Answer:
[tex]\displaystyle T_{1} = m\, g\, \frac{\cos(\theta_{2})}{\sin(\theta_{2})}[/tex].
[tex]\displaystyle T_{2} = \frac{m\, g}{\sin(\theta_{2})}[/tex].
Explanation:
Assuming that the mass is stationary, forces on the mass should be balanced- both in the horizontal component and in the vertical component.
Forces in the vertical component are:
- Weight of the mass: [tex]m\, g[/tex] downward.
- Vertical component of [tex]T_{2}[/tex]: [tex]T_{2}\, \sin(\theta_{2})[/tex] upward.
Since these forces balance each other:
[tex]T_{2}\, \sin(\theta_{2}) = m\, g[/tex].
[tex]\displaystyle T_{2} = \frac{m\, g}{\sin(\theta_{2})}[/tex].
Similarly, forces in the horizontal component are:
- [tex]T_{1}[/tex] (left.)
- Horizontal component of [tex]T_{2}[/tex]: [tex]T_{2}\, \cos(\theta_{2})[/tex] (right.)
Hence:
[tex]\begin{aligned} T_{1} &= T_{2}\, \cos(\theta_{2}) = \frac{m\, g\, \cos(\theta_{2})}{\sin(\theta_{2})}\end{aligned}[/tex].
