A cubical block of mass $m$ and edge $a$ 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.

Consider the free body diagram of the cubical block

The forces are the weight mg, Normal force N and friction fr. Note that the weight that acts about the centre is the resultant or net effect and that the normal force N is the resultant of forces exerted, along the whole surface that touches the incline.

It can be seen that Normal force vector passes through the centre of the block. So, torque due to N

$\tau =$Force$\times \perp r$ distance

$=N\times 0$

$=0$.