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A box of mass 8kg is placed on a rough inclined plane of inclination ${}^{'}θ^{'} .$ Its downward motion can be prevented by applying an upward pull ‘F’. And it can be slide upwards by applying a force ‘2F’, The coefficient of friction between the box and the inclined plane is___

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$\frac{1}{2} \tan θ$

$2 \tan θ$

$3 \tan θ$

$\frac{1}{3} \tan θ$

answer is A.

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Detailed Solution

$F_{u p} = 2 F_{d o w n}$
$m g (\sin θ + μ \cos θ) = 2 m g (\sin g θ - μ \cos θ)$
$\begin{array}{l} \sin θ + μ \cos θ = 2 \sin θ - 2 μ \cos θ \\ 3 μ \cos θ = \sin θ \Rightarrow μ = \frac{1}{3} \tan θ \end{array}$

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