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Difficulty: MediumWheel and Axle

A worker uses a hand-operated winch to vertically lift a bundle of materials weighing 300 pounds300 \text{ pounds}. The winch is driven by a crank handle that has a length of 15 inches15 \text{ inches} from its center of rotation. The cable lifting the load wraps around a drum with a diameter of 5 inches5 \text{ inches}.

Assuming a 100%100\% efficient system, what is the theoretical effort force required on the crank handle to lift the bundle?

Answer: 50 pounds

Answer

50
The system operates as a wheel and axle where the crank handle is the wheel and the drum is the axle. By finding the ratio of their radii (15 inches15 \text{ inches} to 2.5 inches2.5 \text{ inches}), we determine that the mechanical advantage is 66. This means the effort force required is one-sixth of the total load, yielding 50 pounds50 \text{ pounds}.

Step-by-Step Solution

1
Determine the effective radius of both the wheel and the axle.
Wheel radius = 15 inches15 \text{ inches}. Axle radius = 5 inches÷2=2.5 inches5 \text{ inches} \div 2 = 2.5 \text{ inches}.
The formula for mechanical advantage requires consistent measurements, typically the radius of each component.
2
Calculate the Ideal Mechanical Advantage (IMA) of the system.
IMA=152.5=6IMA = \frac{15}{2.5} = 6.
The IMA indicates how many times the simple machine multiplies the input effort force.
3
Calculate the theoretical effort force required.
Feffort=3006=50 poundsF_{effort} = \frac{300}{6} = 50 \text{ pounds}.
In a frictionless (100%100\% efficient) system, the effort force is equal to the load force divided by the mechanical advantage.

Key Concept

Ideal Mechanical Advantage (IMA) of a Wheel and Axle
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Topics

Wheel and Axle

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