1. **State the problem:** We have a ship 140m long with drafts 4.85m aft and 4.25m forward, and LCF 1.9m forward of amidships. 2. **Given data:** - Length $L = 140$ m - Aft draft $D_a = 4.85$ m - Forward draft $D_f = 4.25$ m - LCF (Longitudinal Center of Flotation) is 1.9 m forward of amidships - TPC (Tonnes per cm immersion) = 19.4 t/cm - MCT1cm (Moment to change trim by 1 cm) = 108 t-m - Mass to place = 135 t 3. **Calculate change in trim and mean draft from given drafts:** The trim $T$ is the difference between aft and forward drafts: $$T = D_a - D_f = 4.85 - 4.25 = 0.6\;m = 60\;cm$$ Mean draft $D_m$: $$D_m = \frac{D_a + D_f}{2} = \frac{4.85 + 4.25}{2} = 4.55\;m$$ 4. **Calculate displacement before placing the mass:** Displacement $\Delta$ is related to TPC and mean draft. Assuming linear relation: Since 1 cm immersion corresponds to 19.4 t, immersion in meters corresponds to: $$\Delta = D_m \times 100\;cm/m \times 19.4\;t/cm = 4.55 \times 100 \times 19.4 = 8827\;t$$ 5. **Calculate trim moment before placing the mass:** Trim moment $M_T$ is related to trim and MCT1cm: $$M_T = T(\text{in cm}) \times MCT1cm = 60 \times 108 = 6480\;t\cdot m$$ 6. **Find initial center of gravity (LCG) from amidships:** Trim moment also equals displacement times longitudinal center of gravity (LCG) relative to LCF: $$M_T = \Delta \times (LCG - LCF) \Rightarrow LCG = LCF + \frac{M_T}{\Delta}$$ With LCF 1.9 m forward of amidships, take aft as positive direction (so forward is negative), therefore: $$LCG = -1.9 + \frac{6480}{8827} = -1.9 + 0.734 = -1.166\;m$$ So the center of gravity is 1.166m forward of amidships. 7. **After placing 135 t at position $x$ from amidships, new displacement and moment:** New displacement: $$\Delta' = \Delta + 135 = 8827 + 135 = 8962\;t$$ New moment about amidships: $$M' = \Delta \times LCG + 135 \times x = 8827 \times (-1.166) + 135x = -10300 + 135x$$ 8. **We want to maintain the forward draft at 4.25m, so the trim remains 0.6m, thus same moment and same LCG relative to LCF:** So the new LCG relative to amidships must be: $$LCG' = \frac{M'}{\Delta'} = -1.166$$ Plugging in, $$-1.166 = \frac{-10300 + 135 x}{8962} \Rightarrow -1.166 \times 8962 = -10300 + 135x$$ Calculate left side: $$-1.166 \times 8962 = -10453.65$$ Then: $$-10453.65 = -10300 + 135x \Rightarrow 135x = -10453.65 + 10300 = -153.65$$ So: $$x = \frac{-153.65}{135} = -1.138\;m$$ 9. **Interpretation:** $x$ is negative relative to amidships, meaning the mass of 135t should be placed 1.138 m forward of amidships. **Final answer:** The 135t mass should be placed approximately 1.14 m forward of amidships to maintain the forward draft at 4.25 m.