A wheel is being turned to open a valve on a compressor. The resistance that is applied to the axle of the valve is 235 lb. The diameter of the wheel that is being turned is 20in and an effort force of 45 lb is being applied to the wheel. What is the required diameter of the axle to overcome the resistance force

3.8 in

Explanation:

A wheel and axle machine is a machine in which the wheel rotates around the rod (axle). They rotate together thereby transferring force among each other.

The formula for a wheel and axle is:

Fr * Rr = Fe * Re

Where Fr is the resistance force, Rr is the radius of the resistance force, Fe is the effort force and Re is the radius of the resistance force.

Given that: Fr = 235 lb, Fe = 45 lb, Re = 20 in / 2 = 10 in. Hence:

235 * Rr = 45 * 10

Rr = 45 * 10 / 235

Rr= 1.9 in

Diameter of axle = 2 * 1.9 = 3.8 in

answer: option c.

explanation:

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