A $20$m long, $5$m wide, $2\mathrm{.}5$m tall rectangular, wall$-$sided barge floats at an even$-$keel, zero$-$ trim draft of $1\mathrm{.}25$m in salt water. The vessel$$s KG is $1\mathrm{.}0$m$.$ The barge is subdivided into two $8$m long end compartments and one $4$m long center compartment, as pictured below. For simplicity, the center compartment can be assumed to have a permeability of $1\mathrm{.}0.$

a$.$ What is the initial intact displaced volume, weight, transverse metacentric height, and GMT of the vessel?

b$.$ If the center compartment $($compartment $2)$ is damaged, using the method of lost buoyancy, find the new draft and GMT$.$ Remember in the lost buoyancy method the displacement remains constant, but must be carried by the remaining portion of the hull as the damage compartment is assumed to be entirely removed from both the displaced volume and waterplane area of inertia. As the damage is centered, there is no heel or trim for this problem.

c$.$ Now re$-$solve the problem using the method of added weight. Assume instead of being lost buoyancy, the center compartment is filled with seawater as a tank to the draft you found in $3$b$.$ Calculate the new draft of the vessel, the new displacement $($which rises in the added weight method to include the flooding water$),$ and new GMT $.$ Remember to include the free$-$ surface correction on GMT $,$ and update the vessel's KG for the new weight, KB and BM for the new draft and volumes.

d$.$ Do the drafts match between lost buoyancy and added weight? How about the displacements and GMT $?$ For small angles, the resisting moment of the damaged barge equals M$=*$GMT $*$Sin$(\backslash$phi $).$ Show that even though the GMT and displacement may differ between the lost buoyancy and added mass method, the resisting moment is the same.