# Fluids - A lost shipping container is found resting on the ocean floor

## Homework Statement

A lost shipping container is found resting on the ocean floor and completely submerged. The container is 6.3 m long, 2.1 m wide, and 2.6 m high. Salvage experts attach a spherical balloon to the top of the container and inflate it with air pumped down from the surface. When the balloon's radius is 1.8 m, the shipping container just begins to rise towards the surface. What is the mass of the container? Ignore the mass of the balloon and the air within it. Do not neglect the buoyant force exerted on the shipping container by the water. The density of seawater is 1025 kg/m3.

Container

L = 6.3 m
w = 2.1 m
h = 2.6 m
V = ?

Balloon / Air

r = 1.8 m
p = 1.29 kg / m3
V =
FB =

Water

p = 1025 kg / m3
FB =

## Homework Equations

Archimedes' Principle

FB = Wfluid

Vcontainer = L * w * h

Vballoon = 4/3 $$\pi$$r3

p = m / v

F = ma

## The Attempt at a Solution

First I found the volumes for the container and the balloon.

Vcontainer = L * w * h

Vcontainer = 6.3 * 2.1 * 2.6

Vcontainer = 34.398 m3

Vballoon = 4/3 $$\pi$$r3

Vballoon = 4/3 $$\pi$$1.83

Vballoon = 24.429

I then tried to find the buoyant force of the air.

FB = Wair

FB = mg

FB = pVg

FB = 1.29 * 24.429 * 9.8

FB = 308.83 N

And then for the sea water.

FB = Wwater

FB = mg

FB = pVg

FB = 1025 * 34.398 * 9.8

FB = 345527.91 N

After this, I used newton's 2nd law...

F = ma

m = F / a

m = FBair + FBwater / g

m = 308.83 N + 345527.91 N / 9.80 m/s2

m = 35289 kg ----> 3.5 * 104 kg

But.......this appears to be wrong...Um...thoughts?

Thanks!

Last edited:

mgb_phys
Homework Helper
Check the bouyancy of the balloon

You can ignore the mass of the air - otherwise it's the same calcuation as the container

Hmmm....

The buoyant force equals the weight of the displaced fluid which is the sea water, not the air. So the equation should have been:

FB = Wwater

= mg

= pVballoong

= 1025 * 24.429 * 9.8

= 245389 N

So, using that....

F = mg

m = F / g

m = 245389.30 N + 345527.91 N / 9.80 m/s2

m = 60297.67 kg ----> 6.0 * 104 kg

How is that?

mgb_phys
Homework Helper
Assuming you pressed the right buttons - yes.
You can simplify it a little by ignoring 'g' - since you cancel it out.
You just want the mass of seawater contained in the volume of shipping container and volume of balloon.

Didn't know what you meant by the g's cancelling, but I went back a few steps and tried this...Please excuse pi appearing as an exponent, I have no idea why that happens.

p(4/3 $$\pi$$r3)[STRIKE]g[/STRIKE] + p(L*w*h)[STRIKE]g[/STRIKE] = m[STRIKE]g[/STRIKE]

p(4/3 $$\pi$$r3) + p(L*w*h) = m

1025 kg/[STRIKE]m3[/STRIKE] (24.429 [STRIKE]m3[/STRIKE]) + 1025 kg/[STRIKE]m3[/STRIKE] (34.398 [STRIKE]m3[/STRIKE])

1025 kg (24.429) + 1025 kg (34.398)

25039.725 kg + 35257.95 kg

= 60297.675 kg ---> 6.0 * 104 kg

Do I pass?

mgb_phys