Physics Riddle: The Fate of the Balance

[micromass]

Here is the experiment with the final answer:

https://www.youtube.com/watch?v=b_8LFhakQAk

 

[D H]

If we cut the string on the left container and let the ball float, the whole container doesn't get any heavier ...

It get's lighter. Suppose we put two flasks on the scale, each containing a ping pong ball anchored from below. The scale will balance because we're measuring two identical systems. Now cut the string on the right. The ping pong ball will float, with almost all of the ball out of the water. The air will buoy the part of the ball that is sticking out of the water. The air in the ball (about 4 centigram) will be a part of what is measured on the left. On the right, that air won't register. The scale will tilt down to the left.

 

[D H]

Here is the experiment with the final answer:

https://www.youtube.com/watch?v=b_8LFhakQAk

That analysis is not quite correct. There is a difference between a floating ping pong ball and a submerged one. The submerged ping pong ball system includes the mass of the air inside the ping pong ball. The floating ping pong ball, most of that mass is buoyed by the air. The difference is about 4.3 milligrams.

That said, a lab balance scale that's holding two flasks each containing about a liter of water most likely isn't going to be sensitive to that 4.3 milligram difference.

 

[micromass]

That analysis is not quite correct. There is a difference between a floating ping pong ball and a submerged one. The submerged ping pong ball system includes the mass of the air inside the ping pong ball. The floating ping pong ball, most of that mass is buoyed by the air. The difference is about 4.3 milligrams.

That said, a lab balance scale that's holding two flasks each containing about a liter of water most likely isn't going to be sensitive to that 4.3 milligram difference.

Well, here you go:

https://www.youtube.com/watch?v=7ADBL7_A9qA

 

[chingel]

It get's lighter. Suppose we put two flasks on the scale, each containing a ping pong ball anchored from below. The scale will balance because we're measuring two identical systems. Now cut the string on the right. The ping pong ball will float, with almost all of the ball out of the water. The air will buoy the part of the ball that is sticking out of the water. The air in the ball (about 4 centigram) will be a part of what is measured on the left. On the right, that air won't register. The scale will tilt down to the left.

Wouldn't air also buoy the higher water level that is in the container with the ball submerged? If both of the systems had the same volume they would be buoyed by the same amount, neglecting change of density with height.

 

[D H]

Well, here you go:

The way he's measuring he can't sense a 4.3 centigram difference. 4.3 centigrams is 0.43 milliliters of water. His method of filling the containers with water has to yield an experimental error that is well over that.

Wouldn't air also buoy the higher water level that is in the container with the ball submerged? If both of the systems had the same volume they would be buoyed by the same amount, neglecting change of density with height.

We can eliminate that buoyant force by the air on the containers by putting a small amount of water in the trays and then put the flasks on those trays.

 

[sophiecentaur]

That analysis is not quite correct. There is a difference between a floating ping pong ball and a submerged one. The submerged ping pong ball system includes the mass of the air inside the ping pong ball. The floating ping pong ball, most of that mass is buoyed by the air. The difference is about 4.3 milligrams.

That said, a lab balance scale that's holding two flasks each containing about a liter of water most likely isn't going to be sensitive to that 4.3 milligram difference.

I don't quite go along with that. The volume of the pingpong ball is still displacing the same amount of air, whether it's out in the air or displacing it via the water displacement. So why isn't it experiencing the same amount of upthrust in both positions?

 

[chingel]

We can eliminate that buoyant force by the air on the containers by putting a small amount of water in the trays and then put the flasks on those trays.

I'm not clear on what exactly do you do with the flasks and the water and how would that eliminate the buoyant force? I would think that two systems with equal volume would still experience the same buoyant force if the air density is the same.

 

[AlephZero]

Wouldn't air also buoy the higher water level that is in the container with the ball submerged? If both of the systems had the same volume they would be buoyed by the same amount, neglecting change of density with height.

I think that is right. The fact that the ball is hollow is irrelevant. The experiment would would exactly the same way with a solid ball that floats on water. The issue is whether the ball is rigid or compressible, and the reasonable approximation is that it is rigid. The total volume of water + ball + container is the same on both sides so the air buoyancy forces are equal (unless we are going to include the change in air pressure with altitude!)

We can eliminate that buoyant force by the air on the containers by putting a small amount of water in the trays and then put the flasks on those trays.

Sorry, I don't understand what you are doing there, without a diagram or some equations. Putting the same amount of water on each tray wouldn't seem to affect anything.

 

[D H]

I don't quite go along with that. The volume of the pingpong ball is still displacing the same amount of air, whether it's out in the air or displacing it via the water displacement. So why isn't it experiencing the same amount of upthrust in both positions?

I'm not clear on what exactly do you do with the flasks and the water and how would that eliminate the buoyant force? I would think that two systems with equal volume would still experience the same buoyant force if the air density is the same.

Sorry, I don't understand what you are doing there, without a diagram or some equations. Putting the same amount of water on each tray wouldn't seem to affect anything.

Think of how your toilet works. Toggling the handle lifts the flapper valve off its seat. The flapper valve is less dense than water, so it remains buoyed off the seat until the water level drops below the valve. The valve settles back in place at this point, sealing the tank. The tank starts to fill with water. The pressure from the water seals the valve even more firmly in place.

The flapper valve is less dense than water. So why doesn't the water buoy the valve up, making the toilet run and run and run? The answer is that by sealing itself in place, the flapper valve stops buoyancy in its tracks. Once sealed, the forces on the flapper valve are the weight of the valve, the water pressure from above, and the air pressure from below. The net downward force is stronger than gravity alone. Once the tank fills, that valve is held firmly in place by the pressure differential until the next time the toilet valve is toggled. There is no buoyant force to lift the flapper valve off the seat until then.


I'm doing the same thing here with water in the trays that hold the flasks. The water keeps air from getting under the flask and thereby stops atmospheric buoyancy on the flasks in its tracks.

 

[chingel]

Ok, but if you would deform the shape of the toilet valve, the total force on the valve would change by the buoyancy force of the added piece, no matter what the shape of the added piece is. Consider a cylindrical toilet valve. If you add a 1 cc slice on top as wide as the cylinder, or add a spherical ball of 1 cc, the total force on the valve would change exactly the same. Of course as long as the changed shape doesn't get out of the water or go through another hole in the bottom or things like that.

Edit: here I assumed the water level doesn't change significantly. Both of the deformations would cause the same change in the total force, but it would not be equal to the buoyancy force on the deformation by itself if the water level changes significantly

 

[OmCheeto]

Think of how your toilet works. Toggling the handle lifts the flapper valve off its seat. The flapper valve is less dense than water, so it remains buoyed off the seat until the water level drops below the valve. The valve settles back in place at this point, sealing the tank. The tank starts to fill with water. The pressure from the water seals the valve even more firmly in place.

The flapper valve is less dense than water. So why doesn't the water buoy the valve up, making the toilet run and run and run? The answer is that by sealing itself in place, the flapper valve stops buoyancy in its tracks. Once sealed, the forces on the flapper valve are the weight of the valve, the water pressure from above, and the air pressure from below. The net downward force is stronger than gravity alone. Once the tank fills, that valve is held firmly in place by the pressure differential until the next time the toilet valve is toggled. There is no buoyant force to lift the flapper valve off the seat until then.I'm doing the same thing here with water in the trays that hold the flasks. The water keeps air from getting under the flask and thereby stops atmospheric buoyancy on the flasks in its tracks.

I think you are right about the buoyancy, but am still scratching my head about the value of 4 centigrams. Though I understand now where it comes from.

3.35E-05	m^3 volume of a ping pong ball	
12		N/m^3 of air	
0.000402	Newtons	
102		grams/N	
0.041004	grams of air

I have to go do some shopping. I will pick up some ping pong balls and do the experiment. I will also apparently have to find a couple of dead flies, as wikipedia says that's how many it takes to make 4 centigrams.

The things I do for science...

 

[Creator]

Well, here you go:

https://www.youtube.com/watch?v=7ADBL7_A9qA

Hi Micromass;
Very interesting use of minimal lab equipment...
But was that last comment at the end of the video...something about Buoyancy "doesn't make sense because it gets canceled out by Newton's 3rd law ?? Can you clarify what you meant?

...

 

[A.T.]

You have two identical steel balls hanging on opposite sides of a scale like this:

You have two buckets, one with water and one with glycerin, standing on opposite sides of a scale like this:

Both scales are initially balanced. Then you fully submerge the balls into the buckets without touching the walls.

Does the balance of the scales change? If yes, how?

Spoiler

Then I think the glycerin one will weight more as glycerin is denser than water

But both scales are initially in balance, so the weight of both fluids is the same.

 

[sophiecentaur]

Think of how your toilet works. Toggling the handle lifts the flapper valve off its seat. The flapper valve is less dense than water, so it remains buoyed off the seat until the water level drops below the valve. The valve settles back in place at this point, sealing the tank. The tank starts to fill with water. The pressure from the water seals the valve even more firmly in place.

The flapper valve is less dense than water. So why doesn't the water buoy the valve up, making the toilet run and run and run? The answer is that by sealing itself in place, the flapper valve stops buoyancy in its tracks. Once sealed, the forces on the flapper valve are the weight of the valve, the water pressure from above, and the air pressure from below. The net downward force is stronger than gravity alone. Once the tank fills, that valve is held firmly in place by the pressure differential until the next time the toilet valve is toggled. There is no buoyant force to lift the flapper valve off the seat until then.I'm doing the same thing here with water in the trays that hold the flasks. The water keeps air from getting under the flask and thereby stops atmospheric buoyancy on the flasks in its tracks.

Why should the flapper valve be lighter than water? It could surely be held up by the flow of water through it. I do not know the exact geometry of this because I live in the UK and the classic UK system is different, using a syphon system. However, water flowing up through or down onto this flapper valve (a dynamic thing) can easily provide enough force to counteract with a downwards weight or upwards buoyancy. The dynamic toilet argument can't apply to the ping pong balls because theirs is a static situation.

Any water in the trays will provide its own upthrust. Different from without the water but irrelevant. No amount of upthrust going on in either tray can affect the total weight force on the balance arm. That would require antigrav!

 

[sophiecentaur]

I think people are losing the point in this thread and allowing intuition to take over. If you have the same total masses on each scale pan then the pan will balance, however you arrange them in isolation. (Ball sitting beside the beaker or ball at the bottom of the water in the beaker). If you interact with one side by pushing a ball under the water or partially supporting it (depending on the density of the ball), using external forces, you cannot expect the scales to balance.

 

[russ_watters]

That analysis is not quite correct. There is a difference between a floating ping pong ball and a submerged one. The submerged ping pong ball system includes the mass of the air inside the ping pong ball. The floating ping pong ball, most of that mass is buoyed by the air. The difference is about 4.3 milligrams.

I don't agree either:
If the volume of air displaced by both systems is the same, the buoyant force provided by the air must be the same. In both cases, you have, sitting on the scale, a mass of air (in the ping pong ball), a mass of ping pong ball and a mass of water. And it is buoyed by displacing a volume of air equal to the volume of water and volume of the ping pong ball in both cases.

Whether the ping pong ball is sitting on top of the water or in the water, the volume of ping pong ball + water is the same.

Or, looking at it another way, the tension on the string is internal to the beaker and so it can't affect the force applied to the balance.

 

[rcgldr]

Buoyancy gets canceled out by Newton's 3rd law?

Nothing is accelerating so all forces are Newton third law pairs with no net force on any object. For the left cup, the tape exerts a downwards force onto the ball, and the ball exerts an opposing upwards force on the tape (the source of the upwards force is boyant force, compressing the ball). The tape exerts an upwards force onto the cup, and the cups exerts a downwards onto the tape. The water exerts an upwards buoyant force onto the ball, and the ball exerts a downwards onto the water (the source of the downwards force is the tape). The cup exerts a downwards force onto the scale, the scale exerts an upwards force onto the cup. Gravity is an attractive force between cup and Earth (the Earth towards the cup, the cup towards the earth).

For the right cup, everything is about the same, except that the wire exerts an upwards force on the steel ball, and the steel ball exerts a downwards force onto the wire.

If both balls are the same size, then the buoyant force on each ball is the same, but buoyant forces are not "canceled out" by Newtons 3rd law, instead the upwards buoyant force exerted by the water on the ping pong ball is opposed by the downwards force exerted by the tape (and gravity for a tiny part of the downwards force). For the steel ball, the buoyant force is opposed by (gravity (weight of the steel ball) - the tension in the wire).

Spoiler

The weight of the left cup equals the sum of the weight of the cup + water + ping pong ball (.0027 kg x g (9.80665 m/s^2) = .02648 Newtons) + tape. The weight of the right cup equals the sum of the weight of the cup + water + what would be the weight of the steel ball if the steel ball had the same density as water. Assuming the steel ball is the same size as a ping pong ball, 2 cm radius, then the ball's volume is 33.5 cm^3, and the net downforce exerted by the ball equals the volume of the ball times the density of water (1 g / cm) = .0335 kg x g = .3285 Newtons.

 

[Creator]

The weight of the water for both cups is the same, but the tape exerts an internal upwards force on the cup, reducing the weight of that cup, .

OK; I think you explained what he meant by canceling Newton's force...However, if your statement above is correct ...that the weight of the cup on the left was "REDUCED" (due to the buoyant force pulling up on it),... how come the cup (on the left) didn't move upward BEFORE the steel ball was placed into the cup on the right?
IOWs, If the weight of the left cup was really reduced, it should have arisen (before touching the right cup) because the upward force on the left (according to you) would be equal to the water displaced by the ping pong ball (minus the minor weight of the PP balls). Right?
...

 

[rcgldr]

If your statement above is correct ...that the weight of the cup on the left was "REDUCED" (due to the buoyant force pulling up on it).

I guess there's no need for spoilers now. I corrected my previous post.

To explain the issue with the upwards force exerted by the tape on to the cup, assume the cup is a cylinder (not tapered) and note that the downforce exerted by the water onto the cup equals the pressure at the bottom of the cup times the area of the bottom of the cup. This downforce is related to the height of the water in the cup, not the weight of the water (there is about 600 grams of water). The 2.7 gram ping pong ball has a radius of 2 cm and displaces 33.5 cm^3 of water, or about 33.5 grams of water, so the downforce exerted by the water onto the cup is increased by the equivalent of 33.5 grams of water for a total of 633.5 grams due to the submerged ping pong ball's displacement of water. If an external force was used to submerge the ping pong ball (like a rod with a cupped surface at the bottom), then the weight of the left cup would increase by that equivalent of 33.5 grams of water. However, the ping pong ball is being kept submerged by an internal force pair, downwards on the ball, upwards on the cup, so the net result is a downwards force related to 602.7 grams plus the weight of the tape or whatever keeps the ping pong ball submerged. If including the effect of gravity as part of the system, it's a closed system with no external forces, and the weight of the system equals the sum of the weights of the system's components.

Note that if an external force is used to submerge the ping pong ball or to prevent a steel ball of equal size from sinking, the net downforce on both cups will be 633.5 grams (force) or about 6.21 Newtons. The external force takes care of any residual force not related to the 33.5 grams of displaced water.

A steel ball with a 2 cm radius would have a mass about 268 grams. If the steel ball was supported by an internal force, such as resting on the bottom of the cup, then there are no external forces other than gravity, so again a closed system, and the downforce onto the cup would be 868 grams.

 

[derek10]

But both scales are initially in balance, so the weight of both fluids is the same.

Spoiler

Yes I know, but as glycerin is denser, (I think) that means that a hung submerged ball will exert more force (weight) than the water submerged one. am I correct (probabily no)

 

[rcgldr]

Yes I know, but as glycerol is denser, (I think) that means that a submerged ball will exert more force (weight) than the water submerged one.

True. Again the main issue is the only external force on the left cup is gravity, while the right cup includes a support for the ball. If the left cup also used an external support to keep the ping pong ball submerged as opposed to an internal force to keep the ball submerged (downforce on the ball, upforce on the cup), then the scale would be balanced if the balls were the same size, regardless of the density of the balls.

If there are no external forces other than gravity involved, then the weight will equal the weight of all the components, cup, water, ball, and whatever holds the ball in place.

If there is an external force holding a submerged ball in place, then the weight on the scale will be the weight of the cup + weight of water + weight of water displaced by the ball. The external force will oppose the weight of the ball and the buoyant force.

 

[sophiecentaur]

I don't agree either:
If the volume of air displaced by both systems is the same, the buoyant force provided by the air must be the same. In both cases, you have, sitting on the scale, a mass of air (in the ping pong ball), a mass of ping pong ball and a mass of water. And it is buoyed by displacing a volume of air equal to the volume of water and volume of the ping pong ball in both cases.

Whether the ping pong ball is sitting on top of the water or in the water, the volume of ping pong ball + water is the same.

Or, looking at it another way, the tension on the string is internal to the beaker and so it can't affect the force applied to the balance.

That sums it up nicely. Just one point:
With a deeeeep jar of water, the hydrostatic pressure would be significantly higher and could reduce the volume of the ball and that would reduce the mean density.

 

[Dadface]

Because each ball is held in equilibrium the effective density of each ball,in terms of forces, is the same as the density of the surrounding water. Immersing the balls is equivelent to adding water of volume equal to that of each ball. By Archimedes principle the additional force measured due to the prescence of each ball is the same on both sides and equal to the upthrust.
I think Aleph zero gave the best explanation in post 6.

 

[A.T.]

Spoiler

Yes I know, but as glycerin is denser, (I think) that means that a hung submerged ball will exert more force (weight) than the water submerged one. am I correct (probabily no)

And what about the upper scale, on which the two identical balls hang?

 

[rcgldr]

Because each ball is held in equilibrium the effective density of each ball,in terms of forces, is the same as the density of the surrounding water. Immersing the balls is equivelent to adding water of volume equal to that of each ball. By Archimedes principle the additional force measured due to the prescence of each ball is the same on both sides and equal to the upthrust. I think Aleph zero gave the best explanation in post 6.

The difference is that the ball in the left cup is held in place via an internal force, while the ball in right cup is held in place via an external force.

For the left cup, the buoyant force minus the weight of the 2.7 gram ping pong ball is opposed by internal forces that exert a downward force on the ball and an upward force on the cup. It's a closed system where the only external force is gravity, and the weight of that system is weight of the cup, water, and ball (ignoring whatever is used to hold the ball in place).

For the right cup, the weight of the steel ball minus the buoyant force is opposed by an external force (the wire from above), and there's no internal force that exerts an upward force on the cup. It's an open system, and the weight of the system is the weight of the cup, water, and the weight of water displaced by the ball (ignoring whatever is used to hold the ball in place), and assuming the steel ball is the same size as the 2 cm radius ping pong ball, the right cup system weighs 33.5 grams - 2.7 grams more than the left cup. The right cup system would also be heavier by the same amount if a ping pong ball was held submerged in the water by a rod from above (again an external force).

 

[OmCheeto]

...

I have to go do some shopping. I will pick up some ping pong balls and do the experiment. I will also apparently have to find a couple of dead flies, as wikipedia says that's how many it takes to make 4 centigrams.

The things I do for science...

I did go shopping, but I did not buy the ping pong balls...

Staring at the package of 6 balls, selling for $3.49, I decided that I did not have the facilities to measure the change in either weight nor ball float level.

About an hour ago, I put a dead bathroom vanity light bulb into my vacuum chamber, and could discern no change in eyeballed float level.

My decision to not buy the balls, was vindicated.

ps. I'm surprised ZapperZ hasn't shown up and yelled at us about forgetting about the Casimir effect.

pps. I would do the following experiment : Consider a Ping-Pong ban(sic) floating in a glass of water...

but I'm sure Borek would show up and start talking about partial pressures, and how the increased air pressure would push air molecules into solution, raising the mass of the system...

 

[Orodruin]

And what about the upper scale, on which the two identical balls hang?

I want to play too! :)

Spoiler

Since glycerol has a larger specific gravity, it will result in a larger buoyancy for the ball, so less weight needs to be supported by the string tension. The glycerin side of the upper scale will go up.

Given one of the solutions the other is given by Newton's 3rd law.

 

[Dadface]

The difference is that the ball in the left cup is held in place via an internal force, while the ball in right cup is held in place via an external force.

For the left cup, the buoyant force minus the weight of the 2.7 gram ping pong ball is opposed by internal forces that exert a downward force on the ball and an upward force on the cup. It's a closed system where the only external force is gravity, and the weight of that system is weight of the cup, water, and ball (ignoring whatever is used to hold the ball in place).

For the right cup, the weight of the steel ball minus the buoyant force is opposed by an external force (the wire from above), and there's no internal force that exerts an upward force on the cup. It's an open system, and the weight of the system is the weight of the cup, water, and the weight of water displaced by the ball (ignoring whatever is used to hold the ball in place), and assuming the steel ball is the same size as the 2 cm radius ping pong ball, the right cup system weighs 33.5 grams - 2.7 grams more than the left cup. The right cup system would also be heavier by the same amount if a ping pong ball was held submerged in the water by a rod from above (again an external force).

The increase in force on the right hand side is equal to the weight of water displaced. It seems that we agree on that.

The increase in force on the left hand side depends on how the string is connected. If the string passes through a hole in the bottom of the cup and is held in position externally the increase in force on the left hand side will be the same as that on the right hand side.

If the string is held in place by connecting it to the bottom of the cup there will be an upward force on the cup equal to the tension in the string. In this case the increase in force on the left hand side will be less than that on the right hand side by an amount equal to the weight of the ball.

(I imagined it as the ping pong ball being a bit like a hot air balloon. It's trying to float and lift itself and the cup)
I need to check this. Simplifying assumptions apply.

 

[Doc Al]

I assumed that the string attached to the ping pong ball was also attached to the bottom of the beaker.

Simplify everything. Take the ping pong ball as empty and weightless, and having the same volume as the steel ball. Then the "answer" will be more apparent.

 

[gmax137]

Trying to get clear on what's happening with the ping pong ball. If the string is holding it down, then the string is pulling up on the beaker. But, is it like compressing a spring, and the bottom end of the spring bears against the bottom of the beaker? So the beaker sees the string pulling up, but the spring is pushing down, so there's no net force on the beaker? I don't quite see that as being right.

Is this like the truckload of birds thing, where the driver thinks if he gets them to fly around inside, he can make it over the bridge.

Sorry for the crummy drawing.

 

[Doc Al]

Trying to get clear on what's happening with the ping pong ball. If the string is holding it down, then the string is pulling up on the beaker. But, is it like compressing a spring, and the bottom end of the spring bears against the bottom of the beaker? So the beaker sees the string pulling up, but the spring is pushing down, so there's no net force on the beaker? I don't quite see that as being right.

