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Difficulty: HardLoad Distribution and Balancing

A uniform wooden beam is 1010 feet long and weighs 5050 pounds. It rests horizontally on two sawhorses. Sawhorse A is placed 44 feet from the left end, and Sawhorse B is placed 99 feet from the left end. A 5050-pound bag of cement is placed at the extreme right end of the beam (1010 feet from the left end). Assuming the system is in static equilibrium, what is the upward reaction force exerted by Sawhorse B?

  1. A
    3030 pounds
  2. B
    5050 pounds
  3. C
    6060 pounds
  4. 7070 poundsAnswer

Answer

The upward reaction force exerted by Sawhorse B is 7070 pounds.
For the system to be in static equilibrium, the moments around any pivot point must sum to zero. Choosing Sawhorse A as the pivot: the beam's center of gravity (55 ft mark) is 11 ft right of the pivot, creating 50×1=5050 \times 1 = 50 ft-lbs of clockwise torque. The bag (1010 ft mark) is 66 ft right of the pivot, creating 50×6=30050 \times 6 = 300 ft-lbs of clockwise torque. To balance the total 350350 ft-lbs of clockwise torque, Sawhorse B (located 55 ft right of the pivot) must provide 350/5=70350 / 5 = 70 pounds of upward force.

Step-by-Step Solution

1
Identify the pivot point and the locations of all downward forces.
Using Sawhorse A (at 44 ft) as the pivot. The beam's weight (5050 lbs) acts at its center (55 ft). The cement bag (5050 lbs) acts at the right end (1010 ft).
Choosing one support as the pivot eliminates its unknown force from the equilibrium equation.
2
Calculate the lever arm distances from the pivot for each force.
Beam weight lever arm: 5−4=15 - 4 = 1 ft. Cement bag lever arm: 10−4=610 - 4 = 6 ft. Sawhorse B lever arm: 9−4=59 - 4 = 5 ft.
Torque is determined by multiplying a force by its perpendicular distance from the pivot point.
3
Calculate the total clockwise torque acting on the beam.
Beam torque: 50 lbs×1 ft=50 ft-lbs50 \text{ lbs} \times 1 \text{ ft} = 50 \text{ ft-lbs}. Cement bag torque: 50 lbs×6 ft=300 ft-lbs50 \text{ lbs} \times 6 \text{ ft} = 300 \text{ ft-lbs}. Total clockwise torque = 350 ft-lbs350 \text{ ft-lbs}.
Both the beam's weight and the cement bag push downward on the right side of the pivot, creating clockwise rotation.
4
Set the clockwise torque equal to the counterclockwise torque and solve for Sawhorse B's force.
FB×5 ft=350 ft-lbsF_B \times 5 \text{ ft} = 350 \text{ ft-lbs}, which means FB=350/5=70F_B = 350 / 5 = 70 pounds.
In static equilibrium, the torques must perfectly balance to prevent the beam from rotating.

Key Concept

Rotational Equilibrium and Center of Gravity

Alternative Method

You can verify this by calculating moments around Sawhorse B instead. The beam's center (55 ft) is 44 ft to the left (CCW torque: 50×4=20050 \times 4 = 200). The cement bag (1010 ft) is 11 ft to the right (CW torque: 50×1=5050 \times 1 = 50). Sawhorse A is 55 ft to the left (CW torque: FA×5F_A \times 5). Balancing torques: 200=50+5(FA)→FA=30200 = 50 + 5(F_A) \rightarrow F_A = 30 lbs. Since total downward force is 100100 lbs, FBF_B must be 100−30=70100 - 30 = 70 lbs.
Estimated Time:2m 0s
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Topics

Load Distribution and Balancing

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