Torque

Question

A mechanic applies a 40 N force perpendicular to the free end of a wrench of length 0.25 m in order to loosen a tight bolt. What is the magnitude of the torque produced about the bolt?

RankGuru · Accepted Answer

Torque is the rotational analogue of force and is given by $	au = r F \sin	heta$, where $r$ is the distance from the axis to the point of application and $	heta$ is the angle between the force and the position vector. Since the mechanic pushes perpendicular to the wrench, $	heta = 90^\circ$ and $\sin	heta = 1$, so the entire force is effective in turning the bolt. Then $	au = (0.25	ext{ m})(40	ext{ N})(1) = 10	ext{ N m}$. The value 16 N m has no consistent physical origin and appears to mix unrelated numbers. The value 5 N m would result from halving the lever arm without any justification. The value 100 N m wrongly multiplies the force by 2.5 instead of 0.25, a place-value error. This is the direct NCERT definition of the moment of a force about a point, the quantity that produces angular acceleration. As a unit check, multiplying newtons by metres yields newton-metres, the correct dimension of torque, and the modest lever arm sensibly gives a modest turning effect of 10 N m.

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