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Butoxors [25]
3 years ago
15

A bicycle pedal is pushed straight downwards by a foot with a 11 Newton force. The shaft of the pedal is 20 cm long. If the shaf

t is π/6 radians past horizontal, what is the magnitude of the torque about the point where the shaft is attached to the bicycle?
Mathematics
1 answer:
galben [10]3 years ago
3 0

Answer: 1.1 Nm

Step-by-step explanation:

The formula for torque is T = r · F · sin θ

where θ = π/6, r = 20 cm = 0.2 m, F = 11 N

formula for converting radians to degrees is 1 rad × 180/π = 57.296°

θ = π/6 · 180/π = 30°

Substituting r into the formula for torque, we have

T = 0.2 x 11 x sin 30° = 0.2 x 11 x 0.5

T = 1.1 Nm

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<em>It wasn't clear if you wanted multiplication or composition so I solved both.</em>

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