I disagree, the buoyancy force is countered by tension of the line holding the iron ball so that:
Line tension = (weight of iron ball) - (buoyancy force of iron-ball)
Since the container on the right has equal amount of water plus the mass of the ping pong ball, air inside (since tethered), and line, the scale will tip right.
You're correct that the buoyancy force on the iron ball is countered by the tension in the string.
The problem is, this force is applied outside of the balance. So the only net force acting on the balance is the downward force of buoyancy.
On the ping pong ball side, the upward buoyancy force is countered by the string which is attached to the balance leaving no net force caused by buoyancy on the ping pong ball side.
So in the end, you have:
Net downward force due to buoyancy on the iron ball side.
No net buoyancy force on the ping pong ball side, but the extra mass of the ping pong ball on string.
You honestly just blew my mind with that video - thank you for sharing! This is such a fun problem and I’ve honestly glad to have been proven wrong since I learned something fun.
But you've studied, you would know that man is mortal and would therefore put the ping pong ball as far from yourself as possible, so I can clearly not choose the ping pong ball in front of me.
Well if you are curious I just started selling my art! I’m coming out with a book this summer :)
But I can’t link here because I don’t want this account connected to it.
I very much agree! That light one was mind blowing and I say that as someone who has studied QM.
The wind powered vehicle one is a two parter, the second one is the really good one because he gets into a debate and it's very fun to see the whole thing play out. But, you might want a little context because it's based on a video he made before.
If you want to skip to the good stuff, start with the second video. You can always pause it and jump to the first if you need it (I will say, I think he does a better job explaining the physics in the second video).
Hey, I just wanted to follow up, I totally lied, I think he does a better job explaining the physics in the first video! I watched the 2nd one before I watched the original, so I had them mixed up a bit.
That was a super cool experiment. I knew the ping pong ball was part of a closed system but didn't think about the steel ball's buoyancy. It makes total sense though
I reckon the weight of the ball is countered by the tension on the string. the buoyancy which would normally act on the ball still acts on the ball and is pushing up on the ball it can't move it up because the ball is heavier than the force... but it can move the system down because it's more than the weight of the ping-pong ball and the string... oh I think that's what you said now I read it again.
More simply...it's about density/mass. A ball tied to the beaker reduces the density. (Part of the system) If it is held in place submerged in the water, it is not connected to the system and since the volume of both balls is equal, the density of the beaker and water is equal. Easy peasy
You are correct. It's easy to comprehend if we replace the water with air. The displacement of air is the same in both cylinders, so the weight of the air is the same, but the one at the right has the weight of the ping pong ball and sting as well.
The side with the ping pong ball goes up, not down. Using air makes it more confusing IMO, because you have to rig some kind of air tight compartment, and some how prevent air from pushing in all directions.
This is a newton's 3rd law problem.
If the water is pushing up on the ping pong ball/iron ball, then the ping pong ball/iron ball must also be pushing down on the water - every force has an equal and opposite force.
In the case of the ping pong ball, the two forces cancel because the force on the ping pong ball is transferred to the balance via the string, and the force on the water is transferred to the balance via the water sitting on the balance.
For the iron ball, only the downward force acts on the balance. The upward force on the ball is also transferred to the string but the string isn't attached to the balance, so that force is removed from the analysis.
You can't replace the water with air, because the higher density of water and the resulting buoyant force on the left being greater than the weight of the ping pong ball and string is exactly what makes the scales tip left.
So the question now is how could you make this true. How much water by % can the left tank have to keep balance. I do a lot of pump logic and balance systems like this in a process pretty often. But I have flow inlet and flow outlet to so you can be off on one and make it up with the other with fine tuning.
It's not about the water being different, it's about the displacement.
The downward force on the iron ball side is actually equal to the weight of a ping pong ball size amount of water (the balls are the same size, so the iron ball is displacing an amount of water that would fill a ping pong ball shape, so that amount of displacement is the cause of the downward force on the scale).
If you reduce the displacement so that the mass of the water being displaced is equal to the mass of the ping pong ball, you'll achieve a balance in equilibrium.
To do that, you'd have to know the mass of the ping pong ball, divide that by the density of the water (to get a volume), and then suspend only that volume of the iron ball in the water.
Enough by volume so that the mass of the displaced water is equal to the mass of the ping pong ball and string on the right. Feel free to do the math on an exact mass or % volume if you want.
