A box of mass 50 kg is placed on an inclined plane. When the angle of the plane isincreased to 30º, the box begins to slide downwards. Calculate the coefficient of staticfriction between the plane and the box. Draw the free body diagram.
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At angle θ = 30 deg, the gravitational force is equal to the static friction.
Coefficient of friction = μ.
mass of the box = m
Normal reaction force exerted by the plane on the box = N
Frictional force = μ N
Gravity force along the plane = m g Sinθ = mg/2
The forces perpendicular to the plane are balanced, as the box does not move along that direction (acceleration=0).
So mg Cosθ = N
So Friction = √3/2* μ m g = mg /2
μ = 1/√3
See diagram.
Coefficient of friction = μ.
mass of the box = m
Normal reaction force exerted by the plane on the box = N
Frictional force = μ N
Gravity force along the plane = m g Sinθ = mg/2
The forces perpendicular to the plane are balanced, as the box does not move along that direction (acceleration=0).
So mg Cosθ = N
So Friction = √3/2* μ m g = mg /2
μ = 1/√3
See diagram.
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kvnmurty:
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Answer:If body it rest. Then friction = down ward force which is mgsin30
F=50×10×1/2=250Newtons
Explanation:Friction is equal to 250 Newton
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