A cubical block of mass m and edge `a` slides down a rough inclined plane of inclination with a uniform speed. Find the torque of the normal force ac

The force mg acting on the body has two components mg sin θ and mg cos θ and the body will exert a normal reaction.

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

The only torque will be produced by mg sin θ.

I = F x r(r = a/2)(a = ages of the cube)

i = mgasinθ x a/2

1/2mgasinθ

the torque of the normal force acting on the block about its center is = 1/2mgasinθ