Question

A uniform ladder $3\, m$ long weighing $20 \,kg$ leans against a frictionless wall. Its foot rest on a rough floor $1 \,m$ from the wall. The reaction forces of the wall and floor are

• 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 the ladder.
$\therefore AB = 3\, m$
Its foot $A$ is at distance $1\, m$ from the wall, $\therefore AC = 1 \,m $ and
$BC = \sqrt{\left(AB\right)^{2}-\left(AC\right)^{2}} $
$= \sqrt{\left(3\right)^{2} -\left(1\right)^{2}} $
$ = 2\sqrt{2}m $
The various forces acting on the ladder are
$(i)$ Weight W acting at its centre of gravity $G$.
$(ii)$ Since the wall is frictionless, reaction force $R_1$ of the wall acting perpendicular to the wall.
$(iii)$ 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 translatory equilibrium in the horizontal direction,
$f - R_1 = 0$ or $f = R_1\quad ... (i)$
For translatory equilibrium in the vertical direction,
$N = W = 20 g = 20 \times 10 = 200\,N \quad... (ii)$ For rotational equilibrium, taking moment of the forces about $A$, we get
$R_{1} \left(2\sqrt{2}\right)-W \left(\frac{1}{2}\right) = 0$
$R_{1} = \frac{W}{4\sqrt{2}} = \frac{200}{4\sqrt{2}} $
$ = 25\sqrt{2} N \quad...\left(iii\right)$
From $\left(ii\right), f = R_{1} = 25\sqrt{2} N $
$ R_{2} = \sqrt{N^{2} +f^{2}}$
$= \sqrt{\left(200 N^{2}\right)+\left(25\sqrt{2}N\right)^2} = 203 N$