1. **Problem Statement:** We are given two sets of ship parameters and asked to find the effect of GM (metacentric height) and righting moment at a certain heel angle for the first set, and drafts and effective GM for the second set. --- ### First Set: Delta = 5005 tonnes, L = 120 m, t = 0.25 m (stern), KM = 7.6 m, MCTC = 100 t.m, KG = 6 m, LCF = 5 m abaft amidship **i. Find the effect of GM at the critical instant.** 2. **Formula for GM:** $$GM = KM - KG$$ where KM is the height of the metacenter above keel, KG is the height of the center of gravity above keel. 3. **Calculate GM:** $$GM = 7.6 - 6 = 1.6\,m$$ 4. **Effect of GM:** A positive GM of 1.6 m indicates the ship is stable at the critical instant. --- **ii. Calculate the righting moment at this instant for an angle of heel 5°** 5. **Formula for Righting Moment (RM):** $$RM = \Delta \times GM \times \sin(\theta)$$ where $\Delta$ is displacement, $GM$ is metacentric height, and $\theta$ is heel angle in radians. 6. **Convert heel angle to radians:** $$\theta = 5^\circ = \frac{5 \pi}{180} = 0.0873\,rad$$ 7. **Calculate RM:** $$RM = 5005 \times 1.6 \times \sin(0.0873)$$ $$RM = 5005 \times 1.6 \times 0.0872 = 698.5\,t.m$$ --- ### Second Set: TPC = 20 tonnes, MCTC = 150 t.m, CF = 5 m forward amidships, KM = 9.75 m **i. Calculate the drafts of the vessel at the instants when she is taking the blocks forward and aft.** 8. **Draft change formula:** $$\Delta d = \frac{\text{Moment}}{\text{MCTC}}$$ 9. **Moment at forward blocks:** Moment = $\Delta \times LCF = 5005 \times (-5) = -25025\,t.m$ (negative because forward) 10. **Change in draft forward:** $$\Delta d_f = \frac{-25025}{150} = -166.83\,m$$ 11. **Moment at aft blocks:** Moment = $\Delta \times (L - LCF) = 5005 \times (120 - 5) = 5005 \times 115 = 575575\,t.m$ 12. **Change in draft aft:** $$\Delta d_a = \frac{575575}{150} = 3837.17\,m$$ *Note:* These large values suggest a misunderstanding; normally, moments for draft changes are calculated differently. Usually, the moment arm is relative to the center of flotation (CF), not the entire length. 13. **Correct approach:** Moment about CF: Forward moment arm = distance from CF to forward block = 5 m (given) Aft moment arm = distance from CF to aft block = L - CF = 120 - 5 = 115 m 14. **Calculate change in draft forward:** $$\Delta d_f = \frac{\Delta \times \text{moment arm forward}}{\text{MCTC}} = \frac{5005 \times 5}{150} = 166.83\,m$$ 15. **Calculate change in draft aft:** $$\Delta d_a = \frac{5005 \times 115}{150} = 3837.17\,m$$ *These values are still very large, indicating the problem might expect a different interpretation or smaller scale. Possibly, the problem expects the change in draft per tonne or a different unit. Please verify units or problem context.* --- **ii. The ship’s effective GM at this moment if KG is 7.75 m.** 16. **Calculate GM:** $$GM = KM - KG = 9.75 - 7.75 = 2.0\,m$$ --- **Final answers:** - First set: - GM = 1.6 m - Righting moment at 5° heel = 698.5 t.m - Second set: - Draft changes at forward and aft blocks: 166.83 m and 3837.17 m (verify units) - Effective GM = 2.0 m