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 ELEMENTARY TBEATISE 
 
 STATICS, 
 
 BY GASPARD MONGE. 
 
 ft !J3ingnt{itjinil Sntia nf tljB Slutljnt 
 
 TRANSLATED FROM TUB FRENCH, 
 
 BY WOODS BAKER, A.M. 
 
 OF THE UNITED STATES COAST SURVEY. 
 
 PHILADELPHIA: 
 
 E. C. & J. BIDDLE, No. 6 S. FIFTH STREET. 
 1851. 
 
ENTERED according to Act of Congress, in the year 1851, by 
 
 E. C. & J. BIDDLE, 
 
 in the Clerk s Office of the District Court of the United States, in and for the 
 Eastern District of Pennsylvania. 
 
 STEREOTYPED BY L. JOHNSON AND CO. 
 
 PHILADELPHIA. 
 PRINTED BY T. K. AND P. G. COLLINS. 
 
PREFACE 
 
 AMERICAN EDITION. 
 
 A WANT has long been felt in this country of a good 
 elementary treatise on Theoretical Mechanics. The 
 books on that subject, in the English language, are 
 mostly voluminous, and either adapted to the compre 
 hension of those only who have mastered the various 
 branches of analytical mathematics, or composed chiefly 
 of practical and descriptive details. Hence, an accurate 
 knowledge of the general laws, or grand fundamental 
 truths of mechanics, so important to all men, in this 
 eminently practical age and country, and especially to 
 those who have some one of the useful arts as their 
 daily occupation, has hitherto been attainable only by 
 highly educated men. 
 
 One of the consequences of the want of a familiar 
 acquaintance with the mechanical laws, upon which all 
 machines, of whatever possible kind, must depend, is 
 the large number of failures of inventions occurring 
 every year. The authors of such attempts generally 
 
 3 
 
4 PREFACE. 
 
 have mental ingenuity enough, but, unfortunately, they 
 have not the knowledge necessary to render their con 
 trivances consistent with the laws of Nature, or adapted 
 to attain the proposed ends by the best possible means. 
 Machines, deficient in either of these essential matters, 
 must sooner or later be discarded; and their disap 
 pointed inventors have then to regret the loss of their 
 money and time, which proper information would have 
 prevented. 
 
 To supply this urgent need, in part, the following 
 little book upon Statics, or the science which treats of 
 the equilibrium of forces applied to solid bodies, has 
 been translated. It has long been known and highly 
 admired by those who are familiar with the scientific 
 literature of France ; but to persons who have little or 
 no acquaintance with French authors on such subjects, 
 it may be well to mention, that correct information, so 
 well digested, precise and clear, can be obtained from 
 the literature of no other nation or language. 
 
 From the advertisement prefixed to the seventh 
 French edition of this treatise, which has gone through 
 eight editions in France alone r besides several that have 
 been published at Brussels, the following extracts are 
 made : 
 
 " The first edition of this work, which appeared in 
 1786, was specially intended for young candidates 
 
PREFACE. 5 
 
 for the Navy; now, it is one of the standard books 
 most generally followed. A correct and clear style, 
 rigorous demonstrations, and well connected proposi 
 tions, have long caused it to be preferred for instruction 
 in Statics. It is the first book in which there has been 
 collected all that can be demonstrated in Statics syn 
 thetically. After having studied Euclid s Geometry, 
 this work will be read without difficulty. Being mo 
 delled upon the method of the ancient geometers, it 
 presents very clear ideas upon an abstract science, of 
 which a great number of useful applications are made." 
 
 " A profound knowledge of Statics requires the aid 
 of mathematical analysis, that is to say, of Algebra, and 
 the Differential Calculus ; but it is as important for be 
 ginners to study synthetical statics before analytical 
 statics, as it is fit to precede the study of analytical 
 geometry by that of elementary geometry." 
 
 " The discussion of the elementary machines offers to 
 M. Monge an occasion for showing the truth of a prin 
 ciple, which the author of the Mecanique Analytique 
 (Lagrange) has rendered so productive, and which is 
 known as the principle of virtual velocities." 
 
 The treatise on Statics, of Monge, is a necessary 
 introduction to the work of Poisson, which is a large 
 and thorough analytical treatise, composed for the 
 pupils of the Polytechnic School of Paris. 
 
D PREFACE. 
 
 The translator would here acknowledge his obligation 
 to his friends, R. S. McCulloh, Esq., Professor of Na 
 tural Philosophy in the College of New Jersey, and 
 J. B. Reynolds, Esq., Engineer and Lecturer on Me 
 chanical Philosophy before the Franklin Institute, Phila 
 delphia, for their kindness in giving him valuable advice 
 and assistance in preparing this work for publication. 
 
 Washington, Nov. 1850. 
 
BIOGRAPHICAL NOTICE 
 
 OP 
 
 G A SPAED MONGE. 
 
 IT may be neither uninteresting nor out of place here, to 
 give a slight sketch of the life of the author of this work, 
 who was one of the most distinguished of the eminent men of 
 science in France at the close of the last century. It is 
 selected from two papers,* written by his pupils, MM. Brisson 
 and Dupin, and presented by M. Delambrc in his Analysis of 
 the Labors of the Academy of Sciences during the year 1818. 
 
 Graspard Monge was born in 1746 His progress 
 
 was so meritorious as to procure for him the Professorship of 
 Physics in the College of Lyons, the year after he had com 
 pleted his studies in that institution Having 
 
 gone to Beaune during vacation, he undertook to draw the 
 plan of that city ; for which purpose he was obliged to make 
 the necessary instruments. He presented his labor to the 
 rulers of his native city, who compensated the young author 
 as generously as the limited means of the public purse would 
 
 * Notice Historique sur Gaspard Monge, by M. Brisson ; and Essai Historique sur Us 
 Set-vices et les Travaux Scientifiques de G. Mange, by M. Dupin. Sec the advertisement 
 to the 7th edition of G&ometrie Descriptive, par G. Mongc, Paris, 1847. 
 
 7 
 
8 BIOGRAPHICAL NOTICE. 
 
 permit. A lieutenant-colonel of military engineers, who was 
 then at Beaune, obtained for Monge the position of draughts 
 man and pupil in the school for engineers, and conductors of 
 
 works of fortification As he designed with rare 
 
 ability, his manual talent only was regarded , but he already 
 felt his strength, and could not think, without indignation, of 
 the exclusive esteem accorded to his mechanical skill. He 
 said long afterwards : " I was tempted a thousand times to 
 tear my drawings to pieces through spite at the ado which 
 was made about them, as though I had no ability for produc 
 ing any thing else." The director of the school caused him 
 to make some practical calculations of a particular case of de 
 filement, an operation serving to combine the relief and trace 
 of a fortification, so that the defender may be sheltered from 
 the fire of the assailant. Monge abandoned the mode pre 
 viously followed, and discovered the first geometric and general 
 method that has been given for this important operation. . . 
 .... By applying successively his mathematical talent to 
 different questions of an analogous kind, and generalizing his 
 means of conceiving and operating, he at length succeeded in 
 forming a body of doctrine, which was his Descriptive Ge 
 ometry For more than twenty years he found it 
 
 impossible to get the application of his geometry to draughts 
 of carpentry taught to the corps at Mezieres. He was more 
 fortunate, however, in the application to stone cutting; he 
 carefully followed the methods employed in this art, and im 
 proved them by rendering them more simple by means of his 
 geometry. 
 
 His scientific labors secured him the nomination of Lecturer 
 on Mathematics and Physics, to succeed Nollet and Bossut. 
 He was afterwards appointed the regular professor, when, 
 turning his attention to a series of natural phenomena, he 
 made numerous experiments upon electricity, explained the 
 phenomena relating to capillarity, and was the founder of an 
 
BIOGRAPHICAL NOTICE. 9 
 
 ingenious system of meteorology. He effected the decomposi 
 tion of water, and arrived at this great discovery without 
 having had any knowledge of the researches made a short 
 time before by Lavoisier, Laplace, and Cavendish. He was 
 not satisfied with explaining the theories of science and their 
 application to the students in the lecture rooms ; but liked to 
 conduct his pupils wherever the phenomena of nature and the 
 works of art could render these applications interesting and 
 apparent to the senses. He infused into his pupils his own 
 ardor and enthusiasm, and rendered delightful, observations 
 and investigations, which, had they been viewed only abstractly 
 in the narrow precincts of a lecture-room, would have appeared 
 a wearisome study. 
 
 In 1780, to attract Monge to Paris, he was associated with 
 Bossut as Professor of the course of Hydrodynamics, estab 
 lished by Turgot. In order to reconcile the duties of the 
 two places he had to fulfil, he passed six months of the year 
 at Mezieres, and six months at Paris. The same year he was 
 elected a member of the Academy of Sciences; and, at the 
 death of Bezout, in 1783, was chosen to replace that cele 
 brated examiner for the Navy. The Marquis de Castries fre 
 quently requested Monge to re-write the Elementary Course 
 of Mathematics for the pupils of the Navy ; but he always 
 declined. "Bezout has left," said Monge, "a widow who has 
 no other fortune than the writings of her husband, and I can 
 not think of snatching away the bread from the wife of a 
 man who has rendered such important service to science and 
 the country." The only elementary work which Monge pub 
 lished, was his Treatise on Statics ; and, with the exception 
 of a few passages, where the evidence adduced might be some 
 what more rigorous, the Statics of Monge is a model of logic, 
 clearness, and simplicity. 
 
 At an epoch when the public distress called into the higher 
 ranks all useful and courageous talent to the succor of the 
 
10 BIOGRAPHICAL NOTICE. 
 
 country, which was threatened with an invasion, Monge was 
 appointed Minister of the Navy. He did every thing to retain 
 in France, men eminent for their merit or bravery ; and de 
 scended even to petition, to obtain the continuation of the 
 services of Borda, in which he had the good fortune to suc 
 ceed. He was one of the most active men in scientific labors for 
 the safety of the state. To him is due the construction of the 
 new grinding apparatus in the powder works of Grenelle, and 
 the boring machines on the boats of the Seine. He spent 
 the daytime in giving instructions and directions in the work 
 shops, and the nights in preparing his Treatise on the Art of 
 making Cannons; a work intended as a manual for the di 
 rectors of foundries, and for artizans. 
 
 It was in his course of lectures at the Normal School that 
 he delivered, for the first time, his Lessons on Descriptive 
 Geometry, the secrets of which he had not been able sooner 
 to reveal. Another institution, which preceded the Normal 
 School in the order of conception, but which, being longer in 
 maturing by its authors, followed it in the order of execution, 
 realized in part the hopes that had been conceived in vain of 
 founding the first Universal School opened in France. Monge, 
 by uniting the result of long experience at Mezieres with his 
 original and profound views, arranged the plan of studies, 
 indicated their connection, and proposed practicable means of 
 execution. Out of four hundred pupils, who had first entered 
 the Polytechnic School, fifty of the best scholars were taken 
 to form a preparatory school ; and it was Monge almost alone, 
 who trained them. He remained the whole day amongst them, 
 giving them by turns lessons in Geometry and Analysis ; ex 
 horting, encouraging, and inflaming them, by that ardor, be 
 nevolence, and impetuosity of genius, which caused him to 
 exhibit to his pupils the truths of science with irresistible 
 force and charm. In the evening, Monge commenced his 
 labors anew : he wrote the pages of Analysis which were to 
 
BIOGRAPHICAL NOTICE. 11 
 
 serve for the text of his next lessons ; and on the following 
 day he was with his pupils at the first moment of their assem 
 bling. 
 
 The amiability of Monge was neither the premeditation of 
 the sage, nor even the effect of education : it was an unaffected 
 gentleness of disposition, which he owed to his happy organi 
 zation. He was born to love and admire. He was excessive 
 both in his admiration and his love ; hence he did not always, 
 perhaps, remain within the limits where passionless and cold 
 
 reason would have restrained him As he was the 
 
 father of the pupils in the bosom of the school, he was like 
 wise the friend of the soldier in camp. 
 
 While travelling through Italy, collecting the statues and 
 paintings ceded to France by treaty, Monge had been struck 
 by the strange contrast which the monuments of the Greeks 
 presented to those of the Egyptians, transported to the 
 banks of the Tiber by Augustus and his successors. The 
 comparative character of the antique monuments became the 
 subject of frequent conversation between the conqueror of 
 Italy and the commissioner, who collected for his country 
 the most beautiful of the fruits of victory. Monge con 
 ceived the idea of extending the domain of history beyond the 
 fabulous ages of Greece ; of ascertaining, with the certainty 
 of geometry, what the works of the ancient sages of the East 
 were ; of recovering, by contemplating their monuments, that 
 which had been the processes of their arts, the customs of 
 their public life, the order and majesty of their feasts and 
 ceremonies. 
 
 Charged by the General-in-chief, Napoleon, with bringing 
 the treaty of Campo Formio to the Directory, lie was, shortly 
 afterwards, in the first rank of those who composed the Com 
 mission of Science and Art, which accompanied the Egyptian 
 expedition. He was the first President of the Institute of 
 Egypt, formed upon the model of the Institute of France. He 
 
12 BIOGRAPHICAL NOTICE. 
 
 visited the Pyramids twice, saw the obelisk and the great walls 
 of Heliopolis, and studied the remains of antiquity scattered 
 around Cairo and Alexandria. During a wearisome march in 
 the interior of the desert, he discovered (as he supposed) the 
 cause of that wonderful phenomenon known as the mirage. 
 At the time of the revolt at Cairo, there were only a few de 
 tachments of troops in the city, and the palace of the Institute 
 was guarded only by the scientific corps. It was proposed 
 that they should make their way, sword in hand, to the head 
 quarters ; but Monge and Berthollet, considering that the palace 
 contained books, manuscripts, plans, and antiquities, the fruits 
 of the expedition, maintained that the preservation of this pre 
 cious deposit was the first duty of the scientific corps; and 
 they decided to die, if necessary, in defending this treasure, 
 rather than to .desert it. 
 
 Monge presided over the Commission of Science and Art of 
 Egypt; and by his counsels, contributed greatly to the judi 
 cious conception of the plan, its arrangement, the proportion 
 of the principal parts, and the means of improving the arts of 
 execution. 
 
 He had an inimitable manner of expounding the most ab 
 stract truths, and of rendering them clear by the language of 
 
 action It was, however, only by combating nature 
 
 that he became so excellent a professor : for he spoke with 
 difficulty, indeed almost stammered; his utterance, causing 
 him to drawl some syllables and utter others with too great 
 rapidity. His physiognomy, habitually calm, presented an 
 aspect of meditation; but when he spoke, he appeared sud 
 denly as though a different man : a new fire instantly lighted 
 up his eyes, his features became animated, and his figure in 
 spired 
 
 Enfeebled by age, Monge became the victim of an imagina 
 tion, which, according as the times were adverse or propitious, 
 carried him beyond well founded fears or hopes 
 
BIOGRAPHICAL NOTICE. 13 
 
 The regulations of the service did not permit generous youth, 
 at his funeral, to deposite the palm of gratitude and regret upon 
 the tomb of their first benefactor; but, with the early dawn 
 which followed the day of his obsequies, the pupils silently 
 wended their way to his place of sepulture, and deposited there 
 a branch of oak, to which they suspended a laurel crown. 
 Twenty-three of the former pupils of the Polytechnic School, 
 all residents of the city of Douai, united spontaneously, and 
 decided to write in common to M. Berthollet, begging him to 
 superintend the erection of a monument to be raised at the 
 expense of the old pupils of the Polytechnic School, in honor 
 of Gaspard Monge. 
 
TABLE OF CONTENTS. 
 
 PAGK 
 
 BIOGRAPHICAL NOTICE OF MONGE 7 
 
 DEFINITIONS 15 
 
 CHAP. I. ON THE COMPOSITION AND DECOMPOSITION OF FORCES 17 
 
 II. ON MOMENTS 50 
 
 III. ON CENTRES OF GRAVITY 87 
 
 IV. ON THE EQUILIBRIUM OF MACHINES 118 
 
 ART. I. On the Equilibrium of Forces which 
 act upon each other by means of 
 Cords 120 
 
 ART. II. On the Equilibrium of the Lever 130 
 
 Of Pulleys 141 
 
 Of the Wheel and Axle 153 
 
 Of Cog Wheels ICG 
 
 Of the Jack Screw 1G8 
 
 ART. III. On the Equilibrium of the Inclined 
 
 Plane 169 
 
 Of the Screw 185 
 
 Of the Wedge 193 
 
 NOTE, CONTAINING A DEMONSTRATION OF THE PARALLELOGRAM 
 
 OF FORCES, BY M. CAUCHY 203 
 
 14 
 
AN 
 
 ELEMENTARY TREATISE 
 
 ON 
 
 STATICS. 
 
 DEFINITIONS. 
 
 EVERY thing that is capable of affecting our senses is 
 called a body, or material substance. 
 
 Bodies are divided into solids and fluids. A body is 
 solid when the molecules which compose it are cohesive, 
 and cannot be displaced among each other without 
 effort: the metals, stones, wood, &c., are of this num 
 ber. It is fluid when, on the contrary, all its molecules 
 can be separated with the greatest ease ; such are water, 
 air, &c. 
 
 All bodies are moveable, that is, they may be trans 
 ported from one place to another. A body is said to bo 
 at rest, when all the parts which compose it remain each 
 in the same place ; and it is said to be in motion, when 
 it changes place, or when the parts of which it is com 
 posed pass from one place to another. 
 
 A body at rest cannot enter into motion, and when 
 in motion, cannot change the manner in which it moves, 
 
 15 
 
16 STATICS. 
 
 without the action of some cause, to which has been 
 given in general the name of force or power. 
 
 There are to be considered in a force, 1st, its intm- 
 sity, that is to say, the effort which it makes to move 
 the body, or the particle of the body to which it is ap 
 plied ; 2d, its direction, or the straight line in which it 
 tends to move the point of the body upon which it acts. 
 
 When several forces are applied to the same body, 
 two cases may happen : either these forces counter 
 balance and mutually destroy each other, when they 
 are said to be in equilibrium ; or, by reason of the action 
 of all these forces, the body enters into motion. 
 
 Hence, the term Mechanics is given to that science 
 whose object it is to find the effect which the applica 
 tion of determined forces must produce upon a body. 
 This science is divided into two parts : the first con 
 siders the relations which the forces should have, in in 
 tensity and direction, so as to be in equilibrium, and is 
 called Statics; the second, to which the name Dynamics 
 has been given, determines the manner in which the body 
 moves, when these forces do not entirely destroy each 
 other. 
 
 Each of these parts is again divided into two others, 
 according as the body, to which the forces are supposed 
 to be applied, is solid or fluid. The part of Statics 
 which treats of the equilibrium of forces applied to 
 solid bodies, is named simply Statics, or Statics proper; 
 and Hydrostatics is that which has for its object the 
 equilibrium of forces applied to the different molecules 
 of a fluid body. 
 
 In this Treatise we will consider only the first of 
 these two parts, that is, Statics proper. 
 
IT 
 
 CHAPTER FIRST. 
 
 OF THE COMPOSITION AND DECOMPOSITION OF FORCES. 
 
 1. When a force p, applied to a determined point c 
 of a solid body AB, draws or pushes this body in any 
 direction CF, we may consider this force as though it 
 were immediately applied to any other point D of the 
 body, taken upon the direction of this force. 
 
 For all the points of the body, which are 
 in the straight line CF, can neither approach 
 towards, nor remove from, each other; and 
 none of them can move along this line with 
 out causing all the others to move in the 
 same manner as though the force were im 
 mediately applied to them. 
 
 We are likewise permitted to consider the force p as 
 though it were applied to any other point G, taken beyond 
 the body, upon its direction, provided this point is in 
 variably attached to the body. 
 
 2. Hence it follows that if, upon the direction of the 
 force P, there is found a fixed point D within the body, 
 or an immovable obstacle beyond it, provided in the 
 latter case the obstacle is invariably attached to the 
 body, the force will be destroyed, and the body will re 
 main at rest ; for this force may be regarded as imme 
 diately applied to the fixed point, and its effect will be 
 destroyed by the resistance of this point. 
 
 2* 
 
c 
 b 
 
 18 STATICS. 
 
 xfy- 2. 3. Reciprocally, if the force P, applied to the 
 body AB, is destroyed by the resistance of a 
 single fixed point, this point is found upon 
 the direction of the force : for this point can 
 r . B destroy the effect of the force only by op 
 posing the motion of the point of application 
 o c ; and it cannot prevent this motion, unless 
 it is upon the straight line which the force tends to 
 make the point of application traverse. 
 
 4. A point cannot move in several directions at the 
 same time. 
 
 5. When several forces, differently directed, are ap 
 plied at the same time to the same point, either this 
 point will remain at rest, or it will move in a single di 
 rection, and consequently in the same manner as though 
 it were pushed or drawn by a single force along this di 
 rection and capable of the same effect. 
 
 6. Thus, whatever may be the number and directions 
 of the forces applied at the same time to the same 
 point, there always exists a single force which can move 
 it, or tends to move it in the same manner as all these 
 forces together ; this single force is named the resultant 
 of the former ; and they, with reference to the resultant, 
 are named the component forces. 
 
 The operation, by which we seek the resultant of 
 several given component forces, is named the composi 
 tion of forces ; and that by which we find the compo 
 nents, when the resultant is known, is named the de 
 composition of forces. 
 
 7. Two forces are equal, when, being applied to the 
 same point and directly opposed, they destroy each other 
 and produce an equilibrium. 
 
COMPOSITION OF FORCES. 19 
 
 Reciprocally, when two forces are in equilibrium, they 
 are equal and directly opposed. 
 
 8. For if several forces, differently directed, be ap 
 plied to the same point, it is necessary, in order to 
 put them in equilibrium, or to destroy the effect of their 
 resultant, that a single force, equal to this resultant, 
 should be applied to this point and directly opposed to 
 it, or that several forces be applied, the resultant of 
 which is equal and directly opposed to the resultant of 
 the former. 
 
 9. Reciprocally, when several forces, differently di 
 rected and applied to the same point, are in equilibrium, 
 their resultant is zero ; or what is the same thing, any 
 one of these forces is equal and directly opposed to the 
 resultant of all the others ; or, lastly, the resultant of 
 any number of these forces is equal and directly opposed 
 to the resultant of all the others. 
 
 10. The resultant of two forces, applied to the same 
 point, is in the plane determined by the directions of these 
 forces ; and it is necessarily included in the angle 
 formed by these two forces. 
 
 If the resultant were not in the plane of the two 
 forces, there would be no reason why it should be above 
 the plane rather than below: it cannot be at the same 
 time in two different positions ; hence it is in the plane 
 of the two forces ; moreover, it is included in the angle 
 of the lines along which these forces are directed ; for 
 there is no force tending to move the point into the 
 space adjacent to this angle ; hence it will remain in 
 the angle itself. 
 
 11. A force is the multiple of another force when 
 
20 STATICS. 
 
 it is formed ly the union of several forces equal to the 
 latter. 
 
 Thus, if we apply to the same point and in the same 
 direction several forces equal to each other, and if we 
 take any one of these forces as unity, the multiple force 
 will be expressed by a number equal to that of the 
 added forces. 
 
 As it is always possible to compare numbers with 
 lines, we may represent a force by a straight line taken 
 upon its direction, and its multiple force by another 
 straight line, a multiple of the former ; in Statics, fre 
 quent use is made of this geometrical construction. 
 
 12. Three equal forces, which are applied to the 
 same point 9 and which divide the circumference, of which 
 this point is the centre, into three equal parts, are ne 
 cessarily in equilibrium. 
 
 The point to which these three forces are applied can 
 move only in a single direction ; but the line along 
 which it would move, could be placed in three or six 
 manners perfectly similar with reference to the three 
 forces ; now there is no reason that the motion of the 
 point should take place in one direction rather than in 
 another ; hence it will remain at rest. 
 
 The application of these three equal forces to the 
 same point proves, that there is evidently for this case 
 a resultant of these forces, since any one of the three 
 forces makes an equilibrium with the two others, and it 
 is evident that the direction of one of these forces di 
 vides the angle formed by the two others, into two equal 
 parts ; it is not less evident that the resultant of any 
 two equal forces meeting in a point, divides the angle 
 formed by these two forces into two equal parts. 
 
COMPOSITION OF FORCES. 21 
 
 13. If several forces are applied to the same point 
 along the same line and act in the same direction, it fol- 
 loivs, from proposition No. 11, that their resultant is a 
 single force, equal to their sum, acting along the same 
 line, and in the same direction. 
 
 Hence, to make an equilibrium of all these forces, it 
 is necessary to apply to the same point and in the oppo 
 site direction, a force equal to their sum ; for this force 
 will be equal and directly opposed to their resultant. 
 
 14. From this it follows : 1st. If two unequal forces 
 are applied to the same point in opposite directions, 
 their resultant is in the direction of the greater, and is 
 equal to their difference : for the greater of these two 
 forces may be regarded as composed of two others 
 having the same direction ; one of which is equal to the 
 smaller force, and the other equal to the difference : 
 now the first of these two latter forces is destroyed by 
 the smaller (7) ; then to move the point, there remains 
 only the difference, which is in the direction of the 
 greater. 
 
 2d. If any number of forces are applied to the same 
 point, some being placed in one direction, and the others 
 directly opposed, after having taken the sum of all 
 those acting in one of the two directions, and the sum 
 of all those acting in the contrary direction, the re 
 sultant of all these forces is equal to the difference of 
 the two sums (12), and is in the direction of the greater. 
 
 Hence, to produce an equilibrium of all these forces, 
 it is necessary to apply to the same point, and along 
 the direction of the smaller of the two sums, a force 
 equal to the difference of these sums ; for this force 
 will be equal and directly opposed to their resultant. 
 
22 
 
 STATICS. 
 
 THEOREM. 
 
 15. If at the extremities of an inflexible straight 
 line AB, two equal forces P, Q, be applied along the lines 
 AP, BQ, parallel to each oilier, and which act in the 
 same direction : 
 
 1st. The direction of the resultant R of these tivo 
 forces is parallel to the straight lines AP, BQ, and passes 
 through the middle of AB ; 
 
 2d. This resultant is equal to the sum P-fQ of the 
 two forces. 
 
 DEMONSTRATION. Let another in 
 flexible straight line DE be drawn 
 perpendicular to the directions of the 
 two forces P and Q, and invariably 
 connected with the straight line AB : 
 having prolonged the directions of 
 the two forces, we may suppose that 
 these two forces act (1) at the points 
 D and E ; moreover, we may apply to the same points, 
 the forces p, p , and q, q 9 equal to the forces P and Q, 
 so that the three equal forces P, p, p , meeting at the 
 point D, will divide the circumference, which has its cen 
 tre at the point D, into three equal parts, and so that 
 the three forces Q, q, q, equal to each other and to the 
 first, will divide in like manner the circumference which 
 has its centre at the point E, into three equal parts. 
 
 We have seen (12) that the force P is equal and op 
 posed to the resultant of the couple p, p r ; that the 
 
COMPOSITION OF FORCES. 23 
 
 force Q is equal arid opposed to the resultant of the 
 couple g, q ; hence the forces P and Q have a resultant 
 equal and opposed to that of the system of the two 
 couples g, q r and p, p ; the two forces p f q\ applied to 
 the point F of their direction, have for resultant a third 
 force equal to each of the first two, and acting in the 
 direction CF ; likewise the two forces p and q, applied 
 to the point G in the line of their direction, have for 
 resultant a third force equal to each of them, and di 
 rected along the line GO ; hence, the resultant of the 
 four forces p, p f , q, q r is a single force equal to two of 
 the others, and acting along the line GCF, which divides 
 the line DE into two equal parts ; hence the resultant of 
 the two equal forces P and Q is equal to the sum P-f Q, 
 and divides the line DE, or the line AB, to which these 
 forces are applied, into two equal parts. (See a second 
 demonstration of this theorem, No. 19.) 
 
 COROLLARY I. 
 
 16. Hence, to bring the two forces P and Q into an 
 equilibrium, it is necessary to apply to the middle K of 
 the straight line AB a third force equal to their sum, 
 which will act in a contrary direction, and which di 
 rection will be parallel to the two lines AP, BQ ; for this 
 third force will be equal and directly opposed to their 
 resultant. 
 
 COROLLARY II. 
 
 17. If an inflexible straight line be divided into any 
 number of equal parts, and we apply to all the points 
 
24 STATICS. 
 
 of division, equal and parallel forces, the resultant of 
 all these forces will pass through the middle of the line, 
 in a direction parallel to that of the forces, and will be 
 equal to their sum. 
 
 For, all the partial resultants of these forces, con 
 sidered two and two, and taken at equal distances from 
 the middle point of the line, will pass through this point 
 (15), in the direction of the forces, and each of them 
 will be equal to the sum of the two forces which com 
 pose it ; hence (13) the general resultant will also pass 
 through the middle of the line, in the same direction, 
 and will be equal to the sum of all the partial result 
 ants, that is to say, to the sum of all the component 
 forces. 
 
 THEOREM. 
 
 18. If at the extremities of an inflexible straight 
 line, two unequal forces P and Q be applied, tvhose lines 
 of direction AP, BQ, are parallel to each other, and 
 act in the same direction : 
 
 1st. The resultant R of these two forces is equal to 
 their sum, and its direction is parallel to that of these 
 forces ; 
 
 2d. The point of application c of the resultant di 
 vides the line AB into two parts reciprocally propor 
 tional to the two forces, so that we have 
 
 p : Q : : BC : AC. 
 
COMPOSITION OF FORCES. 25 
 
 DEMONSTRATION. Suppose, in 
 the first place, the two forces P and 
 Q are commensurable ; divide the 
 line AB into two parts AD, DB, pro 
 portional to the forces P, Q ; laying 
 off the line AD from A to E, and 
 the line BD from B to P, the line EF will be double AB ; 
 and since 
 
 p : Q : : AD : DB, 
 we shall have also, 
 
 p : Q : : ED : DF. 
 
 Dividing the lines AD, DB into as many parts as there 
 are units in the forces P and Q, and repeating this di 
 vision upon the lines AE, BF, the whole line EF will be 
 divided into twice as many equal parts as there are 
 units in the sum of the two forces P and Q ; now, the 
 middle of each of these divisions may be considered as 
 the points of application of forces equal to each other 
 
 p_f_ Q 
 
 and each equal to -7, (IT), m being the number of 
 
 units of the sum P+Q; hence the resultant of all these 
 forces will be also the resultant of the two forces P and 
 Q ; but the resultant of equal forces, distributed on each 
 side of the middle of equal divisions of the same line, 
 is equal to the sum of these forces, passes through the 
 middle of this line, and acts in the same direction as 
 the components : hence the resultant of the two forces 
 p and Q is equal to the sum P+Q, passes through the 
 middle point c of the line EF, and acts in the direction 
 of the forces P and Q ; moreover, from the construction 
 of Figure 4, CF= JEF=AB ; then subtracting the com 
 mon part CB, we have AC=BF=BD. For the same reason 
 
 3 
 
26 
 
 STATICS. 
 
 EC=JEF=AB ; hence CB=AE=AD ; hence the middle 
 point c of EF is such that we have P : Q : : CB : CA ; 
 hence this point of application divides the line AB into 
 two parts reciprocally proportional to the forces. 
 
 19. Let us suppose now that the two forces P and Q 
 are incommensurable. Apply to the points A and B, 
 and along the line AB, two equal and opposed forces s, s ; 
 
 the resultant of the two 
 forces P and Q will be the 
 same as the resultant of 
 the four forces P, s, Q, s ; 
 the couple P, s has for re 
 sultant a line Ai included in 
 the angle SAP ; the couple 
 Q, s 7 has for resultant a line 
 directed along BI included in the angle QBS ; supposing 
 these two resultants to be applied to the point of inter 
 section I of their directions, and drawing through the 
 point I a line GH parallel to AB, they will each be de 
 composed into two forces along the directions IH and 
 GH ; the forces in the direction GH being equal to s and 
 s and opposed, they will destroy each other ; the forces 
 in the direction IR combine together and are equal to 
 P+Q; hence, whether the two parallel and unequal 
 forces P and Q be commensurable or incommensurable 
 their resultant is parallel to their direction and equal to 
 their sum.* 
 
 * This demonstration is also true when the two forces P and Q 
 are equal to eacn other ; consequently if is applicable to the theorem 
 of No, 15. 
 
