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Bubble in water tank affects overall weight?

  1. Dec 22, 2014 #1
    1. The problem statement, all variables and given/known data

    Compare the weight in these 3 cases:

    http://imageshack.com/a/img913/4506/hlmnbL.png [Broken]


    2. Relevant equations

    --

    3. The attempt at a solution

    I think the weight is the same in three case, is it correct ?

    http://imageshack.com/a/img538/1879/rjJyK3.png [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 22, 2014 #2

    TSny

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    The title for this thread seems to be meant for another thread.

    In order to receive help, you must show your work or state your reasoning. So, why do you think the weights are the same?
     
  4. Dec 23, 2014 #3
    oops, sorry

    In the second case, there is an upward force but the height of water is bigger so the pressure is more important at bottom.

    In the third case, the conservation of the energy implies the same weight.
     
  5. Dec 23, 2014 #4

    TSny

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    By "upward force" I guess you mean the tension force in the string pulling upward on the bottom of the container. Yes, the downward force from the water pressure at the bottom is greater in #2 compared to #1. But, why do you say that the upward force of the tension is exactly cancelled by the extra dowward force of the water? I don't think that's true.

    Can you elaborate on this?
     
  6. Dec 23, 2014 #5
    Yes, I think it's the same force. With a square surface of 1m*1m with a height of 10m, if g=10 exactly and the density of the fluid =1 (1000 in SI), the force is of 100000N up. The heigh of liquid increases too. If the surface of the container is 2*2, the height increase of 10*1*1/2/2=2.5, the pressure at bottom increases of 0.25*100000*4=100000N, it's the same. I don't take in account the surface of the rope. What do you think it's not the same force ?

    If the weight is not the same, if the container move down at the velocity V constant, this could say the energy that the device can recover is greater (or lower) when a bubble move up. The force works on the distance = V*time.
     
    Last edited: Dec 23, 2014
  7. Dec 23, 2014 #6

    TSny

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    OK. I think you are modelling the bubble of air as having a volume of 10 m3. So, the buoyant force on the bubble is 100,000 N. And, as you have shown, the increase in the force of the water on the bottom of the container due to the increased pressure at the bottom is 100,000 N.

    However, have you taken into account the weight of the bubble of air?

    I'm sorry, I don't understand this. Why would the container move down? I think of the container as resting on a scale to measure the apparent weight of the system. I don't see how energy relates to what the scale reads.
     
  8. Dec 23, 2014 #7
    Yes, like that it seems higher than in the second case. But the volume of the container is higher too and the Archimide's principle with air is higher too, no ? In fact, the exercice don't say if there is air outside or not. If not, you're right.

    I imagine this case for find the weight. At top, I take the container, I place a bubble of air inside, this need an energy I noted E1, I accelerate all the container with the bubbe attached in the bottom of the container. When the container is at the velocity V, I cut the rope, if the weight is higher or lower when the bubble move up this could say the energy recover is higher/lower. The energy E1 is lost in friction I think.
     
  9. Dec 23, 2014 #8

    TSny

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    Yes! Very good. I was neglecting the buoyant force due to the surrounding atmosphere. If the density of the air in the bubble is the same as the density of air in the atmosphere, then the weight of air in the bubble is balanced by the extra buoyant force of the atmosphere on the system when the water level rises due to the insertion of the bubble . So, I think you are right and the apparent weight of the system is the same for picture 1 and picture 2.

    [It's hard to understand how you can tie a string to a bubble of air o_O. I guess we can consider the bubble to be in a very thin shell of material that has negligible thickness and negligible weight.]

    I'm still not getting this energy argument. I understand that you would need to do work to push the bubble down from the surface to where it is attached to the string. So, there is some energy stored that's equal to the work done. Are you imagining the container moving downward at speed V when the rope (or string) is cut? Is V a special velocity or an arbitrary velocity? Can you explain in more detail how energy would be recovered when the string is cut and how this recovered energy is related to the apparent weight of the system?
     
  10. Dec 23, 2014 #9
    Yes.

    Not specific, but if V is high the extra weight works on a big distance.

    I just imagined an external fixed electromagnetic device that can recover energy (efficiency=1 in theory) for keep constant the velocity. If the weight is not the same (+or -) the weight works more or less when the bubble move up. The distance is V*time, and the time is not dependant of V.

    Yes I imagined a small massless container with air inside.
     
  11. Dec 23, 2014 #10

    jbriggs444

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    If you imagine the cord being cut and the bubble rising, do you imagine it rising at a steady rate or at an accelerating rate? [Not a trick question]

    If the bubble is rising in this manner, what does this mean for the net momentum of the water surrounding the bubble? Is it changing or unchanging?

    What about the force of gravity on the water. Is it changing or unchanging?

    Newton's second law can be phrased as F = dp/dt (force is the rate of change of momentum over time). Given all of this, what conclusion can one reach about the force needed to support the tank?
     
  12. Dec 23, 2014 #11

    TSny

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    Yes, that's important.
     
  13. Dec 23, 2014 #12
    I'm not sure if the bubble is accelerating, what is the difference ? The statement of the exercice say only to compare weight. For me it's a "free" bubble like in reality in a container with water. The goal is to know if the weight is the same than in case 1.
     
  14. Dec 23, 2014 #13

    jbriggs444

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    Yes. It matters whether the bubble is accelerating. If the bubble is accelerating, so is the water. If the water is accelerating, its momentum is changing. In order to change its momentum, a net force is required.
     
  15. Dec 23, 2014 #14
    If the bubble is accelerating, the weight is not the same ?
     
  16. Dec 23, 2014 #15

    jbriggs444

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    What definition of "weight" are you using?
     
  17. Dec 23, 2014 #16

    TSny

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    The "apparent" weight is not the same. That is, if you weigh the system on a scale its reading will depend on the acceleration of the bubble. I think your problem is asking you to compare apparent weights.

    Consider a related example of a person in a box resting on a scale. The person holds a massive ball. Compare the reading of the scale for when the ball is held at rest, when the ball is lifted at constant speed, and when the ball is accelerated upward at constant speed.

    The bubble is interesting in that when the bubble rises, some water must move downward to fill the space previously occupied by the bubble.
     

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  18. Dec 23, 2014 #17
    Ok, I understood. But in reality, a free bubble in water (or other fluid) accelerates ?
     
  19. Dec 23, 2014 #18

    TSny

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    When the string is cut there is definitely an initial upward acceleration of the bubble. To see why, compare the buoyant force on the bubble to the weight of the bubble.

    However, due to fluid resistance the bubble quickly reaches a terminal velocity where the acceleration is essentially zero.
     
  20. Dec 23, 2014 #19
    You're right, it's logical. The bubble start with a velocity of 0 and even its velocity is X the bubble need to accelerate during a time dt, correct ? The weight is not the same, correct ? So, during this dt, and if the container moves at V (V is constant by an external fixed device) before cutting the rope, this could say it's possible to recover more energy than the weight. Even dt is very small, V can be very high. I can take V or 2V the time dt is not dependant of V. Where the energy is lost ?
     
  21. Dec 23, 2014 #20

    TSny

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    Correct.
    I'm still not understanding the energy argument.
     
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