As shown in the figure, a block of mass $\sqrt{3}$ kg is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3 \sqrt{3}}$ . The critical force to be applied on the vertical surface as shown at an angle 60⁰with horizontal such that it does not move, will be 3x. The value of x will be............. × 10^–1
$[g = 10 m / s^{2}; \sin 60^{0} = \frac{\sqrt{3}}{2}; \cos 60^{0} = \frac{1}{2}]$

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answer is 33.33.

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Detailed Solution

From equilibrium,

$N = mg + Fsin 60 °$

$\Rightarrow N = \sqrt{3} g + \frac{\sqrt{3} F}{2}$

For no slipping,

$F \cos 60 ° = f \Rightarrow \Rightarrow \frac{F}{2} = μ N = \frac{1}{3 \sqrt{3}} (\sqrt{3} g + \frac{\sqrt{3} F}{2})$

$\Rightarrow \frac{F}{2} = \frac{g}{3} + \frac{F}{6}$

$\Rightarrow F = 10 N$

$\Rightarrow x = 3.33$

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