Calculating the Torque of a Cubical Block on an Inclined Plane

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sachin123
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Homework Statement



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

The Attempt at a Solution


Torque=mgcos[tex]\varphi[/tex] X a/2
mgcos[tex]\varphi[/tex]sin[tex]\varphi[/tex]a/2
Book says:
mgsin[tex]\varphi[/tex]a/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[tex]\varphi[/tex]sin[tex]\varphi[/tex]a/2
and not
mgsin[tex]\varphi[/tex]a/2
Right?