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Buoyancy: Doubtfull and unclear points in theory

  1. Sep 24, 2010 #1
    (volume of body)=(0,1metres)^3
    (weight of body outside the barrel)=10
    (weight of barrel with water without the body)=20
    (weight of barrel with water and the body at the bottom)=20+10=30
    Where's buoyancy? If there is buoyancy in this case, shouldn't the weight of the body inside the barrel be
    (weight)-(weight of displaced water)=10-1=9
    and thus
    (weight of barrel with water and the body at the bottom)=20+9=29?

    By "weight" I mean what the weighing machine under the body or under the barrel shows.

    The weight of the body when it sits at the bottom of the barrel is 10 or 9? If we put one more weighing machine between the body and the bottom of the barrel what does it show? 10 or 9? If it shows 9, then the water+barrel weighs 21 (ignoring the weight of the "in between" weighing machine).

    Buoyancy is prooved by at least two other methods, but regarding when it does not sit on the bottom.
     
    Last edited: Sep 25, 2010
  2. jcsd
  3. Sep 24, 2010 #2
    At the bottom of the barrel, if the body is covered in water, then buoyancy affects the apparent weight.

    If it is not covered by water ( when the adhesion between the material of the body and water is greater then the adhesion between the water and the barrel) i think there is no buoyancy acts on the body.
     
    Last edited: Sep 24, 2010
  4. Sep 25, 2010 #3

    Doc Al

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    I assume that by 'weight' you mean true weight--the gravitational force exerted by the earth.
    The buoyant force is just the net force exerted by the water on the body. Looking at the 'barrel + water + body' system, the buoyant force would be an internal force and cancels out. (If the water exerts an upward force on the body--the buoyant force--then the body also exerts an equal downward force on the water. That's just Newton's 3rd law.) The weight of 'barrel + water + body' will be 30, not 29.

    In any case, buoyant force only affects the apparent weight of the body, not the true weight.

    (FYI, your numbers are a bit off. Given the volume of the body, the mass of the displaced water would be about 1 kg. The weight would be about 10 N. But that doesn't matter.)
    A 'weighing machine' (scale) just measures the force pushing against it, which is the apparent weight, not the true weight (mg). If you could arrange to measure the apparent weight of the body--the force required to support it--you'd get an apparent weight of 9 since the buoyant force helps support it. But the water level rises just enough so that the water pressure increases accordingly. The net force on the bottom of the barrel remains equal to the true weight of 'water + body'.
     
  5. Sep 25, 2010 #4
    (transfered to my next post)
     
    Last edited: Sep 25, 2010
  6. Sep 25, 2010 #5

    Doc Al

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    That logic doesn't hold. Only the apparent weight of the body has changed. If you like, you can consider the apparent weight of the water to have increased by 1, since the water level rises.

    Huh? Again, you are confusing apparent weight with true weight.

    :rolleyes: Give me a break. I'm not the one confused about buoyant force here.
     
  7. Sep 25, 2010 #6

    Doc Al

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    What units are you using for 'weight'? The mass density of water is about 1000 kg/m^3.
     
  8. Sep 25, 2010 #7
    The units of weight I used is what the weighing machine says: (0,1metres)^3 of water ="1".

    Perhaps it's wrong what I said about the "in between" weighing machine, because it also counts weight of water above the body. (I doubt what's wrong here, because an experimental proof of buoyancy when the body is not in contact with the bottom and covered by water above and below, says that the body lost weight equal to the weight of the displaced water, and that weight of the body INCLUDES the weight of water above the body. So what's the weight of the body, the downward force which includes the weight of water above it, or subtracting the weight of the water above it? Anyway.)

    If the weight of the body (meaning, let's say the force the body alone exerts on the bottom) is 9, then the weight of the water+barrel increased from 20 to 21. What is wrong with that conclusion? Or if this conclusion is correct, how is it prooved that it increased to 21?

    I hope you realise that so far you answered nothing at all.
     
    Last edited: Sep 25, 2010
  9. Sep 25, 2010 #8

    Doc Al

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    So your 'weighing machine' uses units of kg.

