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Difficulty: HardLevers and Fulcrums

A worker uses a uniform 6060-inch steel pry bar to lift a 150 lb150\text{ lb} concrete block. The worker places a pivot exactly 2020 inches from the load end of the bar, configuring it as a first-class lever. If the pry bar itself weighs 20 lbs20\text{ lbs}, how much downward vertical force must the worker apply at the opposite end of the bar to perfectly balance the concrete block?

  1. A
    75 lbs75\text{ lbs}
  2. B
    300 lbs300\text{ lbs}
  3. 70 lbs70\text{ lbs}Answer
  4. D
    80 lbs80\text{ lbs}

Answer

The correct force required is 70 lbs70\text{ lbs}.
To achieve equilibrium, the torques around the pivot must balance. The concrete block creates a counter-clockwise torque of 3000 in-lbs3000\text{ in-lbs} (150 lbs×20 inches150\text{ lbs} \times 20\text{ inches}). The pry bar's weight acts at its center (3030 inches from the end), which places it 1010 inches from the pivot on the effort side. This provides an assisting clockwise torque of 200 in-lbs200\text{ in-lbs} (20 lbs×10 inches20\text{ lbs} \times 10\text{ inches}). The effort force is applied at the very end, 4040 inches from the pivot. The equilibrium equation is 3000=200+(F×40)3000 = 200 + (F \times 40), resolving to 2800=F×402800 = F \times 40, which means the required force is 70 lbs70\text{ lbs}.

Step-by-Step Solution

1
Calculate the counter-clockwise torque generated by the concrete block.
150 lbs×20 inches=3000 in-lbs150\text{ lbs} \times 20\text{ inches} = 3000\text{ in-lbs}.
The load is placed 2020 inches from the pivot, creating the primary rotational force to overcome.
2
Determine the location of the pry bar's center of gravity and its resulting torque.
The center of the 6060-inch bar is at 3030 inches. Since the pivot is at 2020 inches, the center of gravity is 1010 inches from the pivot on the effort side. Torque = 20 lbs×10 inches=200 in-lbs20\text{ lbs} \times 10\text{ inches} = 200\text{ in-lbs} (clockwise).
A uniform bar's weight acts at its exact center. Because this center is on the same side of the pivot as the worker's effort, it provides an assisting clockwise torque.
3
Set up the torque equilibrium equation to find the required effort force.
Counter-clockwise Torque = Clockwise Torque
3000=200+(F×40)3000 = 200 + (F \times 40)
For the lever to balance, the sum of opposing torques must be equal. The effort arm is the total length minus the load arm (60−20=4060 - 20 = 40 inches).
4
Solve the equation for the effort force (FF).
2800=F×402800 = F \times 40, which simplifies to F=70 lbsF = 70\text{ lbs}.
Isolating FF gives the exact supplemental force the worker needs to apply to achieve balance.

Key Concept

Torque Equilibrium Involving Lever Mass
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Topics

Levers and Fulcrums

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