Question

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

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

Answer 1

Answer: (A)

Let $AB$ be the ladder.
$\therefore AB=3m$
Its foot $A$ is at distance $1m$ from the wall, $\therefore AC=1m$ 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 $f=R_1...(i)$
For translatory equilibrium in the vertical direction,
$N=W=20g=20\times 10=200N...(ii)$ For rotational equilibrium, taking moment of the forces about $A$, we get
$R_1\left(2\sqrt{2}\right)-W\left(\frac{1}{2}\right)=0$
$R_1=\frac{W}{4\sqrt{2}}=\frac{200}{4\sqrt{2}}$
$=25\sqrt{2}N...\left(iii\right)$
From $\left(ii\right),f=R_1=25\sqrt{2}N$
$R_2=\sqrt{N^2+f^2}$
$=\sqrt{\left(200N^2\right)+{\left(25\sqrt{2}N\right)}^2}=203N$