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

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

    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?

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    Quote Originally Posted by bennyJ View Post
    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?

    Please be complete. Thank you!

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    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.

    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?

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    Quote Originally Posted by bennyJ View Post
    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?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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    This problem seems to want you to use Archimedes' Principle. Since the barrels will be floating, I reasoned that the weight of the displaced water should equal the weight of the object. So, how many barrels are needed to get the level of the lock to the top (i.e., raise the lock's water level 10 cm)? My approach could be wrong but I calculated the volume of the lock based on the information given (200*50*20 = 200,000 meters) then reasoned that raising the lock another 10 cm would require 1,000 cubic meters of water to be displaced. And since "the water has a density of one long ton per cubic meter" that should mean 500 barrels since each barrel displaces 2 long tons of water and it seems like the situation requires that 1,000 cubic long tons be displaced (500*2 = 1,000). This whole approach could be wrong since physics, let alone fluid dynamics, is definitely not my forte. I'm not sure if the barge's weight or volume should be factored in and, if so, how or why. Just as an aside, I reasoned that if each barrel weighs 2 long tons and there are 3,000 barrels that the barrels collectively should weigh 6,000 long tons and, consequently, the barge should weigh 2,000 long tons (8,000 - 6,000 (for barrels) = 2,000 long tons, or barge weight).

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    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?
    Last edited by bennyJ; Yesterday at 06:07 PM.

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