COMPOSITION OF FORCES. 27 
 
 Fi 9- 4 - 20. The resultant R of the two 
 
 forces P and Q, supposed to be in 
 commensurable, passes through a 
 point c in such a manner that we 
 have 
 
 p : Q : : CB :. CA. 
 
 For if it passed through another point H, situated be 
 tween A and c, we should find a force Q applied in the 
 direction Q, which would be such that the resultant of 
 the commensurable forces P and Q would pass through 
 a point K situated between the points c and H ; this 
 force Q would be the fourth term of the proportion 
 
 KB : KA : : P : Q , 
 the lines KB, KA, having for unit of measure a line 
 
 smaller than CH ; but we have > : hence the ra- 
 
 KA CA 
 P . P 
 
 tio is greater than the ratio -. Hence Q is smaller 
 
 than Q : consequently, the resultant of the two forces 
 P and Q will necessarily pass between the points K and 
 Q ; hence it is absurd to suppose that it passes through 
 the point H situated beyond the point K. It might be 
 demonstrated in the same manner that it cannot pass 
 through a point H situated between c and B ; hence it 
 passes necessarily through the point c. 
 
 COROLLARY I. 
 
 21. Hence, in order that equilibrium may ensue be 
 tween the two forces P$ Q, it is necessary to divide the 
 line AB at the point c into two parts reciprocally pro 
 portional to these two forces, and apply to the point c 
 
28 STATICS. 
 
 a third force, equal to the sum P+Q, parallel to them 
 and acting in the opposite direction. 
 
 Remark. 
 
 22. If the ratio of the forces P, Q, and the length of 
 the line AB, -were given in numbers, and it were desired 
 to find the distances of the point c from the points A, 
 B, the proportion 
 
 p : Q : : EC : AC 
 
 could not be employed directly, because in this propor 
 tion we would know only the first two terms ; but it is 
 easy to deduce from it the following, 
 
 P+Q : Q : : BC+AC : AC, 
 which, since BC+AC is equal to AB, becomes 
 
 P+Q : Q : : AB : AC, 
 in which the first three terms are known. 
 
 The distance BC can be found by the proportion 
 
 P + Q : P : : AB : BC, 
 which is likewise deduced from the first. 
 
 COROLLARY II. 
 
 23. When a single force R is applied to a point c of 
 an inflexible straight line, it may always be decomposed 
 into two others P, Q, which, being applied to the two 
 points A, B, given upon the same line, and being di 
 rected parallel to RC, produce the same efiect ; and the 
 intensity of the two forces is found by dividing the 
 force R into two parts reciprocally proportional to the 
 lines AC, CB, by means of the two following proportions : 
 
 AB : BC : : R : P, 
 
 AB : AC : : R : Q, 
 
 in each of which the first three terms are known ; for 
 
COMPOSITION OF FORCES. 29 
 
 the resultant of the two forces p, Q, has the same in 
 tensity, is parallel to, and acts in the same direction as, 
 the force R. 
 
 COROLLARY III. 
 
 24. Since Fig. 6 is the same in all respects as the 
 preceding, if we apply to the point c of the line AB a 
 
 19 ^ force s, equal and directly opposed to 
 
 the resultant of the two forces P, Q, in 
 such manner that we have 
 S=R=P+Q (19), 
 
 the three forces p, Q, s, will be in equi 
 librium, and each of the two forces P, Q, 
 may be regarded as equal and directly 
 opposed to the resultant of the two others. Hence the 
 resultant of the two forces s, Q, which are parallel, and 
 act in contrary directions, is a force p equal and directly 
 opposed to the force P. Now, the force P is equal to 
 the difference of the forces s, Q, and acts in the con 
 trary direction to the greater, s, of these two forces : 
 hence, 1st, the resultant p or p of the two forces S, Q, 
 is equal to their difference s Q, and acts in the direction 
 of the greater and parallel to these two forces. 
 Moreover, we have P+Q, or s : Q : : AB : AC (20). 
 Hence, 2d, the distances of the point of application A 
 of this resultant from the two points c, B, are recipro 
 cally proportional to the forces S, Q. 
 
 Remark. 
 
 25. If the ratios of the two forces s, Q, and the 
 length of the line EC, were given in numbers, and 
 
30 STATICS. 
 
 it were desired to find the distances of the point A 
 from the points B, c, the preceding proportion could not 
 be employed directly, because only the first two terms 
 of this proportion would be known ; but it is easy to 
 deduce the following from it : 
 
 s Q : Q : : AB AC or BC : AC, 
 in which the three first terms are known. 
 
 The distance AB can be found by another proportion, 
 
 s Q : S : : AB AC or BC : AB, 
 which likewise is deduced from the first. 
 
 COROLLARY IV. 
 
 26. If the two forces s, Q, which are parallel, and 
 act in contrary directions, be equal to each other, 1st, 
 their resultant P, which is equal to S Q (24), becomes 
 zero; 2d, in the proportion s Q : Q : : BC : AC, the 
 second term being infinitely great compared with the 
 first which is zero, the fourth term is also infinitely great 
 compared with the third. Hence the point of applica 
 tion A of the resultant P is at an infinite distance from 
 the point c ; hence, to produce an equilibrium between 
 the two forces s, Q, it would be necessary to apply to 
 the inflexible straight line a zero force, whose line of di 
 rection should be at an infinite distance ; which is not 
 absurd, but which cannot be executed. 
 
 We perceive, then, that it is impossible, by means of 
 a single force, to produce an equilibrium between two 
 equal and parallel forces, acting in contrary directions ; 
 but it will be demonstrated (51), that, by means of two 
 forces, an equilibrium can be produced between them in 
 an infinite number of ways. 
 
COMPOSITION OF FORCES. 
 
 31 
 
 PROBLEM. 
 
 27. Any number of parallel forces P, Q, R, s, . . . . 
 which act in the same direction, being applied to the 
 points A, B, c, D, . . . . of given position, and connected 
 together in an invariable manner, to determine the re 
 sultant of all these forces. 
 
 SOLUTION. Consider 
 ing first any two of these 
 forces, such as P and Q, 
 determine their resultant 
 T (18); this resultant will 
 be equal to P+Q; its di 
 rection will be parallel to 
 that of the forces P, Q, 
 and we will find its point 
 of application E, by the 
 following proportion (22) : 
 
 P+Q : Q : : AB : AE. 
 
 In place of the two forces P, Q, substitute their re 
 sultant T ; then having drawn the line EC, determine 
 the resultant v of the two forces T, R ; this resultant V 
 will be also that of the three forces P, Q, R ; its intensity 
 will be T + R or P+Q+R; and find on EC its point of ap 
 plication F by the proportion 
 
 T+R or P+Q+R : R : : EC : EF. 
 
 In place of the three forces P, Q, R, let their re 
 sultant v be substituted, and having drawn the line FD, 
 the resultant x of the two forces v, s, will be found ; 
 this resultant x will be also that of the four forces P, Q, 
 R, s; its magnitude will be v+s, or P+Q+R+S, and 
 
32 STATICS. 
 
 its point of application G upon FD will be found by the 
 proportion 
 
 V+S, or P+Q+R+S : s : : FD : FG. 
 By continuing in this manner, the position of the 
 general resultant of all the forces will be found, what 
 ever their number may be ; and the intensity of this re 
 sultant will be equal to the sum of all these forces. 
 
 COROLLARY I. 
 
 28. Hence, by supposing the point G to be connected 
 with the other points A, B, c, D, .... in an invariable 
 manner, an equilibrium will be produced between all the 
 other forces P, Q, R, s, . . . . by applying to the point G 
 a force parallel to the first, which acts in the contrary 
 direction, and which is equal to their sum 
 
 COROLLARY II. 
 
 29. If among the parallel forces P, Q, R, S, . . . some 
 should act in one direction, and others in the contrary 
 direction, we could determine, first (27), the partial re 
 sultant of all which act in one direction, and then the 
 partial resultant of all which act in the contrary di 
 rection. Thus all the forces would be reduced to two 
 others acting in opposite directions ; and by determining, 
 by the process of No. 24, the resultant of these last 
 two forces, supposed to be unequal, we should have the 
 general resultant, and consequently the force which, 
 being applied in the contrary direction, would produce 
 an equilibrium amon^ all the proposed forces ; if these 
 
COMPOSITION OF FORCES. 33 
 
 forces should be reduced to two parallel and equal 
 forces, we have seen (26), that it is impossible to pro 
 duce an equilibrium between them. 
 
 The general resultant being equal to the difference of 
 the two partial resultants (24), and each of these being 
 equal to the sum of those which compose it (27), it fol 
 lows that the general resultant is equal to the excess 
 of the sum of the forces which act in one direction 
 over the sum of those which act in the opposite di 
 rection. 
 
 COROLLARY III. 
 
 30. If the forces p, Q, R, s, . . . without ceasing to 
 be parallel and without changing in intensity, had 
 another direction and became j9, q, r, s, . . . the result 
 ant t of the first two would also pass through the point 
 E, and would be equal to the sum p + q. Likewise the 
 resultant v of the three forces p, q, r, would pass through 
 the point F, and would be equal to the sum p+q+r. 
 So also the resultant x of the four forces p, q, r, s, 
 would pass through the point G, and would be equal to 
 the sum p-\-q-\-r+s, and so on. Hence the general re 
 sultant of all the forces p, q, r, s, would pass through 
 the same point as the resultant of the first forces 
 P, Q, R, S, 
 
 Hence it appears that, when the intensities and points 
 of application remain the same, the resultant of these 
 forces always passes through a certain identical point, 
 whatever may be their direction ; and the intensity of 
 this resultant is always equal to their sum. 
 
34 STATICS. 
 
 The point through which the resultant of the parallel 
 forces always passes, whatever may be their direction, 
 is named the centre of parallel forces. 
 
 It is easy to see that if the points of application A, 
 B, c, D, . . . of the parallel forces P, Q, R, s, . . . are in 
 the same plane, the centre of these forces is also in this 
 plane ; for this plane contains the line AB, and conse 
 quently the point E of this line, which is the centre of 
 the forces P, Q : it contains also the line EC, and conse 
 quently the centre F of the forces P, Q, R ; it contains 
 the line FD, and consequently the centre G of the forces 
 p, Q, R, s, and so on. 
 
 It may be demonstrated in like manner that, if the 
 points of application are upon the same straight line, 
 the centre of the parallel forces is also upon this line. 
 
 LEMMAS. 
 I. 
 
 31. If a power P be applied to the circumference of 
 a circle movable about its centre A, and 
 in a direction BP tangent to the circum 
 ference, this force tends to turn the 
 circle about its centre, as though it were 
 applied at any other point c, and in a 
 direction CQ tangent to the same cir 
 cumference. 
 
 II. 
 
 A poiver P, applied along the direction of a line on 
 which there is a fixed point, is destroyed by the re 
 sistance of this point. (Nos. 2 and 3). 
 
COMPOSITION OF FORCES. 
 
 35 
 
 THEOREM. 
 
 32. When the directions of two forces P, Q, are con 
 tained in the same plane, and intersect in the same 
 point A, if we lay off in these directions the lines AB, 
 AC, proportional to these forces, so that we have 
 
 p : Q : : AB : AC, 
 
 and complete the parallelogram ABCD, the resultant of 
 these two forces will be in the direction of the diagonal 
 AD of the parallelogram. 
 
 DEMONSTRATION. Suppose, 
 for an instant, that the point 
 D of the diagonal AD is an im 
 movable obstacle ; from this 
 point let fall the perpendicu 
 lars DE DF upon the directions 
 of the two forces ; the trian 
 gles BED, CFD will be similar, 
 because the angles at B and c 
 being equal to A, will be equal 
 to each other, and we shall have 
 
 DC : DB : : DF : DE. 
 Now, we have by supposition 
 
 p : Q : : AB : AC, or : : DC : DB. 
 Hence we shall have 
 
 p : Q : : DE : DE. 
 
36 STATICS. 
 
 From the point r>, as a centre, and with a radius DF, de 
 scribe the arc FG, terminating in the prolongation of ED 
 at G ; then, regarding this arc and the line EG as inflex 
 ible lines, and connected in an invariable manner at the 
 point A, let us conceive the force P to be applied at the 
 point E along its direction, and a force M, equal to the 
 force Q, to be applied to the point G, in a direction 
 parallel to AP, and consequently tangent to the arc 
 FG. This being granted, since M=Q and DF=DG, we will 
 have 
 
 p : M : : DG : DE. 
 
 Hence (18) the resultant of the two parallel forces 
 p, M, will pass through the fixed point D, and will 
 be destroyed by the resistance of this point ; hence 
 these two forces will be in equilibrium around this 
 point. 
 
 Now the force Q, whose direction is tangent to the 
 arc FG, and which we may regard as applied to the point 
 F on its direction, tends to turn this arc in the same 
 manner as the force M (31), and may be substituted for 
 this latter force to counter-balance the force P : hence 
 the two forces P, Q, will also be in equilibrium around 
 the fixed point D ; hence their resultant will be de 
 stroyed by the resistance of this point, and conse 
 quently, (31) the direction of this resultant will pass 
 through the point D. 
 
 But the resultant of the two forces P, Q, should pass 
 through the point of intersection A of their directions 
 (4) ; hence this resultant will be in the direction of the 
 diagonal AD. 
 
COMPOSITION OF FORCES. 
 
 37 
 
 33. ANOTHER DEMONSTRATION.* 
 Suppose that the force Q acts at the 
 point c on its direction, being inva 
 riably fixed at the point A, and 
 that there is applied to the same 
 point c, and in the opposite direc 
 tions CM, CM , two forces M, M , 
 each equal to Q, the effect of the 
 two forces P and Q will be the same 
 as that of the four forces P, Q, M, M , since the last two, 
 M and M , destroy each other, being equal and opposed. 
 Now, these four forces form two couples ; the one, Q and 
 M intersect at the point c ; the other is composed of two 
 parallel and unequal forces P, M , applied to the line 
 AC; the resultant of the two equal forces Q, M, is di 
 vided along the line CK, which divides (12) the angle 
 MCQ into two equal parts ; the resultant HK of the two 
 forces P, M , applied to the points A, c, of the line AC, 
 passes through a point H of this line, so that we 
 have (18) 
 
 p : M , or Q : : HC : HA ; 
 
 moreover, it is parallel to the direction AB of the force 
 p. Hence the point K, the intersection of the two re 
 sultants CK, HK, is a point of the resultant of the four 
 forces P, Q, M, M , and consequently of the first two P 
 and Q. The point K is upon the direction of the di 
 agonal AD of the parallelogram constructed upon AB and 
 AC, as sides ; thus, by construction, the angle HCK is 
 equal to the angle KCD. Now, the angles KCD and CKH 
 
 * This demonstration differs but little from that given by M. 
 Poirisot, m the first edition of his Statique, in the year 1803. 
 
 4 
 
38 STATICS, 
 
 are equal, being alternate internal angles ; hence, in the 
 triangle CHK, the angles K and o are equal ; hence it 
 follows, that the line KII has the same length as the line 
 HC ; but we have the proportion : 
 
 p : Q : : no : HA ; 
 hence we will have 
 
 p : Q : : KH : HA, 
 and since 
 
 p : Q : : CD : AC , 
 
 the ratio of the lines AC and CD is the same as that of 
 the lines AH and UK ; hence the three points A, K, D, are 
 in a straight line, and this line is the diagonal of the 
 parallelogram constructed upon AB and AC as sides. 
 
 COROLLARY I. 
 
 34. If from any point D (Fig. 9), taken upon the 
 direction AD of the resultant of the two forces P, Q, the 
 lines DB, DC be drawn parallel to the directions of these 
 forces, a parallelogram ABCD will be formed, whose sides 
 AB, AC, will be proportional to the forces P, Q, that is to 
 say, we will have : 
 
 p : Q : : AB : AC, or : : DC : DB. 
 
 For if these sides were not proportional to the forces, 
 their resultant would be in the direction of the diagonal 
 of the parallelogram whose sides would be proportional 
 to these forces (32), and not in the direction AD, which 
 would be contrary to the supposition. 
 
 COROLLARY II. 
 
 35. If from any point D (Fig 9), taken upon the 
 direction AD of the resultant of the two forces P, Q, 
 
COMPOSITION OF FORCES. 39 
 
 the perpendiculars, DE, DF, be drawn upon the directions 
 of these two forces, these perpendiculars will be to each 
 other reciprocally as the forces P, Q. 
 For we have just seen (34) that 
 
 p : Q : : DC : DB ; 
 
 and the similar triangles DEE, DCF give 
 DC : DB : : DF : DE ; 
 
 hence 
 
 p : Q : : DF : DE. 
 
 THEOREM. 
 
 36. When the directions of the two forces P, Q, are 
 contained in the same plane, and coincide in a point 
 A, if the lines AB, AC, be laid off on these directions 
 proportional to these forces, so that 
 p : Q : : AB : AC, 
 
 and the parallelogram ABDC be completed, the resultant 
 R of these two forces will be represented in intensity 
 and direction l>y the diagonal AD of the parallelogram ; 
 that is to say, we shall have : 
 
 p : Q : R : : AB : AC : AD. 
 
 Fig. 11. 
 
 DEMONSTRATION. We have 
 
 / \ already seen (32) that the re 
 
 sultant of the two forces P, Q, 
 will be in the direction of the. 
 diagonal AD of the parallelo 
 gram ; it is only necessary to 
 show that its intensity will be 
 represented by this diagonal. 
 
 \ 
 
40 STATICS. 
 
 Let there be applied to the point A, a force s, equal 
 and directly opposed to the resultant R ; this force will 
 be in the direction of the prolongation of the diagonal 
 DA, and the three forces P, Q, S, will be in equilibrium. 
 Hence the force Q will also be equal and directly opposed 
 to the resultant of the two other forces P, S ; and con 
 sequently this last resultant will be in the direction of 
 the prolongation of the line CA. Let CA be laid off on 
 its prolongation from A to H ; draw the line HB, which 
 will be parallel to AD, and consequently to the direction 
 of the force s ; and through the point H draw HK parallel 
 to the direction of the force P : the two forces P, s, will 
 be to each other as the sides AB, AK, of the parallelo 
 gram ABHK (34), that is to say, 
 
 p : s : : AB : AK or IIB. 
 
 Now, by reason of the parallelogram, AD=BII ; more 
 over, the two forces S and n are equal ; hence we have, 
 
 p : R : : AB : AD. 
 But by supposition 
 
 p : Q : : AB : AC. 
 
 Hence, by combining these last proportions, we will have 
 p : Q : R : : AB : AC : AD. 
 
 COROLLARY I. 
 
 37. If the two forces P, Q, be applied to the point A, 
 equilibrium will be produced by applying to the same 
 point a third force in the direction AK, and proportional 
 to the diagonal AD ; for this force will be equal and di 
 rectly opposed to the resultant of the two forces P, Q. 
 
COMPOSITION OF FORCES. 
 
 41 
 
 If the forces P, Q be applied to other points of their 
 direction, equilibrium can be produced by applying to 
 any point of the line AD, and in the direction DA, a force 
 proportional to DA, provided the point of application of 
 this last force is connected in an invariable manner 
 with the points of application of the two forces P, Q. 
 
 Fig. 11. 
 
 COROLLARY II. 
 
 38. A force R, of given in 
 tensity and direction, may be 
 always decomposed into two 
 other forces P, Q, in the direc 
 tion of the given lines AP, AQ, 
 provided these directions and 
 that of the force R are included 
 in the same plane and coincide 
 in the same point A. 
 
 For this purpose, we will represent the force R by a part 
 AD of its direction ; then drawing through the point D 
 the lines DC, DB, parallel to the given directions AP, AQ, 
 we will form a parallelogram ABCD, whose sides AB, AC, 
 will represent the required forces P, Q ; for (36) the re 
 sultant of these two forces will have the same intensity 
 and the same direction as the force R. 
 
 The intensities of the two forces P, Q, may be found 
 by means of the proportions : 
 
 AD : AB : : R : P. 
 
 AD : AC : : R : Q. 
 
42 
 
 STATICS. 
 
 Fig. 11. 
 
 COROLLARY III. 
 
 39. In the triangle ABD 
 (Fig. 11), the sides AB, BD, AD, 
 are proportional to the sines 
 of the angles CAD, DAB, BAG, 
 formed by the directions of the 
 forces P, Q, R ; hence it follows, 
 that we shall have the propor 
 tion 
 
 A . A A 
 
 p : Q : R : : sin (Q.R) : sin (P.R) : sin (P.Q) 
 
 the notation (Q.R), &c., being the angle of the two forces 
 Q and R, &c. 
 
 The same triangle ABD gives 
 
 hence 
 
 COS ABD; 
 
 
 j^2pQ cos (P.Q). 
 
 PROBLEM. 
 
 40. To determine the resultant of any number of 
 forces P, Q, R, S . . . . whose directions, contained or 
 not contained in the same plane, meet in the same 
 point A. 
 
 SOLUTION. Lay off upon the directions of all the 
 forces, from the point A, the lines AB, AC, AD, AE . . . pro 
 portional to their intensities ; then, considering first any 
 two of these forces, as P, Q, complete the parallelogram 
 ABFC, whose diagonal AF will represent in intensity and 
 direction the partial resultant T of these two forces (36). 
 
COMPOSITION OF FORCES. 
 
 43 
 
 Fl ff- 12 - Instead of the forces 
 
 p, Q, take their resultant 
 T, and considering the 
 two forces T, R, complete 
 the parallelogram AFGD, 
 whose diagonal AG will 
 represent in intensity and 
 direction the resultant 
 V of the two forces T, R, 
 which will be that of the 
 three forces P, Q, R. 
 
 In like manner, instead 
 of the forces P, Q, R, take 
 their resultant v, and 
 considering the two forces V, S, complete the parallelo 
 gram AGUE, whose diagonal AH will represent, in intensity 
 and direction, the resultant x of the forces v, s, which 
 will also be that of the four forces P, Q, R, s. 
 
 By continuing in this way, the direction and intensity 
 of the general resultant of all the forces P, Q, R, s . . . 
 will be found, whatever may be their number. 
 
 COROLLARY. 
 
 41. If all the forces P, Q, R, s . . . . be applied to the 
 point of coincidence A of their directions, in order to 
 produce an equilibrium, find first the intensity and di 
 rection of their resultant (40) ; then apply to the point 
 A a force equal and directly opposed to it. But if the 
 forces be applied to other points of their directions, 
 connected together in an invariable manner, an equi 
 librium will be produced by applying to any invariable 
 point in the direction of their resultant, a force equal 
 
44 
 
 STATICS. 
 
 and directly opposed to this resultant, provided the 
 point of application of this force is also connected in an 
 invariable manner with the points of application of the 
 forces p, Q, E, S . . . . 
 
 PROBLEM. 
 
 42. To determine the resultant of any number of 
 forces, whose directions included in the same plane do 
 not meet in the same point, ivhose points of application 
 A, B, c, D . . . . are connected together in an invariable 
 manner, and whose intensities are represented by the 
 
 parts Aa, 
 
 cc, i)d, 
 
 Fig. 13. 
 
 of their directions. 
 
 SOLUTION. Hav 
 ing prolonged the di 
 rections of any two 
 of the forces, for in 
 stance P, Q, until they 
 meet somewhere in a 
 point E, lay off from 
 E to F and from E to 
 G the lines Aa, B#, 
 which will represent 
 these forces ; and 
 complete the paral 
 lelogram EFeG, whose 
 diagonal ~&e will re 
 present in intensity 
 and direction the re 
 sultant T of the two 
 forces P, Q (86). 
 
COMPOSITION OF FORCES. 45 
 
 Instead of the forces p, Q, take their resultant T, and 
 prolong its direction as well as that of the force R until 
 they meet somewhere in a point H ; lay off the line Ee 
 from H to I, the line cc from n to K, and complete the 
 parallelogram HI^K, whose diagonal ilk will represent in 
 intensity and direction the resultant v of the two forces 
 T, R, which will be also that of the three forces P, Q, R. 
 
 In like manner, instead of the three forces P, Q, R, 
 take their resultant v, and prolong its direction, as well 
 as that of the force S, until they meet in a point L ; 
 then laying off from L to M and from L to N the lines 
 H^, vd, which represent the forces V and s, complete 
 the parallelogram LM?N, whose diagonal Ll will represent 
 the resultant x of these two forces, which will also be 
 that of the four forces P, Q, R, s. 
 
 By thus continuing, the intensity and direction of the 
 general resultant of all the proposed forces will be 
 found, whatever may be their number. 
 
 COROLLARY. 
 
 43. Hence, when several forces, directed in the same 
 plane, are applied to points connected together in an 
 invariable manner, these forces always have one result 
 ant : thus it is possible to produce an equilibrium by 
 means of a single force, except in the case where the 
 direction of one of these forces being parallel to that 
 of the resultant of all the others, the force and the re 
 sultant would always be equal to each other and would 
 act in contrary directions ; for we have seen (26) that, 
 in order to produce an equilibrium in that case, it would 
 be necessary to apply a zero force in a line situated at 
 an infinite distance ; which is impracticable. 
 
46 STATICS. 
 
 THEOREM. 
 
 44. If three forces P, Q, R, be represented in intensi 
 ty and direction by the three sides AB, AC, AD, adjacent 
 to the same angle of a parallelopipedon ABFEGD, so 
 that 
 
 p : Q : R : : AB : AC : AD, 
 
 their resultant s will be represented in intensity and di 
 rection by the diagonal AE of the parallelopipedon ad 
 jacent to the same angle, and we shall have 
 p : Q : R : s : : AB : AC : AD : AE. 
 DEMONSTRATION. In the plane ABCF, which contains 
 the directions of the two forces P, Q, draw the diagonal 
 Fig - u - AF ; also draw the diagonal 
 
 DE in the opposite face DHEG : 
 these two diagonals will be 
 equal and parallel ; for the 
 two sides AD, EF of the paral 
 lelopipedon at the extremi- 
 ties of which they terminate 
 are parallel and equal : hence 
 AFED will be a parallelogram. This done, the two 
 forces P, Q, being represented in intensity and direction 
 by the sides AB, AC, of the plane ABFC, which is a paral 
 lelogram, their resultant T will be represented in mag 
 nitude and direction by the diagonal AF (36), and we 
 shall have 
 
 p : Q : T : : AB : AC : AF. 
 
 Likewise the two forces T, R, being represented by the 
 sides AF, AD, of the parallelogram AFED, their resultant 
 
COMPOSITION OF FORCES. 47 
 
 S, which will be also that of the three forces P, Q, R, 
 will be represented by the diagonal AE of the same par 
 allelogram, and we shall have 
 
 T : R : S : : AF : AD : AE. 
 
 Hence, combining the two proportions, we will obtain 
 p : Q : R : S : : AB : AC : AD : AE. 
 
 Now the diagonal AE is also the diagonal of the par- 
 allelopipedon ; hence the resultant of the three forces 
 p, Q, R will be represented in intensity and direction by 
 the diagonal of the parallelopipedon. 
 
 COROLLARY I. 
 
 45. A force S of given intensity and direction can 
 always be decomposed into three other forces P, Q, R, 
 in the direction of the given lines AP, AQ, AR, not in 
 cluded in the same plane, provided these three directions 
 and that of the force S coincide in the same point A. 
 
 Thus, through the three directions, considered two 
 and two, draw the three planes BAG, CAD, DAB ; repre 
 sent the force s by a part AE of its direction ; and 
 through the point E draw three other planes EGDH, 
 EHBF, EFCG, respectively parallel to the first three. 
 These six planes will be the faces of a parallelopipedon, 
 of which AE will be the diagonal, and of which the sides 
 AB, AC, AD, taken upon the three given directions, will 
 represent the intensities of the three required forces P, 
 Q, R ; for (44) the resultant of these three forces will 
 have the same intensity and direction as the force S. 
 
 Or else, draw through the point E three lines paral 
 lel to the directions AP, AQ, AR ; and the parts EF, EH, 
 EG, of these sides, included between the point E and the 
 planes BAG, CAD, DAB, will represent the intensities of 
 
48 STATICS. 
 
 the required forces P, Q, R ; for these lines, being three 
 
 sides of the parallelopipedon, are respectively equal to 
 
 the other sides AB, AC, AD, which are parallel to them. 
 
 COROLLARY II. 
 
 46. When the three forces P, Q, R, are perpendicular 
 to each other, the resultant s is the diagonal of a rectan 
 gular parallelopipedon, whose three sides adjacent to 
 the same summit of an angle are equal to the three 
 forces P, Q, R ; the intensity of this resultant is, in this 
 case, expressed by 
 
 COROLLARY III. 
 
 47. Whatever may be the number of forces P, Q, R, 
 s, . . . applied to the fixed points A, B, c, D, . . . we can 
 always conceive the system of three right lines, perpen 
 dicular to each other, to be transferred to the points of 
 application of the forces so as to occupy positions paral 
 lel to their former positions, and each of these forces to 
 be decomposed into three others, in the directions of the 
 three rectangular lines passing through the point of ap 
 plication ; then all the forces P, Q, R, S, . . . will be 
 decomposed into three systems of forces, so that all the 
 forces of the same system will be reduced to a single 
 force along the same direction ; hence all the forces P, 
 Q, R, s, will have three resultants parallel to the three 
 rectangular lines, fixed and determined in position with 
 reference to these forces. (See No. 53). 
 
COMPOSITION OF FORCES. 49 
 
 COEOLLARY IV. 
 
 48. Call s, s , s" . . . . the forces which act upon a 
 determined point ; and drawing through this point three 
 lines fixed and perpendicular to each other, each of the 
 forces s, s , s", . . . will be decomposed into three others 
 p, q, r, in the direction of the rectangular lines. 
 
 In like manner calling p\ q , r f , the three component 
 forces of the force s ; and p", q" 9 r", the three com 
 ponent forces of the force s", &c. ; the resultant of all 
 the forces S, s , s" will be the diagonal of a rectangular 
 parallelopipedon, whose three sides adjacent to the same 
 angle will be, 
 
 For the first, p+p +p"+ ; 
 
 For the second, q+ q +q"-\-. . . . ; 
 
 For the third, r+r +r"+ 
 
 Hence the expression for this resultant will be 
 
 THEOREM. 
 
 49. Two forces in the direction of lines which do 
 not meet cannot be reduced to a single force equivalent 
 to them. 
 
 DEMONSTRATION. Let P and Q be the two forces 
 whose directions do not meet. If a third force R causes 
 an equilibrium to subsist between them, any two fixed 
 points, one being taken upon the direction of this force 
 R, and the other upon the direction of the force P, will 
 necessarily destroy the force Q : now these two points 
 
50 STATICS. 
 
 cannot be so taken that the line which unites them will 
 not meet the force Q ; hence this force will not be de 
 stroyed ; hence it is absurd to suppose that the two forces 
 p and Q can have a single resultant R. 
 
 THEOREM. 
 
 50. All the forces P, Q, R, s, . . . applied to the 
 points A, B, c, D, . . . joined together in an invariable 
 manner., may in general be reduced to tivo forces in the 
 directions of lines which do not meet. 
 
 DEMONSTRATION. Having extended the lines in the 
 directions of the forces P, Q, R, . . . until they cut a 
 plane having a fixed and determined position with 
 reference to these lines, we may consider the points of 
 intersection as the points of application of the forces ; 
 now each force may be decomposed into two, one situ 
 ated in the plane and the other perpendicular to this 
 plane : all the forces directed in the plane will have one 
 resultant; the forces perpendicular to the plane, and 
 consequently parallel to each other, will have another 
 resultant. In some particular cases, these two resultants 
 meet, and all the proposed forces P, Q, R, S, . . . will be 
 reduced to a single one ; but in general they will not 
 meet : hence we shall have two forces, one situated in a 
 plane assumed arbitrarily, and the other perpendicular 
 to this plane, which will produce an equilibrium with 
 any number of forces P, Q, R, s, . . . . applied to the 
 points A, B, c, D . . . . It is necessary to except from 
 this general conclusion the particular case which we are 
 
COMPOSITION OF FORCES. 51 
 
 about to examine, and which takes place when the forces 
 situated in the plane and the forces perpendicular to 
 the plane, are reduced to one or more couples of equal 
 and parallel forces applied in opposite directions to the 
 same right line. 
 
