Application of Newton's second law to a horizontal pulley

[ElectroMaster88]

Homework Statement

Body A, which has a mass of 3 kg, is on a horizontal surface and is connected by a massless and frictionless string and pulley to body B, which is suspended in the air and has a mass of 2 kg. The coefficient of friction (static and kinetic) between body A and the horizontal surface is 0.2.
A horizontal force F is applied to body A to the left.
A. Calculate what the value of the force F should be in order for the frictional force between body A and the surface to be zero.
B. In what range of values should the force F be in order for the system to be in a static state? There are answers:
A. 20N
B. 14N ≤ F ≤ 26N

Relevant Equations

x=x0+v0t+at^2/2
v=v0+at
f=μ*N
N=mg
Newton's Second Law

I don't understand how to solve these, and I don't understand how an horizontal force can affect the friction force if it's defined by μ*N, and the additional force affect neither of those. I also don't understand how there is a range of possible forces that F can be to make the system static, if only one option can make the horizontal net force 0.
I am clearly missing something and I don't understand what.

 

[PeroK]

I am clearly missing something and I don't understand what.

Static friction is a variable reaction force. It is not a force with a fixed magnitude. The quantity $\mu N$ is the maximum possible force of static fiction. If the box is free from external forces, then the static friction acting on it is zero. If the net external force on the box is less than $\mu N$, then the static friction force will have precisely the same magnitude as the new external force, but in the opposite direction.

 

[ElectroMaster88]

I think I understood:
Because the tension of the string is 20N, to make the net external force on box A 0 (which will make static friction of 0), I need to apply 20N of force in the opposite direction, which is F.
So that's why the answer is 20N.

And because μN is the maximum possible force of static friction (which is in this case 6N), at minimum I need to apply 14N of force to the opposite direction (20-6=14), and at maximum 26N, which will make the same scenario like in the 14N but with switched sides.
So as long as the net external force on the box is less than μN, the system will stay static.
Have I understood it right?