100 



STATICS OF RIGID BODIES 



FIG. 53 



to Oy. Let us suppose that its distance from Oy is b, that of Q 

 being denoted by a. Then taking moments about we have 



(P+ Q)b = Qa, 



, , . I a a I 



so that = = > 



Q P+Q P 



showing that the line of action 

 divides the distance between P and 

 Q in the ratio Q : P. 



Thus we have shown that the 

 resultant of two parallel forces P, 

 Q is a force of magnitude P + Q, 

 parallel to these forces, of which 

 the line of action divides the dis- 

 tance between the lines of action of 

 P and Q, in the ratio Q : P. 



75. Alternative treatment of parallel forces. We can prove directly from 

 68 that parallel forces P, Q will be in equilibrium with a force 



(P + Q) parallel to them and acting along a line which divides the dis- 

 tance between them in the ratio Q : P. 



Take two points A, B on the line of action of P, Q, and a third point 

 C not on the line AB. Then the body, acted on by the forces P, Q, and 



(P + Q), can be kept in equilibrium by the further application of 

 (a) a force JR c at C, perpendicular to ABC ; 



(&) a force R B at B, perpendicular to AB ; 



(c) a force R A at A. 



Thus the system of forces 



P, Q, -(P + Q), R c , R B , R A 

 will be in equilibrium. 



Taking moments about the line AB, we find 

 that R c = 0. Taking moments about A , we find 

 that R B =Q, or else that it acts along BA, in 

 which case it can be absorbed into R A . Resolv- 

 ing perpendicular to the plane of the forces P, Q, 

 we find that R A can have no component perpen- 

 dicular to the plane. Thus the four remaining 

 forces 



P, Q, 

 are all in one plane. 



R 



At 



B 



Fia. 54 