 COROLLARY. 
 
 51. When two forces act along the direction of lines 
 which do not meet, there are an infinite number of sys 
 tems of two forces acting in the direction of other lines 
 which do not meet, whose action is equivalent to that of 
 the first two forces : in fact, any force may be decom 
 posed into two other forces, one perpendicular to a 
 plane assumed at will and the other situated in this 
 plane ; hence any two forces are equivalent to two other 
 forces, one being situated in the plane assumed arbi 
 trarily, and the other perpendicular to this plane. 
 
 PROBLEM. 
 
 52. Two forces being given in the direction of lines 
 which do not meet, to find tivo others equivalent to them, 
 one of which is in the direction of a line having 
 a given position. 
 
 SOLUTION. Let P and Q be the two given forces ; 
 having drawn a plane perpendicular to the line of given 
 position, decompose the forces P and Q into two others 
 P and Q , one having its direction in this plane and the 
 other perpendicular to the same plane ; decompose the 
 force Q , parallel to the given line, into two others q , q f , 
 parallel to Q , of which the one q will pass through the 
 
52 STATICS. 
 
 given line, and the other q" through a point of the force 
 P ; the two forces P , q", meeting in the same point, 
 will be reduced to a single force q ; hence the two forces 
 P and Q will be transformed into two other equivalent 
 forces q, q , the latter of which, </, will pass through a 
 given line. 
 
 53. Examination of a particular case of the compo 
 sition of forces applied to given points A, B, c, D, . . . 
 invariably joined together. 
 
 PROPOSITIONS. 
 
 Having decomposed each of the forces P, Q, R, s, . . . 
 applied to the points A, B, c, D . . . into two others, one 
 situated in a given plane K, and the other perpendicular 
 to this plane, let S and T be the resultants of these two 
 systems of forces ; it may happen that the first system of 
 forces, instead of having a single force S for resultant, 
 will only reduce to a couple of forces +s, s, equal, paral 
 lel, opposite, and applied to the same line ; in this case, 
 the three forces, T, + s, s, will be equivalent to two forces 
 beyond the plane of the couple +s, s: likewise, if 
 instead of a single resultant T, the forces perpendicular 
 to the plane K should reduce to a single couple +, t, 
 (designating by this expression two equal and parallel 
 forces, opposed and applied to the same line), the three 
 forces s,-f t, t, will also be equivalent to two forces, as 
 in the preceding case : finally, if all the forces acting in 
 the plane K and perpendicularly to this plane, should be 
 reduced to two couples +s, s, and -K, , these two 
 couples w r ould compose a single one. 
 
COMPOSITION OF FORCES. 58 
 
 The three propositions which have just been an 
 nounced are included in the two following : 
 
 1st. A force T and a couple +, 8 are equivalent to 
 two forces in the directions of lines which do not meet ; 
 
 2d. Two couples -fs, s and +, t are equivalent 
 to a single couple of a like nature : such for example as 
 -f r, r. 
 
 Demonstration of these two Propositions. 
 
 54. IST PROPOSITION. A force T and a couple +s y s 
 will combine into two forces : thus, the plane of the 
 couple being prolonged intersects the force T in a point 
 which may be regarded as the point of application of 
 the force T ; drawing any line through this point and in 
 the plane of the couple, and regarding this line as fixed 
 with reference to the three forces T,+s, s which are 
 applied to it, decompose the force T into two other 
 parallel ones t, t 9 which will have upon the fixed line 
 the same points of application as the forces -fs, s. 
 The forces , s, meeting in the same point, will have a 
 resultant ; so also will the forces , s : they will have 
 a second resultant ; these two resultants evidently will 
 be equivalent to the three forces T,+S, s. 
 
 55. If the plane of the couple were parallel to the 
 force T, it would be necessary to decompose the forces 
 
 Fig. 13. (.) O f the couple+s and s parallel to T. 
 Suppose this couple is applied to the 
 line AB (Fig. 13 a) perpendicular to 
 this line ; drawing through their points 
 of application A and B, the lines AM, 
 BN, parallel to the direction of the force 
 T, decompose the force +s into two 
 
54 STATICS. 
 
 others in the directions AM and AB, the force s into 
 two others BN and BA ; the forces along the direction 
 AB will be destroyed; the force along the direction AM 
 or BN, and the force T which is parallel to it, will com 
 bine to form a single force : hence, in this case, the three 
 forces T, +,, will reduce to two, one in the plane 
 of the couple +s, s, and the other parallel to this 
 plane. 
 
 56. 2D PROPOSITION. Two couples +, sand -H, , 
 situated in any planes whatever, will combine into a 
 single couple+r, r : thus, the planes of these couples 
 being prolonged will meet in the direction of a line 
 which may be considered as invariably joined to the 
 points of application of the forces which compose the 
 two couples. Let KL (Fig. 13 5) be this line, the in 
 tersection of the two planes LKMN, LZM N , one of which 
 contains the couple -fs, s, applied to the line AB, the 
 other the couple +, t, applied to the line CD ; the 
 directions of the forces +, #>~M> t intersect this 
 line in the points a, b, c, d ; dividing the lines ab y cd 
 Fig. 13, (b.) i n t two equal parts, and marking the 
 middle points / , /, we may transfer 
 the couple -H, t, applied to the line 
 CD, parallel to itself, so that the points 
 /, / will coincide ; we will then have a 
 new couple -M , t . It is necessary, in 
 the first place, to demonstrate that this 
 second couple, composed of forces equal 
 and parallel to those of the first and 
 applied to the line C D equal to CD, will be in equili 
 brium with it, and that in general we cannot change the 
 condition of equilibrium of two couples, by transferring 
 
 M 
 
COMPOSITION OF FORCES. 55 
 
 one of the couples parallel to itself in its plane. Now, 
 this proposition is evident, for the point o being the 
 middle of the line c d, the two forces + ,-, as well as 
 the two forces t 9 -\-t, which act at equal distances 
 from this point o, are in equilibrium ; hence the second 
 couple may be substituted for the first. Now, the force 
 +t f is decomposable into two other parallel forces pass 
 ing through the points a and b; the force ~t also may 
 be decomposed into two others passing through the 
 points b and a; and since c a=d f b, the component 
 forces of these two forces -H and , will be equal, and 
 will differ only in direction : hence, each of the three 
 forces meeting in the point a will be equal to one of the 
 three forces meeting in the point b ; hence, the result 
 ants of the two systems of forces applied to the points 
 a and b will be equal and opposed ; hence it follows 
 that the two couples +s, s and -H, - will be re 
 duced to a single one 4-r, r. 
 
 If the forces +t,t were parallel to the line LK, the 
 couple -fs, s (55) might be changed into another 
 + , s , whose forces would be directed parallel to the 
 line LK, and the four parallel forces -f, + s , , s 
 would be reduced to two equal and opposed forces 
 + (*+* ) and -(+<).* 
 
 * See the theory of couples in the Statique of M. Poinsot, 6th 
 edition, 1834. 
 
56 
 
 CHAPTER SECOND. 
 
 OP MOMENTS. 
 
 57. Two kinds of moments are to be considered. The 
 moment of a force referred to a point is the product of 
 this force multiplied by the perpendicular let fall from 
 the point upon the force ; the moment of a force referred 
 to a plane, is the product of this force multiplied by the 
 distance of its point of application from the plane : this 
 second kind of moment does not change, even though 
 the forces vary in direction ; they differ in this condi 
 tion from the moments of the first kind, which are inde 
 pendent of the points of application of the forces whose 
 direction is constant. 
 
 58. When the moments of several forces, referred to 
 the same point, are considered, this point is named the 
 centre of moments, 
 
 59. Hence it follows, if the intensity of a force be 
 know T n, and its moment referred to a centre or a plane, 
 and if the plane be parallel to the force, we will obtain 
 the distance of the centre, or the plane, from the di 
 rection of the force, by taking the quotient of the 
 moment divided by the force ; if the moment and the 
 distance be known, we shall obtain the intensity of the 
 force by taking the quotient of the moment divided by 
 the distance. 
 
MOMENTS. 
 
 57 
 
 15 - 
 
 COROLLARY. 
 
 60. When the two parallel forces P, Q, act in the same 
 direction, their moments, referred to any point c in the 
 direction of their resultant, are equal. 
 
 For if through the point c the 
 line AB be drawn perpendicular to 
 the direction of the two forces, and 
 terminate in A and B, this line will 
 be divided, at the point c, into two 
 parts reciprocally proportional to 
 ,the forces P, Q (18): that is to say, 
 we shall have 
 
 p : Q : : BC : AC. 
 
 Hence, the product of the extremes being equal to 
 the product of the means, we shall have 
 PXAC=QXBC. 
 
 THEOREM. 
 
 61. If at the extremities A, B of an inflexible right 
 line, two parallel forces P, Q be applied which act in the 
 same direction, and if through the point of application 
 c of their resultant, a right line DB be drawn in any 
 plane, upon which the perpendiculars AD, BE be dropped 
 from A and B, ive shall have 
 
 PXAD = QXBE. 
 
58 
 
 STATICS. 
 
 Fig. 15. DEMONSTRATION. The right an 
 
 gled triangles ADC, EEC, which are 
 similar, because the angles opposite 
 the summit c are equal, give 
 
 EC : AC : : BE : AD. 
 ow we have (18) 
 
 p : Q : : EC : AC. 
 Hence 
 
 p : Q : : BE : AD ; 
 and, the product of the extremes equalling that of the 
 
 means, 
 
 PXAD = QXBE. 
 
 THEOREM. 
 
 Fig. 16. 
 
 
 * 
 
 f AB 
 
 62. Two parallel forces P, Q, 
 which act in the same direc 
 tion^ being applied to the points 
 A, B of an inflexible right line, 
 and through a point F of this 
 line, the line FH being drawn 
 in any plane : 
 
 1st. If the point F be taken 
 upon the prolongation of AB, 
 and if from the points A, B, 
 and from the point of appli 
 
 cation c of the resultant, the perpendiculars AG, BH, ci, 
 
 be dropped upon FIT, we shall have 
 
 11XCI=QXBH + PXAG ; 
 
MOMENTS. 
 
 59 
 
 Fi 9- 17. 2d, If the point F be taken be 
 
 tween A and B, we shall have 
 RXCI=QXBH PXAG. 
 
 DEMONSTRATION. Through the 
 point c, draw DE parallel to FH, 
 intersecting the perpendiculars AD, 
 BH in the points D, E ; we have 
 DG=CI=EII, besides (61), PXAD=QXBE. 
 
 Fig. 16. 
 
 X 
 
 Now, in the first case, the 
 resultant R being equal to the 
 sum of the two forces P, Q 
 (18), we shall have 
 
 R x c i = (Q + P) x ci, 
 or 
 
 But we have GD=AD+AG; 
 hence 
 
 RXCI=QXHE+PXAD+PXAG; 
 or placing the product QXBE, 
 in place of PXAD which is equal to it, we shall have : 
 
 RXci=QXiiE-f QXBE+PXAG; 
 or finally 
 
 RXCI=QXBH + PXAG. 
 
 In the second case we have likewise 
 
 RXCI=QXHE-fPXGD. 
 
 But we have GD=AD AG; hence 
 
 PXAG 
 
60 
 
 STATICS. 
 
 or substituting for PXAD its value QXBE, we shall have 
 
 RXCI=QXHE + QXBE PXAG ; 
 
 and finally, 
 
 RXCI=QXBH PXAG. 
 
 COROLLARY I. 
 
 Fig. 16. 
 
 \ 
 
 V ^ ^ 
 
 JB- **-"--- 
 
 \ 
 
 63. Hence it follows, 1st. 
 When the two points A, B are 
 on the same side of the line 
 FH, the distance ci of the point 
 c of the resultant from this 
 line will be, since R=p-f-Q, 
 
 QXBII-fPXAG 
 
 Cl 
 
 P + Q 
 
 . 
 
 X 
 
 Fi 9- 17 - 2d. When the points A, B are 
 
 placed on different sides of the 
 line FH, this distance will be 
 
 _QXBH PXAG 
 
 P + Q 
 
 In this case the point c will be 
 placed, with reference to the line 
 GH, on the same side as the force Q, for which the pro 
 duct QXBH is the greater. 
 
MOMENTS. 
 
 61 
 
 Fig. 16. 
 
 \ V 
 
 and for Fig. 37, 
 
 COROLLARY II. 
 
 64. If the line FH is per 
 pendicular to AB, the lines 
 AG, BIT, ci will all three be in 
 the direction of AB, and the 
 proposition enunciated in the 
 preceding theorem will still 
 take place. In this case we 
 shall have 
 
 AG=AF, BH=BF, CI=CF. 
 
 Then we shall have for 
 Fig. 16, 
 
 RXCF=QXBF + PXAF ; 
 RXCF=QXBF PXAF. 
 
 As to the distance CF from the 
 point F to the point of applica 
 tion c of the resultant R=P+Q, 
 it will be for Fig. 16, 
 
 _QXBF + PXAF 
 
 P+Q 
 and for Fig. 17, 
 
 _QXBF PXAF 
 P + Q 
 
 THEOREM. 
 
 65. Tivo parallel forces P, Q (Figs. 18 and 19), 
 which act in the same direction, being applied to the 
 points A, B of an inflexible right line, and having drawn 
 a plane MN, through a point F of this line, parallel to 
 their directions : 
 
 <- 
 
 D 
 
 1 
 
 , G 
 / 
 
62 
 
 STATICS. 
 
 1st. If the point F be 
 taken upon the prolongation 
 of AB, the sum of the mo 
 ments of the two forces P, Q, 
 referred to the plane MN, 
 will be equal to the moment 
 of their resultant R : that 
 is to say, if from the points 
 A, B, and from the point of 
 application c of the result 
 ant, the perpendiculars AG, 
 BH, ci, be let fall upon the plane, we shall have 
 
 RXCI=QXBH + PXAG. 
 Fig. 19. 
 
 2d. If the point F be taken be- 
 tiveen A and B, the difference of 
 the moments of the forces P, Q will 
 be equal to the moment of their 
 resultant : that is to say, we shall 
 have 
 
 RXCT=QXEH PXAG. 
 
 DEMONSTRATION. The three lines AG, BH, ci, per 
 pendicular to the same plane MN, are parallel to each 
 other ; moreover they pass through the three points 
 A, B, c, of the same line : hence they are in the same 
 plane drawn through A, B ; hence their feet G, H, I, and 
 the point F are in the same plane. But the four points 
 F, G, H, I, are also in the plane MN ; hence they are in 
 the intersection of two different planes, and conse 
 quently in a straight line. Now draw the line FGIH : 
 
MOMENTS. 03 
 
 it will intersect the lines AG, BH, ci, at right angles ; for 
 it will be in the plane MN to which these lines are per 
 pendicular, and it will pass through their feet. Hence, 
 by considering FGIH as the line FH (Figs 16 and 17) : 
 
 1st. When the point F (Fig. 18) is upon the prolonga 
 tion of AB } we shall have (62) 
 
 RXCT = QXBH + PXAG. 
 
 2d. When the point F (Fig. 19) is between A and B, 
 we shall have 
 
 RXCT = Q + BH PXAG. 
 
 COROLLARY. 
 
 66. Hence, 1st. When the two forces P, Q, (Fig. 18) 
 are on the same side of the plane MN, the distance ci 
 of the plane from the resultant will be equal to the sum 
 of the moments of the forces referred to the plane, di 
 vided by the resultant R, or, what is the same (18), di 
 vided by the sum P+Q of the forces : that is to say, we 
 shall have 
 
 _QXBH + PXAG 
 
 P + Q 
 
 2d. When the plane passes between the directions of 
 the two forces (Fig. 19), this distance will be equal to 
 the difference of the moments divided by the sum of the 
 forces : that is to say, we shall have 
 
 _QXBH PXAG 
 
 CI . 
 
 P + Q 
 
 In the latter case, the resultant will be situated on 
 the same side of the plane MN as the force whose mo 
 ment is the greater. 
 
STATICS. 
 
 THEOREM. 
 
 67. If, to any number of points A, B, c, D, . . . situ 
 ated or not in the same plane, but connected together in 
 an invariable manner, the parallel forces P, Q, R, s, . . . 
 be applied, which act in the same direction, and which 
 are all placed on the same side of any plane MIST parallel 
 to their directions, the sum of the moments of all the 
 forces referred to the plane MN will be equal to the mo 
 ment of their resultant. 
 
 DEMONSTRATION. Draw the line AB, and let E be the 
 point on this line through which the resultant T of the 
 Fig. 20. two forces P, Q pass 
 
 es ; draw the line 
 EC, and let F be the 
 point on this line 
 through which the 
 resultant v of the 
 two forces T, R, 
 passes, which will al 
 so be the resultant 
 of the three forces 
 p, Q, R ; draw FD, 
 and let G be the 
 
 point on this line through which the resultant x of the 
 two forces v, s, passes, which will also be the resultant 
 of the four forces p, Q, R, S ; and so on. Finally, from 
 the points A, B, c, D . . . . and the points E, F, G, . . . 
 let fall upon the plane MN the perpendiculars Aa, B&, cc, 
 vd, . . . . , EC, F/, ag, . . . . 
 
MOMENTS. 65 
 
 Then, the moment of the resultant T will be equal to 
 the sum of the moments of its two components P, Q (65), 
 and we shall have 
 
 In like manner, the moment of the force v will be 
 equal to the sum of the moments of its two components 
 T, K, and we shall have 
 
 Hence, substituting for TXEe its value, we get 
 
 VXF/=PXAa-f QXB#-f RXCtf. 
 
 Likewise the moment of the force x will be equal to 
 the sum of the moments of its two components V, s, 
 which will give 
 
 X X G<7 = V X F/-f S X Dd. 
 
 Hence, substituting for vXF/its value, we shall have 
 
 And so on, whatever may be the number of forces. 
 Hence the moment of any resultant is equal to the sum 
 of the moments of all the components ; hence, &c. 
 
 COROLLARY I. 
 
 68. We have seen (27) that the intensity of the re 
 sultant x of the forces P, Q, R, s, . . . . is equal to the 
 sum p+Q-fR + s .... of these forces: hence the dis 
 tance G(/ of the direction of this resultant from the 
 plane MN is equal to the sum of the moments of all the 
 
66 STATICS. 
 
 forces P, Q, R, s, . . . divided by the sum of all these 
 forces : that is to say, we shall have 
 
 P+Q+R+S 
 
 COROLLARY II. 
 
 69. Hence, if an indefinite plane be drawn on the side 
 on which the forces are placed, parallel to MN and at a 
 distance Gg from it : that is to say, equal to 
 
 P + Q-I-R + S 
 
 this plane will contain the direction of the resultant of 
 all the forces P, Q, R, S . . . . ; for this plane will con 
 tain all the points which, on this side, are distant from 
 the plane MN by the quantity Gg, consequently all the 
 points of the direction of the resultant. 
 
 COROLLARY III. 
 
 70. If the forces P, Q, R, s, . . . be situated on each 
 side of the plane MN, the moment of their result 
 ant, referred to this plane, will be equal to the excess 
 of the sum of the moments of the forces which are situ 
 ated on one side of the plane, over the sum of the mo. 
 ments of the forces which are situated on the other side. 
 
 Thus, let v be the partial resultant of all the forces 
 P, Q . . . which are situated on one side of the plane, 
 whatever their number may be, and E the point of ap- 
 
MOMENTS. 
 
 67 
 
 plication of this force. In like manner, let x be the 
 Fi ff- 21 - partial resultant of 
 
 all the forces R, s, 
 . . . which are situ 
 ated on the other 
 side, and F the 
 point of application 
 of this force. By 
 letting fall upon 
 the plane the per 
 pendiculars Aa, %b, 
 Ee . . . cc, fid, F/, . . 
 we have just seen 
 (67) that we shall 
 have 
 
 and 
 
 Now let Y be the resultant of the two forces v, x, and 
 G its point of application ; this force will be the general 
 resultant of all the forces P, Q, K, s, . . 
 
 This done, the two forces v, x, being situated on each 
 side of the plane MIST, the moment of their resultant is 
 equal to the difference of their moments (65) ; hence, 
 letting fall the perpendicular Gg upon the plane, we 
 shall have 
 
 YXG#=VXEe XXF/. 
 
 Hence, by substituting the values of these last two mo 
 ments, we shall have 
 
STATICS. 
 
 68 
 
 Hence, &c. 
 
 COROLLARY IV. 
 
 71. Hence, in general, in whatever manner several 
 parallel forces P, Q, R, s, . . . acting in the same direc 
 tion, may be situated with reference to a plane MN, 
 parallel to their directions, the distance &g of their re 
 sultant from this plane is equal to the excess of the sum 
 of the moments of the forces situated on one side of the 
 plane, over the sum of the moments of the forces situ 
 ated on the other side, divided by the sum of all the 
 forces : that is to say, we shall have 
 
 _pXAa-f-QXB& . . . . (nxcc+sxvd . . .) > 
 g ~~ P-fQ+R + S 
 
 and this resultant is placed, with reference to the plane 
 MN, on the side on which the sum of the moments is the 
 greater. 
 
 COROLLARY V. 
 
 72. Hence, if on the side of the plane MN, on which 
 the sum of the moments is the greater, we draw an in 
 definite plane parallel to it, which is distant by the 
 quantity G^, or 
 
 P-f Q-f R + S ... 
 
MOMENTS. 
 
 this plane will contain the direction of the resultant of 
 all the forces P, Q, R, S, . . . 
 
 COROLLARY VI. 
 
 Fig. 22. 
 
 . c 
 
 73. If the directions 
 of the forces P, Q, R, S, . . . 
 be all situated in the same 
 plane, perpendicular to 
 the plane MN, the lines 
 Atf, B&, cc, vd . . . . Gg 
 will fall upon the line KL, 
 the intersection of the 
 two planes ; and we shall 
 then also have 
 
 according as the forces are on the same side or on the 
 opposite sides of the line KL. Hence we shall have, in 
 the first case, Fig. 22, 
 
 and, in the second case, Fig. 23, 
 
 P+Q+R+S 
 
70 
 
 STATICS. 
 
 23 - that is to say, when several 
 
 parallel forces, situated in 
 the same plane, act in the 
 same direction, the distance 
 of their resultant from any 
 right line, traced in the 
 same plane and parallel to 
 their directions,is in general 
 equal to the excess of the 
 sum of the moments of the 
 forces situated on one side 
 of the line, over the sum 
 of the moments of the for 
 ces situated on the other 
 side, divided by the sum of the forces. 
 
 PROBLEM. 
 
 74. An indefinite number of parallel forces being 
 given, which act in the same direction and whose points 
 of application may or may not be situated in the same 
 plane, to determine, by means of moments, the direction 
 of the resultant of all these forces. 
 
 SOLUTION. Having drawn at pleasure two different 
 planes ABCD, BCIK, parallel to the directions of the 
 forces, find the distance of the resultant from each of 
 these planes separately (71) ; then draw a plane EFGH 
 parallel to ABCD, at the distance from this latter plane 
 that the resultant is from it, and situated on the side 
 on which the sum of the moments referred to the plane 
 ABCD is the greater ; then this plane EFGH will contain 
 
MOMENTS. 
 
 71 
 
 the direction required (71). Likewise draw a plane 
 % 24 - LMNO parallel to BCIK, 
 
 ~ F 
 
 at a distance from this 
 latter plane equal to that 
 of the resultant from it, 
 and situated on the side 
 on which the sum of the 
 moments referred to the 
 plane BCIK, is the great 
 er ; and this plane will 
 also contain the direction 
 required. Hence the direction of the resultant being 
 both in the plane EFGH, and in the plane LMNO, it will 
 be in the line of intersection PQ of these two planes. 
 
 COROLLARY I. 
 
 75. We have seen (30) that if several parallel forces 
 change in direction, without changing either in intensity 
 or their points of application, and without ceasing to be 
 parallel to each other, their resultant always passes 
 through the same definite point, which has been termed 
 the centre of parallel forces; hence, for the parallel 
 forces which we have just been considering, the centre 
 is placed in the direction PQ of their resultant. 
 
 To find this centre, draw at pleasure a third plane 
 ABKR (Fig. 24), and conceive that all the forces, with 
 out changing either their intensities or points of appli 
 cation, are directed parallel to each other and to the 
 plane ABKR ; find the distance of the resultant of these 
 new forces from this plane (71). This done, if we draw 
 a plane STVX parallel to ABKR, and at a distance from 
 
72 
 
 STATICS. 
 
 this latter plane equal to that we shall have found, this 
 plane will contain the new resultant, and consequently 
 the centre of the forces. Hence the centre of the 
 forces, being found both in the line PQ and in the plane 
 STVX, it will be found in the point of intersection Y of 
 the line and the plane ; or, which is the same thing, 
 this centre will be found at the point of intersection Y 
 of the three planes EFGH, LMNO, STVX. 
 
 COROLLARY II. 
 
 76. If the parallel forces P, Q, R, S, . . . acting in the 
 same direction, be situated in the same plane ; in order 
 Fig. 25. (a) to find the position of their 
 
 ~ resultant, draw in this plane 
 a line LN parallel to the direc 
 tions of the forces, and hav 
 ing let fall upon this line, from 
 all the points of application 
 A, B, c, D, . . . the perpen 
 diculars A#, B&, c<?, i>d . . ., 
 lay off upon a line LM per 
 pendicular to LN, the line L# , 
 equal to (73) 
 
 P+Q+R+S . . . 
 
 and the line </ Y, drawn through the point g parallel to 
 LN, will be the direction of the resultant. 
 
 If all the forces were not on the same side of the line 
 LN, it would be necessary to subtract the moments of 
 the forces situated on the other side, instead of adding 
 them (73). 
 
MOMENTS. 
 
 73 
 
 COROLLARY III. 
 
 77. To find the centre of 
 the parallel forces P, Q, R, s, 
 .. ,, . included in the same 
 plane, conceive these forces, 
 without changing their intensi 
 ties and without ceasing to be 
 applied to the same points 
 A, B, c, D, . . . to be in direc 
 tions parallel to another line, 
 as LM, upon which let fall the 
 perpendiculars Aa , B , c<? , 
 
 Dd f , .... and the distance &g r of this line from the new 
 
 resultant will be (73) 
 
 +QXB/-J-RXCC / 
 
 1 t 
 
 Fig. 25. ( 
 t <l g < 
 
 9. 
 
 : i 
 
 5 N 
 
 
 
 A 
 
 
 B 
 
 
 _ _j G- j 
 
 
 
 
 
 c 
 
 
 
 
 
 
 A 
 
 >4i 
 
 Nfe* /> 
 
 R 
 
 
 P-f Q-fR + S . . . 
 
 Hence, if we lay off upon a line perpendicular to LM, 
 the line *Lg equal to this distance, and through the point 
 g draw g& parallel to LM, this line g& will be the direc 
 tion of the new resultant. Now, the centre of the forces 
 is to be found both upon the direction of the first result 
 ant # Y, and upon that of the second g& ; hence it will 
 be at the point of intersection G of these two directions. 
 
 If all the forces were not on the same side of the line 
 LM, it would be necessary to subtract the moments of 
 those which are situated on the other side, instead of 
 adding them. 
 
74 STATICS. 
 
 COROLLARY IV. 
 
 78. If the points of application A, B, c, D, . . . (Fig. 
 25) are in the same plane to which the directions of the 
 parallel forces P, Q, R, s, . . . are oblique, the centre G 
 of these forces will also be in this plane (30), and its 
 position will be the same as though the directions of the 
 forces were parallel to each other and situated in this 
 plane. Thus, to find in this case the centre of the 
 forces G, draw in the plane any two right lines LN, LM ; 
 then suppose that the forces are in a direction parallel 
 to LN, and find (77) the direction g Y of their resultant 
 on this supposition ; then suppose they are in a direction 
 parallel to LM, and find the direction Gg of their re 
 sultant; the point of intersection G of the two lines 
 <7 Y, g& will be the centre of the forces required. 
 
 The centre being found, if we draw through this 
 point a line parallel to the real directions of the forces 
 p, Q, R, s, . . . this line will be the direction of their 
 resultant. 
 
 COROLLARY V. 
 
 79. Finally, if the points of application a , b f , c , d , . . . 
 (Fig. 25) be upon the same straight line LM oblique to 
 the direction of the forces, the centre g of these forces 
 will be upon this line (30), and its position will be the 
 same as though the directions of the forces were per 
 pendicular to LM. 
 
MOMENTS. 
 
 75 
 
 Hence (76) the distance g ~L of this centre from a 
 given point L upon the line, will be equal to 
 
 L + QX^ L-j-RXc 
 
 P+Q+R + S . . . 
 
 If all the forces are not situated on the same side 
 with reference to the point L, it will be necessary to 
 subtract the moments of those which are situated on the 
 other side, instead of adding them. 
 
 Fig. 26. 
 
 *r 
 
 LEMMA. 
 
 80. When the directions of 
 two forces P, Q, meet in a point 
 A, the moments of these forces, re 
 ferred to any point D in the line 
 of direction of their resultant R, 
 are equal. 
 
 For we have seen (35) that if 
 from the point D the perpendicu 
 lars DB, DC, be dropped upon the 
 directions of the forces, prolonged 
 if necessary, we shall have 
 
 p : Q : : DC : DB. 
 
 Hence, the product of the extremes being equal to 
 that of the means, we shall have 
 
 PXDB=QXDC. 
 
76 
 
 STATICS. 
 
 COROLLARY. 
 
 81. From this it follows that, if the directions of the 
 two forces P, Q meet- in a point A, the moment of any 
 
 Fig. 27. 
 
 one, Q, of these forces, referred to 
 a point D in the direction of the 
 other, will be equal to the moment 
 of their resultant R referred to the 
 same point : that is to say, by let 
 ting fall from the point D, the per 
 pendiculars DB, DC upon the direc 
 tion of the force Q, and upon that 
 of the resultant R, prolonged if ne 
 cessary, we shall have 
 
 QXDB=RXDC. 
 
 For, by applying to the point A a third force S, equal 
 and directly opposed to the resultant R, the three forces 
 p, Q, S, will be in equilibrium ; consequently the force P 
 will be equal and directly opposed to the resultant of 
 the two forces Q, S. Hence, the moments of the two 
 forces Q, S, referred to the point D in the direction of 
 their resultant, will be equal (80) ; hence we shall have 
 
 QXDB=SXDC; 
 
 or, since S=R, 
 
 QXDB=RXDC. 
 
MOMENTS. 
 
 77 
 
 we have, in the first case, 
 
 THEOREM. 
 
 82. When the directions 
 of the two forces P, Q, 
 meet in the same point A, 
 the moment of the result 
 ant E of these forces re 
 ferred to any point D, ta 
 ken in the plane of these 
 directions, is equal to the 
 sum or difference of the 
 moments of the forces P, Q, 
 referred to the same point, ac 
 cording as the point D is with 
 out or within the angle PAQ, 
 formed by the directions of 
 these forces : that is to say, if 
 from the point D we let fall 
 upon these directions, and 
 upon that of the resultant, 
 the perpendiculars DB, DC, DE, 
 
 and, in the second case, 
 
 RXDE = QXDC -PXDB. 
 
 7* 
 
78 STATICS. 
 
 DEMONSTRATION. Draw the line AD, and decompose 
 the force P into two others, p, p , the first in the direc 
 tion AD, and the second in the direction of the force Q. For 
 this purpose (38), represent the force P by the part AF of 
 its direction ; through the point r draw the lines FG, FH, 
 respectively parallel to AQ and AD ; and the two com 
 ponents p, p will be represented by the sides AG, AH 
 of the parallelogram AGFH. 
 
 The point D being upon the direction of the compo 
 nent p, the moment of the other component p , referred 
 to this point, is equal to the moment of their resultant 
 P, and we shall have (81) 
 
 Moreover, taking the two forces p, p instead of the 
 force P, the resultant R, of the two forces P, Q, is also 
 the resultant of the three forces p, p , Q. 
 
