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Zorluk: OrtaLevers and Fulcrums

An automotive brake pedal operates as a second-class lever. The pivot point (fulcrum) is located at the top of the pedal assembly. The pushrod, which connects to the brake master cylinder, is attached 4 inches4 \text{ inches} below the pivot. The driver applies force to the pedal pad, which is located 14 inches14 \text{ inches} below the pivot. If the driver applies a perpendicular force of 50 lbs50 \text{ lbs} to the pedal pad, what is the force, in pounds, transmitted to the pushrod? Assume the system is in equilibrium.

Cevap: 175 lbs

Cevap

175
The brake pedal acts as a second-class lever in equilibrium. According to the principle of moments, the effort torque (50 lbs×14 inches=700 in-lbs50 \text{ lbs} \times 14 \text{ inches} = 700 \text{ in-lbs}) must equal the load torque. Dividing the total torque by the load arm distance (4 inches4 \text{ inches}) gives an output force of 175 lbs175 \text{ lbs}.

Adım Adım Çözüm

1
Identify the lever components and their distances from the fulcrum.
Effort arm = 14 inches14 \text{ inches}, Load arm = 4 inches4 \text{ inches}.
In torque calculations, all distances must be measured directly from the pivot point (fulcrum).
2
Apply the torque balance equation for a lever in equilibrium.
Feffort×deffort=Fload×dloadF_{effort} \times d_{effort} = F_{load} \times d_{load}
To find the unknown force, the applied torque must equal the output torque.
3
Substitute the known values into the equation and solve for the load force.
50×14=Fload×4→700=Fload×4→Fload=175 lbs50 \times 14 = F_{load} \times 4 \rightarrow 700 = F_{load} \times 4 \rightarrow F_{load} = 175 \text{ lbs}.
Isolating the unknown variable gives the final transmitted force.

Anahtar Kavram

Torque Equilibrium and Mechanical Advantage in Second-Class Levers
Tahmini Süre:1m 0s
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Levers and Fulcrums

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