Comparing weights of balloons

[jbriggs444]

Ok, I got to think for a bit I see.

$$ P_{bottom}A - P_{top}A = F_b $$

You seem to be computing the weight of the volume of air that would be present if the jar were a perfect cylinder.

Or the buoyancy of a perfect cylinder in said air. Is $A$ the horizontal area of the jar or of the balloon?

 

[erobz]

You seem to be computing the weight of the volume of air that would be present if the jar were a perfect cylinder.

Or the buoyancy of a perfect cylinder in said air. Is $A$ the horizontal area of the jar or of the balloon?

If there is a helium balloon in the jar will it be given by $\rho g h$?

 

[jbriggs444]

If there is a helium balloon in the jar will it be given by $\rho g h$?

The difference in pressure between top and bottom will be given by $\rho g h$ regardless of the presence of a helium balloon floating in the bell jar, yes.

If the helium balloon is resting at the top of the jar on a contact point of negligible surface area then the total force at the top of the jar will have one contribution from $A P_\text{top}$ and a second contribution from the buoyancy of the balloon minus the weight of the balloon. The support force from the bottom of the jar minus the net force at the top of the jar will turn out to match the weight of the air in the jar plus the weight of the balloon floating in that air. As it must -- that's a second law force balance.

 

[Charles Link]

So are you saying the buoyant force acting on the balloon in the jar becomes larger as the air is evacuated from around it?

The bell jar and its contents which includes the air inside the bell jar is what is getting weighed when the bell jar, totally enclosed, is on the scale. The bell jar does displace a volume of ambient air around it, so the buoyant force from the bell jar becomes part of the picture to be considered.

What the balloon does inside the bell jar is no longer of any bearing in the calculation.

Simply its weight plus the air inside the balloon, plus the weight of the air inside the bell jar, plus the weight of the glass of the bell jar minus the buoyant force of the whole bell jar from the outside air is what the scale will read.

 

[Charles Link]

Perhaps it might be worthwhile at this point to mention why archimedes principle works like it does, that the buoyant force is the weight of the air or fluid that is displaced. What actually occurs is the pressure is just ever so slightly larger as you go further down and thereby the submerged object, when all the external pressure forces are summed, experiences a buoyant force that mathematically is equal to the weight of the displaced air or fluid.

The actual mathematics of this is exact, and follows from a form of Gauss' law.

On the item of a helium balloon inside the jar, the helium balloon does not create any magical upward force of its own. If it is inside the jar, the weight of the balloon plus the helium will add to the weight of the jar.

Helium is less dense than air, so that there is a net upward force on a helium balloon in air when the weight of the helium plus the balloon is less than the buoyant force from the weight of the air that is displaced by the helium balloon.

You may ask what happens if the helium balloon pushes against the underside of the top of the jar? When that occurs, the push comes from the underside of the balloon, and any such pressure/force will be balanced by an equal downward force/pressure against the bottom of the jar (on the inside). That's where it really doesn't matter what happens inside the jar. The scale will see the weight of the contents inside the jar, regardless of what the contents are doing.

 

[erobz]

What does the scale measure in this scenario with the scale completely inside bell jar?

 

[Charles Link]

The weight of the blue egg including its contents of air or fluid minus a buoyant force from the weight of the displaced air inside the bell jar.

 

[erobz]

The weight of the blue egg including its contents of air or fluid minus a buoyant force from the weight of the displaced air.

ok, then when we decrease then amount of air in the jar, the ballon expands the displaced air volume goes up, but the density of the displaced air goes down i.e. there is less air in the same jar. So what is $F_b$ doing, I think its declining in magnitude, because I expect there to be no buoyant force in a perfect vacuum - thus, increasing the readout on the scale?

 

[Charles Link]

ok, then when we decrease then amount of air in the jar, the ballon expands the displaced air volume goes up, but the density of the displaced air goes down i.e. there is less air in the same jar. So what is $F_b$ doing, I think its declining in magnitude, because I expect there to be no buoyant force in a perfect vacuum - thus, increasing the readout on the scale?

What the buoyant force does in that case would be hard to predict, and it would depend upon just how much the balloon expands as the external pressure is lowered. If the balloon expands enough, the buoyant force could increase. If you make a vacuum, yes, then the buoyant force would be zero.

 

[erobz]

What the buoyant force does in that case would be hard to predict, and it would depend upon just how much the balloon expands as the external pressure is lowered. If the balloon expands enough, the buoyant force could increase. If you make a vacuum, yes, then the buoyant force would be zero.