 This being done, in the first case, the two forces Q 
 and p f (Fig. 28), which act along the same line of di 
 rection, are equivalent to a single force equal to their 
 sum Q+y ; thus the force R may be regarded as the re 
 sultant of the two forces p and Q+p ; hence the moment 
 of this resultant referred to the point D, in the direction 
 of the first of these forces, is equal to the moment of 
 the second (81) ; hence we shall have 
 
 RXDE=(Q+J/) DC, 
 or 
 
 QXDC-f//XDC. 
 
MOMENTS. 
 
 79 
 
 Hence, by substituting for the moment p xvc its 
 value, we have 
 
 In the second case, the two forces Q, p f (Fig. 29), 
 which are in the same line, and which act in contrary 
 directions, are equivalent to a single force equal to their 
 difference Q p : now, the moment of this single force, 
 referred to the point D in the direction of the force p, 
 is equal to the moment of the resultant R of these two 
 forces (81) ; and we have 
 
 or 
 
 RXDE=(Q p ) 
 
 RXDE = QXDC 
 
 Hence, substituting for the moment p xvc its value, 
 we have 
 
 RXDE = QXDC PXDB. 
 
 HemarJc I. 
 
 83. This theorem (82) of Statics is a consequence of 
 the following geometrical proposition : 
 
 Fig. 28. (a) If from any point D, taken in the 
 
 plane of a parallelogram AFML, we 
 let fall the perpendiculars DB, DC, 
 DE, upon the sides and the diagonal 
 of this parallelogram, which meet 
 in the same point A, the product 
 
80 
 
 STATICS. 
 
 of the diagonal AM by its perpendicular DE is equal to 
 the sum of the products of the sides multiplied each by 
 its perpendicular DB and DC. Among the known demon 
 strations of this theorem, the following is one of the 
 simplest : The triangles ADF, ADL, ADM, having the same 
 base AD, are to each other as their altitudes F/, L/, MWZ ; 
 but we have Mm F/+LZ; for drawing LK parallel to 
 AD, we have Mm=MK-f Kw=F/-f-L; hence the triangle 
 ADM is equal to the sum of the triangles ADF and ADL : 
 hence 
 
 AMXDE=AFXDB + ALXDC. 
 
 TtemarTc II. 
 
 Fig. 29. 
 
 84. If we suppose the line AD to be inflexible and the 
 point D immoveable ; when this point is placed without 
 the angle PAQ (Fig. 28), the two forces P, Q tend to turn 
 the point A in the same direction around the point D ; 
 and, on the contrary, when the point D is placed within 
 the angle PAQ (Fig. 29), the two forces tend to turn the 
 point A in opposite directions. 
 
MOMENTS. 81 
 
 Hence, if two forces be directed in the same plane, 
 the moment of their resultant, referred to any point 
 taken in this plane, is equal to the sum or difference of 
 ^heir moments referred to the same point, according as 
 these forces tend to turn their point of application 
 around the centre of moments, either in the same or in 
 opposite directions ; and, in all cases, the resultant tends 
 to turn its point of application in the same direction as 
 that of the two forces whose moment is the greater. 
 
 THEOREM. 
 
 85. When the forces P, Q, R, S (Fig. 30), di 
 rected in the same plane, are applied to the points 
 a, b, c, d, connected together in an invariable manner, 
 tending to turn these points in the same direction around 
 another point D taken in this plane, the sum of the mo 
 ments of these forces, referred to the point D, is egual to 
 the moment of their resultant referred to the same point. 
 
 DEMONSTRATION. Let v be the partial resultant of 
 the two forces P, Q ; x that of the two forces V, R, and 
 consequently of the three forces P, Q, R ; Y that of the 
 two forces x, S, and consequently of the four forces P, 
 Q, R, S ; and so on. Then, from the point D let fall upon 
 the directions of the forces and upon those of the par 
 tial resultants V, x, Y, . . . the perpendiculars DE, DF, 
 DG, DH, . . . . DI, DK, DL, . . . This being done, the 
 moment of the resultant v is equal to the sum of the 
 moments of its components P, Q, (82), which gives 
 
 VXDI=PXDE-f QXDF. 
 
In like manner the moment of the resultant X is equal 
 to the sum of the moments of its components v, R, and 
 we have 
 
 or, substituting for the moment VXDI its value, 
 
 XXDK PXDE + QXDF + RXDG. 
 
 Likewise, we have YXDL=XXDK+SXDH, or, substi 
 tuting the value of XXDK, 
 
 YXDL=PXDE + QXDF + RXDG + SXDH ; 
 
MOMENTS. 
 
 83 
 
 and so on, whatever may be the number of forces. 
 Hence the moment of each resultant is equal to the sum 
 of the moments of all its components. 
 
 COROLLARY I. 
 
 86. If the forces P, Q, R, S . . . . do not tend to turn 
 their points of application in the same direction around 
 the centre of the moments D, the moment of their re 
 sultant is equal to the excess of the sum of the moments 
 of the forces which tend to turn in one direction, over 
 the sum of the moments of those which tend to turn in 
 the opposite direction. 
 
 Fig. 31 . 
 
 Thus, let v be the partial resultant of all the forces 
 p, Q, . . . which tend to turn in one direction ; and let 
 X be the partial resultant of all the forces R, S, . . . 
 which tend to turn in the opposite direction ; from the 
 point D let fall upon the directions of the forces and upon 
 those of the two resultants v, x, the perpendiculars DE, 
 DF, . . . DG, DH, . . . DI, DK ; we have just seen (85), 
 that we have 
 
84 
 
 and 
 
 STATICS. 
 
 VXDI=PXDE + QXDF 
 
 XXDK=RXDG + SXDH 
 
 Finally, let Y be the resultant of the two forces v, x, 
 and consequently that of all the forces P, Q, R, s, . . . 
 
 This being established, the moment of the resultant 
 Y referred to the point D is equal to the difference of 
 the moments of its components v, x, which tend to turn 
 in opposite directions (84) : that is to say, by letting fall 
 the perpendicular DL upon its direction, we have 
 
 YXDL=VXDI XXDK. 
 
 Hence, by substituting the values of the two moments, 
 we have 
 
 YXDL=PXDE + QXDF . . . 
 
 COROLLARY II. 
 
 8T. If the directions 
 of the forces P, Q, R, s, . . . 
 in eluded in the same plane, 
 are parallel to each other, 
 the perpendiculars DE, DF, 
 DG, DH, . . . DL let fall 
 from the centre of mo 
 ments D upon these directions and upon that of the re 
 sultant Y, will be in the same straight line ; and the 
 
MOMENTS. 
 
 85 
 
 Fig. 33. 
 
 preceding proportion -will remain just the same, whether 
 all the forces act in the same direction, as in Fig. 32, or 
 whether they act, some in one direction and others 
 in the ODposite direction, as in Figs. 33 and 34. Now, 
 the resultant Y of all these 
 forces is equal to the excess of 
 the sum of those which act in 
 one direction, over the sum of 
 those which act in the opposite 
 direction (29) ; hence the dis 
 tance DL of the centre of mo 
 ments from the direction of 
 the resultant, is equal to the 
 quotient of the excess of the sum 
 of the moments of the forces which 
 tend to turn in one direction, over 
 the sum of the moments of those 
 which tend to turn in the opposite 
 direction, divided by the excess of 
 the forces which act in one direc 
 tion, over the sum of those which 
 act in the contrary direction : thus 
 we have 
 
 Fig. 34. 
 
 DL = 
 
 DL 
 
 PXDE+QXDF . . . (RXDG+SXDH . . .) 
 
 (Fig. 32.) 
 
 PXDE+QXDF . . . (RXDG+SXDH . . .) 
 
 -- 7 - r - -, 
 P + Q . . . (R + S . . .) 
 
 DL _ PXDE+RXDG . . . (QXDF+SXDH . . .) 
 
 p+s . . . (Q+ R . . .) * ig * 
 
86 STATICS. 
 
 In all cases, the resultant acts in the direction in 
 which the sum of the forces is the greater ; and it is 
 placed, with reference to the point D, on the side on 
 which the sum of the moments is the greater. 
 
8T 
 
 CHAPTER THIRD. 
 
 ON CENTRES OF GRAVITY. 
 
 88. THE property by virtue of which bodies, left to 
 themselves, fall towards the earth is named gravitation 
 or gravity. 
 
 All the molecules, of which bodies are composed, have 
 gravitation, and they always have it ; for into whatever 
 number of parts a body is divided, each of these parts 
 continually gravitates, and falls towards the earth when 
 left to itself. 
 
 89. The effort which a body makes to fall, when it is 
 retained or supported by an obstacle which opposes its 
 fall, is called the weight of the body ; this weight may 
 be regarded as the effect of a force which is constantly 
 applied to the body : thus we are accustomed to consider 
 gravity as a force. 
 
 Gravity is not a force rigorously constant for the same 
 molecule ; it varies according to the different positions 
 which this molecule has relatively to the sphere of the 
 earth. 
 
 1st. When the distance of the molecule from the cen 
 tre of the earth changes, its gravity decreases in the 
 same ratio as the square of this distance increases ; be 
 sides, the earth not being perfectly spherical, and the 
 lines drawn from its centre~to the equator being greater 
 
CQ STATICS. 
 
 than those which terminate at the poles, the gravity at 
 the surface of the earth is greater for the same mole 
 cule, when this molecule is placed at the poles, than 
 when it is at the equator, because there the distance of 
 the molecule from the centre of the earth is less. 
 
 2d. The earth turns around its axis, and all the parts 
 composing it perform their revolutions in the same time, 
 that is to say, in about twenty-four hours. The parts 
 of the surface near the equator describe greater circum 
 ferences of circles than those described by the parts 
 near the poles ; their centrifugal force, which likewise 
 is greater, destroys a greater part of the effect of gravi 
 tation, and is a new cause which renders this latter force 
 less at the equator than it is at the poles. 
 
 Thus, rigorously speaking, gravity is variable for 
 the same molecule, when this molecule departs from or 
 approaches the surface of the earth, and when it departs 
 from or approaches the equator : but the distances of the 
 positions in which we are accustomed, in Statics, to 
 consider the same molecule, are so small with reference 
 to the radius of the earth, that the effects of this varia 
 tion are absolutely insensible ; and we may regard 
 gravity as a constant force for the same molecule, what 
 ever its position may be. 
 
 90. The straight line along which a molecule, aban 
 doned to itself, falls to the earth, and which is evidently 
 the direction of gravitation, is named a vertical ; this 
 line is everywhere perpendicular to the surface of the 
 earth, or, more exactly, to the surface of undisturbed 
 water. 
 
 91. A plane is said to be horizontal when it is per 
 pendicular to a vertical. 
 
CENTRE OF GRAVITY. 89 
 
 If the earth were perfectly spherical all the lines of 
 direction of gravitation would meet in the same point, 
 which would be the centre : but the earth not being a per 
 fect sphere, the lines of direction of gravitation for two 
 different molecules may not be in the same plane ; and 
 when they are in the same plane, they meet in the same 
 point. 
 
 However, the molecules in the same body, and those 
 of the different bodies we are accustomed to consider in 
 Statics, are so near each other compared with their 
 distances from the centre of the earth, that the angle 
 formed by the directions of gravitation for any two of 
 them is not sensible, and we may regard all these direc 
 tions as parallel. 
 
 92. We will regard, then, all the molecules of heavy 
 bodies as constantly pushed or drawn towards the earth 
 by forces constant for each of them ; we will suppose 
 that these forces are parallel, and act in the same di 
 rection ; and, consequently, we will be able to apply to 
 them all we have said of the composition, decomposition, 
 and equilibrium of parallel forces. 
 
 Now, when several parallel forces, acting in the same 
 direction, are applied to points invariably connected to 
 gether, we have seen : 1st, that these forces have a re 
 sultant equal to their sum (27); 2d, that the direction 
 of this resultant is that of the components ; 3d, that 
 there exists a centre of forces through which this re- 
 sultant always passes, even though the forces, without 
 changing in intensity and without ceasing to be parallel,, 
 should change in direction (30). 
 
 Hence, 1st, the weights of all the molecules of a solid 
 body have a resultant which constitutes the weight of 
 
90 STATICS. 
 
 the body, and this resultant is equal to the sum of the 
 weights of the molecules ; 2d, the direction of this re 
 sultant, or of the weight of the body, is always parallel 
 to that of gravitation, and consequently vertical; 3d, 
 whatever may be the different positions given to this 
 body, the directions of the resultants for all these posi 
 tions meet in the same point ; for by varying the position 
 of the body, the intensity of the forces, which act upon 
 the molecules, is not altered, and these forces, which 
 only change in direction with reference to the bodies, 
 do not cease to be parallel to each other. 
 
 93. The point through which the direction of the 
 weight of a body always passes, whatever may be its 
 position, is named the centre of gravity. 
 
 94. When several bodies are invariably connected to 
 gether, and we consider their assemblage as though they 
 made but one and the same body, we ordinarily give to 
 this assemblage the name of system. 
 
 95. Every thing that has just been said of a single 
 body may likewise be said of a system of several bodies : 
 that is to say, the weight of the system is equal to the 
 sum of the partial weights of the bodies which compose 
 it ; that the direction of this weight is vertical, and that 
 this direction, whatever may be the position of the sys 
 tem, always passes through the same definite point, 
 which is the centre of gravity of the system. 
 
 COROLLARY I. 
 
 96. We may always regard the weight of a body, 
 or system of several bodies, as a force directed verti- 
 
CENTRE OF GRAVITY. 91 
 
 cally, and applied to the centre of gravity of the body 
 or system : for this weight, which is the resultant of the 
 partial weights of all the molecules which compose the 
 body or system, may be considered as applied to any 
 point of its direction, and consequently to the centre of 
 gravity, which is always upon this direction, whatever 
 may be otherwise the position of the body or the system. 
 
 COROLLARY II. 
 
 97. Hence, we will produce an equilibrium in the 
 action which gravitation exerts upon all the molecules 
 of a body or system of bodies, by applying to the centre 
 of gravity of the body or system, a single force whose 
 direction is vertical, equal to the total weight of the 
 body or system, and which acts in a direction oppo 
 site to that of gravitation. 
 
 Inversely, when a single force produces an equilibrium 
 in the weight of all the molecules of a body or system 
 of bodies, the direction of this force will be vertical, 
 and it will pass through the centre of gravity of the 
 body or system. 
 
 Thus, when a body AB, suspended by a 
 thread ED from a fixed point D, is in equi 
 librium, and the action of gravitation is con 
 sequently destroyed by the resistance of the 
 thread, the direction of this thread will be 
 vertical, and its prolongation will pass 
 through the centre of gravity c of the body. 
 
92 STATICS. 
 
 COROLLARY III. 
 
 98. From this, we deduce a simple man 
 ner of finding, by experiment, the cen 
 tre of gravity of a body of any figure. 
 Thus, if we suspend the same body by a 
 thread, successively by the two different 
 points E, E , and conceive the two directions 
 of the thread to be prolonged into the interior of the 
 body, the point c, in which these two directions inter 
 sect, will be the centre of gravity required. 
 
 Remark. 
 
 99. Since the partial weights of the bodies which 
 compose a system, may be considered as parallel forces, 
 applied to the partial centres of gravity of these bodies, 
 it follows, that when we know the weight of these bodies, 
 and the positions of their partial centres of gravity, we 
 can find the position of the centre of gravity, by the 
 processes which have been given to find the centres of 
 parallel forces, either by means of the principle of the 
 composition of parallel forces, as in No. 28, or by em 
 ploying the consideration of moments, as in Nos. 75, 
 77, 78, and 79 ; we will soon have occasion to give ex 
 amples. 
 
 To find the centre of gravity of a body of any figure 
 whatever, conceive the body to be divided into a certain 
 number of parts, so that we may know the weight of 
 each of them, and the position of its partial centre of 
 gravity ; then, by finding the centre of gravity of the 
 system of all these parts, we will have the required cen 
 tre of gravity of the body. 
 
CENTRE OF GRAVITY. 93 
 
 But when the parts of the body are of the same na 
 ture in all its extent, and when the figure of the body 
 is not very complicated, we may often find its centre of 
 gravity by simpler considerations, and which we are 
 about to employ in order to arrive at results which are 
 used very frequently. 
 
 LEMMA. 
 
 Fig- 86. 100. When a body is considered 
 
 to be composed of parts A, A , B, B , 
 c, c , . . . which, taken two and two, 
 are equal to each other, and so placed 
 that the middle of the lines AA , 
 BB , cc , which join the centres of 
 gravity of the homologous parts, co 
 incide in the same point D, this point, which is the 
 centre of figure of the body, is also its centre of gravity. 
 For the point D is the centre of gravity of each par 
 tial system of two homologous parts ; hence, it is also 
 the centre of gravity of their general system. 
 
 COROLLARY. 
 
 101. By considering lines, surfaces and solids, as 
 composed of parts uniformly heavy, it is evident : 1st, 
 that the centre of gravity of a right line is at the mid 
 dle of its length ; 
 
94 
 
 STATICS. 
 
 2d. The centre of gravity of the area, 
 and that of the contour of a parallelo 
 gram ABCD, are in its centre of figure : 
 that is to say, at the point of intersec- 
 tion E, of its two diagonals AC, BD ; 
 
 D C 
 
 3d. The centre of gravity of the area of a circle, and 
 that of its whole circumference, are at the centre of the 
 circle ; 
 
 4th. The centre of gravity of the whole surface of a 
 parallelopipedon, and that of its solidity, are in its cen 
 tre of figure : that is to say, in the intersection of any 
 two of its four diagonals, or in the middle of one of 
 them; 
 
 5th. The centre of gravity of the convex surface of 
 a right or oblique cylinder, and that of its solidity, are 
 in the middle of the length of its axis ; 
 
 6th. The centre of gravity of the surface of a 
 sphere, and that of its solidity, are at the centre of the 
 sphere. 
 
 PROBLEM. 
 
 102. To find the centre of gravity of the area of any 
 rectilinear triangle ABC. 
 
 Fig. 38. SOLUTION. Having drawn through 
 
 the summit A of one of the angles, 
 and through the middle D of the op 
 posite side, the line AD, if we con 
 ceive the area of the triangle to be 
 divided into an infinite number of 
 elements by lines parallel to BC, the 
 
CENTKE OF GRAVITY. 95 
 
 centre of gravity of each of these elements will be in 
 its middle (101), and consequently upon the line AD; 
 hence the centre of gravity of their system, which will 
 be that of the area of the triangle, will be upon this 
 same line (30). For the same reason, if from the sum 
 mit B of another angle, and through the middle E of 
 the opposite side, we draw a line DE, this second line 
 will contain the centre of gravity : hence this centre 
 will be found both upon the line AD, and upon the line 
 BE ; hence it will be found at the point of intersection 
 F of these two lines. 
 
 COROLLARY I. 
 
 103. If from the summit A of one 
 of the angles of the triangle ABC, 
 and through the middle D of the op 
 posite side, we draw a line AD, and 
 divide this line into three equal parts, 
 the centre of gravity F of the area 
 of the triangle will be upon this line, 
 at the distance of two-thirds from the summit of the 
 angles, or one-third from the opposite side. 
 
 For, if we draw the line DE, this line will be parallel 
 to AB, because the sides BC, AC, are cut proportionally 
 in D and E ; and the triangles ABF, DEF will be similar, 
 because their corresponding angles will be equal ; hence 
 we shall have 
 
 AF : FD : : AB : DE. 
 But the similar triangles ABC, EDC, give 
 
96 STATICS. 
 
 AB : DE : : BC : DC, or : : 2 : 1 (102). 
 Hence we shall have 
 
 AF : FD : : 2 : 1, or AF=2FD, 
 
 hence, 
 
 FD=JAD, and AF= AD. 
 
 COROLLARY II. 
 
 104. If in the plane of a 
 rectilinear triangle ABC we draw 
 any line GI, the perpendicular 
 let fall from the centre of gravi 
 ty F of the area of the triangle 
 upon GI, will be equal to one- 
 third of the sum of the perpen 
 diculars AG+CH+BI let fall 
 from the summits of the angles 
 upon the same line. 
 
 Thus, through the summit A of one of the angles 
 draw the line AM parallel to GI, which will cut in K, L, 
 M, the perpendiculars let fall from the other points ; 
 through the point A, and through the centre of gravity 
 F, draw the line AF, whose prolongation will bisect the 
 opposite side at the point D ; finally, through the point 
 D, draw DN perpendicular to AM : this being done, we 
 shall have 
 
 CK+BM 
 
CENTRE OF GRAVITY. 97 
 
 and the similar triangles AFL, ADN will give 
 
 FL : DN : : AF : AD, or : : 2 : 3 (103). 
 Hence, 
 
 CK+BM 
 
 FL=fDN = g . 
 
 But the lines AG, KH, LO, MI, being equal to each 
 other, we shall have 
 
 LO= 
 
 AG-f-KH-f-MI 
 
 ~3 
 
 Hence, by adding this equation to the preceding, we 
 shall have 
 
 AG + CK + KH + BM-fMI 
 
 that is to say, 
 
 FO 
 
 AG-fCH + BI 
 
 : 3 !* 
 
 From this we deduce 
 another manner of finding 
 the centre of gravity of 
 the area of a rectilinear 
 triangle ABC. Having 
 drawn at pleasure, in the 
 plane of the triangle, two 
 lines GI, GP, and having 
 found the distances FO, 
 FR, of the centre of gravity from each of these lines, 
 
 9 
 
98 
 
 STATICS. 
 
 if we draw the line RV parallel to GI and at the distance 
 FO, this line will contain the required centre of gravity; 
 in like manner if we draw xo parallel to PG and at the 
 distance FR, this second line will contain the centre of 
 gravity ; hence this centre will be found upon the two 
 lines RV, xo ; hence it will be at their point of inter 
 section F. 
 
 Fig. 41. 
 
 PROBLEM. 
 
 105. To find the centre of gravity of the area of a 
 rectilinear polygon ABODE of any number of sides. 
 
 FIRST SOLUTION, by the process of the composition 
 of parallel forces. 
 
 Divide the area of the polygon 
 into triangles by the diagonals 
 AC, AD, . . . drawn from the sum 
 mit of the same angle A, and de 
 termine (102, or 103, or 104), 
 the partial centres of gravity F, 
 G, H of the areas of these trian 
 gles ; then considering these tri 
 angles as weights proportional 
 to their areas and applied to their centres of gravity, 
 join the centres of gravity of the first two triangles ABC, 
 CAD by a line FG, and find upon this line the centre of 
 gravity I of the system of the two triangles, or of the 
 quadrilateral ABCD, by dividing the line FG into two 
 parts reciprocally proportional to the areas of the two 
 
CENTRE OF GRAVITY. 99 
 
 triangles (18,) which may be done by the following pro 
 portion (25) : 
 
 quadrilateral ABCD : triangle CAD : : FG : Fi. 
 
 Through the point I, and through the centre of gravity 
 H of the next triangle, draw the line Hi, upon which 
 find the centre of gravity K of the system of the first 
 three triangles, by dividing this line into two parts re 
 ciprocally proportional to the areas of the quadrilateral 
 ABCD and of the triangle DAE, which may be done by 
 the following proportion : 
 
 pentagon ABCD : triangle DAE : : in : IK. 
 
 + 
 
 By thus continuing, whatever may be the number of 
 triangles, we will find the centre of gravity of their sys 
 tem, and this centre will be that of the area of the pro 
 posed polygon. 
 
 SECOND SOLUTION, taken from the consideration of 
 moments. 
 
 Having divided the area 
 of the polygon, as in the 
 preceding solution, and de 
 termined the partial centres 
 of gravity F, G, H, . . of all 
 the triangles, draw at plea 
 sure in the plane of the poly 
 gon two lines LM, LN, upon 
 . which let fall perpendicu 
 lars from all the centres of gravity F, G, H, . . ; con 
 sider these lines as the intersection of two planes paral- 
 
100 STATICS. 
 
 lei to the direction of gravitation. This being done, the 
 distance of the centre of gravity of the polygon, or of 
 the system of all the triangles, from each of the lines 
 LM, LN, will be equal to the sum of the moments of the 
 triangles referred to each plane, divided by the sum of 
 their areas (77) : thus the distance of this centre from 
 the line LM will be 
 
 ABC X F/drCAD X GU/drDAE X H Jl 
 ABCDE 
 
 and its distance from the line LN will be 
 
 ABC X F/CAD X Gf/ itDAE X H// 
 ABCDE 
 
 Hence, by drawing a line parallel to LM, and at a 
 distance equal to the first of these two distances, this 
 line will contain the centre of gravity of the polygon ; 
 likewise if we draw a line parallel to LN, at a distance 
 equal to the second of these distances, this line will con 
 tain the centre of gravity ; hence the intersection of 
 these two lines will be the required centre of gravity. 
 
 106. If the centres of gravity F, G, II, of the trian 
 gles which compose the area of the polygon, were not 
 all placed on the same side with reference to each of the 
 lines LM, LN, in order to find the distance of the centre 
 of gravity K of the polygon from each of these lines, 
 it would be necessary to subtract the moments of the 
 triangles, whose centres of gravity are situated on the 
 other side of this line, instead of adding them (77). 
 
CENTRE OF GRAVITY. 
 
 101 
 
 PROBLEM. 
 
 107. To find the centre of gravity of the contour of a 
 polygon ABCDE, of any number of sides. 
 
 FIRST SOLUTION, by the process of the composition of 
 parallel forces. 
 
 Fi 9- 43 - Bisect each of the sides 
 
 of the polygon at the 
 points F, G, H, i, K, which 
 will be the partial cen 
 tres of gravity of these 
 sides (101). Then, con 
 sidering all the sides as 
 weights proportional to 
 
 J" * y * * their lengths, find the 
 
 centre of gravity of the system o of any two of them, 
 as AB, EC, by joining their centres of gravity by the line 
 FG, and dividing this line into two parts reciprocally 
 proportional to these sides, which may be done by the 
 following proportion (22) : 
 
 AB+BC : BC : : FG : FO. 
 
 The point being found, draw through this point, and 
 through the middle H of the next side, the line OH, upon 
 which find the centre of gravity P of the system of the 
 three sides, by dividing this line into two parts recipro 
 cally proportional to the side CD and the sum of the first 
 two, AB, BC ; which may be done by the proportion, 
 
 AB+BC+ CD : CD : : on : OP. 
 
 9* 
 
102 STATICS. 
 
 In like manner, drawing the line PI, find the centre 
 of gravity Q of the system of the four sides AB, BC, CD, 
 DE by the proportion, 
 
 AB-f BC-fCD-f DE I DE : : PI : PQ. 
 
 By thus continuing, whatever may be the number of 
 sides of the polygon, find the centre of gravity of their 
 system, and this centre will be that of the contour of the 
 polygon. 
 
 SECOND SOLUTION, taken from the consideration of 
 moments. 
 
 Having bisected each side of the polygon, draw at 
 pleasure the two lines LM, LN, upon each of which let 
 fall perpendiculars from the middle of all the sides. 
 This being done, the distance of the centre of gravity 
 K of the system of all the sides referred to each of the 
 planes, whose lines of intersection are LM, LN, will be 
 equal to the sum of the moments of the sides referred 
 to this plane, divided by the sum of the sides (77); 
 thus, the distance of this centre from the line LM will be 
 
 AB X F/+ BC X Qff + CD X Ilh + DE X I/+ EA X K& 
 
 AB + BC + CD-f DE + EA 
 
 and its distance from the line LN will be 
 
 AB + BC + CD-f DE + EA 
 
 Hence, drawing a line parallel to LM, at a distance 
 equal to the first of these distances, then another line 
 parallel to LN, at a distance equal to the second of these 
 
CENTRE OF GRAVITY. 
 
 103 
 
 distances, the point of intersection of these two lines 
 will be the centre of gravity n of the contour of the 
 polygon. 
 
 Remark. 
 
 108. If the middle points F, G, H, I, K (Fig. 43), 
 of the sides of the polygon, were placed on opposite 
 sides of the lines LM, LN ; in order to find the distance 
 of the centre of gravity R from each of these lines, it 
 would be necessary to subtract the moments of the sides 
 whose middle points were situated on the contrary side, 
 instead of adding them (77). 
 
 PROBLEM. 
 
 109. To find the centre of gravity of the solidity of 
 any triangular pyramid ABCD. 
 
 SOLUTION. Determine the centre 
 of gravity F of the area of one of 
 the faces BCD of the pyramid (103), 
 by drawing through the summit D of 
 one of the angles of this face, and 
 through the middle E of the oppo 
 site side BC, a line DE ; and take 
 upon this line a point F two-thirds 
 of the distance from the summit 
 of the angle or one-third from the 
 base ; then draw the line AF. This being done, if we 
 conceive the pyramid to be divided into an infinite num 
 ber of sections by planes parallel to the face DCB, all 
 
104 STATICS. 
 
 these sections will be similar to this face, and they will 
 be met by the line AF in points, which, being situated 
 upon each of them in the same manner as the point F 
 is in the face BCD, will be the partial centres of gravity 
 of these sections ; hence the centre of gravity of their 
 system, which will be that of the solidity of the pyra 
 mid, will be upon the line AF (30). 
 
 For the same reason, having determined the centre 
 of gravity a of the area of another face ABC, which is 
 done by drawing the line AE, and taking upon this line 
 the part EG=J- AE ; if through this point, and through 
 the summit D of the opposite angle of the pyramid, the 
 line DG be drawn, this line will also contain the centre 
 of gravity of the solidity of the pyramid. 
 
 Hence the lines AF, Da, both containing the centre 
 of gravity of the pyramid, will necessarily intersect in 
 a certain point H ; and the point of intersection of these 
 two lines will be the centre of gravity required. 
 
 Remark I. 
 
 110. It might be demonstrated, independently of the 
 consideration of the centre of gravity of the pyramid, 
 that the lines AF, DG necessarily intersect in one point ; 
 for these lines are in the same plane, which is that of 
 the triangle ADE. 
 
 Remark II. 
 
 111. Any one of the six edges of a triangular pyra 
 mid being cut by four others ; the fifth which does not 
 meet it is called its opposite : if we join the middle 
 point of one of the six edges with that of its opposite 
 
CENTRE OF GRAVITY. 
 
 105 
 
 by a line, it may be demonstrated that the middle of 
 this line is the centre of gravity of the pyramid. (See 
 the Correspondence of the Polytechnic School, tome II, 
 page 1.) 
 
 COROLLARY I. 
 
 112. If from the summits of one of 
 the angles of a triangular pyramid, 
 and through the centre of gravity F 
 of the area of the opposite face BCD, 
 a line AF be drawn, the centre of 
 gravity H of the solidity of the pyra 
 mid will be upon this line, and at 
 one-fourth of the distance from the 
 face, or at three-fourths of the dis 
 tance from the summit of the angle. 
 Draw the line GF, which will be parallel to AD, be 
 cause the lines EA, ED, are cut proportionally in G, F ; 
 the triangles AHD, FHG, whose corresponding angles are 
 equal, will be similar, and will give 
 
 AH : IIF : : AD : GF. 
 But the similar triangles AED, GEF, give 
 
 AD : GF : : ED : EF, or : : 3 : 1 (103) ; 
 hence, we shall have 
 
 AH : HF : : 3 : 1 ; 
 
 that is to say, AH=3HF, and consequently HF=|AF, and 
 AH=?AF. 
 
106 STATICS. 
 
 COROLLARY II. 
 
 113. It might be demonstrated, in a manner analo 
 gous to that of No. 104, that the distance of the centre 
 of gravity of the solidity of a triangular pyramid from 
 any plane, is equal to the fourth of the sum of the dis 
 tances of the summits of the four angles of the pyra 
 mid from the same plane. 
 
 COROLLARY III. 
 
 Fig. 45. 114. The centre of gravity of the 
 
 solidity of a pyramid ABCDEF with 
 any base is upon the line AG, drawn 
 from the summit A to the centre of 
 gravity G of the area of the base, and 
 at a distance of one-fourth of this 
 line from the base or three-fourths 
 from the summit. 
 
