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Buoyant force concept question

  1. Jul 28, 2010 #1
    I have been pondering about this for awhile now. If there are rocks on a boat (that's floating in a water) and one by one the rock is thrown onto the water. Does the water level rise or fall from the initial level?

    I believe that it will rise but am not 100% sure.
     
  2. jcsd
  3. Jul 28, 2010 #2

    ehild

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    Why do you think so? Explain!

    ehild
     
  4. Jul 28, 2010 #3
    my reasoning is because of the forces acting on it....the Fb=density of the fluid * gravity * volume and by taking the rock out, were increasing the Fb so increasing the volume?
     
  5. Jul 29, 2010 #4

    ehild

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    Why do you think that Fb increases if you take the rock out? Fb=Volume of the immersed part of the boat*g*density of the fluid. Does an empty boat immerse deeper in the water or a loaded one?

    ehild
     
  6. Jul 29, 2010 #5
    loaded ones immerses deeper so Fb is less when the rocks are taken out so the volume would be less....?? thus the water level will decrease??
     
  7. Jul 29, 2010 #6

    Dick

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    Think of it this way. When the rock is in the boat, it's displacing a volume of water equal to its weight. When it's in the water it's displacing a volume of water equal to its volume. Think of what happens if the rock is really really dense.
     
  8. Jul 29, 2010 #7

    ehild

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    Yes. Let be the mass of the rock m, that of the boat is M. The boat floats, part of it of volume V1 is immersed in water. Fb = (M+m)g=ρ(water)*g*V1.

    V1=(M+m)/ρ(water), the total occupied volume in the pond is V(water)+V1.*

    You throw the rock into the water. The boat becomes lighter, Fb=Mg=ρ(water)*g*V2,

    V2=M/ρ(water),

    but the rock is on the bottom of the water, (assuming that its density is higher than that of the water) and its volume is

    V(rock) = m/ρ(rock).

    The total occupied volume in the pond is now

    V2+V(water) +V(rock) **

    When is the occupied volume bigger?

    ehild
     
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