Question

A cubical block of mass $M$ and edge $a$, slides down a rough inclined plane of inclination $\theta$ with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude:

• A. $Zero$
• B. $Mga$
• C. $\left(1/2\right)Mgasin\theta$
• D. $Mga\sin \theta$

Answer 1

The correct option is C $\left(1/2\right)Mgasin\theta$

The block moves with constant velocity.

$\Rightarrow f=\mu mg\sin \theta$

The friction produces torque on block .Normal friction produces counter torque.

Normal force does not pass through center.

Torque due to Normal$=\frac{a}{2}\times mg\sin \theta$