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Entrained Water Question!!!

Discussion in 'Mathematics and Physics' started by Iawia, Feb 4, 2016.

  1. Iawia

    Iawia Member

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    Hey all,

    We have been debating for a while on a specific topic and I was wondering maybe if someone here could supply the answer. You guys have pretty much been able to answer everything else I have ever desired to know.

    Here is the problem statement:

    A solid piece of metal with holes is placed in water. In this submerged condition, does the water inside the holes count toward the part center of gravity? Even more perplexing to me, what happens if some of the holes get completely sealed off from the aqueous environment, while others do not? Does this theoretically change its center of buoyancy or center of gravity?

    The second that I think I understand it, I don't.

    -t
     
  2. Les Jones

    Les Jones Well-Known Member

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    If the holes are open then they will not change its centre of gravity. (It will be at the same point in water as it is in air,.) If a hole is blocked offf at both ends then it will provide buoancy. If the two holes of equal volume are an equal distance from the centre of gravity along a straight line then it would not change the centre of gravity but otherwise they would.

    Les.
     
    • Agree Agree x 2
  3. Tony Stewart

    Tony Stewart Well-Known Member Most Helpful Member

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    There exist dynamic and static properties, but for this question, a 1D static gravitation force is the only concern. In this the water has no net effect, whether trapped or free to flow. The net static force is only the difference in density x volume of metal.

    The acceleration could then be considered by a=F/m for the mass of metal,m, ignoring dynamic fluid friction.
     
  4. dave

    Dave New Member

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  5. jpanhalt

    jpanhalt Well-Known Member Most Helpful Member

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    I agree with Les' analysis assuming the reference point is the CG in air,
    John
     
  6. Tony Stewart

    Tony Stewart Well-Known Member Most Helpful Member

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    It depends on the difference in density only and volume occupied therein. Unsealed the water has no net force., and thus only the density and volume of metal are factors. But sealed , it depends on the assumption for sealed with water or air.

    In both cases it is still the difference in density and volume occupied compared to density and volume of water displaced , that determines the gravitation force.

    If air, the then the average density of metal and air times respective volume again relative to the density and volume of water only, determines the net force.
     
  7. Iawia

    Iawia Member

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    So there seems to be an agreement that entrained water does not count toward an objects center of gravity. I am not convinced either way yet myself because if you do a thought experiment where one puts a small rock within the hole, that pebble counts towards its center of gravity. However, according to the statements here water which is also contained within the hole does not.

    Now consider a second thought experiment. Consider the same object in air. Consider heavy moisture in the air. If one was to squeeze all of the moisture out of the airspace occupied by the hole and set it upon the surface of the object, that mass of moisture would count toward the CG, but as particle floating in air it does not contribute to its CG. Now consider that hole sealed off from its environment and the object is heated so that the moisture from the surface is jettisoned into the hole's same airspace, is that not also part of its mass due to its closed nature? I feel there is something fishy about just simply opening or closing an aperture to alter the center of gravity.

    still puzzled. I am building an experiment for it now.
     
  8. atferrari

    atferrari Well-Known Member

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    Having spent enough (!) time on vessels at sea, I think of the center of gravity as a point in space wrt 3 axis. You can have a good balance in the longitudinal sense but the vessel could be completely listed to one side. Your example with a solid piece of metal does not help to perceive what I say below. It would be much easier if your piece was floating (That is a vessel...)

    Depending where they are, sure they are meaningful, not for the water inside them but for the (eventual) lack of balance of the resulting structure. Here, the concept of moment (weight * distance wrt COG) plays a role. As it is submerged you could not notice the lack of balance but it exists.

    If those holes get sealed at both ends, what matters is what you caught inside; water, air or...) Let us say, that most probably, whatever is inside will be "lighter" than the material the body is made of. Whatever is "included" in that body has to be counted as forming part of it, thus affecting balance and buoyancy. Again, like a vessel...

    Eventually you could read about what is the effect of submerging a body into the water. The displaced volume push that body up. That's the reason why a Ch. Officer always will call the whole weight of the vessel, "displacement" (it is given in tons - metric nowadays).

    In case you never heard the question: what would you do to make a cube of 3x3 cm to float in water? I give you the option to choose any tool you could imagine on earth.
     
  9. jpanhalt

    jpanhalt Well-Known Member Most Helpful Member

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    Sorry, I went to bed quite early and didn't see the others' responses until now. The ultimate question is whether the CG changes upon submersion. Here, the question regards the effect of buoyancy or density compared to water. If you consider a hunk of steel on the end of a stick of wood, that "hammer" can be designed so it balances right at the junction between the wood and steel. If you immerse it in water, the wooden handle will rise relative to the head.

    From the standpoint of this thread, the beam is metal, but we all know that you can drill enough holes in metal and seal them off from the water so that that piece of metal will float.
    Thus, one could replace the wood handle of the aforementioned hammer with a metal handle that has a lot of sealed holes.

