i io.] PARALLEL FORCES. 63 



A'l and IV B'. The intersection of these lines, i.e. of the first and last 

 sides of the funicular polygon, gives a point, R, of the resultant of 

 W, W* W\, W* 



Moreover, if the component in A'l be resolved along A 1 1? and the 

 vertical through A 1 , and similarly the component in J3 1 IV along jB'A' 

 and the vertical through J3 1 , the two components along A'' will be 

 equal and opposite, each being equal to the parallel Oo drawn to A'B l 

 in the force polygon. This parallel furnishes, therefore, the magnitudes 

 of the reactions ^ = 01,^ = 50. 



109. Analytically, the resultant of n parallel forces F v F 2 ,.. 

 ... F n , whether in the same plane or not, can be found as follows : 



The resultant of F l and F z is a force F l -{-F 2 situated in the 

 plane (F lt F 2 ), so that F l p l = F 2 ^ 2 (Art. 106), where / x ,/ 2 are the 

 (perpendicular or oblique) distances of the resultant from F 1 

 and Fy respectively. This resultant F l -{-F 2 can now be com- 

 bined with F B to form a resultant F^ F 2 -\- F 3 , whose distances 

 from FI + FI and F B in the plane determined by these two forces 

 are as F s is to /^-f/^. This process can be continued until all. 

 forces have been combined ; the final resultant is 



/5 + F 2 -K,.. +F n . 



Any number of parallel forces are, therefore, in general equiva- 

 lent to a single resultant equal to their algebraic sum. 



110. To find \hz position of this resultant, analytically, let the 

 points of application of the forces F lt F 2 , ... F n be (x^ y^ Z-L), 

 (x v y v ^2)' ( x y ^n)- T ne P omt ^ application of the result- 

 ant F^ + FZ of F^ and F 2 may be taken so as to divide the dis- 

 tance of the points of application of F l and F 2 in the ratio 

 F 2 :F l ; hence, denoting its co-ordinates by x\ y' , z f , we have 



x-x 1 or 



(F l -f- F 2 ) x 1 = 

 and similarly for y' and z l . 



