Comparing weights of balloons

[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.

 

[erobz]

@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.

So, starting from a balloon the size of a baseball, you might need a pretty big bell jar to observe it. Also, what pressure change would this be associated with?

 

[Charles Link]

One other comment though is lowering the outside pressure is not the same thing as blowing up a balloon, where you increase the internal mass considerably. I still need to do further calculations on this.

I do expect there could be a case though where you lower $p_{out}$ to start to cross a threshold, but its very difficult to model mathematically.

 

[Orodruin]

@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.

If you expand the volume by a factor of 10 then the pressure inside the balloon decreases by a factor of 10. This seems like a very very weird balloon.

 

[Orodruin]

but its very difficult to model mathematically.

Which part? The change in potential energy is a fairly simple mathematical model. The strain-stress relationship of rubber outside the purely elastic regime is something else.

 

[erobz]

I'm going to try something "simple" where the pressure differential between the inside of the ballon and outside is some fixed constant ( $P_b-P_j = P'_b - P'_j = C$) It obviously not going to work to vacuum, but is a really bad start? The balloon is to be undergoing adiabatic expansion, and the bell jar air is experiencing a mass change at constant temperature. Does it have a chance, or am I dead in the water?

The buoyant force will be $F_b = \rho'_j V'_b$. It seems like it already could be a bit messy.

If @Orodruin model is already producing this (more or less), I'm not clever enough to tell.

Maybe we should migrate this to a new thread?

 

[erobz]

So I get these starting equations:

For the Jar losing mass at $T$

$P_jV_j = m_jRT$

$\implies \frac{P_jV_j}{P'_jV'_j} = \frac{m_j}{m'_j}$

$\implies P'_j = \frac{P_jV_j}{V_j - (V'_b-V_b) } \frac{m'_j}{m_j} \tag{1}$

So the Pressure in the jar $P'_j$ becomes a function of the mass in the jar $m'_j$, which we can write as some fraction of the initial mass $m_j$, and function of balloon volume $V'_b$.

Then for the balloon we have the adibatic process:

$P_b {V_b}^{k }= P'_b {V'_b}^{k}$ where $k = 1.4$

Then sub in:

$P'_b - P'_j = C$

$\implies P_b {V_b}^{k} = ( C+ P'_j){V'_b}^{k} \tag{2}$

Then you have to solve those (1 and 2) simultaneously.

Thats as far as I feel like going tonight. If I wake up to find it doesn't have legs... so be it.

 

[Orodruin]

If @Orodruin model is already producing this

It won’t because I am not assuming the pressure difference to be constant. Instead, I am basing the argument on when the pressure forces and tension in the balloon are in equilibrium.

 

[Charles Link]

I think in general you might find the buoyant force to decrease on the balloon inside the bell jar as you suck the air out, contrary to what I said in post 44, and also contrary to what @Orodruin may have said in post 59. I had made the error of thinking the balloon has a threshold when you blow one up, but neglected to consider that you are also putting a lot more air into the balloon when you blow one up, as I mentioned in post 72.

From what @Orodruin has in post 69, I think it is possible to see the buoyant force increase for part of the cycle when the air is pumped out, but I think it to be somewhat unlikely.

 

[erobz]

So for $V'_b$ an analytical result looks highly improbable...

$$ \frac{P_b {V_b}^{k} - C {V'_b}^{k}}{{V'_b}^{k}} = P'_j = \frac{P_jV_j}{V_j - (V'_b-V_b) } f $$

Where $f$ is the mass fraction ($m'_j = f m_j$ )

Trying to solve for $V'_b$

$$ C {V'_b}^{k+1} - \left( C( V_j+V_b)+ P_jV_j f \right) {V'_b}^{k} - P_bV_b^{k} V'_b + P_bV_b^{k}( V_j+V_b ) = 0 \tag{3} $$

Then in principle we solve for $V'_b$ as a function of $f$.

From there we employ the Ideal Gas law to get $\rho'_j ( f ) = \frac{P'_j}{RT} = \frac{P_jV_j f }{RT\left( V_j+V_b-V'_b \right)}$

And again $F_b(f) = g\rho'_j(f)V'_b(f)$

So does anyone see some kind of analysis/method that cuts out obtaining the solution to (3) i.e. $V'_b ( f )$ or are am I stuck unable to determine how this function would behave as $f$ declines?

What happens if I say $k=2$, its like a general cubic. Can I say $k= 1$ making it a quadratic? Should/can I check terms in (3) to see if some may be negligible in comparison to others...Are there any cheats were the general function behavior wouldn't change significantly.

 

[Charles Link]

The adiabatic case will always have a lower buoyant force (higher density inside the balloon because the balloon cools with expansion) than the isothermal case, so if you are trying to show the buoyant force decreases when you let the air out, you could solve the isothermal case where $k=1$.

 

[erobz]

Ok, $F_b$ appears (not expansively tested for range of parameters) to be monotonically decreasing as air is evacuated from the bell jar ( balloon expands) under the assumptions of this model (post 79).

Inital parameters:

Functions:

Graphical representation: (I wasn't sure how to reverse the scale in the program)

I'm sure it's a gross simplification based on what has already been discussed (so I'm not sure if I've gained too much), but I couldn't leave it be.