Question

A 4 m long ladder weighing $25kg$ rests with its upper end against a smooth wall and lower end on rough ground. What should be the minimum coefficient of friction between the ground and the ladder for it to be inclined at ${60}^{\circ }$ with the horizontal without slipping? (Take $g=10m/s^2$).

• A. $0.288$
• B. $0.3$
• C. $0.4$
• D. $0.15$

Answer 1

The correct option is A $0.288$

In the figure, AB is a ladder of weight $W$ which acts at its centre of gravity G.
$\angle ABC={60}^{\circ }$
$\angle BAC={30}^{\circ }$
Let
$N_1$ be the reaction of the wall, and
$N_2$ the reaction of the ground. Force of friction $f$
between the ladder and the ground acts along BC. For horizontal
equilibrium,
$f=N_1$ ... (i)
For vertical equilibrium,
$N_2=W$ ... (ii)
Taking moments about B, we get for equilibrium,
$N_1\left(4\cos {30}^{\circ }\right)-W\left(2\cos {60}^{\circ }\right)=0$ ...(iii)
Here, $W=250N$
Solving these three equations, we get
$f=72.17Nand{N}_2=250N$
$\therefore \mu =\frac{f}{N_2}=\frac{72.17}{250}=0.288$