NOTES ON MECHANICS. 



355 



components AB^, &c., of the resultant AD and the resultant in like 

 manner. Then, taking the components perpendicular to the plane of 

 the paper, we have 



CD = C^D^ + CJ)., + . . . . 

 .-. CD .d = C^D^.d + C^D^.d^ .. .. 

 Btit C^D^ . d = B^D^ . p^, &c. = &c. 



.-. BD.p = B,D,.p, + B,D,.p,+ .... 



But BD.p is the moment of AD around 00^. Therefore, &c. 



4. The sum of the moments of two parallel forces is equal to the 

 moment of their resultant around any line. 



Let OX be the given line, and ACB its projection on the plane 

 of the forces ; let the given forces P and Q and their resultant R act 

 at ^, ^, C, respectively. 



Resolve the forces P, .... into Pi, Ps, .... parallel and perpen- 

 dicular, respectively, to OX in the plane A OX, and P^, .... perpen- 

 dicular to A OX. Then, if P4, . . . . denote the resultant of P^, P3, . . . . ; 

 a, b, c,p, q, r, the distances of A, B, C, P4, Q^, R^, respectively, from 

 OX, we have 



P.OA + Q. OB = R.OC. 



Ps.OA + Qs.OB=.Rs.OC. 



Ps . a + Qs . b = Rs . c. 



Pi.p + Qi.q = Ri.r. 



But P^ . p is the moment of P around OX. Therefore, &c. 



5. The centre of parallel forces. 



Let OF be any line, and A^B^, A.jB^, .... the forces whose resultant 

 AB is their sum ; and let A^F^, A.2F2, .... be the distances of the 

 points where the forces act from OF. Drop FF perpendicular to AD. 

 Besolve AB into AD, DB along and perpendicular to OF, respec- 

 tively. Then, taking moments around OF, 



AD . FE = AiDi . F^E^ + 



But AD ■ A^D^ : A^D^ .... =AB : A^B^ : A^B^ .... 

 and FE : F^Ei : F^E^ ....=.AF: A^F^ : A^F^. . . . 



. • . AB . AF -^ A^Bi . A^F^ + 



,„ AiBi . A1F1+ . . . . 



• • ^^= AB ■ 



which is independent of the direction of the forces. Therefore, &c, 



November, 1873. 



