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An object of 6 kg is in the state of limiting equilibrium on an inclined plane having angle of inclination $30^{°}$ . Now if the angle of inclination is increased to $60^{°}$ , then the extra force along the plane to be applied on the object, to keep it in equilibrium will be (take g = 10m/s²)

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3 kg wt

$2 \sqrt{3} kg w t$

$\sqrt{3} k g w t$

$3 \sqrt{3} k g w t$

answer is B.

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Detailed Solution

From first case

$\tan θ = μ_{s}$

That $μ_{s} = \tan 30^{0} = \frac{1}{\sqrt{3}}$

In the second case, the force applied on the block up the plane for equilibrium is

$F = Mg sinθ - μ_{S} gMcosθ$

$= 60 \sin 60^{0} - \frac{1}{\sqrt{3}} (60) \cos 60^{0}$

$30 \sqrt{3} - \frac{30}{\sqrt{3}} = \frac{30 \times 2}{\sqrt{3}} newton$

Or $F = \frac{30 \times 2}{\sqrt{3} \times g} kg wt = 2 \sqrt{3} kg wt$

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