Question 166 of 210

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

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Answer: 3.33

$F\cos 60^{\circ}=\mu N$ or $\dfrac{F}{2}=\dfrac{1}{3\sqrt{3}}N\ldots \left(1\right)$
& $N=\sin 60^{\circ}+\sqrt{3}g\ldots \left(2\right)$
From equation $\left(1\right)$ & $\left(2\right)$
$\dfrac{F}{2}=\dfrac{1}{3\sqrt{3}}\left(\dfrac{F\sqrt{3}}{2}+\sqrt{3}g\right)$
$\Rightarrow F=g=10$ Newton $=3x$
So $x=\dfrac{10}{3}=3.33$