question_answer

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 \[\text{6}0{}^\circ \] with the horizontal without slipping? Take\[\text{g}=\text{1}0\text{ m}{{\text{s}}^{\text{-2}}}\].

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                \[\mu =\frac{2}{4\sqrt{3}}=\frac{1}{2\sqrt{3}}=0.29\]