A traditional water well is operated by a hand crank that raises a bucket of water. The cr… — Armed Services Vocational Aptitude Battery (ASVAB)

A traditional water well is operated by a hand crank that raises a bucket of water. The crank handle has a turning radius of $12 \text{ inches}$ , and it drives a wooden cylinder (the axle) that has a diameter of $8 \text{ inches}$ . If the total weight of the water and bucket is $45 \text{ pounds}$ , what is the ideal effort force required on the handle to lift the bucket at a constant speed?

$15 \text{ pounds}$ Answer
B
$22.5 \text{ pounds}$
C
$30 \text{ pounds}$
D
$135 \text{ pounds}$

Answer

The ideal effort force required is

15 \text{ pounds}

The system operates as a force multiplier because the effort is applied to the larger radius (the crank handle). First, find the axle's radius by dividing its

8 \text{ inch}

diameter by 2, yielding

4 \text{ inches}

. The mechanical advantage is the ratio of the effort radius to the load radius:

12 \text{ inches} / 4 \text{ inches} = 3

. This means the crank multiplies the applied force by 3. To lift a

45\text{-pound}

load, divide the load by the mechanical advantage (

45 / 3

), which results in a required effort force of

15 \text{ pounds}

Step-by-Step Solution

Determine the radius of the axle.

The radius of the axle (

r

) is

4 \text{ inches}

The diameter is given as

8 \text{ inches}

, and radius is half of the diameter (

8 / 2 = 4

Calculate the Ideal Mechanical Advantage (IMA) of the wheel and axle system.

The IMA is

3

The IMA for a wheel and axle where the effort is applied to the wheel is the radius of the wheel divided by the radius of the axle (

IMA = R / r = 12 / 4 = 3

Calculate the required ideal effort force.

The effort force is

15 \text{ pounds}

Effort force equals the load divided by the IMA (

45 / 3 = 15

). This means the crank multiplies the user's force by 3.

Key Concept

Ideal Mechanical Advantage in Wheel and Axle Systems

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