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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 made to slide upwards by applying a force 2 F. The coefficient of friction between the box and the inclined plane is

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$\frac{1}{3} \tan θ$

$3 \tan θ$

$\frac{1}{2} \tan θ$

$2 \tan θ$

answer is A.

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Detailed Solution

$F = m g \sin θ - μ m g \cos θ 2 F = m g \sin θ + μ m g \cos θ$

$(i) o r 2 m g \sin θ - 2 μ m g \cos θ = m g \sin θ + μ m g \cos θ o r$

$μ = \frac{1}{3} \tan θ$

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