Easier said than done. I suppose what one could do is first find the Lagrangian assuming the ladder end touches the wall, solve the equations of motion, find the horizontal component of the momentum. The force will be zero when the horizontal momentum is an extremum (maximum). Such a maximum should exist - if we consider the case where the ladder cannot leave the wall the horizontal component of the momentum starts out as zero, and winds up as zero, but is non-zero in between.
How could you set this up, though? I had a similar problem in my mechanics class this week and I really want to figure it out. The prof said he used energy to solve it (we'll find out how in 2 days) but there has to be a way to solve it with moments and forces, right? I want to figure it out.
The book says the angle at which the ladder loses contact is [tex]sin^{-1}(sin(2/3\theta_{0}))[/tex], where [tex]\theta_{0}[/tex] is the initial angle between the ladder and the ground (I think, I left my book in my car, I'll double check later). Working backward has been so far fruitless.
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