A warehouse worker uses a first-class lever to lift a machinery part weighing 200\text{ lb… — Armed Services Vocational Aptitude Battery (ASVAB)

A warehouse worker uses a first-class lever to lift a machinery part weighing $200\text{ lbs}$ . The fulcrum is placed exactly $2\text{ feet}$ from the machinery part. If the worker applies a downward effort force at a distance of $8\text{ feet}$ from the fulcrum on the opposite side, how much effort force is required to balance the load? (Assume the lever itself is weightless.)

Answer: 50 lbs

Answer

A first-class lever is balanced when the 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 machinery part creates a torque of

200\text{ lbs} \times 2\text{ ft} = 400\text{ ft-lbs}

. To balance this, the effort force applied at

8\text{ ft}

must also create

400\text{ ft-lbs}

of torque. Therefore, the required effort force is

\frac{400}{8} = 50\text{ lbs}

Step-by-Step Solution

Set up the torque balance equation for a lever in equilibrium.

\text{Load} \times \text{Load Arm} = \text{Effort Force} \times \text{Effort Arm}

For a lever to balance, the rotational forces (torques) on both sides of the fulcrum must be equal.

Substitute the given values into the equation.

200 \times 2 = \text{Effort Force} \times 8

The load is

200\text{ lbs}

located

2\text{ ft}

from the fulcrum, and the effort is applied

8\text{ ft}

from the fulcrum.

Calculate the unknown effort force.

400 = \text{Effort Force} \times 8 \implies \text{Effort Force} = \frac{400}{8} = 50

Dividing the total load torque by the effort arm distance gives the required effort force.

Key Concept

The principle of moments (torque balance) dictates that a lever is in equilibrium when the clockwise torque equals the counterclockwise torque.

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