Demonstration of the "Parallelogram of Forces" 225 

 «, /3 and y being connected by the equation 



a + |S+y = 2jr (4.) 



Eliminating P and Q from (1.2. 3.)» R will disappear at 

 the same time, and there results, 



1 — cos 2 \J/ a — cos 2 \p /3 — cos 2 \J/ y ~\ 



— 2 cos\J/a . cos\{//3 . cos\[/y = 0. J " * 



Solve this equation for cos\J/y, 



.*. cos^J/y = — cos\J/a . cos\J//3 + \/(l — cos 2 \[/ a) (1 — cos 2 v|//3) 



= — cos \J/ a . cos \p /3 + sin \p a . sin v[/ /3 



= —cos {\J/a + 4/)3} ; 



.*. (2 w + 1 ) t +^y = ^«±^/3, 



or 4/a + rJ//3 ± 4/y = (2m + 1)tt, . . . . (6.) 



?« (as well as n, p, q and r employed below) being a positive 



2 

 or negative whole number. Again, when a = /3 = y = — 7r, 



3 



we have 



(1. 2. 3.), cos\{/ (-| it J = -1 = cos (2 11 ± -jj 7T, 



•••*(40 = ( 2w± i)* < 7 '> 



Now the left-hand member of (6.) may be written four dif- 

 ferent ways: \J/a + \|//3 + ^y, ^a + ^/3 — \(/y, 4>a — \J//3-t-\|/y, 



or \|/a — vf/ /3 — 4>y; but when \{/a=:\{//3:=\py = 4/( — 7r), 

 the last three give 



4/ f — ttJ = (2»i+l)w, or — (2»i 4- t)», 



which is inconsistent with (7.) ; hence we must have 



<J/a + ^/3 +4/y = (2w+ 1)tt. .... (8.) 

 Differentiate (8.) first with respect to a, and then with re- 

 spect to /3, observing that (4.) -^- = -=% = — 1, 



.■.^'« = 4/'/3 = 4;'y, (9.) 



where, as usual, \J/ ex. = -j- . <p a. Now two of the quantities 



a, |3, y being arbitrary (4.), it is evident that (9.) cannot be sa- 

 tisfied but by \J/ a = constant = c; hence integrating 



4>u = c<x + c x (10.) 



To determine the constants c, c v we observe that when 



