Question

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

• A. zero
• B. $Mga$
• C. $\frac{Mgasin\theta }{2}$
• D. $Mgacos\theta$

Answer 1

The correct option is C $\frac{Mgasin\theta }{2}$

Block moves with constant velocity

$\therefore mg\sin \theta =falsoitdoesnttopple\therefore Z_{N=}{Z}_F=>$

Torque =$F.\frac{a}{2}=\frac{mg\sin \theta a}{2}$