# Thread: Volumetric Displacement Question (3,000 barrels of toxic chemicals)

1. ## Volumetric Displacement Question (3,000 barrels of toxic chemicals)

 A certain lock for raising.

2. Originally Posted by bennyJ
A certain lock for raising and lowering barges from one river level to another is a rectangular parallelepiped 200 meters long, 50 wide, and 20 deep. A barge is floating in the lock that is also a rectangular parallelepiped measuring 80 meters long, 25 wide, and 5 deep. The barge, containing 3,000 barrels of toxic chemicals, displaces 8,000 long tons of water. The water has a density of one long ton per cubic meter. Each barrel is watertight, with a volume of one cubic meter and a weight of two long tons. A group of terrorists render the lock inoperable and attach a time bomb to the side of the barge set to go off in three hours. The barge contains elevators for moving barrels quickly to the deck, but the crew is too shorthanded to roll the heavy barrels up an inclined plane in the time allotted. The deck is only ten centimeters below the top edge of the lock, from which the barrels could be rolled to dry land. If no water is entering or leaving the lock, how many barrels at a minimum would have to be rolled into the water in the lock in order to raise the level of the barge so that its deck would be even with or slightly above the top edge of the lock so that the remaining barrels can be rolled to dry land?
What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

3. When.

4. Originally Posted by bennyJ
When I approached this problem initially, I focused in on a few pieces of critical information and ignored superfluous information in the question. The volume of the lock, especially the depth, the density of the water, and how much water each barrel displaces (calculated by looking at the bolded parts below) I viewed as critical information. I focused on the volume of the lock, especially the depth, because the question wants to know how many barrels have to be dumped overboard to raise the lock's level by 10 centimeters.
Then where did you go - what did you find?

5. This problem seems to want you to use Archimedes' Principle.

6. Is there someone with a bit of a physics background who could lend some insight into this problem or evaluate the approach I took in the previous post? Did anyone else have a difference approach?

7. Anyone? *crickets*

8. Originally Posted by bennyJ
 The water has a density of one long ton per cubic meter. Each barrel is watertight, with a volume of one cubic meter and a weight of two long tons.
The barrels will not float! They are denser than the water. That may change your calculations considerably.

My initial approach would be to find how many cubic meters of water must be added to raise the level by 10 cm. This would be 0.1 m times the area of the water surface. Then you need to add enough barrels to make up that volume.

There may be some fine-tuning required to correct this quick approach. As barrels are removed from the barge, it will float higher, displacing less water (so that more barrels might be needed, but also raising the deck so fewer would be needed!). Either you can think about whether these effects would cancel each other out, or make a more detailed equation expressing the relationship of all the variables. You can easily find how much of the barge is submerged initially, thus determining the actual depth of the water, and so on. I haven't gone through the details.

9. Originally Posted by Dr.Peterson
The barrels will not float! They are denser than the water. That may change your calculations considerably.

My initial approach would be to find how many cubic meters of water must be added to raise the level by 10 cm. This would be 0.1 m times the area of the water surface. Then you need to add enough barrels to make up that volume.

There may be some fine-tuning required to correct this quick approach. As barrels are removed from the barge, it will float higher, displacing less water (so that more barrels might be needed, but also raising the deck so fewer would be needed!). Either you can think about whether these effects would cancel each other out, or make a more detailed equation expressing the relationship of all the variables. You can easily find how much of the barge is submerged initially, thus determining the actual depth of the water, and so on. I haven't gone through the details.
Correct you are. And, yes, there will be some fine tuning required for the solution, it seems. Thanks for the direction.

10. Originally Posted by bennyJ
 A certain lock for raising and lowering barges from one river level to another is a rectangular parallelepiped 200 meters long, 50 wide, and 20 deep. A barge is floating in the lock that is also a rectangular parallelepiped measuring 80 meters long, 25 wide, and 5 deep. The barge, containing 3,000 barrels of toxic chemicals, displaces 8,000 long tons of water. The water has a density of one long ton per cubic meter. Each barrel is watertight, with a volume of one cubic meter and a weight of two long tons. A group of terrorists render the lock inoperable and attach a time bomb to the side of the barge set to go off in three hours. The barge contains elevators for moving barrels quickly to the deck, but the crew is too shorthanded to roll the heavy barrels up an inclined plane in the time allotted. The deck is only ten centimeters below the top edge of the lock, from which the barrels could be rolled to dry land. If no water is entering or leaving the lock, how many barrels at a minimum would have to be rolled into the water in the lock in order to raise the level of the barge so that its deck would be even with or slightly above the top edge of the lock so that the remaining barrels can be rolled to dry land?
200*50 = 10000 m^2

Each barrel thrown in water will actually lower the water level of the lock by 1/10000 m (0.1 mm).

The barge will rise [2/(80*25) =] 1 mm

So they will need to throw 100/.9 = 112 barrels to clear the height ← edited

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