    You realize that your scenario is confusingly presented? So now you want a scale placed directly under the body, which will measure the total force on it including any pressure from the water. And not just the 'weight of the body'. The 'weight of the body' does not include the weight of the water above the body.

    Including the weight of the barrel just confuses the issue. Try this. Put a scale that covers the entire bottom of the barrel. Thus that scale will read the force of whatever is pressing down on it. It will read the total true weight of 'water + body'. To support the body will require 9 units; to support the increased height of water will require +1 additional unit. Thus the additional force required to support the 'water + body' (compare to what is required just to support the water alone) will equal the true weight of the body = 10 units.

    That's the same attitude you presented in the other thread you started on this exact same problem.
     
  10. Sep 25, 2010 #9
    If I got right of what you mean, we place a weighing machine on the surface of the bottom of the barrel. And before the body is dropped into the barrel, it shows that the water weight is 20. Then we let the body sit on the bottom. Now the machine shows 30. And you say that "the force exerted by the body on the machine is 9, therefore the force by the water increased to 21 whereas it was 20". But how do you proove that? I say it is more probable that water's remained 20, and the body's exerted force is 10, and buoyancy disappeared. Why? Because when it sits at the bottom the resistance to the body is solely acted by the bottom, whereas when it does not sit at the bottom, then it sits on the water, and the resistance to the body is acted by the water (which resistance of the water equals buoyancy at least when the body floats). Proove this wrong.
     
    Last edited: Sep 25, 2010
  11. Sep 25, 2010 #10

    Doc Al

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    There are two scenarios to consider. (For simplicity, imagine the body as a cube.)
    (1) The body sits on the bottom of the barrel, but there's a small layer of water underneath it. That's all that's required for the net force from the water surrounding the body to exert an upward buoyant force as usual. (Small protrusions of the barrel surface make contact with the body, support it.)
    (2) The body sits on the bottom of the barrel and makes a perfect, water-tight seal. In that case the net force of the water on the body is now a downward force (since the bottom of the body is not in contact with the water).

    Pick the one you want to discuss. (In ordinary circumstances, scenario #1 is the most likely.) In either scenario, the scale will show 30 units, the total weight of 'water + body'.

    Of course, these two scenarios are the extremes. You can certainly have something in the middle between these two extremes. No matter--the scale still shows 30 units.
     
  12. Sep 25, 2010 #11
    What I want to discuss is the case that it is a sphere and it sits at the bottom. In this case, proove wrong my arguments at my last post.

    When it has not yet sat at the bottom, the weighing machine does not show 30. Does it?!
     
  13. Sep 25, 2010 #12

    Doc Al

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    I don't understand your 'arguments', but if the sphere rests on the bottom then the upward force on the sphere exerted by the bottom will be 9 units. The buoyant force of 1 unit will reduce the apparent weight of the sphere. Once again, the water level rises, increasing the water pressure on the bottom of the barrel; the net force on the bottom due to water pressure will equal 21 units. The total force exerted by the bottom will equal 30 units, the weight of the water + sphere.

    Let's stick to static cases where equilibrium is reached.
     
  14. Sep 25, 2010 #13
    Now that's an interesting answer, but I doubt it's correct:
    You said that the weight of water increased from 20 to 21 because its height rised and the surface of the bottom times the increase of height, equals the volume of water (which weighs 1) displaced by the body. But when we increase the height of water and because of this its weight increases, this happens because also its volume increased. In this case we opened a hole in its previous volume, thus...same weight. I start doubting the obvious now. Why is what I just said wrong?
     
  15. Sep 25, 2010 #14

    rcgldr

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    Not the weight directly, but the downwards force the water exerts on the barrel, which is a function of pressure times the surface area of the bottom of the barrel. The higher the level of the water, the greater the pressure differential between the upper and lower surface of the water. The upper pressure is the same as the air pressure within the barrel, which remains essentially constant since it's just rearranged and it's density is much less than that of the water, so the net result in an increase in pressure at the bottom of the barrel.
     
