Question

A ladder rests in limiting equilibrium against a rough vertical wall and with its foot on rough horizontal ground, the coefficient of friction at both points of contact being 0.5. The ladder is uniform and weighs 300N. Find the angle

Answer 1

The forces equilibrium conditions give

$$\sum F_x=0:\; N_1-F_{f2}=0;\\ \sum F_y=0:\; N_2+F_{f1}-mg=0$$

The friction force is related with normal reaction by relation

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

Hence

$$N_2=\frac{mg}{\mu^2+1},\: N_1=\frac{\mu mg}{\mu^2+1}$$

The momentum equilibrium condition gives

$$F_{f1}L\cos\alpha+N_1L\sin\alpha-mg\frac{L}{2}\cos\alpha=0$$

So

$$\tan\alpha=\frac{mg/2-F_{f1}}{N_1}=\frac{mg/2-\mu N_1}{N_1}$$

$$=\frac{mg}{2N_1}-\mu=\frac{\mu^2+1}{2\mu}-\mu=\frac{1-\mu^2}{2\mu}$$

Finally

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