A cubical block of mass M and edge a slides down a rough inclined plane of inclination e with a uniform velocity. The torque of the normal force on the block about its center has a magnitude

Question

Philoid · Accepted Answer

As, the given surface is rough the force which is translating the block will get shifted towards the bottom of the block.

Also there is no movement in the vertical direction from the perspective of the plane. So, N=mgcosθ and this will act at the center of mass.

The velocity is uniform so the friction force balances mgsinθ. So, the torque due to normal force is (friction)

So, the torque will be,