The problem is as follows:
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?