THE MECHANICAL POWERS. 285 



RULE. Multiply the weight by the radius of the axle 

 and divide the product by the radius of the wheel. 



EXAMPLE. Required, the tractile force necessary to draw 

 2000 Ibs. on a wheel of 2-J- feet radius, and axle of 3 inches 

 radius ? 



SOLUTION. 2000 Ibs., weight, x 3 inches, radius of 

 axle, = 60004-30 inches, radius of wheel, = 200 Ibs. Ans. 



To find the radius required for a wheel to move a given 

 weight by a given force on a given radius of axle. 



RULE. Multiply the weight by the radius of the axle 

 and divide the product by the force. 



EXAMPLE. What radius must a wheel have, the radius 

 of whose axle is 4 inches, to move a weight of 1320 Ibs. by a 

 force of 220 Ibs ? 



SOLUTION. 1320 Ibs., weight, x 4 inches, radius of axle, 

 = 5280 -h220 Ibs., tractile force, = 24 inches. Ans. 



To find the radius of an axle required to move a given 

 weight ~by a given force, on a wheel of a given radius. 



RULE. Multiply the force by the radius of the wheel and 

 divide the product by the weight. 



EXAMPLE. A weight of 1200 Ibs. is to be moved on a 

 wheel of 4 feet radius by a force of 150 Ibs. : What must be 

 the radius of the axle ? 



SOLUTION. 150 Ibs., force, x 48 inches, radius of wheel, = 

 7200-^-1200 Ibs. =6 inches. Ans. 



