VT PARALLEL FORCES AND CENTRE OF GRAVITY 79 



ForceC NY Force AD Force BE 



Distance A 1 > Distance BO Distance CA 

 Or, expressed in words, each force is proportional to the distance 

 the other tiro forces. 



EXAMPLE. A weight of 120 Ibs. is suspended from a light 

 rod (whose weight may be neglected), the ends of which rest 

 upon the shoulders of two men, A and B (Fig. 31). The rod is 6 



FIG. 31. Relation of Parallel Forces to one another. 



feet long, and the weight is 2 feet from the man A, and 4 feet 

 from the man B. What proportion of the weight is borne by 

 each man ? 



From the above we see that 



Amount borne , Amount borne _ B's distance . A's distance 

 by A by B ~~ from weight ' from weight 



= 4:2 = 2. 



So that the man A bears twice as much as the man B, that is, he 

 bears 80 Ibs. and B carries 40 Ibs. 



The Principle of Moments applied to Parallel Forces. It 

 will be evident that the preceding problem resolves itself into 

 one on the moments of forces. Suppose we consider the 

 shoulder of the man A to be a fulcrum about which the rod 

 may turn, then, since the rod is in equilibrium, the moments 

 of the forces tending to turn it are equal and opposite. We 

 may, therefore, write : 



B's exertion x distance from A = weight x distance from A. 

 Or, B's exertion x AB = 120 x 2. 

 That is, B's exertion x 6 = 240 



240 



Therefore, B's exertion = - = 40 Ibs 

 o 



