there- fore, the moment of the power must be equal to the moment of the weight. When tbis one idea is grasped and the position of the fulcrum ascertained the subject- matter is exhausted. EXERCISE L 1. It IS said that the "six simple machines are the lever, the wheel and axle, the pulley, the inclined plane, the wedge, and the screw." Point out that some of these machines are merely modified from others. 2. When does a force act by a machine at a mechanical advantage ? What is meant by mechanical advantage being lost or gained by the intervention of a lever? Ilx- plain under what circumstances either the one result or the other takes place in the case of each kind of lever. 3. There are three classes of levers ; what distinguishes one class from another ? Give familiar examples of each class describing it so as to show clearly that it is an ex- ample. To what class does a wheelbarrow belong ? 4. How does the contrivance of placing the row-locks outside the boat aftect the efforts of the rower ? 5. " We have spoken of only two forces, the power and the weight, as acting upon the lever, yet in reality there must be three forces." What is the third force ? 6. State the conditions for equilibrium of pressures acting on a lever. 7. If a walking-stick be supported near one end by passing that end over the thumb and fourth finger, it is noticeable that the pressure perpendicular to the stick which must be exerted by the fourth finger is greater I BALANCES. 9" when the stick is held horizontally than when it is any slanting position. Explain the fact clearly on mechanical principles, with the help of a diagram. 8. If the forces acting perpendicularly at the extremi- tie;- of the arms of a lever, balance each other, they are inversely as the arms. 9. Show 7vhy a. lever is in equilibrium when the power and the weight are to each other inversely as the perpen- dicular distances of their lines of action from the fulcrum. 10. A man wheels a wheelbarrow along a level road. Point out the conditions which determine how much of the total weight of the load and barrow is supported by the wheel and how much is supported by the man. SECTION II. Balances. 90. Introduction. The lever is a convenient tnstru ment for ascertaining the weight of a body by comparing it with another whose weight is known. The weights of the body to be weighed, and of the counterpoise with which it is compared, are two parallel forces acting in the same direction, and may therefore be made to balance each other by means of a lever of the first kind. Let W be the body to be weighed, and P the counter- poise, then if they do not balance one another, equilibrium may be produced (i) by changing the counterpoise P, (2) changing its position, (3) changing the place of the fulcrum, or (4) changing the inclination of the arms of the lever to the direction of the forces. 91. The Common Balance. The first of these methods is used in the balance with equal arms, usually called the common balance. In this balance the position of the fulcrum and of the points of suspension of the weight and its counterpoise are fixed, and the balance is so constructed, that it will be horizontal when the weight and its counterpoise are equal. For this pur- pose it is necebsary that the arms of the lever should be 9S ELEMENTAR Y STA TICS. equal, and also that the centre of gravity of the instru- ment itself should be in the perpendicular from the ful- crum to the line joining the points of suspension ot the scales. It is m )reover necessary, that the centre of gravity of the balance and its load, should always lie below the fulcrum, in order that, if the instrument were displaced, it may r*^ turn to the horizontal position, for if the centre of gravity lay above the fulcrum, it would have a tendency to upset, and if it coincided with the fulcrum, it would rest indifferently in any position. 92. Qualities of a good Balance. The most im- portant qualities of a good balance are : — 1. Sensibility. The beam should be sensibly deflected from a horizontal position by the smallest difference between the weights in the scale-pans. The definite measure of the sensibility is the angle through which the beam is deflected by a stated percentage of the difference between the loads in the pans. 2. Stability. This means rapidity of oscillation, and consequently speed in the performance of a weighing. It depends mainly upon the depth of the centre of gravity of the whole below the knife-edge, and the length of the beam. 3. Constancy. Successive weighings of the same body must give the same result — all necessary corrections depending on temperature, &c., being allowed for. 93. Remarks. The sensibility of a balance is, in general, of more importance than the stability^ but much depends on the service for which the balance is intended. For weighing heavy goods stability is of more impor- tance. When great accuracy is required, as in chemical balances, sensibility is the quality desired. 94. To determine the true weight of a body by a fal::e bahin. e. Let W be the true weight of the body, and let x and y be the unknown arms of the balance. When W is placed in one scale-pan, let it be balanced by a weight THE COMMON BALANCE. 93 P, and when placed in the other by a weight Q. Then since W and P acting at the arms x and 7, were in equi- librium, we have W^= P^ (i). And since W and Q, acting at the arms x and^, were in equilibrium, we have Wy = Q^ (2). Multiplying(i)and(2), W^jy; = PQA7. Or VV2 = PQ ; .-. VV = \/PQ. That is, the true weight is the square root of the pro- duct of the false weights ; or the true eiglit is a mean proJ>ortional between the two false weights. EXAMPLES. iical fahe 1. If a balance be false, having its arms in the ratio of 15 to i6; find how much per lb. a customer really pays for tea which is sold to him from the longer arm at 75 cents per lb. Since the arms are in the ratio of 15 to 16, i6oz. at the end of the shorter arm will be balanced by 150Z. at the end of the longer arm. The customer will, therefore get 150Z. instead of ilb. and for this he pays at the rate ol 5 cents peroz. Therefore for 16 oz. he will pay 80 cents. 2. A merchant has correct weights but a false balance, that is one having one arm longer than the other j sup- posing that he serves out to each of two customers articles weighing, as indicated by his balance, Wft)s., using for the commodiftes first one scale and then the other ; find whether he gains or loses by the peculiarity of his balance. Let a, b, be the lengths of the arms of the balance, and let P, Q, be the true weights of the commodities, placed at the end of the former and latter arm respec- tively, Then Va = W^, and Q^ = Wo. 11 , i 94 ELEMENTA RY STA TICS, Ileiice P = W — ; a and Q = W a Also 2W = 2W ; . '. P + Q — 2W : a* + b^ = W- - 2^^ ab .(a — b)* W- —^ ; a positive ab quantity. The merchant therefore, loses by his sale to the two customers a weight of substance equal to W (a — b)» ab EXERCISE L 1. How can you determine the true weight of an article by having it weighed in each scale successively of a common balance, that is rendered false, (i) by having its arms unequal, (2) by having one of its scales loaded ? 2. If a straight lever balance, under the action of two weights, in one position, it will balance in every position. What exceptions are there to this ? Why is it not the case in a common pair of scales ? 3. If when a balance is suspended the beam be not horizontal, prove that if the want of horizon tality arise from an inequality in the weight of the scale-pans, the balance may be corrected by putting a weight into the lighter of the two, but that if it arise from a difference of length of the arms the balance cannot be so corrected. 4. What is meant by the sensibility of a balancs ? Other conditions being the same, why is the sensibility less, the greater the weight of the beam ? What are the requisites of a good balance ? I THE COMMON BALANCE. 95 5. A tradesman's balance has arms whose lengths are II in. and 12 in., respectively, and it rests horizontally, when the scales are empty. If he sells to each of two cus- tomers a pound of tea at 55 cents per tt», putting his weights into two different scales for each transaction, find whether he gains or loses, owing to the incorrect- ness of his balance, and how much. 6. A common balance has its arms of unequal length ; a body when placed in one scale-pan appears to weigh 4Bbs., and when placed in the other i6K)S. What is the true weight of the body ? 7. The whole length of the beam of a false balance is 3ft. (jin. ; a certain body placed in one scale appears to weigh 9Kbs , and placed in the other appears to weigh 4tt)s. Find the true weight of the body, and the lengths of the arms of the balance, supposed to be with- out weight. 8. A grocer buys tea wholesale, at the rate of 80 cents per fi)., and in weighing it out to his customers uses a balance, the arms of which are in the ratio of 16 to 15 ; at what price must he profess to sell it per tt)., in order that he may make a profit of 20 per cent ? 9. A tobacconist buys tobacco wholesale, at the rate of 66| cents per fi>., and in weighing it out to his cus- tomers uses a balance, the arms of which are in the ratio of 16 to 15 ; if he profess to sell it at 75 cents per fi)., what profit per cent, does he really make ? 10. One of the arms of a false balance is longer than the other by \ part of the shorter : when used the weight is put into one scale as often as into the other. What will be the gain or loss per cent, to the seller ? 11. The arms of a false balance are to one another as 7 to 8, and the weights are put into one scale as often as into the other ; what will be the gain or loss per cent, to the seller? Twr 96 ELExMENlARY STA TICS, t! :!fi !! I I THE ROMAN STEELYARD. 95- The Roman Steelyard. The Common or Roman Steelyard is another form of tlie balance. It is lepresented in the annexed figure. C G . II UN li 11 rmTO AB is an iron bar turning about the edge C of a tri- angular piece of metal, which works in an aperture of the handle by which the balance is suspended from a point O. At A, a hook or scale is attached for the purpose ct receiving any article whose weight is to be ascertained P is a movable weight which can be 'suspended from any point between C and B. Suppose W to be a body suspended from A, then P must be moved toward C, or towards B, until the beam AB rests in a horizontal posi- tion. Let F be this position of P. The bar is so graduated from C to B that . the reading of F indicates the weight of W. 96. To graduate the Common Sieeiyard, Let the weights W and P be removed. Suppose G to be the centre of gravity of the beam. It W II weight of the beam. The arm CB being by construction heavier than the rest of the beam, will fall when the machine is in this un- loaded state. Let the weight P be placed on the other side of the handle CO at a point D, found by trial, such that the beam is kept in a horizontal position. We shall then have P. CD = W'.CG (i). THE STEELYARD. 97 Now let the beam be loaded with P and W as before; then for equilibrium we must have P. CF + W'.CG = W. AC ; Substituting from ( i ), we have P (CF + CD) = W. AC; DF •••^^ = i'ac <^)- Suppose now we put W, ilb., 2Bbs., 3tt>s., successively, we can calculate in each case the corres- ponding position of F from equation (2). At the point in the arm so determined graduations must be made and marked i, 2, 3, &c. The space between two graduations may be sub- divided into four or more equal parts, and these sub- divisions will indicate with sufficient accuracy, the altera- tion which must be made in W to indicate P placed at any of them. EXERCISE II. 1. Describe the common (Roman) Steelyard, and in- vestigate the method of graduating it. 2. Where must be the centre of gravity of the com- mon steelyard so that any movable weight may be used with it ? 3. If the movable weight for which the steelyard is constructed be tft>., and a tradesman substituted 2Bbs., using the same graduations, show that he defrauds his customers, if the centre of gravity of the steelyard be in the longer arm, and himself if it be in the shorter arm. 4. The movable weight is ifl)., and the weight of the beam is ifc. ; the distance of the point of suspension from the body weighed is 2J inches, and the distance of the centre of gravity of the beam from the body weighed is 3 inches ; find where the movable weight must be placed when a body of 3lbs. is weighed. 5. If the fulcrum divide the beam, supposed uniform, in the ratio of 3 to i, and the weight of the beam be equal to the movable weight, show that the greatest weight which can be weighed is four times the movable weight * 9? ELEMENTAR Y STA TICS, ^ 6. Find what effect is produced on the graduations by increasing the movable weight. ^ 7. Find what effect is produced on the graduations by increasing the density of the material of the beam. THE WHEEL AND AXLE. It 97*. Description of Wheel and Axle. This machine consists of a wheel AB, and a cylinder or axle HH' ; both tuhiing round the same axis. The points C,C', are placed on bearings so as to allow of motion round the common axis. The power is applied at the circumference of the wheel, generally by a cord wrapped round it, as in the figure, and the weight is attached to a cord coiled round the axle in the contrary direction, so that when P descends, W ascends, and vice versa. The second figure represents a vertical section of the machine, by a plane perpendicular to the axis. The power and the weight are, for the sake of simplicity, supposed to act in the same plane. O is the common centre. Draw through O the horizontal line A BOD. A and D will be the points where the cords, connected with the power and weight, leave the wheel and £xle respectively. iisiiii ■^■- THE WHEEL AND AXLE. 99 98. Wheel and Axle a modification of the Lever. Suppose tlic figure to represent the machine at one moment of its action. We rnight, for that moment, suppose the whole machine reduced to a simple lever ABOD, of which O is the fulcrum and A and D the points of application of the power and weight respectively. We should not by this supposition affect in any way the forces at work. Hence we see the wheel and axle is an infinite number of levers, the common fulcrum being O, the arm at which the weight acts being always equal to the radius (OD) of the axle, and that at which the power acts equal to the radius (OA) of the wheel. Hence the wheel and axle is a practical arrangement for continuing the action of a lever as long as may be required, the weight rising all the time. On this account it is called the perpetual lever. 99. To find the ratio of the Power to the Weight in the Wheel and Axle. Take moments about O (fig. 2, Art. 97), and we have P ^ AO = W y. OD, P___Op _ r_ °^ W^ AO" R ' where R is the radius of the wheel, and r the radius of the axle. There will, therefore, be equilibrium when Power: Weight:: radius of Axle: radius of Wheel. If the thickness of the ropes be considerable \^e must add half their thickness to R and r respectively. EXERCISE I. 1. In the wheel and axle what is the difference between axle and axis ? Is the axis fixed or rotatory ? 2. Is there any advantage in having the rope which passes round the wheel thicker than that which passes round the axle ? too ELEMENTARY STATICS. 3. Find the ratio of the power to the weight in the wheel and axle. Explain why the machine has been called i\\^ perpetualln'er. 4. If the rope use be of such a thickness that it becomes necessary to take it into account what would be the ratio between P and W ? 5. Why is the labour of drawing a bucket of water out of a common well generally greater dk-ing the last part of the process than during the first ? 6. What is the greatest weight which can be supported by a power of 4olbs. by means of a wheel and axle, when the diameter of the wheel is 10 times that of the axle? 7. If a power of 35lbs. supports a weight of 84lbs. on a wheel and axle, and the radius of the axle be 5in., what must be the radius of the wheel ? 8. Two men capable of exerting forces of 260 and 3oolbs. respectively, work the handle of a winch and axis. The radius of the axle is 5in. ; what must be the length of the arm of the winch that the men may be ju.'t able to raise a weight of 448olbs. THE PULLEY. III. 100. Def. of Pulley. The Pulley consists of a small wheel which moves freely about an axis, and allows a string to pass over a groove on its surface. The ends of the axis are fixed to a frame called the Block. loi. Def. The Pulley is said to ht fixed ox \no\2\At according as the block is fixed or movable. The wheel is supposed to revolve without friction, and the string to be perfectly flexible. 102. Fixed Pulley. No mechanical advantage is gained by a fixed pulley, for, as the tension in every part of the string is the same (Art 15), if a weight W be suspended at one extremity, an equal weight must be applied at the other to maintain equilibrium ; therefore, in this case, P = W. THE PULLEY. ^^^ wheat has to be raisti fZ"™' ' of th^Tw^r to t'e 1 the upper stories of a building ft m, J ^ ""^ ^.raised by a n,a„ carryi„7i'tfb^; in haT^e" e^ To r^ Tb'-eVars^rrrV^^e^^^^^^^^ %-^t ^ he can raise the wheat 'withltVllgLlwa^S''"' ^ loa ELEMENTARY STATICS, THE PULLEY. 104. The Single Movable Pulley. — To find the ratio of the Power to the Weight in the xingie movable Pulley, Let the weight W be attached t(y ^' the movable pulley B ; and let B ^^T be sustained by the cord ABC, one "^ extremity of which is fastened at A, and the other passing over the fixed pulley C sustains the power P. The tension of the string is the same throughout. Hence we may regard the pulley as acted on by two parallel forces, each equal to P upwards, and by the force W down- wards ; therefore, W = 2P. The pressure on the fixed point A, is equal to P, that is to JW. If the weight of the pulley is too considerable to be omitted, let it be equal to w we have ; then or 2P = W + - ze/ = J (W- -w). 