How far into the water a ship sits (buoyancy, draft)

[bbraves7]

Homework Statement

Consider a large cargo ship, with a steel (density = 7.5g/cc, 7500kg/m^3) hull. If I ignore the mass of the front and back of the ship, how far into the water (height) will the ship sit?

Homework Equations

density = m/v
Force(buoyancy) = (density of fluid)(volume of fluid displaced)(9.8m/s^2)
Area(triangle) = (1/2)(base)(height)

The Attempt at a Solution

I know that an object will float when the force pushing up on it (buoyancy) is equal to the force pushing down (mass*gravity). I know that buoyancy is the weight of the displaced fluid. What I'm not getting is how I find either force without knowing the volume of the ship, which in turns gives me the volume of water displaced. I think the direction I should be going is to get an equation with an unknown volume of both sides, allowing me to cancel them out. I've attached my worksheet, which are basically just a few attempts to get things on paper. I usually find it easier to write down my thoughts as they creep up, but I'm stuck on this one!

Thanks for anyone willing to help!

 

[kuruman]

One way to do this problem is to say that the average density of the ship that is under water is equal to the density of the water. The average density is the mass of the hull (neglecting the air) divided by the total volume of the air plus the hull. The length of the ship doesn't matter and should cancel out eventually.

 

[bbraves7]

Thanks for the reply!

So here is what I've got after your help:

density(water) = average density(ship)

so,

1000kg/m^3 = (mass of hull)/(volume of air plus hull)

then,

1000kg/m^3 = (density of hull * volume of hull)/(volume of air + volume of hull)

1000kg/m^3 = ((7500kg/m^3) * (volume of hull))/(volume of air + volume of hull)

Am I on the right track here?

 

[kuruman]

You are on the right track. Don't forget that (volume of hull) is the the volume of the hull that is under water and depends on how far the hull sinks.

 

[bbraves7]

So how about this:

if (h) = the distance the ship sits in the water

and (L) = the length of the ship

Also, since I want to find the volume of the hull, I think I should be using the formula for finding the volume of a pyramid V = 1/3*base*height*length

1000kg/m^3 = (7500kg/m^3 * 1/3*width*height - h*L) / [(1/3*width*height*L) + (.15*30*2*L)]

1000kg/m^3 = (7500kg/m^3 * 1/3*15.52*28.98 - h*L) / [(1/3*15.22*28.98*L) + (.15*30*2*L)]


1000kg/m^3 = (7500kg/m^3 * 1/3*15.52*28.98 - h*L) / [(1/3*15.22*28.98*L) + (.15*30*2*L)]

then the L would cancel out leaving:

1000kg/m^3 = (7500kg/m^3 * 1/3*15.52*28.98 - h) / [(1/3*15.22*28.98) + (.15*30*2)]


Then when I solve for h, I get 25.3m. Sounds kind of deep to me, as it would only leave about three feet of the ship sticking out of the water. What do you think?

 

[kuruman]

Pyramid is incorrect approach. The volume is the cross-sectional area (triangle) times the length that drops out.

 

[bbraves7]

Ok, then would this be correct:1000kg/m^3 = (7500kg/m^3 * 1/2*base*h*L) / [(1/2*base*height*L) + (.15*30*2*L)]

canceling the L leaves

1000kg/m^3 = (7500kg/m^3 * 1/2*base*h) / [(1/2*base*height) + (.15*30*2)]

plug in the numbers

1000kg/m^3 = (7500kg/m^3 * 1/2*15.52*h) / [(1/2*15.52*28.98) + (.15*30*2)]

which = (roughly) 4.02m

That sounds much more realistic. I think the way I had the (h) set up before put the answer in terms of how much of the hull was sticking out.

 

[kuruman]

Perhaps it is easier to see it this way.

Average density = (mass of hull)/(volume of hull under water)

Mass of hull = (.15*30*2*L) in your expression, that's the numerator.

The denominator should just be the area of two triangles, i.e. 2*(1/2*base*height). The height is the unknown h. What is the base equal to?

 

[bbraves7]

Now I'm really confused! Since I only know the density of the hull, I would have to multiply it by the volume in order to get the mass, right? I thought (.15*30*2) would give me the area of a cross section of the hull?

 

[kuruman]

Sorry, my mistake. I meant to say that the numerator is

Mass of hull = (.15*30*2*L)*density of iron.

You could multiply by the length of the hull (or not); it doesn't matter because it drops out since it also appears in the denominator.

I didn't mean to confuse you.

 

[bbraves7]

Awesome, I really think I have it now. So basically what we are doing by taking the average density is evenly distributing the density of the air(negligible) and hull over the area of the triangle (the ship). And the variable (h) that we solve for is going to give us the height where the density of the triangle equals the density of water. Is that right?

So then I would have

1000 = [7500*(.15*30*2)] / 2(1/2*15.52*h)

h = 4.35m

 

[kuruman]

Can you explain to me where the 15.52 in the denominator came from?

 

[bbraves7]

Oh sorry. That's the base of the triangle. 30*sin(15) = 7.76

7.76*2 = 15.52

 

[kuruman]

It should be the triangle whose base is along the water level not the top the hull. You have calculated the base of triangle of hypotenuse 30 m. You need to use the hypotenuse that is under water only.

Archimedes's principle says

ρ_water*V_{under water} = ρ_Iron*V_hull

so

ρ_water = ρ_Iron*V_hull/V_{under water}

 

[bbraves7]

So the (base) in terms of (h) would be

tan(15) = b/h

b = (h)(tan15)

2b = 2(h)(tan15)

1000 = [7500*(.15*30*2)] / 2[(1/2*((2htan15))*h)]

1000 = 67500 / 2(h^2)(tan15)

1000 = 67500 / 2(h^2).27

solving for h

h = 11.2

 

[kuruman]

That looks about right.

 

[bbraves7]

Wonderful! I have a quick question. What is the difference between approaching this problem from an average density direction, as opposed to a force equation, where:

Force(down) = Force(buoyant)

Are they essentially the same thing, or will you get substantially different answers?

 

[kuruman]

Same thing. You can get the average density formulation from Archimedes's principle when you write down

ρ_fluid*V_submerged=ρ_object*V_object

from which

ρ_fluid=(ρ_object*V_object)/V_submerged

For an object to just barely float, its average density must be the density of the fluid, i.e.

ρ_average=(ρ_object*V_object)/V_submerged

which is what you have.

 

[bbraves7]

That makes sense. Thank you so much for your help, it's actually been quite fun!