Two smooth beads A and B, free to move on a vertical smooth circular wire, are connected by a string. Weights W1, W2, and W are suspended from A, B, and a point C of the string respectively. In equilibrium, A and B are in a horizontal line.

If $ \angle B A C=\alpha \text { and } \angle A B C=\beta \text {, What is the ratio of } \tan \alpha: \tan \beta ? $

Solution:

Resolving forces horizontally and vertically at the points A, B, and C respectively, we get

$ \begin{aligned} &T \cos \alpha=R_{1} \sin \gamma....(i) \\\\ &T_{1} \sin \alpha+W_{1}=R_{1} \cos \gamma...(ii) \\\\ &T_{1} \cos \beta=R_{2} \sin \gamma...(iii) \\\\ &T_{2} \sin \beta+W_{2}=R_{2} \cos \gamma...(iv) \\\\ &T_{1} \cos \alpha=T_{2} \cos \beta...(v) \\\\ &T_{1} \sin \alpha+T_{2} \sin \beta=W......(vi) \\ \end{aligned} $

Using (v), from (i)and (ii), we get,

$ R_{1}=R_{2} \\ $

$\therefore$ From (ii) and (vi), we have,

$ T_{1} \sin \alpha+W_{1}=T_{2} \sin \beta+W_{2} \\ $

or,

$ T_{1} \sin \alpha-T_{2} \sin \beta=W_{2}-W_{1} \\ $

Adding and subtracting (vi) and (vii), we get,

$ 2 T_{1} \sin \alpha=W+W_{2}-W_{1} \\ $ $ 2 T_{2} \sin \beta=W-W_{2}+W_{1} \\ $

Dividing (viii) by (ix), we get,

$ \frac{T_{1}}{T_{2}} \cdot \frac{\sin \alpha}{\sin \beta}=\frac{W-W_{1}+W_{2}}{W+W_{1}-W_{2}} \quad \\ $

or

$ \frac{\cos \beta}{\cos \alpha} \cdot \frac{\sin \alpha}{\sin \beta}=\frac{W-W_{1}+W_{2}}{W+W_{1}-W_{2}} \quad($ from $(\mathrm{v})) \quad \\ $

or

$ \frac{\tan \alpha}{\tan \beta}=\frac{W-W_{1}+W_{2}}{W+W_{1}-W_{2}} \\ $