Answer:
Here,
length of ladder, AB = 4 m.
\[\text{W}=\text{25kg}\] at the
c.g (C) of the ladder, Fig. 3(HT).12.
\[\angle ABO={{60}^{o}}\],
\[\angle BAO={{30}^{o}}\]
Let \[{{R}_{1}}\] be reaction of
the wall, normal to the wall and \[{{R}_{2}}\] be the reaction of the ground normal
to the ground.
Force of friction (/) between
the ladder and the ground acts along BO.
In equilibrium, \[{{R}_{2}}=W\] ?..
(i)
\[f={{R}_{1}}\] ?..
(ii)
Taking moments about B, we get
for 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 \[\frac{f}{{{R}_{2}}}=\frac{BD}{AO}=\frac{BC\,\cos
{{60}^{o}}}{AB\,\sin {{60}^{o}}}\] or \[\mu
=\frac{2}{4\sqrt{3}}=\frac{1}{2\sqrt{3}}=0.29\]
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