# Neutrally Buoyant Bodies

Richard Spiteri
TL;DR Summary
I have a fluid dynamics problem which I have simplified in order to understand a basic concept where part of a body (contraption) is neutrally buoyant while the rest is not.

The problem and questions are posed in the attachment below.
The attachment below describes a tank, hollow pipe and two flexible (balloon like) bags forming one body of weight W_tank.

My two questions are what are the downward forces acting on the submerged body in both cases.

#### Attachments

• SWP Analogue.pdf
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Staff Emeritus

That sounds like a school problem. Is it?

If yes, then our rules require that you show us your attempt at a solution before our helpers offer help.

The mentors may also move this to a homework forum.

Mentor
That sounds like a school problem. Is it?
Nah, he's a retired geophysicist -- from his New Member Introduction thread:

How did you find PF?: Web search

Hello Physics Forum!

I joined this site to peruse questions, learn and perhaps pose a question or two myself.

I am interested in learning a few basics in fluid dynamics.

Richard Spiteri

That sounds like a school problem. Is it?

If yes, then our rules require that you show us your attempt at a solution before our helpers offer help.

The mentors may also move this to a homework forum.
Hello anorlunda,
No, it isn’t a school problem - I am 59 and long out of school.
It is a problem I am pondering since I’ve never seen the physics of a partially immersed (as opposed to ‘partially submerged’) body such as this one. I am also unsure how to treat the ‘balloon-like’ membrane.
thanks and regards,
Richard

Staff Emeritus
OK, not a school problem. But is there any reason why you don't give us your thinking on the problem?

Homework Helper
Gold Member
A helpful truism in problems of this kind is "Water floats in water." In figure A (left) you can ignore the dark blue water inside the contraption. Imagine taking the thing apart and compressing anything that is not water (bags, tube, tank, etc.) into a solid chunk with no cavities inside. If you put that in water, it will experience a force $$F=\rho_{\text{water}}V_{\text{chunk}}~g-m_{\text{chunk}}~g.$$ Here, ##m## is the mass of the chunk and ##V_{\text{chunk}}## is its external volume. You see now that in the original (uncompressed) contraption, the external volume is greater but so is the mass of the object. The equation becomes $$F=\rho_{\text{water}}(V_{\text{chunk}}+V_{\text{water}})~g-(m_{\text{chunk}}+m_{\text{water}})~g.$$ Since ##\rho_{\text{water}}V_{\text{water}}=m_{\text{water}}##, the second equation is the same as the first because we add and subtract the same number. That's the algebraic justification of the aforementioned truism.

In figure B the external volume of the contraption increases by adding air volume ##V_{\text{air}}##. We can again ignore the water and consider the problem of a balloon the skin of which has negligible mass and whose external volume is ##V_{\text{air}}##. If we attach the chunk to this balloon will the thing float or not? Well, the force on it is$$F=\rho_{\text{water}}(V_{\text{chunk}}+V_{\text{air}})~g-(m_{\text{chunk}}+m_{\text{air}})~g.$$If this force is negative, the object will sink, if it is positive, it will float and if it is zero it will be neutrally buoyant.

The take-home message here is that in compound objects, one should compare the average density of the object against the density of the fluid. The average density is the total mass of the object divided by its external volume. If the average density is greater than the density of the surrounding fluid, the object will sink, else it will float. Neutral buoyancy is achieved when the average density is equal to the fluid density. Submarines, for example, regulate their buoyancy by essentially controlling their average density.

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Richard Spiteri and jim mcnamara
Richard Spiteri

That sounds like a school problem. Is it?

If yes, then our rules require that you show us your attempt at a solution before our helpers offer help.

The mentors may also move this to a homework forum.
Hello anorlunda,
No, it isn’t a school problem - I am 59 and long out of school.
It is a problem I am pondering since I’ve never seen the physics of a partially immersed (as opposed to ‘partially submerged’) body such as this one. I am also unsure how to treat the ‘balloon-like’ membrane.
thanks and regards,
OK, not a school problem. But is there any reason why you don't give us your thinking on the problem?
anorlunda,
Yes, because I wanted to hear an unbiased answer since I had one explanation and my father had a different reasoning. We debated it (figure 2 because we both agree on figure 1) and wanted to see what other people think.