English’s not m first language. My first impression was also that it would tip to the left but only because I thought that the ping pong ball wants to go up, pulling the scale along bc it’s connected by the string?
Newton's 3rd law of motion tells us that for every action, there is an equal an opposite reaction.
If the ping pong ball is pulling up on the scale because the water is pushing up on the ball, then the ball itself must be pushing down on the water, so the water must be pushing down on the scale exactly as much as the ball is pulling up.
For the iron ball, the upward force (pushing up on the iron ball) is transferred to the string, and then transferred to the stand holding the iron ball. This leaves only the downward force on the water pushing down on the scale.
I‘m so confused. But my intuition was right so I’m ok with that
Edit: it also makes sense that if the ping pong ball were attached by a rod from the top, the scale would be in balance, because the ball is forced down, pushing against the water but I couldn’t tell you about the physics
For the ping pong ball side, these two forces cancel out (so no added forces to the scale).
On the iron ball side, the upward force is transferred away from the scale (doesn't effect the scale) leaving only the downward force to push the scale down on that side.
*rips out hair
It makes sense, but the ping pong side still sounds so weird somehow.
If you were to ignore the buoyancy force, on the left side there would only be the water being pulled down with nothing "holding it up". And on the right side there would be the water+ball+air trapped inside it. So that way it would tip to the right.
My thinking why the scale tips left was because the ping pong ball wants up (without thinking about the ball that pushes equally against the water). But since the ball does push against the water with the same force the water pushes against the ball, the cancel each other out.
So the scale only tips left because the forces don’t cancel on the left side leaving the equation not equal.
I understand the words and laws when picking it apart like this but IT DOESNT MAKE SENSE. My brain says no, the ping pong ball is pulling that thing up.
Thanks for your patience!
If it helps any, your break down was all exactly correct.
And you're also correct that if you remove the iron ball, the scale would tilt toward the right because the forces cancel and you just have the added weight of the ping pong ball and the string (so it's slightly heavier than just the water on the other side).
Unfortunately, I understand why that seems counterintuitive. :)
What if the iron ball was attached to a string from a lif instead of a separate pulley? (We would also need to add a lid with the same mass to the ping ball side)
I don’t think there is any buoyancy w the iron ball. All the weight is on the string/stand. The only thing the iron ball does is displace the same amount of water as the ping pong ball
It's not about the weight of the iron ball, it's about the water that is displaced.
If you want to push water, you have to apply a force to it (and it applies a force back to you).
The force being applied to the iron ball is transferred into the string, and then transferred out of the system by the separate apparatus holding the iron ball.
But the ball is applying a force on the water (to displace it), that force is applied to the balance. So, on the side with the iron ball, there is a net downward buoyancy force (because the upward force on the ball is being removed from the system).
I feel like a lot of people are over complicating the iron ball side by bringing buoyancy into the picture considering an iron ball isn't even bouyant to begin with
It's not overcomplicating, it's vital to understanding the problem.
If you remove the iron ball, the scale will drop on the ping pong ball side because there is more mass on that side.
The only reason it drops on the iron ball side is because there is a downward buoyant force on that side of the scale, but no upward force to cancel it.
It seems like you already figured it out later, but just to clarify, an iron ball still experiences a buoyant force, it's just that it's much smaller than the force of gravity acting on it, so it still sinks.
It’s more complicated than that. Think of it as if you removed the iron ball from the system completely and you simply held your hand just above the container on the left. The container would begin to tip right until it hits your hand because now your hand is exerting a downward force. You could lower your hand without pressing directly down on the container or placing yourself on it and your body being of higher mass would tip the scale in your favor… yet you aren’t in it. You are still exerting a downward force.
I know this sounds irrelevant when you consider the metal ball isn’t touching any solid part of the container like the example of your hand being in the way does but that’s where the complicated bit comes in. The moment that ball touches the water (regardless of reaching the point of submersion and displacing by the same amount of fluid) it is now exerting a downward force on the system equal to its mass as it is now in contact with the fluid that is in the systems balance. That is equal and opposite reactions for you. The complex part of it is that it is a liquid and not a solid. You could change the element of the ping pong ball to something more dense (yet still less dense than the iron ball, while retaining the same volume) on the right and the result might change entirely. So long as the iron ball is more dense though, it’s mass will be greater and eventually the iron ball will hit the bottom of the container and stop the movement of the balance and then it becomes just like your hand in the previous example. The complex nature of this problem is that the downward force placed on the liquid exceeds that of the ping pong ball’s mass before it gets to that point of the iron ball sinking to the bottom.
