Problem — Torque

Problem

A force $F = 20\;\mathrm{N}$ is applied at a distance $r = 0{,}30\;\mathrm{m}$ from the axis, perpendicular to the radius. Calculate the torque and repeat the calculation if the force forms $60^\circ$ with the radius.

Solution

Perpendicular force ($\theta = 90^\circ$): $\tau = F\,r\sin 90^\circ = 20\cdot 0{,}30\cdot 1 = \ev{6{,}0\;\mathrm{N\,m}}$ Force at $60^\circ$ from the radius: $\tau = F\,r\sin 60^\circ = 20\cdot 0{,}30\cdot 0{,}866 \approx \ev{5{,}2\;\mathrm{N\,m}}$ Only the component perpendicular to the lever arm causes rotation: at $60^\circ$ the torque decreases slightly.

Links

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