How do you calculate the horizontal force created by a load hanging from hinges with a hinged bend (as in a hangar door)?

Question

I am trying to calculate the horizontal force created by a raised hangar door, but for the purpose of this question, we can consider the horizontal force of a 2x4 stud that is connected as shown in the illustration below.

The 2x4 is cut in half and connected with a hinge at the top and a hinge at the middle, and then is supported vertically with a cable (which in the door will pull the door up and down). Not pictured is a track which the 2x4 or door will be connected to that keeps the "bottom" of the door in alignment with the opening.

What I am trying to determine is the method to find the lateral force that is created, or in other words, how much force will the door put on the track when it is open?

I have intentionally not included weights or lengths or angles because I'm looking for a general solution.

Answer 1

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 -> (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) -> 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.

Answer 2

The force is ZERO because in your picture the 2x4 is not attached.

If you want a theoretical answer then have an accurate picture. Seeing the end product might help as there are more than just at rest forces.

Answer 3

First, simplify the problem by assuming:

• The door is at a 90° angle (perfectly horizontal). This creates the maximum load.
• The center of mass is also 90° out from the top hinge (in reality it will be a little lower).
• The center of mass is halfway between the two hinges. (1/4 the length of the fully extended door)

We know that torque = force * distance. We can use weight instead of force because the only force is gravity acting upon the mass--a.k.a. weight. Distance is the distance to the center of mass, which is half of the total distance. And torque in this situation manifests as a force pushing horizontally.

If you choose pounds (weight), feet (length), and foot-pounds (torque force) as your units, everything works out nicely.

So with that, we can find the answer easily with this:

horizontal torque = weight * distance

This gives us the horizontal torque, a.k.a. how much the wheel bearing pushes on the track.

• At a length of 1 foot to the center of mass (i.e. a 4 foot tall door that folds up in half to become 2 foot long, putting the center of mass 1 foot out), the horizontal torque is equal to the weight of the door in pounds.
• At a length of 2 feet, the horizontal torque is double the weight.
• At 3 feet, it's triple the weight.
• And so on.

I find this strategy very intuitive. I used this torque calculator.