Problem — Torque of a 25 cm wrench

Problem

A wrench $25\;\mathrm{cm}$ long is used to tighten a nut. A force of $40\;\mathrm{N}$ is applied perpendicular to the handle, at the end opposite the nut. Calculate the torque. What happens if the force is applied at $30^\circ$ to the handle?

Solution

With the force perpendicular to the arm ($\theta = 90^\circ$, $\sin 90^\circ = 1$) the torque is maximum: $\tau = F\,r = 40\cdot 0{,}25 = \ev{10\;\mathrm{N\cdot m}}$ At $30^\circ$ to the handle only the perpendicular component counts: $\tau = F\,r\sin 30^\circ = 40\cdot 0{,}25\cdot 0{,}5 = \ev{5{,}0\;\mathrm{N\cdot m}}$ The torque is halved: it always pays to push perpendicular to the handle to maximise the turning effect.

Links

Topics: Rotational dynamics Concepts: Torque Skills: Use of trigonometry