First slide

Friction on inclined plane of angle more than angle of repose

Question

A body is sliding down an inclined plane having coefficient of friction 0.5. If the normal reaction is twice that of the resultant downward force along the incline, the angle between the inclined plane and the horizontal is

Moderate

A

15°
B

30°
C

45°
D

60°

Solution

Normal reaction on the inclined plane is $μ m gcos θ$ and resultant downward force $mg (\sin θ - μcos θ)$
Given, $mgcos θ = 2 [mg (\sin θ - μcos θ)]$
Solving by putting $μ = 0.5, θ = 45^{\circ}$

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