I'm somewhat embarrassed to ask this here as the problem seems to be so trivial and we (yes, I have been asking many people around me to pitch in - to no avail) always reach the same conclusion, which however is totally contradicted by reality, so ...

(Spoiler ... I think I've found the difference between reality and the problem description below. I'll add it to the bottom of the question though - no use hiding my mistakes ;-) )

I have a garage door that stays inside of the walls. I have no idea whether this type of door has a particular name in English, so to ensure we have the same understanding I'll add a picture and some explanation:

[![cross-section of the garage door](https://i.stack.imgur.com/YaVY3.jpg)](https://i.stack.imgur.com/YaVY3.jpg)

The garage door is one solid panel (typically of wood or with an internal metal construction) and has at the left and right sides a wheel at the top running inside a fixed (typically via some construction to the ceiling) horizontal rail and at the bottom a wheel running inside a fixed (to the side-wall) vertical rail.

Because of the rigidness of the door, I believe we can reduce this problem to 2D, so basically to the cross-section view given by the attached image. As the wheels have some slack in the rails, they can roll inside them and hence I also think we can assume they won't provide any force in the direction of the rail, only perpendicular to it.

In most installations (as in mine) there will be a counter-weight installed which using cables pulls straight up at the point where the lower wheels are connected to the door. Obviously the counter-weight would be such that the (manual) amount of force needed to open or close the door would be minimal ...

And this is where theory and reality clash. If I call the mass of the door m and that of the counter-weight M, all my attempts indicate that choosing M = m/2 would have this door stand still in any position/angle that it can take, i.e. that it would never further lower or further rise out of itself if put (and shortly held) in that position.

Reality however shows that there is a point where the door is in balance and doesn't move on its own (if fully closed is 0 degrees, then my door seems in balance around 30 degrees - but I guess that might depend on the counter-weight), but that below this point (so more closed), the door further closes automatically and above that point (so more open) the door further opens automatically (and my arms can tell you that the force needed to counter this grows significantly the further you go from that balance point). That seems to indicate that the force needed to counter the door's own weight should (unlike the constant weight of the counter-weight) be a function of the angle the door is at. But as said, this is not what all my attempts to calculate that force result in.

Hence the question: can someone give a formula for the force one would have to supply vertically at the bottom attachment point of such a garage door in function of the weight of the door and the angle it is positioned in? And also explain how you get to that formula. Or otherwise explain why reality and simple mechanics calculations seem to clash here?

Adding some more info as it seems some people think the counter-weight would be something very special:
[![Illustration including counter-weight](https://i.stack.imgur.com/hITGx.jpg)](https://i.stack.imgur.com/hITGx.jpg)

As mentioned a steel cable starts at the bottom wheel at each side of the door, runs over a pulley at the top, so that the cable pulls vertically on that bottom wheel and then via a second pulley to the counter-weight. In my case there is indeed a pulley for each cable attached to the weight and the cable then runs up and is fixed to the ceiling. This just halves the distance the weight needs to travel and the force it applies to the door with its weight, but still that force will be constant over the travel of the weight (and hence the door), right?

Thanks in advance for help saving my sanity.

PS: the border case with the door fully closed also gives a hint that somehow the angle must play a role as it's easy to realize that in this position you will never get the door to open unless you also give it a sideways push. So somehow that angle must have made the F needed to open the door infinite (fully countered by V), right? Of course, that is not the F needed to keep the door in equilibrium, but still it raises that feeling that the angle does play a significant role, no?

==== NORMALITY HAS BEEN RESTORED ====

Or how oversimplification probably caused all my problems above ...
Looking a bit closer at the door, I started to realize that the center of mass of it is actually not in the middle between the wheels. Hence a more correct sketch of the real situation would be:
[![More correct problem description](https://i.stack.imgur.com/Ysr7H.jpg)](https://i.stack.imgur.com/Ysr7H.jpg)
And when I then do a static analysis, I do get a $tan(\alpha)$ in the result and with the counter-weight slightly heavier than half the weight of the door, I also get a point between the extremes where the door would balance.

Conclusion: my original problem description is likely oversimplified in that it assumed the center of mass to be on the axis between the wheels. The further that center of mass moves away from this axis, the more the position of the door plays a role.