$2$: 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 $1\mathrm{.}0.$ Note this is

the in$-$class example problem with different dimensions.

$2$a: What is the initial intact displaced volume, weight, transverse metacentric height, and GM T $,$

of the vessel?

$2$b: If the center compartment $($compartment $2)$ is damaged, using the method of lost

buoyancy, find the new draft and GM T $.$ Remember in the lost buoyancy method the

displacement remains constant, but the 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.

$2$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

$2$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$($$).$ Show that even though the GM T and displacement may differ between the

lost buoyancy and added mass method, the resisting moment is the same.

Dimensions in m

Figure $11\mathrm{.}2$ A simple pontoon$-$intact condition