Question

A uniform ladder of mass $10g$ leans against a smooth vertical wall making an angle of ${53}^o$ with it. the other end rests on a rough horizontal floor. Find the normal force and the frictionless force that the floor exerts on the ladder.

Answer 1

There are four forces acting on the ladder of length L and making 53 degrees with the vertical smooth wall and on a horizontal rough floor.
given weight is 10g=10/1000=0.01kg
1. the weight W=(0.01x9.8) =0.98N acting downwards through the middle of the ladder

2. vertical reaction on the floor acting upwards, equal to W.
These two forces form a couple of magnitudes
W(L/2)sinθ3. Normal reaction N on the wall (horizontal) at the top end of the ladder, equal in magnitude to

4. Frictional force F acting horizontally at the bottom of the ladder.
These two forces form another couple equal to FLcosθSince the ladder is in equilibrium, the two couples must be equal, thus
W(L/2)sinθ = FLcosθ
From which we can solve for F
= (W/2)tanθ

= (0.98/2) tan 53o

= -0.211