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A box of mass m is placed on a rough inclined plane of inclination $θ$ . Its downward motion can be prevented by applying an upward pull F parallel to the inclined plane. The block can be made to just slide upwards by applying force 2F parallel to the inclined plane. The coefficient of friction between the block and inclined plane is

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$\frac{\tan θ}{3}$

$3 \tan θ$

$\frac{\tan θ}{2}$

$2 \tan θ$

answer is A.

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Detailed Solution

$\begin{array}{l} F = mg (\sin θ - μcos θ) \\ F_{up} = 2 F \\ F_{up} = mg (\sin θ + μcos θ) \end{array}$

$(\sin θ + μcos θ) = 2 (\sin θ - μcos θ) 3 μcosθ = sinθ μ = \frac{1}{3} tanθ$

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