1. **State the problem:** We need to find the distance from amidships where a load can be added without changing the draft aft (T_A). 2. **Given data:** - Length of ship, $L = 116$ m - Draft aft, $T_A = 7.3$ m - Draft forward, $T_F = 6.7$ m - Tons per centimeter immersion (TPC), $TPC = 18$ t/cm - Moment to change trim 1 cm, $MCT_{1cm} = 124$ tm - Longitudinal center of flotation (LCF) = 2 m aft of amidships 3. **Calculate initial trim ($ ext{Trim}$):** $$\text{Trim} = T_A - T_F = 7.3 - 6.7 = 0.6 \text{ m (aft)}$$ 4. **Calculate change in trim per ton ($\Delta T_{per\,ton}$):** First convert $MCT_{1cm}$ to $MCT$ per meter: $$MCT = MCT_{1cm} \times 100 = 124 \times 100 = 12400\; \text{tm/m}$$ Then, $$\Delta \text{trim per ton} = \frac{1}{MCT} = \frac{1}{12400} \text{ m/ton}$$ 5. **Determine the weight to change trim by 1 m:** Since $\Delta \text{trim per ton} = \frac{1}{12400}$ m/ton, $$\text{Weight for } 1 m \text{ trim change} = 12400 \text{ tons}$$ 6. **Calculate the total displacement ($\Delta$):** Use average draft: $$T_{avg} = \frac{T_A + T_F}{2} = \frac{7.3 + 6.7}{2} = 7.0 \text{ m}$$ Displacement volume approx = $L \times B \times T_{avg}$ but we lack breadth (B), so use TPC and immersion instead. Calculate total change in draft for all weight (via TPC): Difference in draft between aft and forward and relation to load and moment to change trim. 7. **Use moment to change trim formula:** Adding weight $w$ at distance $x$ from amidships causes a change in trim: $$w \times x = \text{change in trim (m)} \times MCT$$ If draft aft is unchanged, trim change must be zero at stern, so the moment caused by the load does not change the aft draft. To not alter aft draft, the load must be applied to balance the trim shift caused by any other changes. 8. **Calculate moment of displacement about LCF:** The trim moment is given by: $$\text{Trim moment} = \Delta \times (LCF - \text{amidships})$$ Given LCF is 2 m aft, from amidships (0), so LCF = -2 m (aft). 9. **Finding $x$ where load can be added without changing aft draft:** Use the lever principle: $$w \times x = \Delta \times d$$ where $d$ is distance from amidships to LCF (2 m aft), $\Delta$ is total displacement. Also, since draft aft shouldn't change, the load must counterbalance trim moment. 10. **Determine $\Delta$ displacement from TPC:** Displacement in tonnes = change in draft effect over length: Draft difference = $T_A - T_F = 0.6$ m over length $L=116$ m. We do not have breadth $B$ but can calculate displacement change per cm immersion: Total displacement change per cm immersion for ship: $$\text{Displacement} = TPC \times \text{cm immersion}$$ Since TPC is $18$ t/cm, and difference in draft is 60 cm, estimate: $$\Delta = 18 \times 60 = 1080 \text{ tons (approx)}$$ 11. **Calculate $x$ (distance from amidships):** $$w \times x = \Delta \times 2$$ Assuming $w = 1080$ tons (approximate), $$1080 \times x = 1080 \times 2$$ $$x = 2 \text{ m aft (same as LCF)}$$ But this suggests load at LCF; however, the problem asks for distance from amidships where load can be added without changing aft draft. 12. **Considering the problem and the right relationship:** The draft aft will not change if the load is added at a distance $x$ where it creates no moment about the aft. The formula relating change in aft draft $\Delta T_A$ to weight and position is: $$\Delta T_A = \frac{w}{TPC} \times \left(1 - \frac{x}{L} \right)$$ Set $\Delta T_A = 0$ to solve for $x$: $$0 = \frac{w}{TPC} \times \left(1 - \frac{x}{L} \right) \Rightarrow x = L$$ So the load must be added at the very aft to not change the aft draft. 13. **Final conclusion:** The load must be added at the distance of $116$ m (the full length of the ship) from amidships toward the aft to avoid changing the aft draft. **Answer:** The load must be added $58$ m aft from amidships (since amidships is at 0, total length is $116$ m, aft is at $+58$ m from amidships). Thus, **the load added at 58 m aft from amidships will not alter the aft draft.**