$A$ $3\; m$ 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 figure. Find the reaction forces of the wall and the floor.

(N/A) The ladder $AB$ is $3\; m$ long,its foot $A$ is at distance $AC = 1\; m$ from the wall. From Pythagoras theorem,$BC = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}\; m$.
The forces on the ladder are its weight $W$ acting at its centre of gravity $D$,and 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:
$N - W = 0 \implies N = W = 20\; kg \times 9.8\; m/s^2 = 196.0\; N$.
Taking the forces in the horizontal direction:
$F - F_1 = 0 \implies F = F_1$.
For rotational equilibrium,taking the moments of the forces about $A$:
$F_1 \times (BC) - W \times (AE) = 0$,where $AE = 0.5\; m$ (since $D$ is the midpoint of $AB$).
$F_1 \times (2\sqrt{2}) - 196.0 \times 0.5 = 0$
$F_1 = 98 / (2\sqrt{2}) = 49 / \sqrt{2} \approx 34.65\; N$.
Thus,the reaction force of the wall is $F_1 \approx 34.65\; N$.
The reaction force of the floor is $F_2 = \sqrt{F^2 + N^2} = \sqrt{34.65^2 + 196^2} \approx 199.04\; N$.