Question

A box-shaped barge 55 m  10 m  6 m. is floating in fresh water on an even keel at 1.5 m draft. If 1800 tonnes of cargo is now loaded, find the dif-ference in the height of the centre of buoyancy above the keel.

Answer 1

Volume V of the barge = 3300m³. Freshwater density is taken to be 1000kg/m³. Let the mass of the barge be M kg. It displaces an amount of water equal to its own mass. Density=mass/volume, so 1000=M/volume, making volume v of the water displaced M/1000 m³. The base area of the barge=550m². Draft=v/(base area)=(M/1000)/550=M/550000=1.5m, M=825000kg=825 tonnes.

The mass is increased by 1800t, making the total mass 2625t. The draft is proportional to the mass. So 825t produces a draft of 1.5m, 2625t produces a draft of 2625/825×1.5=105×1.5/33=105/22 m.

So the draft difference is 105/22-1.5=36/11=3.27m approx.