 Conceive the pyramid to be divi 
 ded into an infinite number of sec 
 tions by planes parallel to the base : 
 all these sections will be similar to the base, and the 
 point where each of them is intersected by the line AG 
 will be situated upon this section in the same manner as 
 the point G is upon the base ; consequently this point 
 will be the centre of gravity of the section : hence the 
 centres of gravity of all the sections will be upon the 
 line AG ; hence the centre of gravity of their system, 
 
CENTKE OP GRAVITY. 107 
 
 which is that of the solidity of the pyramid, will be 
 also upon this line (30). 
 
 Moreover, let the base be divided into triangles by 
 the diagonals BE, BD, and conceive that through these 
 diagonals and through the summit A, the planes ABE, 
 ABD be drawn, which will divide the proposed pyra 
 mid into as many triangular pyramids as there are tri 
 angles in the base ; then through the centres of gravity 
 H, I, K of the triangular bases, draw the lines AH, AI, 
 AK ; finally, let the points L, M, N be taken upon these 
 lines, upon each of them at the distance of one-fourth 
 of its length from the base ; these points will be the 
 centres of gravity of the triangular pyramids (112). 
 This being done, the points L, M, N, which will divide 
 proportionally the lines AH, AI, AK, drawn from the sum 
 mit of the pyramid upon the base, will be in the same 
 plane parallel to the base ; hence the centre of gravity 
 of the system of triangular pyramids, that is to say, 
 the centre of gravity of the solidity of the proposed 
 pyramid, will be in this same plane ; hence the centre 
 of gravity, being found both in this plane and in the 
 line AG, will be at the point of their intersection o. 
 
 Now, the line AG will be cut by the plane LMN in parts 
 proportional to the divisions of the lines AH, AI, AK ; 
 hence the centre of gravity o of the solidity of the 
 pyramid will be placed upon AG, at one-fourth of this 
 line from the base, or three-fourths from the summit. 
 
108 
 
 STATICS. 
 
 COROLLARY IV. 
 
 115. The centre of gravity of the solidity of a cone 
 of any base is upon the line drawn from the summit to 
 the centre of gravity of the base, and at the distance 
 of one-fourth of this line from the base, or at three- 
 fourths from the summit ; for this solid may be consid 
 ered a pyramid whose base has an infinite number of 
 sides. 
 
 PROBLEM. 
 
 116. To find the centre of gravity of the area of a 
 section made in the hull of a vessel ly a horizontal 
 plane. 
 
 SOLUTION. Let CEuhec be 
 the proposed section, AB the 
 line of intersection of the 
 plane of this section with the 
 vertical plane drawn through 
 the keel of the vessel. It is 
 evident, since the whole section is symmetrical on each 
 side of the line AB, the required centre of gravity K 
 will be upon this line ; thus, to construct this point, it 
 will suffice to know its distance AK from a line Cc, drawn 
 through a given point perpendicular to AB. 
 
 For this purpose, let the line AB be divided by the 
 perpendiculars or ordinates vd, EC, F/j . . . into a suffi 
 ciently great number of equal parts, so that the arcs CD, 
 DE, EF, .... included between two adjoining perpendi- 
 
CENTRE OF GRAVITY. 109 
 
 culars, may be regarded as right lines, which will divide 
 the area of the section into trapeziums ; then let each 
 of these trapeziums be divided into triangles, by means 
 of the diagonals cd, DC, E/, . . . . This being done, if 
 we take the sum of the moments of all the triangles re 
 ferred to the vertical plane passing through the line cc 9 
 and divide this sum by the sum of the areas of the tri 
 angles, the quotient will be the distance required AK 
 (77). Now, each triangle may be considered as having 
 for base one of the perpendiculars, and for height the 
 common distance from each other of two consecutive 
 perpendiculars ; hence the area of each triangle will be 
 equal to the half of the product of the ordinate which 
 serves for base, multiplied by the common distance. 
 For example, the area of the triangle DEe will be equal 
 to the half of the product EeXLM ; that of the triangle 
 vde will be the half of DC?XLM, and so of the others. 
 Moreover, the distance of the centre of gravity of each 
 of the triangles from the plane cc, will be equal to one- 
 third of the sum of the distances of the summits of its 
 three angles from the same plane (104) : for example, 
 the distance of the centre of gravity of the triangle 
 DEe from the plane cc will be one-third of AL+AM+AM, 
 and so of the others. 
 
 Hence it will be easy to have the sum of the areas of 
 all the triangles, and the sum of the moments of these 
 areas referred to the plane cc ; and by dividing the 
 second of these two sums by the first, we shall have the 
 required distance of the centre of gravity K from the 
 line cc. 
 
 The preceding solution is not rigorous, because the 
 
 parts CD, DE, . . ., cd, de, . . ., of the sides of the sec- 
 
 10 
 
110 STATICS. 
 
 tion arc not right lines, as we have supposed ; but it is 
 evident that the result will approach exactness as much 
 more as these parts are smaller : that is to say, as the 
 number of perpendiculars are greater. 
 
 117. The operation just described is susceptible of 
 some reduction. Thus, according to the preceding, the 
 area of the triangle 
 
 cc 
 
 That of CDcZ=ALX, 
 
 That of vde=ALX~, 
 2 
 
 That of DEe=ALX , 
 
 JH 
 
 That of E>/=ALX??, 
 A 
 
 That of EF/=ALX^; 
 a 
 
 and so on with the others. By adding all these pro 
 ducts together, we see that their sum is equal to the 
 product of the common factor AL, multiplied by half the 
 sum of the two extreme perpendiculars and the sum of 
 all the others. 
 
CENTRE OF GRAVITY. Ill 
 
 As to the moments of these triangles referred to the 
 plane C, we have 
 
 That of ccd=ALX x-- 
 
 mi c 7 D ^ 2AL 
 
 That of CD<i=ALX x ^-, 
 2 o 
 
 mi T>d 4AL 
 
 That of Dd p e=ALX- 6 -x-^-, 
 
 That Of DE=ALX X 
 
 2 3 
 That of Eef=ALX-r 
 
 That of EF/=ALX-X-^-, 
 
 and so on ; in which we see that the number which mul 
 tiplies AL, in the moment of the last triangle, is always 
 equal to three times the number of the intervals minus 
 unity ; or, which is the same thing, to three times the 
 number of the last perpendicular, less 4. By adding 
 together all these moments, we find their sum equal to 
 the product of the common factor ALXAL, multiplied by 
 the sum composed of one-sixth of the first perpendicu 
 lar, one sixth of the last, multiplied by three times the 
 number of perpendiculars less 4, then of the second 
 perpendicular, double the third, three times the fourth, 
 . . . and so on. 
 
112 STATICS. 
 
 Now the sum of the moments and that of the areas 
 having the common factor AL, their quotient will also be 
 the same if we suppose this factor to be in both terms 
 of the division ; hence, to obtain the distance of the 
 centre of gravity K from one of the extreme ordinates 
 c<?, it is necessary , 1st, to take one-sixth of the first or- 
 dinate Co ; one-sixth of the last nh, multiplied by three 
 times the number of ordinates., less 4 ; then the second 
 ordinate, double the third., three times the fourth, . . . 
 and so on ; which will give the first sum : 2d, to the 
 half of the two extreme ordinates add all the interme 
 diate ordinates ; this will give the second sum : 3d, di 
 vide the first of these two sums by the second, and multi 
 ply the quotient by the common interval of the ordinates. 
 
 PROBLEM. 
 
 118. To find the centre of gravity of the volume of 
 the submerged part of the hull of a vessel. 
 
 SOLUTION. We will suppose that the vessel, being 
 afloat, has its keel horizontal, and that the vertical plane 
 drawn through the keel divides the volume of the hull 
 into two perfectly symmetrical parts. This being done, 
 the centre of gravity of the part submerged will be in 
 this plane, and the question will be reduced to find the 
 distances of this point from two lines of known position 
 in the vertical plane. 
 
CENTRE OF GRAVITY. 
 
 Fig. 47. 
 
 113 
 
 B I 
 
 ~~x_ _^- 7i 
 
 J 
 
 ^ 
 
 E ( 
 
 c / 
 
 \ 
 
 
 P 
 
 / 
 
 3 
 
 -~^^f 
 
 
 / 
 
 M 
 
 \ 
 
 N 
 
 / 
 
 Let ABCDF be the section of a vessel through the ver 
 tical plane CD, its keel, and conceive the plane of flota 
 tion, or the section made in the vessel at the level of the 
 water, to be represented by the line Rb parallel to the 
 keel. Let the interval of the two lines B&, CD, be divi 
 ded into a certain number of equal parts BE , B B", B"B " 
 . . . ., and through each point of division suppose there 
 
 are horizontal sections represented by B 5 , B"^" 
 
 In like manner let the line B&, from the point B of the 
 stern-post, be divided into equal parts BF, FF , F F". . . . ; 
 and through each point of division imagine vertical 
 planes to be arranged perpendicular to the keel, and 
 represented by the lines BB ", F/, F^. . . . ; the sub 
 merged part of the hull will be divided into rectangular 
 prisms, whose sides will be perpendicular to the vertical 
 plane drawn through the keel, and which will be termi 
 nated on both sides at the surface of the vessel. (It is 
 necessary that the divisions of the lines BB ", Eb should 
 be so small that the part of the surface of the vessel, 
 which terminates each prism, may be regarded as a 
 plane). Finally, let each rectangular prism, represented 
 by its base LMNP, be divided into two triangular prisms, 
 by a diagonal plane, represented by MP. 
 
 This being done, 1st, each triangular prism will always 
 be divided into three pyramids of the same base as the 
 
 10* 
 
114 STATICS. 
 
 prism (Legendre s Geometry, Book VI.), and each of 
 which will have for height one of the sides of the prism : 
 hence, if, by actual measurement of the vessel, we get 
 the length of all the sides, it will be easy to find the 
 solidity of each pyramid, by multiplying the area of the 
 common base, LMP, by one-third of the side, which 
 measures the height of the pyramid ; and by taking 
 the sum of all these solidities, we will obtain that of the 
 submerged part of the hull. 2d. The moment of a tri 
 angular pyramid referred to a plane, being equal to the 
 product of the solidity of the pyramid, multiplied by 
 one-fourth of the sum of the distances of the summits 
 of its four angles from this plane (113), it will be easy 
 to find the moment of each pyramid referred to the ver 
 tical plane BB " or to the horizontal plane CD ; because 
 the distances of the summits of these angles from each 
 of these planes are known ; and by taking the sum of 
 all these moments, we shall have the moment of the 
 submerged part of the hull. 
 
 This being done, the quotient of the sum of the mo 
 ments referred to the vertical plane BB ", divided by the 
 sum of the solidities, will be the distance KX of the re 
 quired centre of gravity from the vertical BB " ; in like 
 manner the quotient of the sum of the moments re 
 ferred to the horizontal plane CD, divided by the sum of 
 the solidities, will be the distance KY of the same point 
 from the keel. We shall have, therefore, the distances 
 of the centre of gravity from two lines of known posi 
 tion in the vertical plane drawn through the keel, and 
 consequently the position of this point will be deter 
 mined. 
 
CENTRE OF GRAVITY. 115 
 
 The preceding solution is not rigorous ; because the 
 surface of the vessel being curved, the part of this sur 
 face which terminates each triangular prism cannot be 
 regarded as a plane, as we have supposed ; but the re 
 sult will approach exactness as much more as the num 
 ber of divisions, both in the direction of the height of 
 the vessel, and in that of its length, are greater. 
 
 119. The operation just described is susceptible of 
 some reduction; and by reasoning as in No. 116 we find, 
 that to get the distance KX of the centre of gravity of 
 the submerged part of the hull from the vertical plane 
 BB ", it is necessary, 1st, for each horizontal section, to take 
 one-sixth of the first ordinate which is in the plane BB ", 
 one-sixth of the last, multiplied l>y three times the num 
 ber of ordinates contained in the section, less 4 ; then the 
 second ordinate, double the third, three times the fourth, 
 . . ., which will form a partial sum for each section; 
 then add together the half of the first of these sums, the 
 half of the last, and all the intermediate ones, which 
 will form a dividend ; 2d, to one-fourth of the four or 
 dinates placed at the angles of the rectangle EW s ", add 
 one-half of all which are upon the contour of this rect 
 angle, and the whole of all those in the interior, which 
 will form a divisor ; 3d, divide the dividend by the di 
 visor, and multiply the quotient by the interval BF pa 
 rallel to the distance required KX. 
 
 To find the distance KY of the centre of gravity from 
 the horizontal plane drawn through the keel, it is ne 
 cessary to operate upon the vertical sections as upon 
 the horizontal sections in the preceding case : that is to 
 say, 1st, for each vertical section, take one-sixth of the 
 lower ordinates, one-sixth of that which is in the plane 
 
116 STATICS. 
 
 of flotation, multiplied by three times the number of 
 ordinates of the section, less 4 : then the second ordinate 
 from the bottom, double the third, three times the fourth, 
 . . . tuhich ivill form for each section a partial sum ; 
 then add together the half of the first of these sums, the 
 half of the last, and all the intermediate ones, which will 
 form a dividend ; 2d, divide this dividend by the same 
 divisor as in the preceding case, and multiply the 
 quotient by the interval BB parallel to the distance 
 sought KY. 
 
 Remark. 
 
 120. In the preceding problem, the only object is to 
 find the centre of gravity of the volume of the submerged 
 part of the hull, or, which is the same thing, the volume 
 of water displaced by the vessel. But if it were re 
 quired to find the centre of gravity of the vessel itself 
 either laden or unladen : that is to say, to find the dis 
 tances of this point from the horizontal plane drawn 
 through the keel, and from the vertical plane perpen 
 dicular to the keel, it would be necessary to take, with 
 reference to each of these planes, the sum of the mo 
 ments of all the parts which compose the vessel and its 
 load, and then to divide each of these sums by the total 
 weight of the vessel and its load ; observing, in taking 
 the moments, to multiply, not the volume, but the weight 
 of each part, by the distance of the partial centre of 
 gravity of this part from the plane to which the moments 
 are referred ; and the quotients of these divisions would 
 be the distances required. 
 
CENTRE OF GRAVITY. 117 
 
 It will be easy to find, at least in a manner sufficiently 
 near, the partial centre of gravity of each of the parts 
 of the vessel and its load ; because we may always de 
 compose this part into parallelopipedons, cylinders, py 
 ramids, or other solids of which we have given the 
 means of finding the centres of gravity. 
 
118 
 
 CHAPTER FOURTH. 
 
 ON THE EQUILIBRIUM OF MACHINES. 
 
 121. Every instrument intended to transmit the ac 
 tion of a determined force, to a point which is not found 
 upon its direction, so that this force may move a body 
 to which it is not immediately applied, and move it in 
 a direction different from its own, is called a machine. 
 
 122. In general, we are not able to change the direc 
 tion of a force, but by decomposing this force into two 
 others ; one of which is directed toward a fixed point, 
 which destroys the force by its resistance, and the other 
 acts in a new direction : this latter force, which is the 
 only one that can produce any effect, is always a com 
 ponent of the first ; and may be either smaller or great 
 er than it is, according to circumstances. By changing, 
 in this manner, the directions and intensities of the 
 forces, we may, by the aid of a machine and the points 
 of support which it presents, produce an equilibrium 
 between two unequal forces which are not directly 
 opposed. 
 
 123. The force, whose direction is to be changed by 
 employing a machine, is ordinarily named a power, and 
 the term resistance is applied to the body it has to 
 
MACHINES. 119 
 
 move, or to the force with which it has to produce an 
 equilibrium by means of the machine. 
 
 124. We propose here to find only the ratios w T hich 
 should subsist between the power and the resistance ap 
 plied to the same machine, in order that, with regard to 
 their directions, they may be in equilibrium. We will 
 leave friction out of consideration : that is to say, the 
 difficulties which the different parts of the machine may 
 experience in slipping or rolling upon each other ; 
 and we will suppose that cords, when they enter into 
 the composition of the machine, are perfectly flexible. 
 Thus, having given to a power the intensity which is 
 proper for the condition of equilibrium in this supposi 
 tion, it would not suffice to augment it by a small quan 
 tity to destroy the equilibrium and put the machine in 
 motion ; but we must first apply the whole quantity re 
 quired to overcome the obstacles just mentioned ; and 
 then a slight increase will produce motion. 
 
 125. Although the number of machines is very great, 
 we may regard them all as composed of three simple 
 machines, which are : cords, the lever, and the inclined 
 plane. We will content ourselves with presenting the 
 theories of these three machines, and of those which 
 are immediately derived from them ; it will then be 
 easy, by simple applications, to find the ratio of the 
 power to the resistance, for the condition of equilibrium, 
 in every machine, however complicated it may be. 
 
120 
 
 STATICS. 
 
 AKTICLE I, 
 
 On the equilibrium of forces which act upon each other 
 by means of cords. 
 
 126. We will suppose that the cords are without 
 weight ; and because the property they have of trans 
 mitting force, is independent of their size, we will sup 
 pose them to be reduced to their axes, and regard them 
 Fi 9- 4a as straight, flexible, and in- 
 
 extensible lines. Taking 
 this for granted, let us con 
 sider, first, the case of equi 
 librium between three forces 
 p, Q, R, acting upon each 
 other by means of three 
 cords united together by a 
 knot A. 
 
 1st. The three forces P, 
 Q, R, cannot be in equili 
 brium, unless the three di 
 rections, and consequently 
 the cords by means of which they transmit their actions, 
 are in the same plane (10). 
 
 2d. If we represent any two of these forces, for ex 
 ample, the forces P, Q, by the parts AC, AD of their di 
 rections, and upon these lines as adjacent sides construct 
 the parallelogram ACBD, the diagonal AB will represent 
 in intensity and direction the resultant of these two 
 forces (36) ; now the three forces being in equilibrium, 
 the force R must be equal and directly opposed to the 
 resultant of the two others ; hence the direction of the 
 
CORDS. 
 
 121 
 
 force R will be in the prolongation of BA, and its in 
 tensity will be represented by this diagonal : thus we 
 shall have 
 
 p : Q : R : : AC : AD : AB ; 
 
 or because the sides AD, BC of the parallelogram are 
 equal to each other, the three forces P, Q, R will be to 
 each other as the sides of the triangle ABC. 
 
 127. The angles of the triangle ABC, being given by 
 the directions of the forces P, Q, R, and the magnitudes 
 of its sides being proportional to the intensities of these 
 forces, it follows that, of the six things to be considered 
 in the equilibrium in question, namely, the directions 
 of the forces and their intensities, any three being 
 given, we can find the three others in all cases in which, 
 of the six things to be considered in the triangle ABC, 
 namely, the angles and the sides, three being given, we 
 can determine the three others. 
 
 For example, when the three forces P, Q, R are known, 
 we can find the angles which the cords make with each 
 other when in equilibrium, by constructing the triangle 
 ABC, the sides of which are proportional to these forces. 
 But when the directions are given, we can know only 
 the ratios of the three forces ; because in the triangle 
 ABC, the knowledge of the three angles determines only 
 the ratios of the sides, but does not determine their 
 magnitudes. Thus, it will be necessary, besides, to know 
 the magnitude of one of the three sides P, Q, R, in order 
 to find that of the two others, by means of the propor 
 tional series : 
 
 P : Q : R : : AC : BC : AB. 
 
 11 
 
122 STATICS. 
 
 Remark I. 
 
 49- 128. If the three cords be 
 
 united by a slip-knot ; for ex 
 ample, if the cord PAQ passes 
 | / through a ring attached to 
 / the extremity of the cord RA, 
 
 ^ the conditions just enuncia 
 
 ted are not sufficient to es 
 tablish equilibrium : it is ne 
 cessary, in addition, that the 
 angles PAB, QAB, formed by 
 the two parts of the cord and 
 by the prolongation AB of the direction of the other 
 cord, should be equal ; for it is evident that, otherwise, 
 the ring A would slip upon this cord toward the greater 
 of the two angles. 
 
 Remark II. 
 
 129. What has just been said contains the whole 
 theory of equilibrium betwee*n three powers applied to 
 cords united in the same knot ; but we have supposed 
 the construction of the parallelogram ACBD, we may 
 however enunciate this theory independently of all con 
 struction. 
 
 Thus, in every triangle ABC, the sides are proportional 
 to the sines of the opposite angles : that is to say, we 
 have 
 
 AC : BC : AB : : sin ABC : sin BAG : sin ACB. 
 
CORDS. 123 
 
 Now, the sines of these angles are respectively the 
 same as those of their supplements RAQ, RAP, PAQ; 
 hence we have also 
 
 AC : BC : AB : : sin RAQ : sin RAP : sin PAQ, 
 and consequently 
 
 p : Q : R : : sin RAQ : sin RAP : sin PAQ. 
 
 That is to say, when the three powers, which act by 
 cords upon the same knot, are in equilibrium, each of 
 them is as the sine of the angle formed ly the, directions 
 of the other two. 
 
 COROLLARY I. 
 
 - 50 - 130. If the cords AP, AQ, 
 
 instead of being drawn by 
 two powers, be attached to 
 two fixed points at P and Q, 
 and we represent the force 
 R by the diagonal AB of the 
 parallelogram ABCD, the two 
 sides AC, AD will represent 
 the tensions of these two cords, or the efforts which 
 they exert upon the fixed points in the line of their di 
 rections. 
 
 COROLLARY II. 
 
 131. When the angle PAQ is very large, the sides AC, 
 AD of the parallelogram are very great compared with 
 the diagonal AB, and consequently the effort which the 
 
124 
 
 STATICS. 
 
 power R exerts upon the fixed points P, Q, to make them 
 approach each other, is very great compared with this 
 power. Hence, by means of cords, we can put a 
 moderate power in the condition of producing a very 
 great effect. 
 
 COROLLARY III. 
 
 132. In the case of equilibrium, however small the 
 force R may be, the diagonal AB, which represents it, is 
 not zero, and the three points c, A, D, are not in a straight 
 line ; hence, by supposing that a cord PAQ without weight 
 is stretched in a line by two forces P, Q, the smallest 
 force R, applied to A, will bend it at this point and cause 
 it to make an angle PAQ. Thus, it is rigorously impossi 
 ble to stretch a heavy cord in a straight line, unless it 
 be vertical ; for the weight of the parts which compose 
 it may be regarded as forces applied to this cord, and 
 which must necessarily deflect it from a straight line. 
 
 COROLLARY IV. 
 
 Fig. 51. 
 
 133. If any number 
 of powers P, Q, R, s, 
 T, . . . act upon each 
 other by cords join 
 ed together three by 
 three in the same knot, 
 it is easy, from what 
 precedes, to find the 
 ratios which these pow- 
 
CORDS. 125 
 
 ers should have with each other as to their directions, 
 so as to be in equilibrium. For the general equilibrium 
 cannot take place, unless, 1st, the three powers joined 
 together in each knot are in equilibrium with each other ; 
 2d, each of the cords AB, BC, which join two knots, are 
 equally stretched in the two directions. Hence, by 
 naming v, x, the tensions of the two cords AB, BC, we 
 will have (129), by reason of the equilibrium at the 
 knot A, 
 
 p : Q : : sin QAB : sin PAB, 
 p : v : : sin QAB : sin PAQ ; 
 
 by reason of the equilibrium at the knot B, 
 
 v : R : : sin RBC : sin ABC, 
 v : x : : sin RBC : sin ABR ; 
 
 by reason of the equilibrium at the knot c, 
 
 x : T : : sin SCT : sm BCS. 
 
 And by continuing these proportions, we will find the 
 ratio of any two of these powers, and the ratio of one 
 of them to the tension v, x of any cord whatever. 
 
 For example, by multiplying in order the 2d propor 
 tion and the 3d, we find 
 
 p : R : : sin QABXsin RBC : sin PAQ x sin ABC; 
 
 by multiplying the 2d and 4th, 
 11* 
 
126 STATICS. 
 
 p : x : : sin QABXsin RBC : sin PAQXsin ABR ; 
 by multiplying the 2d, 4th and 5th, 
 
 p : S : : sin QABXsin RBC X sin SCT 
 : sin PAQXsin ABR x sin BCT; 
 
 by multiplying the 2d, 4th and 6th, 
 
 p : T : : sin QABXsin RBC x sin SCT 
 : sin PAQXsin ABR x sin BCS, 
 
 and so on. 
 
 It also follows from this, that the three cords united 
 by the same knot are in the same plane (126), although 
 those which are united at two knots may be in different 
 planes. 
 
 COROLLARY V. 
 
 134. If the forces Q, R, s, 
 be weights suspended by the 
 knots A, B, c, to the same 
 cord EABCF, and this cord 
 be retained at its extremities 
 by two fixed points E, F : 
 
 1st. The main co^d and the 
 cords of the weights Q, R, s, 
 are in the same vertical plane ; 
 
CORDS. 127 
 
 for the two parts EA, AB of the cord are in the vertical 
 plane drawn through the cord AQ ; in like manner, the 
 two parts AB, BC are in the vertical plane drawn through 
 BR : now these two vertical planes pass through the 
 same line AB, and coincide ; hence the parts EA, AB, BC 
 of the cord, and the directions of the cords AQ, BR, are 
 in the same vertical plane. In the same manner it may 
 be demonstrated that the part CF of the cord and the 
 direction CS are in this same plane, and so on. 
 
 2d. The tensions of the two extreme parts of the cord 
 are to each other reciprocally as the sines of the angles 
 which these parts make with the vertical ; for the angles 
 QAB, ABR, are supplements of each other, and have the 
 same sine : it is the same with the angles RBC, BCS, and 
 so of the rest ; hence, by neglecting the common factors 
 in the proportion which gives the ratio of the two ex 
 treme tensions P, T (133), we have 
 
 E : T : : sin SCT : sin PAQ. 
 
 3d. The vertical HI, drawn through the point of inter 
 section G of the prolongations of the two extreme parts 
 of the cord, passes through the centre of gravity of the 
 system of all the weights Q, R, s, . . . ; for the two ex 
 treme parts being in the same plane, their tensions have 
 a resultant whose direction passes through the point G ; 
 moreover, these tensions supporting the system of 
 weights Q, R, S, . . . their resultant should be vertical, 
 and pass through the centre of gravity of these weights ; 
 hence the point a and the centre of gravity of the 
 weights Q, R, s, are in the same vertical. 
 
128 
 
 STATICS. 
 
 COROLLARY VI. 
 
 135. When a heavy cord 
 EHF is suspended in equilibri 
 um to the two fixed points 
 E, F, we may consider its axis 
 as a thread without weight, 
 charged with a weight dis 
 tributed throughout its whole 
 extent : hence 1st, this axis is 
 in the vertical plane drawn through the two points of 
 suspension; 2d, if the directions of the two extreme 
 elements of this axis be prolonged to EG, FG, and through 
 the point of intersection G we draw the vertical IH, the 
 tensions of these two elements are to each other recipro 
 cally as the sines of the angles which these elements 
 make with the vertical : that is to say, by naming P and 
 T these tensions, we have 
 
 p : T 
 
 sin IGF : sin IGE 
 
 3d, the centre of gravity K of the cord is in the ver 
 tical IH. 
 
 Finally, by considering the total weight z of the cord 
 as a force applied to the point G of its direction, we will 
 find the efforts which the cord makes upon the two 
 points of support E, F, along the directions EG, FG, by 
 decomposing the force z into two others which act in 
 these directions ; and we shall have (129) 
 
 z : P : T : : sm EGF : sm IGF : sin IGE. 
 
COEDS. 129 
 
 Remark. 
 
 136. So far we have supposed that there were only 
 three cords united at each knot, because, if the cords as 
 sembled at the same knot are greater in number and 
 included in the same plane, it is not sufficient to know 
 their directions in order to find what ratios the applied 
 powers should have so as to be in equilibrium : that is 
 to say, these ratios may vary in an infinite number of 
 ways, without the forces ceasing to be in equilibrium. 
 
 Thus, whatever may be the number of powers directed 
 in the same plane, it suffices, in order to be in equili 
 brium about the same knot, that the resultant of any 
 two of them is equal and directly opposed to the re 
 sultant of all the others ; hence all these forces, except 
 two, being taken at pleasure, which determine the in 
 tensity and direction of the resultant, we can find the 
 intensities of the latter two forces which will make an 
 equilibrium with this resultant. 
 
 However, when four cords, joined at the same knot, 
 are not in the same plane, their directions being given, 
 the ratios of the intensities which the applied forces 
 must have, to produce an equilibrium, are determined : 
 for we have seen (44), that these forces must be to each 
 other as the diagonal, and the adjacent sides of the 
 parallelopipedon constructed on their lines of direction. 
 But when the forces are not directed in the same plane, 
 and their number is greater than four, the ratios of the 
 forces are no longer determined by the knowledge of 
 the direction of the cords. 
 
130 
 
 STATICS. 
 
 X 
 
 ARTICLE II. 
 
 On the Equilibrium of the Lever. 
 
 Fig- 64. 137. The lever is an 
 
 inflexible rod ACB (Fig. 
 54), or CAB (Fig. 55), 
 either straight or curved 
 and moveable around 
 one of its points c, ren 
 dered fixed by means of 
 any obstacle ; and this 
 obstacle is termed the 
 point of support or ful 
 crum. 
 
 138. First, supposing 
 the lever to be without 
 weight, and that it cannot 
 in any manner slip upon 
 its fulcrum, let P, Q be two 
 powers applied, either im 
 mediately or by means of 
 the cords AP, BQ, to the 
 two points A, B of a lever. 
 If we consider the resist 
 ance of the point c as the effect of a third force R ap 
 plied to the lever at this point, we have seen, in order 
 that equilibrium may subsist between these three forces, 
 1st, their directions must be included in the same plane, 
 and meet in the point D (10) ; 2d, the forces P, Q must 
 be to each other reciprocally as the perpendiculars CE, 
 
 Fig. 65. 
 
THE LEVER. 131 
 
 CF, let fall from the fulcrum upon their directions (35) : 
 that is to say, we must have 
 
 p : Q : : CF : CE ; 
 
 3d, if we lay off from the point D the lines DL, DM, 
 upon the directions of the forces P, Q, proportional to 
 the intensities of these forces, and finish the parallelo 
 gram DLMN ; the diagonal DN will represent in intensity 
 and direction the force E, and consequently the resist 
 ance of the fulcrum (35) ; thus we will have 
 
 p : Q : E : : DL : DM or NL : DN, 
 or (129), 
 
 p : Q : R : : sin QBE, : sin PDE : sin PDQ. 
 
 COEOLLAEY I. 
 
 139. If we leave out of consideration the resistance 
 of the fulcrum : that is to say, if we suppose this point 
 to be capable of an indefinite resistance, it is necessary, 
 in order that the two powers P, Q may be in equilibrium 
 around this point by means of the lever, 1st, their di 
 rections and the fulcrum should be in the same plane ; 
 2d, the two forces P, Q should tend to turn the lever 
 around the fulcrum c in opposite directions, and their 
 moments, referred to this point, should be equal : that 
 is to say, we should have (80) 
 
 PXCE=QXCF. 
 
132 STATICS. 
 
 COROLLARY II. 
 
 140. Hence it appears, 1st, that however small the 
 power Q may be, we may always, by means of a lever, 
 put it in equilibrium around a fulcrum c, with another 
 power P of given intensity and direction ; for the di 
 rection of the force P being known, the distance CE of 
 this direction from the fulcrum will be known, and we 
 will know the moment PXCE : hence it will be sufficient 
 to arrange it so that the moment QXCF of the power, is 
 equal to the preceding : that is to say, to direct this 
 power in such a manner that its distance CF from the 
 
 P X CE 
 
 fulcrum is equal to , and that it tends to turn the 
 
 lever in the opposite direction to the force P. 
 
 2d. If the distance CF of the direction of the force 
 Q from the fulcrum is known, we will find the intensity 
 which this force must have in order to produce an equi 
 librium with the force P, by dividing the moment of this 
 latter force by the distance CF : that is to say, we shall 
 have 
 
 PXCE 
 
 CF 
 
 COROLLARY III. 
 
