ANGULAR CRANK. 



by the length of the line M N. Let L N be 

 horizontal, and the angle L N M therefore a 

 right angle. 



Now, the force applied vertically at M is not 

 applied in the direction of turning, which is the 

 tangential direction M O. The whole force 

 represented by the length of M N is therefore 

 not employed in turning. 



To find how much is so employed, complete 

 the parallelogram P 0, whose diagonal is M N, 

 and one of whose sides, M O, is in the tangential 

 direction that is, the direction of turning at 

 the point M, and therefore at right angles to 

 L M. Then, by the parallelogram of forces, it 

 is seen that the actual force applied by the 

 foot, of magnitude and direction M N, is com- 

 pounded at the point M of two forces, one of 

 direction and magnitude M P, wasted in com- 

 pressing the crank L M, and the other of mag- 

 nitude and direction M O. This latter, there- 

 fore, is the portion of the foot's pressure which 

 is employed at M in turning the crank, with the 

 advantage for turning of the longer leverage. 



This, of course, gives as much turning effect 

 as a greater force employed with the shorter 

 leverage at the point N, How much greater 

 would the latter have to be to produce the same 

 turning effect as the force M O applied at M ? 

 Obviously, as much greater as the lever L N is 

 less than L M. Since the triangles are right- 

 angled, and have one of their acute angles in 



109 



