Question

As shown in the figure, a block of mass $\sqrt{3}\text{kg}$ 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 ${60}^{\circ }$ with the horizontal such that it does not move will be $3x$. The value of $x$ will be _______. $($up to two decimal places$)$

$[g=10{\text{m/s}}^2;\sin {60}^{\circ }=\frac{\sqrt{3}}{2};\cos {60}^{\circ }=\frac{1}{2}]$

Answer 1

From equilibrium,
$N=mg+F\sin {60}^{\circ }$

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

For no slipping,

$F\cos {60}^{\circ }=f$

$\Rightarrow \frac{F}{2}=\mu N=\frac{1}{3\sqrt{3}}(\sqrt{3}g+\frac{\sqrt{3}F}{2})$

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

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

$\Rightarrow \frac{F}{3}=\frac{10}{3}$

$\Rightarrow F=10\text{N}=3x\text{N}$

$\Rightarrow x=3.33$