Calculating the Torque of a Cubical Block on an Inclined Plane

AI Thread Summary
The discussion focuses on calculating the torque of a cubical block sliding down an inclined plane at a uniform velocity. The initial attempt at a solution suggests that the torque is mgcos(φ) × a/2, but the book indicates that gravity does not contribute to torque since it acts at the center of mass. Instead, it emphasizes that friction creates torque, which the normal force must counterbalance. The correct expression for torque, according to the book, is mgcos(φ)sin(φ)a/2. This highlights the importance of understanding the roles of gravity and friction in torque calculations.
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Homework Statement



cubical block of mass m ,edge a,slides down the inclined plane of inclination \varphi with uniform velocity.
torque of normal reaction on the block about its centre is?

The Attempt at a Solution


Torque=mgcos\varphi X a/2
mgcos\varphisin\varphia/2
Book says:
mgsin\varphia/2
 
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Think about it this way: gravity acts on the center of mass, so gravity contributes no torque. Friction does, and the normal force has to balance the torque that friction creates.
 
ideasrule said:
Think about it this way: gravity acts on the center of mass, so gravity contributes no torque. Friction does, and the normal force has to balance the torque that friction creates.

So that should make the answer mgcos\varphisin\varphia/2
and not
mgsin\varphia/2
Right?
 
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