105. First System of Pulleys. To find the ratio of the Power to the Weighty in a systetn of Pulley Sy in which each Pulley hangs by a separate string. The annexed figure represents a system of pulleys in which the string that pas.ses round one pulley has one of its ends fastened to a fixed point, and the other attached to the block of the next pulley ; the weight is applied to THE PULLEY, 103 the block of the lowest pulley, and the power to the extremity ol the first string. i^ cr 10 me R N i\r H AB = I tension of BC = jr W But „ AP = P; ^ .-. P = JW. Thus, when there are three movable pulleys, W 23 are'", ^4^ pTe^f " '"'' '^ ^''°"''' '^^ *''^" '"-« or 2« 2»» P = W. I04 ELEMENTARY STA TICS. io6. Pulleys supposed heavy. If the weight ol the pulleys be taken into account, let the weight of the three pulleys, beginning with C, be a/j, Wg, Wg, respectively; the tension in BC = — (VV + k'j), N AB = ^(W + a/i)+ --w^. But N AP =-2T(^ + «'i) + 7i«'2 +-7^«- AP = P; W w \P = ,_i^!^^!^. 2* 2* 2* 2 and similarly for any number of pulleys. If the weight of each pulley be the same and equal to «/, we have 23 P = W + 7f/(23— l). or P — w = — r (W — u>) ; and similarly, if n be the number of pulleys, P W — — j^(VV— 7£/). EXAMPLES. 1. In a system of pulleys, where each pulley hangs by a separate siring, the number of pulleys is 5. What weight will a power of s&s. support ? Here VV = 2* x 5 — i6oft)s. 2. In the above system, \%. is supported by loz. ; find the number of pulleys. 160Z. = 2** X loz., where n is the number of pulleys, or 2" = 16 ; ..•.«= 4. It of hree ely; W,. ito by hat )Z. jys, THE PULLEY, los hi4;es?are^,""7' whose weights, heglnning with the a^cSnt?oV'fi''."^ 6Ibs., respectively, are arranged Here P = 462 .^ 462 + 6^ 2* 52. 2* 2 + —- + 23 _3. a = -ill 16 4. In the case of the single movable pulley will the mechanical advantage be increased or diminished bv taking mto account the weight of the pulleyT ^ When the weight of the pulley is neglected ' W -p =2. But if the weight of the Pulley be taken into account 2 P - W+w, or _W _ a, p -• 2 p-» which is less than 2, and, therefore, the mechanical advantage is dirrinished EXERCISE I. 1. In the single movable pulley, is any mechanica 2. In a system of four pulleys, where each pulley 4. If a weight of rft». is supported by loz what is thp I|:i H: I io6 ELEMENTARY STATICS. 5. The mechanical advantage is equal to 16 ; how many puPeys are there ? 6. If three movable pulleys, the weights of which are 2, 4, and 8 oz. respectively, beginning at the lowest, be arranged as in the first system, what is the least force that will raise a weight of io4lbs. ? 7. With three movable pulleys arranged according to the first system, each of which weighs 8oz., what weight can be supported by a force of iR). ? 8. In a system of pulleys, in which each pulley hangs by a separate string, there are three pulleys of equal weights ; the weight attached to the lowest is 32S)s., and the power is 11 lbs. ; find the weight of each pulley. 9. In a system of five pulleys, in which each pulley hangs by a separate string, the tension of the string attache'd to the lowest block but one is 8o5>s. ; find the weight supported and the power exerted. 10. If a man weighing i6olbs. supports a weight equal to his own, and there are three pulleys, find his pressure on the floor on which he stands. 11. In the system of pulleyf? in which each pulley hangs by a separate string, a platform is suspended from the lowest block ; what force must a man who weighs i8oRs., standing on the platform and pulling downwards on the string which passes round the highest pulley and over the pulley fixed to the beam, exert to sustain him- self? 107. Second System of Pulleys lopid the ratio of the Fouier to the Weighty in a system of PutleySy in which the same string passes round all the Pulleys. In this system there are two blocks, each containing a number of pulleys. The lowest block is movable, the upper one fixed. The pulleys are of such relative sizes that the different portions of the string between them are all parallel. This system will be understood from the annexed figure, and is known as the Second System of Pulleys. II THE PULLEY. 107 Suppose there are four pulleys, two in each block. Let W denote the weight which is suspended from the lower block, and let P denote the power which acts vertically downwards at one end of the string. The tension of the string is the same throughout and is equal to P. We may, therefore, regard the lower block as acted on by four parallel forces, each equal to P up- wards, and the force W downwards. Therefore W = 4P. In a similar manner it may be shown that if there are n strings at the lower block, W = «P. If the weight of the lower pulleys and block be w, the total weight supported will be W + «', therefore W + Z£/ = «P. EXERCISE II. uimg , the sizes ■mare 1. In the system of pulleys where the same string passes round each pulley, if the number of strings at the lower block be six, what limit must be put to the weight of the lower block, so that any mechanical advantage may be gained by this system of pulleys ? 2. In the system of pulleys where the same string passes round each pulley, there are 7 pulleys at the lower block. What weight will a power of 2Bt)s. support ? 3. If there are twelve strings at the lower block in a system of pulley in which the same string passes round all the pulleys, find the weight which a power of loBbs. will support, the weight of the pulleys l Mng neglected. w I'f io8 ELEMENTARY STATICS. 4. Tf there are 6 strings at the lower block, in the Second System of pulleys, find the greatest weight which a man weighing i4olbs. can possibly raise. 5. In the Second System of pulleys the mechanical advantage is expressed by the number 10; how many movable pulleys are there? 6. Find the number of strings in the lower block in order that a power of 40Z. may support a weight of 4B)S. 7. A man weighing i68tt)s. supports a weight equal to half his own ; if there are 7 strings at the lower block, find his pressure on the floor on which he stands. 8. Find what weight can be supported by a power of loBbs., if there are 3 pulleys at the lower block, and the weight of the lower block is three times the power. 9. In a system of pulleys where the same string passes round each pulley, if there are four pulleys at the lower block, two of which weigh ifi). each, and the other two half a pound each, what weight will a power of 5tt)s. support ? 10. In a system of pulleys in which the same string passes round any number of pulleys, if the weight of the pulleys be regarded, under what circumstances will the mechanical advantage be reduced to nothing. 1 1. If six strings pass round the lower block and a man weighing i4oIbs. supports himself by standing on a platform suspended from the lower block, and holding the rope that passes round the pulleys ; find the tension of the rope. 12. A man sitting upon a board suspended from a single movable pulley pulls downwards at one end of a rope which passes under the movable pulley and over a pulley fixed to a beam overhead, the other end of the rope being fixed to the same beam. What is the small- est proportion of his whole weight with which the man roust pull in order to raise himself? 13. With what force would he require to pull upwards, if the rope, before coming to his hand, passed under a pulley fixed to the ground, as well as round the other two pulleys ? THE PULLEY. 109 108. Third System of Pulleys. To find the ratio 0/ the Po7vtr to the Weighty in a system of Pulleys in which all the strings are attached to the weight. In this system the cord which passes over any one pulley has one extremity fastened to the block to which the weight is attached, while the other is attached to the block of the next pulley. The highest pulley is iixed This system is represented in the annexed figure and is kn^wn as the Third System of Pulleys. lom a of a Iver a >f the jmall- man ^ards, kder a othei ^ / ^ y f V ^ ly ^ rA..>i.Q.»..;(A;T^pi X 4 w Suppose there are four pulleys. Let W denote the weight to which all the strings are fastened, and let P be the power which acts vertically downwards at the end of the string which passes over the lowest pulley. The tension of the string which passes over the lowest pulley is P ; the tension of the string which passes over the next pulley is 2P ; the tension of the string which passes over the next pulley is twice this or 2" P ; the tension of the string whiqh passes over the highest pulley no ELEMENTARY STA TICS. is twice this, that is, 2 'P. But the sum of all these tensions must equal W, since the strings support VV; therefore W == P + 2P+ 2^? + 2^? = P(l +2+22 + 23) =r P(24-l). Similarly, if « be the number of pulleys, W=: ? (2»*— i). 109. General Method of finding the relation of P to W in Pulleys. Begin with the end of the system at which the Power acts. The tension of the string which supports P will be equal to P throughout; against each of the parallel portions of the string write P. Proceed to the next string, and find what its tension is by observing how many strings, each having the tension P, produce it; write the expression for its tension against each parallel portion ; and so on with the next string. When the tension of each string of the system has been written down, it is easy to see how many of them support VV, and by adding their tensions together we have the rela- tion between P and W required. i EXAMPLES. 5. Find the power necessary to sustain a weight of looBbs. with three movable pulleys arranged according to the Third System, the weights of the pulleys being 80Z., 60Z., and 40Z., respectively. Let /j, /j, /g, t^ be the tensions of the strings begin- ning with lowest pulley. Then A = P, t^ = 2P + 8, /g =4P+ 16 + 6, t^ = 8P + 32 + 12 + 4. The sum of the tensions equals the weights j therefore i5P + 78=i6oo, or P = 101460Z. = 6'34K)S. these THE PULLEY, EXERCISE III. XIX 1. If It IS required to raise a weight equal to three times the power by a system of pulleys in which all the strings are attached to the weight, how many strings must there be ? -^ 6 2. A power of 2oIbs. supports a weiglit of 388Ibs by means of four pulleys of equal weights, arranged accord- mg to the third system ; find the weight of each pulley. 3. It m the third system there are 5 pulleys and the weights of the four movable pulleys, commencing with the lowest, are 2, 3, 4 and slbs., find the weight which will be sustained by a power of i2ll>s. 4- What weight will be supported by a power of i2lbs acting by means of a system of pulleys in which eacli cord is fastened to the weight, when there are five mov- able pulleys each weighing i J pounds ? 5. In the third system if there are eight pulleys, find the ratio that the weight of each pulley must bear to the weight supported, in order that the latter may be just supported by the weight of the pulleys alone. 6. What advantage does the third system of pulleys possess over the first or second system ? IIS ELEMENTARY STATICS. THE INCLINED PLANE. IV. 110. Introduction. The Inclined Plane is a rigid plane inclined at any angle to the horizon. The Plane is supposed to be perfectly smooth and per- fectly rigid^ so as to be capable of counteracting any force which acts upon it in a direction perpendicular to its surface. When, therefore, a body is supported on an Inclined Plane by a force directly applied to it, the case is that of a body kept at rest by three forces, its own weighty acting vertically downwards, the power applied, and the resistance ot the plane, acting at right angles to its surface. 111. Definitions. In the annexed figure AB i« called th<; length of the Plane; AC >|^ is called the base; and B C, which is perpendicular to A C, is called the height of the Plane. The an- gle B A C is called the angle ot in- clination of the Plane. 1 1 2. Remarks. It may seem A that the Inclined Plane can scarcely be called one of the Mechanical Powers. - It is not obvious that it can be usefully employed like the others. But it will be seen that if we have to raise a body, we may draw it up an Inclined Plane by means of a Power which is less than the Weight of the body. &«^' THE INCLINED PLANE, SECTION I. 113 The Power acting parallel to the Plane. 113. To find the. ratio of the Potver to the Weight when there is equilibrium on a smooth Inclined Fiane tfu power acting parallel to the plane. ' R JP Tne body is kept at rest by three forces, P actin? paralle to the plane, W, the weight of the body, acting vertically downwards, and R the reaction of the plane In the vertical line through the centre of gravity of the ^u^' ^^^^a"^ !^"^^ ^^' *^^ ^^"g*h of the plane. Through d draw be parallel to the direction of P meet- uig the direction of R produced backwards. ' In the triangles abc, ABC, we have the side ab equal to the side A13, the right angle acb equal to the right angle ACB, and the angle bac equal to the angle BAC each being the complement of the angle Aab, therefore* A oJ^"f^^^^ ^^^ ^^ ^^"^^ *" ^^^^ ^^spec' to the triangle ABC (Euc. I. 26), and its sides are respectively parallel to the directions of the three forces, P, W, R, and therefore proportional to their magnitudes. Hence, by the Triangle of Forces, we have W a^ "" AB ~ length "* "T* In other words, Ui V n' Hh ■! 114 ElEAlENTARy STATICS, If in the inclined plane the power be applied parallel *.o the plane, the Power is to the Weight as the height qf the flane is to its length. P_ __ _^ __ BC __ height " ~ AC "" Also, R ao base Ex. I. A weight, W, is supported on an inclined plane, the inclination of which to the horizon is 30** , find the Power and the pressure on the plane. In the preceding figure let the angle BAG = 30**. Then if AB := 2, BC will = 1, and AC = V3- W " 2 Hence And w p = — . P _ 1 R ^ V3 /.R =P/3 W EXERCISE I. 1. A force of lolbs. sustains a weight upon an inclined plane rising 2 in 5 ; the force acts parallel to the plane ; required the weight.* 2. A railway train weighing 67,2ooIbs. is drawn up an inclined plane rising ift. in 60, by means of a rope and a stationary engine \ find what number of Rs., at least, the rope should be able to support. 3. What power will be necessary to keep a weight of Sottas, in equilibrium when the length of the plane is 6ft. and the height of the plane ift. 6 in. *An inclined plane is said to rise 2 in 5 ; if when its height is 2, tts length is 5. THE INCLINED PLANE, "5 4. If a weight of 1 ot)S. is placed on an inclined plnne whose base is i6ft. and height i2it., and is attached by a string to an equal weight hanging over the top of the plane, find how much must be added to the weight on the plane that there may be equilibrium. 5. The length of an inclined plane is 5ft. and the height 3ft. ; find into what two parts a weight of i04fi)s. must be divided so that one part hanging over the top of the plane, niay balance the other resting on the plane. 6. What is the vertical height of an mclined plane when the length is aft. 6in. and a body weighing 96ft)s. is kept in equilibrium by a power of 2oBbs. ? 7. What is the leng'.h of an inclined plane when the vertical height is 3ft., the weight of the body 56Jbs., and the power i6fcs. ? 8. If a force ff 4olbs., acting parallel to the length, sustains a weight of i;61bs. on i.n inclined plane whose base is 340ft., find the height and length of the plane. 9. What weight is that which it would require the same exertion to lift as to sustain a weight of 4R)s. upon a plane inclined at an angle of 30" to the horizon. 10. What force will sustain a weight of 561bs., on a plane inclined at an angle of 45* to the horizon ? Find also the pressure on the plane. 11. Two planes, having the same height, are placed back to back, and two weights of 7 ounces and 10 ounces respectively, connected by a string passing o\^r the summit, are in equilibrium upon them ; find the ratio of the lengths of the planes. 12. Two inclined planes, of the same height, one of which is 8ft. long, and the other 5ft., are placed so as to slope in opposite directions and so that their summits coincide. A weight of 20 ounces rests on the shorter plane, and is connected by a string passing over a pulley at the common summit of the two planes with a weight resting upon the longer plane; how great must this weight be to prevent motion ? is 2, Ii6 ELEAfENTARY STATICS, it ■"" r SECTION II. The Power acting parallel to the base. 114. To find the ratio of the Power to the Weighty lit^en there is equilibrium on a smooth Inclined Plane^ tk', powei' acting parallel to the base. The body is kept at rest by three forces, P acting parallel to the base of the plane, W acting vertically downwards, and R the reaction of the plane. In the vertical line through the centre of gravity of the body take ac equal AC, to which the force is parallel j through c draw ah parallel to P, meeting the direction of R produced backwards. In the triangles acb^ ACB, we have the side ac in the one, equal to the side AC in the other, the right angle acb equal to the right angle ACB, and the angle cab equal to the angle CAB, each being the complement of. the angle Aas. acting horizontally ; what is the inclination of the plane ? 7. The angle of a plane is 45° ; what weight can be supported by a horizontal force of 3 ounces and a force of 4 ounces parallel to the plane, both acting togetl.er ? 8. A weight of 2oBbs. is supported by a power of i2lbs. acting along the plane ; show that if it were required to support the same weight on the same plane by a power acting horizontally, the power must be increased in the ratio of 5 to 4, while the pressure on the plane will be increased in the ratio of 25 to z6. *i< ii8 £LEM£ArAJi Y STA TICS. 9. If a power which will support a weight when acting l)arallel to the plane be half that which will do so acting horizontally, find the inclinatif n of the plane. 10. If R be the pressure on the plane when the power acts horizontally, and R' when it acts parallel to le plane, show that RR =s W». THE WEDGE, 119 THE WEDGK V. 115. Introduction.— The Wedge is a solid triang- ular prism, formed of some hard material such as steel. Its section made perpendicular to its length is generally an isosceles triangle. Its use is to separate the parts of a body ; this is effected by introducing its edge between them and then applying a force to its back sufficient for the purpose. 1 16. To find the ratio of the power to the resistafueinan isoscdes wedge. Let ABC be the section of an isosceles wedge intro- duced into the cleft DFE. The resistance on each side of the wedge will be the same. We may call this resis- tance ^W, since the whole resistance corresponds to what we have in other machines called W. Since the wedge is smooth its action must be perpendicular to its surface ; the forces ^W, therefore, act perpendicularly to the sides of the wedge. Let the power P act at the * point H, the centre of the back of the wedge. The directions of these three forces when produced will meet in a point G (art. 34). They may therefore be considered as three forces in equilibrium acting upon a I Il'„l( m 1 20 ELEMENTARY STATICS. point. The sides of the triangle ABC are respectively perpendicular to the directions of these forces, and there- fore proportional to them (art. 32) ; hence we have P AB or }W W BC JAB^ BC' that is, the power is io the total resisianee as half the back of the weitge is to the side of the wedge. 117. Remark. — The power usually employed with the wedge is not a pressure, but a blow, and as the pro- portion between a blow and a pressure cannot be well defined, the theory for calculating the power of the wedge is of little practical importance 7 HE SCREIK X2I THE SCRE^A-. VI. tt8. Introduction. This mechanical power is a combination of the lever and the incHned plane. It consists of a cylinder with a continuous projeccing thread wound round its surflice. This cylinder n^orks in a block m which there is a groove correspouding to the projecting thread; or the cylinder may ha/e a groove and the block a projecting thread. The thread or groove is inclined at every point at the same angle to the axis of the cylinder. The thread of the screw may be conceived to be gen crated by wrapping an inclined plane round a cylinder. Let a piece of paper be cut in the shape of aright-angled triangleABC; and let the side BC be placed on a pencil, or some other cylindrical sur- face, so as to be parallel to the axis of the cylinder. Let the paper be wrapped round the cylinder, the AB will mark out a continuous di scending path, which will be th thread of the screw. The screw is generally worked by means of a bar o- arm inserted into the head of the cylinder perpendicular to its axis. One revolution of the cylinder will cause the screw to be pushed through the block or nut a dis- tance equal to that between two threads of the screw, measured parallel to the axis of the cylinder. if Ml ■ if s I 122 ELEMEN TAR Y STA TICS, 119. To find the ratio of the Power to the Weight in the Screw. Suppose the screw to be vertical and used for raising a weight W. Perpendicular- ly from the axis of the screw extends an arnn, at the extremity of which the power P acts, P being at riiiht angles both to this arm and to the axis of the screw. The machine is kept in equilibrium by W acting vertically down- wards, P acting horizontal- tally, and the reaction oi the groove at points whcit the thread is in contact with it. The groove being supposed smooth, these re- actions are at right angles to it. Let now a portion of the thread be conceived to be unwrapped from the cylinder; , let A B, the hypothenuse of a right-angled triangle ABC, be such a portion, BC being the vertical distance betvveen con- tiguous threads, and AC the circumference of the cylinder. Let Rj be the reaction of the groove at any point a. Construct the triangle abc equal in every respect to ABC. Then the vertical and horizontal components of Rj are BC I.e. R ^ ^AB' Ri. AB* Suppose the other reactions R2, resolved ; then, since the two angles to one another form, R3 similarly sets of forces at riglit parately, systems in equi- a r t I t] w is A Th'E SCREW. 123 libriurn, W will be supported by the vertical, and P will have "^ ^^'^ horizontal components. Hence, we AC W = ^(R. +R. + R,.. )...,(,, Taking moments about the axis, we have BC AB Dividing (i) by (2) we have _W _ AC^ P.« - BCTr ' W AC -jT = — . « -f BC ^•^ = — C^^+^a+Ra )'-....(2), 2TCa circumference of Power distance between contiguous threads' .n^*'/^ o"^ ^^'^^"^ ^^ *^^ '^'"^^ ^^' inclined at an angle ot 30 to a transverse section of the cylinder whose radms is 9 inches the length of the lever whTch turns i68oBbT'' ^'^"^ ^^'' ^"^ ^^^' P^^^^^ ^^^^ «"^tah the'Xlyhnd^^:""'""" "' " '""'""' ^^^''^" ^' then if BAC = 3o^ BC will = d; and since the radius IS 9 inches the circumference AC = 2n,^. *If c be the circumference of a circle, then-^ ^^-dheCreek 314159. zr P) ? ' if t.i« 124 ELEMENTARY STA TICS. Therefore EC = *< /I. w Hence 1680 2^.48 ~ 16' therefore p 1680^3 16 = 105^/3 — i8i.865tt)s. EXl'lRCISE I. 1. Describe the common screw and explain its prin- ciple. Does friction assist the power or the weight ? How is the efficiency of a screw affected by increasing (i) the length of the arm, (2) the diameter of the screw, (3) the distance between the threads ? 2. The distance between the contiguous threads of a screw is 2in., and the arm at which P acts is 20 in. ; determine the ratio of P to W. 3. The circumference of the circle described by a lever working a screw is 2ft. pin., the space between the threads is J of an inch. If a force of i3lbs. be applied to the lever what will be the pressure of the screw ? 4. A carriage weighing ii2ott>s. is raised by a screw. The threads of the screw are \ of an inch apart, and the lever is 16 inches. What power is applied to work the lever? . 5. The diameter of the circle described by the power is 2in., and the distance between the threads is J of an inch ; find the mechanical advantage. 6. The length of the arm of the power is 15 in. find the distance between two consecutive threads of the screw that the mechanical advantage may be 30. THE SCREW, «25 7. What must be the distance between the threads of a screw so that a pouer of 281bs., applied at the whth^l''llt'/K''^"'^ ^'5^'!" *^« '^^^^^« o^a screw Which IS Tvorked by an arm /, when the power applied to Its extremity ,s an mh part of the pressure on the^screir! i! I.i' » CHAPTER IX. VIRTUAL VELOCITIES. As applied to Machines. T20. Introduction. There is another principle from which the theory of equilibrium in machines has been derived, but which is neither so evident in itself nor capable of such an easy demonstration as the principle of the composition and resolution of forces, and the principle of moments. It has, however, the advantage of being much more conveniently applied to questions of equilibrium of complicated machinery. It js called the Principle of Virtual Velocities. 121. Def of Virtual Velocity. If a machine is in equilibrium under the action of a power P and a weight fV, and we suppose it to receive any displacement, con- sistent with the connection of its various pat ts, then the spaces described by P and PV, cs *imated in their respective direc-tions, are called the Virtual Velocities of the power and 'iveis,ht. 12 2. Remarks. The word virtual is used to indi- cate that the motion referred to, does not really take place, but is only supposed to take {)lace. The word velocities is used because we may conceive the points of application of P and W to move into their new positions in the sunie time, and then the lengths of their paths described will be proportional to their velocities. mmim VIRTUAL VELOCITIES, 127 123. Principle of Virtual Velocities. The Principle of Virtual Velocities asserts that in any machine the Power multiplied by its virtual velocity is equal to the Weight multiplied by its virtual velocity. This important principle is only a particular case of a far more general one which we cannot here examine ; nor can we even give a general proof of the principle as stated above. All that we shall be able to do is to prove that the principle holds good in those C£\?es of equili- brium which we have already established by other methods. 124. To apply the Principle of Virtual Velocities to find the ratio of the Power to the Weight in the lever. O B W |/i>' indi- take word Ints of sitions paths Suppose the lever to be a straight lever AB, having its arms AO=dr, and B0=^, and to be acted on bv forces P and W perpendicular to the arms. Let the lever AB be turned vertically through a right angle into the position A'B', P and W remaining const- ant in direction ; then, the virtual velocity of P is OA' or a, and I. If W is OB' or d. By the Principle of Virtual Velocities we have P X tf = W X ^, the condition of equilibrium established by the principle of moments. w taS ELEMENTARY STATICS, In a similar nianner the principle may be applied to find tH ratio of the power to the weight in a lever of lecond or third order. Hence, the Principle of Virtual Velocr^^s holds in the case of the Lever. nf The Wheel and Axle. Conceive the ma- chine, lo make one complete turn ; then, th h virtual velocity of P =: circumference of wheel, H II W = It axle ; and the Principle of Virtual Velocities asserts that P > circumference of wheel = W x circumference of axle. liivide both sides of this equation by 2n', and P X radius of wheel = W x radius of axle, tbA result obtained in Article 99, and hence the Prin- ci^«4e holds in the case of the Wheel and Axle. 126. The Pulley. In applying the Principle of V xtual Velocities to pulleys, we suppose the weight to be r;ised through a small space, which space will be its v'ftual velocity, and the corresponding space through f 'hich the point of application of P must be moved will he the virtual velocity of P. 127. The Single Movable Pulley. In the figure Article 104, suppose W raised through i inch, the string on either side of the pulley B will be shortened by the same quantity ; consequently P will descend through 2 inches ; hence, the virtual velocity of P = 2 inches, II M W = I inch \ and the Principle of Virtual Velocities asserts that P X 2 = W X I, the condition found in Article 1 04. • 128. The First System of Pulleys. In the figure Article 105, let VV be raised through i inch, then the pulley C rises through 1 inch, the pulley B, through 2 inches, the pulley A through 2x2, or 2* inches, and \ toe VIRTUAL VELOCITIES, 129 so on ; hence, if there were n movable pulleys, the «th l)ulley will rise through 2 inches, and P will descend through 2" inches ; hence, W's virtual velocity = i in., P's n II = 2** in. ; and the Principle of Virtual Velocities asserts that P X 2« = Wx I, which is the result obtained in Article 105, and, there- fore, the Principle holds in the case of first system of pulleys. 129. The Second System of Pulleys. Tn the figure Article 107, let W be raised through i inch, then if there were // strings between the two blocks, each of them will be shortened by i inch ; consequently P will descend through n inches : hence, W's virtual velocity = i in., P's II II = « in. ; and by the Principle of Virtual Velocities we have P X « = W X I the condition obtained in Article 107. The Principle is, therefore, true in the second system of pulleys. 130. The Third System of Pulleys. In the figure Article 108, let W be raised through i inch ; then the highest movable pulley descends through i inch. The next pulley descends through 2 inches in conse- quence of the descent of the pulley above it, and through 1 inch in addition, in consequence of the ascent of the weight; hence on the whole its descent will be (2 + i) inches. The next pulley descends through twice this, in consequence of the descent of the pulley above it, and through I inch besides in consequence of the ascent of the weight; therefore its whole descent will be j2(2 + i) + ij inches, that is (2^ + 2 + 1) inches. In a similar manner it may be shown that P descends through (2^ + 22 + 2 + 1) inches, hence, the virtual velocity ofP = (23 + 22 + 2 + 1) inches = (2* — i) inches, mod H W = I inch; IK IJO ELEMENTARY STATICS, f and by the Principle of Vir i al '/elocities we have p X {2^ — i)=. W < I, the result obtained in Article lo '. 131. The Inclined Plane. In the figure Article 113, let W be at A and suppose it drawn up to B ; t^^n P's displacement in its own direction is the length 01 Plane. Now W's direction is vertical, and its dispL ce- ment in that direction is evidently the height of the plane ; hence P's virtual velocity = AB, W's II = BC; and the Principle of Virtual Velocities assert that P X AB = W X AC, the condition obtained in Article 113 ; hence the Prin- ciple holds in the case of the inclined plane, when the power acts parallel to the plane. In a similar manner it may be shown that the Principle holds when '? power acts parallel to the base. 132. The Screw. If the arm on which P acts be made to describe a complete revolution, the weight will be raised or depressed through a space equal to the vertical distance between two threads of the screw; hence, P's virtual velocity = circumfer. of circle described by P, W's II = vertical dist. between two threads ; and the Principle of Virtual Velocities asserts that P X circumference of circle described by P = W x dis- tance between two threads, which is the result obtained in Article 119. The Principle, therefore, holds in case of the screw. 133. Remarks. We have thus shown that the Princi- ple of Virtual Velocitie%^^ holds good in all the simple machines, since it always leads to correct results ; we may, therefore, safely conclude that it holds good in any com- bination of these machines. We have, therefore, a princi- ple of great beauty, and very easy of application which VIRTUAL VELOCITIES. 131 P, Lds ; I ici- ^ple lay, )m- ici- ^ich brings all cases of equilibrium in machines under one general law, and enables us to find the ratio of the power to the weight in machines of the most complica- ted construction. If P be always supposed to describe the same space in the same time, it will easily be seen that " what is gained in power in any of the above machines is lost in time." This is sometimes stated as a mechanical prin- ciple. EXERCISE I. . I. Explain the term virtual velocity. State the Prin- ciple of Virtual Velocities, and show that it holds good in a lever of the third order. 2. State the relation between two weights v*hicii bal- ance one another on a given wheel and axle. Prove that the Principle of Virtual Velocities is true in this instance. 3. Show that the Prin- 'ple of Virtual Velocities holds in the case of a body 1. equilibrium on a smooth in- clined plane, the power aci s., rises 5 J ft. What is the smallest force that can be em- ployed to turn the wheel ? H^K 133 ELEMEHTAR Y STA TICS. I. • . llil n. 8. In the first system of pulleys the distance of the highest pulley from the fixed end of the string which passes round it is i6ft. ; find the greatest height through which the weight can be raised, there being four pulleys. 9. In the second system of pulleys, how much cord would be required for a man to raise himse'*' 30ft., there being three pulleys in the lower block ? 10. Employ the Principle of Virtual Velocities to find the weight which ilb. will support by means of two blocks of pulleys, of which the upper one is fixed, and the lower one, weighmg ilb., is movable, each block containing 4 pulleys, and the portions of string between the blocks being all vertical. 11. A weight of 2538)8. is supported on an inclined plane rising Soft, in a length of 137ft. by two equal forces one being horizontal and the other parallel to the plane ; find the forces. 12. Two weights hang over a pulley fixed to the summit of a smooth inclined plane, on which one weight is supported, and for every 3 inches that one is made to descend, the other rises 2 inches ; find the ratio of the weights and the length of the plane, the height being I ft. 6in. 13. If a power of ift describes a revolution of 3 feet whilst the screw moves through ^in., v*^hat pressure will be produced ? CHAPTER X. 1^' Hi I.V FRICTION. 134. Def. of Friction. In the preceding investigations we have supposed that all surfaces are snaooth. Prac- tically this is not the case. When we attempt to make the surface of a body move upon that of another, with which it is kept in contact by pressure, there is, in general, a resistance to motion ; this resistance is called friction. 135. Direction in which Friction acts. Friction at any point of a surface acts in the direction exactly contrary to that in which motion would occur if the surface were smooth. Thus, if a body on an inclined plane, under the action of any force, is on the point of ascendingj the force of friction acts downwards ; but if the body is on the point of descending^ the action of friction is upwards. Hence a given weight may be sus- tained by a less power than is required independently of friction, but requires a greater power to move it. 136. Laws of Friction. The following laws, which have been established by experiment, apply to a body which is on the point of moving on the surface of ano- ther body, and is only prevented from doing so by fric- tion. I. The friction between two surfaces of the same kind is in direct trotortion to the pressure between them. 134 ELEMENTARY STATICS. 1 II. The amount of fridion is independent of tfi€ extent of the surface in contact, so long as tJn pressure remains the same, 137. Coefificient of Friction. Let R be the per- pendicular pressure between two surfaces, F the friction, then by the first Law, tlie ratio of F to R is a constant number for two given substances. This number is called the coefficient of friction and is denoted by A (the letter ;;/ of the Greek alphabet), that is F -— - = //, and therefore F = /< R. R 1 38. To find the coefficient of friction between Itvo given substances. Let one of the substances form an inclined plane AB, movable about the point A, and let a block of the other substance, having a plane base, be placed upon it. Let the plane AB be elevated till the body is on the point of sliding down it. The body is then supported on the plane by the three forces F, W, and R ; hence we havo • F BC , -^=^^, -Wart. 137). The value of this coe^cient has been found by ex- periment for various substances ; thus, for metals ©n metals it is about .17 ; for wood on wood it has been found to be about .33. The angle BAG is called the angle of repose, and some- time the /imitin^q angie offrictiofi. FRICTION. EXAMPLES. '35 I. A body is jtist on the point of sliding on a rough frktbn ^' "'^' ^ '" ^ ^"""^^^ ""^ ^ ' ^""^ ^^^ coefficient^ of In_the preceding figure let AB=:s and BC=3, then F w^ ^^i^T, ''J'^P^ ^' '^'^ °^ '^^ P^^^e by the forces 1*, w, and R; hence we have j;^ __ BC _ 3 K "" AC ~ T "" '"■ The coefficient of friction is, therefore, ~ 4 2. What force, acUng paraUel to the plane, will be re- ^Zl^Tt^l ^'^^"^'^ ^'^^^ ^f ti'^b^r "P ^ Pjane in- n ?