Contest question The sinking oil barge. Help

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The discussion revolves around a physics contest question involving a sinking oil barge being towed by a tugboat. Participants are tasked with determining how long it will take for the barge to sink as oil leaks out and water flows in. Key points include the need to calculate the overall density of the barge as oil is lost, ensuring it matches the density of seawater at the moment of sinking. The final answer provided in the discussion is 118 kilometers, with participants collaborating to derive the necessary formulas and calculations. The conversation highlights the importance of understanding specific gravity and density in solving the problem.
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Contest question! The sinking oil barge. Help!

OK. Here it goes!

A tugboat is towing an oil barge in the sea. You may consider the barge to be a large steel container, with an empty weight of 3000 tons, completely filled with 50000 tons of oil. The density of steel is six times that of sea water, and the density of oil is three-quarters that of sea water. An industrial saboteur from certain pipe-line interests opens some drain and vent valves. This let's oil leak out at a constant rate of one ton per second, and water flows into keep the barge full. How long will the resulting oil slick be if a speed of ten kilometres per hour is maintained until the barge sinks?(Answers in km)

I know the density of the barge must be equal to the density of water when it submerges. I also know that the weight of the barge when it sinks is equal to the oil left inside the barge and the water in. This question is something related to specific gravity which is new to me and I'm having problems with.
 
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Hi Chi. I didn't realize. I'm new here. I read the link. You should know that I relly spent a lot of time but specific gravity and densities confuse me. This type of question is very repetitive ( the idea...) and comes up often. Can you really help? Thanks.
 
It's a rather silly, stretched out problem. Luckily the hard part has been given to you: the rate at which the oil leaks out.

You need to find out how long in time you have until the boat's overall average density becomes equal to that of water.

As oil leaks out, it is assumed that an equal VOLUME of water comes in.

Derive a formula for the overall density of the ship as a function of time.
 
Thanks a lot Chi. Can you please have patience on me and show me how it's done?
 
Sure, but you'll have to show some effort.

Start by finding the formula for density, and expand it to find the starting density of the entire ship.

Since densities are given in fractions of the density of water, use some variable, such as D as the density of water, call the density of the steel, 6D, and of oil 3D/4. Find the volume of the steel, and of the oil to get the volume of the entire ship (in terms of an integer times "D").

And figure out how many seconds it takes to increase mass by one ton.

And I come and go from this forum, so I may not be back for some time.
 
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Is the contest over?
 
I did that but now what? We have to know how much oil is inside when it sinks. For the buoyant force on the volume of steel when starts to sink:

B = 3000/6 g = 500 g

So now there must be some quantity of oil which is supporting the rest of the mass of the steel,

(((otherwise the ship would go down immediately, but it doesn't, at the very moment it starts to sink. That's why at this moment the ship and the water must have the same density.)))

which is 2500 tons of steel. Here is where I get stuck. How much of oil supports a quantity of steel, knowing their respective densities? The final ANSWER on the book is 118 km.
 
Volume of Steel = 3000/6D = 500/D
Volume of oil = 50000*4/3 /D = 66666.667/D

TOTAL VOLUME = 67166.6667 tons/D

The second part I'm not sure. The contest is definitely not over, hehe, at least for me. Some more help? ...
 
  • #10
fisico said:
Volume of Steel = 3000/6D = 500/D
Volume of oil = 50000*4/3 /D = 66666.667/D

TOTAL VOLUME = 67166.6667 tons/D

...

The final ANSWER on the book is 118 km.
Yep, I got 188 km (phew!).

You can just round the volume to 67170 T/D). Now you have volume and total mass, find a statement for the total density of the ship.

What's the prize for the contest?
 
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  • #11
All right! About the price, ah, ah, ah ... wow I didn't think of that one either, hehe.

So:

TOTAL DENSITY = (6 + 3/4)D

?
 
  • #12
fisico said:
All right! About the price, ah, ah, ah ... wow I didn't think of that one either, hehe.

So:

TOTAL DENSITY = (6 + 3/4)D

?
no. get the total density by taking the total mass and putting it over the volume you just figured out. You will get a fraction times D. The boat will float until this fraction = 1D.

You got to get it yourself from this point. You're 3/4 of the way there and you know the answer. If you can't get it, I'm sorry, then you should not win this contest.
 
  • #13
OH, I got it. It all looks so much easier now, hehe. Thanks a lot.
 
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