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Difficulty: MediumLevers and Fulcrums

A carpenter uses a claw hammer to pull a stubborn nail out of a board. The point where the hammer head rests on the board acts as the fulcrum. The distance from the fulcrum to the nail is 1.5 inches1.5\text{ inches}, and the carpenter grips the handle 15 inches15\text{ inches} away from the fulcrum. If the carpenter applies a 25-lb25\text{-lb} pull at the handle, what is the maximum nail resistance that can be overcome, and what class of lever is being used?

  1. A
    275 lbs275\text{ lbs}; second-class lever
  2. B
    250 lbs250\text{ lbs}; second-class lever
  3. 250 lbs250\text{ lbs}; first-class leverAnswer
  4. D
    2.5 lbs2.5\text{ lbs}; third-class lever

Answer

The maximum resistance is 250 lbs250\text{ lbs}, and the system acts as a first-class lever.
The correct answer accurately identifies the claw hammer as a first-class lever because the fulcrum is located between the applied effort and the load. Applying the principle of moments (torque balance), the effort force multiplied by the effort arm (25×15=37525 \times 15 = 375) must equal the load force multiplied by the load arm. Dividing 375375 by 1.51.5 yields a maximum resistance of 250 lbs250\text{ lbs}.

Step-by-Step Solution

1
Determine the relative positions of the fulcrum, effort, and load to identify the lever class.
The fulcrum (where the hammer rests on the board) is located between the effort (the hand pulling the handle) and the load (the nail). This arrangement defines a first-class lever.
Knowing the correct lever class confirms how the forces interact around the pivot point.
2
Set up the torque balance equation using the principle of moments.
Effort Force×Effort Arm=Load Force×Load Arm\text{Effort Force} \times \text{Effort Arm} = \text{Load Force} \times \text{Load Arm}
For the lever to overcome the nail's resistance, the torque applied by the carpenter must at least equal the opposing torque of the nail.
3
Substitute the given values into the equation and solve for the load force.
25×15=Load Force×1.5→375=Load Force×1.5→Load Force=375/1.5=250 lbs25 \times 15 = \text{Load Force} \times 1.5 \rightarrow 375 = \text{Load Force} \times 1.5 \rightarrow \text{Load Force} = 375 / 1.5 = 250\text{ lbs}.
This calculation yields the maximum resistance force the lever can lift under the applied effort.

Key Concept

Torque balance and lever classification
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Topics

Levers and Fulcrums

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