Buoyant Force of a ship and water

AI Thread Summary
A ship floating in fresh water displaces a greater volume of water compared to when it is floating in sea water due to the difference in density. The buoyant force equation indicates that the volume of water displaced is proportional to the depth of the ship below the surface. Since sea water is denser than fresh water, the depth of the ship in sea water (ds) is less than in fresh water (df). This relationship leads to the conclusion that ds is less than df, confirming that the ship floats higher in sea water. The discussion emphasizes the importance of understanding buoyancy and the relationship between volume displaced and depth.
dab353
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Homework Statement


Consider a ship that is floating in fresh water. The bottom of the ship is a depth of (df) below the surface. If the same ship is floating in sea water, the bottom of the ship (ds) below the surface. Given that the density of sea water is greater than the density of fresh water, which one of the following statements is correct?

The correct answer is ds<df


Homework Equations



Buyont Force = B=pVg

The Attempt at a Solution



B=(pf)(Vf)g=(ps)(Vs)(g)

Vf>Vs ---> How was this simplified to give the correct answer. I am confused.
df>ds
 
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dab353 said:
Vf>Vs ---> How was this simplified to give the correct answer. I am confused.
df>ds
How does volume of water displaced relate to depth below the surface?
 
Volume relates to depth below the surface in the equation (Pressure under water: P=ρ•g•h ; where ρ is the density of water m/V and h is the depth of water)
 
dab353 said:
Volume relates to depth below the surface in the equation (Pressure under water: P=ρ•g•h ; where ρ is the density of water m/V and h is the depth of water)
That equation describes the pressure beneath the surface of a fluid. What I asked for was much simpler. If a ship displaces a greater volume (Vf > Vs, say), what can you say about the depth of its bottom beneath the surface? (It's an easy question--don't over think it.)
 
That they are proportional. I am still confused as to how from the Buoyant force equation we were able to come up with the result of Vf being greater then Vs.
 
dab353 said:
That they are proportional. I am still confused as to how from the Buoyant force equation we were able to come up with the result of Vf being greater then Vs.
Ah, now I understand your question.

I think you understood this:
(pf)(Vf)g=(ps)(Vs)(g)

Now just rearrange as ratios, so that (pf)/(ps) = ?

How? Divide both sides by g, then by Vf, then by ps.
 
So if the density of Ps is greater then Pf, and not knowing by how much; how exactly would I set up the ratio? Also the ratio of Vs and Vf --> Would it be something like this? [(Pf)/(Ps)] [(Vf)/(Vs)]=
 
dab353 said:
So if the density of Ps is greater then Pf, and not knowing by how much; how exactly would I set up the ratio?
Start with the equation I gave in the last post. (Which was from your first post.)

Then do each of the three steps I outline.

1) Divide both sides by g
... and so on.
 
I am more of a visual learner as to where what goes. Still confused, but i'll ask someone from my department at school. Thanks!
 
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