1. **State the problem:** We need to find the length of cord AC so that an 8-kg lamp is suspended in equilibrium at point A. The spring AB has an undeformed length $l_{AB} = 0.4$ m and stiffness $k_{AB} = 300$ N/m. The spring is stretched from B to A, and cord AC makes a 30° angle with the horizontal. 2. **Identify forces and geometry:** - Weight of lamp: $W = mg = 8 \times 9.81 = 78.48$ N downward. - Spring force $F_{AB} = k_{AB} \times \Delta l$, where $\Delta l$ is the extension of the spring from its natural length. - Cord AC tension $T$ acts along AC at 30° above horizontal. 3. **Set coordinate system and equilibrium:** At point A, forces must balance: - Horizontal: $T \cos 30^\circ = F_{AB}$ - Vertical: $T \sin 30^\circ = W$ 4. **Solve for tension $T$ from vertical equilibrium:** $$T = \frac{W}{\sin 30^\circ} = \frac{78.48}{0.5} = 156.96 \text{ N}$$ 5. **Find spring force $F_{AB}$ from horizontal equilibrium:** $$F_{AB} = T \cos 30^\circ = 156.96 \times \frac{\sqrt{3}}{2} = 135.95 \text{ N}$$ 6. **Calculate spring extension $\Delta l$:** $$\Delta l = \frac{F_{AB}}{k_{AB}} = \frac{135.95}{300} = 0.4532 \text{ m}$$ 7. **Find current length of spring AB:** $$l_{AB} = l_{AB,0} + \Delta l = 0.4 + 0.4532 = 0.8532 \text{ m}$$ 8. **Determine length of cord AC:** From geometry, horizontal distance between B and A is 2 m (given). Since spring AB is horizontal, point A is 0.8532 m from B horizontally. The horizontal distance from A to C is: $$AC_x = 2 - 0.8532 = 1.1468 \text{ m}$$ Cord AC makes 30° with horizontal, so length of AC is: $$l_{AC} = \frac{AC_x}{\cos 30^\circ} = \frac{1.1468}{\sqrt{3}/2} = 1.323 \text{ m}$$ **Final answer:** The required length of cord AC is approximately $\boxed{1.32}$ m.