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Archimedes principle

  1. Dec 25, 2015 #1
    1. The problem statement, all variables and given/known data
    35m^3 of a boat's volume is submerged under water. What does the boat weigh? The boat is 14 metres long.

    2. Relevant equations
    pVg=mg

    3. The attempt at a solution
    pVg=mg
    pV=m
    m=pV=1000*35

    This is what my teacher said, but I don't understand it. The submerged volume of the boat is equal to the volume of the displaced water. So by calculating m we are getting the water's mass, which is not equal to the boats mass. or is it?
     
  2. jcsd
  3. Dec 25, 2015 #2
    The "m" that appears in your force balance equation is the mass of the boat. So, of course whatever you calculate for "m" has to be the boat's mass. You are probably a little confused because [itex]\rho_{w} V[/itex] is also the mass of the water that has been displaced.
    But if you trust your mathematical manipulations, that result is exactly the implication of Archimedes Principle: that for an object partially submerged (floating) in a fluid, the mass of fluid displaced is equivalent to the mass of the object itself.
     
  4. Dec 25, 2015 #3
    this principle is confusing me so much :(

    "Archimedes' principle indicates that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces."

    Doesn't that mean that m is the waters mass?
     
  5. Dec 25, 2015 #4
    Archimedes' Principle is essentially
    [tex]F_{up} = m_{w}g = \rho_{w} V g[/tex]

    The first equation you have up there is actually not so much a statement of Archimedes' Principle but rather a force balance equation, that the upthrust acting on the boat is equal to the weight of the boat:
    [tex]F_{up} = F_{g} = m_{boat}g[/tex]
     
  6. Dec 25, 2015 #5
    So the boat's mass is equal to the waters mass and the waters volume is equal to the boats volume that is under water?
     
  7. Dec 25, 2015 #6
    Yes, it would be strange if it wasn't so, wouldn't it?
    In equilibrium, yes. That is what applying Archimedes' Principle tells us.
     
  8. Dec 25, 2015 #7
    oh now I get it!!!! what if the object is kinda floating but still not. it is fully under water but it is not touching the ground/bottom. The boyant force is still equal to mg if the ibject is at rest rght?
     
  9. Dec 25, 2015 #8
    Well, technically yes. Although such a system is not a stable equilibrium.
     
  10. Dec 25, 2015 #9

    HallsofIvy

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    This situation, an object floating so that it is completely underwater but not touching the bottom, can happen only if the object has exactly the same density as water.
     
  11. Dec 26, 2015 #10
    This is only true if the object has uniform density. If the object is hollow, for example, then no. The situation that Drizzy is describing is called "neutral buoyancy," and is attained when the weight of the displaced water is exactly equal to the weight of the fully submerged object.

    Chet
     
  12. Dec 27, 2015 #11

    HallsofIvy

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    If the density of the object is NOT uniform, we have to talk about its average density and what I said is still true.
     
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