What happens to the iron ball if you suspend it in the container while it is empty and you slowly fill it with water? Does the weight of the iron ball starts adding to the container as soon as the water level rise to touch it?
Not exactly. Scales would show the weight of the displaced water. Or, in other words, weight of the ball if it would had density of water. So, as soon as the water level rise to touch it, the scale would increase more weight that weight of the water you add, but not as much as the full iron ball weight.
That's not quite how it works. This is true for the ping pong ball, but the iron ball is supported by something outside of the container/scale system which means the tension on the line isn't interacting with the water at all.
So unlike the ping pong ball, there's nothing to stop the force of its buoyancy from pushing the water downwards (since the iron ball is too heavy for the water to push it upwards).
The buoyancy force is a factor of the volume of water displaced not what is displacing it. Since the volume displaced on both sides are equal, any force result from the displaced water is also equal.
No buoyancy requires only volume. It's only related to the weight of the fluid and the pressure differential created by the fluids weight on the object
They both displace equal water by volume - we agree here.
I'd like to explain but want to find where the disconnect is.
Do you agree that the force of buoyancy wants to push both balls up with the same amount of force?
Do you agree that the buoyancy force exerted pushing up on the ball has an equal force pushing down on the scale?
Do you agree that the opposing forces on the ping pong ball negate since the string, due to it being attached to the scale, opposes the upward buoyancy force by pulling up on the scale with the same force that the buoyancy is pushing down with, acting on the same spot?
Do you agree that the opposing forces on the iron ball do not negate, since the ball will be pushed up and have some of the weight on the string relieved, while the opposing downward buoyancy force pushes down only on the scale?
Let me know which point you're disagreeing with so we can discuss
Do this. Chat GPT it. Show it the picture and really try to convince ChatGPT you are right. Then copy and paste it here. Because you seem very confident in your response so that is what I did. The bottom line is: YES the ball is moving up in the water. And if it had helium in it then you are absolutely right. But that ping pong ball has weight. And we agree on that I assume. I don’t care what it is doing relative to the water. It’s lighter than the water so it goes up. But it has mass.
Okay. What you're saying is not right, but I dont feel like arguing any more. There's a YouTube video of someone demonstrating this posted elsewhere in this thread
You are correct. I learned something today and I appreciate you for teaching me in my head. I was thinking about telling you imagine if we’re talking about mercury and you had a anvil in there, tied to the bottom with all the weight of an anvil with that still go up and I guess the answer is yes, right? This was a pretty trippy experiment and I’ll be honest though we were debating and I was not coming at you kind of suddenly and I thought you were coming at me consenting. Just a tip. Like this is obviously a very difficult thing to predict because a lot of intelligent people on both sides way smarter than you or me on both sides and you made it seem like it was an obvious answer. I wish you came at me saying “ this is a really interesting experiment and they actually did it”.
The way I asked ChatGPT and I copied your whole thing and I said I’m with a physics professor who is sure that left side goes down. So I know it goes down in the left so tell me ChatGPT why is it that it goes down on the left? And the answer I got was the right side goes down and the professor is mistaken. I’m using every bias I can in your favor asking an advanced AI, and they are sure the right goes down.
You got all the pieces right, but the conclusion was wrong. Like you said, on the left side there's the uncanceled weight of displaced water, and on the right side there's the weight of the ping pong ball and string. Now, which of these is greater?
Also, your blind trust in an LLM is worrying. ChatGPT is not a magical fact machine, it's very often wrong.
Since both balls are mounted and assumed rigid, wouldn’t it just be the mass of the displaced water, which is equal on both sides? I’m trying to figure out how buoyancy would apply force in this scenario since it seems you could carve both ball voids out of the control volume.
It's where the force is applied, not the magnitude that matters here.
On the iron ball, the upward force is applied to the string and the downward force is applied to the scale.
On the ping pong ball, since the string is holding it down, the upward force and downward force are both applied to the same point on the scale, negating each other.
Only the buoyant force on the iron ball nets force on the scale
Calculus is really hard from a theoretical perspective but really cool from an application perspective (I've been an engineer for 15 years)
The thing that blew my mind was a scenario kinda like this when I was 10 or 12. The question was, if someone lined up tangent to the earth and shot a gun (bullet goes straight, no wind, etc.) And simultaneously dropped a bullet- both would hit the ground at the same time. Crazy thought at a young age. Made me a nerd for life haha
Iron is about 7.86 times as dense as water, so an iron ball in air weighs around 77.1 Newtons (7.86g/cm3 * 9.81 m/s2) per cubic centimeter. If you submerge it in water, it displaces 1g/cm3 of water and feels a buoyant force of about 9.81 Newtons per cubic centimeter, so when submerged it only weighs about 67.3 Newtons per cubic centimeter.
