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Q.

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

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a

90°

b

30°

c

60°

d

45°

answer is D.

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Detailed Solution

N = 2 (mg \sin θ - μ mg \cos θ)

$mg \cos θ = 2 mg (\sin θ - \frac{\cos θ}{2})$

$\cos θ = 2 \sin θ - \cos θ$

$2 \sin θ = 2 \cos θ$

$\tan θ = 1$

$θ = 45 °$

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