A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in Fig.7.27. Find the reaction forces of the wall and the floor.

Explanation

Length of ladder `AB=3m`
The distance of foot of ladder fro the floor `AC=1m`
`therefore` Right angle `DeltaACB" "BC=sqrt(AB^(2)-AC^(2))`
`=sqrt((3)^(2)-(1)^(2))`
`=sqrt(8)`
`=2sqrt(2)m`
The forces on the ladder are its weight W acting at its centre of gravity D.
The reaction forces `F_(1)andF_(2)` are acting on wall and floor.
Force `F_(2)` is resolved into two components are normal reaction N and the force of friction F parallel to AC.
which prevents the ladder from sliding away from the wall and therefore directed toward the wall.
For translational equilibrium
`N-W=0` (in vertical diretion)
`thereforeN=W...(1)`
Taking the forces in horizontal direction
`f-F_(1)=0`
`therefore F_(1)=f...(2)`
For rotational equilibrium, taking moment of forces about A
`F_(1)(BC)-W(AE)=0`
`therefore F_(1)(2sqrt(2))-W((1)/(2))=0`
`therefore 2sqrt(2)F_(1)=(W)/(2)`
`therefore F_(1)=(W)/(4sqrt(2))....(3)`
Now `W=20g [becauseW=mg]`
`=20xx9.8`
`=196N`
`therefore` From eqn. (1),
`N=196N`
From eqn. (3),
`F_(1)=(196)/(4sqrt(2))`
`therefore F_(1)=34.65N`
`therefore` From eqn. (2),
`f=34.65N`
Now in `DeltaANF_(2)`,
`AF_(2)=sqrt(AN^(2)+NF_(2)^(2))`
`=(sqrt(N^(2)+f^(2)))`
`=sqrt((196)^(2)+(34.65)^(2))`
`=sqrt(38416+1200.62)`
`=sqrt(39616.62)`
`=199.039`
`therefore AF_(2)~~199N`
Suppose force `AF_(2)` makes an angle `alpha` with horizontal
`therefore "In " DeltaANF_(2)`,
`tanalpha=(N)/(f)`
`=(196)/(34.65)`
`=5.6565`
`therefore alpha=tan^(-1)(5.6565)`
`=79^(@)56`.
`therefore alpha~~80^(@)`