Question

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

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

Answer 1

The correct option is B $0.29$

In figure $AB$ is a ladder of weight w which acts at its center of gravity G

$\angle ABC={60}^{\circ }$

$\therefore \angle BAC={30}^{\circ }$

Let $N_1$ be the reaction of the wall, and $N_2$ reaction of the ground . Force of friction f between ladder and the ground act along BC. For horizontal equalibrium

$f=N_1$ ...(i)

For vertical equalivrium

$N_2=w$ ...(ii)

taking moment about B we get for equalibrium

$N_1\left(4\cos {30}^{\circ }\right)-w\left(2\cos {60}^{\circ }\right)=0$ ...(iii)

Here $w=250N$

Solving these three equations, we get

$f=72.17N$

and $N_2=250N$

$\therefore \mu =\frac{f}{N_2}=\frac{72.17}{250}=0.288$