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Q.

A box 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 inclined plane is

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a

\frac{1}{2} \tan θ

b

\frac{1}{3} \tan θ

c

2 \tan θ

d

3 \tan θ

answer is A.

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Detailed Solution

F = mg \sin θ - f

$2F = mgsin θ + f$

$2 (mg \sin θ - f) = mg \sin θ + f$

$2mgsin θ - 2f = mgsin θ + f$

$mgsin θ = 3 μ mgcos θ$

$\tan θ = 3 μ$

$μ = \frac{1}{3} \tan θ$

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