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° with horizontal such that it does not move, will be $$3x$$. The value of $$x$$ will ______

Let the magnitude of the applied force be $$F$$. The force acts at $$60^{\circ}$$ below the horizontal, so its components are:

• Horizontal component (towards the right): $$F_h = F\cos 60^{\circ} = \dfrac{F}{2}$$
• Vertical component (downwards): $$F_v = F\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}\,F$$

Because the vertical component is downward, it increases the normal reaction $$N$$ exerted by the floor on the block. Taking upward as positive, the forces in the vertical direction are

$$N - mg - F_v = 0 \;\;\Rightarrow\;\; N = mg + F_v$$

Substituting $$F_v$$:

$$N = mg + \dfrac{\sqrt{3}}{2}\,F$$ $$-(1)$$

The limiting (maximum) static friction is given by

$$f_{\text{max}} = \mu N$$ $$-(2)$$

For the block to be on the verge of sliding, the horizontal component of the applied force must equal this limiting friction:

$$F_h = f_{\text{max}}$$ $$\dfrac{F}{2} = \mu N$$ $$-(3)$$

Insert $$N$$ from $$(1)$$ into $$(3)$$:

$$\dfrac{F}{2} = \mu\left(mg + \dfrac{\sqrt{3}}{2}\,F\right)$$

Collect the $$F$$ terms on the left:

$$\dfrac{F}{2} - \mu\,\dfrac{\sqrt{3}}{2}\,F = \mu mg$$

$$F\left(\dfrac{1}{2} - \dfrac{\mu\sqrt{3}}{2}\right) = \mu mg$$

Multiply numerator and denominator by 2 to isolate $$F$$:

$$F\left(1 - \mu\sqrt{3}\right) = 2\mu mg$$

$$F = \dfrac{2\mu mg}{1 - \mu\sqrt{3}}$$ $$-(4)$$

Insert the given numerical values $$m = \sqrt{3}\text{ kg}, \quad \mu = \dfrac{1}{3\sqrt{3}}, \quad g = 10 \text{ m s}^{-2}$$:

First evaluate $$\mu\sqrt{3}$$: $$\mu\sqrt{3} = \dfrac{1}{3\sqrt{3}}\times\sqrt{3} = \dfrac{1}{3}$$

Now substitute into $$(4)$$:

$$F = \dfrac{2\left(\dfrac{1}{3\sqrt{3}}\right)\left(\sqrt{3}\right)(10)}{1 - \dfrac{1}{3}} = \dfrac{2\left(\dfrac{1}{3}\right)(10)}{\dfrac{2}{3}}$$

$$F = \dfrac{\dfrac{20}{3}}{\dfrac{2}{3}} = 10 \text{ N}$$

The problem states that this critical force equals $$3x$$, so

$$3x = 10 \;\;\Rightarrow\;\; x = \dfrac{10}{3} \approx 3.33$$

Therefore, the required value of $$x$$ is $$\mathbf{3.33}$$.