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Difficulty: MediumLoad Distribution and Balancing

A uniform steel bar is 1010 feet long and weighs 200200 pounds. It is placed on a pivot located exactly 22 feet from its left end. To keep the bar perfectly horizontal, a downward force must be applied to the extreme left end. What is the magnitude of this downward force?

  1. A
    100100 pounds
  2. B
    120120 pounds
  3. 300300 poundsAnswer
  4. D
    500500 pounds

Answer

The required downward force is 300300 pounds.
To maintain a horizontal position, the torques around the pivot must balance perfectly. Because the bar is uniform, its entire 200200-pound weight acts at its center of gravity, which is 55 feet from the end. The pivot is located 22 feet from the left end, meaning the center of gravity is exactly 33 feet to the right of the pivot (5−2=35 - 2 = 3). This creates a clockwise torque of 600600 lb-ft (200200 lbs ×\times 33 ft). To balance this rotation, a downward force applied at the left end (which is 22 feet from the pivot) must create an equal 600600 lb-ft counter-clockwise torque. Dividing the required 600600 lb-ft by the 22-foot distance gives a required downward force of 300300 pounds.

Step-by-Step Solution

1
Determine the location of the center of gravity (CG).
The CG is exactly in the middle of the 1010-foot bar, which is 55 feet from either end.
A uniform object's weight is evenly distributed, so its total weight acts geometrically at its center.
2
Calculate the distances from the pivot point.
The left end is 22 feet from the pivot. The CG is 33 feet from the pivot (55 feet - 22 feet).
Torque is calculated using the distance from the exact pivot (fulcrum) to the point where the force acts.
3
Calculate the clockwise torque created by the bar's weight.
200200 lbs ×\times 33 ft = 600600 lb-ft.
The weight of the bar pulls down on the right side of the pivot, creating a clockwise rotational tendency.
4
Calculate the required counter-clockwise force at the left end.
Force ×\times 22 ft = 600600 lb-ft, which means Force = 300300 lbs.
For the bar to remain horizontal (in equilibrium), the counter-clockwise torque must exactly balance the clockwise torque.

Key Concept

Principle of Moments (Torque Equilibrium)
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Topics

Load Distribution and Balancing

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