# Buoyant Force of a ship and water

## 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 Attempt at a Solution

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

df>ds

Doc Al
Mentor
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)

Doc Al
Mentor
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.

Doc Al
Mentor
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)]=

Doc Al
Mentor
So if the density of Ps is greater then Pf, and not knowing by how much; how exactly would I set up the ratio?