1. **Problem statement:** A uniform rod AB of length 120 cm and weight W N is supported at points A and C. The reaction at A, $R_A$, is $\frac{2}{3}$ of the reaction at C, $R_C$. We need to find the distance AC such that the rod is in horizontal equilibrium. 2. **Knowns and unknowns:** - Length of rod, $AB = 120$ cm - Weight, $W$ acts at the center of the rod (at 60 cm from A) - $R_A = \frac{2}{3} R_C$ - Distance $AC = x$ (unknown) 3. **Equilibrium conditions:** For the rod to be in equilibrium: - Sum of vertical forces = 0 - Sum of moments about any point = 0 4. **Sum of vertical forces:** $$ R_A + R_C = W $$ 5. **Using the relation between reactions:** $$ R_A = \frac{2}{3} R_C \implies \frac{2}{3} R_C + R_C = W \implies \frac{5}{3} R_C = W \implies R_C = \frac{3}{5} W $$ Then, $$ R_A = \frac{2}{3} \times \frac{3}{5} W = \frac{2}{5} W $$ 6. **Sum of moments about A:** Taking moments about point A (counterclockwise positive): - Moment due to $R_C$ at distance $x$ is $R_C \times x$ - Moment due to weight $W$ at center (60 cm) is $W \times 60$ For equilibrium: $$ R_C \times x = W \times 60 $$ Substitute $R_C = \frac{3}{5} W$: $$ \frac{3}{5} W \times x = W \times 60 $$ Divide both sides by $W$: $$ \frac{3}{5} x = 60 $$ Solve for $x$: $$ x = \frac{60 \times 5}{3} = 100 \text{ cm} $$ 7. **Answer:** The distance $AC$ is 100 cm. **Final answer:** $\boxed{100 \text{ cm}}$ (option c)