Question

A 4 metre long ladder weighing 25 kg 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? 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

Answer: (B)

0.29

Answer 2

Here, length of ladder, $AB = 4 m$
$W$ = 25 kg at the C.G. (C) of the ladder, as shown in the figure.
$? ABO = 60?, ? BAO = 30?$
Let $R_1$ be reaction of the wall, normal of the wall and $R_2$ be the reaction of the ground normal to the ground.

Force of friction $(f )$ between the ladder and the ground acts along $BO$.
In equilibrium, $R_2 = W$ ...(i)
$f = R_1$ .....(ii)
Taking moments about $B$, at equilibrium,
$R_2 \times 0 - W \times BD + R_1 \times AO = 0$
Using (i) and (ii), we get
$- R_2 \times BD + f \times AO = 0$
or $\frac{f}{R_2} = \frac{BD}{AO} = \frac{BC \, \cos \, 60^\circ}{AB \, \sin \, 60^\circ}$
or $\mu = \frac{2}{4\sqrt{3}} = \frac{1}{2\sqrt{3}} = 0.29$