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$3\; m$ long ladder wetghing $20 kg$ leans on a frictionless wall. Its feet rest on the floor $1\; m$ from the wall as shown in Figure Find the reaction forces of the wall and the floor.
Solution
The ladder AB is 3 m long, its foot $A$ is at distance $A C=1$ 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 $F_{1}$, 1 s perpendicular to the wall, since the wail 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. $N-W=0$
Taking the forces in the horizontal direction.
$F-F_{1}=0$
For rotational equilibrium, taking the moments of the forces about $A$
$2 \sqrt{2} F_{1}-(1 / 2) W=0$
Now $\quad W=20 g =20 \times 9.8 N =196.0 N$
$N=196.0 N$
$F_{1}=W / 4 \sqrt{2}=196.0 / 4 \sqrt{2}=34.6 N$
$F=F_{1}=34.6 N$
$F_{2}=\sqrt{F^{2}+N^{2}}=199.0 N$
The force $F_{2}$ makes an angle $\alpha$ with the horizontal.
$\tan \alpha=N / F=4 \sqrt{2}, \quad \alpha=\tan ^{-1}(4 \sqrt{2}) \approx 80^{\circ}$
