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Metallic coin over wooden block in glass of water - buoyancy

  1. Nov 29, 2014 #1
    1. The problem statement, all variables and given/known data
    So there is some water inside a container; the height of water inside the container is l. I placed a wooden block on the water and it's floating to some height x, on top of the block is a metallic coin (see the diagram below).

    What will happen if I throw that coin inside the water? Will the Height x and l change? Can we derive an expression for this change? I tried doing it, but failed.

    cdtEi.png


    2. Relevant equations
    Fb=gρhA


    3. The attempt at a solution

    Here is what I've tried Let's assume that our wooden block has mass m1 Volume v1height h1, height underneath the water as y and let's assume mass of Metallic coin as m2 Now total mass on block will be equal to (in which buoyant force is actin upon) : m = m2 + m1 and let's also say that total height h1 = x + y

    Principle of buoyancy of a liquid is Give as : Fb=gρhA
    Simplifying :
    Fb=g(m/v1)(x+y)A ... {1}

    Now, archimedes principle is given as F=gρhA
    F=g(ρf - ρb)hA // ρb is Density of the Body

    And then I am not sure what to do next .. Can anyone help me out?
     
  2. jcsd
  3. Nov 29, 2014 #2

    haruspex

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    Where ρ, h and A are what?
     
  4. Nov 29, 2014 #3
    A - Area
    ρ - density
    h -
     
  5. Nov 29, 2014 #4

    haruspex

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    Sure, but of what?
     
  6. Nov 29, 2014 #5
    Area of Wooden Block, ρb --> Density of Body
    ρf --> Density of Liquid (water)
    h is of Wooden block too
     
  7. Nov 29, 2014 #6

    haruspex

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    OK, but you previously posted
    If ρ is the density of the liquid, why do you have m/v1 there (where you have defined m = m1+m2)?
    If h is the total height of the block, then Ah would be the volume of the block. If you multiply that by the density of the liquid, what mass will that give you?
     
  8. Nov 30, 2014 #7

    haruspex

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    Let me try to explain again:
    This is wrong. The buoyant force does not depend in any way on the mass or density of the object nor on any part of its volume that is not immersed. It depends on the density of the fluid and the volume of the object that is submerged (and gravity, of course).
    The principle of buoyancy is Archimedes' principle. It applies independently of whether the set-up is static and of whether the object is fully or partly immersed. Your other equation should come from the fact that the block+coin float.
     
  9. Nov 30, 2014 #8
    Didnt Realize it, Thanks .. Gonna rederive it
     
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