FACTIONS FOR A SINGLE FORCE. 57 



being of equal length, the resultant of the two loads evidently 

 acts at d and is 2 P, so that 



= 2P (check). 



Graphically the solution would bo exactly like that of Fig. 

 32, using the resultant 2P acting at d. 



Let Fig. 34 represent a ladder 

 supported by a wall at B. If the 

 friction between the wall and the 

 ladder be neglected, R 2 will be 

 horizontal and the line of R^ must 

 pass through C because R 2 , P, and R i 

 are in equilibrium. The magnitudes 

 of R l and R 2 may be found from a force triangle. 



This may also be solved by passing a string polygon through 

 the one known point A of R : as shown in Fig. 35. 



.1, j<- . . t f . . 



1_ .] 



Fig. 36. 



Fig, 35. 



The case of Fig. 36 is the same 

 as that of Fig. 31. If R 2 were hori- 

 zontal, it would be the same as that 

 of Fig. 34 and R would be in- 

 clined. 



Algebraically, Fig. 34, from 

 5 vert, comps. = 0, R 1 cos ft = P, 

 and from 2 hor. comps. = 0, 

 R t sin fi-R 2 , and from 2 moms. 

 = with center at A, R 2 XLtau a 



