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Sink or float?

  1. Oct 26, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider an object that floats in water but sinks in oil. When the object floats in a glass of water, half of the object is submerged. We now slowly pour oil into glass so it completely covers the object.. Compared to water level, does the object move up, move down, or stay in same place.


    2. Relevant equations

    I think i understand that for the object to float in water, the bouyant force exceeds the density of the water.. or that the object is less dense than water, but more dense than oil.

    I know that for an object to float it has to displace more water than its weight. However, when oil is added to the top of the water, does it prevent the object from displacing the water- in effect causing it to sink?
     
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  3. Oct 26, 2007 #2

    Astronuc

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    Assuming that the object is floating in water, and it is solid with half the object above the water, it must be less dense than water. If the oil covers it, the object is denser than oil.

    But then does the object move up, or down, or remains half submerged in water (and oil)? What happens to the water with oil on top? Does the water change density?
     
  4. Oct 26, 2007 #3
    no the density of the water is unchanged.. the oil just sits on top without changing the set up too much.. but would the oil produce a downward force on the block, pushing it down? b/c the force exerted by the water originally only opposed gravity*weight of block
     
  5. Oct 27, 2007 #4

    stewartcs

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    There isn't a net force change. Pascal's principle says that the pressure is transmitted undiminished in all directions so the pressure due to the oil on top of the water would be transmitted through the water and to the bottom of the object as well. So the increases would cancel out essentially (or least I think so).

    It sounds like the only forces are the buoyant force pushing up and the gravitational force (weight of the object) pulling down. Since the object was floating (half in and half out) originally, and there was no net change in force, the object should remain in the same position.

    At least that seems reasonable to me!
     
    Last edited: Oct 27, 2007
  6. Oct 27, 2007 #5

    Shooting Star

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    To those who have seen my last posting, I am deleting it because I misunderstood the question.
     
    Last edited: Oct 27, 2007
  7. Oct 27, 2007 #6

    Shooting Star

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    The object will rise up with respect to the water level.

    Suppose, instead of oil, water had been added. Then the object would rise up to maintain the same relationship with the present water level, which means it'd rise up with respect to the original level.

    Now suppose that the oil is almost as dense as water, but just slightly less. What'll happen now?

    The OP should figure out the rest by himself. The math is not difficult, remembering that if any portion of an object is submerged in a liquid, there is a bouyant force.
     
  8. Oct 27, 2007 #7
     
  9. Oct 27, 2007 #8

    Shooting Star

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    Think of it this way. The buoyant force of water was there initially, because the body displaced water. Now it is displacing some oil too, and so there is a buoyant force from the oil. Thus, there is an extra force upward, which lifts it out of the water a bit until the body reaches a position when again the two buoyant forces add up to the weight.

    It is not floating on the surface of the oil, but rising a bit wrt the interface between the water and the oil.

    In my earlier post, I had tried to give you a hint for a continuity argument. Think about that, too.
     
  10. Oct 27, 2007 #9

    Shooting Star

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    The pressure difference between the top and the bottom of an object is essentially the buoyant force. The object is half out of water. So, when the object is covered in oil, there’s a pressure difference between the top of the body and where the water surface intersects the body, due to the pressure of the oil. That results in the buoyant force due to oil only. Both the pressures are transmitted to all parts according to Pascal’s law.
     
  11. Oct 27, 2007 #10

    stewartcs

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    If the object, let say a sphere, is floating in water (halfway submerged), then oil is added on top of the water and covers the sphere, the oil will float on top of the water since it is less dense. I don't see how any extra net buoyant force would be applied to the bottom of the sphere to make it rise since the oil is floating on the water which means only the top half of the sphere is covered by the oil (the hydrostatic pressure due to the oil would be "pointing" perpendicular to the surface of the top half of the sphere only, and not the bottom). The net hydrostatic pressure at the bottom of the object is still the same since the extra head pressure that is transmitted through the water due to Pascal's principle is negated by the downward pressure of the oil acting on the top half of the sphere. So, I don't see it being able to rise or sink.

    Maybe I'm not thinking of this correctly, so please correct me if I'm wrong.
     
  12. Oct 27, 2007 #11

    stewartcs

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    The reason an object floats is due to the net hydrostatic pressure at the bottom of the object being greater than the downward force of the object due to gravity...i.e. it's weight is less than or equal to the buoyant force.
     
  13. Oct 27, 2007 #12

    Shooting Star

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    Hi stewartcs,

    By your logic, nothing will change even if we pour water instead of oil. But that is not the case, as is obvious.

    Please note that the pressure at the interface of water and oil, due to oil, is more than the pressure of the oil at the top, and cannot be negated, like you have said.

    Use Archimedes’ Principle directly. As I have said, any part of a body which is immersed in a liquid will face an upward buoyant force, which, as we all know, is equal to the weight of the liquid displaced.

    The lesser the density of the oil, the lesser the effect will be on the body.
     
  14. Oct 28, 2007 #13

    stewartcs

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    Yep, I agree now. I was assuming the pressure differential would be 0, but it's not like you pointed out (since the object is NOT completely submerged in the water - only partially). The net force on the bottom would be greater. After this discussion I thought about it with a square that was half in and half out of the water, with the oil rising around it, but not on top of it yet, the hydrostatic pressure would be increasing at the bottom of the square (since it is transmitted through the water), resulting in a net increase in force at the bottom. This then lead me to realize that at any infinitesimal change (of hydrostatic pressure due to the oil being added) would always result in a larger force at the bottom since the rate of change of the pressure on the bottom of any object would be larger than at the top.
     
    Last edited: Oct 28, 2007
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