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Concept of Archimedes

  1. Mar 24, 2007 #1
    i just don't understand why an object will experience buoyant force equal to the weight displaced when it is partially or wholly immersed in a fluid.I don't want to memorize this,i want to understand and get the idea.But i just don't seem to understand.
     
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  3. Mar 24, 2007 #2

    HallsofIvy

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    If it didn't experience a force exactly equal to its weight, it would have a resultant force and accelerate either up or down, wouldn't it?

    If a weight is floating in water, whether on top or completely submerged, its weight must be completely offset by the bouyant force.
     
  4. Mar 24, 2007 #3

    arildno

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    Perhaps you'll understand it in this way:

    1. Why doesn't a portion of the fluid fall down from its position into the Earth?
    Answer:
    The immediate reason is that the rest of the fluid keeps it where it is, by exerting a force upon it equal to the portion's own weight.
    This is the buoyancy force acting upon the fluid element.
    2. Now, secretly substitute that fluid portion with an object of the same shape, but of possibly different material composition.
    Now, ask yourself:
    Why would the rest of the fluid suddenly stop exerting its force or alter it as a result of this substitution?
    Only if the inserted object somehow altered the fluid chemistry (say, by inducing the formation of a film that can support tensile stress), or say, the local electro-magnetic field should we expect such changes in the buoyancy force to occur. A more prosaic way of changing the pressure distribution would be if the object was moving, rather than at rest.

    however, if none of this happens, the buoyancy force exerted upon the object should remain unaffected, which is what Archimedes' principle states.
     
  5. Mar 26, 2007 #4
    Can i relate this to Newton's law?
     
  6. Mar 26, 2007 #5

    russ_watters

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    Which one...?
     
  7. Mar 26, 2007 #6
    The third law
     
  8. Mar 26, 2007 #7

    russ_watters

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    It certainly applies, but I don't know that it tells us anything useful for how buoyancy works.
     
  9. Mar 26, 2007 #8
    The buoyant force comes from the difference in pressures exerted by the fluid in the top and the bottom of the object.

    Take a vertical volume with cross section S and height L. If the top is at a deep Lt the vertical component of the force due to the fluid pressure will be rho*g*Lt*S where rho is the density and g the acceleration of gravity. The upward force in the bottom will be rho*g*(Lt+L)*S.

    The net upward force will be then: rho*g*Lt*S. This is the weight of the volume if it was filled with the fluid. As you can divide any object in a lot of vertical volumes. The total upward force will be the sum of all individual forces, and is equal to the weight of a volume of fluid equal to the volume of the object. That is the displaced fluid.

    If the object floats, each individual vertical volume feels a force equal to the weight of fluid from the bottom to the surface of the fluid.
     
  10. Mar 26, 2007 #9
    I'm slow at this principle,and can only understand through this law
     
  11. Mar 27, 2007 #10
    Think of it this way. Consider an object floating in the water that displaces x liters of water, where the diaplced water weighs y newtons. Without the object floating there, there will be x liters of water floating in place of the object, right? It is being supported by the water underneath this 'would be displaced' water, with an upward force of y newtons. Now, with the object there, the upward force beneath the floating object must equal the weight of the water displaced, and for the object to be in equilibrium (no net acceleration) the upward buoyant force must equal the weight of the object. This is a hand waving explanation of archimedes' principle, Fbouyant = Wdisplaced.

    ok, it looks like this was already explained by others in this thread.
     
    Last edited: Mar 27, 2007
  12. May 18, 2007 #11
    sorry for bumping this thread, but i would like to ask, since a force is experienced on an object submerged in the water (by the water), by Newton's third law, an equal and opposite reaction force should be experienced on the water by the object. So how is that balanced by the water system?
     
  13. May 18, 2007 #12

    Doc Al

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    The water pushes on the object--and the object pushes back on the water. It has to push back, to keep the water from crushing it. By Newton's 3rd law, the force that the water exerts on the object is equal and opposite to the force the object exerts on the water. (If I missed your question, ask again.)

    Also realize that when the object is submerged the water level rises, increasing the pressure it exerts at a given height above the bottom.
     
  14. May 18, 2007 #13
    Yes I do get the part where there is a force by the object on the water, but the point is that I got sort of confused if a scale is put below this cylinder of water + object. How would the scale read? With or without the object submerged?

    Sorry for ignorance, but I've lost touch with my academic work for quite a few months (which proved to have significantly lowered my ability to think) as I'm serving the army.
     
  15. May 18, 2007 #14

    Doc Al

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    Imagine a bucket of water on a scale: The scale will read the weight of bucket + water.

    Now float an object in the water: The scale will read the weight of bucket + water + object.

    Is that your question?
     
  16. May 18, 2007 #15
    I thought the weight of the object will be supported by the upthrust on the object by the water, and hence the scale reading isn't it just water+bucket?

    and I am aware of the missing force which is the force on the water by the object and it should reflect in the scale reading
     
    Last edited: May 18, 2007
  17. May 18, 2007 #16

    Doc Al

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    The water pushes up on the object with a force equal to the object's weight. Accordingly, per Newton's 3rd law, the object pushes down on the water with an equal force. That force on the water is transmitted to the bucket and thus the scale.
     
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