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 $60^\circ$ with horizontal such that it does not move, will be $3x$. The value of $x$ will be ______ $\left[g = 10\ \text{m s}^{-2};\ \sin60^\circ = \frac{\sqrt{3}}{2};\ \cos60^\circ = \frac{1}{2}\right]$
Numerical value type. Enter your answer.
Answer: 3.33
The force $F$ pushes on the vertical face, directed into the block and downward at $60^\circ$ to the horizontal.
Normal reaction: $N = mg + F\sin60^\circ = 10\sqrt{3} + \dfrac{\sqrt{3}}{2}F$.
At the critical (limiting) condition, the horizontal component equals limiting friction:
$$F\cos60^\circ = \mu N \Rightarrow \frac{F}{2} = \frac{1}{3\sqrt{3}}\left(10\sqrt{3} + \frac{\sqrt{3}}{2}F\right) = \frac{10}{3} + \frac{F}{6}$$
$$\frac{F}{3} = \frac{10}{3} \Rightarrow F = 10\ \text{N}$$
$3x = 10 \Rightarrow x = 3.33$
Solution by Sreeraj P, M.Sc Physics