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. $mgasin\theta$
• D. $\frac{mgasin\theta }{2}$

Answer 1

The correct option is D $\frac{mgasin\theta }{2}$

$f=mg\sin \theta$

$\tau =f\ast \frac{a}{2}$

$\tau =mg\sin \theta \ast \frac{a}{2}$

$\tau =\frac{mga\sin \theta }{2}$