Bubble in water tank affects overall weight?

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Homework Statement



Compare the weight in these 3 cases:

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


Homework Equations



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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]
 
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  • #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?
 
  • #3
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The title for this thread seems to be meant for another thread.
oops, sorry

So, why do you think the weights are the same?

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.
 
  • #4
TSny
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In the second case, there is an upward force but the height of water is bigger so the pressure is more important at bottom.

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.

In the third case, the conservation of the energy implies the same weight.

Can you elaborate on this?
 
  • #5
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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.
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 ?

Can you elaborate on this?
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.
 
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  • #6
TSny
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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 ?

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?

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.

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.
 
  • #7
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However, have you taken into account the weight of the bubble of air?
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.

Why would the container move down? I
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.
 
  • #8
TSny
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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 ?

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 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.

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?
 
  • #9
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Are you imagining the container moving downward at speed V when the rope (or string) is cut?
Yes.

Is V a special velocity or an arbitrary velocity?
Not specific, but if V is high the extra weight works on a big distance.

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?
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.

[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.]
Yes I imagined a small massless container with air inside.
 
  • #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?
 
  • #11
TSny
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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]

Yes, that's important.
 
  • #12
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do you imagine it rising at a steady rate or at an accelerating rate?
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.
 
  • #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.
 
  • #14
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If the bubble is accelerating, the weight is not the same ?
 
  • #15
jbriggs444
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What definition of "weight" are you using?
 
  • #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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  • #17
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Ok, I understood. But in reality, a free bubble in water (or other fluid) accelerates ?
 
  • #18
TSny
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Ok, I understood. But in reality, a free bubble in water (or other fluid) accelerates ?
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.
 
  • #19
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When the string is cut there is definitely an initial upward acceleration of the bubble.
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 ?
 
  • #20
TSny
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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 ?
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 ?

I'm still not understanding the energy argument.
 
  • #21
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I'm still not understanding the energy argument.
I compare the energy from a container that move down without a bubble and a container with it when the bubble move up. I give the same cycle for two container:

1/ I move up 2 containers without bubble
2/ In one container I place a bubble with a rope
3/ I move down (without friction) 2 containers
4/ When containers reach the velocity V I recover the energy from an external device (electromagnetic or other), V is constant
5/ I cut the rope, during dt one container has more weight, the additionnal energy is V*dt*F with F the force from the extra weight

The energy from the extra weight is V*dt*F, F and dt are independant of V (V is constant), and V can be high. The other container without a bubble recover less energy. It's for that I thought the weight was the same.



http://imageshack.com/a/img904/8712/Js5EM4.png [Broken]
 
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  • #22
jbriggs444
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Not only must the bubble accelerate and move. It must also stop or disappear. Those events have transient effects on the apparent weight of the container. Where does the energy argument account for that? In addition, the water will have moved relative to its nominal V. Where does the energy argument account for that?
 
  • #23
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It must also stop or disappear.
But I can recover all kinetic energy before the bubble reach the surface, no ? I stop in a very short time (in theory) the device.

In addition, the water will have moved relative to its nominal V
True, but this energy is small and it is not dependant of V, no ?
 
  • #24
jbriggs444
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Sorry, I'm with TSny. I do not understand the energy argument at all. You have multiple containers of different types with and without cords and bubbles moving at high velocities and stopping instantly to avoid the transient effects that you want to ignore in an effort to explain the transient effects that you care about. It does not make for a convincing argument.
 
  • #25
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You have multiple containers of different types with and without cords and bubbles moving at high velocities and stopping instantly to avoid the transient effects that you want to ignore in an effort to explain the transient effects that you care about. It's does not make for a convincing argument.
I compared the energy of 2 containers, one with a bubble one without a bubble, it's all. If one has more weight when it moves down, for me the additionnal weight gives an additionnal energy dependant of V. The energy must be the same for 2 containers, no ?

Why I can't stop all the device in theory for compare the energy ?

that you want to ignore

Because I don't understand all dynamic effect of the bubble, I try to understand.
 
