Abicycle pedal is pushed straight downwards by a foot with a 16 newton force. the shaft of the pedal is 20 cm long. if the shaft is π/5 radians past horizontal, what is the magnitude of the torque about the point where the shaft is attached to the bicycle?

2.56 N*m

Explanation:

Torque is force times distance, with the force being perpendicular to the distance.

If the force and distance are not perpendicular, a projection of one of the two must be used instead.

We will use the projection of the length of the shaft upon a perpendicular of the force applied. This will have a magnitude of:

Lh = L * cos(a) (I name it Lh because it is a horizontal projection in this case)

Lh = 0.2 * cos(π/5) = 0.16 m

Then:

T = f * d

T = 16 * 0.16 = 2.56 N*m

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