A mechanic uses a manual screwdriver to remove a tightly secured machine screw. The handle… — Armed Services Vocational Aptitude Battery (ASVAB)

A mechanic uses a manual screwdriver to remove a tightly secured machine screw. The handle of the screwdriver has a diameter of $1.5 \text{ inches}$ , and the metal shaft has a diameter of $0.25 \text{ inches}$ . If the mechanic applies a rotational effort force of $12 \text{ pounds}$ to the outer edge of the handle, how much output force is exerted by the shaft onto the screw? (Assume $100\%$ efficiency.)

$72 \text{ pounds}$ Answer
B
$36 \text{ pounds}$
C
$18 \text{ pounds}$
D
$2 \text{ pounds}$

Answer

72 \text{ pounds}

In a wheel and axle system where effort is applied to the wheel (the handle), the force is multiplied by the ratio of the wheel's diameter to the axle's diameter. The ratio of

1.5 \text{ inches}

0.25 \text{ inches}

provides a mechanical advantage of

6

. Multiplying the

12 \text{ pounds}

of effort by

6

yields an output force of

72 \text{ pounds}

Step-by-Step Solution

Identify the components of the wheel and axle system.

The handle acts as the wheel (Diameter =

1.5 \text{ in}

), and the shaft acts as the axle (Diameter =

0.25 \text{ in}

Establishing which component is the wheel and which is the axle determines how force is multiplied.

Calculate the Ideal Mechanical Advantage (IMA).

IMA = \frac{1.5}{0.25} = 6

The IMA for a wheel and axle is found by dividing the diameter of the wheel by the diameter of the axle.

Calculate the total output force applied to the screw.

\text{Output Force} = \text{Input Force} \times IMA = 12 \text{ lbs} \times 6 = 72 \text{ lbs}

The system multiplies the input effort force by the mechanical advantage to produce the final output force.

Key Concept

Ideal Mechanical Advantage of a Wheel and Axle

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