# 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 good way to approach this problem is to think about the potential energy of the window vs. the height to which it has been pulled up; the force needed to hoist the cable is the derivative of this potential energy with respect to height. Due to the folding of the window, its center of mass moves upwards at half the speed of the cable, meaning you effectively get a factor of two in mechanical advantage. The weight on the cable is 100 lbs. – Yly Aug 30 '17 at 23:28
• You say $200\;lbs$ but the sketch says $190\;lbs$. – John Alexiou Aug 31 '17 at 0:11
• @ja72 - Good catch. I've been throwing around a nice round number to describe it but the actual combined weight is 190 pounds! – BasileSoftware Aug 31 '17 at 16:49
• @Yly - Much appreciated! This will be great help in my calculations. – BasileSoftware Aug 31 '17 at 16:50

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.