In a problem related to static equilibrium, there is a ladder leaning against the wall supposed to rotate about a point. Why are we not assuming it to slide down in a way similar to that in the real case?

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    $\begingroup$ If the ladder is in static equilibrium it will not rotate. The sum of the moments and sum of the forces about any point are zero. If the ladder slides down, static equilibrium requirements are not being met. Hopefully in your practical life ladders aren't typically sliding down while you're on them. $\endgroup$
    – Bob D
    Commented Apr 5, 2019 at 13:41

1 Answer 1


1) when it slids down, the bottom must move also, and can be messy to write down the relation between the two motions

2) Any istantaneous motion of a rigid body in a plane can be described as a rotation around some point (this point go to infinity for rectilinear motion)

3) with the rotation you can write easily the ladder motion

Drawing of falling ladder

  • $\begingroup$ The OP submitted a new question putting numbers on it. It says the wall is smooth. Otherwise, you're right that with friction on the wall it gets messy because you will have redundant vertical reactions (wall friction and floor reaction) which makes the problem statically indeterminant. $\endgroup$
    – Bob D
    Commented Apr 5, 2019 at 14:25
  • $\begingroup$ I agree, to solve the static problem the ladder doesn't move, so we do not need any rotation and we can solve it just by sum of forces. I guess the rotation comes from textbooks that continue by describing the falling motion, how much it accelerates, for which the rotation description is useful $\endgroup$
    – patta
    Commented Apr 5, 2019 at 14:32

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