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. $25\sqrt{2}N,203N$
• 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 ladder.
$\therefore AB=3m$
Its foot $A$ is at distance $1m$ from the wall,
$\therefore AC=1m$
$BC=\sqrt{(AB{)}^2-(AC{)}^2)}=\sqrt{(3{)}^2-(1{)}^2}=2\sqrt{2}m$
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 translation equilibrium in the horizontal direction,
$f-R_1=0$ or $f=R_1$
For translation equilibrium in the vertical direction,
$N=W=20g=20\times 10=200N$
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 $\left(ii\right),f=R_1=25\sqrt{2}N$
$R_2=\sqrt{N^2+f^2}=\sqrt{{\left(200N\right)}^2+{\left(25\sqrt{2}N\right)}^2}=203N$