^.f K^- """.T ^^^" ^"«^« ^f^'^^ the coefficient of friction being .63, and the weight of the block 5 cwt. ? SC^s, Since tke body is on the point of moving up the ihe plane, friction acts downwards (art nc) The body is kept in equilibrium on the piane by the forces P, R, friction which is equal to Tr. and he weight, 5 cwt. ; of these forces P and ;.R a^e equ^vallnt to the single force P-^^R acting up the plane, °' 136 ELEMENTAHY STA 7/CS. Since the angle BAG = 60**, the sides AB, BC, AC, of the right-angled triangle ABC, are in the ration of 2, I and V 3 ; hence we have R 2 I .('). and — = — (2). 52 Substitute the value of R from (2) in (i), and for /n substitute its value .63, and we have P— .6 ^xi- = 5V3 2 .-. P = 4 (1/3 + -(^3) ^m f ■ - ; J i| = -?- (1.732 + .63) 2 = 5.9 CWt. EXERCISE I. 1. What is meant by frirtion, and by the angle of repose ? 2. What is meant by the coefficient of friction ? How is it related to the limiting angle of resistance ? 3. What are the laws of friction ? How may the co- efficient of friction for different substances be determ- ined ? 4. What are the forces acting on a body, which stands at rest on the side of a hill ? 5. A ladder AB has one end A on a rough horizontal road, and the other end B, against a rough vertical wall ; what are the forces acting on it ? 6. If the coefficient of friction is — , show that the in- 3 clination of plane is 30**. 7. A body rests without support on a plane inclined to the horizon at an angle of 45° ; what is the coefficient of friction between the body and the plane? IV FRICTTOir, «37 lands /all le m- td to It of 8. A plane rises 5 in 13 ; find the greatest weight which can be sustained upon it by a power of 2oIbs. acting along the plane, the coefficient of friction being J-. 9. Find the least force which will sustain a weight of ioK)s. upon a plane rising 7 in 25 when the force acts along the plane, and the coefficient of friction is J-. 10. Find the least force which will draw a weight of iofi)s. up a plane rising 1 1 in 1 22, when the force acts along the plane, and the coefficient of friction is ■^^. 11. A body placed on a horizontal plane requires a horizorital force equal to its own weight to overcome the friction ; supposing the plane gradually tilted up, find at what angle the body will begin to slide. Ex. 3. A ladder rests against a vertical wall to which it is inclined at an angle of 45"; the centre of gravity of the ladder is \ the length from the foot. The coefficient of friction for the ladder and plane is J, and and for the ladder and wall J. If a man whose weight is half the weight of the ladder ascends it, find to what height he will go before the ladder begins to slide. j Let AB, the ladder, = 3^ ft. in length. II AC = ic, be the height to which the man has ascended before the ladder begins to slide. The weight of the ladder, W, acts at D, its centre of gravity, AD bein^j equal to a. \ 138 ELEMENTARY STATICS. When the man has ascended to E, the ladder is kept in equilibrium by the forces as indicated in the figure. Since the forces are in equilibrium, we have and therefore R' = — R, 3 W R + JR'=: — - + Wj R' = 2 6W 13 .(I). Take moments about A, and we have — X — + VV X — = R' X A- + — R' X ±-. 2 ^2 y'2 ;/2 4 ^2 or [■Wii=R'x — ^ — 6W 13 4 iSa , from (I) ; 19 19 19 13 39 39 the length of the ladder. of EXERCISE II. 1. A uniform ladder resting at an angle of 45** between a smooth vertical wall, aiid a rough horizontal plane, is just supported by the friction of the ground ; find the coefficient of friction. 2. A ladder inclined at an angle of 60"" to the horizon rests between a rough pavement and a smooth wall of a house. Show that if the ladder begins to slide when a man has ascended so that his centre of gravity is half way up, then the coefficient of friction between the foot of the ladder and the pavement is ^y'3. 3. A ladder 10 feet long, rests with one end against a smooth vertical wall, and the other on the ground, the coefficient of friction being 5 ; find how high a man. FRICTION. >39 of whose weight is three times that of the ladder may ascend before the ladder begins to slip, the foot of the ladder being 6 feet from the wall. 4. A beam 21ft. long rests with its ends upon a hori- zontal and a vertical plane, respectively ; find how far the lower end may be drawn out from the wall before the beam will slip ; the coefficients of friction being \ and J, respectively. 5. A uniform ladder is placed between a rough hori- zontal plane and a rough vertical wall, at an angle of 45^ the coefficient of friction between the ladder and the ground being \ ; a man whose weight is half that of the ladder ascends ; find what the coefficient of friction must be between the ladder and the wall, that when the man reaches the top of the ladder it may just begin to slip. 6. A ladder which is divided by its centre of gravity into segments of loft. and 20ft., is placed against the side of a house at an angle of 30^ ; find the highest round upon which a weight of 4cwt. can be suspended ; the weight of the ladder being 3cwt., and the coefficient of friction between the ladder and the wall, and also be- tween the ladder and the horizontal pavement which supports its heavier end being —. 45" mtal nd ; lizon lof a pn a half Ifoot 1st a the lan. CHAPTER XI. The application of Similar Triangles to the solution of statical problems. Introduction. Some of the questions set to candi- dates for first class certificates require a knowledge of a few propositions in Euclid B. VI. The following are the most important; : — If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those sides pro- duced proportionally ; and if the sides, or the sides pro- duced, be cut proportionally, the straight line which Joins the points of section, shall be parallel to the remaining side of the triangle. Euc. B. vi. 2. Similar Triangles. Two triangles are said to be similar when they are equiangular. The sides about the equal angles of similar triangles are proportionals ; and those which are opposite to the equal angles are antecedents or consequents of the ratios. (Euc. B. vi. 4). EXAMPLES. I. A. uniform beam AB, 12 feet long and weighing 4ott)s., rests with one end A at the bottom of a vertical wall, and a point C, 2ft. from the other end,is connected by a string CD to a point D, 8ft. above A; find the reaction at A, and the tension of the string. APPLICA TION OF SIMILAR TRIANGLES, 141 The beam is stf^ported by three forces — 1. T, the tension of the string along CD. 2. W acting vertically through G, 6ft. from A, its direction meeting CD in E. 3. The reaction R at A along AE. Since G is the middle point of AB, AG=6, and GC=4 ; DC is, therefore, cut in the point E in the ratio of 6 to 4 (Euc. vi. 2). Hence, DE = — x 6 = 3. 6 feet. 10 And DE» +AD8 = AE^ ; /. AE = 8.77. From the triangle of forces, we have T DE or W 40 .-. T AD' 8 ' i8fi)s. Similarly R = 43.81bs. 2. A rod 8 feet long, the weight of which has not to be considered, is placed across a smooth horizontal rail, and rests with one end against a smooth vertical wall, the distance of which from the rail is i foot ; the other end of the rod bears a weight of i2lbs. Find the posi- tion of equilibrium and the pressure on the rail. The three forces acting on the rod AB, are the weight i2ft)s., at B verti- cal, the reaction of the wall at A hori- zontal, and the pressure on the rail C perpendicular to AB. Since the di- rections of the first two meet in E, the third must be along EC. Hence the position of equilibrium is such that EC IS perpendicular to AB. I » ir Mi 14a ELEMENTARY STATICS. h"' ■ Draw CD horizontal. Then the triangles ACD, ACE, ABE, are similar ; hence we have (Euc. vi. 4) CD CA AE CA ~ AE " AB' therefore CA2 = AE.CD — AE, since CD = i. And AE« = CA.AB = 8CA, since AB = 8. therefore CA = 2, and AE = 4. Therefore the angle CAE = 60° (art. 28). Again the triangle EAB has its sides perpendicular to the directions of the three forces acting on the rod, and therefore proportional to them (art, 32). Hence if P be the pressure on the rail, we have P 8 3. A weight pended from a is attached to rests at D on 12 4 ■ ,*. P = 24lbs. W, of 34lbs, sus- string MD, which a fixed point M, an inclined plane CA, whose base AB is 39 inches ; and its vertical height CB, 26 inches. The position of D is such that BD is at right angles to CA ; and the point M is in BC produced, CM being 8 inches. Prove that the tension of the string MD is 2oK)s. ; and find the reaction of the plane on the weight. First Class Exam.f July^ 1^73' The weight W, is kept at rest by the three forces — 1. The tension of the; string T, acting along DM. 2. The weight W, acting vertically downwards. • 3. The reaction R, at right angles to the plane and therefore in BD produced. Hence the triangle BDM has its sides parallel to the three forces and is, therefore, proportional to them. I t ^ c E Since AB == 39 Since the triangi ' liave we lia or Also or SIMILAR THIAA and BC = 26, • ^^'D is similar 5C_CB CB - AC' DC GLES, AC »43 31/13. to the triangle ACB 26 DC 2^ ^3i/i3 4i/i3- Bp_ AB JBC"~^AC' BD 26 BD = 39 ^ZV^Z 6^13. Draw DE parallel to ABT snnilar to the 'J'he triang/e CDE AI30 And tmngle CAB; hence behave PE_ AB ^ I^C -AC' .*. DE =12. CE__CB ED - AB ^ •. CE = 8 DM2 = DE2+EM> IS DM 122 + i62 20. From the triangle offerees, we have or And or i= DM W MB y 20 < T = 20. -^ = -22 W BM' • ■ 3^ 34 R V13 34 ' 144 ELEMENTARY STAl ICS. ■ % i t. M < I .-' >' We might have obtained the preceding results without using similar triangles. I'ind AC as before. Then AC X BD =AB X CB, since each proJuclisdoubletriangle ABC; .-. BD -- 6^13. Again, in the triangle BCD, we have DC« = BC2 ~ BD8 i .'. DC = 4i/i3- And DExBC — BDxDC, each product being double triangle BDC; .-. DE= 12. Also EC* = DC* ~ DE« ; .-. EC = 8. And DM = 20, as before. EXERCISE I. 1. If two forces P and W sustain each other on the arms of a bent lever PCW, whose fulcrum is C, and act in directions PA, WA, which form the sides of an isos- celes triangle PAW ; show that if AC be joined, and pro- duced to meet PW in F, P : W : : FW : FP. 2. If a weight be suspended from one extremity of a rod movable about the other extremity A, which remains fixed, and a siring of given length be attached to any point B in the rod, and also to a fixed point C above A m the same vertical line with it, then the tension of the siring varies inversely as the distance AB. 3. Three forces act on a particle, their d' parallel to the three perpendiculars di angles of a triangle to the opposite sides, a i thei; nitudes inversely proportional to these perpendi' ilars ; show that the three forces are in equilibrium. 4. A uniform beam AB, whose weight is W, and length 6 feet, rests on a vertical prop CD equal to 3 feet ; the other end A is on the horizontal plane AD, and is prevented from sliding by a string DA equal to 4 feet ; find the tension of this string. o he .ag- wimmm SIMILAR TRIANGLES. 145 and 'eet ; id is feet; 5. Find the centre ofgravityof a conical shell con- tained between two riglit circular conical surfaces, hav- ing the same axis, the outer diameter of the shell being 8 inches, the inner diameter 6 inches, and the height of the whole 1 2 inches. 6. A uniform beam AB whose weight is 5oIbs., rests with one end B on the top of a vertical wall, the height of which is half the length of the beam, the other end A is on the horizontal plane Ai), and is prevented from sliding by a siring DA equal to j of A R • find the ten- sion of the string. ^^ 7. A 9-pounder gun, weigh- ing i3cwt. 2qrs. is suspended at the extremity B of a spar AB, which is supported by a stay BC, fixed at C ; find the tension of the stay and the pressure on the sgar when AB = 13 feet, AC = 10 feet, and AD, the horizontal distance of the gun from A, = 5 feet. 8. A gun, to be lowered over the parapet of a fortress, is suspended from the extre- mity C of a spar AC, 10 feet long, held in its position by a slay BC 15 feet in length, the distance from A to B being 10 feet; find the ten- sion of BC and the thrust upon AC, when AB is horizontal, and the weight of the gnn33cwt. 9. Two weights, connected by a string passing over the common vertex of two inclined planes, balance each other ; if they are set in motion, show that their centre of gravity will move in a horizontal straight line. ID. A ladder 10 feet long, rests with one end against 1 smooth vertical wall, and the other on the ground, the coefficient of friction being .1; ; find how high a man, whose weight is three times that of the ladder, may ascend before the ladder begins to slip, the foot of the ladder being 6 feet from the wall. ii [ff-1 \ i I m CHAPTER XII. Examination Papers in Statics set to Candi- dates for First and Second Class Certificates. 1871. SPECIAL EXAMINATION FOR COUNTY INSPECTORS. 1. State the principle of the Pardlelogram of Forces. What is the magnitude of the resultant of five forces acting on a particle, and represented respectively in magnitude and direction by the sides of a regular penta- gon taken in order ? 2. Draw a system of pulleys in which the relation be- tween the power and the weight is expressed by the formula W = P x 2** , ;/ being the number of movable pulleys. (2) Should the weight of the pulleys be taken into account, what is the relation between W and P, when there are four movable pulleys in the system, and W the weight of Cuch is — ? IS 3. Enunciate the principle of Virtual Velocities ; and assuming the principle, apply it to find the relation be- tween the power and the weight in the screw. 4. On a plane inclined to the horizon at an angle 01 39", a weight W is supported by a power P, acting parallel to the plane. Show that a power equal to 2P, acting parallel to the base, would be required to sustain W on a plane inclined to the horizon at an angle of 45". EXAMINA TION PA PERS, HI 5. One end of a uniform beam whose weight is loolbs., rests on a horizontal plane, and in a vertical plane pas- sing through the beam. Find the reaction of the ground against the beam, and the amount of friction. £iEGOND CLASS CERTIFICATES. July 1C71. 1. In a straight lever CBA, a weight W of lolb. acts at B, a distance of 5 feet from the fulcrum A ; and the Power P is applied at C, on the same side of the ful- crum as W, but in an upward direction. If AC be 1 2 J feet, what is P ? And what is the pressure on the ful- crum ? 2. Assuming the principle of Virtual Velocities, de- duce therefrom the relation that must subsist between the Power and the Weight in the lever, in order that there may be equilibrium ? 3 At points A, B, C, which are in the same straight line, v/eights of 81bs., i2lbs., and 2olbs., respectively, are placed. If AB be 12J feet, and BC be 5 feet, it is required to find the centre of gravity of the three weights. 4. (a) In a system of pulleys where each pulley hangs by a separate string, what power will sustain a weight of I o4tt)S., there being two movable pulleys in the system? ij?) If the weight of each of the pulleys be ijlbs., ijy what power will the weight of io4lbs. be sustained, ihe weight of the pulleys being taken into account ? 5. A weight of 8olbs. is sustained on an inclined plane by a power of 6oIbs. acting parallel to the base. How many feet does the plane rise in the hundred ? 6. (a) One end of a uniform beam rests on the ground, the otlier being supported in the hands of a man who exerts a pressure which is at right angles to the length of the beam, and in a vertical plane passing through the beam. Mention the different forces by which the beam is kept at rest. (J?) If the weight of the beam be 2ooflbs., and its inclination to the ground 60", what f?rce does the man cxerf? I' I 148 ELEMEMTARY STATICS. m %- \ ■■>!'' h t" FIRST CLASS CERTIFICATES. JULV 187 1. 1. (a) Enunciate the principle of Virtual Velocities. {d) Assuming the principle, deduce the relation that must subsist betvreen the Power and the Weight in the inclined plane, in order that there may be equili- brium ; pointing out precisely the quantities which repre- sent the virtual velocities of the Power and the Weight respectively ; tb^ Power being supposed to act parallel to the plane. 2. In a straight lever, the lengths CA, CB, measured to the left of the fulcrum C, are 6 and 10 feet respect- ively ; and CD, CF, measured to the right are 5 and 8 feet respectively. Mention any weights which, sus- pended from A, B, D, E, respectively, will keep the lev("r at rest. 3. A balance has one of its arms 6 inches long and the other 6^ inches. A package of tea is weighed out to a purchaser in the one scale against the weight of 2K)s. in the second scale ; and another package is weighed out td'Hum in the second scale against a weight of zlbs. in the first scale. How much tea does he get altogether? 4. A weight W is just sustained on an inclined plane by two forces, one of plbs. acting parallel to the base and toward the plane; the other of 2ilbs. acting parallel to the plane, and in an upward direction. The base of ])lane being 40 feet, and its perpendicular height 30 ieet, what is W ? SECOND CLASS CERTIFICATES. December 1871. 1. Describe the Wheel and Axle, stating the relation between the power and the weight necessary in order to equilibrium. 2. Describe the Screw, stating the relations between the power and the weight, necessary in order to equili- brium. 3. Show that the principle of Virtual Velocities holds in the Screw. EXAMINA TION^PAPERS. 