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

The ladder AB is 3 m long, its foot A is at distance `AC = 1 m` from the wall. From Pythagoras theorem, `BC = 2 sqrt(2) m`. The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces `F_(1) and F_(2)` of the wall and the floor respectively. Force `F_(1)` is perpendicular to the wall, since the wall is frictionless. Force `F_(2)` is resolved into two components, the normal reaction N and the force of friction F. Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall.
For translational equilibrium, taking the forces in the vertical direction,
`N – W = 0` (i)
Taking the forces in the horizontal direction, `F – F_(1) = 0` (ii)
For rotational equilibrium, taking the moments of the forces about A,
`2sqrt(2)F_(1)-(1//2)W=0` (iii)
Now `W = 20 g = 20 × 9.8 N = 196.0 N`
From (i) `N = 196.0 N`
From (iii) `F_(1)=W//4sqrt(2)=196.0//4sqrt(2)=34.6N`
From (ii) `F=F_(1)=34.6N`
`F_(2)=sqrt(F^(2)+N^(2))=199.0N`
The force `F_(2)` makes an angle `α` with the horizontal,
`tanalpha=N//F=4sqrt(2),alpha=tan^(-1)(4//sqrt(2))~~80^(@)`