Solution of a Functional Equation. 69 



forces are supposed to tend to diminish the distances p, q, r, and 

 by (4) we get P(^/?+a'^g+R(^r=0, (5). 



Now since a. b, a' ^ Z>', are each invariable (because they be- 

 long to fixed points), we have <5p = — 8x^ dq = — 8y^ 8r = 

 {x — a']8x-\-{y-b'Yy 



'- — , and by substituting these values (5) he- 

 comes -VSx-GiSy^^'^^^'Sidx -{-'^^ R«y2/=0, or (^^R-p) 



ly-h' \ 



ox-\-[ — — R — Q,)Jy=iO; hence, since Sx, ^y are arbitrary and 



rp q/ qj J/ 



independent of each other, we must have ' R - P=0, R 



-a=o, or p=^^'R,a=~^R, .-. p- + a^ = R% (S), 



{x-a'Y-\-{y-h'y a y~b' 



smce " = 1> also p-= _ ^ , (9) ; hence from (8) 



and (9) it is evident that if two forces are represented in quantity 

 and directions by the two sides of a rectangle, their resultant is 

 represented in quantity and direction by the diagonal which pass- 

 es through the angle formed by the two sides that represent the 

 forces. 



For other applications of (4) we shall refer to the Mecanique 

 Analytique of Lagrange, and the Mecanique Celeste of La Place, 

 especially to the first volume of the former work. 



Art. YL — Solution of a Functional Equation, which has been 

 employed by Poisson in demonstrating the parallelogram of 

 forces ; by George R. Perkins, A. M. 



PoissoN, in his able Traite de Mecanique, (see second edition, 

 Yol. I, p. 44 et seq.) has given a beautiful demonstration of the 

 parallelogram of forces. He makes his demonstration rest upon 

 the determination of (px, so as to satisfy the condition, 

 cpx(pz—(p{x-^z)-\-(f{x—z) (1). 



He says, cpx—2.CQS. ax, will satisfy (1), and he further says, 

 that no other value of (px will satisfy it ; but he does not show 

 how he determined this value of (px ; but seems to have obtained 

 it by induction. Neither does he show why there may not be 

 other values of <px, which will satisfy (1). 



