# Homework Help: Buoyancy very easy question

1. Sep 16, 2010

### luckis11

Ballon with density 1,2kgr/m^3 (including its cover, thus it contains a gas lighter than air), of a volume of 1m^3, placed in water.

Taking in account buoyancy, but ignoring water resistance, what is the acceleration with which it would move upwards?

2. Sep 16, 2010

### rock.freak667

If you draw a free body diagram, there are only two forces acting on it, what are they?

3. Sep 16, 2010

### luckis11

I have found an answer but want to see whether it agrees with yours, and I do not want to prejudice you.

4. Sep 16, 2010

### JaredJames

It won't prejudice us, part of the rules is that you must show some effort in attempting the question.

Perhaps show some working or just give your answer and I'll give you my own workings or answer in return.

Don't take this badly, but it sounds as if you simply want to be handed the answer for you to copy.

Last edited: Sep 16, 2010
5. Sep 16, 2010

### luckis11

Let's say that I have almost written a book on this subject just by trying to figure out your logic, and I disagree with your illogical solution. Now are you going to give me your solution?

6. Sep 16, 2010

### JaredJames

I'm sorry I don't understand you. This sounds like a homework question.

The rules of PF are that you must show effort in working out the solution.

I am saying that I won't give you an answer without seeing at least some work on your part, because I think you are going to copy my solution and use it as your own. Nothing illogical about my reasoning there.

From the global guidelines:
(I've highlighted 'textbook style exercises' because this is in the form of a question asked in a textbook.)

Last edited: Sep 16, 2010
7. Sep 17, 2010

### luckis11

Do they all use Buoyancy Mass and True Mass? Because there is another solution too.

8. Sep 17, 2010

### JaredJames

I've given you the rules, it's up to you if you follow them.

Give both solutions if it makes you feel better, it really doesn't matter to me. But I can't and won't help you if you don't provide some evidence of effort on your part (as per the PF rules).

9. Sep 20, 2010

### luckis11

Another question (I have done my homework, but this...):

Bottom of the sea. An air bubble comes off the ground and becomes a hemisphere. It raises or not? Obviously yes. But a metal-coated hollow hemisphere with almost the density of the air (including the metal coat), it raises or not?

Before you answer, take in mind that they (all?) do not subtract buoyancy at such body shapes, and particular case, "because the water cannot push them upwards" or "because there is no below and above water pressure difference" etc, which seems true if the metal hemisphere does not rise.

Correct me if you are sure enough I am wrong in any of the above thoughts.

10. Sep 20, 2010

### JaredJames

Why would the metal sphere not rise if it's density is less than that of the water?

If the object neither rises or falls, it is neutrally buoyant. Which means it's mass = mass it displaces. If you have a metal sphere whose overall density is that of air, it is the equivalent of having a bubble of air, so it will rise (float).

The shape of a submerged body is irrelevant, as long as it displaces a mass of water = to that of it's own mass, it will rise (float). If it does not, it will sink. If the amount of mass displaced = mass of object, it is neutrally buoyant and will neither rise nor sink.

11. Sep 22, 2010

### luckis11

If you read the hydrostatics interperetation of buoyancy, you will see that they imply that the metal hollow hemisphere will not rise (when in contact with the bottom of the sea). Since the air bubble does rise, thιs theory (that says that none of them will rise) is wrong (?). But also the opposite theory (that says that both will rise) cannot be correct if the metal hollow hemisphere does not rise (?). So, my question now is not what physics are saying, but what experiments show: Does the metal hollow hemisphere rise or not?

The questionmarks are because there might be some conditions that the theories do not imply, e.g. the (seeming) theory that says that only the medium-body densities define the phenomenon and not the pressure underneath says that only when the whole body is surrounded by the same medium. So the theoretical answer is not ... known. Please tell me the experiments.

Last edited: Sep 22, 2010
12. Sep 22, 2010

### JaredJames

Does the hydrostatics interpretation of buoyancy say this? Or are you simply assuming it says this?

If you want to do a simple experiment, you know a helium balloon will rise, so fill a balloon with helium and sit it on the floor. If it doesn't rise then your assumption is correct, if it does rise (I think you'll find it will), then your assumption is wrong.

As long as the mass of the metal sphere plus contents = mass of air bubble and as such the metal spheres overall density = density of air bubble, then it will rise.

I refer you to this link:
http://www.anzcp.org/CCP/Physics&Chem/Hydrostatic%20Pressure%20&%20Buoyancy%20force.htm [Broken]

If the force F is greater than the downwards force (weight of the object), then the object will rise.

Depth is the key here, not the amount of fluid below the object.

EDIT: Also the wiki article - http://en.wikipedia.org/wiki/Buoyancy

From the equations there, you can see that buoyancy is the result of differing density. Nothing to do with "pressure underneath".

Last edited by a moderator: May 4, 2017
13. Sep 22, 2010

### luckis11

Since you do not know the experimental answer, I am asking YOU if the hydrostatics are saying this. Are YOU certain it's not saying this? I have done a lot of work ending up seeing that it's saying this. Have a start:
http://glasstone.blogspot.com/2006/03/analytical-mathematics-for-physical.html

MY assumption is wrong? What assumption? I said that the bubble RISES, and I asked whether the metal-coated hemipshere does rise too, or not. Again, YOU are making assumptions that they both rise.

