A 60\text{-pound} crate is placed on one end of a first-class lever, exactly 4\text{ feet}… — Armed Services Vocational Aptitude Battery (ASVAB)

A $60\text{-pound}$ crate is placed on one end of a first-class lever, exactly $4\text{ feet}$ from the fulcrum. What downward effort force must be applied at a distance of $8\text{ feet}$ on the opposite side of the fulcrum to perfectly balance the lever?

Answer: 30 lbs

Answer

30\text{ lbs}

A lever is perfectly balanced when the turning force (torque) on one side of the fulcrum equals the torque on the other side. Torque is calculated by multiplying the force by its distance from the fulcrum. The crate creates a torque of

240\text{ ft-lbs}

(

60\text{ lbs} \times 4\text{ ft}

). Because the effort is being applied twice as far from the fulcrum (

8\text{ ft}

), it only needs to be half the force to generate the same

240\text{ ft-lbs}

of torque. Therefore,

240 \div 8 = 30\text{ lbs}

Step-by-Step Solution

Set up the torque balance equation.

F_1 \times d_1 = F_2 \times d_2

For a lever to be balanced, the torque (force times distance) on the left side must equal the torque on the right side.

Substitute the known values into the equation.

60 \times 4 = F_2 \times 8

The crate applies a

60\text{ lb}

force at a distance of

4\text{ ft}

, and the effort force is applied at a distance of

8\text{ ft}

Solve for the unknown effort force (

F_2

240 = 8 \times F_2 \rightarrow F_2 = 30

Dividing both sides of the equation by

8

isolates

F_2

, revealing the required balancing force.

Key Concept

Torque equilibrium and mechanical advantage in a first-class lever

Estimated Time:45s

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