Physics Riddle: The Fate of the Balance

[micromass]

What will happen to the balance:

 

[Nathanael]

I assume the balls are equal volume?

Spoiler

I think the balance will tip to the right, only because the string on ping pong ball is pulling up on the container. If the ping pong ball were held in place (so it didn't float up) from the outside (like on the right side) then my answer would be that the scale will not move.

My reasoning is that the difference in weight (between the balls) is irrelevant because the excess weight of the steel ball is counteracted by the string holding it up. Both of the balls only push down on the water with the bouyant force proportional to their volume, so their mass is irrelevant.

But since there is a force of tension upwards on the left container, it must push down on the scale less, thus the scale tips to the right.

 

[Doc Al]

Ain't saying.

(Good one, micro! )

 

[Vanadium 50]

What will happen

The steel ball will rust.

 

[TurtleMeister]

I'll give it a try. I'm not certain of my answer though. Assuming we're supposed to say what will happen after insertion of the balls: The scale will tilt down on the side of the ping pong ball. (I haven't read the spoiler)

 

[AlephZero]

Spoiler

The weight of both containers is the same. The depth of fluid in both containers is the same, so the hydrostatic pressure acting on both scale pans is the same. The only difference is the upwards force from the string on the left hand pan.

 

[TurtleMeister]

I forgot to give the reason for my answer.

Spoiler

The balance scale measures a difference in mass. After insertion of the balls, the mass on the left plate has changed but the mass on the right plate has not changed (the steel ball is supported by the external hanger). Buoyancy is irrelevant because it's the same for both.

 

[ZetaOfThree]

Spoiler

At first I thought the scale tips to the left, but now I think it tips to the right. Assume the containers have the same mass so forget about them. Also assume the balls have the same volume. The plate on the left feels the weight of the ping pong ball plus the water. The plate on the right feels the weight of the water plus the buoyant force from the steel ball (the rest of the force is taken care of by the string so that it stays put in the water). The buoyant force from the steel ball is the weight of a ball of the same volume as the steel ball, but made of water. So since a ball made of water is heavier than the ping pong ball, the force on the right plate is greater, and the scale tips to the right.

 

[D H]

Way to go Zeta!

There apparently are a lot of people here who think they can pick themselves up by their bootstraps.

 

[davenn]

Way to go Zeta!

There apparently are a lot of people here who think they can pick themselves up by their bootstraps.

LOL ... I would have been another

instead I second the vote for V50's reply haha

Dave

 

[Vanadium 50]

Good job, Zeta!

 

[rcgldr]

Spoiler

It might have been interesting to also include a cup filled with water and no ball in it, then ask to compare the weights of all 3 cups, but that might have given away the answer. It could also be noted that the cup with the ping pong ball has lower average density, and no external forces (other than gravity) are exerted onto the ping pong ball.

 

[BobKat]

It boils down to this question: Does the upward force caused by the air inside the ping pong ball counterbalance the added weight of the ping pong ball itself and the string which holds it. Both balls displace the same amount of water (presumably) and the ball on the right adds no weight to the right side. I suspect the scale stays even, because the air lifts the ball and string - BUT the string is attached to the bottom of the glass, so I don't think it can lift the glass by its own bootstrap (string).

 

[BobKat]

Guess I don't understand how the steel ball contributes the weight of its volume in water to the weight on the right...

 

[OmCheeto]

It boils down to this question: Does the upward force caused by the air inside the ping pong ball counterbalance the added weight of the ping pong ball itself and the string which holds it. Both balls displace the same amount of water (presumably) and the ball on the right adds no weight to the right side. I suspect the scale stays even, because the air lifts the ball and string - BUT the string is attached to the bottom of the glass, so I don't think it can lift the glass by its own bootstrap (string).

I just did the experiment. Zeta is correct. Although, I replaced the two balls with a plastic tool pick container, and the string with a rubber band. 9 oz vs 12 oz.

