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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?`
one year ago

Arun
23513 Points
```							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.
```
one year ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions