Buoyancy very easy question

[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?

 

[rock.freak667]

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

 

[luckis11]

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

 

[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.

 

[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?

 

[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:

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

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

 

[luckis11]

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

 

[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).

Perhaps there is someone else here who will help you, and blindly give you the answer.

 

[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.

 

[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.

 

[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.

 

[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

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".

 

[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.

Your link says:
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.

 

[rock.freak667]

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.

Your link says:
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!

What I have grasped from the (somewhere implied) theory, 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

Downward force acting = weight = mg

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

In a simple case, V=Area*height.

 

[luckis11]

That's what I said.

 

[JaredJames]

Since you do not know the experimental answer

What experiment? I gave you an experiment to try and I told you exactly what the result would be.

I am asking YOU if the hydrostatics are saying this. Are YOU certain it's not saying this?

The hydrostatics are not saying this. I am CERTAIN it is not saying this so long as the conditions I outlined are followed.

I have done a lot of work ending up seeing that it's saying this.

Saying what? That a metal sphere won't float (conditions observed)? I'd like to see this work.

Have a start:
http://glasstone.blogspot.com/2006/03/analytical-mathematics-for-physical.html

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.

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.

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.

Your link says:
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 thory says? The weight of the displaced fluid is AhD, where h is the height...of the body!

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.

 

[JaredJames]

That's what I said.

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

 

[luckis11]

That nuclear article was saying that the balloon 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?

 

[JaredJames]

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

 

[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.

 

[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?

 

[JaredJames]

One more important question, which is very easy to be prooven experimentally. Just to make sure: You have a solid which weighs 1kg, and a barrel containing some water, and water plus barrel is weighting 2kg. Now we place the 1 kg of the solid body in the barrel, and it goes to the bottom. The barrel now weighs 3kg, correct?

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.

I don't mean to sound hostile, but you are asking questions easily found and explained in the wiki article on buoyancy. Aside from adding to our post counts, the discussion in this thread is unnecessary after I posted that link. Please, please, please read that article. It has some good explanations, it will help you. If you are not prepared to read, you can't expect everyone to do it for you. If you are reading, then post what it is you are having trouble with and I will do my best to help.

 

[luckis11]

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

 

[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.

 

[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?

 

[JaredJames]

Since (as you say) it did not lose weight equal to the weight of water it displaced,
i.e.
good - (weight of body outside the barrel)=10kg
good - (weight of barrel with water without the body)=20kg
good - (weight of barrel with water and the body at the bottom)=20+10=30kg
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=9kg? But you said it is not
(weight of barrel with water and the body at the bottom)=20+9=29kg
And if there's no buoyancy in this case, then remember you were insiting that there is. So where is it?

OK, firstly, kg = mass. N = weight. I know it's picky, but it is an important distinction. mass is a scalar and weight is a vector.

Remember, for every action there is an equal and opposite reaction. So if the object floats, the water is pushing upwards with a buoyant force = to 98.1N (10 * 9.81). This force is also present in the opposite direction (from the weight of the object) and so this force is exerted downwards. The NET force is zero.

If the object sinks, it rests on the bottom. The object is pushing down on the bottom with a force of 98.1N and the bottom has a reactive force against it of 98.1N (thanks to Newtons third law).

The section I have made bold:
If the object displaces 0.1m^3 of water within the barrel, the water level will rise slightly, if it overflows from the barrel the weight of the barrel will change. If it does not overflow then the weight of the water and barrel does not change. Can you see this?

Regardless of whether the object floats or not, its mass does not change and its weight does not change. The only reason the barrel / water weight will change is if some of the barrel / water is lost at some point (from overflow etc).

You are confusing the buoyant mass of the object. If an object is neutrally buoyant it will appear to be weightless in the surrounding fluid (it will neither rise nor fall and so 'float' - think of an airship or a submarine). Again, it's in the wiki if you read it.

 

[luckis11]

You are answering my questions by "we have answered your questions". Well, wikipedia does not answer them and everything is "meant".

"buoyant mass" is supposed to answer my questions? Mass=~(number of nucleons)(mass of each nucleon). So, since the number of nucleons does not change, I am supposed to figure out how and why the mass of each nucleon changes? Some kind of relativity that explains the nature of mass ... somehow? Why should it change? I say it remains the same, why am I wrong?

 

[JaredJames]

You are answering my questions by "we have answered your questions". Well, wikipedia does not answer them and everything is "meant".

"buoyant mass" is supposed to answer my questions? Mass=~(number of nucleons)(mass of each nucleon). So, since the number of nucleons does not change, I am supposed to figure out how and why the mass of each nucleon changes? Some kind of relativity that explains the nature of mass ... somehow? Why should it change? I say it remains the same, why am I wrong?

The mass doesn't change, at all. I have said this over and over. So does the wiki article.

The reason an object feels lighter in water is because of buoyancy. It doesn't become lighter. The buoyant force upwards makes the object feel lighter, but there is an equal and opposite reaction downwards from the weight of the object. NET force = 0.