So then it's fair to say that there is something complicated occurring if you think it won't necessarily monotonically decrease toward zero. I'm a little bit -

For a "rigid balloon" (I expect) it decreases monotonically as the air is evacuated. What is it about the rubber?

 

[Charles Link]

So then its fair to say that there is something complicated occurring if you think it won't necessarily monotonically decrease toward zero. I'm a little bit -

You've greatly changed the scenario from what @kuruman has with his isolated bell jar sitting on top of the scale. His case is simple. His balloon is inside an isolated bell jar. It doesn't matter for his case how the balloon responds.

When you have a balloon, basically in ambient air on a scale, and you change the air pressure, (e.g. lowering it) the balloon could expand an unpredictable amount. That makes for a case that doesn't have the same simple answer for what the buoyant force will do. If the balloon suddenly expands to a much larger volume, the larger volume could more than offset the lower density of the ambient air.

 

[erobz]

You've greatly changed the scenario from what @kuruman has with his isolated bell jar sitting on top of the scale. His case is simple. His balloon is inside an isolated bell jar. It doesn't matter for his case how the balloon responds.

When you have a balloon, basically in ambient air on a scale, and you change the air pressure, (e.g. lowering it) the balloon could expand an unpredictable amount. That makes for a case that doesn't have the same simple answer for what the buoyant force will do. If the balloon suddenly expands to a much larger volume, the larger volume could more than offset the lower density of the ambient air.

Have I though? So…that whole assembly is now on a scale of its own. The scale reading inside is unpredictable, and the scale reading outside is predictable. That’s kind of interesting isn’t it?

 

[Charles Link]

You could have the balloon inside the bell jar be a helium balloon that originally floats in the bell jar. It doesn't matter at all to the scale outside what the helium balloon is doing. It will read the weight of the contents of the bell jar, minus the buoyant force of the entire bell jar from the outside air.

The scale inside the jar will read zero though if the helium balloon is floating in the air or rises to the top and pushes against the top of the bell jar. If you then empty the air out of the jar, the scale outside the jar will read less, but the scale inside will now have a helium balloon resting on it and it will read its weight. (Note: there is zero buoyant force inside the jar). The outside scale reading went down, and the inside scale reading went up. Yes, it's very interesting. :)

 

[Thamska]

The answer was given by the teacher, which the OP didn't buy. I pointed out OP's misapplication of th 3rd law in post #12.

Let's say you have a bowl of water on a scale. Then you submerge your hand in the bowl without touching the bowl so the hand is freely in the water not moving. This will result in the scale registrering more weight since there will be additional force from somewhere on the scale. I just thought that the additional force was a opposite force from the buoyancy force on my hand, but it seems you think I am wrong?

If you apply this to the balloons then you could say the scale readings will both be the same? Their mass will always be the same.

 

[Charles Link]

@erobz addition to post 48: The scenario mentioned at the bottom of this post with the helium balloon also applies to @kuruman 's condom demonstration of post 14. The outer scale will show less weight, but a scale inside the bell jar would show an increase as the air inside the bell jar is removed.

to answer @Thamska of post 49, your first paragraph is correct, but I don't see how you get the result of the second paragraph from that, and the second paragraph is incorrect if air is present to supply a buoyant force in the scale readings to the balloons.

Note: Even the bowl of water (no hand in it) appears to weigh less than it really does because there is a buoyant force from the outside air. The scale would show a slight increase if you were to remove the air.

 

[Orodruin]

Let's say you have a bowl of water on a scale. Then you submerge your hand in the bowl without touching the bowl so the hand is freely in the water not moving. This will result in the scale registrering more weight since there will be additional force from somewhere on the scale. I just thought that the additional force was a opposite force from the buoyancy force on my hand, but it seems you think I am wrong?

No, that's fine.

If you apply this to the balloons then you could say the scale readings will both be the same? Their mass will always be the same.

This is not fine. The mass is the same but the buoyancy is not so the force pressing down on the scale is greater as, in equilibrium, it is equal to the weight of the balloon minus the buoyancy of the balloon. The reading on the scale is based on the force pressing down on the scale, not on the actual weight of the object being measured.

 

[Steve4Physics]

Let's say you have a bowl of water on a scale. Then you submerge your hand in the bowl without touching the bowl so the hand is freely in the water not moving. This will result in the scale registrering more weight since there will be additional force from somewhere on the scale. I just thought that the additional force was a opposite force from the buoyancy force on my hand, but it seems you think I am wrong?

