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The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If co-efficient of friction between the body and inclined plane is $\frac{1}{2 \sqrt{3}}$ , the angle of inclined plane is

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$60^{°}$

$45^{°}$

$30^{°}$

$15^{°}$

answer is C.

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Detailed Solution

To hold block stationary

$F_{1} = m g \sin θ - μ m g \cos θ$

Force required to move block up

$F_{2} = m g \sin θ + μ m g \cos θ$

Given $F_{2} = 3 F_{1}$

$\Rightarrow m g (\sin θ + μ \cos θ) = 3 m g (\sin θ - μ \cos θ)$

$2 \sin θ = μ \cos θ$

$\tan θ = 2 μ$

$\Rightarrow θ = 30^{\circ}$

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