Question

A cubical block of mass M and edge a slides down a rough inclined plane of inclination θ with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude
(a) zero
(b) Mga
(c) Mga sinθ
(d) 12 Mga sinθ.

Answer 1

A cubical block of mass M and edge a slides down a rough inclined plane of inclination θ with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude 12 Mga sinθ.

Explanation:

Step 1:

Let N be the block's normal reaction.

It is evident from the block's free body diagram that the forces N and $\mathrm{mg} \cos \theta$ pass through the same side. There will therefore be no torque due to N and $\mathrm{mg} \cos \theta$ as much as possible.

$\therefore \vec{\tau}=\vec{F} \times \vec{r}$

Where  

$\vec{\tau}$ is torque

F is the force

Step 2:              

Since the cube edge is a, $r=\frac{a}{2}$

hence, we get :

$\tau=m g \sin \theta \times \frac{a}{2}$

$\tau=\frac{1}{2} m g a \sin \theta$