If you apply this to the balloons then you could say the scale readings will both be the same? Their mass will always be the same.

Hi @Thamska. Maybe this will help.

You are not comparing like-with-like.

For the ‘hand in the bowl of water’:
- the water exerts an upthrust (buoyant force) on the hand (Archimedes);
- therefore the hand exerts an equal magnitude ‘downthust’ on the water (N3L)
This entire ‘downthrust’ is transmitted to the scale – increasing its reading.

However, for the Post #1 balloons, it is the atmosphere that provides the upthrust. Any change in the atmosphere's 'downthrust' is (in principle) spread-out over the entire surface of the earth, not concentrated on the scale.

Edit. Minor changes to (hopefully) improve clarity.

 

[jbriggs444]

For a "rigid balloon" (I expect) it decreases monotonically as the air is evacuated. What is it about the rubber?

This is rather unrelated to the thread topic, but the expansion of a rubber balloon is more complicated than one might imagine.

As the radius of the balloon increases, the curvature of the surface decreases. So it takes less pressure differential to keep it inflated. This is the primary effect that makes it hard to start blowing up an air balloon but easier once you have it started.

As the radius of the balloon increases, the tension (per linear section across the tension direction) in the balloon skin increases. As in Hooke's law. This acts to decrease the effect of the decreased curvature.

There may be a departure from Hooke's law due to the decreasing thickness of the balloon material during expansion.

There will be a departure from Hooke's law if we are measuring tension per unit cross sectional width because that width is increasing.

There will be a departure from Hooke's law due to the nature of rubber. It reaches its elastic limit and sustains increased tension without much further expansion. Then it pops.

 

[Charles Link]

I could not follow @Steve4Physics 's explanation in post 52. The problem is that the buoyant force will be less for the (cold) balloon of smaller volume, and thereby it will read more on the scale than the (warmer ) larger balloon. (Both balloons have the same mass).

Edit: I think I see it now. He's trying to explain why the buoyant force doesn't push right back on the scale, negating its effect. The buoyant force is an upward force on the balloon, and is larger with a balloon of larger volume, reducing the scale reading. If you instead had someone standing on the earth creating an upward force, the Newton's action reaction force would be back on the earth and not on the scale. Thereby the Newton action reaction from the buoyant force does not go to the scale. Yes, maybe that's where the OP was missing something...

 

[kuruman]

Let's say you have a bowl of water on a scale. Then you submerge your hand in the bowl without touching the bowl so the hand is freely in the water not moving. This will result in the scale registrering more weight since there will be additional force from somewhere on the scale. I just thought that the additional force was a opposite force from the buoyancy force on m

No, that part is correct. The buoyant force exerted by the water on the hand has a 3rd law counterpart, an equal and opposite force exerted by the hand on the water which increases the reading on the scale. However, it is instructive to write down some equations and see how all this is put together.

The figure on the right shows
(A) A glass with water on a scale. A brick is suspended by a string over the glass.
(B) The brick is lowered until it is fully submerged. The water level in the glass rises by $\Delta h$.
(C) The string is cut and the brick sinks to the bottom.
What does the scale read in each case? We can ignore the weight of the glass; it can always be added to the readings derived below.

Case (A)
Clearly, the reading on the scale is the weight of the water.
Note that if the height of the water in the glass is $h$ and the area of the glass is $A$, we can write the reading on the scale $$R_{\text A}=m_{\text{water}}g=\rho_{\text{water}}(Ah)g=(\rho_{\text{water}}gh)A=p_{\text{bot.}}^{\text{(A)}}A$$ where $p_{\text{bot.}}^{\text{(A)}}$ is the gauge pressure of the water at the bottom of the glass. So we can conclude that the contribution of the water to the scale reading is the water pressure at the bottom times the area.

Case (B)
When the brick is submerged, the pressure at the bottom increases, $p_{\text{bot.}}^{\text{(B)}}=p_{\text{bot.}}^{\text{(A)}}+\rho_{\text{water}}g\Delta h.$ The reading on the scale is now $$R_{\text B}=(p_{\text{bot.}}^{\text{(A)}}+\rho_{\text{water}}g\Delta h)A=m_{\text{water}}g+\rho_{\text{water}}(A \Delta h)g. $$ Note that the second term in the above expression is the weight of the water displaced by the brick, i.e. the buoyant force $BF=\rho_{\text{water}}(A \Delta h)g~$ on the brick. In other words, the reading on the scale is increased by an amount equal to the buoyant force.