I hope this helps and it is in line with the policies of this forum

Richard

Richard Spiteri
A helpful truism in problems of this kind is "Water floats in water." In figure A (left) you can ignore the dark blue water inside the contraption. Imagine taking the thing apart and compressing anything that is not water (bags, tube, tank, etc.) into a solid chunk with no cavities inside. If you put that in water, it will experience a force $$F=\rho_{\text{water}}V_{\text{chunk}}~g-m_{\text{chunk}}~g.$$ Here, ##m## is the mass of the chunk and ##V_{\text{chunk}}## is its external volume. You see now that in the original (uncompressed) contraption, the external volume is greater but so is the mass of the object. The equation becomes $$F=\rho_{\text{water}}(V_{\text{chunk}}+V_{\text{water}})~g-(m_{\text{chunk}}+m_{\text{water}})~g.$$ Since ##\rho_{\text{water}}V_{\text{water}}=m_{\text{water}}##, the second equation is the same as the first because we add and subtract the same number. That's the algebraic justification of the aforementioned truism.

In figure B the external volume of the contraption increases by adding air volume ##V_{\text{air}}##. We can again ignore the water and consider the problem of a balloon the skin of which has negligible mass and whose external volume is ##V_{\text{air}}##. If we attach the chunk to this balloon will the thing float or not? Well, the force on it is$$F=\rho_{\text{water}}(V_{\text{chunk}}+V_{\text{air}})~g-(m_{\text{chunk}}+m_{\text{air}})~g.$$If this force is negative, the object will sink, if it is positive, it will float and if it is zero it will be neutrally buoyant.

The take-home message here is that in compound objects, one should compare the average density of the object against the density of the fluid. The average density is the total mass of the object divided by its external volume. If the average density is greater than the density of the surrounding fluid, the object will sink, else it will float. Neutral buoyancy is achieved when the average density is equal to the fluid density. Submarines, for example, regulate their buoyancy by essentially controlling their average density.
Thank you, kuruman!
Your explanation is excellent and easy to follow. My friend and I were debating whether or not you need to ‘lift’ the weight of the portion of the water inside the air chamber.
You clearly state that you do not lift that chunk of water.
Richard

berkeman and kuruman
Richard Spiteri
Thank you, kuruman!
Your explanation is excellent and easy to follow. My friend and I were debating whether or not you need to ‘lift’ the weight of the portion of the water inside the air chamber.
You clearly state that you do not lift that chunk of water.
Richard

I need to re-emphasize that the red solid plates are fixed and cannot move. Hence the red and green balloons (or air bubbles) anchored to these plates are also fixed and do not rise with the tank and hollow tube).

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Richard Spiteri
My thoughts are that the tank and portions of the hollow (water-filled) tube of weights Wtank, Wtube2 and Wtube3 in Figure A are neutrally buoyant (hence we do not use W*<anything>).

When wrapped with the weightless air tank in Figure B, the tank and portions of the tubes that are wrapped contribute to the total buoyancy of the Air container as well as to the weight. Since the air container is fastened to the frame, however, that upwards buoyancy is zero when calculating the forces on the entire object because it is countered by the tension on the lower plate. Likewise, the weight (were it not negligible) would be taken up by the tension in the upper anchor points.

Therefore I think that only the lower bag, upper bag, tube 1 and tube 4 contribute towards Vchunk and hence buoyancy but W*tank and W*tube2 and W*tube3 must be added to mchunk meaning that they are no longer neutrally buoyant.

Is my logic of apportioning the single hollow tube into tube 1 and tube 4 that provide buoyancy and another sector of tube 2 and tube 3 that doesn't correct?

Also, is my logic where tube 2, tank and tube 3 are now dead weight in air correct?

Where did all that energy (buoyant force) that was present in the tank, tube 2 and tube 3 in Figure A go?

