Question

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

Answer 1

The ladder AB is $3$m long, its foot A is at distance AC$=1$m from the wall. From Pythagons 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,
$N-W=0$ $\left(i\right)$
Taking the forces in the horizontal direction,
$F-F_1=0$ $\left(ii\right)$
For rotational equilibrium, taking the moments of the forces about A,
$2\sqrt{2}{F}_1-\left(1/2\right)W=0$ $\left(iii\right)$
Now $W=20g=20\times 9.8N=196.0N$
From $\left(i\right)N=196.0$
From $\left(iii\right)F_1=W/4\sqrt{2}=196.0/4\sqrt{2}=34.6N$
From $\left(ii\right)F=F_1=34.6N$
$F_2=\sqrt{F^2+N^2}=199.0N$
The force $F_2$ makes an angle $\alpha$ with the horizontal,
$\tan \alpha =N/F=4\sqrt{2},\alpha ={\tan }^{-1}\left(4\sqrt{2}\right)={80}^o$