Answer to Question #140497 in Quantum Mechanics for Blessings

Question

A ladder rests in limiting equilibrium against a rough vertical wall and with it foot on rough horizontal ground,the coefficient of friction at both points of contact been 0.5.the ladder is uniform and weights 300N.find the angle which the angle of the ladder with horizontal ground?

Answer 1

Solution

The forces equilibrium conditions give

$$\sum F_x=0:\; N'-F_{f2}=0;\\ \sum F_y=0:\; N''+F_{f1}-mg=0$$

The friction force is related with normal reaction by relation

$$F_{f1}=\mu N_1;\: F_{f2}=\mu N_2$$

Hence

$$N''=\frac{mg}{\mu^2+1},\: N'=\frac{\mu mg}{\mu^2+1}$$

The momentum equilibrium condition gives

$$F_{f1}L\cos\alpha+N'L\sin\alpha-mg\frac{L}{2}\cos\alpha=0$$

So

$$\tan\alpha=\frac{mg/2-F_{f1}}{N'}=\frac{mg/2-\mu N'}{N'}$$ $$=\frac{mg}{2N'}-\mu=\frac{\mu^2+1}{2\mu}-\mu=\frac{1-\mu^2}{2\mu}$$

Finally

$$\alpha=\tan^{-1}\left(\frac{1-\mu^2}{2\mu}\right)=\tan^{-1}\left(\frac{1-0.5^2}{2\times 0.5}\right)=37^{\circ}$$