1. **Problem Statement:** Calculate the tension in cable BC given that the tension in cable EF is 5.0 kip and the weight of the beam is 3.5 kip. 2. **Setup and Known Values:** - Tension in cable EF, $T_{EF} = 5.0$ kip - Weight of beam, $W_{beam} = 3.5$ kip - Length of beam AB = 12 ft - Vertical distances and cable attachment points as described. 3. **Assumptions and Approach:** - The beam is in static equilibrium. - Sum of forces and moments must be zero. - We will take moments about point A to find $T_{BC}$. 4. **Calculate moments about point A:** - Moment due to beam weight acts at midpoint of beam (6 ft from A): $$M_{beam} = W_{beam} \times 6 = 3.5 \times 6 = 21 \text{ kip-ft}$$ - Moment due to tension in cable EF: Cable EF is attached at point E, 12 ft from A horizontally. Vertical component of $T_{EF}$ creates moment: $$M_{EF} = T_{EF} \times 12 = 5.0 \times 12 = 60 \text{ kip-ft}$$ - Let $T_{BC}$ be the tension in cable BC. Cable BC is attached at point B (12 ft from A horizontally) and goes up 3 ft vertically to point C. The horizontal distance for moment arm is 12 ft. 5. **Sum of moments about A for equilibrium:** $$\sum M_A = 0 = M_{BC} - M_{EF} - M_{beam}$$ 6. **Express moment from cable BC:** Cable BC tension acts along cable BC, which is vertical distance 3 ft and horizontal 0 ft (since it goes up vertically from B to C). The tension force has a vertical component only, so the moment arm is horizontal distance 0, but since cable BC is vertical, the tension force creates a moment about A equal to: Actually, cable BC tension force acts vertically at point B, so moment arm is horizontal distance from A to B = 12 ft. Therefore, $$M_{BC} = T_{BC} \times 3 \text{ ft (vertical component)}$$ But this is incorrect because the tension force is along cable BC, which is vertical, so the moment arm is horizontal distance from A to B = 12 ft. So moment from $T_{BC}$ is: $$M_{BC} = T_{BC} \times 12$$ 7. **Set up moment equilibrium equation:** $$T_{BC} \times 12 = 60 + 21 = 81$$ 8. **Solve for $T_{BC}$:** $$T_{BC} = \frac{81}{12} = 6.75 \text{ kip}$$ **Final answer:** The tension in cable BC is $\boxed{6.75}$ kip.