80 ELEMENTARY GENERAL SCIENCE CHAP, 



The force exerted by A may be found in a similar way by 

 taking the moments about the shoulder of B considered as a 

 fulcrum. In this case we have 



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

 Or, A's exertion x 6 = 120 x 4. 



Therefore, A's exertion = - ~ = 80 Ibs. 



We may also consider the moments of forces acting at the 

 point C, from which the weight is suspended. Then we have 



A's exertion x AC = B's exertion x BC 



Or, A's exertion x 2 = B's exertion x 4 



B's exertion x 4 

 Therefore, As exertion = - 



2 

 = B's exertion x 2 



If B's exertion is denoted by jc, then A's exertion is expressed 

 by 2x. The total weight supported is 120 Ibs. ; so that 



A's exertion + B's exertion = 120 Ibs. 

 Or 2x + x = 120 ,, 

 That is 3x = 120 ,, 



120 



Therefore x = = 40 Ibs. 



= B's exertion. 



We have thus taken moments about three different points and 

 have found that they all give the same results. As a matter of 

 fact, it does not matter what point is considered as the fulcrum 

 when forces act upon a body in one plane, in the manner here 

 considered. Under these conditions the rule already given for 

 the moments of forces holds good, and the moments may be taken 

 about any point in the plane in which they act. As it is im- 



FIG. 32. To illustrate the Principle of Moments applied to Parallel Forces. 



portant that the student should thoroughly understand this, we 

 give another example : 



