CHAP, n.] DIFFERENT POINTS OF APPLICATION. 27 



and Si S\. But from the force polygon we see at once that the 

 resultant of S and S' is given in direction and intensity by 

 C C f , and this is also the resultant of S t and S\. The line join- 

 ing g and It, must therefore be parallel to C C'. For the second 

 force P 2 we can show similarly that the line joining k and I is 

 parallel to C C'. But k is a common point of both lines hence 

 g k and I lie in the same straight line parallel to C C'. 



[NOTE. The pure geometric proof is as follows : The two 

 complete quadrilaterals 1 C' C and g k a' a have jive pairs of 

 oomatponding sides parallel, viz., 1 and a a', 1 C' and a' k, 

 C and a g, o C' and a' g, 1 G' and a k hence the sixth pair 

 are also parallel, viz., rJ C' and g k. In like manner for 1 2 

 C C' and I kb' b and so on.~\ 



We can make use of this principle in order from one given 

 equilibrium polygon S a b c d S 4 and pole, to construct another, 

 the direction of C C' being known. For this purpose, having 

 assumed the position of the first string S' we draw through its 

 intersection g with S a line M N parallel to C C'. The next 

 string must therefore pass through the intersection a' of S' and 

 P! and through the point k, of intersection of the second string 

 of the first polygon and the line M N. It is therefore deter- 

 mined. The next side must pass through b' and I, and 

 so on. 



\JN~ote. Observe that the intersections r and r' of the first and 

 last lines of both polygons must lie in a straight line parallel to 

 4:, the direction of the resultant.] 



27. 9Iean polygon of equilibrium. Since the pole may 

 have any position, let us suppose it situated in one of the angles 

 of the force polygon. It is evident that the first line of the 

 corresponding equilibrium polygon, then coincides with the first 

 force. If now the pole be taken at the beginning of the first 

 force in the force polygon, then the first side of the correspond- 

 ing equilibrium polygon will coincide with the first force, and 

 the last line will be the resultant itself in proper position. 



Take for instance, the pole at o in the force polygon, Fig. 15 

 (#), PL 4. The first side S reduces to zero. The next S x coin- 

 cides with 1. In (b) therefore P x is the first side of the equi- 

 librium polygon. The next side S 2 corresponds with S 2 in (a). 

 Thus we obtain the polygon a b c d 0, the last side of which S 7 , 

 is the resultant itself. That is, S 2 is the resultant of P x and P.J, 

 S 3 of P w , S 4 of P 14 and so on. Every line in the polygon then 



