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Hydrostatics & Buoyancy

  1. Jan 15, 2007 #1
    Hi everyone, I am having some trouble understanding hydrostatics and buoyancy as taught in my textbook

    1) "In a static fluid, the pressure is ISOTROPIC (same in all directions)" <----what does it mean by "same"? Same magnitude in all directions at a single point? Or same magnitude in the entire fluid at every depth?

    2) "A boat will float if the weight of the siplaced water EQUALS the weight of te boat." <----is this a correct statement? Because on another page in my textbook it says that an object floats on the surface if it weighs LESS than the fluid it displaces, which is right?
    The following links to the actual page of my textbook...
    [​IMG]

    3) "An ice cube is floating in a glass of water that is filled entirely to the brim. When the ice cube melts, the water level will
    a) fall
    b) stay the same, right at the brim
    or c) rise, causing the water to spill"
    The correct answer is b, but I don't understand the physcis behind it...can somebody explain this to me, please?

    I hope that someone can help me out! Thank you for explaning!:smile:
     
  2. jcsd
  3. Jan 15, 2007 #2

    cristo

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    Isotropic means that the pressure has equal magnitudes at all directions. It does not mean that it is equal at all depths.

    They're both right, but are saying (slightly) different things. The first comment says that a boat floats if the displaced water weighs the same as the boat. The second is saying that it floats if the object weighs less than the water displaced. However, what the second does not mention, is that the object will rise up in the water (e.g. half in, half out) such that the weight of the volume displaced is equal to weight of the object. This is what the first comment is saying.

    Well, using the above argument, the ice cube will have to displace enough water to support its weight in order to float. The volume of this is V=m/d, where m is the mass of the ice cube, d is the density of water. When the ice melts, it turns into water. The volume of water will also equal V. Hence the water level remaisn the same.
     
    Last edited: Jan 15, 2007
  4. Jan 15, 2007 #3

    berkeman

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    Yeah, that melting ice cube thing used to always bother me. But then I finally understood that the cube's density changes when it melts, increasing from the density of ice to that of water. As christo says, the weight of the ice cube is the weight of the water it displaces, not the weight of a cube of water the same size as the ice cube. So when it melts, it melts into a box the size of the submerged part of the ice cube, so there is no increase in the volume of the water.
     
  5. Jan 16, 2007 #4
    "the weight of the (whole) ice cube is (equal to) the weight of the water it displaces" <----but if they are equal, wouldn't the ice cube be underwater instead of floating on the surface (e.g. half in, half out)?
     
  6. Jan 16, 2007 #5
    2) If the displaced water weighs the same as the boat, can the boat be floating on the surface? (e.g. 60% of the boat is submerged) If so, how can this be? Because from the textbook page I uploaded, for the neutrally buoyant case, I saw the whole object underwater "hanging" in the middle of it...
     
  7. Jan 16, 2007 #6

    cristo

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    Yes, the boat would float on the surface. Imagine we had an object whos mass we could change remotely. At the beginning of the experiment, the object weighs less than the total water it's volume can displace, and so it is floating on the surface. As we increse the mass of the object the object moves down, in order to stay afloat. Why is this? Well, in order for the object to float, the weight of the object must equal the weight of the displaced water, and so the object must displace a greater volume of water, in order for the weight of this water to equal the weight of the object. As we increase the mass of the object further, it will carry on dropping down in the water. Now, when the object's mass is increased so that its density is equal to the density of water (i.e. the weight of the object is equal to the weight of the water displaced by the total volume of the object) then it will sit, just under the surface, like in the neutrally buoyant case.

    Does this make sense?
     
    Last edited: Jan 16, 2007
  8. Jan 16, 2007 #7
    I see!

    So is it true that, if an object's weight is equal to the weight of the displaced fluid, then no part of the object can be above the water level?

    And is it true that, if part of an object is above the water level and part of it is submerged, then the obejct's weight MUST be less than (not equal to) the weight of the displaced fluid?
     
  9. Jan 16, 2007 #8

    cristo

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    Well, a slight correction, as the object will displace a volume of fluid equal to the volume of the object underwater.

    Your statement can be rephrased: if the weight of the object equals the weight of the maximum volume of water it can displace, then no part can be above water level.

    If part of the object is not submerged, then the weight of the object must be less than the weight of the maximum volume the object can displace.

    Do you see the difference here, between the two sets of statements?
     
  10. Jan 16, 2007 #9

    cristo

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    The weight of the ice cube is equal to the weight of the water is displaces, since it is floating! However, note that the volume of water that the ice cube displaces is not equal to the volume of the ice cube; it is only equal to the volume of the ice cube which is submerged.
     
  11. Jan 18, 2007 #10
    Thanks everyone, I will have to think about it...

    And I have another question that is puzzling me:

    4) Assume that air is an ideal fluid (incompressible, irrotational, non-viscous, and has a steady state), why is it windier near the base of a tall building than somewhere far from the building? How would the height of the building affect the wind speed near the ground?

    What principles/equations/concepts can I use to explain this? I have absolutely no clue...
     
  12. Jan 18, 2007 #11
    Think about a water, since it is incompressible and air is not. The priciple is continuity of flow, i.e. for steady state flow mass in = mass out along any portion of the pipe or reach. So since the building reduces the area available for flow, to remain in steady state, the velocity has to increase.
     
  13. Jan 19, 2007 #12
    One of the four assumptions of the "ideal fluid" is "incompressible", then how can the air (which is assumed to be an ideal fluid in this case) be compressed to a smaller area?

    And what do you mean by "steady state"? My textbook says that steady state means that the fluid velocity at each point in the fluid is constant...I am not too sure what it means now...
     
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