149 4. A body, situated at A, one of the angular points of , the square ABCD, whose diagonal is AC, is acted on by- three forces, represented in magnitude and dirction by the lines AB, AC, AD, respectively. Draw a line repre- senting in magnitude and direction, a fourth force which, in connection with the three above-mentioned, would keep the body at rest. 5. A uniform straight lever ABC, whose fulcrum is B ; weighs lefts. The arm AB is 12 feet; the arm BC, 8 feet. Weights of 8 and i2lbs. are suspended from A and C respectively. Find the additional weight which must be suspended from C, in order to keep the lever at rest ; the weight of the lever being taken into account. FIRST CLASS CERTIFICATES. December 187 i. 1. In a system of pulleys, each of which hangs by a separate string, apply the principle of Virtual Velocities to determine the relation between the power and the weight when there is equilibrium. 2. In the previous question should each pulley weigh 2lbs., and should the weight of the pulleys be taken into account, what power would be required to sustain a weight of 4ofi)s., there being two movable pulleys in the system ? 3. ABCD is a square ; and, at a point E, in the diagonal AC, is placed a particle to which are attached strings that pass over smooth pulleys at the corners, A, B, C, D, of the table, and support weights represented in magnitude by the lines CE, DE, AE, BE, respectively. Show that the particle wiil be kept at rest NORMAL SCHOOL 1871. 1. A uniform straight lever ABC, whose fulcrum is B, has the arm AB 10 feet, and BC 8 feet. A weight of 2oS)S. hangs from the middle point of the lever, and another of 3oIbs. from the extremity A. What weight must be suspended from C to produce equilibrium ? If :. If 'SO ELEMEN7AR Y STA TJCS, 2. Describe the Differential Wheel and The radius of .the axles are 4 J and 5 inches, respectively. What power will susain a weight of 2ooolbs. ? 3. What weight will be supported on an inclined plane by a power of 6oIbs. acting parallel to the base ; the height of the plane being 7 feet, and the base 17 feet ? 4. Define the centre of gravity. How can it be shown that every system of particles he , a centre of gravity ? Where is the centre of gravity of two heavy particles? In the straight line AD take AB = 3 ft., BC = i ft , CD = I ft. ; and place at B, C, D, weights of 3lbs., 2lbs , lib., respectively. Then if d be the distance of the centre of gravity of these three weights from A prove that (1 + 2 + 3) ^ = (3«-o2) + (32- r 2) + {z^-2^). 5. A uniform beam AB, 20 feet long, and weighing 9ooIbs., rests in a horizontal position on equal upri'^ht props CA and DB. It is loaded at a point E, 5 feet from A; with a weight of ioo!bs. ; find the pressure on each prop. SECOND CLASS CERTIFICATES. July 1872. T. Describe the two kinds of levers, and give two examples of each kind. 2. In a bent lever, AFR, whose fulcrum is F, the arms AF and FB are straight, and inclined to one ano- ther at an angle AFB, which is such, that the perpendi- cular let fall from A on BF produced is equal to two- fifths of AF. Also FB = 2FA. When FB is horizontal, find the force, which, acting vertically at A, will balance a force of I oolt)s. act, g vertically at B. 3. Describe the Screw. 4. The base, AC, of an inclined plnne, AB, is 8ft. ; and the vertical height BC, 6 feet. A weight W, of loolbs. is sustained on AB by a force P, acting ])arallel to AB. Find the reaction of the plane against W (or the pres- sure of W on the plane in a direction perpendicular to the plane). "rn * EXAMINA TION PAPERS. 151 5, The triangle ABC has the angle ACB a right angle ; AC = 3 feet, and BC = 8 feet. From A let the straight line AD be drawn to D, the middle point ol BC ; and suppose that a particle at A is acted on by lliree forces represented in magnitude and direction by the lines AB, AD, AC respectively. Find the length of the line which represents the sum of the resolved parts in a direction perpendicular to AC ; and deduce the magni- tude of the resultant of the three forces. 6. In sliding friction what is the relation between friction and pressure? How does friction depend on the extent of surface in contact, and on the velocity of motion ? FIST CLASS CETIFICATES. July 1S72. 1. A straight lever ACB, whose fulcrum is C, is kept at rest by two forces P and Q, the former acting at A in the direction AP, at the latter at B in the direction BQ. The lines PA and QB produced meet in a point D. Show that, if the forces P and Q be represented in magnitude by the lines DA and DB respectively, CA is equal to CB. 2. On the same base AB, and on the same side of it, are described the equilateral triangle ABC, and the isosceles triangle ABD, which is right-angled at B. A })article at A is acted on by two forces, one in the direc- tion AC, the other in the direction DA (not AD) ; and these forces are represented in magnitude by lines which are 2 feet and 4/2 feet in length, respectively. Prove that their resultant is at right angles to AB. SECOND CLASS CERTIFICATES. July 1873. I. Tlie mechanical advantages of a machine, on which a weight W is sustained by a power P, being W understood to be the rati 3 -p-; and the machine being said to gain, or to lose, advantage, according as this i IJ2 Bt.EMtLNTAkY SfATtCS, m '/ If AS % \ ratio is greater or less than unity ; describe two kinds of straight levers, the one of which always gains, and the other always loses advantage. When mechanical advantage is lost, does the machine serve any good purpose ? If so, what ? Illustrate by a familiar example. 2. When a door is turned round its hinges by a force applied horizontally at the handle, what is the resistance which the force applied overcomes? If the weight of the door be increased, is the force required to turn the door increased ? If so, how does the increase of a vertical force render necessary the increase of a horizotidaliorct} 3. " A number of pulleys are attached to the same block, which supports a weight, aud the same string passes round all the pulleys." Draw a diagram repre- senting such a system, with three pulleys in the block ; and state (without proof) the mechanical advantage. 4. Enunciate the principle of Virtual Velocities ; and show how it holds good in the system of pulleys referred to in the previous question. 5. What is friction ? Inquire whether friction acts to the advantage or to the disadvantage of the Power, ^rj/, when the Power is on the point of raising the Weight ; and secondly, when the Power is just preventing the Weight from descend- ing. What is the coefficient of friction ? FIRST GLASS CERTIFICATES. July 1873. 1. A particle at E, a point on a square surface ABCI), is kept at rest by four forces, which are repre- sented in direction by the lines EA, EB, EC, ED, res- pectively ; and in magnitude by EA, 2EB, 3EC, 4KD, respectively. Prove that the point E is equally distant from one pair of the opposite sides of the square ; and that its distances from the other pair of opposite sides are in the proportion of 7 to 3. 2. Let BA, BC, represent a double inclined plane ; the length of BA being 25 feet, and "hat of BC 40 feet ; AC being horizontAl. K weight P, resting on AB, is r. EXAMINA TION PAPERS. 153 balanced by a weight W, resting on BC ; P and W being connected by a string, which passes over a fixed pulley at B, and the two portions of which are parallel to AB and BC respectively. If W be loolbs., find P. Inquire whether the tension of the string can be found from what is given. 3. Show that the principle of Virtual Velocities holds good in the case of equilibrium described in the prev- ious question. 4. A weight W, of 34lbs., suspended from a string ML), which is attached to a fixed point M, rests at D on the inclined plane CA, whose base AB is 39 inches; and its vertical height CB, 26 inches. The position of D is such that AD is at right angles to CA ; and the point M is in BC produced, CM being 8 inches. Prove that the tension of the string MD is 2ott)s. ; and find the reaction of the plane on the weight. 5. A uniform heavy rod AB, 14 feet in length, weighing 56tt>s., and having a weight W, which is also 561bs., suspended from a poii.t E in the rod, is sus- tained in a horizontal position by strings AD and BC, whidi are attached to fixed points D and C in the same horizontal line DC. The length of AD is 15 feet ; of BC, 13 feet; and the perpendicular distance between the parallel lines DC and AB is 12 feet. Show that the point E, from which the weight is suspended, dWidcs the rod in the ratio of 3 : 11. NORMAL SCHOOL. FINAL EXAMINATION. June 1873. 1. Three smooth vertical posts are fixed at the angles of an equilateral triangle and a cord is passed round them, to each end of which a force of loolbs. is applied ; find the pressure on each post. 2. A beam, AB has one fjnd attached to a hinge A, and the other end attached to a cord BC, one end of which is tied to a peg. The weight of the beam is ^olbs., i tS4 £.LEMENTaR V STA TICS, and may be supposed to act at its middle point. The beam and cord make angles of 60^ on opposite sides of the vertical. Find the tension of the cord. 3. Define the centre of gravity of a body, and deter- mine the position of the centre of gravity of a triangular board. If the centre of gravity of a four-sided figure coincide with one of its angular poirtts, show that the distance of this point and the opposite angular point from the line joining the other two angular points are as i to 2. 4. A uniform iron rod, a foot of which weighs ijibs., rests on a fulcrum two feet from one end : find what weight, suspended from that end, will keep it at rest when the pressure on the fuffcrum is i5oIbs. 5. An equilateral triangle is placed upon an inclined plane, its lowest angle being fixed : find how high the plane may be elevated before the triangle rolls. 6. In a system of pulleys, in which each pulley hangs by a separate string, there are three pulleys of equal weights ; the weight attached to the lowest is 32lbs., and the power is i ilbs. ; find the weight of each pulley. SECOND CLASS CERTIFICATES. December 1873. 1. Describe the wheel and axle, and state the me- chanical advantage of the machine. Show that the wheel and axle, in the state of equilibrium, is identical with the lever. 2. ACB is a straight lever, whose fulcrum is C. Find the weight which, acting at A, is balanced by 15 J oz. at B, and, acting at B, is balanced by i6j| oz. at A. 3. A straight lever ACB, whose fulcrum is C, is kept at rest in a horizontal position by two forces, one a weight suspended from B, and the other the tension of a string which is attached to A, and runs in the direction AE. If AE be equal to AC, and the perpendicular dis- tance of E from AB be equal to BC, find the relation between the weight suspended from B and the tension of the string EA. EXAMINA TION PAPERS, m 4. Draw a diagram representing a system of pulleys (3 moveable pullies in the system), in which the me- W chanical advantage (that is, -p ) is 8. (/;) If W = i61bs., state the tension of the string pas- sing round the middle pulley. 5. Let ABCD be a square. Bisect BC in E and CD in ¥. Join AE and AF. If a particle A be acted on by two forces, represented in magnitude and direction by the lines AE and AF respectively, find the numerical value of the length of a line representing the magnitude of their resultant, the sides of the square being unity. FIRST CLASS CERTIFICATES. December 1873- 1. Let ABCD be a parallelogram such that each of the four sides is equal to the diagonal AC. Bisect AC in E. A particle at A being acted on by three forces represented in magnitude by AD, (i + 2|/J)AE, and ^{ I + |/J) AB, respectively, and in direction by AD, EA (not AK), and AB, respectively, prove that their resultant is in the direction of the diagonal of a square described on AC. 2. A uniform heavy rod, AC (which may be con- sidered as a line), is in equilibrium, in a horizontal posi- tion, with one end C resting on a smooth inclined plane ED, and the other end A connected by a string AB with a fixed point B. Prove that the tension of the string and the reaction of the plane on the rod, are equal to one another. 3. Let BCD be a triangular board, forming two oppo- site inclined planes, BD being horizontal. Suppose a perfectly flexible uniform heavy cord ABCDE, weighing i4K)s., to pass over it, the portions BA and ED hanging vertically. Let thejengths of AB, BC, CD, and DE, be 6 ft., 8 ft., 6 ft., 8 ft., respectively. The cord being supposed capable of moving freely in virtue of its own weight, inquire what weight (if any) must be attached to either extremity (A or E as the case may be) in order that equilibrium may subsist. ^mmmm 156 ELEMENT A R Y STA TICS. SECOND CLASS CERTIFICATES. July 1874. X. ACB is a bent lever, the fulcrum being C. The arms AC and CB are straight ; and the angle formed by BC and AC produced is f of a right angle. If a weight of lolbs. acting at A balance a weight of 8R)S. at B when the arm AC is horizontal, find what weight at A will balance a weight of Slbs. at B, when BC is horizontal. 2. {a) Enunciate the principle of Virtual Velocities. (/») Assuming the principle, apply it to determine the relation between the Power and the Weight in the Wheel and Axlo. 3. {a) Draw a diagram representing a system of pulleys (three movable pulkys in the system) in which a separate string passes round each of the pulleys. {I)) Apply the principle of Virtual Velocities to determine the relation between the Power and Weight in this system of pulleys. 4. {a) Enunciate the princ'ple of the Parallellogram of Forces. (b) If a particle at A be acted on by two forces represented in magnitude and direction by the lines AB and AC respectively, CAB being a right angle ; if the length of AB be 5 feet, and the length of the resultant of the two forces be 13 feet ; find the length of AC. FIRST CLASS. 1874. r. A uniform heavy rod AB, whose weight is 2ofi>s., is kept at rest in a horizontal position by four forces (in addition to its own weight) : a force of lolbs., acting at B in a direction BD at right angles to AB ; a force of lolbs. acting at A in a direction AE which is such that EAC is J of a right angle ; a force m acting vertically at C ; and a force n actmg horizontally at B. If AC = 2CB, find ;// & //. 2. On AB, an inclined plane, whose base is AC, and which has the angle BAC equal to f of a right angle, a heavy body is kept at rest by two equal forces, the one 3 ■' EXAMINA TION PAPERS. 157 lorces lolbs., IB; a |ch is Icting latB. and [le, a one acting in the direction of AB, and the other in a direc- tion parallel to AC and towards the plane. Prove that the reaction of the plane on the body is equal to the weight of the body. 3. Let ABCD, a uniform heavy square, be suspended by a string EB from a point E ; and from A and C let weights P and Q be suspended by strings AP and CQ. Prove that if P exceed the weight of the square by 3Q, the direction of the string EB will bisect the line drawn from A to the centre of the square. SECOND CLASS CERTIFICATES. July 1875. 1. A straight lever ACB, without weight, the fulcrum being C, is in equilibrium, in a horizontal position, under the influence of two weights, namely, P acting at A, and W at B. If AC = 2k ^eet, and BC = 4f feet, and if the pressure on the fulcrum be 24lbs., find P and W. 2. Assuming the Principle of Virtual Velocities, apply (W\ p Jofthe Screw, the Power being supposed to act in a plane at right angles to the direction of the screw. 3. A weight W is kept at rest on an inclined plane by a power A acting parallel to the base. Does such a machine ever act at a mechanical disadvantage ? If so, when ? Illustrate by an example. 4. If a pupil should say that in the case of a body at rest on an inclined plane under the influence of a Power P acting parallel to the plane, the weight of the body and the force P, being neither equal nor directly oppo site, cannot possibly counterbalance one another ; and should ask what force, additional to these, acts on the body so as to keep it at rest ; what would you reply ? 5. What is meant by the Resultant of a number of forces acting at a point ? Draw any lines AB, BC, CD ; and let a particle at A be acted on by forces parallel to the lines AB, BC, CD, taken in order, and represented by them in magnitude. 158 ELEMENTARY ST A TICS. Prove (assuming the principle of the Parallelogram of Forces) that the resultant of these three forces is repre- sented in directioi and magnitude by AD. 6, ABC is an equilateral triangle, of which the side is one foot. A parlicle at A is acted on by a force repre- sented in magnitude and direction by AB. Let the force be resolved into two forces, one in a direction parallel to BC, the other in a direction perpendicular to BC. P'ind the lengths of the lines representing these forces respectively. FIRST CLASS CERTIFICATES. July 1875, 1. A particle at A, a point in the straight line FAE, is at rest under the influence of three forees, namely, a force of lib. acting in the direction AC, which makes the angle FAC equal to one third of a right angle ; a force of ;// lbs., in the direction AB, which is at right angles to AC, and makes BAE two-thirds of a right angle ; and a force of « lbs., in the direction AD, making the angle DAE one-half of a right angle. Find the rela- tion between m and //. ' • 2. Let AB be a uniform heavy beam, resting with one end B against a wall, and the other end A on the ground. If the reaction of the wall on the beam, and the friction at B, be together equal to the reaction of the ground on the beam at A, compare the distance of A from the wall with the height of B above the ground. 3. Let ABC be a uniform straight rod, in a horizontal position ; AB being 6| feet ; and BC, 3| feet. In DB, a straight line drawn at right angles to AC in a vertical plane, take the point D above the rod, and let DB be 4^ feet. Suppose the rod to be acted on" by two forces besides its own weight, namely, a force of 61b3. acting at A in the direction AD, and one of 8lbs. acting at C in the direction CD. If the rod weigh loflbs., enquire whe- ther it be in equilibrium. If it be not in equilibrium, specify any force or forces, which, in conjunction with those acting on it, will produce equilibrium. (// may be assumed that A CD is a right angle). EXAMIAA TION PAFE/iS. 159 3. Let ABC be a horizontal line , and let BE and BD represent two inclined planes lying towards opposite sides, the angle ABE being two-thirds of a right angle, and the angle CBD being one-third of a right angle. • Prove, that, if a uniform heavy rod FG, lying on the planes, the extremity F on BD, and G on BE, be in equilibrium, BF is half the length of the rod. NORMAL SCHOOL. FIRST GLASS CERTIFICATES. 