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A block of mass 8 kg 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 made to slide upwards by applying a force 2F. The coefficient of friction between the box and the 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

$F = mg \sin θ - mg \cos θ$

$2 F = mg \sin θ + μ mg \cos θ$

$2 (mg \sin θ - μ mg \cos θ) = mg \sin θ + μ mg \cos θ$ $\Rightarrow mg \sin θ = 3 μ mg \cos θ$

$μ = \frac{\sin θ}{3 \times \cos θ} = \frac{Tan θ}{3}$

$μ = \frac{\tan θ}{3}$

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