346 Parallelogram of Forces. 



pose that we have n"z'=x' , (2) ; then, since the directions of R 

 and S, make equal angles with that of z' , each of which is express- 



ed by ^ which equals the angle made by the direction of z' , with 



that of X, we shall evidently have n"^ and n'"^, for the values of 

 R and S, when resolved in the direction of z' , v^'hose sum must 

 equals;', .'. n'"(R4-S)=2;'; hence (2) becomes n" - (K-\-^)-=x' ^ 



Again, by resolving S in the direction of a:, we have by what has 

 been proved, S cos. a = the value of S when resolved in the direction 

 oi X, and by adding R, which acts in the direction of x, we shall 

 have S cos. a-j-R =3^', since by the nature of the components and 

 their resultant, the resultant when resolved in any direction, must 

 equal the sum of the components when resolved in the same direc- 

 tion ; hence by substituting the value oi x' in (3) we get S cos. a+ 

 R=w/'-(R-}-S), or since R = S, we have 2w'''-=l+ cos. a, (4), 



a a 



but by trig. 1-j-cos. a=2 cos.^ -? .'. (4) becomes %"=■ cos. ^» (5). 



a 

 By (5)j we have n"z'=z' cos. ^ =z' resolved in the direction of a;; 



and in the same way for a force z", whose direction makes the an- 



a a . a 



gle g = oi with that of x, we shall have z" cos. ^ = z" resolved 



2 



in the direction of a?, and soon ; hence, if Z denotes any force whose 



a . 

 direction makes the angle ^^ with the direction of x, where m rep- 

 resents any positive integer, and if X denotes the value of Z, when 



resolved in the direction of x, we shall get Z cos. ^^ = X, or by 



restoring the value of a, we have Z cos. \ ^ — / = X, (6). 



Hence, the force z, whose direction makes the angle ^ with that of 

 X, is easily resolved in the direction of x ; for since the integers m 

 and n, are arbitrary, they may evidently be taken such, that 



P 



4 + 2n P 



shall differ from ^ by an angle less than any given angle, 



