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Difficulty: HardWheel and Axle

A 180 lb180 \text{ lb} load is suspended by a cable wrapped around a solid cylindrical axle. The cable hangs freely from the right side of the axle. A steering handwheel with a radius of 12 inches12 \text{ inches} is rigidly attached to this axle. The axle has a diameter of 8 inches8 \text{ inches}. To lift the load, a worker applies a tangential force to the outer edge of the handwheel. Viewed from the front, what is the required direction of rotation and the ideal effort force needed to lift the load at a constant speed?

  1. Counterclockwise; 60 lbs60 \text{ lbs}Answer
  2. B
    Clockwise; 60 lbs60 \text{ lbs}
  3. C
    Counterclockwise; 120 lbs120 \text{ lbs}
  4. D
    Counterclockwise; 540 lbs540 \text{ lbs}

Answer

Counterclockwise; 60 lbs60 \text{ lbs}
The correct answer accurately determines that raising a load from a cable hanging off the right side of a cylinder requires counterclockwise rotation. Furthermore, it accurately calculates the mechanical advantage by dividing the handwheel radius (12 inches12 \text{ inches}) by the axle radius (4 inches4 \text{ inches}), yielding an IMA of 33. Dividing the 180 lb180 \text{ lb} load by this IMA gives a required effort of 60 lbs60 \text{ lbs}.

Step-by-Step Solution

1
Determine the required direction of rotation.
Counterclockwise rotation.
Since the cable hangs freely from the right side of the axle, pulling the right side upwards (winding the cable) requires turning the axle to the left, which is a counterclockwise direction.
2
Determine the working radii of the wheel and the axle.
Wheel radius (RR) = 12 inches12 \text{ inches}; Axle radius (rr) = 4 inches4 \text{ inches}.
The handwheel's radius is given as 12 inches12 \text{ inches}. The axle's diameter is 8 inches8 \text{ inches}, so its radius is half of that (8/2=4 inches8 / 2 = 4 \text{ inches}).
3
Calculate the Ideal Mechanical Advantage (IMA).
IMA=3\text{IMA} = 3
For a wheel and axle system where effort is applied to the wheel, IMA=R/r=12/4=3\text{IMA} = R / r = 12 / 4 = 3.
4
Calculate the required ideal effort force.
Effort=60 lbs\text{Effort} = 60 \text{ lbs}
Effort=Load/IMA=180/3=60 lbs\text{Effort} = \text{Load} / \text{IMA} = 180 / 3 = 60 \text{ lbs}.

Key Concept

Wheel and Axle Mechanical Advantage and Rotational Kinematics
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Topics

Wheel and Axle

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