18 FORCES IN THE SAME PLANE. [CHAP. II. 



and b 1, C 1 ai\d 1 2, are parallel each to each, therefore the 

 sixth pair 2 and a b must also be parallel ; b is therefore a, 

 point of the resultant passing through a, parallel to 2.] 



17. The above Construction holds good equally well Tor 

 Parallel Forces. By means of it we find in PI. 3, Fig. 9 (a) 

 and (/;) and Fig. 10 (a) and (&), the resultant of a pair of paral- 

 lel forces, in the first case, both acting in the same direction ; 

 in the second, in opposite directions. 



In both cases we have simply to choose a pole C, and draw 

 S S t and S 2 . Then taking any point c in the line of direc- 

 tion of P 1? as a point of application for S , draw through this 

 point Si, thus finding d, the point of application for S 2 . S 

 and 83 prolonged, intersect upon the resultant, whose intensity, 

 direction, and position thus become fully known. 



18. Property of the Point b. It is plain that thus a point 

 of intersection b, through which the resultant must pass, can 

 always be found, provided S and S 2 do not fall together in 

 the force polygon, or intersect without the limits of the draw- 

 ing. By properly choosing the position of the pole C, this can 

 always be avoided if the points 2 and in the force -polygon do 

 not themselves coincide, i.e., if the force polygon does not close. 



The point 5, Figs. 8, 9, and 10, which by reason of the arbi- 

 trary position of the pole may lie anywhere upon the resultant, 

 has a remarkable property. If we draw a line m n through 

 this point parallel to S l5 and let fall from it perpendiculars p t 

 and p 2 upon P t and P 2 , then in all three cases, and therefore 

 generally, the triangle c in b is similar to C 1, and d b n is 

 similar to 1 C 2. Hence we have the proportions 



01:lCIIcm:ra&, and 

 1 C : 12 : : n b : n d. 



From these proportions we find 



01:12* *. cmxnbimbxnd. 



Now the triangles c m b and d n b have the same height 

 above the base m n ; the bases m b and b n are therefore pro- 

 portional to their areas. But their areas are equal to half their 

 sides cm and nd multiplied byj?i and^ 2 respectively. Hence 

 we have from the above proportion, since cm = nd, 



01:12 : : n d x p z ' n & x P\ or 

 01:12 ::p 2 :pi. 

 orP t : P 2 : \p*\p. 

 That is, the perpendiculars let fall from any point of the 



