A uniform ladder of mass 10 kg leans against a smooth vertical wall making an angle `53^(@)` with it. The other end rests on a rough horizontal floor. Find the normal force and the frictional force that the floor exerts on the ladder.

Explanation

The ladder is in equilibrium.
`therefore N_(1)=f` and `N_(2)=W`
Taking torque about .B.
`N_(1)(AO)=W(CB)`.
or `N_(1)(AB) cos 53 = W((AB)/(2))sin 53` or `N_(1)=(2)/(3)W`
and `N_(2)=W=10xx9.8=98 N`.
The frictional force is `f=N_(1)=(2)/(3)W=65 N`