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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 coefficient 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

$F_{up} = 3 F_{down}$

$mg (\sin θ + μ_{k} \cos θ) = 3 mg (\sin θ - μ_{k} \cos θ)$ $4 μ_{k} \cos θ = 2 \sin θ$

$4 μ_{k} = 2 \tan θ$

$2. \frac{1}{2 \sqrt{3}} = \tan θ$

$\tan θ = \frac{1}{\sqrt{3}}$

$θ = 30 °$

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