At this point it is worth considering the external forces on the water plus brick system. We have already taken account of the water. For the brick, we have $$T-m_{\text{brick}}g+BF=0 \implies T=m_{\text{brick}}g-BF.$$ Case (C)
When the brick sits at the bottom, the water plus brick system is at rest just like in case (C) and the water level is the same. For that to be true, the tension $T$ that kept the brick at rest must be replaced by an equal normal force provided by the bottom of the glass. To keep the bottom of the glass at rest, the normal force provided by the scale must increase by $T$. Thus, the reading on the scale will be $$R_{\text C}=R_{\text B}+T=(m_{\text{water}}g+BF)+(m_{\text{brick}}g-BF)=(m_{\text{water}}+m_{\text{brick}})g.$$ The reading on the scale is just the sum of the weights of the contents of the glass. There is no buoyant force that reduces the contribution of the brick.

 

[Thamska]

I think I understand it now. I just had my physics exam today and it went well. Thanks for the answers!

 

[kuruman]

The scale reading inside is unpredictable . . .

In what sense unpredictable? If you know the parameters on which it depends, then it can be predicted. From the FBD of the balloon on the right we have $$N+BF-m_{\text{in}}g=0 \implies N=m_{\text{in}}g-BF.$$ The normal force is the reading of the scale $R$. In terms of the densities of the air inside and the air outside the balloon and the volume of the balloon, we have $$\begin{align} & R=\rho_{\text{in}}V_{\text{bal}}g-\rho_{\text{out}}V_{\text{bal}}g\nonumber \\
& R=\left(1-\frac{\rho_{\text{out}}}{{\rho_{\text{in}}}} \right){m_{\text{in}}g}.
\end{align}$$ The last expression assumes that the scale has been tared to subtract the weight of the balloon. We write the density as $\rho=\dfrac{N\mu}{V}$ where N is the number of air molecules and $\mu$ the mass of a single molecule. Then the ideal gas law becomes $$p=\frac{\rho}{\mu}kT.$$Because $p_{\text{in}}>p_{\text{out}}$ (otherwise the balloon would not be inflated) it follows that $\rho_{\text{in}}>\rho_{\text{out}}.$ In that case equation (1) predicts that the reading on the scale will always be less than the weight of the gas inside the balloon as air is sucked out of the bell jar. To get an exact number, one would have to know the density ratio.

 

[Charles Link]

View attachment 356851In what sense unpredictable? $$Because $p_{\text{in}}>p_{\text{out}}$ (otherwise the balloon would not be inflated) it follows that $\rho_{\text{in}}>\rho_{\text{out}}.$

One minor correction : If the air inside the balloon is of a lighter atom or molecule, such as helium (He) or hydrogen ($H_2$). Then the scale reading can actually be negative, i.e. if you attached the balloon to the scale with a string, it would pull upward.

 

[Orodruin]

The point is that, without more information on the condom, you cannot determine $\rho_{\rm in}$. The condom volume - and therefore the density of the air inside - will depend on the elastic properties of the condom. The important thing to know if you want to know if the reading increases or decreases as you suck the air out is not knowing the density ratio - it is knowing how the density ratio changes as you change $\rho_{\rm out}$ and to know that you need to do some modelling.

Essentially, taking a spherical condom and defining $p_{\rm out}/p_{\rm in} = x$ I believe you would get something like$$
x= \frac{Rp_{\rm out}}{{R p_{\rm out} +{\sigma(R)}}}
$$ where $\sigma(R)$ is the dependence of the surface tension as a function of size $R$. If $\sigma(R)$ grows linearly with $R$ then $x$ would decrease with $p_{\rm out}$, meaning the scale reading would increase as you suck the air out. However, I do not think a condom has a linear $\sigma(R)$ ... In the limit of constant $\sigma(R)$ you would find $x \propto R^3$ and the reading would decrease when you suck the air out.

Note: These were just quick back-of-the-envelope considerations and I am tired so I may be completely outlandish here ...

 

[kuruman]

One minor correction : If the air inside the balloon is of a lighter atom or molecule, such as helium (He) or hydrogen ($H_2$). Then the scale reading can actually be negative, i.e. if you attached the balloon to the scale with a string, it would pull upward.

I followed the original question where air is both inside and outside.

Essentially, taking a spherical condom . . .

I think that a cylindrical condom would be a better fit to your model.

Anyway, thank you both for reinforcing my original point that which way the scale reading will tip is not unpredictable but depends on the specific parameters of the situation.

