How Deep Will a Barge Sit When Moving from Saltwater to Freshwater?

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SUMMARY

The discussion centers on calculating the depth a barge will sit in freshwater after being in saltwater. The barge measures 3 x 20 m and initially sits 0.8 m deep in saltwater, displacing a volume of 48 m³. The key equations used are buoyancy (B = density_displaced * V_displaced * g) and the equilibrium condition (B - mg = 0). Participants clarify that the density of saltwater should be used for initial calculations, and the mass of the barge equals the mass of the displaced water, which is crucial for determining the new depth in freshwater.

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Homework Statement


A barge is loaded in a saltwater harbor and measures 3 x 20 m, sitting 0.8 m deep in the harbor. If this barge moves to freshwater, how deep will it sit?

Homework Equations


B = densitydisplaced * Vdisplaced * g
B - mg = 0

The Attempt at a Solution


So I began by solving for B - mg, getting to m = 1000 * 48 to find out the mass of the water displaced. This itself confuses me, however, because I 1) wasn't sure whether to use the density of saltwater or seawater and 2) don't really know what to do with this information. Can someone help walk me through it?
 
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In what context was the voume displaced 48m3? In that context, which density applies?
eglaud said:
m = 1000 * 48 to find out the mass of the water displaced.
In the equation you quote:
eglaud said:
B - mg = 0
of what is m the mass?
 
The barge sits 0.8 meters deep and has an area of 60, which gives a volume of 48. I don't know what density applies, we never got a density for the barge. m is the mass of the barge, which equals the mass of the water displaced I think?
 
I just found out how to do the problem, you take the bouyancy of the salt water situation and set it equal to the freshwater, and use A * h for the volume. However, is it possible to solve this problem with my method?
 
eglaud said:
The barge sits 0.8 meters deep and has an area of 60, which gives a volume of 48. I don't know what density applies, we never got a density for the barge. m is the mass of the barge, which equals the mass of the water displaced I think?
I don't think you understood my questions.
The context for the 48 m3 displacement was the saltwater harbor, so the displaced water was salt. That tells you which density to use.
you wrote that you did not know what to do with information, but you correctly stated that it told you the mass of the barge.
You can then calculate the volume if fresh water that would provide the necessary buoyant force.
This is not different from the method in post #4. That method only seems shorter because your description left out a step. Instead of going weightsaltwater displaced=weightbarge=weightfreshwater displaced you left out the middle part and just wrote weightsaltwater displaced=weightfreshwater displaced
 

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