`A` rod `AB` rests with the end `A` on rough horizontal ground and the end `B` against a smooth vertical wall. The rod is uniform and of weight w. If the rod is in equilibrium in the position shown in figure.Find
(a)frictional force shown at `A`
(b) normal reaction at `A`
(c) normal reaction at `B`.

Explanation

a. The free body digram of the ladder is shown in figure.
Note that all the three forces acting on the ladder have to pas through a common point `O`, otherwise it cannot be in equilibrium.

The total reactiion force `F` from the horizontal surface is inclined at an angle `alpha(gttheta)` to the horizontal. The horizontal component of this force is friction force and its vertical component is the normal reaction applied by the ground on the ladder.
b. Applying the conditions of equilibrium, we get
`SigmaF_(x)=0implies-N=0`
`implies f=N` ............i
`Sigma F_(y)=0impliesN_(1)-Mg=0`
`implies N_(1)=Mg`...........ii
Taking torque about centre of the rod `C, Sigmatau_(C)=0`
`(N_(1))L/2costheta-(f)L/2sintheta-NL/2sintheta=0`.......iii
Substituting the value of `N_(1)` and from eqs i and ii in eqn iii we get
`MgL/2costheta-NLsintheta=0`
`N=(Mg)/(2tantheta)=1/2mgcottheta=f`........iv
Thus, the net force applied by ground on the ladder is
`F=sqrt(N_(1)^(2)+f^(2))`
`F=1/2Mgsqrt(4+cot^(2)theta)`
Alternative method:
As ladder is in equilibrium, the net torques about `A` will be zero. `sum tau_(A)=0` taking torque about end `A` of the rod.
`Mg(L/2costheta)-N(Lsintheta)=0`
`N=(Mg)/(2tantheta)=f`
It is the same result as we get by taking torque about `C`.