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Question

Difficulty: HardLoad Distribution and Balancing

A single-axle utility trailer has its axle located 88 feet behind the hitch. The empty trailer weighs 600600 pounds, and its center of gravity is 55 feet behind the hitch. A machine weighing 1,2001,200 pounds is loaded onto the trailer so that its center of gravity is 66 feet behind the hitch. When the trailer is parked horizontally, what is the downward force, in pounds, exerted on the hitch?

Answer: 525 pounds

Answer

525
The correct answer is derived by treating the trailer as a lever in static equilibrium. Taking the sum of torques around the axle, the total rotational force from the empty trailer and the loaded machine is 42004200 lb-ft. Dividing this by the 88-foot lever arm to the hitch yields exactly 525525 pounds of force.

Step-by-Step Solution

1
Identify a pivot point and determine all relevant distances.
Using the axle as the pivot point. Distance from axle to trailer CG is 3 ft (8−58 - 5). Distance from axle to machine CG is 2 ft (8−68 - 6). Distance from axle to hitch is 8 ft.
Establishing a pivot point allows us to use the principle of moments (torques) to find the unknown reaction force.
2
Calculate the total counter-clockwise torque around the axle.
Trailer torque = 600 lbs×3 ft=1800 lb-ft600 \text{ lbs} \times 3 \text{ ft} = 1800 \text{ lb-ft}. Machine torque = 1200 lbs×2 ft=2400 lb-ft1200 \text{ lbs} \times 2 \text{ ft} = 2400 \text{ lb-ft}. Total torque = 4200 lb-ft4200 \text{ lb-ft}.
The weight of the trailer and the machine both push down, attempting to rotate the system around the axle.
3
Set up the static equilibrium equation.
4200 lb-ft=Fhitch×8 ft4200 \text{ lb-ft} = F_{hitch} \times 8 \text{ ft}.
For the trailer to remain stationary, the clockwise torque provided by the hitch must perfectly balance the total counter-clockwise torque.
4
Solve for the force at the hitch (FhitchF_{hitch}).
Fhitch=42008=525 lbsF_{hitch} = \frac{4200}{8} = 525 \text{ lbs}.
Dividing the total torque by the distance to the hitch gives the required balancing force.

Key Concept

Static Equilibrium and Torque Distribution

Alternative Method

You can also solve this by setting the hitch as the pivot point. The downward torque would be (600×5)+(1200×6)=3000+7200=10200(600 \times 5) + (1200 \times 6) = 3000 + 7200 = 10200 lb-ft. The upward force from the axle balances this: Faxle×8=10200F_{axle} \times 8 = 10200, so Faxle=1275F_{axle} = 1275 lbs. Since the total weight is 18001800 lbs (600+1200600 + 1200), the remaining force supported by the hitch is 1800−1275=5251800 - 1275 = 525 lbs.
Estimated Time:2m 0s
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Topics

Load Distribution and Balancing

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