How do I find the minimum force to be applied to a block sliding in a incline such as it moves with constant velocity?

Question

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?

Answer 1

Hint.

Calling $\beta = \frac{\pi}{2}-\alpha$ and projecting the forces along the incline and assuming the movement positive upwards,

$$ F\sin\beta - m g\sin\alpha = \mu\left(F\cos\beta+m g\cos\alpha\right) $$

Answer 2

Free Body Diagram of the mass 'm'

From the vertical (with respect to the slope) forces, we get:
N = mg$\cos θ$ + F$\sin θ$

So, by balancing the horizontal (with respect to the slope) forces, we get:

mg$\sin θ$ + f = F$\cos θ$
→ mg$\sin θ$ + μN = F$\cos θ$
→ mg$\sin θ$ + μ(mg$\cos θ$ + F$\sin θ$) = F$\cos θ$
→ mg$\sin θ$ + μmg$\cos θ$ + μF$\sin θ$ = F$\cos θ$
→ mg$\sin θ$ + μmg$\cos θ$ = F($\cos θ$ - μ$\sin θ$)
→ mg($\sin θ$ + μ$\cosθ$) = F($\cos θ$ - $μ\sin θ$)

Dividing both sides by $\cos θ$, we get:
→ mg($\tan θ$ + μ) = F(1 - μ$\tan θ$)
∴ F = mg$\frac{\tan θ + μ}{1 - μ\tan θ}$

Now, substituting θ = 30°, m = 2 kg, μ = 0.3 and g = 10 ms², we get:
F = $2*10\frac{\frac {1}{\sqrt{3}} + 0.3}{1 - 0.3*\frac{1}{\sqrt{3}}}$
→ $20\frac{0.57 + 0.3}{1 - 0.57*0.3}$
→ $20\frac{0.87}{0.829}$
→ $20*1.04$
→ $20.8$
~ 21 N