A cubical block of mass M and edge length a slides down a rough inclined plane of inclination heta at a uniform velocity. Determine the magnitude of the torque produced by the normal force on the block about its centre.

Step-by-step solution

Since the cubical block moves down the inclined plane with uniform velocity, its translational and rotational accelerations are zero. For translational equilibrium along the inclined plane, the frictional force f balances the component of weight: f = Mg For rotational equilibrium about the centre of the block, the net torque must be zero. The forces acting on the block are weight Mg , normal force N , and friction f . The weight Mg acts through the centre, so its torque about the centre is zero. The frictional forc