What is the maximum value of the force $F$ such that the block shown in the arrangement, does not move?

Step 1: Identify the given parameters.

The physical parameters of the block and the surface are given.

$$m = \sqrt{3}\text{ kg}$$ $$\mu = \frac{1}{2\sqrt{3}}$$ $$\theta = 60^\circ$$ $$g = 10\text{ m/s}^2$$

Step 2: Resolve the force $F$ into components.

The force $F$ is directed downwards and to the right at an angle of $60^\circ$ to the horizontal.

$$F_x = F \cos 60^\circ = \frac{F}{2}$$ $$F_y = F \sin 60^\circ = \frac{\sqrt{3}}{2}F$$

Step 3: Set up the vertical force balance.

The normal force $N$ balances both the weight of the block and the downward component of the applied force.

$$N = mg + F \sin 60^\circ$$ $$mg = 10\sqrt{3}\text{ N}$$ $$N = 10\sqrt{3} + \frac{\sqrt{3}}{2}F$$

Step 4: Apply the condition for impending motion.

At the verge of slipping, the horizontal driving force equals the maximum static friction force.

$$F \cos 60^\circ = \mu N$$ $$F \cos 60^\circ = \mu \left( mg + F \sin 60^\circ \right)$$

Step 5: Solve for the maximum force $F$.

Substitute the known values into the equilibrium equation.

$$\frac{F}{2} = \frac{1}{2\sqrt{3}}\left(10\sqrt{3} + \frac{\sqrt{3}}{2}F\right)$$ $$\frac{F}{2} = 5 + \frac{F}{4}$$ $$\frac{F}{4} = 5$$ $$F = 20\text{ N}$$

Answer: The maximum value of the force $F$ is $20\text{ N}$.