The figure shows a block over an incline. Find the minimum force that must be applied to the block so that the body of mass $m=2\,kg$ such as that body moves with constant velocity upwards in the incline. It is known that the coefficient of friction between the surfaces is $\mu = 0.3$ and the angle of the incline is $\alpha = 30 ^{\circ}$.

The alternatives given on my book are as follows:

$\begin{array}{ll}

1.&21\,N\\

2.&23\,N\\

3.&18\,N\\

4.&20\,N\\

5.&2.2\,N\\

\end{array}$

I really need help with this problem. Initially I thought that I should decompose the force and weight. Which I assumed that from the figure the force is parallel to the floor which is the base of the incline.

By doing this and considering that the coefficient of friction (which I assumed that is static) this would be translated as follows:

$F\cos\alpha - \mu N = 0$

The normal or the reaction from the incline I found it using this logic:

$N- mg \cos\alpha - F\sin\alpha = 0$

$N= mg \cos\alpha + F\sin\alpha$

Inserting this in the above equation:

$F\cos\alpha - \mu \left(mg \cos\alpha + F\sin\alpha\right) = 0$

$F\cos\alpha - \mu mg \cos\alpha - \mu F\sin\alpha = 0$

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

$F=\frac{\mu mg \cos\alpha}{\cos\alpha - \mu \sin\alpha}$

Therefore by inserting there the given information would become into:

$F=\frac{\frac{3}{10} (2\times 10) \cos 30^{\circ}}{\cos 30^{\circ} - \frac{3}{10} \sin 30^{\circ}}$

$F=\frac{\frac{6\sqrt {3}}{2}}{\cos 30^{\circ} - \frac{3}{10} \sin 30^{\circ}}$

$F=\frac{\frac{6\sqrt {3}}{2}}{\frac{\sqrt{3}}{2} - \frac{3}{10} \times \frac{1}{2}}$

Here's where simplification becomes ugly:

$F=\frac{3\sqrt{3}}{\frac{10\sqrt{3}-3}{20}}$

$F=\frac{60\sqrt{3}}{10\sqrt{3}-3} \approx 27.25$

Therefore in the end I obtain that value for the force. But it is nowhere near to the answers. Can somebody help me with this?. What could I had done wrong?. How could I simplify this?. Can somebody offer a FBD help for this problem? :help: