A cargo barge is loaded in a saltwater harbor for a trip up a freshwater river. If the rectangular barge is 3.0 m by 20.0m and sits 0.80 m deep in the harbor, how deep will it sit in the river? Thank you

### 4 Answers

Solution :

w(salt water) = density of salt water = 1.025

w(river water) = density of river water = 1.0

A = surface area of barge

d = submerged depth

A.d.w(salt water) = A.d.w(river water)

(3 x 20) x (0.80) x (1.025) = (3 x 20) x d x (1.0)

d = 0.82 meter (Answer) – submerged depth in river

A very short cut solution.

You need the density of the river water and the saltwater to do this calculation.

However let’s go ahead with the some densites I got off the net.

I took the density of sea water to be 1030 Kg/m^3

http://hypertextbook.com/facts/2002/EdwardLaValley…

and the density of riverwater to be 1000 Kg/m^3

Now the amount of saltwater displaced by the barge is 3*20*0.8m^3

which is 48 cubic meters. With the volume and density we can calculate the weight of the displaced liquid.

48*1030=49440 Kg.

Now the weight of the displaced liquid is equal to the upthrust or the force acting to keep the barge afloat.

Now the density of the river water is less, so to generate the same amount of upthrust the barge must sink deaper.

49440=volume of displaced RIVER water *density of RIVER Water

49440=3*20*x*1000

x=0.824 meters

Displacement (volume) * density = constant. Also area = constant. Thus depth * density = constant.

The average near-surface seawater density I’ve seen most often is 1025 kg/m^3 (ref.).

So 1025*0.8 = 1000*x ==> x = 0.82 m.

deeper….because of the water density change? cant say how much, you would need to know the two densities.