You can imagine the string being replaced by a massless spring.

First start out with just the beaker of water. That water presses down on the bottom of the beaker with some force equal to pressure x area.

Now submerge the ping pong ball under the water, attaching it to the bottom via that spring. The water level rises, increasing the water pressure on the bottom of the beaker. The buoyant force pushes up on the ping pong ball, stretching the spring, which pulls up on the bottom of the beaker. The upward pull of the spring (or string) exactly balances the additional downward force due the increased depth of water.

 

[D H]

I don't agree either:
If the volume of air displaced by both systems is the same, the buoyant force provided by the air must be the same. In both cases, you have, sitting on the scale, a mass of air (in the ping pong ball), a mass of ping pong ball and a mass of water. And it is buoyed by displacing a volume of air equal to the volume of water and volume of the ping pong ball in both cases.

I think you are wrong with regard to buoyancy. Buoyancy is not a fundamental force. It's a consequence of a pressure gradient (and pressure in turn is not a fundamental force). It can be defeated.


However, that's not the question here. The question is, to paraphrase a year old song, "What does the scale say?" In that regard, I'm most definitely wrong. I'll look at two extremes, a scale whose trays have tiny little bumps of negligible area that raise the flask a tiny bit above the tray, versus a perfectly flat tray combined with a flask with a perfectly flat bottom so no air can get underneath and cause buoyancy.

I'll assume a flask with thin vertical walls and a thin bottom. In the first case, the scale will detect the atmospheric pressure at the bottom of the flask plus the apparent weight of the flask, including buoyancy: $W_{\text{tot}} = AP_{\text{bottom}} + (mg - A(P_{\text{bottom}} - P_{\text{top}}))$, where $A$ is the area of the interior of the flask, and $P_{\text{bottom}}$ and $P_{\text{top}}$ are the atmospheric pressure at the bottom and top of the flask. After canceling common terms, this reduces to $W_{\text{tot}} = mg + AP_{\text{top}}$. In the second case, the scale will detect the weight of the contents of the flask plus the weight of atmosphere acting on the top of the flask: $W_{\text{tot}} = mg + AP_{\text{top}}$. That's the same result as the first case! Whether or not buoyancy acts on the flask and it's contents is irrelevant. What is relevant is the mass of the contents of the flask and the pressure at the top.

This contribution from atmospheric pressure is important in comparing what a scale can sense with regard to an intact versus crushed ping pong ball. The top of the liquid will be slightly higher in the case of the intact ping pong ball suspended within the water compared to a ping pong ball floating atop an equal amount of water. So what does the scale say?

The trivial explanation is to assume that density and gravity don't change that much over the sub-centimeter level. With this assumption, the hydrostatic equilibrium condition, $\frac{dp}{dz}=-\rho g$, says that pressure decreases linearly with increased height. The non-trivial explanation, which involves the lapse rate for a non-ideal gas, simplifies to the trivial explanation in the case of centimeter-level changes. (I've wasted too much time playing with the math.) This pressure change is important.

Bottom line: If the pressure inside the ping pong ball is close to local atmospheric pressure, there is essentially no difference in the weight sensed by the scale between a submerged versus floating ping pong ball.

 

[Vanadium 50]

You can imagine the string being replaced by a massless spring.

I'd go further. I'd replace the string with a post. And then the post with a tube. And then a tube that is open at both ends. Then replace the whole system with glass, so you have a funny shaped beaker. And then with an ordinary beaker with the same volume of water.

At each step you have the same weight.

 

[phinds]

I'd go further. I'd replace the string with a post. And then the post with a tube. And then a tube that is open at both ends. Then replace the whole system with glass, so you have a funny shaped beaker. And then with an ordinary beaker with the same volume of water.

At each step you have the same weight.

Only If water and glass weigh the same, which they likely don't.