1. **State the problem:** A uniform rod AB of length 4 m and weight 40 N rests on supports at A and B. A weight of 60 N is placed at point C where AC = 3 m. We need to find the magnitudes of the reactions at supports A ($R_A$) and B ($R_B$) when the rod is in horizontal equilibrium. 2. **Set up the equilibrium conditions:** Since the rod is in equilibrium, the sum of vertical forces and the sum of moments about any point must be zero. - Sum of vertical forces: $$R_A + R_B = 40 + 60 = 100$$ - Sum of moments about A (taking clockwise moments as positive): $$R_B \times 4 = 40 \times 2 + 60 \times 3$$ Here, 40 N acts at the midpoint (2 m from A), and 60 N acts at 3 m from A. 3. **Calculate moments:** $$R_B \times 4 = 80 + 180 = 260$$ $$R_B = \frac{260}{4} = 65$$ 4. **Calculate $R_A$ using vertical forces sum:** $$R_A + 65 = 100$$ $$R_A = 100 - 65 = 35$$ 5. **Conclusion:** The reactions at supports are: $$R_A = 35\,N, \quad R_B = 65\,N$$ This corresponds to option (c).