Question

A cubical block of mass $m$ and edge ${}^{\prime }{a}^{\prime }$ slides down a rough inclined plane of inclination $\theta$ with a uniform speed. Find the torque of the normal force acting on the block about its centre.

Answer 1

The weight of the block $mg$ acting on the body has two components $mg\sin \theta$ and $mg\cos \theta$ and the body will exert a normal reaction as shown in the fig.

Consider that the normal force acting on the block is $R$

Since $R$ and $mg\cos \theta$ pass through the centre of the cube, there will be no torque due to $R$ and $mg\cos \theta$.

The only torque will be produced by $mg\sin \theta$.
$\therefore \tau =F\times r$ $(r=a/2)\&(a=edgeofthecube)$

$\Rightarrow \tau =mg\sin \theta \times a/2$
$=1/2mga\sin \theta$.