1. Statement of the problem: A ship of length $L = 110\text{ m}$ floats on an even keel with draft $4.85\text{ m}$ and its LCF is $1.0\text{ m}$ forward of amidships. 2. A load of $30\text{ t}$ is moved forward until the forward draft becomes $5.05\text{ m}$. 3. Find the distance the weight was moved given $\text{MCT}_{1\text{ cm}} = 64\text{ t-m}$. 1. Compute the change in forward draft. The forward draft increases from $4.85\text{ m}$ to $5.05\text{ m}$, so $$\Delta T_f = 5.05 - 4.85 = 0.20\text{ m} = 20\text{ cm}$$ 2. Find the distance from the LCF to the forward perpendicular. Half the ship length is $L/2 = 55\text{ m}$, so the distance from the LCF (1.0 m forward of amidships) to the forward perpendicular is $$d_f = 55 - 1.0 = 54\text{ m}$$ 3. Relate the change in forward draft to the change in trim. Trimming about the LCF gives the relation $$\Delta T_f = \Delta T\cdot\dfrac{d_f}{L}$$ where $\Delta T$ is the total change in trim (forward draft minus aft draft). 4. Solve for the change in trim $\Delta T$. $$\Delta T = \Delta T_f\cdot\dfrac{L}{d_f}$$ Substituting numbers, $$\Delta T = 0.20\times\dfrac{110}{54} = 0.407407\text{ m} = 40.7407\text{ cm}$$ 5. Compute the trimming moment required. Moment to change trim is $M = \text{MCT}_{1\text{ cm}}\times \Delta T(\text{cm})$, so $$M = 64\times 40.7407 = 2607.111\text{ t-m}$$ 6. Relate the trimming moment to the moved weight and solve for distance. Moving weight $w=30\text{ t}$ a distance $x$ produces a trimming moment $w\,x$, so $$x = \dfrac{M}{w}$$ Thus $$x = \dfrac{2607.111}{30} = 86.9037\text{ m}$$ 7. Final answer. Therefore the weight was moved approximately $86.9\text{ m}$ forward.