How do you calculate the horizontal force created by a load hanging from hinges with a hinged bend (as in a hangar door)?
Assume: The two halves of the door are of equal length and weight. (L and W)
Let Theta be the angle between vertical, and the upper portion, measured from the opening. Thus, when the door is closed, theta is 0, and when open all the way (impossible in real life) theta is 90
Constraint:  The door will form an isosceles triangle at all times when closing. (two equal length sides)
The X component of the length will be L sin( theta).
H will be 2L cos(Theta)
Let R be the reaction of interest.

Take the sum of the torques about P (hinge pin at top)  The forces P and T disappear as the moment arm is zero.
Sum Torque = 0 -&gt; (L/2 sin(theta) * 2 W)(clockwise+) - R 2L cos(theta)(counterclockwise, -ve)
            = LW sin(theta) - 2RL cos(theta)
            = W sin(theta) - 2R cos(theta)
            2R cos(theta) = W sin(theta)
            R = W tan(theta)/2

Consider the results -  if the door is closed, theta is zero and tan(theta) is zero.
If the door is fully open, tan(90) -&gt; infinity and the universe ends.
If the door is open 89 degrees, then the force at R is 57/2 W.
At 85 degrees, it's 11/2 W.
So you'll need to restrict the door opening with a stop in the track somewhere to keep things within reason.   The greatest forces will occur at the top of the track.