As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle $$30°$$, with horizontal. For $$\mu_s = 0.25$$, the block will just start to move for the value of $$F$$: [Given $$g = 10$$ m$$\cdot$$s$$^{-2}$$]

Given: mass = 10 kg, force F at 30° to horizontal, $$\mu_s = 0.25$$, $$g = 10$$ m/s².

The normal force: $$N = mg - F\sin 30° = 100 - \frac{F}{2}$$

For the block to just start moving, the horizontal component of F equals the maximum static friction:

$$ F\cos 30° = \mu_s N $$

$$ F \times \frac{\sqrt{3}}{2} = 0.25\left(100 - \frac{F}{2}\right) $$

$$ \frac{\sqrt{3}}{2}F = 25 - \frac{F}{8} $$

$$ F\left(\frac{\sqrt{3}}{2} + \frac{1}{8}\right) = 25 $$

$$ F \times \frac{4\sqrt{3} + 1}{8} = 25 $$

$$ F = \frac{200}{4\sqrt{3} + 1} = \frac{200}{7.928} \approx 25.2 \text{ N} $$

Therefore, $$F \approx 25.2$$ N.