Buoyancy problems always have me scratching my head, and mopping up in the kitchen afterwards.

 

[OmCheeto]

Guess I don't understand how the steel ball contributes the weight of its volume in water to the weight on the right...

Replace the steel ball on the right, with a ping pong ball. Instead of a string, smoosh the ping pong ball down with your fingers. I actually did this, but my ping pong ball had too little volume to indicate a weight change on my cheesy little 16 ounce dollar store diet scale. Hence my toothpick container upgrade.

 

[rcgldr]

Guess I don't understand how the steel ball contributes the weight of its volume in water to the weight on the right...

Spoiler

Gravity exerts a downwards force onto the steel ball equal to the steel balls weight, which in turn exerts some of it's weight onto the water, and some onto the wire connected to the steel ball. This force is opposed by an upwards buoyant force exerted by what would the the weight of the water displaced by the steel ball. The remaining component of downwards force is opposed by the external (to the cup) upwards force from the wire connected to the steel ball. So the net downwards force exerted by the steel ball onto the water is equal in magnitude to the upwards buoyant force (Newton third law pair of forces), which is equal to what would be the weight of water displaced by the steel ball.

 

[BruceW]

you can think of it like a variation of energy problem, too.

Spoiler

If the right side moves down by $\delta y$ (i.e. positive $\delta y$ means right side goes down), then the change in GPE on the right is equal to $- (V_{water} + V_{ball}) \rho_{water} g \delta y$ This is because all the water moves down by $\delta y$ plus some small amount to account for the fact that the steel ball has moved upwards relative to it. The GPE of the steel ball itself does not change. Also, the left side will move upwards, so the change in GPE on the left side is: $V_{water} \rho_{water} g \delta y + \rho_{pingpong} V_{ball} g \delta y$ This is simply because everything on the left side moves upwards by $\delta y$ (including the ping pong ball). So, the total change in GPE is $(\rho_{pingpong}-\rho_{water}) V_{ball} g \delta y$ Which is negative, since the density of the ping pong ball is less than that of water, so in conclusion, if the right side moves down and left moves up, the total GPE will decrease, so that is what will happen.

 

[Doc Al]

For those still struggling:

Spoiler

If the string pulling up on the steel ball supported the full weight of the ball, then both sides would be balanced. But it doesn't. (The buoyant force partially supports the ball.) Thus the right side is heavier.

 

[chingel]

Spoiler

I think it is easiest to think of it in terms of external forces on the containers, since their weights are the same. The pressure of the water depends only the the height of the water level, which is the same for both of them, but the left container has an additional string pulling it up.
I think we have to be careful not to say that the steel ball's weight somehow channels through the water, because the water pressure is the same in both containers.

 

[Gerinski]

Spoiler

I think it is easiest to think of it in terms of external forces on the containers, since their weights are the same. The pressure of the water depends only the the height of the water level, which is the same for both of them, but the left container has an additional string pulling it up.
I think we have to be careful not to say that the steel ball's weight somehow channels through the water, because the water pressure is the same in both containers.

"easiest to think of it..."? I did not understand one word of what you said. And their weights (including the balls in them) are not the same.

 

[chingel]

"easiest to think of it..."? I did not understand one word of what you said. And their weights (including the balls in them) are not the same.

View the container itself as a system, viewing everything inside and outside of it as external. The containers themselves are the same, they weigh the same. Water pressure on both of the containers is exactly the same everywhere, but the left container has an extra string attached to it pulling it up, therefore it will go up.

 

[BruceW]

that is a nice way to think about it. Maybe not very intuitive though.

edit: It is probably the most inventive, of the ones I've seen so far :)

 

[derek10]

I still don't get it:

Spoiler

Why does the steel ball have buoyancy and the ping pong one doesn't? is it because the steel one is more massive than the other?

(BTW I though than the balance would lean to the left due to the extra weight of the ping pong ball)

 

[Doc Al]

I still don't get it:

Spoiler

Why does the steel ball have buoyancy and the ping pong one doesn't? is it because the steel one is more massive than the other?

