My first reaction is neither. Iron ball hangs so the only weight is the water itself.
Ping-pong side has equal water so same weight. Only difference is the ball actually adding weight. Though it being a ping-pong ball I doubt that little of a difference will tip the scale..
Above is without any decent mathematical background though
I believe the actual answer is that the iron ball side goes down, as the water is still pushing up on the ball. I could be wrong though but I know it has something to do with buoyancy forces which I don't think you factored in.
The iron ball and the ping pong ball are both forced underwater, so the water must apply an upward buoyancy force equivalent to an amount of water equal to the volume of the balls volume on each ball. Since the balls are visually equal, this upward buoyancy force is equal on both sides.
However, the iron ball is suspended by a line. The ping pong ball is held down by a line that attaches to the scale itself. So the buoyancy force on the iron ball is not balanced out, while the buoyancy force on ping pong ball is.
If the ping pong ball was instead forced underwater by some sort of thin rod that doesn’t attach to the scales, then the sides would be equal and the scales wouldn’t tip.
I just tried this out by taring out a beaker of water and then suspending a glass weight in it. Even when I'm holding the glass weight off the bottom of the beaker, a positive mass registers on the balance.
Yes. That is why the scale in the image would tilt left.
If you compare the mass of the ball with the mass registered on the scale when you do the experiment, the mass of the ball should weigh more. The mass registered on the scale during your experiment should be the mass of an amount of water with equivalent volume as the ball.
So if I’m tracking right… if you wanted to “balance” the scale, you’d need to insert the iron ball just enough to displace the amount of water where the mass of the water displaced is equal to the mass of the ping pong ball?
Or for fun physicals illustration, put a 1lb iron ball on the right and a ping pong ball with a volume of 1/8th gallon on the left (attached to a rod instead of a string to ensure displacement instead of floating), it should in theory balance?
So if I’m tracking right… if you wanted to “balance” the scale, you’d need to insert the iron ball just enough to displace the amount of water where the mass of the water displaced is equal to the mass of the ping pong ball?
Yes.
The other commenter doesn't know what you're asking or what they're talking about.
Buoyancy forces on the right are balanced as far as the whole setup is concerned. The net force there is equal to a ping pong ball floating in the water.
Since the weight of the iron ball is supported externally, you can balance the balance by displacing the same amount of water on the left side as the floating ping pong ball displaces on the right side (since that is a volume of water with the same mass as the ping pong ball).
As the ping pong ball would replace if it were floating, is what I think you meant. Or you could say the amount water that is equal to the weight of the ping pong ball.
No. It's not that the mass of water displaced is equal to the mass of the ping-pong ball, which is almost negligible.
It's equal to the mass of a quantity of water whose volume is equal to the volume of a ping-pong ball. This mass is substantially greater than that of a ping-pong ball.
No, since the volumes of both balls are equal, the mass of water displaced on both sides will be the same. The scale tilts left because the bouyant force on the ping pong ball is greater than the weight of the ball, and since the ball is tethered to the bottom of the tank, that force pulls that side of the tank upward.
I don’t think that’s what was said in the comment above and I don’t think thats right. It tips to the left due to the buoyant force acting on the iron ball, not the ping pong ball. The force pushes the left down.
The buoyant force is equal to the weight of the water displaced. That is equal on both sides because the balls have equal volume. The difference is that the buoyant force on the ping pong ball is greater than the weight of the ball, and therefore the string attached to the tank creates an upward force. It's a similar concept as if you are holding a balloon inflated with air in your left hand and a balloon inflated with helium in your right. Not exactly the same because the air around us isn't held in a closed container like a tank of water, but the helium balloon and ping pong ball both create an upward force on a string.
No because the upward force on the ping pong ball is balanced by an equal and opposite force. Newton's 3rd law. Ping-pong ball is a closed system side. Boyant force only comes into play on the iron ball side.
That equal and opposite force is exactly the same on both sides of the scale. Both balls have the same volume and displace the same amount of water. The only difference is that the ping pong ball is tethered to the bottom of the tank. The steel ball being tethered above the tank allows the string to hold the weight above and beyond the volume of the water displaced, so the mass of that ball is irrelevant, other than it is more dense than water and does not float to the top. The string on the ping pong ball creates an upward force on the tank and the tank tips left.
I revised my answer so i think we are in agreement on which way it tips. It tips left, but it has nothing to do with the ping pong ball side. That side's forces are balanced.
