A cubical block of mass M and edge a slides down a rough inclined plane of inclination $\mathrm{\theta }$ with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude

The block travels at a constant speed.

$\mathrm{f}=\mathrm{\mu mgsin\theta }$

Torque is produced on the block by friction.

Since the block is not rotating, net torque should be zero. This means a counter torque is produced by normal reaction.

therefore, the normal force doesnot pass through the centre.

Torque due to norma l force = $\frac{\mathrm{a}}{2}\times \mathrm{mgsin\theta }$