MECHANICS. 



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forces, the first 

 iv.*nltant of any number uf equal ones applied to a body at 



ir.j n line. '1'he number may bo either 

 \ hall consider each separately. First, tako odd ; 

 mil let it be -even, as in Kit,'. 1-- Now, supposing each to bo 

 i ml, if we tako tho middln one, whieh is eviilently at tho 

 miilille of t lie lino A B, we find that 

 there are threo pounds on cither 

 it net inn in pairs at equal 

 <li-Unccs from M. Tho resultant 

 of tho nearest pair gives, as proved 

 F1 K< 12. above, two pounds at M ; tho next 



pair also give two, and so does tho 



third. Those mako six pounds of resultant at M, which, with tho 

 single one already there, are seven pounds tho sum of all tho 

 forces for resultant. Wore tho number thirteen tho conclusion 

 AM. 11 l.l bo the same. There would be six on either side of tho 

 .i.i.Mln one, and you would have a resultant of thirteen pounds ; 

 and the same holds good of any other odd number yon select, 

 bo it large or small. 



Now, suppose wo have an even number of such forces, say 

 six, as in Fig. 13, counting them from either end towards tho 

 middle, there will be no middle pound ; and tho middle of tho 

 line A B will bo in tho middle of tho space between the middle 

 pair of forces. What have we then ? The irside pair gives two 



pounds at M, so does tho next 

 outside, and so tho next; and 

 there are evidently thus six 

 pounds of resultant at tho cen- 

 ^ tro of A B. Tako any other even 



Fig. 13. number, and tho result is the 



same ; and thus, for both odd 



and even numbers, we arrive at this conclusion : The resultant 

 of any number of equal parallel forces acting on a body at equal 

 distances along a line, is equal to their sum, and bisects the 

 line joining tho points of application of the extreme forces. 



An instance of this is the working of a hand fire-engine. 

 Suppose seven men at tho lever on either side, that is, fourteen 

 hands on each lever ; supposing the men to be equally arranged 

 and of equal strength, this makes fourteen equal forces applied 

 at equal distances, tho resultant of which is the muscular power 

 of seven acting at the centre on cither side. 



Now we shall, without difficulty, find the resultant of two 

 \vnequal parallel forces. As before, let A and B be their points 

 of application, and let us first suppose that they act in the same 

 direction. Measuring the forces by pounds, or ounces, or even 

 grains, there are three cases which may occur. The number, 

 eay of pounds, in the forces may be both even, or both odd, or 

 one oddandtho other even. 1. We shall take "both even" first, 

 and, for simplification sake, let them be six at A and four at B. 

 Divide now the line A B into ten equal parts, that is, into as many 

 parts as four and six together make. Extci .1 also A B on either 

 side, as represented (Fig 14) by the dotted lines, and measure 

 off on the extensions any number of portions you please, each 



equal to one of the 



""X subdivisions of A B. 

 Beginning at A, sup- 

 pose you apply apound 

 force at the end of the 

 first subdivision to the 

 Fi S- !* right, another pound 



at the end of tho third, 



another at that of the fifth, and so on until you come to B. 

 You will find then that there will bo a pound at the end of the 

 first division from B. Put pounds now at the end of the first 

 division from A on tho dotted line, on the third, and on the fifth, 

 and do the same on tho dotted line from B, on the first and third. 

 Count all the pounds you have ; they are ten, five inside and five 

 outside. Calling the points occupied by the extreme pounds 

 f and Q, the resultant of these ten, so distributed at equal 

 distances, must pass through the middle, M, of P Q, and be ten 

 pounds, by the principle last established. But if we take 

 separately the three outside and the three inside A, they make six 

 pounds acting at A. Also tho two pair on either side of B make 

 four pounds at B. The ten pounds at M must therefore produce 

 the same effect on the body as the six at A and the four at B, 

 jnd therefore must be the resultant of these forces ; that is to 

 aay, tho resultant is tho sum of the components. 



