310 On the Parallelogram of Forces. 



i 

 Suppose then, that x is changed to x-\-x', but that y is the same 



as before ; let z' be the resultant of these forces, 6' the angle which 



its direction makes with that of x ; then it is manifest that d 7 is Z$, 



and that b -&'= the angle formed by the directions of z* and x. Put 



d — 6'=Vj (a), then by resolving z' in the direction of z by (1), we 



shall have 2^(1?)= the resultant resolved in the direction of z; but 



x'y(i)=x' resolved in the same direction, and by adding z, we have 



the components reduced to the same direction; hence z'$(v)~z-{> 



Put -=m, y=n,^t^=m', l-=n',(c),thenby (4'),m 2 +n 2 = l> 



^ <2 -2 2 



m /a +n' 3 = l, (rf); by (c),a?=mr, x+a? 7 =7nV, y — «r=wV; hence 



we shall have 2/ = — , x =m'z' — x — )z, but by (1)>- 



n f \ ri I z 



*»=?(£), hence 2:-fx / ^) = (n / (l — i» 2 )-|- nram')— — (since 1 — m" 



n 7 



n a ,) = (n'ra -{-mm') — ^wim'+wtt')*' ; hence by (6), we shall have 



(p(t?) = w/7i / +ft7i / , (e). It is evident that 6 and d 7 are independent of 

 each other, and that m, n are functions of 6 without 7 , and that mf, 



w 7 are functions of &' only; hence by putting ^ — <P 7 (y)> and tak- 



ing the partial differentials of (e) relative to 6 and d', we shall have ?/(tf)X 



tf-u dm . ,{//* ,, . flu dm 1 . da f , , / x dv t dv 

 = m _ + n <p (•») X — —m •,— + n-— ;butby(a), — = l>-r, 



1, .-.^)=m^+n^= - L^+n^l (/). Now by (. 



</j </j \ <// di' J 



dn _ _m dm dn' _ wi' c?m 



(/) 



we have by reduction, — —rt = c !l!L—n / * (g). Since the first mem- 



J dt dr KqJ 



ber of (g) is a function of 6 only, and the second of &\ it is evident 

 that each member = a const. ; .".let — k denote the constant, and 



since by (d) n = V 1 - m 2 , we shall have by multiplying by d6 9 



kd) y whose integral gives m =-= cos. (&4 + c), c = the 



z 



dm 



1 



correction ; hence (as before,) we shall have #=* cos. Q. 