 141. The effort or load, which the fulcrum c sustains, 
 being equal to the resultant of the two forces P, Q, we 
 will find this load by means of the following proportion : 
 
THE LEVER. 133 
 
 p : Q : R : : sin QDR : sin PDR : sin PDQ : 
 that is to say, we shall have 
 
 _PXsin PDQ 
 sin QDR 
 
 or, 
 
 _QXsin PDQ 
 sin PDR 
 
 COROLLARY IV. 
 
 142. Hence, if the fulcrum does not possess indefinite 
 resistance, in order that it may not be moved, and that 
 equilibrium may subsist, it is necessary that the resist 
 ance in the direction CD, should be equal to the result 
 ant of the two forces P, Q : that is to say, equal to 
 
 PXsin PDQ 
 
 sn QDR 
 
 , . , . ,, QXsin PDQ 
 
 or, which is the same, to -. - . 
 
 sm PDR 
 
 COROLLARY V. 
 
 143. In general, of the six things to be considered 
 in the equilibrium of the lever, namely the intensities 
 and directions of the two powers p, Q, and those of the 
 load on the fulcrum, any three being given, the other 
 
 three may be determined in all cases; or of the six 
 
 12 
 
134 
 
 STATICS. 
 
 analogous things which may be considered in the tri 
 angle DLN, namely, the sides and the angles, three being 
 given, we can determine the others. 
 
 Remark I. 
 
 144. If the lever can slip upon the fulcrum, the con 
 ditions, which have just been given, are not sufficient to 
 keep the lever stable, and to produce an equilibrium ; 
 it is necessary, besides, that the direction DC of the load 
 on the fulcrum should be perpendicular to the surface 
 of the lever at the point c ; for, if this direction were 
 oblique, the lever would have a tendency to slip towards 
 the side of the greater angle, and in fact would slip, 
 whenever this tendency should be greater than the fric 
 tion upon the fulcrum which opposes this effect, as we 
 will demonstrate in treating of the inclined plane. 
 
 Remark II. 
 
 145. What has just been stated contains the whole 
 theory of the equilibrium of two powers applied to a 
 lever, considered to be without weight, and retained by 
 a fulcrum ; we will now make application of it to a few 
 simple cases. 
 
 Fiff- 56- If the directions of the pow 
 
 ers P, Q, applied to a lever, are 
 parallel to each other : for ex 
 ample, if there are two weights 
 suspended at the points A, B, 
 the load, which the fulcrum c 
 sustains, is equal to their sum 
 
THE LEVER. 135 
 
 P+Q, and the two perpendiculars CE, CF, let fall from 
 the fulcrum upon their directions, are in a straight line. 
 Hence, if the lever be straight, the triangles ACE, BCF 
 will be similar, and give 
 
 CF : CE : : CB : CA. 
 
 Hence we shall have, in the case of equilibrium, 
 p : Q : : CB : CA : 
 
 that is to say, the weights P, Q will be to each other re 
 ciprocally as the arms of the lever. 
 
 Thus, the intensity and the lever arm of a resistance 
 p being given, 1st, the lever arm, which must be given 
 to a power Q in order to produce an equilibrium with it, 
 will be 
 
 PXCA 
 CB= . 
 
 2d. The intensity of the power necessary to be ap 
 plied to the given point B, in order to produce an equi 
 librium with it, will be 
 
 _PXCA 
 
 ~~* 
 
 Finally, if the two weights P, Q, and the length AB 
 of the lever, be given, we will find the fulcrum c, around 
 which these weights will be in equilibrium, by dividing 
 the lever AB into two parts reciprocally proportional to 
 the two weights. 
 
136 STATICS. 
 
 Remark III. 
 
 146. When there are more than two powers applied 
 to the same lever, it is not sufficient to know their di 
 rections, and the position of the fulcrum, in order to 
 determine the ratios which they should have to each 
 other so as to be in equilibrium ; but, as equilibrium 
 cannot take place between several forces about a ful 
 crum, unless the resultants of all the forces are destroyed 
 by the resistance of this point, it is evident, in this 
 case, that the conditions of equilibrium are reduced to 
 the two following : 1st, all the forces must have a single 
 resultant ; 2d, the direction of this resultant must pass 
 through the fulcrum. 
 
 COROLLARY. 
 
 147. If the directions of all the forces are included 
 in the same plane, these forces have necessarily a single 
 resultant (43), and the first condition is fulfilled ; hence 
 it suffices for the equilibrium, that the direction of this 
 resultant passes through the fulcrum, or, which amounts 
 to the same, the sum of the moments of the forces, 
 which tend to turn the lever in one direction around the 
 fulcrum, is equal to the sum of the moments of those 
 which tend to turn it in the opposite direction. 
 
 Remark IV. 
 
 148. So far we have abstracted weight from the 
 lever ; but if this weight enter into consideration, it 
 must be regarded as a new force, applied to the centre 
 
THE LEVER. 137 
 
 of gravity of the lever in a vertical direction ; and in 
 the case of equilibrium, the conditions just stated sub 
 sist between all the forces, including that under con 
 sideration. 
 
 Let P, Q be two weights sus 
 pended from a heavy lever AB, 
 and in equilibrium about the 
 fulcrum c : we will consider the* 
 weight of the lever as a third 
 weight S, suspended from the 
 centre of gravity of the lever, 
 and the sum of the moments of the two weights Q, S, 
 referred to the fulcrum c, will be equal to the moment 
 of the weight P : that is to say, we shall have 
 
 QXCB+SXCG=PXCA; 
 
 or, subtracting from these two equal quantities the mo 
 ment of the lever SXCG, 
 
 QXCB=PXCA SXCG. 
 
 Thus, knowing the length and the weight of the lever, 
 the position of its centre of gravity, that of the fulcrum 
 and one of the two weights P, Q, it will always be pos 
 sible to get the other weight ; for we shall have 
 
 _QXCB-fSXCG 
 CA 9 
 
 and 
 
 PXCA SXCG 
 
 Q= 
 
 CB 
 
 12* 
 
138 
 
 STATICS. 
 
 The load at the fulcrum is evidently equal to the sum 
 of the weights p-fQ+s. 
 
 F ^- 58< 149. But, if the weight P, 
 
 suspended from the heavy 
 lever AB, be retained in equi 
 librium about the fulcrum c, by 
 means of a vertical power Q, 
 and directed upwards, the mo 
 ment of the force Q, which 
 tends to turn the lever in one 
 direction, will be equal to the 
 sum of the moments of the 
 weights P, s, which tend to 
 turn it in the opposite direc 
 tion, and we shall have 
 
 QXCB=PXCA + SXCG, 
 
 or, subtracting the moment of the lever, 
 
 QXCB SXCQ=PXCA. 
 
 Thus the intensities, which the two forces P, Q must 
 have in order to be in equilibrium, will be 
 
 _ 
 
 QXCB SXCG 
 ~CA 
 
 and 
 
 _ 
 
 PXCA+sxca 
 
 CB 
 
 and the weight upon the fulcrum p+S Q. 
 
THE LEVER. 
 
 139 
 
 O tlOLLARY. 
 
 Fig. 57. 150. Hence it appears by 
 
 regarding the weight P as a re 
 sistance, and the force Q as a 
 power which has to bring the 
 weight into equilibrium or to 
 overcome it, by means of the 
 lever AB, the weight of this 
 lever is a force which may either increase or diminish 
 the power, according as this weight tends to turn the 
 lever in the same direction as the power, or in the oppo 
 site direction. For example, in the case of Fig. 57, 
 the weight of the lever increases the power Q ; and by 
 prolonging the lever arm CB of this power, it would be 
 placed in the condition of producing an equilibrium with 
 a greater resistance, for two reasons : 1st, because its 
 Fig- 58. moments would be thus aug- 
 
 ^ mented; 2d, because the 
 weight S of the lever would 
 be augmented, which alone 
 would produce an equilibrium 
 with a greater part of the 
 resistance. But, in the case 
 of Fig. 58, the weight of 
 the lever diminishes the pow 
 er Q, and we cannot increase 
 the length of the lever arm 
 *" CB without at the same time 
 
 augmenting its weight S, which forms part of the resist 
 ance ; thus, in order that there may be in this case an 
 
 I 
 
 A 
 v 
 
* "* ^ j* 
 
 " 
 
 140 STATICS. 
 
 advantage in prolonging the arm of the lever, it is 
 necessary that the moment of this prolongation should 
 be less than the resulting increase in the moment of the 
 power,. 
 
 THEOREM. 
 
 151. Two powers p, 
 Q, api^lied to a lever 
 AB, and in equilibrium 
 about a fulcrum c, are 
 t each other recipro- 
 cally as tlie spaces 
 which these powers tra- 
 verse in the line of 
 their directions , if the 
 equilibrium be disturbed infinitesimally . 
 
 DEMONSTRATION. From the fulcrum c let fall the 
 perpendiculars CE, CF upon the directions of the powers ; 
 and instead of the straight lever AB, let us consider the 
 bent lever ECF ; at the extremities E, F of which con 
 ceive the powers P, Q to be applied ; then suppose, by 
 virtue of a derangement in the equilibrium, that the 
 bent lever EOF takes the infinitely near position ecf. 
 This being done, the small arcs EC, F/, will be the spaces 
 which the powers P, Q would traverse by virtue of this 
 derangement. Now, the angle ECF of the bent lever 
 being invariable, the two angles EC<?, FC/ are equal, and 
 we have 
 
THE LEVER. 
 CF : CE : : F/ : Ee ; 
 
 moreover, because of the equilibrium, we have (35) 
 
 p : Q : : CF : CE ; 
 hence we have also 
 
 141 
 
 We will have occasion to show, in a subsequent part, 
 that an analogous proposition occurs in cases of equi 
 librium for all other machines. 
 
 On Pulleys. 
 
 Fig. 60. 
 
 I. 
 
 152. A pulley is a 
 wheel having a groove on 
 its circumference to re 
 ceive a cord PGDHQ, and 
 is traversed at its centre 
 by an axle E, upon which 
 it turns in a sheath or 
 block EL. 
 
142 
 
 STATICS. 
 
 II. 
 
 Fig. 60. 153. Suppose the axis 
 
 of the pulley being fixed, 
 two forces p, Q are ap 
 plied to the extremities 
 of the cord, and this cord 
 being perfectly flexible 
 exerts no friction upon 
 the groove of the pulley, 
 so that it may slide free 
 ly on this rim. What 
 ever the figure of the 
 pulley may be in other 
 respects, that is to say, 
 whether the arc GDH, em 
 braced by the cord, be circular or not, it is evident that 
 the two forces cannot be reciprocally in equilibrium 
 unless they are equal ; for if they were unequal, the 
 greater would overcome the smaller by causing it to 
 slide in the groove of the pulley. 
 
 Upon the same supposition, the pulley, having no 
 other fixed point than its centre, and being drawn by 
 the two forces P, Q, cannot remain at rest, unless the 
 resultant of these two forces is directed towards the 
 centre, and is destroyed by the resistance of this point. 
 Hence, having prolonged the directions PG, QH of the 
 two cords, until they meet in some point A, and taking 
 upon these directions the equal lines AB, AC to repre 
 sent the forces P, Q, if the parallelogram ABDC be com 
 pleted, the diagonal AD, .which will represent the resultant 
 of these two forces, must pass through the fixed point E. 
 
PULLEYS. 
 
 143 
 
 Fig. 61. 
 
 Now, when the figure 
 is circular, this last con 
 dition is always fulfilled : 
 thus, the triangle ABD 
 being isosceles, the angle 
 BAD is equal to the angle 
 BDA, and consequently to 
 the angle DAG ; hence the 
 diagonal AD divides the 
 angle BAG into two equal 
 parts. But, if we draw 
 the line EA, this line di 
 vides the same angle into 
 two equal parts ; for if, 
 through the centre E, we draw, to the points of contact 
 of the cords, the radii EG, EH, perpendicular to the di 
 rections of these cords, the two rectangular triangles 
 EGA, EHA will be in all respects equal, and we shall have 
 the angle EAG of the one, equal to the angle EAH of the 
 other. Hence the line EA, and the diagonal DA, will 
 have the same direction. 
 
 Hence, the centre of a pulley being fixed and its 
 figure circular, it is necessary that the two forces P, Q, 
 to be in equilibrium, should be equal, and that the pulley 
 remain at rest, at the same time, about its axis. 
 
 The load which the axis of the pulley sustains is evi 
 dently equal to the resultant of the two forces P, Q; 
 hence, if we name R this load, we shall have (36) 
 
 p : Q : R : : AB : AC : AD. 
 
144 
 
 STATICS. 
 
 Fig. 60. 
 
 Finally, draw the cord 
 GH of the arc embraced 
 by the rope ; the two tri 
 angles GHE, ABD will be 
 similar, because they will 
 have their sides perpen 
 dicular each to each ; and 
 we shall have 
 
 AB : AC : AD : : GE : EH : GH; 
 
 hence we shall have 
 
 p : Q : n : : GE : EH : GH. 
 
 III. 
 
 154. If the axis of the pulley is not absolutely fixed, 
 but retained simply by the power s, by means of the 
 block EL and the cord LS ; in order that this axis may 
 be at rest, and the three forces P, Q, S in equilibrium, 
 it is necessary that the force S should be equal and di 
 rectly opposed to the load which the axis sustains. 
 Hence, 1st, the direction of this force should coincide 
 with the line EA ; 2d, its intensity should be equal to 
 the resultant n of the two forces P, Q, and we shall 
 have 
 
 p : Q : S : : GE : EH : GH. 
 
PULLEYS. 
 
 145 
 
 Thus, when two forces P, Q, applied to a rope em 
 bracing a pulley, are in equilibrium with each other, and 
 with a third force S applied to the axis of the pulley, 
 1st, the two forces P, Q are equal to each other ; 2d, the 
 direction of the force S bisects the angle formed by the 
 directions of the two others ; 3d, each of the two forces 
 P and Q is to the third force S, as the radius of the pul 
 ley is to the chord of the arc embraced by the rope. 
 
 COROLLARY I. 
 
 Fig. 60. 155. We observe that 
 
 if the cord, fastened to 
 the block, instead of being 
 drawn by a force S, is at 
 tached to a fixed point of 
 indefinite resistance, and 
 if it be proposed, by em 
 ploying the pulley, only 
 to place the two forces 
 P, Q in equilibrium, or to 
 overcome a resistance P, 
 by means of a power Q, 
 the pulley does not assist 
 the power : it has no 
 
 other effect than to change the direction of this force, 
 
 without altering its intensity. 
 
 is 
 
146 
 
 STATICS. 
 
 ir. 61. 
 
 But if one of the ex 
 tremities of the cord, 
 which embraces the pul 
 ley, is attached to a fixed 
 point P, and, by using 
 the pulley, we design to 
 place a power Q in equi 
 librium with a resistance 
 S, attached to the block, 
 the pulley assists the 
 power, which is always 
 less than the resistance ; 
 for we have 
 
 Q : S : : EH : GH, 
 
 COROLLARY II. 
 
 Fig, 63. 
 
 156. When the directions 
 of two parts PG, QH of the cord 
 are parallel to each other, and 
 consequently to that of the 
 cord of the block, the chord 
 GH becomes a diameter, and 
 is double the radius ; the pro 
 portional series of No. 153 
 then becomes 
 
 p : Q : S : : 1 : 1 : 2, 
 
 that is to say, the force S, or the load of the axis of the 
 pulley, is equal to the sum of the two powers P, Q, or to 
 
PULLEYS. 147 
 
 double one of them. Thus, in the case of Fig. 63, the 
 power Q, which, by means of the pulley and the point 
 of support P, will produce an equilibrium with the re 
 sistance S, will be only half of this resistance. 
 
 IV. 
 
 157. A pulley is said to be immoveable when its 
 block is attached to a fixed point and the power and re 
 sistance are applied to the cord which embraces it ; but 
 when the resistance is attached to the block, and the 
 pulley has to move with it, as in Figs. 61, 63, the pulley 
 is said" to be movedble. 
 
 This being established, let there be any number 
 of moveable pulleys, (Fig. 64,) and considered without 
 weight ; let the first bear a weight P suspended to its 
 hook, and embraced by a cord, one of whose extremi 
 ties is attached to the fixed point D, and the other is 
 applied to the block of the second pulley ; let this pul 
 ley be embraced by another cord, one of whose extremi 
 ties is fixed to the point H, and the other attached to 
 the block of the next pulley ; let this third pulley be 
 embraced by a third cord fixed on one side to M, and 
 drawn on the other by a power Q ; and so on, if the 
 number of pulleys be greater. Finally, supposing the 
 whole system to be in equilibrium, draw the radii and 
 the subtending chords of the pulleys as is shown in the 
 figure. We may consider the equilibrium of the first 
 pulley A as though this pulley were alone ; and represent 
 ing by X the tension of the cord BX, we shall have (155) 
 
 p : x : : BC : BA. 
 
148 
 
 STATICS. 
 Fig. 64. 
 
 For the same reason, calling Y the tension of the cord 
 FT, -we shall have 
 
 x : Y : : FG : EF. 
 
 We shall have, in like manner, for the third pulley, 
 Y : Q : : KL : IK ; 
 
 and so on, whatever may be the number of pulleys. 
 Hence, by multiplying in order all these proportions, 
 we shall have 
 
 p : Q : : BCXFQXKL : BAXEFXIK: 
 
PULLEYS. 
 
 149 
 
 that is to say, the resistance is to the power as the 
 product of the subtending chords is to the product of 
 the radii. 
 
 COROLLARY. 
 
 158. When all the ropes CD, GH, LM, . . are parallel, 
 the subtending chords will be diameters, and the pre 
 ceding proportion will become 
 
 p : Q :: 2x2x2 : Ixlxl, or : : 8 : 1 ; 
 
 Fig. 65. 
 
 from which it is evident, that, 
 in this case, the resistance is to 
 the power as the number 2, 
 raised to a power indicated by 
 the number of moveable pul 
 leys, is to unity. 
 
 Thus, by suitably increasing 
 the number of moveable pulleys, 
 we may put a moderate force in 
 equilibrium with a very great 
 resistance. For example, with 
 three pulleys, and by means of 
 the fixed points D, II, M, the 
 power produces an equilibrium 
 with a resistance eight times 
 greater than itself. 
 
 However advantageous this 
 disposition of moveable pulleys 
 may at first appear, it is rarely 
 
 13* 
 
150 STATICS. 
 
 employed ; because, in order to make the first pulley A 
 traverse a certain space, it is necessary that the second 
 should traverse a double space, and the third a quadru 
 ple ; and so on, which requires too much room ; and 
 thus, most generally, muffles are used instead. 
 
 V. 
 
 159. The term muffle is applied to a system of several 
 pulleys assembled in the same block and upon separate 
 axes, as in Figs. 66, 67, or upon the same axis, as in 
 Fig. 68. A fixed muffle and a moveable muffle are 
 always employed at the same time ; and all the pulleys 
 of the two muffles are embraced by the same cord, one 
 of whose extremities is attached to one of the two 
 muffles, and the other extremity is drawn by the power ; 
 and the resistance is suspended to the moveable muffle. 
 
 We may give different diameters to the pulleys, and 
 arrange them in such a manner that all the parts of the 
 cord, which go from one muffle to the other, may be 
 parallel to each other, as in Figs. 66, 67 ; but this ar 
 rangement increases the extent of the muffles. They 
 may be reduced to a volume much smaller and more 
 convenient, by mounting in each of them all the pulleys 
 upon the same axis, as in Fig. 68. By this means, the 
 cords which are on one side of the muffles are not 
 parallel to those on the other side ; but, when the 
 distance of the muffles is considerable, the departure 
 from parallelism is very small, and it may be regarded 
 as insensible. 
 
PULLEYS. 
 
 151 
 
 Fig. 
 
 Fig. 68. 
 
 160. By considering, then, the cords of the muffles 
 as parallel to each other, and abstracting all weight 
 from the whole machine, let Q be a power in equilibrium 
 with the resistance P, suspended from the block of the 
 moveable muffle. Equilibrium cannot exist throughout 
 the whole machine, unless it occur in each particular 
 pulley, and the two parts of the cord which embraces 
 this pulley are equally stretched (154). Thus, the ten 
 sions of the two cords QA, BC are equal to each other ; 
 so likewise are those of the two cords BC, DE, and those 
 
152 
 
 STATICS. 
 
 A 
 
 Fig. 6G. O f the cords DE, FG, and so on, what- 
 
 || ever may be the number of the cords ; 
 
 hence all the cords which go from one 
 5 muffle to the other are equally stretch 
 ed. Now the sum of these tensions 
 produces an equilibrium with the re 
 sistance P, and is equal to it ; or, what 
 is the same, the tension of one of the 
 cords, multiplied by their number, is 
 equal to the resistance ; hence the ten 
 sion of one of the cords, or the power 
 Q, is the quotient of the resistance p di 
 vided by the number of cords which go 
 from one muffle to the other. 
 
 Hence it is evident, that, in the case 
 of Fig. 66, where the extremity of the 
 cord is attached to the fixed muffle, the 
 power Q should be one-sixth of the re 
 sistance to be in equilibrium with it, and 
 in the case of Fig. 67, where the ex 
 tremity of the cord is attached to the 
 moveable muffle, it should be one-fifth, 
 
 because there is one pulley and one cord less. 
 
 If it be desired to introduce the consideration of the 
 
 weight of the moveable muffle, it may be regarded as 
 
 constituting part of the resistance. 
 
WHEEL AND AXLE. 
 
 153 
 
 Of the Wheel and Axle. 
 
 I. 
 
 161. The wheel and axle, windlass, or capstan, is a 
 machine consisting of a cylinder moveable upon an axis, 
 and of a cord, which, having one of its extremities 
 wound around the cylinder, while a power Q causes it 
 to turn, overcomes a resistance P attached to its other 
 extremity. The cylinder is furnished at its two ends 
 with trunnions A, B, bearing upon supports, which are 
 named boxes, and by means of which it turns freely on 
 its axis. 
 
 Fig. 69. 
 
 162. There are several modes of applying the power 
 to this machine, to communicate the movement of rota 
 tion to the cylinder. 
 
154 STATICS. 
 
 1st. We may unite solidly with the cylinder and upon 
 the same axis, a wheel, whose circumference, having a 
 groove in it like that of a pulley, is surrounded by a 
 second cord. This cord, being drawn by the power, 
 causes both the wheel and the cylinder to turn upon 
 their common axis. This first arrangement, to which we 
 will refer all the others, is rarely employed, because it 
 requires the cord of the wheel to be very long, while 
 the space to be traversed by the resistance is incon 
 siderable. 
 
 2d. The rim of a wheel, furnished with spokes placed 
 equally far apart, and to which men s hands are applied, 
 furnishes them with the means of making use of part 
 of their weight in turning the machine. 
 
 3d. In other cases, instead of the wheel, there is 
 mounted upon the cylinder a large hollow drum in which 
 men or animals can walk ; and then by their feet they 
 cause both the drum and the cylinder to turn. 
 
 4th. Sometimes, in place of making use of the wheel 
 and the drum, the cylinder is traversed by bars perpen 
 dicular to its axis, and at the extremities of which men 
 act by the force of their muscles and part of their 
 weight. 
 
 5th. Finally, most frequently, and when the resist 
 ance is not very great, there are adapted to the ex 
 tremities of the cylinder one or two cranks, which men 
 turn by employing the force of their arms. 
 
 163. The names of this machine vary with its object, 
 and even with its position. Ordinarily it is named wind 
 lass, and wheel and axle, when the cylinder is horizontal ; 
 and capstan, when the cylinder is vertical, and horizon 
 tal bars are used to put it in motion. 
 
WHEEL AND AXLE. 
 
 155 
 
 It is evident that the different modes in which the 
 power may be applied to the cylinder of the wheel and 
 axle, can all be referred in theory to the first we have 
 described; for, whatever may be the direction of the 
 power, when it is directed in a plane perpendicular to 
 the axis of the cylinder, we may always conceive it to 
 be applied to a wheel whose circumference would be 
 tangent to the direction of this power. Thus we will 
 suppose that VXYZ being the cylinder of the machine, 
 whose axis ACB is perpendicular to the plane of the 
 wheel JiDD Jc, 1st, the power Q is applied to the circum 
 ference of this wheel in any direction DQ, contained in 
 the plane of the wheel, and tangent to the circumference 
 of the radius CD, at a given point D ; 2d, that the di 
 rection KP of the resistance is tangent at K to the sur 
 face of the cylinder, and situated in a plane parallel to 
 that of the wheel. Lastly, to render the conception 
 clear, we will suppose that the axis AB of the cylinder 
 is horizontal, and consequently that the direction KP of 
 
156 STATICS. 
 
 the resistance is vertical. This being granted, two 
 questions present themselves to be solved : the first is 
 to find the ratio which the power Q and the resistance P 
 should have to produce an equilibrium ; the second is to 
 find the weight which the supports of the two pivots 
 A, B, sustain. 
 
 II. 
 
 164. To solve the first of these two questions, let us 
 conceive a horizontal plane KMEN to pass through the 
 axis AICB of the cylinder : this plane will pass through 
 the point K, where the direction of the resistance touches 
 the surface of the cylinder ; moreover, it will intersect 
 the plane of the wheel in a horizontal line ME, which 
 will pass through the centre c, and it will meet the di 
 rection of the power Q in some point E. Prolong the 
 line ME to ES, and through the point E draw the vertical 
 ER; the three lines EQ, ER, ES being included in the 
 same plane, which is that of the wheel, we can decom 
 pose the power Q into two forces R, S, directed along EG, 
 EH. For this purpose, we will represent this power by 
 the parts EF of its direction ; and completing the paral 
 lelogram EGFH, we will have 
 
 Q : S : : EF : EH, 
 
 Q : R : : EF : EG, or FH ; 
 
 or, since, by drawing the radius CD, the two rectangular 
 triangles CDE, EHF will be similar, and give 
 
 EF : FH : EH : : CE : CD : DE 
 
we shall have 
 
 WHEEL AND AXLE. 
 
 Q : S : : CE : DE, 
 Q : E : : CE : CD. 
 
 15T 
 
 Thus, in place of the power Q, we can take the two 
 forces E, S, whose values 
 
 QXCD 
 E=- , 
 
 CE 
 
 QXDE 
 
 are known, since the direction of the force Q being 
 given, all parts of the triangle CDE are known. 
 Fig. 69. 
 
 Now, of the two forces E, S, the latter being directed 
 towards the axis of the cylinder which is immoveable, 
 it may be regarded as immediately applied to the point 
 
158 STATICS. 
 
 c, and destroyed by the resistance of the axis ; this 
 force, then, cannot contribute in any manner to the 
 motion of rotation of the cylinder, and it has no other 
 effect than that of pressing the pivots against their 
 supports. Hence, there remains only the force R which 
 can be employed to produce an equilibrium with the re 
 sistance P. 
 
 Now, in the horizontal plane KMEN draw the radius 
 of the cylinder KI, which will be perpendicular to the 
 axis and parallel to ME ; also draw the line KE which 
 will intersect the axis in some point L. This being 
 done, the point L, being in the axis, may be regarded as 
 immoveable, and the line KE may be taken for an inflexi 
 ble rod, retained by a fulcrum L, at the extremities of 
 which are applied the two forces R, P. Now the direc 
 tions of these two forces are both vertical, and conse 
 quently parallel ; hence in order that equilibrium may 
 subsist between them, it is necessary that they should 
 be reciprocally proportional to their arms LE, LK, or 
 that we should have 
 
 R : p : : KL : LE. 
 But the similar rectangular triangles KIL, LCE, give 
 
 KL : LE : : KI : CE ; 
 
 hence we shall have 
 
 R : p : : KI : CE. 
 
WHEEL AND AXLE. 159 
 
 Hence by multiplying in order, this proportion and 
 the following 
 
 Q : R : : CE : CD, 
 
 which we found above, we shall have, in case of equi 
 librium, 
 
 * 
 
 Q : p : : KI : CD ; 
 
 that is to say, the power will be to the resistance as the 
 radius of the cylinder is to the radius of the wheel; 
 which is the answer to the first question. 
 
 By making the product of the extremes equal to that 
 of the means in the above proportion, we have 
 
 QXCD = PXKI. 
 
 The line CE meets the circumference of the wheel at 
 the point D ; imagining that there is through this point 
 a force Q equal to Q directed along the tangent D Q , we 
 shall have (31) 
 
 QXCD=Q XCD / =PXKI ; 
 
 from which it appears, that in the case of equilibrium, 
 the moments of the power Q , equal to Q, and of the re 
 sistance P, both taken with reference to the vertical 
 plane passing through the axis of the cylinder, are equal 
 to each other. 
 
160 
 
 STATICS. 
 
 III. 
 
 165. The pressure which the trunnions exert against 
 their points of support is evidently the effect only of 
 the forces P, Q, and of the weight T of the machine, 
 which may be considered as united at its centre of 
 gravity g, where, by taking the two- forces R, S, instead 
 of the power Q, these pressures are produced by the four 
 forces P, E, s, T. 
 
 Fig. 69. 
 
 These forces are all known ; for, 1st, the resistance 
 p and the weight T of the machine are given imme 
 diately ; 2d, the two other forces R and s, whose values 
 we have found to be, in general, 
 
 QXCD 
 
 B=*5?, 
 
 CD 
 
WHEEL AND AXLE. 161 
 
 become, in the case of equilibrium where we have 
 QXCD=PXKI, 
 
 s= 
 
 PXKI 
 R= , 
 
 CE 
 
 PXKIXDE 
 
 CDXCE : 
 
 and containing only known quantities, since the direc 
 tion of the power Q being given, we can find all the 
 parts of the right-angled triangle CDE. 
 
 Now the two forces P, R, whose directions are vertical, 
 and which are in equilibrium about the point L, exert 
 upon this point of the axis a weight whose direction is 
 vertical, and which is equal to their sum P+R. More 
 over, this weight P+R, being sustained by the two points 
 of support at A and B, must be regarded as the resultant 
 of the two vertical pressures which it exerts at these 
 points ; and we will find each of these pressures, by 
 dividing the resultant P+R into two parts reciprocally 
 proportional to the distances of the point L from the 
 two supports. Let x be the pressure exerted at the 
 point A, and x that exerted at the point B ; we will find 
 these two pressures by the following proportions : 
 
 AB : BL : : P+R : x, 
 AB : AL : : P+R : x . 
 
 In like manner the weight T of the whole machine, 
 supposed to be united at the centre of gravity g, may 
 be regarded as the resultant of the vertical pressures 
 
102 
 
 STATICS. 
 
 which it produces upon the two points of support : and 
 we will find these pressures by dividing the weight T 
 into two parts reciprocally proportional to the distances 
 
 A #> # B - 
 
 Then let Y and Y be the pressures which result re 
 spectively at the points A and B; we will find these 
 pressures by the two proportions which follow : 
 
 AB : B# : : T : Y, 
 AB : Ag : : T : Y . 
 
 Lastly, the horizontal force s, applied to the point c 
 of the axis, produces horizontal pressures upon the two 
 supports, directed perpendicularly to the axis AB, and 
 of which it is the resultant : we will find likewise these 
 pressures by dividing the force s into parts reciprocally 
 proportional to the lines AC, CB. Then let z, z be the 
 horizontal pressures, produced respectively upon the 
 
WHEEL AND AXLE. 163 
 
 points A, B : we will find these pressures by the pro 
 portions : 
 
 AB : BC : : S : z, 
 AB : AC : : S : z . 
 
 Thus the point of support A sustains the two vertical 
 pressures x, Y, and the horizontal pressure which acts 
 in the direction of the force S. In like manner the point 
 B sustains the vertical pressures x , Y , and the horizon 
 tal pressure z . Hence, by compounding, for each of 
 these points, the forces which act upon it, we will find 
 the intensity and direction of their resultant ; and we 
 shall have the intensity of the resistance of which it 
 must be capable, as well as the direction in which it 
 must resist, so as not to yield to the united efforts of the 
 resistance P, the power Q, and the weight T of the ma 
 chine ; which is the object of the second question. 
 
 IV. 
 