  16. Sep 25, 2010 #15
    I answer to that my previous post. How did your last post prooved wrong my previous argument?
     
    Last edited: Sep 25, 2010
  17. Sep 25, 2010 #16

    rcgldr

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    Another way to look at this is that the water weighs the same as before, but now the sphere exerts a downwards force onto the water (due to gravity), which is transmitted through the water (via increased pressure at the bottom of the barrel), increasing the total force the water exerts on the bottom of the barrel by the weight of the sphere.
     
  18. Sep 26, 2010 #17
    So you want to deny the existence of buoyancy because it leads to the same reading of total weight ? since we already know that buoyancy exists when body is immersed in water ( in normal case) you should state a solid reason why the buoyancy disappears when the body is at the bottom as i did.
    Your argument (although lack of clarity) does show that it is possible for the body to be unaffected by buoyancy but only at that special position and it has a condition (that you didn't state). The condition makes the theory valid. End of story.
     
  19. Sep 26, 2010 #18

    Doc Al

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    :rofl:
    The true weight of the water doesn't increase, but the height of its surface does. That causes an increase of water pressure at the bottom of the barrel.
    Why would the volume of the water increase?
    If I understand what you mean by 'opened a hole', you are referring to the fact that we haven't changed the amount of water, just displaced it by an amount equal to the volume of the sphere. That's certainly true. But we have changed the water pressure at the bottom.
     
    Last edited: Sep 26, 2010
  20. Oct 1, 2010 #19
    A rising bubble does not accelerate, but deccelerates before it reaches terminal speed. Does anybody KNOW whether the other rising objects accelerate or deccelerate before reaching terminal speed?

    I guess your answer regaring objects with density close to that of water, will be that they accelerate. What about a buoy and a ping-pong ball?
     
    Last edited: Oct 1, 2010
  21. Oct 2, 2010 #20
    If the water the body have displaced spills from the container then you can "say" that the body lost that weight and see the problem from your point of view.
    And every body accelerates if it start with speed under its terminal velocity, and if you shoot it with higher speed it decelerates. I dont think that bubbles first accelerate to some speed higher than their terminal speed and then decelerate to that speed.
    Except maybe there is something with the decreasing of the pressure so they become bigger and their terminal speed decreases.
     
    Last edited: Oct 2, 2010
  22. Oct 2, 2010 #21
    See that they deccelerate instead of accelerate before reaching terminal speed:
    http://74.6.238.254/search/srpcache...5&icp=1&.intl=us&sig=R0HLt_Q_RwxvVm_bHJ6P2A--
    So what about larger balls filled with air and rising in the sea?

    Another question: When you press a volleyball under the surface of the sea, the speed of the ball increases or decreases when it comes out of the water? I.e.
    1.) The average speed (while being) above water is more or less than the average speed below water?
    2.) The average speed above water is more than its final speed below water (?!)
    3.)...(lol, what should the question be?)
     
    Last edited by a moderator: Apr 25, 2017
  23. Oct 2, 2010 #22
    Yes, as I imagined. As the bubble is going in the lower pressure zone it expands so its surface (the radius) is getting bigger. And the terminal velocity depends on the radius of the sphere (inversely proportional to the radius or inversely proportional to the square of the radius depending on the regime).So yes it could be decelerating, but this is happening only when the bodies are compressible.
    And the ball is increasing its speed, if not it wont move, it will stay at the place you had left under water (the speed goes from 0 to some final speed, depending on the depth the ball is put).
    When the ball exits the water it decelerates, of course.
     
    Last edited: Oct 2, 2010
  24. Oct 5, 2010 #23
    When you press a volleyball under the surface of the sea, when it comes out of the water continues to move upwards. Regarding while moving upwards above water in the air:
    1.)Its average speed above water is more or less than its average speed below water?
    2.)Its average speed above water is more or less than its final speed below water?
    3.)Its average speed above water regarding the first half distance is more or less than its average speed below water?
    4.)Its average speed above water regarding the first half distance is more or less than its final speed below water?
     
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