A uniform ladder, $AB$, is leaning against a smooth vertical wall on rough horizontal ground at an angle of $70°$ to the horizontal. The ladder has length $8\ \rm m$, and is held in equilibrium by a frictional force of magnitude $60\ \rm N$ acting horizontally at $B$, as shown in the diagram. Write down the magnitude of the normal.
The normal reaction makes both A and B rotate, so I don't understand how we can ignore it by taking the moment at B OR A?
I have 60*8cos20 on the LHS (clockwise) = X*9.8*4sin20 (anti) but then I remember the normal force and it feels like I should add everything from the LHS to the RHS and I don't understand why that's wrong.
If I'm taking the moment at B, the 60N to the left frictional force still makes point -A rotate clockwise. and the Normal force, makes point A rotate anti-clockwise, so I still can't ignore it?
Similarly, there have been other seesaw questions, where I am told if I take the moment at the pivot then I can ignore the reaction at the pivot. Sure, but isn't there still a normal reaction from the seesaw back up into the (particle) weights on top of them?
Please can someone explain the simple thing I must be doing incorrectly, thanks!