A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10N. The coefficient of static friction between the block and the plane is: [Take $$g = 10$$ m/s$$^2$$]

$$N = mg\cos30^\circ$$

$$f_{\text{max}} = \mu N = \mu mg\cos30^\circ$$

Case 1: Maximum force down the incline before slipping ($$F_1 = 2\text{ N}$$), friction acts up the incline:

$$F_1 + mg\sin30^\circ = f_{\text{max}}$$

$$2 + mg\sin30^\circ = \mu mg\cos30^\circ \quad \text{--- (1)}$$

Case 2: Maximum force up the incline before slipping ($$F_2 = 10\text{ N}$$), friction acts down the incline:

$$F_2 = mg\sin30^\circ + f_{\text{max}}$$

$$10 = mg\sin30^\circ + \mu mg\cos30^\circ \quad \text{--- (2)}$$

Subtracting (1) from (2): $$10 - 2 = 2mg\sin30^\circ \implies 8 = 2mg\left(\frac{1}{2}\right) \implies mg = 8\text{ N}$$

$$2 + 8\left(\frac{1}{2}\right) = \mu (8)\left(\frac{\sqrt{3}}{2}\right)$$

$$2 + 4 = 4\sqrt{3}\mu \implies 6 = 4\sqrt{3}\mu$$

$$\mu = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}$$