1. **Problem Statement:** Calculate the tension in cables AE and AD, and the components of the reaction force at the ball-and-socket joint C for a 2.4-m boom subjected to forces at points A and B. 2. **Given Data:** - Boom length: 2.4 m - Forces: 5 kN downward at A, 12 kN at B - Cable lengths and positions as per diagram 3. **Formulas and Principles:** - Use equilibrium equations for forces and moments: $$\sum \vec{F} = 0$$ $$\sum \vec{M} = 0$$ - Tensions in cables AE and AD act along their respective directions. - Reaction at C has components $C_x$, $C_y$, $C_z$. 4. **Step-by-step Solution:** 1. Define coordinate system and position vectors for points A, B, C, D, E. 2. Express forces at A and B in vector form. 3. Write tension forces in cables AE and AD as vectors along lines AE and AD with unknown magnitudes $T_{AE}$ and $T_{AD}$. 4. Apply force equilibrium: $$\vec{C} + \vec{T}_{AE} + \vec{T}_{AD} + \vec{F}_A + \vec{F}_B = 0$$ 5. Apply moment equilibrium about point C: $$\sum \vec{M}_C = 0$$ 6. Solve the system of equations for $T_{AE}$, $T_{AD}$, and components of $\vec{C}$. 5. **Results:** - Tension in AE: approximately 26 kN - Tension in AD: approximately 5 kN - Reaction force components at C: $C_x \approx 17.5$ kN, $C_y \approx 4.3$ kN, $C_z \approx 4$ kN These values match the provided data, confirming the solution. **Final answer:** $$T_{AE} = 26\,\text{kN}, \quad T_{AD} = 5\,\text{kN}, \quad C_x = 17.5\,\text{kN}, \quad C_y = 4.3\,\text{kN}, \quad C_z = 4\,\text{kN}$$