To solve for the tensions in the strings, we resolve the forces horizontally and vertically.
1. Resolve horizontally:
\(T_1 \cos 35^{\circ} = T_2 \cos 40^{\circ}\)
2. Resolve vertically:
\(T_1 \sin 35^{\circ} + T_2 \sin 40^{\circ} = 2.4g\)
where \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\).
3. Solve the system of equations:
From the horizontal resolution, express \(T_2\) in terms of \(T_1\):
\(T_2 = \frac{T_1 \cos 35^{\circ}}{\cos 40^{\circ}}\)
Substitute \(T_2\) in the vertical resolution equation:
\(T_1 \sin 35^{\circ} + \left(\frac{T_1 \cos 35^{\circ}}{\cos 40^{\circ}}\right) \sin 40^{\circ} = 2.4 \times 9.8\)
Simplify and solve for \(T_1\):
\(T_1 (\sin 35^{\circ} + \frac{\cos 35^{\circ} \sin 40^{\circ}}{\cos 40^{\circ}}) = 23.52\)
\(T_1 = \frac{23.52}{\sin 35^{\circ} + \frac{\cos 35^{\circ} \sin 40^{\circ}}{\cos 40^{\circ}}} \approx 20.4 \text{ N}\)
4. Substitute \(T_1\) back to find \(T_2\):
\(T_2 = \frac{20.4 \cos 35^{\circ}}{\cos 40^{\circ}} \approx 19.0 \text{ N}\)
