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I need to mount a heavy panel against a wall. The panel will be first mounted on two legs which go from the top of the panel to the floor. These legs will transfer most of the panel's weight to the floor, and will also be bolted to the wall to prevent the panel falling forward. Since there are only two legs (behind the panel), there will be some force exerted on the bolts and thus on the wall as the panel tries to fall forward.

How do I calculate this horizontal force exerted on the wall?

Thanks!

Here's a drawing: enter image description here

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    $\begingroup$ Didn't see the image (it is blocked in my country), but why would the panel try to fall forwards? You say it's already weighing on the legs, so there must be negligible force on the wall, which is probably only caused by the compression of the panel or the legs.. Do you need to know the maximum force on the wall (which is the tensile strength of the bolts)? $\endgroup$ – Gürkan Çetin Oct 25 '17 at 18:48
  • $\begingroup$ No, the panel only has 2 legs, which are bolted to the wall. The panel is 12 inches deep and weighs 150 lbs. Thus as it sits on the legs it will tend to pivot out and away from the wall (if it was not bolted to the wall it would fall over). This is why I included the drawing as it illustrates this clearly. $\endgroup$ – Ryan Griggs Oct 25 '17 at 21:44
  • $\begingroup$ If I were trying to hold the panel upright on its legs, this would require some pulling tension to keep it from falling/pivoting forward. This is the force I'm trying to calculate. $\endgroup$ – Ryan Griggs Oct 25 '17 at 21:50
  • $\begingroup$ Ok, thanks for clarifying. I had thought it was to be standing like a poster. It's clear now. $\endgroup$ – Gürkan Çetin Oct 26 '17 at 4:15
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As an order of magnitude approximation, do a moment balance. Assume the weight is in the center of the box, so 6 inch from the wall. Suppose the wall force is at a single point half way up (54 inches from the floor).

 force = 150lb * 6 / 54 ~= 17 lb.

In reality that force will be distributed among (I imagine) 6 or 8 fasteners. Size them for 20lb each should be no problem and give plenty of margin.

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