1. **State the problem:** A non-uniform rod AB is 10 m long with a weight of 200 N acting at point C. The rod is supported at ends A and B. The reaction force at support A is 80 N. We need to find the distance AC such that the rod is in equilibrium. 2. **Relevant formulas and rules:** - For equilibrium, the sum of vertical forces must be zero: $$R_A + R_B = W$$ where $R_A$ and $R_B$ are reactions at supports A and B, and $W$ is the weight. - The sum of moments about any point must be zero. Taking moments about A: $$R_B \times AB = W \times AC$$ 3. **Given values:** - Length of rod, $AB = 10$ m - Weight, $W = 200$ N - Reaction at A, $R_A = 80$ N 4. **Calculate reaction at B:** $$R_B = W - R_A = 200 - 80 = 120 \text{ N}$$ 5. **Apply moment equilibrium about A:** $$R_B \times AB = W \times AC$$ Substitute values: $$120 \times 10 = 200 \times AC$$ $$1200 = 200 \times AC$$ 6. **Solve for AC:** $$AC = \frac{1200}{200} = 6 \text{ m}$$ **Final answer:** $$\boxed{6 \text{ m}}$$ The correct choice is (b) AC = 6 m.