  • #26
jbriggs444
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Let me try to decode the Rube Goldberg machine. Possibly we have a language difficulty.

1/ I move up 2 containers without bubble
2/ In one container I place a bubble with a rope
3/ I move down (without friction) 2 containers
4/ When containers reach the velocity V I recover the energy from an external device (electromagnetic or other), V is constant
5/ I cut the rope, during dt one container has more weight, the additionnal energy is V*dt*F with F the force from the extra weight
1. So you start with 2 empty containers at the top of a track.
2. In one container you add a bubble and a rope. [Which requires energy].
3. You allow the containers to free fall until they achieve velocity V. This is energy-neutral. Potential energy is being converted to kinetic energy.
4. You allow the containers to continue moving downward at a constant velocity, harvesting gravitational potential energy as it is being released.
5. You cut the rope in the one container. The bubble begins to rise. You do not calculate whether this increases the effective weight of the container or decreases it. But you accept the proposition that it changes the effective weight of the one container.

You argue that this violates conservation of energy because you can make the energy discrepancy between the two containers arbitrarily large by increasing V.

However, the fallacy becomes clear...

The container with the bubble is still moving at velocity V. But the water in that container is not moving at velocity V. It is moving at a different velocity and, as a result, has a different kinetic energy than the water in the bubble-free container. The faster V is, the larger this kinetic energy discrepancy becomes. This is enough to exactly cancel the effect of increasing V. Instead of magnifying the effect that you were trying to analyze, the increase in V did not alter the energy balance at all.
 
  • #27
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You argue that this violates conservation of energy
At start, I imagined this cycle for argue that the weight is constant...I don't imagined it was so complex.

It is moving at a different velocity
If the bubble moves up, the water moves down in the container, so its velocity is higher than V ?
 
  • #28
jbriggs444
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If the bubble moves up, the water moves down in the container, so its velocity is higher than V ?
Yes.

If its water velocity is v then its kinetic energy is ##\frac{m(v+V)^2}{2}=\frac{mV^2}{2}+mVv+\frac{v^2}{2}##. The ##mVv## is the part that scales with V and ruins things for this line of argument.
 
  • #29
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I don't understand, if the velocity of the water is higher when the bubble moves up the kinetic energy is higher, no ? Could explain with the difference with one container without a bubble please ?
 
  • #30
jbriggs444
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The kinetic energy is higher and the effective weight is lower. So the force applied is lower so the energy recovered is lower, but the distance moved by the COM is greater so the pseudo-work is greater. The water is moving relative to the container so there is some work done by the container on the water. We don't have an explicit term for the energy going into turbulence. There is the velocity squared term and the associated gravitational potential energy to deal with. Lots of terms with lots of different scaling factors. You conjured the scenario to try to eliminate a calculation. If it does not simplify things then it has failed to fulfil its purpose.
 
  • #31
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I would like to think with forces and the distance only. The work is the force by the distance, if the bubble accelerates and moves up when the container move down from A to B with a constant velocity: the force F works from A to B. Then if the weight is lower or higher (it was logical it is higher if the bubble accelerates to up, the water accelerates to the bottom), the work from the weight is lower/higher too, Fd. The center of mass can move, but forces are applied to the bottom of the container, not in the center of mass. So, if I want to calculate the work from F I must draw another force ?

http://imageshack.com/a/img905/5034/IebquJ.png [Broken]
 
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  • #32
jbriggs444
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Force F times distance d is the work done by the container on the generator/restraining mechanism. Yes. That force is reduced because the water is accelerating downward with respect to the container.

Energy is conserved -- the work done by the generator/restraining mechanism on the container is the negation of this quantity of work.

Then you can consider the work done by the container on the water and the work done by the water on the container. That part is more difficult. Water is not rigid. It can absorb mechanical energy. The work done across the interface between water and container (computed by multiplying the force supporting the water by the motion of the container) is not equal to the pseudo-work (computed by multiplying the force supporting the water by the motion of the water's center of mass). Energy is still conserved. The difference manifests as turbulent energy in the water.
 

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