A body mass m is placed on a rough inclined plane of inclination θ. Its downward motion can be prevented by applying an upward pull F parallel to the inclined plane. The block can be made to just slide upwards by applying force 2F parallel to inclined plane. Then coefficient of friction between the block and inclined plane is
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11th
Physics
Laws of Motion
Introduction to Friction
A box of mass 8 kg is place...
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A box of mass 8 kg is placed on a rough inclined plane of inclined θ. Its downward motion can be prevented by applying an upward pully F and it can be made to slide upwards by applying a force 2F. The coefficient of friction between the box and the inclined plane is
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For downward direction
Force equation in perpendicular direction is given as
N=mgcosθ
f
′
=μmgcosθ
Force equation parallel to incline is
f+f
′
=mgsinθ
f+μmgcosθ=mgsinθ
f=mgsinθ−μmgcosθ..............(1)
For upward direction
Force equation in perpendicular direction is given as
N=mgcosθ
f
′
=μN
f
′
=μmgcosθ
Force equation parallel to incline is
2f=mgsinθ+f
′
2f=mgsinθ+μmgcosθ............(2)
From equation (1),equation (2) becomes
2mgsinθ−2μmgcosθ=mgsinθ+μmgcosθ
mgsinθ=3μmgcosθ
tanθ=3μ
μ=
3
tanθ
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