Question

A cubical block of mass m and edge a slides down a rough inclined plane of inclination $\theta$ with a uniform speed. Find the torque of the normal force acting on the block about its centre.

• A. Zero
• B. $-\frac{1}{2}mg \ a \cos \theta$
• C. $-\frac{1}{2}mg \ a \sin \theta$
• D. $\frac{mga}{2}$

Answer 1

Answer: (C)

$-\frac{1}{2}mg \ a \sin \theta$

Answer 2

The block slides at uniform speed, implying both translational and rotational equilibrium.

1. Translational equilibrium perpendicular to the incline gives normal force $N = mg \cos \theta$.
2. Translational equilibrium parallel to the incline gives kinetic friction $f_k = mg \sin \theta$.
3. Rotational equilibrium about the block's center of mass (CM):
   • Weight's torque about CM is zero.
   • Friction force $f_k$ acts at the base, distance $a/2$ from CM. Its torque is $\tau_{f_k} = f_k (a/2)$ (counter-clockwise).
   • Normal force $N$ acts at the base, at some distance $x$ from the CM's vertical line. Its torque is $\tau_N = -N x$ (clockwise).
   • For equilibrium, $\tau_{f_k} + \tau_N = 0 \implies f_k (a/2) - N x = 0 \implies N x = f_k (a/2)$.
4. The torque of the normal force is $\tau_N = -N x = -f_k (a/2)$.
5. Substitute $f_k = mg \sin \theta$: $\tau_N = -\frac{1}{2} mg a \sin \theta$.