If I have a situation like this
Where my blue thing is wall and the yellow thing is the ground. If both were to be smooth then they will exert a contact force only normal to them which I have depicted as $ N_1 and N_3 $. Now let's calculate the torques :-
$\tau _1 $ = $W \times 5 ~cos 60 ~~~~~ = W \times \frac{5} {2} $
$\tau _2 $ = $ N_3 \times 10 ~ sin60 ~~~~~ = N_3\times \frac{5\sqrt{3}} {1} $
I have taken the pivot point to be the point of contact of ground. By the right hand rule the $\tau_1 $will go into the page and $\tau_2$ will come out of the page . Now, for rotational equilibrium $$ \tau _1 - \tau _2 = 0 $$ $$ N_3 ~5\sqrt3 - W~ \frac{5}{2} = 0$$
$$ N_3 \sqrt3 = W/2 $$ $$ N_3 = \frac {W} {2 \sqrt{3}}$$ Well that just means that if $N_3 $ is $\frac{1} {2\sqrt{3} }$ of W then our ladder wouldn't rotate. SO much clear this far.
But $$ \sum F_x \neq 0 $$ because we have just $N_2$ as a horizontal force and nothing else to compensate for it. Okay, well nothing is compensating it but what it can do , all it can do is cause a motion horizontally but the ladder is pivoted at bottom and we have found if $N_3 = \frac{W} {2\sqrt{3}} $ then there would be no rotation at all about the bottom point.
But I have read that if both wall and ground were to be frictionless then the ladder would slip. What is slipping? Is it a translational motion or a rotation? As far as I can see slipping is kind of rotation. How the ladder will slip if if we have managed to do rotational equilibrium?
Thank you. Any help will be much appreciated.