1. **Problem statement:** A uniform rod AB of length 6 m and weight 120 N is supported at points C and D. The distances are AC = 1 m and AD = 5 m. An external weight of 80 N is placed on support D. We need to find the reaction forces at supports C and D when the rod is in horizontal equilibrium. 2. **Known data:** - Length of rod, $L = 6$ m - Weight of rod, $W = 120$ N (uniformly distributed) - External weight at D, $W_D = 80$ N - Positions: $AC = 1$ m, $AD = 5$ m 3. **Concepts and formulas:** - The rod is in static equilibrium, so the sum of vertical forces and moments must be zero. - Let $R_C$ and $R_D$ be the reaction forces at supports C and D respectively. Equilibrium conditions: $$\sum F_y = 0 \implies R_C + R_D = W + W_D = 120 + 80 = 200$$ Taking moments about point C (counterclockwise positive): $$\sum M_C = 0$$ The weight of the rod acts at its center, which is at 3 m from A, so distance from C is $3 - 1 = 2$ m. The external weight acts at D, which is 4 m from C ($5 - 1 = 4$ m). Moment equation: $$R_D \times 4 - 120 \times 2 - 80 \times 4 = 0$$ 4. **Calculations:** $$4 R_D = 120 \times 2 + 80 \times 4 = 240 + 320 = 560$$ $$R_D = \frac{560}{4} = 140$$ Using vertical force equilibrium: $$R_C + 140 = 200 \implies R_C = 60$$ 5. **Answer:** The reactions are: $$R_C = 60 \text{ N}, \quad R_D = 140 \text{ N}$$ This corresponds to option (c).