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 $$g = 10 \text{ m} \cdot \text{s}^{-2}$$]

Lets write the equation for the block in both along and perpendicular direction of the incline

along the direction of the incline

$$F\cos\ \theta\ =\mu\ _sN$$

equation perpendicular to the incline

$$N+F\sin\ \theta\ =mg$$

$$N=mg-F\sin\ \theta\ $$

Now substituting N in the above equation

$$F\cos\ \theta\ =\mu\ _s\left(mg-F\sin\ \theta\right)\ $$

$$F\cos\ \theta\ =0.25\left(100-F\sin\ \theta\right)\ $$

$$F\cos\ \theta\ +0.25F\sin\ \theta\ =25$$

$$F\times\ \frac{\sqrt{\ 3}}{2}\ +0.25F\times\ \frac{1}{2}=25$$

$$F\times\ \sqrt{\ 3}\ +0.25F=50$$

$$1.732F\ +0.25F=50$$

$$1.982F\ =50\longrightarrow\ F=\frac{50}{1.982}$$

$$\ F=25.2N$$