Homework Helper
Gold Member
My thoughts are that the tank and portions of the hollow (water-filled) tube of weights Wtank, Wtube2 and Wtube3 in Figure A are neutrally buoyant (hence we do not use W*<anything>).

When wrapped with the weightless air tank in Figure B, the tank and portions of the tubes that are wrapped contribute to the total buoyancy of the Air container as well as to the weight. Since the air container is fastened to the frame, however, that upwards buoyancy is zero when calculating the forces on the entire object because it is countered by the tension on the lower plate. Likewise, the weight (were it not negligible) would be taken up by the tension in the upper anchor points.

Therefore I think that only the lower bag, upper bag, tube 1 and tube 4 contribute towards Vchunk and hence buoyancy but W*tank and W*tube2 and W*tube3 must be added to mchunk meaning that they are no longer neutrally buoyant.

Is my logic of apportioning the single hollow tube into tube 1 and tube 4 that provide buoyancy and another sector of tube 2 and tube 3 that doesn't correct?

Also, is my logic where tube 2, tank and tube 3 are now dead weight in air correct?

Where did all that energy (buoyant force) that was present in the tank, tube 2 and tube 3 in Figure A go?
I am not sure what tubes 1, 2, 3, and 4 are because they are not labeled in the figure. The idea is simple: Whatever displaces water contributes to the buoyant force, whatever does not displace water, does not contribute to the buoyant force. Stated differently: Weigh your contraption in air. Fill to the brim a tank large enough to hold the contraption with water. Put the contraption in the tank. Collect all the water that spilled and weigh it. There are only two possibilities:
1. The contraption sinks. When this happens, the weight of the spilled (displaced) water (and hence the buoyant force) will be less than the weight of the contraption.
2. The contraption floats. When this happens, the weight of the spilled (displaced) water (and hence the buoyant force) will be equal to the weight of the contraption. If you push the contraption deeper into the water with your finger and hold it there, more water will spill out. The weight of that additional water is equal to the force of your pushing finger because now the buoyant force has to compensate for two forces, the gravitational attraction and the finger force. Note that both of these are external forces, i.e. generated by agents external to the contraption. Internal forces, i.e. forces exerted by one part of the contraption on another, do not count because, according to Newton's third law, they come in pairs and cancel each other when you add all the forces to get the net force acting on the object.

Richard Spiteri
Thank you, kuruman, as always!

The definition of tubes 1, 2, 3 and 4 are found to the right of Figure A and B (with Wtubex and W*tubex separated by dashed lines clearly labelled for x = 1, 2, 3, and 4).

Your answer makes sense to me if the air balloon enveloping the tank and parts of the central tube was left to float and sink on its own. Given that I fastened the air balloon to the two unmovable solid plates (in red), I am either confusing myself or it is more complicated than what you explain.

Using my definitions where W*tank, W*tube2 and W*tube3 are the water-filled weights of the part of the contraption that is enveloped in air (Figure B), I agree with your logic of calculating Vchunk to arrive at a weight of water displaced to calculate buoyancy. I also agree with the elegant algebraic solution of either including the volume of water trapped and then subtracting its weight; or else simply to ignore the trapped water from both the upward buoyancy and downward weight. This clearly answers my first question ("Is my logic of apportioning the single hollow ...and tube 3 that doesn't, correct?"). Thank you for that - your answer helped me grasp this better.

I am still unclear on the answer to my second question about dead weight in air - i.e., the downward gravitational force of the tank, tube 2, and tube 3 enveloped in air in Figures B and C? Do we have to add its weight W*tank, W*tube2 and W*tube3 (i.e., the weight of the tank, tube 2 and tube 3 plus contained water) or do we just add Wtank, Wtube2 and Wtube3 (i.e., the material without water)?

I hope this is clear - we can switch to PF mail conversations if this thread is getting too unfocussed.

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Using my definitions where W*tank, W*tube2 and W*tube3 are the water-filled weights of the part of the contraption that is enveloped in air (Figure B), I agree with your logic of calculating Vchunk to arrive at a weight of water displaced to calculate buoyancy. I also agree with the elegant algebraic solution of either including the volume of water trapped and then subtracting its weight; or else simply to ignore the trapped water from both the upward buoyancy and downward weight. This clearly answers my first question ("Is my logic of apportioning the single hollow ...and tube 3 that doesn't, correct?"). Thank you for that - your answer helped me grasp this better.