J y/ i i\i ~t i i 

 ii ^ 



4 



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now the number of Hubdivuion* on either side, 

 from M to A an. I n. There are four on the aide of A and KU on 

 B'H aide that u to say, the resultant cuU the line A B in the 

 ..in of tho number* 4 and 0, with this peculiarity, how- 

 evor, that the imaller number in on the aide of the greater forue 

 Thin w what we might expect, for the resultant ought twturaUy 

 tu t'-n.l toward* the greater, on account of iU preponderance. 

 When a line in cut in thin way, the smaller portion being on the 

 side of the greater number of pounds, it M said to be oat 

 inversely as the two numbers that in, in the contrary order. 



2. Now let us take the cam of two odd numbers ; let them be 

 9 and 7. It is evident that if we put another 9 pounds at A, 

 and 7 at B, the resultant of thin second 9 and 7 should in every 

 respect agree and coincide with that of tho first, and that the 

 resultant of the four should be the sum of two nines and two 

 sevens. But the double 9 at A is 18 pounds, and the double 7 

 at B 14 pounds. The case, therefore, becomes one of even num- 

 bers, and the line A B, as proved above, must be cut by the 

 resultant in the inverse proportion of 18 to 14. But to divide 

 a line no that there may be 18 parts one side and 14 on the 

 other becomes, by throwing every two of the subdivisions into 

 one, the same thing as dividing it so that 9 may be on one side 

 and 7 on the other. In this case then, also, A B is divided 

 inversely as tho forces. 



3. When the number i are one odd and the other even, say 4 

 and 7, the result is the same. By doubling each force yon get 

 8 and 14 pounds, both even numbers ; the line A B is divided by 

 tho resultant inversely as 14 to 8, which is the same as 7 to 4 

 inversely as the forces. 



We have supposed in all these cases that the forces con- 

 tained an exact round number of pounds ; but what should wo 

 do if there were fractions of a pound in either or in both ? I 

 say, reduce the forces to ounces, and work by round numbers in 

 ounces. If there were fractions of ounces, work in grains. 

 Tou can thus still secure round numbers, and the above proofs 

 will hold good. But what are* 

 you to do if there are fractions 

 of grains ? Work them by 

 tenths, or hundredth, or thou- 

 sandth parts of grains, or by even 

 far smaller fractions, and you 

 will still have round numbers, 

 and you can say that the result- 

 ant cuts A B inversely as these 

 numbers, however great they be, 

 and therefore inversely as the forces. To trouble you about 

 smaller fractions would only get you into a cloud of metaphysics 

 for no practical purpose. 



I have proved this important principle only for particular 

 even numbers, 6 and 4, but you will find that the reasoning will 

 be the same whatever be the even numbers you choose. The 

 rule simply is to divide the line A B into as many equal parts as 

 there are pounds in both forces, and then to distribute all the 

 pounds at A in two batches on either side of that point, and to 

 do the same at B with the pounds there acting, observing to 

 place the pounds as you go from A or B in any direction, at the 

 first, third, fifth, and so forth, points of division. 



You are now in a position to find the resultant of three or 

 more parallel forces acting, say, at the 

 points A, B, c, D, as in Fig. 15. First 

 join A with B, and cut it inversely as the 

 forces which are there applied ; next join the 

 point x so found with c, and cut the join- 

 ing line at Y inversely as the sum of the two 

 first forces to that at c; join this again 

 with D, and cut it inversely as the three 

 first forces to that of D ; and so proceed 

 until you have exhausted all the forces. 

 The point z last found is that through 

 which the resultant of aM passes, and is 

 called the centre of parallel forces. 



Suppose, for example, that the centre was required in the caso 

 of parallel forces of 1, 2, 3, and 4 pounds applied to the four 

 corners of a square board, A, B, c, D (Fig. 16). First divide A s 

 into three parts, and take two next to A and one to B. The 

 point x so found is the parallel centre for these two forces. 

 Join x now with c, and cut X c into six parts (the sum of 1, 2, 

 and 3), and take three next to c and three to x. The centre T 



v 



Fig. 15. 



Fig. 16. 



