Here is the problem:

The figure shows a block above a slope. Find the minimum force to apply on the block for the mass body $ m = 2 , kg $ such that this body moves with a constant speed upward in the inclination. We know that the coefficient of friction between the surfaces is $ mu = 0.3 $ and the angle of inclination is $ alpha = 30 ^ { circ} $.

The alternatives given on my book are:

$ 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 I should break down the strength and weight. Which I supposed that from the figure the force is parallel to the floor which is the base of the inclination.

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

$ F cos alpha – mu N = 0 $

The normal or the tilt reaction I found using this logic:

$ N- mg cos alpha – F sin alpha = $ 0

$ N = mg cos alpha + F sin alpha $

Insert this into the equation above:

$ 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, inserting the given information would become:

$ 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 is where simplification becomes ugly:

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

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

So in the end, I get this value for strength. But it is far from the answer. Can any one help me with that ?. What could I have done wrong? How could I simplify that? Can any one offer FBD help to solve this problem?