A uniform 10-foot steel pry bar weighing 40 pounds is used to lift a rock. A fulcrum is pl… — Armed Services Vocational Aptitude Battery (ASVAB)

A uniform $10$ -foot steel pry bar weighing $40$ pounds is used to lift a rock. A fulcrum is placed exactly $2$ feet from the load end of the bar. A $480$ -pound downward force from the rock acts exactly at the load end. How much downward effort force must be applied at the opposite end of the pry bar to keep it perfectly balanced and horizontal?

A
84 lbs
105 lbsAnswer
C
120 lbs
D
135 lbs

Answer

105 lbs

The system is balanced when clockwise and counter-clockwise torques are equal. The rock creates

960 \text{ lb-ft}

of counter-clockwise torque (

480 \times 2

). The bar's weight creates

120 \text{ lb-ft}

of clockwise torque (

40 \times 3

, since its center is

3 \text{ ft}

from the fulcrum). The remaining clockwise torque must be provided by the effort force applied at

8 \text{ ft}

from the fulcrum. Therefore,

960 = 120 + 8E

, which solves to

E = 105 \text{ lbs}

Step-by-Step Solution

Calculate the counter-clockwise torque created by the rock.

480 \text{ lbs} \times 2 \text{ ft} = 960 \text{ lb-ft}

Torque is force multiplied by the distance to the fulcrum. The rock acts

2

feet from the fulcrum.

Determine the location of the pry bar's center of gravity relative to the fulcrum.

The center is

5 \text{ ft}

from either end. The fulcrum is at

2 \text{ ft}

. The distance from the fulcrum to the center of gravity is

5 - 2 = 3 \text{ ft}

on the effort side.

A uniform bar's weight acts at its exact center point.

Calculate the clockwise torque created by the pry bar's own weight.

40 \text{ lbs} \times 3 \text{ ft} = 120 \text{ lb-ft}

The bar's weight assists the effort force because its center of gravity is on the same side of the fulcrum as the applied effort.

Set up the torque balance equation to find the required effort force.

960 \text{ lb-ft} = 120 \text{ lb-ft} + (E \times 8 \text{ ft})

For the lever to be balanced, total counter-clockwise torque must equal total clockwise torque. The effort is applied

10 - 2 = 8 \text{ ft}

from the fulcrum.

Solve for the effort force (E).

8E = 960 - 120 \Rightarrow 8E = 840 \Rightarrow E = 105 \text{ lbs}

Isolating E gives the exact downward force required at the end of the bar.

Key Concept

To balance a lever, the sum of all clockwise torques must equal the sum of all counter-clockwise torques. For heavy levers, the object's own weight acting at its center of gravity must be included in the calculation.

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