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A cubic 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 cubic 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?

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1 Answers

Arun
25750 Points
5 years ago
The forces on the block are the gravitational force, the normal reaction and friction. If you resolve gravitational force along the plane and perpendicular to it, you get normal reaction = mg cosθ and hence friction = μmg cosθ 

Since the block has a constant velocity friction is equal to the component of weight along the plane. So friction = mg sinθ 

Now, since the block slides, the resultant torque on it about its centre is zero. The friction produces a torque r x F = (mga sinθ)/2 in one direction. Since gravity acts through the centre of mass, which is in this case the centre, it produces no torque. Hence the only other force which can produce a torque is the normal force, which produces a torque of equal magnitude, i.e., (mga sinθ) /2 in the opposite direction.

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