    Of course, this thought experiment looks at extremes, but I think Les' explanation still holds in lesser extreme cases. The principle is the same. The SG can shift in the unsymmetrical situation.

    John
     
  10. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    No, so long as ALL holes contain water and the object is fully submerged. The buoyancy provided by the water will act upon the metal displacing water equally. In other words, no different that that buoyancy (however slight) provided by the air it was in, prior to being submerged.
    Two answers:
    1. No, so long as no holes contain any air (or other flotation).
    2. Yes. As in the first example, if certain portions of the submerged piece of metal contain flotation (air pockets), that will alter the CG

    Regardless of the shape and configuration, and so long as NO flotation is created by that shape and configuration, this (Archimedes Principle) holds true:
    upload_2016-2-5_12-38-30.png
    Overall, the object weighs less if submerged, BUT the weight distribution within the fluid remains the same.
     
  11. Iawia

    Iawia Member

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    I work for a company who deals with entrained water within vessels so it does have some definite benefits to knowing the truth behind this.

    Here's the deal. Follow me really quick on one extreme scenario.

    Let's say you have an cylindrical object made of a material called water. There is a large hole in the object off its center axes. The location of CG is: if the hole is on the left, CG will be located in the right plane.

    Now you 'fill' that hole with water. Does the material added to the hole count toward cg? I assume everyone here would say yes, and the CG would shift to back to center.

    Now place the object in water as well. According to folks statements it will have to shift back to the right plane due to the fact that the entrained water in the hole contributes nothing. Hmm.

    Now suppose we were to alter the material properties only of the object (not the contents of the hole) by changing the water density until it becomes a solid. Clearly, changing the density of an object will change the overall CG for an unsymmetric part, now where is CG and does the entrained water contribute to its location? I am trying to demonstrate that physical principles and natural laws should not change merely because material properties shift.

    To be clear certainly the resultant 'wet weight' (weight - buoyancy) force will not change however, I am not certain the CG remains unaffected. Hence, forces at play are not in question, only CG.

    -t
     
    Last edited: Feb 5, 2016
  12. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    My emphsis.
    OK. Although the original material was metal.
    OK.
    Here's where you make a conceptual leap that (you'll excuse me) " doesn't hold water". The object, all of which is water, has a CG in the middle. The so-called "entrained water" is no different that the rest of the object, and this entire object is IN water. Nothing is contained or has a different density and, therefore, there is no, nor can there be, a CG.
    The fallacy of your argument is that you are assuming that water in a hole of a cylinder contributes to the weight of that cylinder if submerged in water. It does not.The "entrained" water is, and can only be, part of the material (the water) in which the cylinder is submerged, i.e., it can only contribute to the buoyancy side of the equation.
    Again, my emphasis.
    Now, if the cylinder is removed from the water, with the hole still filled with the entrained water and within a vessel, then yes, the object's CG will be altered. That water is now additional Mass that has to be included in the Total Mass of the object (the vessel) and, as such, will most definitely alter the CG, in or out of the water.

    Which will, in turn, be reflected in the alignment of the vessel's CG with respect to the horizontal planes (pitch and yaw).

    You have too many variables and terms, all being mushed together without regard to context, to have proven such a statement.
     
  13. strantor

    strantor Active Member

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    A picture is worth a thousand words and an experiment is worth a thousand thought experiments.

    Take an empty mason jar and fashion a long counterbalance handle/arm for it out of clothes hanger. Find the CG, hang the apparatus from a string from the CG. Dunk it in your bathtub and make sure the jar fills. Observe CG.

    Now remove it, cap the jar, find the new CG, and Dunk it again.

    If you can show that the two experiments yield the same result and/or if you can show that dunking the open jar apparatus changes the GC, I'll send you a whole $0.47 via paypal.
     
  14. strantor

    strantor Active Member

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    I suggested the experiment because I couldn't think of a better way to explain it than what's already been explained, but let me take another stab at it.

    The "weight" of water per depth is typically expressed in Atmospheres (ATM), and there's a reason for that.
    1atm = 14.7PSI = 1.013 bar = 760 Torr.
    Notice these are all pressure units, not weight units.

    An object is surrounded by an atmosphere. In air, the atmosphere is ... air.
    In water, the atmosphere is ... water.
    Water is the atmosphere, and the "Weight" of the atmosphere acts equally on all surfaces of the object.
    See Pascal's law:
    Yeah, surrounding an object with water puts the weight of the water on top of it, but it puts the exact same weight pushing up from bottom of too, cancelling out.
    If your object has a void in it, say a hole drilled through, the "weight" of the water is inside that hole, but it's pushing UP on the top of the hole as much as it's pushing down on the bottom of the hole.

