A cubical block of mass M and edge a slides down a rough inclined plane of inclination 0 with a uniform velocity.
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The block moves down the incline with uniform speed and therefore the net force along the incline is zero.
Thus,
f−mg sinθ=0
f=mg sinθ
The gravitational force due to symmetry about the center produces no torque.
Since the only force acting on the block are gravitational force mg, normal force N ang frictional force F,the net torque is provided only by the later two forces.
Therefore,
τnet=τF+τN
But the block does not roll and therefore the net torque is zero.
τF+τN=0
τN=−τF
τN=−F(a/2)
τN=−1/2mga sinθ
Thus the torque due to normal force is in clockwise direction.
Thus,
f−mg sinθ=0
f=mg sinθ
The gravitational force due to symmetry about the center produces no torque.
Since the only force acting on the block are gravitational force mg, normal force N ang frictional force F,the net torque is provided only by the later two forces.
Therefore,
τnet=τF+τN
But the block does not roll and therefore the net torque is zero.
τF+τN=0
τN=−τF
τN=−F(a/2)
τN=−1/2mga sinθ
Thus the torque due to normal force is in clockwise direction.
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24
hope it helps ........
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