 

[Orodruin]

which way the scale reading will tip is not unpredictable but depends on the specific parameters of the situation

I think ”unpredictable” or not depends on what you are given. In the OP’s case they are not given the elastic properties of condom rubber, which makes the way of tipping unpredictable given the information. Of course, the actual setup in the OP (as well as the setup weighing the full contraption) is predictable.

 

[Orodruin]

I think that a cylindrical condom would be a better fit to your model.

Also, speak for yourself!

I am a theoretical physicist - everything is a sphere!

 

[erobz]

@kuruman I was just trying to follow along with this notion proposed by @Charles Link that the buoyant force on the balloon might not be decreasing at all stages between the initial state and vacuum in the bell jar due to the complexity of the balloon expansion.

 

[kuruman]

@kuruman I was just trying to follow along with this notion proposed by @Charles Link that the buoyant force on the balloon might not be decreasing at all stages between the initial state and vacuum in the bell jar due to the complexity of the balloon expansion.

I understand. Thank you for clarifying this point.

 

[jbriggs444]

The point is that, without more information on the condom, you cannot determine $\rho_{\rm in}$. The condom volume - and therefore the density of the air inside - will depend on the elastic properties of the condom. The important thing to know if you want to know if the reading increases or decreases as you suck the air out is not knowing the density ratio - it is knowing how the density ratio changes as you change $\rho_{\rm out}$ and to know that you need to do some modelling.

The situation could be even worse. For some elastic properties, the equilibrium volume of the balloon might be multi-valued (more than one equilibrium state for a given pressure). Or there might be no stable equilibrium at all.

The volume of the balloon might not be a function of pressure alone. It might be a function of pressure history. This is not as esoteric a possibility as one might think. I claim that an ordinary rubber balloon is multi-stable in this sense. This might be usable with a pneumatic digital computer.

 

[Charles Link]

@kuruman I was just trying to follow along with this notion proposed by @Charles Link that the buoyant force on the balloon might not be decreasing at all stages between the initial state and vacuum in the bell jar due to the complexity of the balloon expansion.

@Orodruin explains this very well with his computations in post 59. You might not have seen this post, because you didn't give it a "like" yet.

 

[Charles Link]

@erobz Reading it more carefully, I think @Orodruin has a typo in post 59, and it should read $x=\rho_{out} / \rho_{in}$, (instead of $p$'s.) He's referring to @kuruman 's calculations of post 57. See also his post 61.

Edit: I need to look at it more closely though...If the gases are the same, and the temperature are the same, then the $p$ is proportional to $\rho$.... But I think in general he's trying to show it for the $\rho$ ratio.

It should also read $\sigma \, \alpha \, p_{out} / R^3$, instead of $x$. ? I need to study this further. Right now I'm just guessing...

 

[Orodruin]

@erobz Reading it more carefully, I think @Orodruin has a typo in post 59, and it should read $x=\rho_{out} / \rho_{in}$, (instead of $p$'s.) He's referring to @kuruman 's calculations of post 57. See also his post 61.

Not really, I used the pressure for finding the equilibrium of the radius. However, the ideal gas law gets you $p \propto \rho$ so the pressure ratio is equal to the density ratio. I am assuming the gas inside is the same as the outside.

 

[Orodruin]

It should also read $\sigma \, \alpha \, R^3$, instead of $x$.

It should not. That case is for $\sigma$ constant (which is kind of unphysical). In essence I base it on finding the minimum in the total potential energy from pressure and tension, which becomes $\sigma(R) = \Delta p R = R(p_{\rm in} - p_{\rm out})$. Solving for $x$ gives you the expression in #61.

If you assume constant $\sigma$ and use that $p_{\rm in} R^3$ is also constant and therefore $p_{\rm in} = \kappa/R^3$, then$$
\sigma = \frac{\kappa}{R^2} - p_{\rm out} R
$$ and therefore $$
x = \frac{p_{\rm out}R}{p_{\rm out} R + \sigma} = \frac{p_{\rm out}R}{\kappa/R^2} = \frac{p_{\rm out} R^3}{\kappa}
$$
It now strikes me that I missed the explicit $p_{\rm out}$ dependence in the numerator, which is of course also important. To determine the actual behaviour would require some additional analysis of a third degree polynomial to get $R$ as a function of $p_{\rm out}$, which I am not overly keen to do at the moment.

 

[Charles Link]

@Orodruin I also have tried to model the balloon mathematically, and it seems to be rather difficult. What is apparent though, that if you decrease $p_{out}$ a little, and the balloon suddenly expands by a factor of ten or more in volume, that you will see an increase in the buoyant force.