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A bicycle pedal is pushed by a foot with a 60-N force which made 75 degrees with the shaft of the pedal. The pedal hand is 12 cm long. Find the magnitude of the torque about the center of the pedal

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Mistadijex

Answer: 6.95Nm

Step-by-step explanation:

A torque of a circular rotating body is calculated by multiplying the force acting on the centre with it's lever arm. If this force acts at an angle like the one we have in the question, we derive our resultant force and use it in our calculations.

F = 60N

Angle = 75°

Sin 75 = 0.9659

Lever arm = 12cm = 0.12m.

Deriving the resultant torque from the given angle becomes:

T = 60 * 0.12 * sin75 = 6.95Nm.

boffeemadrid

The torque at the center of the pedal is required.

The  magnitude of the torque about the center of the pedal is 6.95 N.

F = Force = 60 N

[tex]\theta[/tex] = Angle between pedal and shaft = [tex]75^{\circ}[/tex]

r = Radius = 12 cm

Torque is given by

[tex]\tau=rF\sin\theta[/tex]

[tex]\Rightarrow \tau=0.12\times 60\times \sin75^{\circ}[/tex]

[tex]\Rightarrow \tau=6.95\ \text{Nm}[/tex]

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