Question

. A box-shaped barge 75 m  6 m  4 m displaces 180 tonnes when light. If 360 tonnes of iron are loaded while the barge is floating in fresh water, find her final draft and reserve buoyancy.

Answer 1

Volume V of the barge=75×6×4=1800m³. Fresh water density=1000kg/m³.

Displacement=180 tonnes=180000kg. 

Density=mass/volume=180000/volume=1000, so volume=180m³=V/10.

Assuming that 4m is the height of the barge, then the base area is 6×75=450m². The draft of the empty barge would be 180/450=0.4m. The total mass of the barge with a load of 360t of iron=360+180=540t=540000kg. This is the mass of freshwater displaced so the corresponding volume is: v=540000/1000=540m³. If we divide this by the area of the base we get 540/450=1.2m, giving us the draft. (3 times the weight (compared with the unladen barge) produces 3 times the draft.)