1875. 1. ABCD is a square. In CD a point E is taken, and in CB a point F, such that each of the angles EAC and FAC is one third of a right angle. Find the mag- nitude of four forces acting on a particle at A, namely, a force of 3 lbs. in each of the directions AD and AB, and a force of y(6)lbs. in each of the directions EA and FA (twt AE and AF). 2. A uniform straight rod AB, with one end A resting on the ground, has its other end B connected by a siring with a fixed point C in the vertical line CA. The rod being in equilibrium under its own weight, the tension (T) of the string, friction (F), and the reaction (R) of the ground on the rod, prove that if ABC be an equi- lateral triangle, * is a mean proportional between T + F andT— F. 3. Let AC be a uniform straight rod, in a horizontal position ; B, its middle point ; D, a point above the rod, in the vertical plane passing through the rod ; and DE, the perpendicular let fall from D on AC. If the weight of the rod be represented in magnitude by the straight line DE, prove that the rod will be in equilibrium under the influence of the following forces, namely, its own weight, a force acting at A in the direction AD and re- presented in magnitude by ^f- a force acting at C in the direction CD and represented in magnitude by °^' and a force acting in the direction of the rod and represented in magnitude by EB. IS^ i6o ELEMENTARY STAIJZS, NORMAL SCHOOL. SECOND CLASS CERTIFICATES. June 1875. 1. Give two familiar examples of a lever working at a mechanical disadvantage, and point out the use of such a machine. 2. If a straight lever be in equilibrium in a iiorizontal positioii under the influence of two weights, show that it will be in equilibrium under the influence of the same weights applied at the same points, in every position. 3. In the straight lever ABDE, let AB, BD, DE, be equal distances of one foot ; and let the lever (supposed to be without weight) be in equilibrium round the ful- crum (^, under the influence of a weight of 3lbs. applied at A, of stbs., applied at B, of 7 lbs. applied at D, and of 9tbs. applied at E. Find the position of the fulcrum C. 4. Assuming the principle of the Parallelogram of Forces, deduce the corollary, that, if the directions of three forces acting at a point be parallel to the sides of a triangle taken in order, and their magnitudes be pro- portioned to the sides, they will keep the point at rest. 5. A particle at A is acted on by" a force in the direc- tion AB. Suppose this force resolved into two forces, one acting in a direction AC, and the other in a direc- tion AD at I'ght -angles to AC, the angles BAC and ]iAD being each less than a right angle. If the resolved part of the force in the direction AC be W of die whole force acting on the particle, inquire what part of that whole force the resolved part in the direction AD shall be. 6. {a) Dra\v a diagram representing a system of two movable pulleys, the last pulley supporting the weight, and the free portions of the strings being all vertical ; and state the mechanical advan iage y~^ when tlie weigl^.ts of the pulleys are not taken into account. {b) Let the weights of the pulleys be taken into account, the weight of each bein^ Q. Inquire, then, EXAMINATION PAPERS. i6l shall If two leight, Itical ; tlie then, what must be the relation between P and Q in order that no mechanical advantage may be either gained or lost by the machine. SECOND CLASS CERTIFICATES. July 1876. 1. What particulars are required to be known in order to specify a force ? What are the conditions of equili- brium of three forces acting at a point ? What of three forces acting on a rigid bcdy ? 2. Forces of jlb., 4lt)s., and 6lfes., respectively, act on a particle, the force of 4tb3. being inchned at an angle of 60° to each of the others, find the magnitude and di- rection of'" pir resultant. 3. {a) What is meant by the moment of a force with respect to a point ? State the principle of moments. {b) A uniform rod a foot of which weighs 3lbs., rests on a fulcrum two feet from one end, what weight suspended from that end will keep it horizontal when the pressure on the fulcrum is 3oofi)S. ? 4. A circular plate of two feet radius has a circular hole of eight inches radius cut in it, find the distance of the centre of gravity from the centre of the plate, if the centre of the plate and hole are twelve inches apart. 5. n cylinders of the same height //, the radii of which are equal to r^, ra, r 3» - r, n» respectively, stand one upon another with their axes in the same straight line ; find the hei'^ht of their common centre of gravity above the base of the first. . FIRST CLASS CERTIFICATES. July 1876. I. Enunciate the Triangle of Forces, and by means of it deduce the Principle of Moments. Find the resultant of three forces acting in consecutive directions round a triangle, and represented respectively by its sides. 1 63 ELEMENTARY STATICS, lit 2. A lever -without weight is c feet in length, and from its ends a weight is supported by two strings in length a and d feet respectively. Find the ratio of the lengths of the arms, if there be equilibrium when the lever is hori- zontal. 3. A piece of uniform wire is bent into the form of a triangle ; find the position of its centre of gravity. 4. A swing-gate weighing 96lfes. rests on a hinge y^, and against a frictionless turning-point £, four feet directly beneath A. Find the strain on the hinge and the pressure on the point, given that the centre of gravity of the gate is 4 ft. 7 in. from AB. What will be the strain and the pressure if a boy weighing 108 lbs. stands on the gate 6 ft. from AB i » INTERMEDIATE EXAMINATION. June 1876. 1. State the principle of the Parallelogram of Forces. Explain the meaning of the terms employed in your state- ments. Shew that if four forces acting on a point be represented by the sides of a rectangle taken in order, they will be in equilibrium. 2. Apply the Triangle of Forces to find the least hori- zontal force necessary to draw a wheel four feet in diameter and weighing 10 cwt. over, an obstacle the height of vvhich is six inches, situated c*i the horizontal plane on which the wheel rests. 3. Define the moment of a force with respect to a given point. A uniform beam A B, whose weight is 100 lbs. and length 50 feet, rests with one end (A) on a horizontal plane A C, and the other end against a vertical wall C B. If a siring C A, equal in length to C B, prevents the beam from sliding, find the tension of the string. 4. In the system of pulleys in which each pulley hangs by a separate string, a platform is suspended from the lowest block ; what force must a man who weighs 180 pounds,''standing on the platform, exert to sustain him- self when there are three movable pulleys. »':-v7V. EXAMINA TION PAPERS, 1^3 5. State the principle of virtual velocities, and apply it to find the relation between the power and the weight in the inclined plane. INTERMEDIATE EXAMINATION. December 1876. 1. A uniform straight bar, 18 feet long, has weights of 12 snd 20 pounds at the two ends. How heavy must the bar be, in order that it may rest on a pivot 10 feet from one of its extremities ? 2. A bent lever has arms of equal length, making an angle of 120''. Required the ratio of the weights at the ends of the arms when the lever is '1 equilibrium with one arm horizontal. 3. {a) Statetheprincipleofthe Parallelogram of Forces. (/;) A unit of force is resolved into two forces, F and G acting in directions at right angles to one another. If the magnitude of F be \, what is the magnitude of G? 4. {a) State the principle of Virtual Velocities. (<{') Draw a diagram exhibiting the system of pulleys in which the same string passes round all the pulleys and the parts of it between the pulleys are parallel. {l) Apply the principle of Virtual Velocities to determine the ratio between the Power and the Weight in til is system. "■■T APPENDIX I. Duchayla's Proof of the Parallelogram of Force. In Chapter II. we explained the meaning of the l)ioposition known as the " Parallelogram of Forces," and deduced its truth by means of expe/iment. We shall now establish the truth of the proposition upon evidence of the same kind as that on which the Theorems of Euclid rest. The demonstration is divided into th.ree parts : (i) To find the direction of the resultant when the forces are commensurable. (2) To find the direction of the resultant when the forces are incommensurable. (3). To find the magnitude of the resultant. 7. To demonstrate the Parallelogram of Forces^ so far as relates to the direction of the resultant y the forces being commensurable. (i) We assume that if two equal forces act on a particle the direction of the resultant will bisect the angle between the directions of the forces. This seems obvious, for no reason can be adduced for its inclining towards one of the forces which would not equally apply to make it incline towards the other. (2) Suppose that the proposition is true for two forces / and w, inclined at any angle ; and also for two forces / and ;/, inclined at the same angl^ . we shall that it will be true for two forces / aad m -f- //, inclined at the same angle. ^i^ APPENDIX, 165 % Let A be the particle on which the forces act. Draw AB, AD representing the forces in and/ respectively in magnitude and direction. Produce A B, and make EC prrportional to the force n in the same ratio as AB is to the force m. Then since a force may be transferred to any point of it J direction, BC will represent the force n in magnitude and direction. Complete the parallelogram ABED, and draw the diagonal A E. Then by hypothesis, A E represents the resultant (R suppose) of/ and in in direction; and the point of application of this resultan may may be sup- posed to be at E, a point in its direction rigidly connected with A (art. 15). Now since the force R acting at A is equivalent to two / and m, when at E it may evidently be resolved back again 10 the same forces acting parallel to their original directions. Produce DE to L, and BE to H. Then instead of the forces/ and in at A, we have a force m acting along L vat E, and a force/ acting along EH at E. We may again change the points of application of these forces so thdt ;// may be supposed to act at F and / at B. Complete the paralellogram EBCF and draw the diagonal B F. Then, by hypothesis, B F represents in direction the resultant of the two forces p and n acting i66 APPENDIX. at B. And this resultant may be supposed to act at F a point in its direction, instead of at A. At F it may be resolved again into the forces/ and n. We have now removed the forces / and /;/ + ;/ which acted at A to F. Therefore F must be a point in the resultant. Whence the direction of the resultant of jp and ;« + « is along the diagonal of the parallelogram constructed on the lines representing them. Now, we know the proposition to be true for the direc- tion of the resultant of two equal forces / and /. Since therefore it is true for / and /, and p and /, by the preceding it is true for / and / +/ or 2 /. And since it is true for/ and /, and/ and 2/, it is true for / and 2/ +/ or 3/, and so on ; therefore generally it is true for/ and r/, where r is a whole number. And again, since it is true for r p and/, it is true for r p and 2/, and so on by similar successive deductions it may be proved to be true generally ol r p and j/, where ^ is a whole number ; therefore it is true for all commensurable forces, i.e. for all forces the ratio of whose magnitudes can be expressed by the ratio of two whole numbers. Remarks. — The preceding demonstration generally seems obscure to students who meet with it for the first time. This results from the somewhat unusual foi-m of the proof The student will notice that the demonstra- tion consists of two parts. In the first part it is shown that if the principle is true in two cases, viz., with regard to the pair of forces/ and ;// and the pair / and ;/, it must also hold good in a third case, viz., in regard to the pair of forces p and m + n ; this part of the proof is purely hypothetical, as much so as m the case of a d-^'monstration by reduction to an absurdity. The second part of the proof takes up the argument, but as a matter of fact the proposition is true in two certain cases ; therefore it must be true in a third case, therefore in a fourth case, and so on. APPENDIX. 167 (2) To demonstrate the Parallelogram of Forces so far as relotes to the direction of the resultant^ the forces being incommensurable. Let AB, AC represent in magnitude and direction two of tlie incommensurable forces. c Complete the parallelogram ACDB. Draw AD. Then we have to show that AD represents in direction the resultant. If it do not, let any other line as AE represent that direction. Divide AC into a number of equal parts, such as that each is less than ED. Mark off along CD parts, each equal to one of these. Then one division will evidently fall between E and D, at F suppose. Join A F, and draw EK, FL parallel to AC. The forces represented by AC, AL are commensur- able. Therefore their resultant will be in the direction of A F, and we may suppose this resultant to be substituted for them. Then the resultant of the forces represented by AC 9sA AB is equivalent to the resultant of two forces, one acting in the direction AF, the other represented by LB, and which may therefore be supposed to act at A. in the direction AB : and this resultant must lie withtn the angle BAF. I! i6S APPENDIX. ^:~^" ':%^ But, by hypothesis, it acts in the direction AL, without the same angle, which is absurd. In like manner it may be shown that no direction but AD can be that of the resultant of the forces represented by AB, AC. Whence the proposition is true for all forces as far as the direction of resultant is concerned. (3 ) To demonstrate that the Parallelogram of Forces holds also with respect to tlie magnilude of the resultant. Let A C, A B represent in magnitude and direction two forces P and Q. Complete the Parallelogram B C. Then by what has already been jvoved, the diagonal A D will be the direction of the resultant of P and Q. Produce A D backwards towards K. Then if we suppose a force equal in magnitude to the resultant of P and Q to act along A K, this force and P and Q will be in equilibrium But if any number of forces be in equilibrium any one is equal and opposite to the resultant of all the rest. Thefore the force P must be equal and opposite to the resultant of the force acting along A K and Q. Produce C A to F, and make A F equal to A C. Then AF represents in magnitude and direction the resultant of Q and the force alonff A K. ;1 APPES'DJX. 169 i Join F B, and draw F G parallel to A B. Then since FAC is parallel to BD, and FA=AC =BD, therefore also FB is equal and parallel to AD (Euc I. 33). Now AF represents in magnitude and direction the resultant of Q and the force acting along DK. But the forces represented in magnitude and direction by the lines AG, AB, have a resultant acting along AF. Therefore AG must represent in magnitude the force which we supposed to act along AK. But this force is equal and opposite to the resultant of P and Q Therefore the magnitude of the resultant ot P and O = AG=FB = AD. ^ That is, the magnitude of the resultant of P and Q is represented by the diagonal of the parallelogram, of which the sides represent the magnitude of the Forces PandQ Therefore, if two forces, &c., Q E. D. RESULTS, HINTS, Szc, FOR THE EXKRCISES. Chaptkr I. Exercise I. Page 6. (4) A force is mensured numerically by the weight ii would sustain if acting vertically upwards. The most common unit of weight employed is ilb. Hence in the absence of other information, I should understand by a force of 10 or a force of P, a force of such magnitud';; that it would support iolt)s. or Plt)3. respectively it applied to the weight in the manner stated. (8) 8Ibs. PARALLELOGRAM OF FORCES. Chapter IL Exercise J. Page 12. (2) i5lbs. in the same direction as the forces. (3) lit), in the direction of the greater force. (4) II oz. acting in the direction of the latter forces, (8) islbs. (9) i3lbs. (10), if P be smaller force R = P^io. (11) 6lfe-?. (12) 3olb^. (13) ictbs. (14) the latter. fi5) 27lbs. and i2ott),. (16) i.2lb3. and i.6tt>s. 17) 129.72)3. and 129.5103. \l I: r ISES. gilt ii most n the by a itud'^ Iv u sas. ces. rce '1 ^ RESULTS, HINTS, &*C. Exercise II. Page 16. 171 (i) 6ilbs. (2) 7fl)s. (3) 7lbs. (4) soslbs. (5) Let the force along each ratter be P, then we have two equal forces, each P, making an angle of 60° . with each other, and having for resultant qoILs. ; hence P = S^V3', (6) 4.63^)8. (7) 3i.9olbs. (9) This is the same as to fmd the resultant of two equal forces, each equal to 61bs., making an angle of 60" with each other. The resultant is 6v^3lbs. (10) First, we have given the resultant, W, of two equal forces, acting at an angle of 60°, to find the forces. If P be P P one of the equal forces, W^ = (P + — )s + ( — 1/3)2, or w r= — . The pressure on B or C is resultant of two 1/3 equal forces, each = — , acting at an ancle of 120°; V3 resultant is, therefore, equal to either of the forces, that W is pressure on B or C = — . Pressure on -A is result- V3 W ant of two equal forces, — , acting at angle of 60' ; pres- sure on A is, therefore, \V. Exercise III. Page 17. (3) The resultant is equal to one of the forces. (4) 120°. (5) Euc. 1. 20, (6). If p be the lesser force, resultant will be 2P, and greater will be P^3 ; therefore sides of triangle are in ratio of i, 2, v3 ; hence angle between resultant and lesser force is 60", and betweeo resultant and greater 30°. Or thus, bisect resultant and join point of bisection with right angle. In any right-angled triangle the line joing the point of bisection of the hypo- thenuse and the right angle is half the hypothenuse. The angle between lesser fc. and resultant, is, .