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Difficulty: HardLoad Distribution and Balancing

A uniform wooden plank is 1616 feet long and weighs 120120 pounds. It rests horizontally on two supports. The left support is positioned at the extreme left end of the plank, and the right support is positioned 1212 feet away from the left end. If a 6060-pound weight is placed on the extreme right end of the plank, how much upward force must the right support provide to maintain equilibrium?

  1. A
    9090 pounds
  2. B
    120120 pounds
  3. 160160 poundsAnswer
  4. D
    180180 pounds

Answer

160160 pounds
The system is in equilibrium, meaning the torques must balance. Taking the left support as the pivot, the downward forces create a clockwise torque: the plank's weight (120120 lbs) acts at its center (88 ft away) creating 960 lb-ft960 \text{ lb-ft} of torque, and the box (6060 lbs) acts at the end (1616 ft away) creating another 960 lb-ft960 \text{ lb-ft}. The total downward torque is 1920 lb-ft1920 \text{ lb-ft}. The right support is located 1212 feet away and must push upward to create an equal counter-clockwise torque. Therefore, 12×Force=192012 \times \text{Force} = 1920, which means the required force is 160160 pounds.

Step-by-Step Solution

1
Set the left support as the pivot point (fulcrum).
All distances for torque calculations will be measured from the 00-foot mark.
Choosing one of the unknown supports as the pivot eliminates its force from the torque equation, allowing us to solve directly for the other support.
2
Calculate the total clockwise torque acting on the plank.
Plank torque = 120 lbs×8 ft=960 lb-ft120 \text{ lbs} \times 8 \text{ ft} = 960 \text{ lb-ft}. Weight torque = 60 lbs×16 ft=960 lb-ft60 \text{ lbs} \times 16 \text{ ft} = 960 \text{ lb-ft}. Total clockwise torque = 1920 lb-ft1920 \text{ lb-ft}.
The uniform plank's weight acts at its center of gravity (88 feet from the left), and the external weight acts at the extreme right end (1616 feet from the left). Both create a downward, clockwise rotation around the left support.
3
Equate the clockwise torque to the counter-clockwise torque to solve for the right support's force.
Fright×12 ft=1920 lb-ftF_{right} \times 12 \text{ ft} = 1920 \text{ lb-ft}. Dividing by 1212 gives Fright=160 lbsF_{right} = 160 \text{ lbs}.
For the system to remain in static equilibrium, the upward (counter-clockwise) torque provided by the right support must exactly balance the downward (clockwise) torques.

Key Concept

Static Equilibrium and Principle of Moments
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Topics

Load Distribution and Balancing

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