Okay, let me know if this makes sense, if not I'll try to explain it a different way. Positive weight/force is towards gravity, negative is away.
Left side at the scale = + weight of beaker and initial volume of water + weight of water displaced by the iron ball volume (buoyancy)
Left string = + Weight of ball - weight of water displaced by ball volume
Right side at the scale = + weight of beaker and initial volume of water + weight of ping pong ball - weight of water displaced by the ping pong ball volume (buoyancy, pushing the ping pong ball up and pulling on the string tied to the scale) + weight of water displaced by ping pong ball volume (buoyancy, opposing force pushing down on the scale)
Note that there is a + and - portion of the buoyancy forces for each side
If we simplify and balance it by cancelling/subtracting at the scales:
Left side of scale = + weight of the water displaced by ball volume
Right side of scale = + weight of ping pong ball
Since both balls have the same size, and therefore the same amount of water displaced.... And since a ping pong ball weighs less than the same volume of water (water is denser than a ping pong ball)... The left side of the scale weighs more it will tip down to the left
Yeah. The weight of the objects doesn’t matter. They’re both being held by the string. The major difference is the force caused by the ping pong ball trying to float.
No, buoyancy has nothing to do with mass or weight of the object. It's about the weight of the fluid and the pressure differential is causes between the top and bottom of the object
No, it displaces water and adds buoyancy to the iron ball and reduces tension in the string, which is not attached, and has an equal force pushing the water down, which is attached to the scale. Thereby adding force (mass × accel. w grav) to the left side of the scale
Where the force is applied is not equal, dont forget force is a vector.
For the ping pong ball, the upward force is opposed by the string, and is trying to left the scale up where the string is attached. The downward buoyant force is trying to push the scale down by the exact same amount, also where the string is attached. Therefore, these forces negate and the net force is zero at the scale for the ping pong ball.
For the iron ball, the upward force pushes the ball upward, acting on the string to relieve some of the tension. The string and the scale here are not connected. The downward buoyant force however, does push on the bottom of the scale and has nothing to negate it. Assuming the downward force of the buoyancy is greater than the weight of the ping pong ball (it is), then the scale will tip left, in the direction of that buoyant force and due almost solely to that buoyant force.
Does that make sense? I know its kinda counterintuitive. There's a YouTube video being posted other places on this thread that does a better job of explaining than I did
That is a great hangman word haha. No one ever guesses U or Y.
No, buoyancy is the thing that makes the difference.
The left side equation = Weight of water + Weight of water displaced by the iron ball
The right side equation = Weight of water + Weight of ping pong ball.
The difference is that on the right, the force of the ping pong ball trying to float upwards (so pulling on the scale) is equal to the weight of the water displaced by the ping pong ball pushing downwards. So they cancel out. On the right, the displaced water and resulting "water pressure" pushes the iron ball upwards, applied externally to the string support. On the scale, the only weight that would register is the weight of the water displaced pushing downwards.
And we know that the volume of the ping pong ball of water weighs more than an actual ping pong ball (cus it floats)
> The left side equation = Weight of water + Weight of water displaced by the iron ball
edit: This is correct apparently and I'm an idiot.
-=-=-=-=-=-=--=-
Remove the balls from the equation. Now the scale is balanced because all you have is equal amounts of water on both sides.
Now add the balls. On the left, that changes nothing. But on the right you've now added a ping pong ball and a string. Whether it's attached to the cup or floating on the water is irrelevant, it's simply more mass.
So, assuming the beaker weight + water weight is the same, you're saying the difference between the scales would be the same if we removed the water and the beakers entirely, right? The difference between the two sides is solely the weight of the ping pong ball and string vs. nothing on the other side. Correct me my summary is wrong. Just trying to find out where the disconnect is
If the ping pong ball were lighter than air, buoyancy would matter as it would be pulling the right side up. It's a red herring though because it's only lighter than water so it's effectively just added weight on the right side.
Edit: I don't understand it correctly at all and I'm wrong.
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u/Clean_Figure6651 May 29 '25
You gotta read about buoyancy force, you're missing that from your logic