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Homework Help: Calculating the height of the water column

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A solid cubical block with a side of 11.0 cm floats in mercury so that 5.48 cm of the side of the block is submerged in the mercury. Water is then poured on the system until the block floats completely submerged. calculate the height of the water column that was added. The density of mercury is 13590 kg/m^3

    2. Relevant equations

    3. The attempt at a solution

    I do not know if this makes sense without a specific height of the container given. Does this problem make sense at all?
  2. jcsd
  3. Jun 5, 2013 #2
    Yes it does. Well, you have to assume that the container is sufficiently tall such that within the parameters of the problem, the fluid does not overflow, but that isn't too much of a problem and can be safely ignored.

    You need to be able to compute the buoyancy forces acting on the block. If you need to read up more, I suggest looking up a general physics textbook.
  4. Jun 5, 2013 #3

    could you explain a little more about the problem please?
  5. Jun 5, 2013 #4


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    Read the problem statement carefully.

    1. The block is a cube with sides of 11 cm.
    2. When the block is floating in pure mercury (density 13590 kg/m^3), 5.48 cm of the block is submerged.
    3. Water is now poured on top of the mercury until the block becomes totally submerged.

    What is the height of the water above the surface of the mercury?

    Can you find the weight of the cube? Hint: use Archimedes Principle
    Knowing the weight of the cube, find out the depth of water such that the cube floats with its top surface submerged.
  6. Jun 5, 2013 #5
    Initially, the block is partially submerged in mercury. The mercury exerts an upthrust force on the block, which balances the weight of the block.
    Can you find the weight of the block?

    Then, water is added to the system. Water floats on mercury because it is less dense. So now, the lower part of the block will be in mercury and the upper part in water. Once again, the upthrust is equal to the weight of the block.
    Can you find the new upthrust?
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