1. **State the problem:** We have a system of rods and weights in equilibrium. We need to find the weights of objects A, B, and C given the distances from the pivot points and the known weight of 6.0 N. 2. **Recall the principle of moments:** For a system in equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments about any pivot point. 3. **Identify pivot points and distances:** - Object B hangs 3.0 cm from pivot S2. - Rod S2 is 5.0 cm long and 2.0 cm from pivot S1. - Rod S1 is 6.0 cm long to the right and holds object C 8.0 cm from pivot S3. - A 6.0 N weight hangs 4.0 cm from the pivot between rods S3 and object A at 8.0 cm. 4. **Calculate moments about pivot S1:** Let weights be $W_A$, $W_B$, and $W_C$. Sum of moments about S1: $$W_B \times 3.0 = 6.0 \times 4.0 + W_A \times 8.0 + W_C \times 6.0$$ 5. **Calculate moments about pivot S2:** $$W_B \times 3.0 = 6.0 \times 4.0 + W_A \times 8.0$$ 6. **Calculate moments about pivot S3:** $$6.0 \times 4.0 = W_C \times 6.0$$ 7. **Solve for $W_C$ from pivot S3:** $$6.0 \times 4.0 = W_C \times 6.0$$ $$24.0 = W_C \times 6.0$$ $$W_C = \frac{24.0}{6.0} = 4.0$$ 8. **Substitute $W_C$ into the equation from step 4:** $$W_B \times 3.0 = 6.0 \times 4.0 + W_A \times 8.0 + 4.0 \times 6.0$$ $$W_B \times 3.0 = 24.0 + 8.0 W_A + 24.0$$ $$W_B \times 3.0 = 48.0 + 8.0 W_A$$ 9. **Assuming equilibrium and symmetry, solve for $W_B$ and $W_A$ by isolating variables or using additional information if available.** Since the problem does not provide more equations, we can express $W_B$ in terms of $W_A$: $$W_B = \frac{48.0 + 8.0 W_A}{3.0} = 16.0 + \frac{8}{3} W_A$$ **Final answers:** - $W_C = 4.0$ - $W_B = 16.0 + \frac{8}{3} W_A$ - $W_A$ remains unknown without additional data. This completes the calculation based on the given data.