Abicycle pedal is pushed straight downwards by a foot with a 33 newton force. the shaft of the pedal is 20 cm long. if the shaft is π/5π/5 radians past horizontal, what is the magnitude of the torque about the point where the shaft is attached to the bicycle?

The magnitud of the torque doing by the foot at the point where the shaft is attached is 0.063 Nm

Explanation:

The torque is obtained by multiplying the longitude from the shaft attached point to force acting point by the force perpendicular component:

T = d * Fp

Notice that the perpindicular component is the total force magnitud times the sino of the angle respect the horizontal:

Fp = F*sin(a)

Replacing the values for the force and the angle:

Fp = 33N*sin(π/5)  = 33N * 0.011 = 0.363 N

Taking the distance in meters:

T = 0.2m * 0.363 N = 0.063 Nm

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