1. **State the problem:** We have two pipes weighing 43 kg and 28 kg supported at points A and B by two cords AB and AD and a spring CB. The system is in equilibrium with angles \(\theta=37^\circ\) and \(\phi=24^\circ\). We need to find the tensions in cables AB and AD. 2. **Identify forces:** - Weight at A: \(W_A = 43 \times 9.81 = 421.83\,\text{N}\) downward. - Weight at B: \(W_B = 28 \times 9.81 = 274.68\,\text{N}\) downward. - Tension in AB: \(T_{AB}\) along cable AB at angle \(\phi=24^\circ\) from horizontal. - Tension in AD: \(T_{AD}\) along cable AD (angle not given, assume horizontal for simplicity or solve symbolically). - Spring force in CB: \(F_{CB}\) at angle \(\theta=37^\circ\) from vertical. 3. **Set coordinate system:** Let horizontal be x-axis and vertical be y-axis. 4. **Write equilibrium equations:** At point A, sum of forces in x and y directions must be zero. - Horizontal (x): $$ T_{AB} \cos(24^\circ) + T_{AD} = F_{CB} \sin(37^\circ) $$ - Vertical (y): $$ T_{AB} \sin(24^\circ) + W_A = F_{CB} \cos(37^\circ) $$ At point B, vertical equilibrium: $$ F_{CB} \cos(37^\circ) = W_B + T_{AB} \sin(24^\circ) $$ But since spring CB connects C and B, and tensions are internal, we focus on A's equilibrium. 5. **Solve for \(T_{AB}\) and \(T_{AD}\):** From vertical equilibrium at A: $$ F_{CB} = \frac{T_{AB} \sin(24^\circ) + W_A}{\cos(37^\circ)} $$ Substitute into horizontal equilibrium: $$ T_{AB} \cos(24^\circ) + T_{AD} = \left(\frac{T_{AB} \sin(24^\circ) + W_A}{\cos(37^\circ)}\right) \sin(37^\circ) $$ Simplify: $$ T_{AB} \cos(24^\circ) + T_{AD} = (T_{AB} \sin(24^\circ) + W_A) \tan(37^\circ) $$ Rearranged: $$ T_{AD} = (T_{AB} \sin(24^\circ) + W_A) \tan(37^\circ) - T_{AB} \cos(24^\circ) $$ 6. **Numerical values:** - \(\sin(24^\circ) \approx 0.4067\) - \(\cos(24^\circ) \approx 0.9135\) - \(\tan(37^\circ) \approx 0.7536\) - \(W_A = 421.83\,\text{N}\) Substitute: $$ T_{AD} = (T_{AB} \times 0.4067 + 421.83) \times 0.7536 - T_{AB} \times 0.9135 $$ $$ T_{AD} = 0.3063 T_{AB} + 318.1 - 0.9135 T_{AB} = -0.6072 T_{AB} + 318.1 $$ 7. **Additional relation needed:** To find numeric values, we need the angle of AD or another equation. Assuming AD is horizontal (angle 0°), tension is purely horizontal. 8. **Assuming equilibrium at B:** At B, vertical forces: $$ F_{CB} \cos(37^\circ) = W_B $$ $$ F_{CB} = \frac{W_B}{\cos(37^\circ)} = \frac{274.68}{0.7986} = 343.9\,\text{N} $$ 9. **From step 5 vertical equilibrium at A:** $$ F_{CB} = \frac{T_{AB} \sin(24^\circ) + W_A}{\cos(37^\circ)} $$ Substitute \(F_{CB} = 343.9\): $$ 343.9 = \frac{T_{AB} \times 0.4067 + 421.83}{0.7986} $$ Multiply both sides: $$ 343.9 \times 0.7986 = T_{AB} \times 0.4067 + 421.83 $$ $$ 274.68 = 0.4067 T_{AB} + 421.83 $$ $$ 0.4067 T_{AB} = 274.68 - 421.83 = -147.15 $$ $$ T_{AB} = \frac{-147.15}{0.4067} = -361.8\,\text{N} $$ Negative tension is not physical, so assumption about AD angle or forces may be incorrect or more info is needed. **Summary:** - Without angle of AD or more info, exact tensions cannot be found. - Method involves writing equilibrium equations and solving simultaneously. **Final answer:** Tensions depend on unknown angle of AD; more data needed for numeric solution.