A 3 m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in the following figure, Find the reaction forces of the wall and the floor.

Explanation

Here, AB=3m
AC=1m
`BC=sqrt(AB^(2)-AC^(2))=sqrt(3^(2)-1^(2))=2sqrt2m`
Force acting on the ladder:
1. Weight acting at centre of gravity D
2. Reaction force `F_(1)` of the wall at B
3. Reaction force `F_(2)` of the floor at A, which can be resolved into two components: normal reaction N and force of friction F. The frictional force F prevents the ladder from sliding away from the wall and is thus directed towards the wall.
For translational equilibrium in vertical direction,
N=W
For translational equilibrium in horizontal direction,
`F=F_(1)`
For rotational equilibrium, considering the moments about point A, we have
`F_(1)xxBC=WxxAE" "[N,F_(2)` and F will have zero moment about A]
`impliesF_(1)xx2sqrt2=Wxx(1)/(2)`
Given, `W=20g=20xx9.8=196N`
`:.F_(1)=(196)/(4sqrt2)=34.6N`
N=W =196 N
`F=F_(1)=34.6N`
`F_(2)=sqrt((F)^(2)+N^(2))=sqrt(34.6^(2)+196^(2))=199N`
If `F_(2)` makes and angle `alpha` with F then
`tanalpha=(N)/(F)=4sqrt2=5.657`
`impliesalpha~=80`