(BTW I though than the balance would lean to the left due to the extra weight of the ping pong ball)

The buoyant force on each ball is the same. Buoyant force depends only on the volume of fluid displaced, not the mass of the ball itself.

Extra weight of the ping pong ball?

 

[derek10]

The buoyant force on each ball is the same. Buoyant force depends only on the volume of fluid displaced, not the mass of the ball itself.

Extra weight of the ping pong ball?

Spoiler

Yeah because if I interpreted the OP diagram properly it is tied into the container right? so it has a extra weight exerted into the container even if it's floating while the steel one is hanged so it doesn't weight anymore (only displaces water?)

What I don't understand is why does the steel ball exert a weight equivalent to a water ball of the same size but the ping pong doesn't (per Zeta's explanation, if I understood it well)

 

[A.T.]

Water pressure on both of the containers is exactly the same everywhere, but the left container has an extra string attached to it pulling it up, therefore it will go up.

That was my first idea too. And it seems like the simplest argument so far to me.

 

[A.T.]

Here is a similar one:

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?

 

[derek10]

Here is a similar one:

Spoiler

No it won't change, both sides of the scale have the same mass/weight

Thats the first thing I thought so I'm pretty sure I'm wrong

 

[Gerinski]

View the container itself as a system, viewing everything inside and outside of it as external. The containers themselves are the same, they weigh the same. Water pressure on both of the containers is exactly the same everywhere, but the left container has an extra string attached to it pulling it up, therefore it will go up.

I don't thing that's correct, the string in the left container is not pulling it up, as someone said "nobody can lift himself by pulling his bootstraps".

I see the buoyancy argument but as Derek I do not see why only the right side should have the weight of the steel ball's buoyancy and not the left side the same buoyancy by the volume of water displaced by the ping-pong ball.

 

[Doc Al]

I see the buoyancy argument but as Derek I do not see why only the right side should have the weight of the steel ball's buoyancy and not the left side the same buoyancy by the volume of water displaced by the ping-pong ball.

As both balls have the same volume and are completely submerged, they both have the same upward buoyant force.

 

[Doc Al]

Spoiler

No it won't change, both sides of the scale have the same mass/weight

Thats the first thing I thought so I'm pretty sure I'm wrong

The balls are held in some fully submerged position, not just dropped and resting on the bottom. So other forces are involved.

 

[derek10]

The balls are held in some fully submerged position, not just dropped and resting on the bottom. So other forces are involved.

Ok thanks I haven't read the "without touching the walls" part properly so I suppose they are hung too :)

Spoiler

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

Probably wrong too, but I often think too quickly in these cases.

 

[chingel]

I don't thing that's correct, the string in the left container is not pulling it up, as someone said "nobody can lift himself by pulling his bootstraps".

I see the buoyancy argument but as Derek I do not see why only the right side should have the weight of the steel ball's buoyancy and not the left side the same buoyancy by the volume of water displaced by the ping-pong ball.

If we cut the string on the left container and let the ball float, the whole container doesn't get any heavier and so you can say in that sense that it can't pull itself up with the string. However, this is not the situation the original problem compares it with, because the other container has the same water level, but no string attached to the container.

Consider only the container, it has water pressure pushing it down and a string pulling it up, it doesn't care where is the water pressure or the string coming from, or what is pulling the string.

Regarding the other one with glycerin, using the same analysis, only the liquid pressure is acting on the containers, the change of height of the liquid level is the same for both of them, so with the denser liquid the pressure on the bottom will increase more, making it go down.

 

[Gerinski]

As both balls have the same volume and are completely submerged, they both have the same upward buoyant force.

That's weird to hear from you as one of your previous posts says:

"If the string pulling up on the steel ball supported the full weight of the ball, then both sides would be balanced. But it doesn't. (The buoyant force partially supports the ball.) Thus the right side is heavier."

 

[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!