Does normal contact force exert torque?

[Sharp2]

I came across a question in which a cubical body of mass 'm', edge 'a' is slipping down an inclination with constant velocity and torque due to normal contact force is to be found about the centre of the block.
As friction will exert an anticlockwise torque= 0.5amgsinx, where 'x' is the inclination, I was told that the normal contact force will exert an equal and opposite torque to keep the body in rotational equilibrium.
But doesn't normal contact force pass through the centre of the body? Then how can it possibly exert a torque?

 

[arildno]

But doesn't normal contact force pass through the centre of the body?

Nope.

The normal force is called "normal" because it acts along a line perpendicular to the surface, i.e, along the normal. This line need not pass through the bdy's center of mass (or, for that matter, its geometric center).

The normal force "prevents" collision&interpenetration of two touching bodies and will, therefore, arrange itself in such a manner (in terms of magnitude&position where it acts) so that these collisions won't happen.

This will always be a success, until the internal stress in one (or both) surface(s) becomes too great, and a rupture/fracture occurs.

 

[Sharp2]

Thanks! I came across another question in which the normal force acts along the corner of the cube (to find the condition for toppling and sliding)! Thank you. =)

 

[Selwyn]

normal can exert torque
consider a case in which a disc is kept on an inclined plane

 

[a.ratnaparkhi]

The torque due to normal reaction is the vector product of the normal reaction it self &the perpendicular distance between point of contact and center of mass.