A depot worker must slide a 1200\text{ lb} steel safe up a straight ramp to a storage plat… — Armed Services Vocational Aptitude Battery (ASVAB)

A depot worker must slide a $1200\text{ lb}$ steel safe up a straight ramp to a storage platform that is $3\text{ feet}$ above the main floor. The worker uses a mechanical winch that can pull with a maximum force of $240\text{ lbs}$ . Assuming a frictionless surface, what is the minimum ramp length required to move the safe?

Answer: 15 feet

Answer

The mechanical advantage required to move a

1200\text{ lb}

load using only

240\text{ lbs}

of force is

1200 / 240 = 5

. Because the Ideal Mechanical Advantage of an inclined plane is its length divided by its height, a ramp with an IMA of

5

lifting a load

3\text{ feet}

vertically must be

5 \times 3 = 15\text{ feet}

long.

Step-by-Step Solution

Calculate the required Ideal Mechanical Advantage (IMA).

IMA = 5

The mechanical advantage is the ratio of the load force to the maximum effort force (

1200 / 240 = 5

Set up the equation for the mechanical advantage of an inclined plane.

5 = Length / 3

For an inclined plane, IMA is equal to the length of the slope divided by the vertical height.

Solve for the ramp length.

Length = 15 feet

Multiplying the IMA by the height (

5 \times 3

) gives the minimum required ramp length to achieve that mechanical advantage.

Key Concept

Ideal Mechanical Advantage (IMA) of an Inclined Plane

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