I'm sorry to see that Phinds is worn out a little. My impression is that's partially caused by the medium we are working with: it's easy to trigger misunderstandings with such indirect communication. Can't be helped at the moment. I am completely convinced of Phinds good will and expertise. I am also convinced of postfans sincerity, curiosity and courage in tackling a challenging set of AAPT contest questions, so I'll go along a bit more.
Going back to square 1 (post #1?) with this exercise (usually going under the tags ladder - wall, e.g.
here -- check out post #4 for some relevant equations !), I think a lot of hassle would have been avoided with the picture (good for 1k words, they say) you drew ! (Can you still show it ?)
Things start well in post #1with ##2mg## down acting at the center of mass of the whole contraption -- awkward place, but probably cancels anyway. Easier to deal with if we use symmetry and look at one leg only: ##mg## halfway the ladder. (You can see I'm too lazy to carry the 2 around, so I use m for the mass of one leg only).
[edit:} Can't do that: too confusing.
Things start well in post #1 with ##mg\ ## down acting at the center of mass of the whole contraption -- awkward place, but probably cancels anyway.
Easier to deal with if we use symmetry and look at one leg only: ##{1\over 2}mg\ ## halfway the ladder. Straight down.
Then a second force "normal force ##mg\; \cos\left ( (180^\circ - \theta)/2 \right )## up (relative to the surface of the leg) " -- here you can see how useful a picture would have been ! As well as a bit of basic trig: ##\cos (\pi/2 - \alpha) = \sin\alpha##. Apparently we are projecting ##mg## (
That is the actual normal force) on a line perpendicular to the leg. For the right leg would point a bit up and to the right. Project that onto the floor and presto: there is a horizontal component where there was no horizontal component before. No good.
Would be wiser here to not lump the two legs together. One normal force is ##{1\over 2}mg## pointing straight up at the point the right leg is on the floor. You can guess the second normal force is ##{1\over 2}mg## pointing straight up at the point the left leg is on the floor. Together they nicely add up to ##mg## pointing straight up and this is compensating gravity: the stand won't sink into the floor, nor will it float up. Both normal forces are equal, which is good too: the thing won't play dog and lift one leg :) or start rotating in a vertical plane any other way.
Third force in post #1: ##\mu N## inwards. Let me not be corny and go on about lumping together. There is ##\mu {1\over 2}mg## pointing to the left at the point the right leg is on the floor, etc. Left and right add up to 0. Everybody happy -- if that were all. And it isn't, as the bloke who looked at each leg separately has noticed already: the legs lean against each other and a fourth force comes in the picture. Two actually: left on right and vice versa.
Now we can have a sidebar about the direction of these forces -- acting at the hinge, which is frictionless, thank goodness.
Is it purely horizontal, or is it in the direction the tip would want to go (i.e. perpendicular to the leg), so that the legs don't just lean against one another, but also ON one another. I'm in favor of the latter (thinking of one heavy leg and one light leg). I hope someone else reads this far and is knowledgeable enough to settle this.
The good news is it doesn't matter, because of the symmetry: both normal forces at the bottom will remain at ##{1\over 2}mg## . And at the top we only need the horizontal component -- to compensate ##\mu {1\over 2}mg## at the bottom.
As they say: one picture... The above isn't far short of a thousand words (and cost me a lot of time).
Back to work: four forces per leg, dragged together in such a way that the force balance is in proper order. One more condition for equilibrium to fulfil. My bet is the ##{1\over 2}mg\ ## divides out :) and the L too.
