Question

A cubical block of mass $m$ and edge $a$ slides down a rough inclined plane of inclination $\theta$ with a uniform speed. Determine the torque of the normal force acting on the block about its center.

Answer 1

$\Rightarrow f=mg\sin \theta$

As fractional force nol pass through the center, it produces torque $=\frac{1}{2}mg\sin \theta .$

To overcome the torque, normal reaction does not pass through center and produces a torque of $\frac{1}{2}mg\sin \theta .$

Hence, the answer is $\frac{1}{2}mg\sin \theta .$