    Now you seal off the hole, water can't get in, you change the model. Now you've created a new atmosphere, inside an existing atmosphere. Since the new atmosphere has a lower density, it isn't a cut & dried issue of pressure; NOW weights come into play. Your new atmosphere of fixed volume and fixed pressure has a fixed weight.
     
  15. Ratchit

    Ratchit Well-Known Member

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    The metal will weigh less submerged by the amount of water displaced. Removing metal above the surface will lighten the load, Submerging the altered metal will lighten the load by the same ratio as before. If the metal entraps a multitude of vacuum tubes, then it will be lighter when submerged because water will be displaced by the vacuum tubes. If many of the vacuum tubes are displaced from the surface center of gravity, the submerged CG will change. If some of the vacuum tubes lose their seal and water enters, they lose their buoyancy and the CG changes again. Any questions?

    Ratch
     
  16. Ratchit

    Ratchit Well-Known Member

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    Why a small rock? Why not any size rock that fits in the metal? Is the metal displacing any solid rock material like it is water? Seems to me the metal is supporting the weight of the rock in water. The rock is a drag on the metal, and will change the CG.

    Opening or closing the chamber where the displacement takes place will not change anything unless the content of the displacement changes when the chamber is opened or closed.

    Ratch
     
  17. Ratchit

    Ratchit Well-Known Member

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    The water in the "hole" displaces the same amount of water contained in the hole for a net weight change of zero. Therefore, the CG will not change when submerged.

    Impossible and conflicting conditions. For instance, balsa wood has an extremely low density, but is a solid. Mercury has a very high density, and is not a solid. Cranking up the density does not make a solid.

    How does an overall CG differ from a regular CG? Unsymmetric parts have a CG, too. Changing density of a object will not change its CG.

    That's right. They change because their displacements and force application positions change.

    As I said before, it all depends on force application positions.

    Ratch
     
  18. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    This is ridiculous, strantor. No longer interested in your nonsense.

    Ratch - Beware! Utter waste of time here.
     
  19. strantor

    strantor Active Member

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    Care to elaborate? I wasn't trying to be nonsensical so I'm a little baffled by your response. What is so ridiculous about the experiment?

    EDIT: I can't help but wonder if maybe you're mistaking me to be OP/TS? This is not my thread. The post you quoted was my first reply to this thread.
     
    Last edited: Feb 5, 2016
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  20. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    You're right, strantor. My sincere apologies :(. You were not nonsensical nor was your experiment ridiculous. Please forgive me.

    I mistook your post for that of Iawia. His posts were rapidly looking like that of a troll: no responses refuting ours, just a stream of changed argument components, materials and, well, nonsense. I had allowed myself to get angry rather than just dropping out of the thread.

    That blinded me to the obvious merits of your post.

    CBB
     
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  21. Iawia

    Iawia Member

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    ???

    Some folks are misconstruing the original premise. Where is CG for an object that has holes in it and is placed in water? By applying water as the object's material is only for conceptual purposes. Obviously no one can make an object out of water and put a hole in it. However, it can be used to demonstrate that CG is affected by entrained water. I also never said wet weight is affected as I have already computed this fact (weight does actually change, however this effect is cancelled by the same change in buoyancy). Thus, all of the wet weight calculations in the world can not prove whether CG is affected, hence the employment of thought experiments. Dynamically, of course all the entrained water does effect the term 'm' in F = ma, because generally speaking, all of those particles are being accelerated. Statically, I wanted to further define the location of CG and CB.

    CONCLUSION FOUND: I have determined the answer to my question by conducting an experiment. Very similar to what strantor had suggested, I precisely machined an assembly which held sealed vessels at each end. A string held the assembly from the middle such that if both ends were sealed and filled with water, the device should be balanced and have nearly zero moment. I computed the torque that would be provided if entrained water did not count toward CB/CG and determined roughly ~1 in-lbf can be observed using a 3' assembly using 12 fl. oz water bottles. The assembly with both ends sealed performed as expected and did not rotate.

    Dunking the assembly in water with both ends seal is the reference case so that if there was any issue with imperfections in machining, material densities, string placement etc., the moment would be observable and have to be offset. It did not require an offset. Next, if a vessel was to be unsealed, we will see a shift in CB/CG IF entrained water DOES NOT count toward CG/CB, it can be removed from equation and the CG/CB will shift to the opposite side of where the jar was unsealed. Pls note that a moment can be observed because CB shifts at a different rate than CG therefore will produce a moment upon the assembly. The CB/CG separation is ~1.31"

    Result: No change occurred, no additional moment was observed, hence, CG/CB is affected by entrained water.

    And to cowboybob, I am not sure what the hell a troll is anyway or what is suppose to be inferred by your comment. Thanks to all those who had commented and contributed to this discussion.

    -t
     
    Last edited: Feb 8, 2016

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