A body of mass 4kg is in limiting equilibrium over an inclined plane of inclination $30^{\circ}$ . If the inclination is made $60^{\circ}$ , the minimum force required to prevent the body from sliding down is $(g = 10 m s^{- 2})$

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$\frac{80}{\sqrt{3}} N$

$40 \sqrt{3} N$

$\frac{40}{\sqrt{3}} N$

80 N

answer is C.

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Detailed Solution

$F = mg (sinθ - μcosθ)$ , $μ = \tan 30^{0}$

$F = 4 \times 10 (\sin 60 ° - \tan 30 ° . \cos 60 °)$

$F = 40 (\frac{\sqrt{3}}{2} - \frac{1}{2 \sqrt{3}}) = \frac{40}{2} (\frac{3 - 1}{\sqrt{3}})$

$F = \frac{40}{\sqrt{3}}$

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