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Difficulty: MediumLevers and Fulcrums

An automotive technician is inspecting a suspended brake pedal assembly in a vehicle. The pedal pivots on a stationary pin located at the very top. The master cylinder pushrod (which provides the load resistance) is connected to the pedal 3 inches3\text{ inches} directly below the pivot pin. The center of the driver's foot pad (where the effort force is applied) is located 15 inches15\text{ inches} directly below the pivot pin. If a driver applies exactly 60 lbs60\text{ lbs} of horizontal force to the foot pad, what is the resulting horizontal force applied to the pushrod, and what class of lever does this assembly represent?

  1. 300 lbs300\text{ lbs}; Second-class leverAnswer
  2. B
    300 lbs300\text{ lbs}; Third-class lever
  3. C
    240 lbs240\text{ lbs}; First-class lever
  4. D
    12 lbs12\text{ lbs}; Third-class lever

Answer

The assembly represents a second-class lever, and the resulting force is 300 lbs300\text{ lbs}.
The brake pedal assembly features a fulcrum at the top pin, the load connection in the middle, and the effort application at the bottom. The Fulcrum-Load-Effort (F-L-E) arrangement defines a second-class lever. To find the output force, we use the law of levers (Torque = Force × Distance from fulcrum). The input torque is 60 lbs×15 inches=900 in-lbs60\text{ lbs} \times 15\text{ inches} = 900\text{ in-lbs}. To maintain equilibrium, the output torque must also be 900 in-lbs900\text{ in-lbs}. Dividing this by the load distance of 3 inches3\text{ inches} yields an output force of 300 lbs300\text{ lbs}.

Step-by-Step Solution

1
Identify the arrangement of the fulcrum, load, and effort to determine the lever class.
The pivot (fulcrum) is at one end, the pushrod (load) is in the middle, and the foot pad (effort) is at the other end. This F-L-E configuration is a second-class lever.
Correctly classifying the lever establishes the proper distances for the torque equation.
2
Set up the torque balance equation for the lever: Feffort×deffort=Fload×dloadF_{effort} \times d_{effort} = F_{load} \times d_{load}.
60 lbs×15 inches=Fload×3 inches60\text{ lbs} \times 15\text{ inches} = F_{load} \times 3\text{ inches}.
In equilibrium, the torque input from the effort must equal the torque output against the load.
3
Solve the algebraic equation for the unknown load force (FloadF_{load}).
900=3×Fload900 = 3 \times F_{load}, which simplifies to Fload=300 lbsF_{load} = 300\text{ lbs}.
Isolating the variable gives the exact force applied to the master cylinder pushrod.

Key Concept

Torque Balance and Lever Classification
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Topics

Levers and Fulcrums

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