226 Demonstration of the Parallelogram of Forces. 



a. = and 8 = y m k, the forces all act in the same straight 

 line, P in one direction and Q and R in the opposite ; hence 

 P = Q + R: and we derive from (1. 2. 3.), 



cos \J/ = — 1 == cos (2 p + 1 ) 7r and cos \j/ k = 1 = cos 2 q 7r, 



,\ vj/ = (2^? + 1 ) 7r and vf/ w = 2 <? 7r ; 

 hence (10.) gives 

 c x = ( c 2p + 1) w 9 and c n + Cj = 2 # 7r, or c = 2 (</ — p — 1) 



+ 1 = 2 r + 1 (suppose), 

 .*. (10.) becomes \J/a = (2r + 1) a + (27? + 1)tt, 

 and 



cos\{/a = cos {(2r+l)a + (2/>+l)*} = — cos(2r + I) a. (11.) 

 Moreover, if 



« ss -, and 8 = y — it — . — . 



2r+1 » ^ / 2/- + 1 2' 



we shall have (11.), 



cos\J/« = 1 and cos\{//3 == cosv[/y = 0, 

 .-. (1.2. 3.) Q= R and P = 0; 



7T 



hence the equal forces Q, R acting at an angle -, ba- 



j o o 2 r + 1 



lance each other; but this is impossible, unless r = 0, when 

 they act in the same straight line in opposite directions. Make 

 therefore r = in (11.), and we have finally, 



cos4'a=— cos a (12.) 



The equations (1, 2, 3.) therefore become, 



R = — Pcos/3 — Qcosa (13.) 



Q = — Pcosy — Rcos« (14.) 



P = — Qcosy- Rcos/3 (15.) 



Multiply these equations by R, Q and P respectively, and 

 we readily get 



R 2 = P* + Q 2 + 2PQcos 7 . . . . (16.) 



Eliminate P, and then Q, from any two of the equations 

 (13. 14-. 15.), and reduce by means of (4.), 



.,X= "« =-«-.. .... (,7.) 



sin a sin 8 sin y 



After having (by any means) obtained the form of 4>a 

 (= cosvj/a), the formula (17.) may be found in the following 

 manner: — 



Produce RO to R', draw G OH perpendicular to RO R', 



