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.6.27. Find the reaction forces of the wall and the floor.
Example-(6.9)
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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 $F_1$ 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 \quad\quad\text{(i)}
$
Taking the forces in the horizontal direction,
$
F-F_1=0 \quad\quad\text{(ii)}
$
For rotational equilibrium, taking the moments of the forces about A,
$
2 \sqrt{2} F_1-(1 / 2) W=0 \quad\quad\text{(iii)}
$
Now $\quad W=20 g =20 \quad 9.8 N =196.0 N$
From (i) $N=196.0 N$
From (iii) $F_1=W / 4 \sqrt{2}=196.0 / 4 \sqrt{2}=34.6 N$
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=4 \sqrt{2}, \quad \alpha=\tan ^{-1}(4 \sqrt{2}) \approx 80^{\circ}
$
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