A cubical block of mass m and edge L slides down a rough inclined plane of inclination θ = 30° (with horizontal) with a uniform speed. Torque of the normal force acting on the block about the centre is mgL/n such that n = ...
Physics · Mixed Chapter
JEE Mains

Question

A cubical block of mass m and edge L slides down a rough inclined plane of inclination θ = 30° (with horizontal) with a uniform speed. Torque of the normal force acting on the block about the centre is mgL/n such that n = ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Uniform speed along the slope requires zero net downslope force, so friction force f=mg sin30°=mg/2 directed upslope. Step 2: About the cube's centre, gravity creates no torque. Friction acts on the contact face a perpendicular distance L/2 from the centre, producing torque τ_f=f(L/2)=mgL/4. Step 3: No angular acceleration means the torque of the normal-force resultant balances friction torque. Hence |τ_N|=mgL/4=mgL/n ⇒ n=4.