I am still unclear on the answer to my second question about dead weight in air - i.e., the downward gravitational force of the tank, tube 2, and tube 3 enveloped in air in Figures B and C? Do we have to add its weight W*tank, W*tube2 and W*tube3 (i.e., the weight of the tank, tube 2 and tube 3 plus contained water) or do we just add Wtank, Wtube2 and Wtube3 (i.e., the material without water)?
I don't see the difference between your first and second question. The answer that @kuruman gave you is general, and applies in the second case as well. You can include or ignore the weight and volume of the trapped water, just be consistent about it.

Richard Spiteri
Homework Helper
Gold Member
I'd rather not switch to private messaging. This is a thread that is still open and all PF users must have the chance to read it, offer their comments and profit from it.

There are two total forces acting on the contraption because there are two agents that interact with it, each in its own way. The Earth and the surrounding water. Here is the breakdown of the constituents of these forces. This applies to both figures A, B and C.

Earth
To get the total force exerted on the contraption by the Earth, you must add
(a) the weight of all the water contained in all parts of the contraption
(b) the weight of the skin of all the bags that form the boundary with the surrounding water
(c) the weight of the tubes, tank, platforms, etc. that form the rigid structure.
These are the weights measured in air and are the sum total of all downward force, i.e. the total downward force in all figures A, B and C. This is the force with which the Earth attracts the contraption and is the same regardless of whether the contraption is in water or not.

Fluid.
The total upward force is the buoyant force and is equal to the weight of the displaced water. As we discussed earlier, the contraption will float or sink depending on how the variable weight of the displaced water compares with the fixed weight of the contraption. I hope this answers your question.

I started writing this before @A.T. posted, but I will leave it as is anyway because it is more detailed.

Richard Spiteri
Richard Spiteri
I don't see the difference between your first and second question. The answer that @kuruman gave you is general, and applies in the second case as well. You can include or ignore the weight and volume of the trapped water, just be consistent about it.
Thanks, @A.T.,

After I sent my last message, I reflected further and saw that I am confusing myself unnecessarily. I just need to treat this as you and @kuruman said - an upward buoyancy force and a downwards gravitational force. If one component of the contraption is held static (like my fixed air-filled balloon), then I must treat that as a external force as @kuruman said.

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Richard Spiteri
I'd rather not switch to private messaging. This is a thread that is still open and all PF users must have the chance to read it, offer their comments and profit from it.

There are two total forces acting on the contraption because there are two agents that interact with it, each in its own way. The Earth and the surrounding water. Here is the breakdown of the constituents of these forces. This applies to both figures A, B and C.

Earth
To get the total force exerted on the contraption by the Earth, you must add
(a) the weight of all the water contained in all parts of the contraption
(b) the weight of the skin of all the bags that form the boundary with the surrounding water
(c) the weight of the tubes, tank, platforms, etc. that form the rigid structure.
These are the weights measured in air and are the sum total of all downward force, i.e. the total downward force in all figures A, B and C. This is the force with which the Earth attracts the contraption and is the same regardless of whether the contraption is in water or not.

Fluid.
The total upward force is the buoyant force and is equal to the weight of the displaced water. As we discussed earlier, the contraption will float or sink depending on how the variable weight of the displaced water compares with the fixed weight of the contraption. I hope this answers your question.

I started writing this before @A.T. posted, but I will leave it as is anyway because it is more detailed.
Thanks once again, @kuruman,

It is clear to me now and I like your explanation. I can calculate the upwards buoyancy force as you suggested in your very first answer using Vchunk and mchunk as long as I am consistent about whether I include the trapped water or not; then treat the downwards gravitational force as mentioned above; and, if I have an external force holding the balloon fixed, I must treat that separately against the downwards force only. It is clear and I see how I can apply it to any complex body as you mentioned in your first answer. Thank you @kuruman and @A.T.!

A.T. and kuruman