Free-body diagram: Spring-loaded cylinder sliding on a metal rod

[Steve4Physics]

How would you describe the direction of the normal force on the cylinder? Is it always vertically upwards) oppsite of gravity or does the direction of the normal force vary with the horizontal displacement x?

The normal force of the rod on the cylinder ($\vec N$) is (by definition of 'normal') perpendicular to the contact surfaces. So in this problem ,$\vec N$ must act vertically up, or vertically down, or be zero.

The cylinder never has a vertical acceleration, so the net vertical (y) force on the cylinder is always zero: $\vec W + \vec {F_y} + \vec N = 0$

Can you see under what conditions:
$\vec N$ acts vertically up?
$\vec N$ acts vertically down?
$\vec N$ = 0?

Also, have you covered simple harmonic motion (SHM)? This could help you understand how your rod/cylinder/spring system will behave.

Edit. Of course I should simply have asked what @Orodruin asked in Post #30!

 

[Lnewqban]

You have represented the N forces correctly oriented in your diagrams, only that the origin of the vector could be better located at the top surface of the hole in the cylinder.
That is where the bar is excerting a normal force on the cylinder in order to counteract gravity and the normal component of the spring force.

 

[ymnoklan]

Now I am wondering about the equilibrium points of the system. I find that they must be at x = -0.4, x = 0 and x = 0.4 (by finding where there would be 0 force in the horizontal direction), but I struggle characterising which would be stable and which would be unstable. What does this even mean?

 

[Steve4Physics]

Now I am wondering about the equilibrium points of the system. I find that they must be at x = -0.4, x = 0 and x = 0.4 (by finding where there would be 0 force in the horizontal direction), but I struggle characterising which would be stable and which would be unstable. What does this even mean?

If x=-0.4m the cylinder will not be entirely on the rod (assuming the Post #6 diagram shows x=0.4m). So presumably you are meant to assume the rod is longer than shown.

Equilibrium states can be stable or unstable. What is the difference? Have you looked-up these terms?

By the way, you never answered my Post #31 questions relating to the direction of $\vec N$.

 

[Orodruin]

but I struggle characterising which would be stable and which would be unstable. What does this even mean?

For this, I suggest working with the potential of the system. A (un)stable equilibrium corresponds to a minimum (maximum) of the potential.

Alternatively, find how the force looks close to each equilibrium. If it points towards (away) from the equilibrium, it is a (un)stable equilibrium.