16k t '6f 



LEVERAGE 27 



29. Moment of Forces. All problems in leverage may 

 be solved by arithmetic and without using a model. 



Suppose that two weights are balanced as in Fig. 9 at the 

 distances shown therein. u ______ u . ___ t|j __ a . __ ^ 



As 11 times 18 equals 12 Pt j^ 



times I>y 2 (198) it fol- 



lows that the weight of one 



side times its distance from FlG . 9. The Moment of Forces. 



the fulcrum is equal to the 



weight on the other side times its distance from the fulcrum. 



W X D = W X D' 



This rule always holds true for all classes of levers. If, 

 therefore, the amount of both weights and one distance are 

 known, the other distance can always be found; or if any 

 three of the four quantities are known, the fourth can always 

 be found. As an example, if we know all but the 16J/2 M )8 - 

 in Fig. 9 we can find this figure in the following way: 



18 X 11 



= Ifii^ Ibs. 

 12 



In all classes of levers the weight or force times its perpen- 

 dicular distance from the fulcrum is called the moment. 



Thus in the above problem, 12 X 16^ is one moment and 

 18 X 11 the other. As another example: What force will balance 

 a weight of 100 Ibs., 12 in. from a fulcrum located at the short end 

 of a lever? The long end of the lever is 24 in. in length. 



100 X 12 = moment of acting force 

 W X 24 = moment of resisting force 



But, when a lever is balanced, the moments of forces are equal, 

 according to the rule explained above. 



