Question

A $4m$ 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 the ladder for it to be inclined at ${60}^{\circ }$ with the horizontal without slipping? (Take $g=10m{s}^{-2}$)

• A. $0.19$
• B. $0.29$
• C. $0.39$
• D. $0.49$

Answer 1

The correct option is B $0.29$

In the figure, $AB$ is a ladder of weight $W$ which acts at its centre of gravity $G$.

$\angle ABC={60}^o$

$\angle BAC={30}^o$

$N_1$ Let 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 with $BC$.

For horizontal equilibrium,

$f=N_1$........(1)

For vertical equilibrium,

$N_2=W$..............(2)

Taking moments about $B$, we get for equilibrium,

$N_1\left(4cos{30}^o\right)-W\left(2cos{60}^o\right)=0$..................(3)

Here, $W=250N$

Solving these three equations we get

$f=72.17N$ and $N_2=250N$

$\therefore \mu =\frac{f}{N_2}=\frac{72.17}{250}=0.29$