 166. Hitherto we have regarded the cords as infinitely 
 fine threads ; but, when the weight P is suspended by 
 the cord KP, the line of direction of this weight does 
 not coincide with the axis of the cord ; and in the case 
 where the cord by wrapping around the cylinder does 
 not change in figure, its axis is always at a distance 
 from the surface of the cylinder equal to the semi-di 
 ameter of the cord. Thus, by reason of its thickness, 
 the cord is in the same condition as though, being in 
 finitely fine and reduced to its axis, it were wrapped 
 
164 
 
 STATICS. 
 
 upon a cylinder whose radius was greater than the radius 
 of the cylinder of the machine, by a quantity equal to 
 the semi-diameter of the cord. It is the same with the 
 cord of the wheel, -which, by reason of its thickness, 
 may be regarded as a mathematical line wrapped upon 
 a wheel whose radius is greater than that of the wheel 
 of the machine, by a quantity equal to the semi-diameter 
 of this cord. Hence, in all the relations we have just 
 found, it is necessary to augment the radius of the 
 cylinder and that of the wheel by quantities respectively 
 equal to the semi-diameters of the cords which envelope 
 them. Thus, for example, in the case of equilibrium of 
 the wheel and axle, the power Q is to the resistance P as 
 the radius of the cylinder, augmented by the radius of 
 the cord KP, is to the radius of the wheel, augmented by 
 the radius of the cord DQ. 
 
 Fig. 70. 
 
 V. 
 
 167. If several wheels and axles 
 are so arranged that the cord BQ of 
 the first wheel, being drawn by a 
 power Q, the cord CB of the cylin 
 der of this wheel, instead of being at 
 tached immediately to the resistance, 
 is wound around the second wheel ; 
 the cord FH of the cylinder of the 
 second is likewise wound around the 
 wheel of the third ; and lastly, the cord 
 IP of the cylinder of the last is applied 
 to the resistance P ; the power and the 
 
WHEEL AND AXLE. 165 
 
 resistance will not be in equilibrium, unless there is also 
 equilibrium between the two forces which act upon each 
 separate wheel. Thus, by naming K the tension of the 
 cord CE, and L that of the cord FH, in the case of 
 general equilibrium we have : 1st, by reason of the 
 equilibrium of the first wheel and axle (164), 
 
 Q : K : : CA : AB ; 
 2d, by reason of the equilibrium of the second, 
 
 K : L : : DF : DE ; 
 3d, by reason of the equilibrium of the third, 
 
 L : P : : GI : GH ; 
 
 and so on, whatever may be the number of machines. 
 Hence, by multiplying all these proportions in order, we 
 have 
 
 Q : p : : CAXDFXGI : ABXDEXGH ; 
 
 that is to say, the power is to the resistance as the pro 
 duct of the radii of the cylinders is to the product of 
 the radii of the wheels. 
 
 For example, if the radius of the wheel of each is 
 four times the radius of its cylinder, in the case of 
 equilibrium of all three wheels and axles, we have 
 
 Q : p : : Ixlxl : 4x4x4, or : : 1 : 64. 
 
166 
 
 STATICS. 
 
 Hence we see, that by multiplying in this manner the 
 number of machines we can put moderate powers in a 
 state of equilibrium with very great resistances; but 
 this arrangement is almost never employed, because it 
 requires too great a length of cord in the first wheels, 
 while the space which the resistance has to traverse is 
 inconsiderable. 
 
 VI. 
 
 168. When we wish to profit by 
 the advantages of this arrange 
 ment, 1st, the cords are suppressed 
 which transmit the motion from one 
 wheel and axle to another ; 2d, on 
 the circumference of each wheel, 
 teeth are placed equally distant 
 from each other; 3d, on the ar 
 bor of each of the toothed-wheels, 
 another wheel similarly toothed, 
 of a smaller diameter, and which is 
 called the pinion, is fixed solidly; 4th, lastly, the 
 whole system is so arranged that the teeth of each 
 pinion interlock with the teeth of the following wheel. 
 By this means one wheel cannot turn upon its axis, with 
 out its pinion communicating motion to the wheel with 
 which its teeth are engaged, and causing it to turn upon 
 its axis; and the number of revolutions which each 
 pinion can cause the following wheel to make, is un 
 limited. The system is then in the same condition as 
 though the pinion, considered as the cylinder of a wheel 
 
COG WHEELS. 167 
 
 and axle, and the wheel with which it interlocks, were 
 embraced by the same cord as in the preceding case. 
 
 Hence, when a power Q, applied to the circumference 
 of the first wheel, is in equilibrium with a resistance P, 
 applied to the circumference of the last pinion, the 
 power is to the resistance as the product of the radii 
 of the pinions is to the product of the radii of the 
 wheels. 
 
 VII. 
 
 169. Toothed wheels are employed in a very great 
 number of machines, principally in mills and in the 
 works of time pieces. Their immediate object is to 
 communicate to a cylinder or arbor a movement of ro 
 tation about its axis, by the aid of the rotary motion of 
 another arbor. For this purpose it is not necessary that 
 the axes of the two arbors should be parallel, as we 
 have heretofore supposed ; it is sufficient that they are 
 in the same plane. 
 
 Fig. 72. 170. When the axes of the two arbors 
 
 are at right angles, the teeth ordinarily 
 are placed perpendicularly to the plane 
 of the wheel, as may be seen in Fig. 72 ; 
 then they can interlock with those of the 
 pinion, or with the staves of the trundle 
 AB, which produces the effect of a pinion : 
 by this interlocking, the teeth of the 
 wheel are forced to slide upon the staves 
 in the direction of the axis of the trundle, which also 
 causes friction. 
 
168 
 
 STATICS. 
 
 171. In general, whatever 
 may be the angle BAG, which the 
 axes of the two arbors make, 
 % so that the movement of rota 
 tion of the one is communicated 
 to the other by means of two 
 toothed wheels DEFG, EHIF, and 
 the teeth do not slide upon each 
 other in the direction of the axes, it is necessary that 
 these two wheels, called leveled wheels, should be trunca 
 ted sections of two cones DAE, EAH, which have the 
 same summit A, and whose axes coincide with those of 
 the arbors ; moreover, the teeth of the two wheels should 
 be terminated by conic surfaces which have for their 
 summit the common point A. 
 
 VIII. 
 
 172. The jack-screw is also a machine which may be 
 referred to the wheel and axle. 
 
 The simple jack-screw is composed of 
 a bar of iron AB, furnished with teeth 
 upon one of its sides, and moveable in 
 the direction of its length within a box 
 DE. The teeth of the bar interlock with 
 those of a pinion c, which is turned 
 upon an axis by means of a crank F. 
 The teeth of the pinion carry along those 
 of the bar, and cause the weight to rise, which rests 
 upon the head A of the bar, or which is raised by the 
 hook B. This machine, evidently, is nothing else than 
 a wheel and axle, and, in the case of equilibrium, by 
 
THE INCLINED PLANE. 169 
 
 supposing the direction of the power to be perpendicu 
 lar to the arm of the crank, the power is to the resist 
 ance as the radius of the pinion is to the arm of the 
 crank. 
 
 In the compound jack-screw, the teeth of the first 
 pinion interlock with those of a second toothed wheel, 
 and the teeth of the pinion of this wheel interlock with 
 those of the bar. By this means the power is placed in 
 a state of equilibrium with a greater resistance. This 
 case is related to that of toothed wheels, and we will 
 not dilate any more upon that subject. 
 
 173. When two powers are in equilibrium by means 
 of a pulley or wheel and axle, it is very evident that 
 we may consider them as though they were in equili 
 brium at the extremities of a lever whose fulcrum is in 
 the axis of the cylinder or the pulley ; hence (151), 
 these powers are to each other reciprocally, as the 
 spaces which they would traverse along their directions, 
 if the equilibrium were disturbed infinitesimally. 
 
 ARTICLE III. 
 On the Equilibrium of the Inclined Plane. 
 
 174. A plane is said to be inclined, 
 when it makes an angle with a horizon 
 tal plane ABEF, and this angle is not a 
 right angle. 
 
 175. If through any point H, taken 
 upon the line of intersection AB of 
 the inclined plane and the horizontal 
 
170 STATICS. 
 
 plane, two perpendiculars be drawn to this line, the one 
 HK in the horizontal plane, and the other HI in the in 
 clined plane, the angle IHK, formed by these two per 
 pendiculars, is the measure of the inclination of the 
 plane. The plane of the angle IHK, which passes 
 through two lines perpendicular to AB, is perpendicular 
 to the line AB ; hence it is perpendicular to the two 
 planes ABCD and ABEF, which intersect in this line; 
 hence this plane is at the same time vertical and per 
 pendicular to the inclined plane. 
 
 176. Reciprocally, every plane which is at the same 
 time vertical and perpendicular to the inclined plane, is 
 perpendicular to the intersection AB of the inclined 
 plane with the horizontal plane ; hence, the lines HI and 
 HK along which it intersects these two last planes, are 
 also perpendicular to AB, and form between them an 
 angle IHK, which is the measure of the inclination of the 
 plane ABCD with reference to the horizontal plane. 
 
 177. If through the point I there be drawn a vertical 
 line IL, this line will not leave the plane of the angle 
 IHK ; it will meet the horizontal line HK to which it will 
 be perpendicular, and it will form a rectangular triangle 
 IHL. The hypotenuse of this triangle is named the 
 length of the inclined plane ; the side IL is its height, 
 and the other side HL is its base. 
 
 178. When a body, which, in a single point Q, touches 
 an immoveable plane ABCD, is pushed by a single force 
 p, whose direction PQ, 1st, passes through the point of 
 
THE INCLINED PLANE. 
 
 171 
 
 Fig. 76. contact Q, 2d, is perpendicular 
 
 to the plane, this body remains 
 at rest. 
 
 Thus, we can conceive the 
 force P to be applied to the point 
 Q of its direction. Now this di 
 rection being perpendicular to 
 the plane, and consequently to 
 all the lines QR, QS, QT, which we can draw in the plane 
 through the point Q, it is similarly disposed with 
 reference to all these lines ; there is then no reason why 
 the point Q should move along one of the lines rather 
 than along any other; hence this point, and conse 
 quently the whole body, will remain at rest. 
 
 Fig. 77. 179. Both the conditions just men 
 
 tioned are necessary to the rest of the 
 body : for, 1st, if the direction n 
 of the force does not pass through 
 the point of contact, nothing prevents 
 the point a of the body which is upon 
 this direction from approaching the 
 plane, and the body would be put in motion. 2d. If 
 the direction PQ of the force passes through the point 
 Fig. 78. O f contact, and is not perpendicular 
 
 to the plane, by conceiving this force 
 to be still applied to the point Q, let 
 its direction be prolonged to QR, and 
 through the point Q draw the line QS 
 perpendicular to the plane ; through 
 the two lines QR, QS draw a plane, 
 which will intersect the first plane 
 ABCD somewhere in a line Hi, which will pass through 
 
p 
 
 a 
 
 172 STATICS. 
 
 the point Q. This being done, if we represent the force 
 p by the part QR of its direction, and complete the pa 
 rallelogram QTRS, instead of the force P we may take 
 the two forces represented by QS and QT. Now the 
 force QS, which passes through the point of contact, and 
 whose direction is perpendicular to the plane ABCD, will 
 be destroyed by the resistance of this plane (178) ; but 
 nothing will oppose the action of the force QT, whose 
 direction is parallel to the plane ; the point Q will then 
 move in the direction QII, and the body will not be at 
 rest. 
 
 Fig. 78. 180. Hence it follows, that, when a 
 
 body, affected by the simple action 
 x" \ of its gravity, remains in equilibrium 
 
 upon a plane ABCD, which it touches 
 in a single point Q, 1st, the centre 
 of gravity P of this body, and the 
 point f contact Q, are in the same 
 A~~ vertical line; 2d, the plane ABCD is 
 
 horizontal; for, since the weight of the body can be 
 regarded as a single force applied to its centre of gravity, 
 the body cannot be at rest, unless the direction of this 
 force passes through the point of contact and is perpen 
 dicular to the plane upon which the body is sustained. 
 Fig. 79. 181. It follows again, that 
 
 when a body, affected by the 
 single action of gravitation, is 
 sustained upon an inclined plane 
 ABCD by a single point Q, and 
 this point is found in the verti 
 cal drawn through the centre of gravity, this body must 
 tend to slip upon the plane ; and the direction QH along 
 
THE INCLINED PLANE. 
 
 173 
 
 Fig. 80. 
 
 which the point Q tends to move, is the intersection of 
 the plane ABCD with the plane IHK, which is at the 
 same time vertical and perpendicular to the inclined 
 plane. 
 
 182. What has just 
 been said of a body 
 pushed by a single force 
 against a plane, must ap 
 ply to a body pushed 
 against a curved surface 
 AQB, which it touches only 
 in the single point Q : that 
 is to say, this body cannot 
 be at rest, unless the direction of the force which pushes 
 it passes through the point Q, and is perpendicular to 
 the curved surface at this point ; for this body may be 
 considered as sustained upon the plane DE tangent to 
 the curved surface at the point Q. 
 
 183. We see, then, that when a lever may slide upon 
 its fulcrum, it is not sufficient for its remaining at rest, 
 that the resultant of the two powers which are applied 
 to this lever are directed towards the fulcrum ; it is be 
 sides necessary that the direction of this resultant 
 should be perpendicular to the surface of the lever at 
 the point where it touches the fulcrum. 
 
 II. 
 
 184. When a body, pushed by a single force P 
 against an immoveable plane ABCD, is sustained upon 
 this plane by a definite base VXYZ, if the direction PQ 
 
 15* 
 
174 STATICS. 
 
 of the force meets the base somewhere in a point Q, and 
 if it is at the same time perpendicular to the plane, the 
 body is at rest ; for we have seen (178) that if the base 
 Fi 9- 81. were reduced to the single 
 
 point Q, rest would take place ; 
 it is evident that the other 
 points of support, which the 
 base presents, cannot disturb 
 it. 
 
 We will demonstrate, as in 
 No. 179, that both these con 
 ditions are necessary to the 
 body s remaining at rest : the effect of the first is to 
 prevent the body from turning upon one of the sides of 
 its base ; the effect of the second is to prevent it from 
 slipping upon the plane ABCD. 
 
 185. If the body, instead of being supported upon 
 the plane by a continuous base, simply touches it by 
 several points separated from each other, we may regard 
 these points as the summits of the angles of a polygonal 
 base ; and the body is at rest when the direction of the 
 force, which pushes it against the plane, is perpendicu 
 lar to this plane, and at the same time passes through 
 the interior of the polygon. 
 
 Thus, since the weight of a body may be regarded as 
 a vertical force applied to its centre of gravity P, we 
 see, 1st, that a body, which rests with its base upon a 
 horizontal plane ABCD, cannot be stable, unless the ver 
 tical PQ, drawn through the centre of gravity, meets 
 some point Q of the base ; 2d, if the body rests upon 
 the plane by a certain number of points of support, 
 
c 
 
 - U| 
 
 THE INCLINED PLANE. 175 
 
 u, x, Y, . . . it cannot be stable, 
 unless the vertical PQ, drawn 
 through its centre of gravity 
 F, passes through a point 
 Q taken in the interior of the 
 polygon UXY, which may be 
 formed by joining the exterior 
 points of support by the lines 
 ux, XY, YU. 
 
 III. 
 
 186. Hitherto we have supposed that the body, sup 
 ported upon a plane, was pushed by a single force ; but it 
 is evident that, if the body is pushed by several forces 
 at the same time, it cannot be at rest, unless the result 
 ant of all these forces satisfies the preceding conditions ; 
 that is to say, unless the direction of this resultant is 
 perpendicular to the plane, and passes through the base 
 of the body. There is then in this case a third con 
 dition necessary to the rest of the body, and this condi 
 tion is, that all the forces which act upon it must have 
 one resultant. 
 
 The whole theory of the equilibrium of a body, pushed 
 by as many forces as we please to assume, and supported 
 upon a single resisting plane, consists in the search for 
 the directions and intensities which the forces must have, 
 so that the three conditions just mentioned may be ful 
 filled. We will content ourselves with developing it for 
 a few simple cases, and principally for that in which the 
 body is pushed by two forces. 
 
176 
 
 STATICS. 
 
 IV. 
 
 Fi 9- 81. 187. Let LUZ be a body sup 
 
 ported by a base UXYZ upon a 
 resisting plane ABCD, and so 
 licited at the same time by two 
 forces R, s. According to the 
 preceding remarks, in order 
 that the body may be at rest, 
 it is necessary, 1st, that the 
 two forces R, s should have 
 one resultant : now, two forces cannot have one result 
 ant, unless their directions are in the same plane, (10) ; 
 hence, 1st, the directions of the two forces R, S must be 
 included in the same plane, and intersect in a certain 
 point N. 
 
 2d. It is necessary that the direction PNQ of the re 
 sultant of the two forces R, s, should be perpendicular 
 to the plane ABCD : now the resultant of the two forces 
 is always contained in the plane drawn through their 
 directions ; hence, 2d, the plane in which the two forces 
 R, s are directed, should be perpendicular to the plane 
 ABCD. 
 
 3d. It is necessary that the direction of the resultant 
 should pass through a point Q of the base. 
 
 188. From this it follows, that if one of the two 
 forces, for example the force R, is the weight of the 
 body which may be considered as applied to the centre 
 of gravity M, and whose direction MR is vertical, the 
 body cannot be at rest upon an inclined plane ABCD, 
 unless the direction NS of the other force is contained 
 
THE INCLINED PLANE. 
 
 177 
 
 in a vertical plane, drawn through the centre of gravity 
 M, perpendicularly to the inclined plane, and, moreover, 
 that the direction PQ of the resultant of the two forces 
 is perpendicular to the inclined plane, and passes through 
 a point Q of the base of the body. 
 
 Now all these conditions, relative to the directions of 
 the forces, being supposed to be filled, we will seek the 
 relations which the two forces E, s, and the weight P 
 of the plane, have for each other in the case of equi 
 librium. 
 
 Fig. 83. 
 
 V. 
 
 189. Let LXU be a body sup 
 ported by its base ux upon a re 
 sisting plane HI, and kept at rest 
 upon this plane by the two forces 
 E, S. Having prolonged the di 
 rections of the two forces until 
 they meet in a point N, draw 
 B. through this point the line NP per 
 
 pendicular to the plane Hi. We have seen that this 
 line will be the direction of the resultant of the two 
 forces E, s. Hence, if we represent this resultant by 
 the part NE of its direction, and if, by drawing through 
 the point E the lines EG, EF, parallel to the directions 
 of the forces E, s, we complete the parallelogram NFEG, 
 the sides NF, NG will represent the intensities of the 
 forces E, s. Hence, by naming P the weight on the 
 plane which is equal to the resultant, we shall have 
 
 E : s : P : : NF : NG or FE : NE. 
 
178 STATICS. 
 
 In order to have the ratios of these three forces ex 
 pressed in quantities independent of the construction 
 of the parallelogram NFEG-, we will remark, that in the 
 triangle NEF the sides are to each other in the ratio of 
 the sines of the opposite angles, or we shall have 
 
 NF : FE : EN : : sin NEF : sin FNE : sin NFE. 
 Hence we shall have 
 
 E : s : P : : sin NEF : sin FNE : sin NFE. 
 
 Now these three angles are those which form with each 
 other the directions of the three forces E, s, P ; hence 
 these forces are to each other as the sine of the angle 
 which the directions of the other two forms. 
 
 We see then that, of the six things 
 which may be considered in this 
 equilibrium, and which are, the di 
 rections of the three forces and their 
 intensities, any three being given, we 
 can determine the other three, in all 
 cases where, of the six things which 
 we consider in the triangle NEF, namely, the angles and 
 the sides, the three analogous ones being given, we can 
 determine the three others. 
 
 190. If one of the forces, for example the force E, 
 is the weight of the body whose direction is vertical, 
 and passes through the centre of gravity M, and the di 
 rection of the force S, which retains the body in equi 
 librium upon the inclined plane, is parallel to this plane, 
 draw the base KH and the height KI of the inclined 
 
THE INCLINED PLANE. 
 
 179 
 
 plane ; the right-angled triangles NFE, IHK will be simi 
 lar, because the angles NFE, HIK, whose sides are paral 
 lel each to each, will be equal, and we shall have 
 
 NF : FE : EN : : HI : IK : KH ; 
 hence we shall also have 
 
 R : s : P : : HI : IK : KH. 
 
 Thus, in this case, the weight of the body is to the 
 force which holds it in equilibrium, as the length of the 
 inclined plane is to its height. 
 
 Fi 9- 85 - 191. By supposing still that 
 
 s the force R is the weight of the 
 body, if the direction of the 
 force S is horizontal, and con 
 sequently parallel to the base 
 HK of the inclined plane, the 
 right-angled triangles NFE, HKI are again similar ; be 
 cause the sides of the one will be perpendicular to the 
 sides of the other, and we shall have 
 
 NF : FE : EN : : HK : KI : IH, 
 
 we shall also have 
 
 R : s : P 
 
 HK : KI : m. 
 
 Hence, the weight of the body is to the force which 
 holds it in equilibrium, as the base of the inclined plane 
 is to its height. 
 
180 
 
 STATICS. 
 
 VI. 
 
 192. Let us now consider 
 the equilibrium of a body sup 
 ported by two inclined planes. 
 Let M be a body subjected to the 
 single action of gravity, and re 
 tained by the two inclined planes 
 ABCD, ABEF, which intersect some 
 where in the line AB. Let H, I be the points at which 
 the body touches the two planes, and GR the vertical 
 line drawn through its centre of gravity, and which will 
 consequently be the direction of its weight. It is evi 
 dent that this body cannot be at rest, unless its weight 
 R can be decomposed into two other forces P, Q, applied 
 to the same body, and which are destroyed by the re 
 sistance of the two planes : or, what is the same thing, 
 unless the directions of the two forces P, Q pass through 
 the points of support I, H, and are perpendicular each 
 to the corresponding inclined plane. Now the direction 
 of a force and those of its two components are contained 
 in the same plane, and necessarily meet in the same 
 point ; hence, in order that the body M may be at rest 
 between the two inclined planes, it is necessary that the 
 perpendiculars IG, HG, drawn through the points of sup 
 port I, H to the two inclined planes, should be in the 
 same plane with the vertical drawn through the centre 
 of gravity of the body, and intersect this vertical in the 
 same point G. 
 
THE INCLINED PLANE. 181 
 
 193. Hence it follows, that, in order that a body M 
 may be at rest between two inclined planes, inde 
 pendently of the position of the body, the planes should 
 satisfy the condition that the line AB of their intersec 
 tion should be horizontal. Thus, the plane IGH, which 
 must contain the vertical GR, and the perpendiculars 
 IG, HG to the two inclined planes, is at the same time 
 vertical and perpendicular to these two planes ; hence, 
 reciprocally, the two inclined planes must be perpen 
 dicular to the vertical plane IGH ; hence the line AB 
 of their intersection should be perpendicular to this 
 same plane, and consequently horizontal. 
 
 87. 194. These conditions, 
 
 which have for object the 
 respective positions of the 
 body and of the two in 
 clined planes, being sup 
 posed to be fulfilled, in 
 order to find the ratio of 
 the weight R of the body 
 to the weights P, Q, which the two planes support, we 
 will remark that the plane IGH, vertical and perpendicu 
 lar to the two inclined planes containing the angles 
 which these two planes form with the horizon, includes 
 all that relates to the question, and we may be content 
 to consider it alone, as in Fig. 87. Hence, let AD, AF 
 be the intersections of the vertical plane IGH with the 
 two inclined planes ; these lines form with the horizon 
 tal line vz, or with any other horizontal line XY, angles 
 which will measure the inclinations of the two planes. 
 This being granted, if we represent the weight of the 
 
 16 
 
182 STATICS. 
 
 body by the part GR of its direction, and complete the 
 parallelogram GPQR, we shall have 
 
 R : p : Q : : GR : RQ : QG. 
 
 Now the triangles GQR, XAY, whose sides are perpendicu 
 lar each to each, give 
 
 GR : RQ : QG : : XY : YA : AX ; 
 hence we shall also have 
 
 R : p : Q : : XY : YA : AX ; 
 
 or, lastly, because the sides of the triangle XAY are pro 
 portional to the sines of the opposite angles, we shall 
 have 
 
 R : p : Q : : sin YAX : sin AXY : sin XYA, 
 
 that is to say, by representing the weight of the body 
 by the sine of the angle which the two inclined planes 
 form with each other, the weights which these two 
 planes support are to each other reciprocally as the 
 sines of the angles which these planes form with the 
 horizon. 
 
THE INCLINED PLANE. 183 
 
 VII. 
 
 195. Lastly, if a body is 
 sustained by three points 
 A, B, c, upon three inclined 
 planes, it is evident that 
 this body cannot be at rest 
 unless its weight p, whose 
 direction DP is vertical, and 
 passes through the centre of gravity of the body, can 
 be decomposed into three other forces Q, E, s, which are 
 destroyed by the resistances of the inclined planesj 
 that is to say, unless the directions of the three forces 
 Q, n, s pass through the three points of support, and 
 are each perpendicular to the corresponding inclined 
 plane. 
 
 196. In order that the body may be at rest, it is 
 necessary that the weight P should be susceptible of 
 being decomposed into two forces Q, x; the first of 
 which, Q, being directed towards one of the points of 
 support A, perpendicularly to the inclined plane which 
 passes through this point, the other force x may itself 
 be decomposed into two others B, s, in the directions 
 of the two other points of support B, c, perpendicular 
 to the other inclined planes. 
 
 Hence we see, in this case, it is not necessary that 
 the perpendiculars to the inclined planes, drawn through 
 the points of contact A, B, c, should all three meet in 
 the same point, nor even that they should all three meet 
 the vertical drawn through the centre of gravity of the 
 body. 
 
184 STATICS. 
 
 THEOREM. 
 
 197. When a body without gravity, supported upon 
 an inclined plane EC by a single point C, is in equili 
 brium between two poivers P, Q, these powers are to each 
 other reciprocally as the spaces which they ivould tra 
 verse in the line of their directions, if the equilibrium 
 were disturbed infinitesimally. 
 
 Fig- 89. DEMONSTRATION. The two 
 
 -g powers P, Q being in equili 
 brium, their resultant should 
 B be perpendicular to the inclined 
 plane, and pass through the 
 point of support c (186) ; it 
 follows from this that the di- 
 rections of these powers should 
 coincide in a certain point G, and that the line GO 
 should be perpendicular to the inclined plane. This 
 being established, from the point c draw upon the di 
 rections of the powers P, Q, the perpendiculars CE, OF : 
 it is evident that the angle EOF, supposed to be invaria 
 ble, may be considered as a bent lever, at the extremi 
 ties of which are applied the two powers in equilibrium 
 about the point of support c : hence we can demonstrate, 
 as in No. 151, that the powers are to each other recipro 
 cally as the spaces which they would traverse in the line 
 of their directions, if the equilibrium were disturbed 
 infinitesimally. 
 
THE SCREW. 
 
 185 
 
 On the Screw. 
 
 198. If we conceive a cylin 
 der ABCD to be enveloped by 
 a thread AGHIK . . . ., and so 
 disposed that the angles FLO, 
 FMP, FNQ . . . ., formed by 
 the direction of the thread 
 with the lines drawn upon the 
 surface of the cylinder paral 
 lel to the axis, are equal to each other, the curve which 
 the thread traces upon the surface of the cylinder is 
 named a helix. 
 
 199. Hence it follows, that if we develope the surface 
 of the cylinder, and extend it upon a plane, as we see 
 in abed, 1st, the developement ah or lik of one revolu 
 tion of the helix will be a straight line ; because the 
 angles which this line will form with all such lines as ef, 
 drawn parallel to the side ad, will be equal to eact 
 other. 2d. This developement ah of a revolution of 
 the helix will be the hypotenuse of a right-angled tri 
 angle abh , whose base ab will be equal to the circum 
 ference of the base of the cylinder, and whose height 
 bh will be equal to the distance of the revolution which 
 we consider, from the revolution which follows it. 3d. 
 All the hypotenuses ah , hJc being parallel to each 
 other, the right-angled triangles abh , hh k . . ., will be 
 equal and similar, and will have equal heights. Hence 
 the intervals LM, MN, . . . between two consecutive revo 
 lutions of the helix, considered upon the surface of the 
 cylinder, are everywhere equal to each other. 
 
 16* 
 
186 
 
 STATICS. 
 
 Fig. 92. 
 
 200. This being established, the screw may be con 
 sidered as a right cylinder, enveloped by a projecting 
 fillet, adhering and wound as a helix upon the surface 
 of the cylinder. In the wooden screw, the form of the 
 fillet is such, that if it be cut by a plane drawn through 
 the axis of the cylinder, its section is most frequently 
 an isosceles triangle, as may be seen in Fig. 92 ; but 
 in large iron screws which are made with care, the sec 
 tion of the fillet is rectangular, as in Fig. 91. The 
 constant interval AB, which is found between two con 
 secutive revolutions of the fillet, is named the pitch of 
 the screw, or the helical interval. 
 
 201. The piece MN, into which the screw enters, is 
 named the nut. Its cavity is invested with another pro 
 jecting fillet, adhering, and wound likewise as a helix ; 
 and whose figure is such that it exactly fills the inter 
 vals left by the fillets of the screw. Thus, the screw 
 can turn in its nut, but it cannot do so without moving 
 in the direction of its axis ; and for one entire revolu 
 tion, it moves in the direction of the axis by a quantity 
 equal to the pitch of the screw. 
 
THE SCREW. 187 
 
 202. Sometimes the screw is fixed, and the nut 
 moves around it ; then, for each revolution, the nut 
 is carried upon the screw by a quantity equal to the 
 interval. 
 
 203. The screw may serve to elevate weights or over 
 come resistances ; but it is employed most generally 
 when great pressures are proposed to be exerted. For 
 this purpose we apply a power Q to the extremity of a 
 bar which traverses the head of the screw, Fig. 92, or the 
 nut, Fig. 91, according as it is the one or the other of 
 these two pieces which is moveable ; and this power, by 
 causing the piece to turn to which it is applied, makes 
 the head of the screw advance towards the nut, or re 
 ciprocally. The objects to be compressed are ranged 
 between two plates ; one of these plates is fixed, the 
 other is pressed by the movable piece, which can ad 
 vance only by reducing the volume of the objects. 
 
 We now propose, disregarding the friction, to find the 
 ratio of the power Q to the resistance P, which is in 
 equilibrium with it by being opposed to the motion of 
 the moveable piece, along a direction parallel to the 
 axis of the screw ; and, because the effect is absolutely 
 the same, whether the screw turns in its nut, or the nut 
 turns upon the screw, it will be sufficient to examine the 
 latter case. 
 
 204. The screw being fixed and in a vertical position, 
 we will conceive the nut to be left to the action of gravi 
 ty, and even, if we please, that it is charged with an 
 
 ,- additional weight ; it is evident that it will descend by 
 turning, and that it will traverse all the interior fillets 
 of the screw, by sliding over them as over inclined 
 surfaces. It is also evident that we oppose this effect 
 
188 
 
 STATICS. 
 
 by preventing the nut from 
 turning around the screw, and 
 consequently by applying to 
 the extremity of the bar FV a 
 power Q which is directed per 
 pendicularly to this bar, and 
 in a plane perpendicular to 
 the axis of the screw. 
 
 205. Suppose, for an in 
 stant, that the nut rests upon 
 the surface of the fillet of the screw only in a single 
 point ; this point, during the motion of the nut, will de 
 scribe a helix, whose pitch will be the same as that of the 
 screw, and which we may conceive to be traced upon 
 the surface of a cylinder, whose radius would be equal 
 to the distance of the describing point from the axis of 
 the screw. Let ABCD be the cylinder, EFGHI the helix 
 
 in question, and M the 
 point of the nut which 
 describes it. Let XY be 
 the tangent of the helix 
 at the point M ; through 
 a point Y of this tangent 
 draw the vertical YZ, equal 
 to the pitch of the helix, and in the vertical plane XYZ of 
 the horizontal line xz : it is clear, that this last line will 
 be equal to the circumference of the base of the cylin 
 der ABCD (199), or to the circumference of the circle 
 whose radius is CK ; and we will represent it by circ. CK. 
 This being established, the point M, which we may 
 .regard as charged with the whole weight P of the screw, 
 will be supported upon the helix as though it were upon 
 
TIIE SCREW. 189 
 
 the inclined plane XY ; thus, in order to hold it in equi 
 librium by means of a force R, which should be imme 
 diately applied to it, and which should be directed pa 
 rallel to xz, it is necessary (191) that this force R should 
 be to the weight P, as the height of the inclined plane 
 is to its base, or that we should have, 
 
 R : p : : YZ : circ. CK. 
 
