Question

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

Answer 1

Given in the question .

Total number of force acting on the block is three.

Weight= mg

Friction= F

Normal force= R

Refer to the Attachment

Here, Force mg acting on the two component i.e mg. cosθ and mg.sinθ

Now,

∵ R and mg.cosθ passing by the cube centre, hence no torque will be working, only torque is produced by mg.sinθ

Thertefore,

i = F× r

i = mgsinθ × (a/2) (r = a/2)

i = $\frac{1}{2} mga sin\theta$

Hope it Helps :-)