Composition and Decomposition of Forces. 35 



66. We may thus derive a very simple method of obtaining 

 the position and magnitude of the resultant of any number of 

 forces, when they all act in the same plane. 



Let us suppose, in the first place, that they are all parallel ; 

 and, not to make the problem more complicated than is neces- 

 sary, let us suppose that there are only three forces ; it will be 

 easily inferred how we are to proceed in case of a greater 

 number. 



Accordingly, let there be the three known forces p, 9, r, the Fig. 21. 

 two first being directed the same way, and the third having a 

 contrary direction. Having drawn arbitrarily any line FABC, 

 perpendicularly to the directions Ap, B q, &c., we will suppose 

 that D is the point through which the resultant g is to pass. 

 Then, having taken at pleasure a point F in FABC, we shall have, 

 according to what has been demonstrated, 63 



p X FA + q x FB r x FC = p X FD. 

 Now the distances FA, FB, FC, and the forces p, q, r, being 

 known, it will be easy to deduce from the above equation, the 

 value of the distance FD, through which the resultant would 

 pass, if the value of this resultant p were known. In order to 

 find it, we take another point (p in AF produced, and by proceed- 

 ing as above, we have 



p X (pA + q X <f>B r X <pC= Q X <pD. 



If from this second equation we subtract the first, recollecting 

 that 



9>A FA = tpF, (pB FB = <pF, 



we shall have 



p X <pF + q X <pF r X <pF = p X 0>F; 

 that is, the whole being divided by (pF, 

 P + q r = Q. 



If we examine the process now pursued, we shall see that it 

 does not depend in any degree upon the number of forces, but 

 that it is applicable, whatever this number may be. We must infer 

 therefore, that the resultant of any number of parallel forces is 

 equal to the sum of those which act in one direction minus the 

 sum of those which act in the opposite direction. 