 But if, instead of a force R immediately applied to the 
 point M, we employ a force Q, whose direction is parallel 
 to that of the first, and which acts at the extremity of 
 a bar M /, it is necessary that this force should exert 
 upon the point M the same effect as the force R, and for 
 that purpose, these forces should be to each other re 
 ciprocally as their distances from the axis of the cylin 
 der ; that is to say, we should have, 
 
 Q : R : : M/ : M /; 
 
 or, since the circumferences of circles are to each other 
 as their radii, we should have, 
 
 Q : R : : circ. M/ : circ. m f. 
 
 Then, by multiplying together in order this proportion 
 and the first, we shall have, since circ. CK=circ. M/, 
 
 Q : p : : YZ : circ. M /; 
 
 that is to say, the power which retains the nut in equi 
 librium, will be to the weight of the nut, as the pitch of 
 
190 STATICS. 
 
 the screw is to the circumference of the circle which the 
 power tends to describe.* 
 
 206. Since the distance of the point M from the axis 
 of the screw does not enter into this proportion, it fol 
 lows that, whatever may be this distance, the ratio of 
 the weight p of the nut to the power Q, which is in 
 equilibrium with it, is always the same, provided this 
 power is always applied to the same point. 
 
 207. If the fillet of the nut is supported upon that 
 of the screw by several points unequally distant from 
 the axis of the screw, as it generally happens, the total 
 weight of the nut can be regarded as divided into par- 
 
 tial weights, each applied to 
 one of the points of support. 
 Now the partial power ap 
 plied to the point v, and which 
 makes an equilibrium with one 
 of these partial weights, is to 
 this weight, in the constant 
 ratio of the pitch of the screw, 
 to the circumference which 
 the power tends to describe. 
 Hence the sum of the partial weights, or the total weight 
 of the nut, is to the sum of the partial powers, or to the 
 total power Q, in the same ratio. 
 
 208. From this it follows : 1st. The force necessary 
 to be applied to the nut parallel to the axis of the screw, 
 to produce an equilibrium with the power Q, which tends 
 
 * This relation is independent of the form of the fillet of the 
 screw ; because the only question is to prevent the moveable piece 
 from turning around a line which is supposed to be fixed, No. 203. 
 
THE SCREW. 
 
 191 
 
 to turn the nut, should be to this power, as the circum 
 ference of the circle which this power tends to describe, 
 is to the pitch of the screw ; 
 
 2d. For the same screw, the effect of the power Q is 
 as much greater as this power is applied at a greater 
 distance from the axis of the screw ; 
 
 3d. For two different screws, the power being applied 
 at the same distance from the axis, its effect is as much 
 more considerable as the height of the pitch is less ; that 
 is to say, the closer the fillets of the screw are together, 
 the greater is the effect of the power for compressing 
 in the direction of the axis. 
 
 II. 
 
 209. The screw is 
 sometimes employed for 
 communicating to a 
 toothed wheel a motion 
 of rotation upon its 
 axis. For this purpose, 
 having given to the screw 
 a pitch DE, equal to one 
 of the divisions of the 
 toothed wheel, it is so 
 arranged that its axis is 
 in the plane of the wheel, and its fillet catches in the 
 teeth. This being done, when a power Q turns the screw 
 upon its axis, by means of a crank FG, the fillet carries 
 along the teeth, which follow each other, and it turns 
 the wheel in spite of the resistance P, which opposes 
 its rotary motion. 
 
192 STATICS. 
 
 When the screw is employed for this purpose, it is 
 named the endless screw. 
 
 To find the ratio of the power Q to the resistance P 
 in the case of equilibrium, suppose the resistance is 
 suspended to a weight by a cord which envelopes the 
 arbor of the wheel. By virtue of this weight, the tooth 
 of the wheel presses the fillet of the screw parallel to 
 the axis HF ; and if we name R this pressure, we have 
 (164), 
 
 p : R : : AC : AB. 
 
 Now we may regard the pressure of the tooth as that 
 which a nut, pushed by a force R parallel to the axis 
 of the screw, would exert ; we have, in the case of equi 
 librium, 
 
 R : Q : : circ. FG : DE ; 
 
 hence, by multiplying together the two proportions in 
 order, we have, 
 
 p : Q : : ACXcirc. Fa : ABXDE ; 
 
 that is to say, the resistance is to the power, as the pro 
 duct of the radius of the wheel by the circumference 
 which the crank describes, is to the product of the 
 radius of the arbor of the wheel by the pitch of the 
 
 screw. 
 
THE WEDGE. 
 
 193 
 
 On the Wedge. 
 
 210. The wedge is a triangular prism 
 ABCDEF, which is introduced by its sharp 
 edge EF into a crack, to split or sepa 
 rate the two parts of a body. It is 
 also made use of for exerting great 
 pressures or to stretch cords. 
 
 Knives, hatchets, punches, and, in 
 general, all cutting and penetrating in 
 struments may be considered as wedges. 
 
 The face ABCD, upon which we strike the wedge to 
 sink it, and which receives the action of the power, is 
 named the heel of the wedge; the side EF, by which the 
 wedge commences to penetrate, is named the Hade; and 
 the name of sides are given to the faces AFED, BFEC, by 
 which it compresses the bodies which it has to separate ; 
 or, since we are accustomed to represent the wedge by 
 its triangular profile ABF, the base AB of the triangle 
 is called the heel of the wedge ; AF and BF are its sides. 
 Fig. 96. 211. We are accustomed to suppose 
 
 that the direction of the power is per 
 pendicular to the heel of the wedge ; 
 because, generally, the wedge is slink by 
 striking upon the heel with a hammer, 
 or with any other object which has no 
 connection with it, and which, in this 
 case, if the direction CD of the shock is 
 not perpendicular to the surface of the 
 heel, the action is naturally decomposed into two other 
 forces, CH, CE ; the first of which, being parallel to the 
 
194 
 
 STATICS. 
 
 heel of the wedge, can have no other effect than that 
 of making the hammer slip, and the second being per 
 pendicular to the face AB, is the only one which is trans 
 mitted to the wedge, and conduces to the effect which 
 is to be produced. But if the power were applied to 
 the wedge by means of a cord, whose point of attach 
 ment could not slip, then, by considering this power, it 
 would be necessary to take account of its direction. 
 
 I. 
 
 212. Let c, D, be two 
 points separated by a 
 wedge ABF, and retained 
 by a cord CD, which is 
 attached to them, and 
 which opposes their sepa- 
 
 j V "|\ * /T\j " rat i n ? suppose, more- 
 *> A ""^- over, that these points 
 
 are supported against a 
 resisting plane which 
 prevents them from moving in a direction perpendicu 
 lar to the cord, and that a power P, applied perpendicu 
 larly to the heel AB of the wedge, makes an effort to 
 separate them and move them in the direction of the 
 length of this cord. This being established, it is re 
 quired to determine, 1st, the tension which results in 
 the cord CD ; 2d, the force necessary to be applied to 
 one of the two points c, D, to prevent them from moving 
 in the direction of the cord; 3d, the pressure which 
 each of these points exerts upon the plane which resists 
 them. 
 
THE WEDGE. 195 
 
 It must be remarked first, that, if the direction of the 
 power P is not such that it may be decomposed into two 
 others Q, R, whose directions pass through the points i 
 c, D, and are perpendicular to the sides AF, BF, the ( 
 wedge will turn between the two points c, D until this , 
 condition is fulfilled, and then only will the power P j 
 produce all its effect. We will suppose, moreover, that, 
 having drawn through the points c, D the lines CE, DE* 
 perpendicular to the sides of the wedge, the point of 
 intersection E of these two lines is in the direction of 
 the power P. 
 
 This being the case, the force P will be decomposed 
 into two other forces Q, R, directed along EC, ED ; and, 
 if we represent this force by the part EX of its direc 
 tion, and finish the parallelogram EZXY, we shall have, 
 
 p : Q : R : : EX : EY : EZ or YX ; 
 
 or, since the two triangles EXY and ABF, whose sides are 
 perpendicular each to each, are similar, we shall have, 
 
 p : Q : R : : AB : AF : BF, 
 and, consequently, 
 
 PXAF PXBF 
 
 Q= , R= , 
 
 AB AB 
 
 The point c not being able to move in the direction 
 ECH, because of the resistance of the plane upon which 
 it is supported, the force Q, which is applied to it, will 
 be decomposed into two others ; one of which, in the 
 
196 
 
 STATICS. 
 
 direction of the line ci, perpendicular to the cord, will 
 be destroyed by the resistance of the plane, and the 
 other, in the direction of the prolongation of DC, will 
 be employed in stretching the cord. Thus, by making 
 CH=EY, and completing the rectangle CGHI, the two 
 components of the force Q will be represented by ci and 
 CG ; and we shall have, 
 
 Q : force ci : force CG : : CH : ci : CG, 
 and, consequently, 
 
 force ci 
 
 QXCI 
 
 QXCG 
 
 force CG= : 
 
 CH 
 
 or, substituting for Q its value previously found, we shall 
 have, 
 
 PXAFXCI 
 
 force ci= 
 
 ABXCH 
 
 PXAFXCG 
 
 ABXCH 
 
 In like manner, if, 
 upon the prolongation 
 of ED, we make DL=EZ, 
 and complete the rect 
 angle DKLM, whose side 
 DK is upon the prolonga 
 tion of CD, and whose 
 side DM is perpendicular 
 to CD, the force E will 
 be decomposed into two 
 
THE WEDGE. 197 
 
 others DM, DK, the first of which will be destroyed by 
 the resistance of the plane, and the second will be 
 wholly employed in acting upon the cord ; and we will 
 find, likewise, 
 
 RXDM PXBFXDM 
 
 force DM= = , 
 
 force DK= 
 
 DL ABXDL 
 
 RXDK PXBFXDK 
 
 DL ABXDL 
 
 Thus the cord CD will be drawn in one direction by the 
 force CG and in the contrary direction by the force DK. 
 
 Now, when a cord is drawn in contrary directions by 
 two unequal forces, the tension which it suffers is always 
 equal to the smaller of these two forces ; for, when the 
 two forces are equal, one of them is the measure of the 
 tension of the cord; and when they are unequal, the 
 excess of the greater over the smaller, not being coun 
 ter-balanced by anything, does not contribute to stretch 
 the cord, and has no other effect than to draw it along 
 in the direction of its length. 
 
 Hence, 1st, the tension of the cord CD will be equal 
 to the smallest of the two forces CG, DK. 
 
 2d. The cord will be drawn along in the direction of 
 its length and in that of the greater of the two forces 
 CG, DK ; so that, in order to oppose this motion, it will 
 be necessary to apply to one of the two points c, D a 
 force equal to the difference of these two forces, and 
 directly opposed to the greater. 
 
 3d. The pressures exerted by the two points c, D 
 upon the plain which retains them, will be equal ; the 
 first to the force ci, the second to the force DM. 
 
 17* 
 
198 STATICS. 
 
 II. 
 
 fr- 98 - 213. If the sides AF, BF 
 
 p | of the wedge be equal, the 
 
 heel AB is parallel to the 
 cord which retains the two 
 points c, D ; and, at the same 
 time, if the direction of the 
 force P is perpendicular to 
 the middle of AB, 1st, the 
 wedge will not turn, because the lines CE, DE, drawn 
 through the two points of support perpendicular to the 
 sides of the wedge, will meet in a point E of the di 
 rection of the power. 2d. Both sides being perfectly 
 alike, the forces CG, DK will be equal to each other, and 
 each of them will be the measure of the tension of the 
 cord CD. 3d. By letting fall from the point F the per 
 pendicular FN upon the heel of the wedge, the two tri 
 angles CGH, BNF, whose sides will be perpendicular each 
 to each, will be similar, and give, 
 
 CH : CG : : BF : FN. 
 Now we have, 
 
 Q : force CG : : CH : CG, 
 
 and, consequently, 
 
 Q : force CG : : BF : FN ; 
 
 But, we have, also, 
 
 p : Q : : AB : BF. 
 
I 
 
 THE WEDGE. 199 
 
 Hence, by multiplying together these two proportions 
 in order, we shall have 
 
 p : force CG : : AB : FN ; 
 
 that is to say, in this case, the power p will be to the 
 tension of the cord CD, as the heel of the wedge is to 
 its height. 
 
 We will not make any application of the theory of 
 the wedge to the use which may be made of this in 
 strument for splitting bodies ; because, under such cir 
 cumstances, the resistance which has to be overcome is 
 always unknown, and it is useless to know the ratio of 
 this resistance to the power which is in equilibrium 
 with it. 
 
 LEMMA. 
 
 214. If from the summits B, c of two of the angles 
 of a triangle, the perpendiculars BE, CD be dropped upon 
 the opposite sides, these perpendiculars will be to each 
 other reciprocally as the sides upon which they are 
 dropped ; that is to say, we shall have, 
 
 BE : CD : : AB : AC. 
 
 DEMONSTRATION. By con 
 sidering AB as the base of 
 the triangle, the perpen 
 dicular CD will be its height, 
 and the surface of the tri- 
 
 i TI -1 AI> X GD -r- 
 
 angle will be - = . In 
 
200 STATICS. 
 
 like manner, by taking AC for the base, the surface will 
 
 be - ; hence, we shall have the two equal products 
 & 
 
 ACXBE=ABXCD, which will give the proportion, 
 BE : CD : : AB : AC. 
 
 THEOREM. 
 
 215. When the two powers P, Q are applied to the 
 faces AB, AC of a wedge, the third face of which EC is 
 supported upon a resisting plane MN, and these two 
 powers are in equilibrium through the resistance of the 
 plane, they are to each other reciprocally as the spaces 
 which they traverse in the line of their directions, if the 
 equilibrium be disturbed infinitesimally. 
 Fig. 100. 
 
 DEMONSTRATION. Since the powers P, Q are in equi 
 librium, their directions are perpendicular to the faces 
 of the wedge to which they are applied (179). Now 
 suppose, by virtue of a derangement in the equilibrium, 
 the wedge slips upon the resisting plane, and takes the 
 infinitely near position abc ; and the direction QE is pro 
 longed until it meets ac in e : it is evident that the 
 
THE WEDGE. 201 
 
 small lines Ee, Dd will be the spaces which the powers 
 will have traversed in the lines of their directions. 
 
 Lastly, draw Aa ; prolong CA, ba until they intersect 
 somewhere in F, and from the points A, a let fall the 
 perpendiculars AH, aa ; we will have evidently Ee=Ga 
 and j)d=AH. 
 
 This being determined, the powers P, Q being in equi 
 librium, they are to each other as the sides of the 
 wedge to which they are applied ; then we shall have 
 (212), 
 
 p : Q : : AB : AC, 
 and, because of the similar triangles ABC, FA, 
 
 p : Q : : F# : FA ; 
 hence (214), we shall have, 
 
 p : Q : : a& : AH 
 and, consequently, 
 
 p : Q : : Ee : Dd. 
 
 216. We have seen that the analogous proposition 
 takes place in the equilibrium of all the machines which 
 we have considered. By following the same process, it 
 could be demonstrated directly that, when two powers 
 are in equilibrium by means of points of support which 
 any machine presents, they are to each other reciprocally 
 as the spaces which they would traverse in the lines of 
 their directions, if the equilibrium were infinitesinr.ally 
 
202 STATICS. 
 
 disturbed. By means of this proposition, it will be 
 easy to find in practice the relation which should subsist 
 between a power and a resistance applied to a proposed 
 machine, in order that these forces should be in equi 
 librium, leaving out of consideration friction and other 
 obstacles to motion. 
 
203 
 
 NOTE. 
 
 A NEW DEMONSTRATION OF THE PARAL 
 LELOGRAM OF FORCES. 
 
 BY M. CAUCHY.* 
 
 IF we suppose, as in general, a force to be repre 
 sented by a length laid off from its point of application 
 along the direction in which it acts, the resistance R of 
 two forces P, Q, simultaneously applied to a material 
 point (A), will be represented in intensity and direction 
 by the diagonal of the parallelogram constructed upon 
 these two forces. This proposition has already been 
 demonstrated in several manners. But, among the 
 demonstrations which have been given, some require the 
 consideration of new material points connected with the 
 point (A) by rigid and invariable lines ; others are 
 founded upon the use of the differential calculus, or of 
 derived functions; others again are deduced from the 
 relations which exist between the cosines of certain 
 angles. I am here about to demonstrate the same propo 
 sition without recurring to these different considerations, 
 
 * This demonstration is extracted from the work -which M. Cauchy 
 publishes in numbers under the name of Exercises de Mathematiques. 
 
204 STATICS. 
 
 and, in order to attain my object, I will establish suc 
 cessively several lemmas, which may be enunciated as 
 follows. 
 
 LEMMA I. 
 
 If we designate by R the resultant of the two forces 
 p, Q, simultaneously applied to the point (a), and by x 
 any number, the resultant of two forces, equal to the pro 
 ducts P#, Qx, and directed along the same lines as the 
 forces P, Q, will be represented by the product RZ, and 
 directed along the same line as the force R. 
 
 DEMONSTRATION. In the first place let #= , m and 
 n representing any two whole numbers. We shall have 
 
 _ p Q 
 
 n n 
 
 Besides, we may consider the component P, or m , as 
 
 n 
 
 produced by the addition of several forces equal to -, 
 
 n 
 
 and the component Q, or m - , as produced by the addi- 
 
 n 
 
 tion of as many forces equal to . From this it is easy 
 
 to conclude, that the resultant of the forces P#, Q#, will 
 both be produced by the addition of several forces 
 
 p o 
 
 equal to the resultant of - and -. Moreover, it is evi- 
 n n 
 
 dent that the first two resultants are to the last, as the 
 
NOTE. 205 
 
 numbers n and m are to unity. Hence, the second re 
 sultant will be equivalent to the first multiplied by the 
 
 997 
 
 ratio =Xj that is to say, to the product nx. 
 
 % 
 
 Let us suppose in the second place that the number 
 x is irrational. Then we can vary the whole numbers 
 
 / /77 
 
 m and n in such a way that the fraction will converge 
 
 7l/ 
 
 to the limit x; and it is evident that, in this case, the 
 resultant of the forces , , always directed along 
 
 the same line, and always equal to , will tend more 
 
 n 
 
 and more to coincide, in intensity and direction, with the 
 resultant of the forces PX, Q#. Hence, this last result 
 ant will be directed along the same line as the force R, 
 and it will have for measure the limit of the product 
 
 , that is to say, the product ~RX. 
 n 
 
 w 
 
 COROLLARY. 
 
 A 
 
 If we designate by the notation (p, Q) the angle in 
 cluded between the directions of the two forces P and Q 
 
 A A 
 
 (p, R), (Q, R) will be the angles included between the di 
 rections of the components P and Q and their resultant 
 R. This being fixed, if we make successively R#=P, 
 
 P Q 
 R#=Q, or, what amounts to the same, #=-, #=-, we 
 
 R R 
 
 18 
 
206 STATICS. 
 
 ill conclude from the preceding lemma, 1st, that the 
 
 p2 pQ 
 
 force P may be replaced by two components and , 
 
 Ei 11 
 
 A A 
 
 which form with it angles equal to (P, R) and (Q, K) ; 
 
 2d, that the force Q may be replaced by two com- 
 
 O 2 PQ 
 
 ponents and , which form with it the angles (Q, R) 
 
 R II 
 
 A 
 and (P, R). 
 
 LEMMA II. 
 
 The resultant R of two forces P, Q, which intersect at 
 right angles* is represented in intensity by the diagonal 
 of the triangle constructed upon the two components, so 
 that we have, 
 
 DEMONSTRATION. Let us conceive the force P to be 
 replaced by the two above-mentioned components, that 
 
 "P "PO 
 
 is to say, by two forces and , which form with 
 
 R R 
 
 it the angles (P, R) and (Q, R). Let us conceive also that 
 
 Q 2 i PQ 
 the force Q is replaced by two components - and , 
 
 A A 
 
 which form with it the angles (Q, R) and (P, R). We 
 
NOTE. 207 
 
 p2 Q2 
 
 can suppose that the forces , are directed along the 
 
 R R 
 
 same line as the resultant R, and then the two forces 
 
 PO 
 equivalent to will each form with the direction of R 
 
 an angle equal to (P, Q). Hence, they will form between 
 
 A 
 them an angle equal to double (P, Q). Hence, since the 
 
 A 
 angle (P, Q) is a right angle, by hypothesis, the forces 
 
 PQ 
 equivalent to will be equal, but directly opposed. 
 
 
 
 Consequently, they will be in equilibrium ; and for the 
 
 p2 Q2 
 
 forces , , directed along the same line, we can sub- 
 R n 
 
 stitute only the forces P, Q, or their resultant R. Hence, 
 we shall have the equation, 
 
 P 2 , Q 2 
 
 R= + , 
 
 R R 
 
 from which formula (1) is immediately deduced. 
 This demonstration is due to Daniel Bernoulli. 
 
 LEMMA III. 
 
 The resultant R of the two forces P, Q, which intersect 
 at right angles, is represented, not only in intensity, as 
 we have proved above, but also in direction, by the diago 
 nal of the rectangle constructed upon the two components. 
 
 DEMONSTRATION. This proposition is evident in the 
 case where the forces P, Q are equal to each other. 
 
208 STATICS. 
 
 Then the resultant R should necessarily divide the angle 
 
 A 
 (p, Q) into two equal parts, and we have, by virtue of 
 
 Lemma II, 
 
 R 2 =2p 2 , or R=pv/2. 
 
 It is also easy to demonstrate, Lemma III, in the case 
 where we suppose Q 2 =2p 2 , or Q=p^2. Thus, we will 
 consider three forces, equal to P, directed along three 
 lines which are perpendicular to each other. These 
 three forces will be represented by three sides of a cube 
 which meet in the same summit. Moreover, the result 
 ant of two of these forces being equal to p^2, and di 
 rected along the diagonal of one of the forces of the 
 cube, the resultant Q of the three forces will of necessity 
 be included in the whole plane, which will contain one 
 of the forces P and the diagonal of the square con 
 structed upon the other two. Now, there are three 
 planes of this kind, and these three planes intersect in 
 the diagonal of the cube. Hence, the resultant of the 
 three forces P, or what is the same, the resultant of the 
 forces P and P^S, which intersect at right angles, will 
 be directed along the diagonal of the cube, which is at 
 the same time the diagonal of the rectangle constructed 
 upon the forces P and p^2. 
 
 It might be proved, precisely in the same manner, 
 that if we designate by m a whole number, and suppose 
 Lemma III to be demonstrated in the case w r here we 
 have Q p^w, the resultant of three forces, respect 
 ively equivalent to 
 
 p, P, 
 
NOTE. 209 
 
 and represented by three lines perpendicular to each 
 other, will be directed along the diagonal of the rect 
 angular parallelopipedon, which will have these same 
 lines for sides. From this we conclude, in the admitted 
 hypothesis, that Lemma III will also subsist, if we take 
 for Q the resultant of the forces P and P^w, that is to 
 say, if we make Q=p^m+l. Besides, Lemma III is 
 evident, when we have Q=P, or, what is the same, m=1. 
 Hence, this lemma will also subsist, if we take 
 Q=P\/l-f l==pv/2, or Q=pN / 2+l=p> / 3, etc., 
 
 or, in general, Q=P^/W, m being any whole number. 
 
 Let us conceive now, that m and n designate two 
 whole numbers ; and construct a rectangular parallelo 
 pipedon, which has for its sides three lines representing 
 the three forces 
 
 The resultant of these three forces will evidently be 
 contained; 1st, in the plane which includes the force 
 p^n and the diagonal p^m+1 of the rectangle con 
 structed upon the forces P, p^m, 2d, in the plane 
 which includes the force p^m and the diagonal 
 of the rectangle constructed upon the forces P, 
 Hence, this resultant will be directed along the diagonal 
 of the parallelopipedon ; and the plane, which contains 
 the same resultant with the force P, will intersect the 
 plane of the two forces p^m, P^n along the diagonal 
 of the rectangle constructed upon these two forces. 
 Hence the resultant of the forces ?*/m, p-v/w, which 
 evidently should be comprised in the plane in question, 
 
 18* 
 
210 STATICS. 
 
 will be directed along this last diagonal. Hence Lemma 
 III will subsist, when we replaee the forces P and Q 
 by two forces equal to Q^w, P^n, that is to say; by 
 two forces whose squares are to each other in the ratio 
 of m to n. Hence, Lemma III will also subsist between 
 the forces P and Q, if we suppose 
 
 ; =pj~- 
 
 Now, let Q=Payx designating any number whatever. 
 We can vary the whole numbers m and n, in such a 
 
 manner that the ratio will converge to the limit ar, 
 n 
 
 and it is evident, that, in this case, the resultant of the 
 forces P and P*J , directed along two lines perpendicu 
 lar to each other, will tend more and more to coincide, 
 in intensity and direction, on the one hand with the re 
 sultant of the forces P, PX, and, on the other, with the 
 diagonal of the rectangle constructed upon these two 
 forces. Hence, the resultant of the forces P, ?x will 
 be represented by the diagonal in question. 
 
 COROLLARY I. 
 
 If the force R be represented by the length AB laid 
 off from its point of application (A) upon the line along 
 which it acts, and if we draw through the point (A) two 
 axes perpendicular to each other, we may substitute for 
 the force R the two forces represented in intensity and 
 
NOTE. 211 
 
 direction by the projections of the line AB upon the 
 two axes. 
 
 COROLLARY II. 
 
 Let us conceive now, that two forces P, Q, being applied 
 to the same point (A), and represented by two lines AB, 
 AC, which form any angle between them, to be traced in 
 the plane of these two forces, two axes, one of which 
 coincides with the diagonal of the parallelogram to 
 which they serve as sides, and the other perpendicular 
 to this diagonal. We can substitute for the two forces 
 p, Q, the four forces represented in intensity and di 
 rection by the projections of the lines AB, AC upon the 
 two axes. Now, of these four forces, two, being directly 
 opposed, will be in equilibrium. The other two, directed 
 along the diagonal of the parallelogram, will be added 
 together, and will give, for their sum, a force represented 
 in intensity and direction by this same diagonal. Hence 
 we may enunciate the following proposition : 
 
 THEOREM. 
 
 The resultant R of two forces P, Q simultaneously ap- 
 
 : plied to a material point (A) and directed in any manner 
 
 whatever, is represented in intensity and direction by 
 
 the diagonal of the parallelogram constructed upon these 
 
 two forces. 
 
212 STATICS. 
 
 COROLLARY I. 
 
 As the diagonal R of the parallelogram constructed 
 upon the two forces P, Q is at the same time the third 
 side of the triangle, which is formed by drawing through 
 the extremity of the first force a line equal and parallel 
 to the second, and as the angle in this triangle, opposite 
 
 A 
 to the side R, is the supplement of the angle (P, Q), we 
 
 have necessarily, by virtue of a known formula in 
 trigonometry, 
 
 A 
 
 R2 = p2-f Q2 + 2PQ COS (P, Q) (2). 
 
 COROLLARY II. 
 
 In the case where the forces P, Q become equal to 
 each other, their resultant R is represented in intensity 
 and direction by the diagonal of the lozenge constructed 
 upon these same forces. Then the formula (2) re 
 duces to 
 
 A 
 
 R 2 =2P 2 fl + COS(P, Q)J. (3) 
 
 A 
 Moreover, by supposing (P, R)=0, we will find, in the 
 
 A 
 present case, (P, Q)=0; and as we have, in general, 
 
 cos 2^=2 cos 2 ol, (4) 
 
 the equation (3) will give R=2p cos 0, or, what is the 
 same, 
 
 A 
 R=2p cos (P, R). (5) 
 
NOTE. 213 
 
 For the rest, we may be directly assured that the 
 second member of the formula (5) represents the di 
 agonal of the lozenge constructed upon the two forces 
 equal to P. 
 
 It is easy to demonstrate the theorem of page 211, 
 for the case in which the forces P, Q have any ratio to 
 each other, when we have once established this theorem 
 for the case where we have Q=P, that is to say, when 
 formula (5) is established. Now, we can give a direct 
 demonstration of this formula, which, in fact, is deduced 
 from equation (4), but which appears to merit special 
 remark. I will explain it in a few words. 
 
 Let us admit that formula (5) is verified for the case 
 A 
 
 where we have (P, R) *, * designating either a right 
 angle or an acute angle. I say that it will still subsist, 
 if we suppose 
 
 (p,n)=|, or(p,R)=?-|. 
 
 A 
 Now, in these two hypotheses, the angle (P, Q), included 
 
 between the directions of the two equal forces P, Q, 
 will be equivalent to one of the angles *, rtt; and we 
 can prove by reasoning as in Lemma II, that we can 
 substitute in the system of the two forces P, Q, or for 
 
 p2 
 
 their resultant R, four components equal to , among 
 
 which, two will be directed along the same line and in 
 the same direction as the force R, while the other two 
 will each form with the resultant R an angle equivalent 
 
 A 
 to (P, R), that is to say, to * or to * *. Now, since we 
 
214 STATICS. 
 
 suppose formula (5) to be verified in the case where we 
 
 A 
 have (P, R)=*, the last two components evidently can 
 
 p2 
 
 be replaced by a single force equal to 2 cos *, and 
 
 R 
 
 situated in the direction of the resultant R, or in the 
 opposite direction. Consequently, we will find defini 
 tively, 
 
 R=2-2- cos *=2- (Idbcos *) ; 
 R R R v 
 
 or, 
 
 R 2 =2P 2 (1COS *), (6) 
 
 the sign being necessarily reduced to the sign+, in 
 the case where (P, R)=O? and to the sign in the case 
 
 where (P, R)=^ -. Since we will also get the formula 
 (4) by placing successively 
 
 1+cos *=2 cos 2 5, 1 cos *=2 cos 2 , 
 equation (6) will give in the first case 
 
 R=2 cos o, 
 and, in the second, 
 
 7i if 
 R = COS 
 
 2 
 Hence, if equation (4) subsists when we attribute to the 
 
 angle (P, R) the value t, it will also subsist when we 
 
NOTE. 215 
 
 -. ! 
 
 2 i 
 
 attribute to the same angle one of the values ~, 
 
 Now, this equation is verified when we suppose (P, R)=5, 
 
 jj 
 
 since we have evidently in this hypothesis R=o, and 
 
 A 
 cos (P, R) =0. Hence, it will be equally true if we 
 
 suppose 
 
 and, consequently, if we take 
 
 / A N 1 * n A i - 3^ 
 
 < P > B ) = 2 4=8 or ( p > R )=2 (*-3=-s 
 
 Hence it will also be true, if we attribute to the angle 
 
 A 
 (p, E) one of the values, 
 
 1 7t _ * 1 07t _ 3rt -lx 3rtv _ 5rt 
 
 2 8~16 2 8""l6 2^~8"^"~16 
 
 By continuing in the same manner, it may be proved 
 
 A 
 that formula (5) generally has place when the angle (P, R) 
 
 2?w4-l 
 receives a value of the form - -, n representing any 
 
 whole number, and 2m +1 an odd number less than 2 n . 
 If we now represent by an acute angle taken at will, 
 we can vary the whole numbers m and n in such a man 
 
 ner that the ratio = will indefinitely approach the 
 
216 STATICS. 
 
 limit e ; and the resultant n will tend more and more 
 to coincide, on the one hand, with a force equivalent to 
 2p cos 0, and, on the other hand, with the resultant of 
 two forces equal to P, which would form between them 
 an angle double of o. Hence, this last resultant will 
 be represented in magnitude by 2p cos 0, and will also 
 verify formula (5). 
 
 THE END. 
 
 STEREOTYPED BY L. JOHNSON AND CO. 
 THILADELrHIA. 
 

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