A 4 kg wedge is placed on a rough horizontal surface. A 2 kg block is gently kept on the frictionless inclined surface of the wedge. If the wedge is on the verge of slipping, the coefficient of static friction between the wedge and the horizontal surface is

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$\frac{1}{\sqrt{3}}$

$\frac{1}{3 \sqrt{3}}$

$\frac{\sqrt{3}}{2}$

$\frac{1}{3}$

answer is B.

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Detailed Solution

For the block, $N = 2 x 10 x \cos 60^{o} N = 10 N$

For the wedge,

$f_{\max} = Nsin 60^{o} = 10 x \frac{\sqrt{3}}{2} N = 5 \sqrt{3} N$

$N' = 40 + Ncos 60^{o} = (40 + 10 x \frac{1}{2}) N = 45 N$

$Now, f_{\max} = μN' = 45 μ$

$∴ 45 μ = 5 \sqrt{3} \Rightarrow μ = \frac{5 \sqrt{3}}{45} = \frac{1}{3 \sqrt{3}}$

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