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

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Homework Help Overview

The problem involves a barge transitioning from saltwater to freshwater and seeks to determine how deep it will sit in the freshwater. The context is rooted in buoyancy principles and the relationship between displaced water volume and density.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of buoyancy and the relevant densities for saltwater and freshwater. Questions arise about the mass of the barge and the volume of water displaced, as well as the appropriate density to use in calculations.

Discussion Status

Some participants have provided insights into the relationship between the buoyancy of the barge in saltwater and the conditions in freshwater. There is an ongoing exploration of the assumptions regarding density and volume displacement, with no explicit consensus reached yet.

Contextual Notes

Participants note the lack of provided density for the barge and the confusion surrounding which density to apply in the calculations. The discussion reflects uncertainty about the implications of the displaced volume in different water types.

eglaud

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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