In your answer you keep saying the object gets lighter, and you even gave some maths which I highlighted in the post above and explained why it is wrong. I have not at any point, said something loses mass when placed in a fluid.

You bring up nucleons here, but I have already shown in a previous post that unless the density or volume of the object changes, the mass does not change.

 

[luckis11]

“The mass does not change” when they make it of two kinds instead of keeping it the same?

“Because of buoyancy the object feels lighter but it does not become lighter”?

Next time you’ll tell me that 2+2=5.

Why nobody else responds here? The system of the forum is that a group of somehow connected teachers respond to students and when one teacher takes on a question the others are supposed to agree? Can I change my teacher please?

 

[JaredJames]

There is confusion here between mass and weight. The mass of a an object in a fluid does not change. The weight of the object does not change since weight = mass x gravity. Neither the mass nor the gravity change and so the weight remains constant.

The buoyancy force counteracts the weight, so if you were to weigh it, it would register as lighter than it really is. You would have to compensate for the buoyancy to gain the correct weight of the object.

However, in your example of a barrel of water with an object placed in it, the overall weight of the system is added A + B = C because it is a closed system and the NET force = 0.

 

[Doc Al]

“Because of buoyancy the object feels lighter but it does not become lighter”? It DOES become lighter.

Don't confuse actual or 'true' weight--the force of gravity on an object--with the object's apparent weight--a measure of the force needed to support the object. Obviously dunking an object into a barrel of water won't change the Earth's gravitational pull on it. What does change is the amount of force required to support it--it feels lighter. If the buoyant force happened to equal the true weight of the object, the object would feel 'weightless'--it would just float without needed any support.

 

[fizzickschick]

OMG, entirely overcomplicated. The assumption that there is no pressure difference at the top and bottom of the object didn't really get addressed. There's always a pressure difference, unless you have a dimensionless object. The difference in pressure is what causes the buoyant force, we all agree. So if the object does not rise, does this mean there is no pressure difference, or is there somthing else going on? Namely, the weight of the object! A generic metal hemisphere could rise or sink, depending on the overall density, but the one stated in the original question is neutrally buoyant. Was the original contributor to this thread confused about how this is possible? A simple thought experiment would do the trick. Imagine a block of lead sunk at the bottom of a container. Then the buoyant force is less than the weight of the block. Now remove a piece of lead from the inside of the block and replace it with air. If the air bubble is small enough, the block will still be sunk. But if enough lead is replaced with air, eventually the weight of the block + air will be less than the buoyant force (weight of fluid displaced, rho v g). There. Why is this not simple to explain?

 

[fizzickschick]

Also, I think luckis has a few questions that could be answered by a referal to Newton's 2nd Law and suggesting he draw an FBD.

 

[JaredJames]

OMG, entirely overcomplicated. The assumption that there is no pressure difference at the top and bottom of the object didn't really get addressed. There's always a pressure difference, unless you have a dimensionless object. The difference in pressure is what causes the buoyant force, we all agree. So if the object does not rise, does this mean there is no pressure difference, or is there something else going on? Namely, the weight of the object!

I've discussed the weight and buoyant force over and over and the OP is getting confused with actual weight and buoyant weight (the weight the object feels).

A generic metal hemisphere could rise or sink, depending on the overall density, but the one stated in the original question is neutrally buoyant.

Could you show me where it states neutrally buoyant? His OP regarding the hemispheres tells us:

a metal-coated hollow hemisphere with almost the density of the air (including the metal coat), it raises or not?

If the overall density is virtually that of air, it will definitely rise (it has been pointed out to me that with an air tight seal (perfect seal) between the perfect hemisphere and the bottom of a tank there would only be pressure acting downwards and so the hemisphere would sit on the bottom despite density. How you would get a perfect seal with a metal hemisphere I don't know and would be interested in evidence showing such an example in real life.)

Was the original contributor to this thread confused about how this is possible? A simple thought experiment would do the trick. Imagine a block of lead sunk at the bottom of a container. Then the buoyant force is less than the weight of the block. Now remove a piece of lead from the inside of the block and replace it with air. If the air bubble is small enough, the block will still be sunk. But if enough lead is replaced with air, eventually the weight of the block + air will be less than the buoyant force (weight of fluid displaced, rho v g). There. Why is this not simple to explain?

As I explained, if the overall density of an object is less than that of the displaced water, it floats (as per his example). If it is more than the displaced water, it sinks. If it is equal, it is neutrally buoyant. Nothing difficult there.

The problems arise in the previous posts when the response from the OP were constantly contradicting over numerous lines. He would make a statement in line one and then contradict it a few lines later. Not easy to work with.

I'd also point out that certain posts here have been edited after my responses given and so now don't fit well with what I wrote in response.

Also, I think luckis has a few questions that could be answered by a referal to Newton's 2nd Law and suggesting he draw an FBD.

I refer you to post number 2.