Question

What is the reaction forces of the wall and floor on a uniform ladder, $3m$ long weighing $20kg$ leaning against a frictionless wall with its foot resting on a rough floor $1m$ from the wall.

• A. $25\sqrt{2}N, \, 203 \, N$
• B. $50\sqrt{2} \, N, \, 230 \, N$
• C. $203 \, N, \, 25\sqrt{2} \, N$
• D. $230 \, N, \, 50\sqrt{2} \, N \, $

Answer 1

Answer: (A)

Let $AB$ be ladder.
$\therefore AB=3 \, m$
Its foot $A$ is at distance $1 \, m$ from the wall,
$\therefore \, AC=1 \, m$
$B C=\sqrt{\left.(A B)^{2}-(A C)^{2}\right)}=\sqrt{(3)^{2}-(1)^{2}}=2 \sqrt{2} m$
The various forces acting on the ladder are:
$\left(\right.i\left.\right)$ Weight $W$ acting at its centre of gravity $G$ .
$\left(\right.ii\left.\right)$ Since the wall is frictionless, reaction force $R_{1}$ of the wall acting perpendicular to the wall.
$\left(\right.iii\left.\right)$ Reaction force $R_{2}$ of the floor. This force can be resolved into two components, the normal reaction $N$ and the force of friction $f$ .
For translation equilibrium in the horizontal direction,
$f-R_{1}=0$ or $f=R_{1}$
For translation equilibrium in the vertical direction,
$N=W=20g=20\times 10=200 \, N$
For rotational equilibrium, taking moment of the forces about $A$ , we get
$R_{1}\left(2 \sqrt{2}\right)-W\left(\frac{1}{2}\right)=0$
$\Rightarrow R_{1}=\frac{W}{4 \sqrt{2}}=\frac{200}{4 \sqrt{2}}=25\sqrt{2}N$
From $\left(i i\right), \, f=R_{1}=25\sqrt{2} \, N$
$R_{2}=\sqrt{N^{2} + f^{2}}=\sqrt{\left(200 \, N\right)^{2} + \left(25 \sqrt{2} N\right)^{2}}=203 \, N$