A single-axle utility trailer has a total loaded weight of 2,000 pounds. The loaded traile… — Armed Services Vocational Aptitude Battery (ASVAB)

A single-axle utility trailer has a total loaded weight of $2,000$ pounds. The loaded trailer's center of gravity is located $3$ feet forward of the axle. If the trailer hitch is located $12$ feet forward of the axle, what is the downward force exerted on the hitch when the trailer is horizontally parked?

$500$ poundsAnswer
B
$1,000$ pounds
C
$1,500$ pounds
D
$8,000$ pounds

Answer

500

pounds

In static equilibrium, the sum of torques around any pivot point must be zero. By treating the axle as the pivot, the weight of the trailer (

2,000

pounds) acts at a distance of

3

feet from the pivot, creating a torque of

6,000

ft-lbs. To balance this, the upward force from the hitch (located

12

feet away) must equal

6,000

ft-lbs divided by

12

feet, which yields

500

pounds.

Step-by-Step Solution

Identify the pivot point and the forces acting on the system.

The axle acts as the pivot point (fulcrum). The two forces creating torque around the axle are the trailer's weight (

2,000

lbs) pushing down at the center of gravity, and the hitch support pushing up.

Setting the axle as the pivot eliminates its unknown support force from the initial torque equation.

Calculate the torque (moment) created by the trailer's weight around the axle.

Torque = Force × Distance =

2,000\text{ lbs} \times 3\text{ ft} = 6,000\text{ lb-ft}

This is the rotational force that the hitch must counterbalance to keep the trailer horizontal.

Set up the equilibrium equation to solve for the force at the hitch.

F_{\text{hitch}} \times 12\text{ ft} = 6,000\text{ lb-ft}

In static equilibrium, the clockwise torque must equal the counterclockwise torque.

Solve for the hitch force.

F_{\text{hitch}} = 6,000 \div 12 = 500\text{ pounds}

Dividing the required balancing torque by the lever arm distance gives the necessary support force.

Key Concept

Static Equilibrium and Torque (Moments)

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