-. an angle of an equil. triangle. IMAGE EVALUATION TEST TARGET (MT-3) / o 1.0 I.I 1.25 (50 1^ M M U III 1.6 <^ '/# /i // VI e. ^V'«>*' > ^ / \ Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 S. ■i^ V ^ 4 ^ \ ^1.^ 6^ % ri::^ % I7S RESULTS, HINTS, &-C. (7) A force DA. (8) ^2 at C, parallel to DB. (10) 6^2. (12) The straight lines drawn from the angles ot a triangle to the points of bisection of the opposite sides meet at the same point and each of the bisecting lines is trisected at that point (art. 78) TRIANGLE OF FORCES. Chapter II. Exercise I. Page 21. (i) Euc. I, 20. (2) Arts. 29and3r. (5) Euc. I, 48. (6) The side which does not pass through the point, represents the third force in direction, but not in line of action. (7) Construct triangle having its sides 5, 6, 7. Produce side 7 backwards through intersection of 5, 7, and make produced part equal to side produced. Through intersection of 5, 7, draw a line parallel and and equal to 6. The lines 5, 6, 7, acting at angular point will •represent forces in direction and magnitude. (8^ The forces make angles of 60** with each other. (9) 2P making an angle 60" with P. (10) makes an angle of 30° with greater force, see Ex. 6, page 18. (11) I : |/2. Exercise II. Page 24. (i) 8ott)S. and 6oIbs. (2) Slbs. and 6ibs. (3) Pro- duce side 24 backwards and const, right-angled triangle having side making angle 30° with vertical for hypothen- use. The three sides of this triangle are parallel to the 3 fcs., which keep point at which wt. is attached at rest, and, thei^fore, proportional to them. Let hypothenuse — 2, &c., and horizontal tension is 8^3, the other i6y'3. (4) The sides of the triangle whose sides are 10, 6, 8, are parallel to the three forces which keep point B at [rest, and, therefore, proportional to them. Tension is 26f cwt i I f: .5i* KESULTS, mXTS, 6-C. 173 (5) Let vertical through C meet line AB in D. Then AB will be bisected in D. Through D draw DE paral- lel to BC ; AC is bisected in the point E. The straight line drawn through point of bisection of one side of a triangle parallel to the base bisects the other side. (This may each be proved by drawing through point of bisection of one side of triangle lines parallel to other two sides. Show that the two triangles thus formed are equal (Euc. i. 26). Then since opposite sides of parallellograms are equal, we have the required result). The sides of the triangle CDE are parallel to the three forces which keep point C at rest, and there- fore proportional to them. CD = 12, AD = 35, .'. AC = 37; EC = HC = &c. The weights are equal to tension of string = 154 Jibs. (6) 17 : 8. (7)3K>s. !( I»! Exercise III. Page 27. (i) The sphere is kept at rest by reactions of planes and its own wt, which may be supposed to act at its centre. These three fcs. must, therefore, pass through centre (art 34). Show that the angles which line of action of wt. makes with reactions at centre are 60" and 30". Triangle formed by one of the reactions, weight of sphere, and plane, will have its side parallel to three forces which keep sphere at rest &c. Pressures are 100^3 and looBis. respectively. (2) Through top of// draw line parallel to F. Triangle formed will have its side parallel to the three force F, VV, and reaction of // which passes through centre. (3) Produce the wt. backwards till it meets the ten- sion ; then reaction of hinge must pass through this point, tension = 5^2H)s. (4) solbs. (5) Produce wt. backwards till it meets reaction of wall ; join this point with A. The triangle of which this line is hypothenuse has its sides parallel to the three forces which keep rod at rest. If perpendicular of triangle be i, base is ^, and, therefore, hypothenuse 4^/5. Hence W : R :: i : Ji/5 .•.!< = ?V5- i 174 HESULTSy HINTS, &'C. (6) Beam is kept at rest by three forces, W acting at its middle point, T, tension of string, and reaction of wall. These three meet in string CB, at a point E. BC is bisected in E, (note Ex. 5, Exercise II). From B draw BD perpendicular to wall ; then CA = AD, since EA is drawn through E parallel to BD. Let BD = y, CA = .r, then AD will also = x. 52 = y^ +{2x)^ \ 38 = v^+^c^. Therefore^ =>^- The sider, of triangle CBD are parallel to three forces which ke';p beam at rest, .\ T : W :; 5 : Sy'J or T =^^-^. o RESOLUTION OF FORCES. Chapter IV. Exercise I. Page 34. (i) 1.7 and i.35v'3- (2) Draw one side at right angles to 12, and equal to 25; complete parallelogram having 25 for one side and 12 for diagonal ; the other side will be 27.73. (?) Resolve tension of cord into two parts, one indirection of keel, and other perpendi- cular to it. The latter tends to turn the bow towards the land, and this tendency will be counteracted by the pressure on the rudder, if the rudder be turned from the land. (8) 5o-/3lbs. and 5otb. (9) If P = given force, the first force = P, the second Py'2. (10) 40y'3lbs. and 4olbs. (11) 52lbs. (12) 25lbs. (13) Resolve in di- rection offeree 100 and at right angles to it. R makes an angle of 90^ with 100, and is = ■/3. (14) Resolving the 80 and 100 horizontally we have 40 and 50\/3, respectively. The angle between these forces is 30^", and their resultant is found by the paral- lelogram of :orces to be i22.8Ibs. RESULTS, IILVTS, &-C. Exercise II. Page 39. 175 (l) 2lb. (2) I2v^3lbs. (3) 5^2lbs. (4) ^3: I. (5) 8V3tt>s. and i6V3^fes. (6) i : V3- (?) Resolve the 20 and 20^3 along and at right angles to the direc- tion of the force 40, and we have R = 80 acting in a direction opposite to the force of 4olh3. (8) Resolving along the plane and at right angles to it we have ^^W = P + JP,orP=-^^^-lX. R = iW + iPV3 ^ 3 = W. (9) Resolve along plane, iWv3 = 2Por P - J\\V3. M R, and R + ^Pv3 - ^PV3 + JVV (ro) Let t = tension in AC, t' = tension in BC. Resolve horizontally, and — = /'. H vertically II V2 = W; , , ^, . ••• i = WV2, and /' =W. (11) Ihe tension of the string must be the same throughout. Resolve ?o and W along each plane, and equate results, and we have W = 10V6 =24 Jibs, nearly. PARALLEL FORCES. Chapter V. Exercise I. Page 46. (i) 70iV^s. and 50/^Ibs. (2) 388)3. and 1148)3. (3) 6 jV ft. from one end. (4) 1 2 m. (S) Resolve i2ft>s. into forces of 4fi)s. and 8B)s. acting at the extremity of the line bisecting its width ; resolve then into 4, 4, ; 2, 2, actaog at its corners. (6) olbs. at A, 62)s. at B, 4tbs. at C, sibs. at D. ■HI MMB asifc^ 176 RESULTS, HINTS, cShC. (7) Resolve the i5oIbs. into two parallel forces, one of softs, acting at angular point, and another of loolbs. acting at middle point of side of triangle. The Toolbi. can be resolved into two equal parallel forces of 5olb.s. each acting at angular points. Hence, pressure on each prop is 5oft)s. (8) Pressure on A = i4lbv, B = 61b3., C = lolbi., D = 3ofts. (9) No. The weight may be resolved into 4 equal parts acting on each leg of the table. (10) Art. 49. (11) If ;c = the distance from 2ft)3., then, by article 46 we have 2X ~ $ {x — I J), or :«; = 2^ ; that is, the hand must be placed ift. from the middle of the sash or 6in. from the broken cord. Ot, we might proceed as follows : — Resolve the 5^)3. into two parallel forces of 2^ft3. each, acting vertically downwards, 2lbs. of one of the forces is balanced by the tension of the unbroken cord which is 2ft5. The problem then is, to find the position of the resultant of forces, ^ft. and 2\Vbi, 3 feet apirt, which gives the same result as above. MOMENTS. Chapter VI I. Exercise I. Page 56. (i) Arts. 51, 52. (2) Art. 53. (3) Art. 55, second case, and art. 54. {4) Art. 58 ; take moments about pt. of intersection of t'vo of the forces, then moment of these two will be ztro ; moment of third must also be zero if the forces are in equilibrium, and must, therefore, pass through same point. (5) Let W be wt. of bundle, b its distance from his shoulder: P, the pressure of hand, a its distance from shoulder ; then, The left-hand side being constant, if a diminishes P increases, and therefore P + VV, or the pressure increases. HESULTS, HINTS, &>€. ill Regarding the man and things carried as one body, we see that the change only affects the mutual actions at the head and shoulder, and not the pressure on the ground. (7) Art. 57. (8) When three forces are in equilibrium each is equal and opposite to the resultant of the other two, then Art. 56. (9) Arts. 59 and 61. Exercise II. Page 60. (i) 7rds. (2) If the pressure on one prop is 8Ibs., the pressure on the other will be 481bs. Let x = dis- tance at which 56Ibs. must be placed; take moments about that point, and we have 48 ;c = 8 (7 — a:) ; .'. .r = I ft. from one end. (3) 20 inches from one end. (4) Distance from first man must equal J of pole. (5) i inch from fulcrum on side of 7lbs. (6) Take moments about fulcrum, and distance = 1.74 ft. (7) 8 inches from smaller force. (8) Each man pushes back the boat with his feet with a force exactly equal to that which he applies to the handle of the oar, />„ 5oibs. ; but he presses it forimrd also with the force which his oar produces on the row- lock; if P be that force, then P x 7^ = 50 x 10, or P = 66flbs. Each man, therefore, presses forward the boat with a pressure equal to 66f — 50 = 16 Jibs. The resultant force propelling the boat is i6|lbs. x 8= 133?, lbs. (9) 7n>s. (10) ioV3lbs. (11) P : Q :: i : 2. (12) 1": 2. (13) P = 3olbs. ; pressure on the fulcrum = 36.61 ; produce horizontal and vertical forces till they meet, and join their point of intersection with the fulcrum ; the pressure on the fulcrum will act along this line (art. 34). Exercise III. Page 65. (i) 32a>s. (2) I. Reaction of wall = 83. 23565. ; 2 Tension of string, = 93.232)3. ; 3. Reaction of plane = II2ft)S. (4) Produce reactions of wall and ground till they meet. Take moment about this point, and force will = ioV3tt>s. 178 RESULTS, HINTS, dr-C. (5) Reaction of wall is equal to horizontal force = W. Take moments about foot of ladder, and required dis- tance = 4 J ft. (6) The weight of each beam, acting at its centre, may be resolved into 5olbs. acting at A, and 5oIbs. acting at C. Take moments about C, and thrust =5oIbs. (7) Since the ladder lies on the wall, the re- action of the wall will be at right angles to the ladder. (Art. 62, III). Produce reactions of wall and horizontal plane till they meet. Take mo- ments about points of intersection ; then if P be the pressure on the peg, P x 4V2 = 120 x J x — r-, or P = v 2 255)s. To find reaction of wall, take moments about foot of ladder, and reaction = 2 5 V2. Exercise IV. Page 67. (i) From D draw perpendicular /, p', on each of equal sides a \ then a-Kp + a^-p' = 2 Area triangle ABC = tf X P. »'. p+P' = P. If F be one of the equal forces. / X F-f/ x^F = P X F = const. (2) Sum of perpendiculars from any point within an equilateral triangle = perpendicular from angular point on opposite side. (4) The line drawn from the middle point of the hypothenuse to the right angle is half the hypothenuse. (5) Since the force acting along the rope is constant, its moment or turning power will be greatest when the perpendicular on it from the foot of the tree is greatest. The perpendicular will be greatest when the point of the tree to which the rope is attached is equal to the distance of the man from the foot of the tree. This can easily be shown by a simple geometrical construction. It is equivalent to saying that if a rectangle and a square have equal diagonals, the perpendicular from an angulai pt. of the square on 'ts diagonal is greater than corres- ponding perpendicular in rectangle. Or, we may reason as follows : HESULTS, HINTS, &>C\ ,79 Letrtf = length ot each side of right-angled triangle when sides are equal ;/ = corresponding perpendicular : / = length of rope. Let b. c, be sides when unequal, and/ = correspond- ing perpendicular ; / will be the same in each case. n Then a a pi ; be = p' /, or ~ = -4 ; also be p ' b^+e^ =/s,aiid2«« = /2;.-. 2rz^ = b^-^e^. Hence b^ + e 2 be 1 = —r ; but the sum of the squares of two quantities is greater than twice their product. Iherefore/ is greater than/. CENTRE OF GRAVITY. Chapter VII. Exercise I. Page *J2. (i) 12 inches from 2off)s. (2) Between the 2lbs. and 3tts., and 8 inches from the latter. (3) 10 inches from the i2lbs. (4) 6 inches. (5) 2 inches. (6) 6ir>s. (7) 9 feet. (8) Let W be wt. of rod acting at its C. G. lake moments about edge of table and W x i = i x 2 or W = 2lbs. (9) 8J inches from the ylbs. (10) The 6oIbs. acts at C. G. Each man bears 3oibs. Let x — distance of other man from C. G. Take moments about C. G., and 30 ^ = 30 x 8, or x - 8ft. from C. G. or 4ft. from other end. (11) nft. from smaller end. (12) VVx I = 3x4, or W = 12 cwt. (12) If or = distance of C. G. from top, then 6a: = 4 X 3.39 + 2x3, or .r = 3.26 ; that is the C. G. is 3.26 inches from the top. (13) Articles 50 and 65. (14) Art. 65. Exercise IL Page "jt. (i) Articles 69 and 70. (2) Art. 75. (3) Art. 73. (4) Art. 72. (5) Arts. 73 and 74. (6) The vertical line through C. G. falls within the base. (7) When at greatest height vertical through C. G. falls at end of i8o RESULTS, HINTS, 6-^ diameter of base. If slant hci-^ht be 5 and height 4, diameter of base must be 3 ; but diameter of base is 60 ; therefore tlie height is Soft. (8) 2V3. (9) Art. 76. (10) The C. G. is raised, and returns to its original position. (11) When it rests on two pts. if it receive a small displacement, the C. G. will descend and cannot return to its original position. (r2) To upset a body in a state of stable equilibrium we must raise its C. G. The solid cylinder will be the heavier, cceieris paribus, and will therefore require a greater force to upset it. (13) If there were only two feet, the equilibrium would be unstable. (14) Because the base is so small that it is practically impossible to place the pin so that tiie line of direction may fall within it. But even if it were so placed it would not stand for an instant, for its alti- tude is so great in proportion to the base, that the slightest force would be sufficient to upset it. (15) The C. G. is higher in the first case, and there- fore the equilibrium is more unstable. (17) 4 inches. (18) It is in stable equilibrium when its C. G. is directly under its geometrical centre. It is in a position of unstable equilibrium when its C. (j. is vertically above its geometrical centre. (20) 60°. (21) The C. G. is in the plumb line which, therefore, bisects the hypothenuse, and is also equal half of it. The acute angles are 60° and 30". The ratio of the sides are as \/3 : i. Exercise III. Page 85. (i) Let wt. of 2lbs, be suspended at C, in the triangle ABC. Bisect AB in D : join DC, and bisect it in E. E is the C. G. (3) Let a = length of either of the equal sides. The C. G. of squares will be at right angle. Take moments about this point and (za^ a' a' asl2 a ' -J2' C. G. from vertex. .'. X = - , wjjere .v is distance of 15 HE^ULTS, IffNTS, ^C. tSi (5) 15111. from side corresponding to wts. 3, 4 ; i2in. from side corresponding to wts. 4, 6. From centre, therefore, Vs inches. (6) 'J^zfrom centre. (7) On diameter passing through both centres ; to left of centre of a large circle at a dis- tance = j^^.-^-., (8) Draw diameter passing through vertex of triangle and bisecting opposite side. Greatest wt. will be at extremity of this diameter. Take moments about point of bisection of side by diameter, and wt. will be found to be equal tc jvt. of table. ■ (9) 4o5.61hs. (10) Let W be wt. acting at C. G. of slab. Resoive W into J\V at vertex, and ^VV at pt. of bisection of the base. Again, resolve J\V into JVV, & |W, at extremities of base. Exercise III. Page 86. (12) Let a, b, be the sides, then the hypothenuse will be \/(^" +b^). Weights at angular points will h?. ma^, ;////-, ;//(«- -^-b'^). Take moments about right angle, and if A.- be distance of C. G. from right angle X = ab V(«2+^2^' Or, find C. G. of weights ?na^, mb"^ ; then C. G. of whole will be at middle point of line joining this point with right angle. (13) The weights of the beams act at their middle points, and are proportional to their lengths. Since hypothenuse is 20, each of equal sides will be ~ or . V 2 10 V 2. The resultant of these two is 2o\f2 acting at middle point of perpendicular. Take moments about right angle, and (2o+-2o\/2) x — 20 x 10 + 201/2 x 5, or X — 51/2. mmm mg>ri TS M i8a JiESULTS, HINTS, &*C. (14) Let a — side of square. Take moments about centre of square, and {a* ->r^a*) x = \a^ y^ \ a .'. x = y^ff rtf, the distance of centre of square measured on line joining this centre with vertex of triangle. (16) 10 cent pieces. (17) Centre of sphere of which hemispherical end forms a part. (18) Art. 79 ',^ {hy — //g) from base. (20) Art. 72. MECHANICAL POWERS. Chapter VIIL BALANCES. Sec. II. Exercise L Page 94. (5) The tradesman loses jj-jfts. or^Y? cents. (6) 83)s. (7) True weight = 61bs. l^et a be shorter arm, b longer arm; then a-\-b — 3! ft. ; also 6a = 4^. Substitute these value and b = zii. 9in, a — ift. 6in. (8) 90 cents. (9) 20 per cent. (10) When lii is put into the scale of shorter arm, let / be the weight of the goods put into scale-pan of A.7V longer arm ; then $p = 47^', or/ = ^-- . Also, let ^ from shorter arm balance zv from longer, then ^q =■ 5^/, or 4 In latter case seller's loss will be ^ — w ■= In former case seller's gain will be w — / = w 4 w Therefore seller's loss is 7V w w = — in two sales. 20 4 5 or - — - in 100 sales. Therefore seller's loss = — = 2j •20 20 w per cent. Or gijice he loses — on two sales, he will 20 w l(jse — on one sale, that is 2^ per cent, as before. 40 (11) ift>. RESULTS, HINTS, ^C, 153 THE STEELYARD. Exercise II. Page g;. (4) 7 inches from point of suspension. (6) The point from which the graduation begins is brought nearer forthe'rC c "'P'"''°" ^' ^^^ ^^'' P°^°' ^ '' ^^k^" THE WHEEL AND AXLE, Exercise I. Page 99. 5 . vv .. r +/ . K-f /. (5) The rope coils on itself and THE PULLEY, Exercise I. Page los (2) 18). 12 oz. (3) 4 pulleys. (4) 4 pulleys. (8( «Ibs. (9) The weight supported is i6olbs., and tne power exerted is I sflis. (10) i4oIbs. (11) The number of pulleys has been omitted in the question The system consists of three movable pulleys. 