1. **Problem Statement:** Determine the tension in each segment of the cable ($F_{BA}$, $F_{BC}$, $F_{CD}$) and the total length of the cable. 2. **Equations of Equilibrium:** At joint B: $$\sum F_x = 0: F_{BC} \cos \theta - F_{BA} \frac{4}{\sqrt{65}} = 0 \quad [1]$$ $$\sum F_y = 0: F_{BA} \frac{7}{\sqrt{65}} - F_{BC} \sin \theta - 50 = 0 \quad [2]$$ At joint C: $$\sum F_x = 0: F_{CD} \cos \phi - F_{BC} \cos \theta = 0 \quad [3]$$ $$\sum F_y = 0: F_{BC} \sin \theta + F_{CD} \sin \phi - 100 = 0 \quad [4]$$ 3. **Geometry and Angles:** $$\sin \theta = \frac{y}{\sqrt{y^2 + 25}}, \quad \cos \theta = \frac{5}{\sqrt{y^2 + 25}}$$ $$\sin \phi = \frac{3 + y}{\sqrt{y^2 + 6y + 18}}, \quad \cos \phi = \frac{3}{\sqrt{y^2 + 6y + 18}}$$ 4. **Substitution and Solution:** Substitute the trigonometric expressions into the equilibrium equations [1]-[4] and solve the system for $F_{BA}$, $F_{BC}$, $F_{CD}$, and $y$. Given solution: $$F_{BC} = 46.7, \quad F_{BA} = 83.0, \quad F_{CD} = 88.1, \quad y = 2.679$$ 5. **Total Cable Length:** Calculate the length of each segment: $$l = \sqrt{7^2 + 4^2} + \sqrt{5^2 + 2.679^2} + \sqrt{3^2 + (2.679 + 3)^2}$$ Calculate each term: $$\sqrt{49 + 16} = \sqrt{65} \approx 8.062$$ $$\sqrt{25 + 7.176} = \sqrt{32.176} \approx 5.672$$ $$\sqrt{9 + 32.176} = \sqrt{41.176} \approx 6.466$$ Sum: $$l = 8.062 + 5.672 + 6.466 = 20.2$$ **Final answers:** - Tensions: $F_{BA} = 83.0$, $F_{BC} = 46.7$, $F_{CD} = 88.1$ (all in lb) - Total cable length: $20.2$ ft