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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 ...


-1

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 ...


3

Understand the constraints of the existing framing. I'd have probably kept the new vertical 2x4 you added 3.5" shorter (along with the plywood spacer you added), and made the header stick out 3.5" on that side to rest on top of that new 2x4. You'd essentially be using the existing vertical 2x4 as the king stud and your new 2x4 as the jack stud, ...


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