1 hen with three pulleys a power of ift. would sustain a weight of 81bs iSolbs. must, therefore, be divided into two parts in the ratio of i : 8. The power will be 2olbs. Exercise II Page 107. (i) w must be less than 5P. (2) 28fts. (3) 1 20ms. (4) 84oIbs. (5) 5 pulleys. (6) 16 strings (7) i561bs. (8) 3olbs. (9)37fi>s. ^ ^ (10) Since «P = W + 7^, .-. « _ i5L = :^ W _ P P ' P ~° when -^ - —. = o or «P = «,. That is, when the Zffi ""[^^ ^°'^^'' ^^^^^ '^ ^^"^^ ^^ the power multi- plied by the number of strings at the lower block. r i ! • 1^4 I^ESCLTS, I/rXTS, &'C. (n) Since there are six strings at tlie lower block, the weight must be six times the power ; we have, tlierefore to divide the i4olt)s. into 7 parts, 6 of which win be weight and i power. The tension will be 2oK>s. (12) I ol his weight. (12) A force equal to his own weight. Exercise ill. Pasfe iii. (i) The number of s'^-'^ts must be two. (2) 81t)s. (3) 44olbs. (4) 84i.5lhs. (5) Let w be the weiglit of each pulley. Proceed as in example 5, and put P — o ; then w : W :: I : 247. (6) In the Third Systenj the weights of the pulleys act 7iit/i the power, in the other two systems against it. THE INCLINED PLANE. Exercise I Page Z14. (l) 25lbs. (2) II2olt)S. (3) I2^1bs. (4) 6§lb.s. (5) Let X be the part hanging over the top of the plane, then 104 — x will be the part resting X on it Hence = | ; x = 39, and the other is 6q. 1 04 — X (6) 6 J- inches. (7) loj feet. (8) Let x be the X 40 height and y the length, then — = -— , or "jx = ^y. Also / — x'^ = 340, ory 49 = 340 ; hence >• 26 42, and X = 18.91. (9) 2lbs. (10) 28V2, and 28\/2. (12) Let // be the height of the planes, and /, / their lengths ; let P be the tension of the string that supports the weights. Then -- = y-, and ~r"~J ;oi'''i^'"g. -7 = A (12) 32 oz. / RESULTS, HINTS, ^C. 185 Exercise II. Page 117 (i) 20^/3. (2) 15^41. (3) 56ft>. (4) 12V3. (5) 4. (6) 30^ (7) 8-651^3. (9) Oo\ THE SCREW. Exercise I. Page 124. « (i) Article 135. From equation expressing the rela' tion between the power and the weight we see that i^ th<^ arm be increased the efficiency is increased. (2) I : 62.83. (3) i7i61bs. (4) i.Slbs. (5) 25.T32. (6) 3.1416. (7) .4398 inches. (8) '^^ n VIRTUAL VELOCITIES. Chapter IX. Exercise I. Page 131. (4) 3o!bs. (5) ^ inch, (6) 6 inches. (7) 3!!^ -. (8) I ft. (9) I Soft. (lo) 7lb5. (11) If P be one of tlie equal forces, then P X 137 + P X 105 r= 253 X 88 .'. P = 92ft)S. (13) 2 : 3; height 27 inches. (14) T44lbs. FRICTION. Chapter X. Exercise I. Page 136. (i) Arts. 134, 138. (2) 137. If a point be taken in either of the sides containing the angle and a perpendi- cular be let fall on the other side, the perpendicular of the right-angled triangle thus formed divided by the base is the co-efficient of friction. (4) Friction acting up the hill, the weight of the body and the reaction of the hill. (6) If the co-efficient of friction be-r-, the perpendicular is i and base V3, and, V3 therefore, hypotlicriusc 2, hence inclination =30^. i86 RESULTS, HINTS, 6-C (7) In this case the body is on the point of moving down the plane, and, therefore, friction acts with the / V 20 + iR power (art. 135). Hence — ^ — = -/j, and R= W |«^; Hence W = i3oIbs. (9) The friction acts with the power. Sohition same as last question ; P = 6.4 oz. (10) See Ex. 2. slbs. (ii) From the first part of the question we have F = W, that is //W = W, or /^ = I ; but /z = perp. divided by base. The perp! is, t'.ierefore, equal to base j hence angle = 45°. Exercise II. Page 138. (i) Take moments about the intersection of the two reactions. >u = J. (2) Let R == reaction of pavement, and let W be weight of man and ladder both acting at same point. Take moments about intersection of two reactions, and ^ R X V3 = W X ^ ; but W = R .-. ;, = JV3. (3) See example 3. 7!^ feet. (4) The ladder is kept at rest by R', reaction of verti- cal plane, R, reaction of horizontal plane, and friction R T, R' — ; R+ -- 4 6 = W. Take moments about foot of ladder, and R' |/(2i'» -x^) + -I^a: = — = ^R' X — . 6 26 2 /. x-=- 9.71 ft. from foot of ladder. (5) Let R, R' be reactions of plane and wall, respectively; |R, //R', friction of plane and wall W W respectively. Resolve W into — , — acting at top and foot of ladder. Then we have R' = f R, R+yuR' = Take moments about foot of ladder, and if distance of ladder from wall be «, we have u^a^-^a = Wa =(i} R' + |//R)a; JR', and JR respectively. Then R' 3W RESULTS, HINTS, 6fC. 187 (6) Same notation as in last. R'= — -, R+ -— = V3 V3 3 + 4- Take moments about foot of ladder, and we have 15R' -^ w: '^ 15V3 = Z ^ 5^3 + -^'^Z X 4 .-. ^ = 18.75, t^^at is, the weight cannot be suspended on a round higher than 18.75 ^^et from foot of ladder. APPLICATION OF SIMILAR TRIANGLES. Chapter XI. Exercise I. Page 144. (i) Draw FL, FM parallel to PA and WA respec- tively. Smce the lever is kept in equilibrium by the forces P, W, and the reaction at C, the resultant of P aud VV must pass through C ; hence AC is its direction, and by the parallellogram of forces, P and W must be prDportional to AM and AL, respectively, or P : VV :: FL : FM. Now, by construction PMF and FLW are both isosceles triangles similar to each other therefore FL : FM :: FW : FP, or P : W :: FW : FP. (2) Let E be end of rod from which W is hung, and T tension of string CB. The system is in equilibrium under action of forces, W acting vertically downwards at E, T acting at B in direction BC, and reaction (R) of hinge at A. Since last must be equal and opposite resultant of T and VV acting at D where they meet, AD must be its direction. Hence triangle ADC has its sides parallel and therefore proportional to the three forces, which acting at D keep the whole system in equilibrium, therefore T: W :: CD . AC; also from similar triangles CBA, EBD, CD : AE :: CB : AB, naaa i88 RESULTS, HINTS, C-^C. ■ ! ! I i CD .-. T = W — = AC AC W AE.CB = W AB ' AE.CB I AC 'AB' that is, T varies as ~ because W, AE, BC, aud AC are all constant. (4) Take moments about intersection of reactions and/'X3 =-- 2|W;.-. /=:4W. (5) Since the heights are proportional to the diame- ters of the bases, the height of the cavity is to 12 as 6 • 8 • hence it is equal to 9 inches. The volume of a cone the height of which is //, and radius r is tc r'-A. Hence volume of solid contained by outer conical sur- face IS 7r.64, and that contained by the inner surface is ^.27. Therefore volume of shell is n.^l- The C G of whole cone is 3 in. from base; that of cavity d in irom base. Hence ' ?r.64 X 3 = ;r.27 X 2J + 7r.37 X ^ ; n n ca 1 *?, f "^ ^'^^ inches, the distance of the C. Lr. of the shell from the base. (6) Let length of beam be 6a ■ then height will be xa, and length of string will be 4a ; evidently i ft. of beam IS resting on wall. Take moments about intersection of reactions and tension of string is found to be 14 4lbs (7). Smce AB ^ 13 and AD = 6, BD = 12 /also , ~AyV' ■'•^'■ougli A draw AE parallel to BD 'J'o" ^]^ '' ^^ '■'' ^'^ • ^^ •' therefore aIe =. 8. Also' f A V •■• ^ ; 'V tJ^erefore, EB = V41. The sides of the triangle ABE are parallel to the three forces vvhicii keep the point C at rest, and are therefore pro- partional to them. Hence, tension of BC : 13J :; \/4i :8 .-. tension of BC = fJV4i = 10.805 cwt. • Again, Pressure oh spar : 13^ :: 13 : 8 .-. Pressure on spar = ^ ^^ 'L== 3, .^33 cwt RESULTS, HINTS, &>€. 189 Or thus : From A draw AF perpendicular to AB. Then AF X 3^41 =10x12, each being double triangle ABC. . 40 .-. J\i< = -----. Let / = tension of stay. Take •v/41 moments about A, and / x -^°- = i3i x 5; .-. / = — as before. 16 From C draw CG perpendicular to BA produced Then 13XCG = lox 12, or CG = ^3- I^et / be pressure on spar. Take moment about C, and p x y^ = 13* X i5> or/ = -—-—---, as before. o It will also be a profitable exercise to solve by resolv- ing vertically and horizontally. (8) 6.276 cvvt. ; 37.4192. (9) Let /, /, be the lengths of the two planes. Let P, Q, be the two weights at a distance from the common vertex of the planes = x and y respectively, where x+y =: c, say. Through vertex of planes draw hue parallel to base ; take moments al out this line, and (P + Q)x —Vp + Qq-, where/ and q are perpendiculars from weights on parallel line through vertex. lih be height of planes, then — = X and q = hy h i P^ hx V Resolve both weights along the plane, and P--- = Q --. Hence we have (p+Q). = pi;f+Qi!: = p4?+pi^ = P-y(.v+7)=P~ = con- stant. .\ X ^ constant; that is, the centre of gravity is mdependent of the position of the weights, and is always at a constant distance from the parallel line through the vertex, and must therefore be in a straight Ime. (10) 7| feet 190 RESULTS, HINTS, ^C, Chapter XII. EXAMINATION PAPERS. Special Examination for Inspectors. Page 146. (i) o. (2) Art. 105 ; P : W :: I : 8. (3) Arts. 121, 123, 132, (4) In the first case we find 2P = W; in second case we find the Power equal to the Weight. .'. 2P would be required to sustain W. (5) The following clause of this question has been omitted: — "And the other end is sustained by a force of 4oll>s. acting at right angles to the length of the beam, and in a vertical plane passing through the beam." Let a = length of beam, and x = distance of lower end of beam from direction of wt. Take moments about foot of ladder, and 40^5 = loox, ox x •=z ^a. .*. base of right-angled triangle formed by the ladder is \a, and height will be |«. Hence sides of right-angled triangle are in the ratio of 3, 4, 5. Produce reaction of ground till it meets direction of force 4ott)s. If x be distance from pt. of intersection to horizontal plane, by similar triangles — = J, .*. ^=y. Take moments about intersection of reaction and direc- tion of 40, and Fxy = 100x2 .*. F = 32. Take moments about top of ladder, and Rx4=:32x3 + 100 x a. .'. R = 74. Second Class Certificates. Page 147. (i) 4lbs. and 6flis. (2) Art. 124. (3) The- Centre ot Gravity is at the point B. (4) a 26, ^ 27. (5) 60ft. of height for looft. of length. (6) (a) Weight of beam, acting vertically downwards at its middle point ; the pressure exerted by the man acting as described in the question ; the reaction of the ground against the end of the beam which rests on the gjound, acting vertically upwards ; and friction, acting horizontally, (b) 5o]bs. RESULTS, HINTS, &>€. First Class, July I871. Page 14a 191 (i) (a) Art. 123. (d) 131. (2) 5 at A, 4 at B, 6 at D, 5 at E. (3) 4^5lbs- (4) 47. Second Class, December 1871. Page 148. (i) Arts. 97 and 99. (2) Arts. 118 and 119. (3) Art. 132. (4) The forces are equivalent to 2 AC. The required line will be equal to 2AC^nd a direction CA. (5) 2jn>s. First Class, December 1871. Page 149. (i) Art. 128. (2) i4lbr,. Normal School, 1871. (i) 4olb5. (2) ioofl>s. (3) i454tt)s. (4) Arts. 66 and 69. (5) On CA, 525lbs. and on DB, 475^3. Second Class, July 1^2. Page 150. (i) Art. 88. (2) Take moments about fulcrum and force = . (3) Art. 118. (4) 8oIbs. (5) Length of line representing resolved parts is 12ft., and magnitude of resultant is 13ft. (6) Art. 136. The friction is independent of the velocity when there is sliding motion. First Class, July 1872. Page 151. (i) From C draw CE, CF, perpendiculars on AD, BD respectively. Since the lever is at rest, the moments about C must be equal in magnitude and opposite in direction ; that is AD.CF - BD.CF; .'. triangle ACD =-■■ triangle BCD ; .-. AC = CB. (2) Resolve each force vertically and horizontally. The horizontal components are i, i, acting in opposite directions and, therefore, neutralizing each other. The sum of the vertical components is the resultant and is at right angles to AB. 193. RESULTS, HLVrS, 6*C Second Class, July 1873 Page 151. (i) Art. 88. (2) {a) Friction, art. 135. (b) Art. 136. (3) Art. 107. (4) Arts. 121, 123, and 129. (5) Arts. 134, i35» a»d 137. First Class, July 1873 Page 152. (1) Resolve tlie forces vertically and horizontally. (2) See note to question 11, page 40, and question 12, page ii5;P = 62ilbs (3) Let 62J be at A, and suppose it drawn up to L ; then 100 will descend to a point D, 25ft, from B. Through D draw DE parallel to AC. Then BE : 25 :: // : 40 or BE = ^h. The Principle of Virtual Velocities asserts that (i2\. force, and will, therefore, be equally inclined to the lines of action of the 61bs. and 4lbs forces. (3) Arts. 52 and 58. {d) 24oB>s. (4) See Ex. 4, page 84. I J inches from centre. 196 RESULTS, HINTS, 6fC. (5) The volume is proportional to the weight and may, therefore, be used for it. Take moments about base using volume for weight, and we have 3 r «+r,« + r,« + + rn •• First Class, July 1876. Page 161. (i) The enunciations of the Triangle of Forces given in articles 29 and 32 may be still further extended as follows : — If three forces, the magnitudes of which are proportional to the sides BC, CA, AB, of a triangle ABC, and of which the directions are either parallel to these sides or respectively inclined to them at equal angles, act upon a point they will produce equilibrium, and conversely. Three forces acting in consecutive directions round a triangle ABC (AC horizontal) are equivalent to a couple (Art. 49). To show this, through A draw AD equal and parallel to CB, and in it introduce a pair of balancing forces, each equal to CB. Of the five forces, three AC, AD and BA, are in equilibrium, and may be removed; there are then left two forces CB and DA, equal, parallel, and in contrary directions which consti- tute a couple. (2) The direction of the weight produced must pass through the fulcrum (art. 59). 'J'he question then becomes a simple geometrical one, viz., having given the three sides of a triangle to find the segments njade by a perpendicular from the vertex on the base. II x and^ be the segments, then ' — ^-a -_y 5: {x-^y) (x—y) = c (x—y) a'^^b'' = X X — V = and x-^y and X = fl2+.-«~^» 2C ff = a^+c^ -a' 2C X :y ::<2"+'8l!i has not been introduced into the Hinph School, Ingersoll, as a Text Book, but much of its contents has been brought before tlie notice of the students in the form of lectures. '* I have carefully examined the work, and I have no hesitation in pronounc- ing it superior to anvthing yet presented on the subject of English. It is especially adapted to High School work. I shall be gratified to learn that it is placed on the list of authorised Text Books.*' IT ri.EMl>0 8 ANALYSIS OF THE EN'OLISII LANGUAGE. II. M. H1CK8, MA., Head Mrtsitr, If. S., Trenton 'I hivfc u^ed the work and think it as good aa any I have seen on thfi subject." 8. :e p. C. McGREOon, D.A , Heatf .Vasfer, H.S., Almonte. "I have already introducjd FUminjj's Analysis into our IIf(fh School here. ,In the senior clasi I And it an excellent work : (ar tx.veding any jrramiiiar we have yet had." o j o A Slsci,AiR, M.A., Head Muster, IfJS., lyhidsor. " Fleming's Analysis is U8c«l by the teachers of the Windsor High School in giving notes on Grammar. &c. We hai^e not u id it as a text boik yet. since it is not authorized. The work is certainly very go^d." JAMK8 TORSBILL, B.A., Head Master, H. S., Clinton. " I winsider it an excellent book of reference, and as such It is In th« bands of some of the pupiU of the senior class." « D. McBridk, B, A., Head Master, H. S., Port Perry. "I have used 'Fleming's Analysis* in the senior grammar class in this school, and I consider it an excellent work." J. R. WiGllTMAN, M.A., Head Master, H. S., Newcastle. ♦ •» "I^/^'"® ****• n"'"'' vf ""y.^'ehly. The list of examination questions appended to It I find especially useful. ' n "« P. A. SwiTZER, B. A., Head Master, H. $., Oakt'ille. ••Our English Master uses Fleming's Analysis as a book of reference. He would like to see it allowed as a text book." Rev. F. F. Mac.nab, B A., Head Master, H. S.,-Carlton Place. "1 regard it as the very best I have seen on that most important department list o?text*book" • *°" "° hesitation in introducing it if on the authorized James Mills, M.A., Head Master Brant/ord Collegiate Institute. -et for Sl^h tchoS^ovV^ *"^'"^ *^'* ^ '*'"''^''" '' °"' "' *•*" ^^ ^"^^^ '^« <=*" S, Woods, M.A , Rector, Kingston Collegiate Institute. "It fills a place too long unoccupied, by giving in small space, more genuin* i?n aruaiTteT'""*" ^"^'"® *"*' ***''®'^ Vt'^rawy text book with which I J. Campbell, M.A., Had Master, H. S., Napane; " ■■«» High School purposes, such a work is just what is needed." fLKMING'S ANALYSIS O? THE ENOLIiH LANOUAGF, Wm. Coc;jra\f, D.D., PresUient of Fncnlty Young Ladies' College, Brantford, " \Vc have need ' Fleming's Analysis ' in the Brantford Ladies' Collejpe for t!" pas) two yeirs with >,'roiit stti^ritction. We reifard it as admirably adapted for th;L i-enior cliWfses nf oiir l)i^^her schools and collotfcs. Your Ann have conferred a favour tif>on the frie'idh of etluu.ition in our laud, by the Issue of a Canadian edition of this useful text book." Wm. Tassik, LL.D., Gall Coiiegiaie Itisiitutt. \ " I am convinced thut from the extent of ground gone over, and the amount of |» nforination contained in it, and that from sources from which many are precluded. \. that it will ho a valuable acquisition tj students qrcnerally, and especially to Public tichool T^ar.Iifjrs." • James IIuours, P. 5. Inspector, Toronto. '• I regard It as a very able and exhaustive work. The explanations of the principles of Kn^lish GianiiMar are clear and simple, and ought to do s great deal towards removiujj the mystciies and difficulties which some treatises have thrown around this important subject. The numerous exercises which it contains, render it of (jreat value to teachers." GtoBB, May 1 "For High School purposes and for use in the higher classes of our Publ'-o Schools, Mr. Fleming's IJook is unq\iestionably the very best at present availabl No other volumu yet pulilished contains in so small a space so much valuable in« ter. Wbll arranged and nitelligibly expressed as this one does. It is not surprising, therefore, that it lus midt; ius way nipidly into favor in the schools of Canada, or that a publiinly necessary papers, and, • to which the the etymolo- Iditional con- r educational \x and High are claimed, ihridge Local Etymological