1. Problem 16: Calculate weights A, B, and C assuming rods have negligible mass and system is in equilibrium. 2. Use the principle of moments: For equilibrium, sum of clockwise moments = sum of anticlockwise moments about any pivot. 3. Assign pivot at point where forces act and write moment equations for each segment S1, S2, S3. 4. Let weights be $W_A$, $W_B$, $W_C$. Given force at S3 is 6.0 N. 5. Write moment balance equations: - Taking moments about the left end, $$W_B \times (2.0 + 3.0) + W_C \times (2.0 + 3.0 + 4.0) = 6.0 \times (2.0 + 3.0 + 4.0 + 6.0) + W_A \times (2.0 + 3.0 + 4.0 + 6.0 + 5.0 + 8.0)$$ 6. Calculate distances: - $W_B$ at 5.0 cm, - $W_C$ at 9.0 cm, - 6.0 N force at 15.0 cm, - $W_A$ at 20.0 cm. 7. Substitute and solve for $W_A$, $W_B$, $W_C$ using equilibrium conditions and sum of vertical forces = 0. --- 8. Problem 17: A uniform meter rule pivoted at 34 cm balances with 64 g mass at 4 cm. Find mass of meter rule. 9. Use principle of moments about pivot: $$\text{Mass} \times (\text{distance of center of mass from pivot}) = 64 \times (34 - 4)$$ 10. Center of mass of uniform rule is at 50 cm mark, distance from pivot = $|50 - 34| = 16$ cm. 11. Let mass of rule be $M$ grams: $$M \times 16 = 64 \times 30$$ 12. Solve for $M$: $$M = \frac{64 \times 30}{16} = 120$$ grams. --- 13. Problem 18: Calculate total moment about P due to two forces on bar PQ. 14. Forces: 16 N at 30° and 5 N horizontal, both at 50 cm from P. 15. Moment = Force $\times$ perpendicular distance. 16. For 16 N force, perpendicular component = $16 \times \sin 30^\circ = 8$ N. 17. For 5 N force, perpendicular component = $5 \times \cos 0^\circ = 5$ N (since horizontal force acts perpendicular to bar). 18. Total moment about P: $$8 \times 0.5 + 5 \times 0.5 = 4 + 2.5 = 6.5$$ Nm. 19. Check directions and sum moments accordingly; if both cause rotation in same direction, add moments. 20. Final total moment about P is 6.5 Nm.