Art. 47. REACTIONS FOR ANY NUMBER OF LOADS. 



Fig. 41. 

 first method is the better. By the law of the lever 



the location of the resultant of 

 the loads thus reducing the prob- 

 lem to that of finding reactions 

 for a single load. 



In Fig. 41, the location of 

 the resultant of the three loads 

 is not apparent, therefore the 



2 = -JL. 20 + 11- 15 + 10 = 

 #1 + # 2 = 22f + 22 = 45 = 20 + 15 + 10; check. 

 Applying the equations of equilibrium directly, 

 20 7^ = 20X14+15X9+10X4 =-455 from which 



20 R 2 = 10X16+15X11+20X6=- 445 from which 



Graphically, the problem is to divide the equilibrant of the 

 three loads into two parts, R^ and E 2 , such that when acting at 

 A and B respectively, they will balance the loads. 



Fig. 42. 



The force polygon (Fig. 42) abcda gives the equilibrant da. 

 To find the point n, draw the equilibrium polygon DFHKED 

 and On parallel to ED. The ray 5 corresponds with the string 



D E -the closing line. The force polygon abcdna closes, and 



dn~R 2 , and na = R 1 . B lf 1, and 5 are in equilibrium at D, 

 also 72 2 , 5, and 4 at E. 



The student should prove that there is equilibrium as was 

 done in Art. 37. 



