Forces on bifolding door with horizontal hinge I'm a programmer who's been put in charge of some motor control problems and I need your help, so please forgive me if the question is silly.
The problem in question (see inserted picture) is basically whether or not I should factor in the entire mass of the (bifold with horizontal hinge) door into the lifting load of the hoist, or if some of that load will be absorbed by the building (we can disregard any forces absorbed by the building structure here):

So if the total weight of both halves of the door is 200 pounds, should I treat the holding torque of the hoist as 200 pounds * Gravity?
If not, how is the total force on the pulley line calculated from the 200 pounds of total door weight?
 A: You need to analyse the forces acting on each body separately. What is important is the location of each center of mass. I am approaching this as a static (or quasi static) problem because the dynamics are way too complex for this forum.

I am not assuming the two leafs are of the exact same size, in order to accommodate the pivot-to-rope clearance $s$.
The angle $\theta$ is such that $s = 2 b \sin \theta - 2 c \sin \varphi$, or $$\theta = \sin^{-1} \left( \frac{ 2c \sin \varphi + s}{2 b} \right) $$
The vertical distance of the rope attachment to the door pivot is
$$ h = r + 2 b \cos \theta + 2 c \cos \varphi $$
The sum of the forces along the x and y axes are 
$$ \begin{aligned}
  A_x &= 0 \\
  A_y-\tfrac{W}{2}-\tfrac{W}{2}+T & = 0 
\end{aligned} \Rightarrow  \begin{aligned}
  A_x &= 0 \\
  A_y & = W-T
\end{aligned}$$
Now we need to balance the moments. Using the pivot as the origin the rotational balance is
$$ s\, T + c \cos\varphi \tfrac{W}{2} + (b \cos \theta-s) \tfrac{W}{2} = 0
\Rightarrow T = \frac{W}{2} \left(1 - \frac{h-r}{2 \,s} \right) < \frac{W}{2}
 $$
So as the $h$ distance gets smaller, the tension gets less and less.
