You push perpendicular to a door with a force of 40 N at a distance of 0.800 m from the hinges. What torque do you exert about the hinges?
A50.0 N m
B0.020 N m
C32.0 N m
D3.20 N m
Answer & Solution
Correct answer: C. 32.0 N m
1. Use tau = r F sin(theta) with r = 0.800 m, F = 40 N and theta = 90 degrees.
2. sin(90 degrees) = 1, so the whole force is effective and tau = (0.800 m)(40 N)(1).
3. That gives tau = 32.0 N m about the hinges.
4. Dividing rather than multiplying by the distance gives 50.0 N m, a common slip.
_Source: OpenStax College Physics (CC BY 4.0), Ch 9 "Statics and Torque", section 9.2 The Second Condition for Equilibrium_
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