The Buoyant Force [the force diving an object (eg air bubble) upward]:
F = AhD
Force [F] is equal to the area [A] times the height [h] (or depth) times the density of the fluid

Now...is this what...the theory says? The weight of the displaced fluid is AhD, where h is the height...of the body and not of the depth.

What I have grasped from the (somewhere implied) hydrostatics equation with buoyancy, is that Buoyant Force is the (upward force of the fluid towards the down surface of the body)-(downward force of the fluid towards the up surface of the body above)=AhD, where each of these fluid forces is (pressure)A.

Last edited: Sep 22, 2010
14. Sep 22, 2010

### rock.freak667

Downward force acting = weight = mg

Upward force acting = Upthrust = Buoyancy = weight of fluid displaced = ρVg

In a simple case, V=Area*height.

15. Sep 22, 2010

### luckis11

That's what I said.

16. Sep 22, 2010

### JaredJames

What experiment? I gave you an experiment to try and I told you exactly what the result would be.
The hydrostatics are not saying this. I am CERTAIN it is not saying this so long as the conditions I outlined are followed.
Saying what? That a metal sphere won't float (conditions observed)? I'd like to see this work.
I'm slightly concerned you linked me to an article on nuclear weapons. What is wrong with the wikipedia article I link you to. The wiki article explains buoyancy a lot more clearly than this 'history of buoyancy' in your nuclear dossier.

There's no assumption on my part. I told you very clearly. If the metal sphere has equal mass and as such overall density of the air bubble, it will react identically to the air bubble. Period.

That makes no sense at all. Their calcs omit gravity for the reason outlined below. The weight of the displaced fluid = volume of fluid displaced x density x gravity. This must equal the weight of the object = volume of object x density x gravity.

However, as gravity is a constant, you can remove this and compare the mass of the displaced fluid with the mass of the object. If you want to quote forces (weight and buoyancy) gravity must be included.

EDIT:
When submerged, a buoyant force > weight = rise. buoyant force < weight = sink. buoyant force = weight = neutrally buoyant (neither rise nor sink).
When above water, a ship needs to be neutrally buoyant.

Last edited: Sep 22, 2010
17. Sep 22, 2010

### JaredJames

No, you omitted gravity with AhD so it isn't weight, just displaced mass.

Last edited: Sep 22, 2010
18. Sep 22, 2010

### luckis11

That nuclear article was saying that the ballon must be pushed underneath. Fortetabouti, check this one, it's better:
http://physics.valpo.edu/courses/p111/lectures/lecture21.pdf
(at the chapter 15)
As you see, he did not subtract buoyancy regaring the air that the body displaced above the sea level. And ... it is NOT a simplification as the results are almost the same (I suggest have some doubts if you think that he meant this simplification). So...WHY do you think he did not subtract it?

19. Sep 22, 2010

### JaredJames

Which slide are we talking about here? That's a long power point and I don't intend reading it all.

20. Sep 22, 2010

### JaredJames

It is a simplification. You clearly didn't read the wiki article I linked. Here is a quote directly from it:

"Air's density is very small compared to most solids and liquids. For this reason, the weight of an object in air is approximately the same as its true weight in a vacuum. The buoyancy of air is neglected for most objects during a measurement in air because the error is usually insignificant (typically less than 0.1% except for objects of very low average density such as a balloon or light foam)."

The force exerted by the atmosphere on a body such as your cork in the slides, is negligible when compared to the others forces involved.

Last edited: Sep 22, 2010
21. Sep 22, 2010

### luckis11

One more important question, which is very easy to be prooven experimentally. Just to make sure: You have a solid which weighs 100kg, and a barrel containing some water, and water plus barrel is weighting 200kg. Now we place the 1 kg of the solid body in the barrel, and it goes to the bottom. The barrel now weighs 300kg, correct? No buoyancy subtracted from the weight of 300kg, correct?

Last edited: Sep 22, 2010
22. Sep 22, 2010

### JaredJames

Whether it goes to the bottom or not, the barrels weight increases by the weight of the object placed in it. In this case, 1kg.

Again, in the wiki article, if you read it fully, there is a good section on buoyant mass.

Last edited: Sep 22, 2010
23. Sep 22, 2010

### luckis11

Since the barrel weighs the sum of the two weights, then how did the solid lose weight when placed in the barrel?

24. Sep 22, 2010

### JaredJames

How did the solid lose weight? In what way did it lose weight?

To lose weight, W = mg, either the mass would reduce or gravity would reduce. Neither of those situations occur and as such it doesn't lose weight.

mass is in kg not weight - that is in newtons. So the mass of all is 3kg, the weight of the barrel is 29.4N. Either way, no weight is lost or gained.

Last edited: Sep 23, 2010
25. Sep 23, 2010

### luckis11

Since (as you say) it did not lose weight equal to the weight of water it displaced,
i.e.
(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
then where's the buoyancy? E.g. when the volume of the body is (0,1metres)^3, then shouldn't the weight of the body inside the barrel be (weight)-(weight of displaced water)=10-1=9? But you said it is not
(weight of barrel with water and the body at the bottom)=20+9=29
And if there's no buoyancy in this case, then remember you were insiting that there is. So where is it?

Last edited: Sep 23, 2010