A `3m` along ladder weighing `20 kg` leans on a frictionless wall. Its feet rest on the floor `1 m` from the wall. Find the rection 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=2sqrt(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 components, the normal reaction N and the force of friction F. Not 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
Taking the forces in the horizontal direction,
`F-F_(1)=0`
For rotational equilibrium, taking the moments of the forces about A,
`2sqrt(2) F_(1)-(1//2)W=0`
Now `W=20g=20xx9.8N=196.0 N`
From (i) N=196.0 N
From (iii) `F_(1)=W//4sqrt(2)=196.0//4sqrt(2)=34.6 N`
From (ii) `F=F_(1)=34.6 N`
`F_(2)=sqrt(F^(2)+N^(2))=199.0 N`
The force `F_(2)` makes an angle `alpha` with the horizontal.
`tan alpha=N//F=4sqrt(2), alpha=tan^(-1)(4sqrt(2))~~80^(@)`.