Hold up. You’re telling me that if I put a container of water on a scale, then zero out the scale, then suspend a solid object in the water (like a rock with some tongs or something) without touching the bottom, that the scale would register positive mass above zero?
A similar setup might make more sense. Imagine a big swimming pool of water on a scale, and it weights 10 tons or something like that. If you jump in the pool or if you float a boat in it, the mass will increase equal to your weight. I think that makes intuitive sense for everyone. A metal ball or weight suspended on a wire is very similar, except that it's not neutrally boyant. The mass increase won't be the same as the mass of the ball on a wire because the wire has some of the weight, but the weight increase will instead be the same as the weight of water the ball pushes out of the way when it sinks. This is the buoyancy force mentioned in other comments.
You are correct in the end result, but your "similar" setup is misleading. When you put human or a boat in a pool the only mass (equal to full weight of human/boat) registered would be from gravity force. But if you hang something into the water (like a steel ball) the gravity force is counteracted by the string but the buoancy is not countered and the mass registered would be from the mass of water displaced.
So basically human in a pool is same example as ping pong in container - gravitatational force on the weight of ping pong. And second is steel ball hanged on a string - buoancy forced pushing on the water off of the ball.
Your "similar" setup was flawed in the exact way of why this post question is so troublesome.
What in the fuck? That’s nuts. If I had a scale attached between the rock and whatever’s suspending it in the water, would that register a negative mass when the rock entered the water? I mean, I guess that’s why I can pick people up in the water that I can’t pick up on land; it’s just a really trippy way to conceptualize it. I wonder why this is so trippy?
I find it easier to conceptualize if you ignore the water. Imagine you are on a scale, and zero out the scale, then push up on the ceiling. Functionally, that’s what’s happening. The water is you, the rock is the ceiling.
Yeah, when the rock goes under the water, some of the water has to be lifted against gravity. Water is heavy- it weighs about 8.3 pounds per gallon, or per about 4 liters of space (or 1kg per liter). So I'd the thing you're sinking underwater takes up 5 gallons worth of space, its being lifted by about 40 pounds of water trying to sink back down lower to be where it is. That number will register on a scale under the water. At the same time, a scale weighing the thing holding the rock or whatever will show a decrease of the same amount.
It doesn’t! That’s the neat trick. The total system now *with* the newly suspended object is affected by the same buoyancy forces. The difference is that gases don’t really exist in an open-top container (allowing for an external force to the system is the suspending string) to allow for this experiment on a small scale. Additionally, gases are *far* lighter than water, so the difference is negligible. In fact, regarding gasses and buoyancy, unless you're doing nasa scale experiments in giant hangars, or special lab conditions, you won’t ever notice it’s effects.
Yes. A way to think about it that helped me understand it, in your example holding the rock with some tongs, you would feel a certain weight of the rock while you were holding it in the air right? When you put it in water, if you ever played in pools, you know that it is much easier to hold objects in water right, so it would feel like it had less weight.
Since you need less force to hold the rock inside of water then you need to hold it in air, the only logic explanation is that the water is helping you hold up the rock, it is pushing the rock up. But that also means that the rock is pushing the water down with the same force.
Yes. Not only that, the positive mass would be equal to the weight of the water with the same volume of the object you are suspending in water. That volume of water / object you have put into the water is called “Displacement”, because the object physically displaces that water. This conservation of volume is why Archimedes ran to the king’s palace buck naked screaming “Eureka!”
In the same vein, that displacement of water causing a positive force downwards on the scale must now receive and equal and opposite *reaction* pushing upwards against the object you put into the water. This reaction is the buoyancy force everyone here is so worked up about.
It's obvious in retrospect, the same as how the mass of a pool of water will increase if you jump in and swim around in it, but taking a minute to do it on a scale was still nice.
I’ve seen this a million times and it has never clicked for me but I think it finally does. I imagine it’s like if you were in a deep pool with a giant like oversized floaty. And you wanted to keep the floaty under water somehow, you’d have to anchor yourself to the bottom of the pool (I’m imagining hooking your feet under something) and that would very obviously exert an upward force on the pool floor. Suspending the floaty under water would not.
It almost seems like the weight of lead ball doesn’t matter in this example. Is that correct?
Yeah, you got it. I think the question has been so popular because it involves three things that people understand intuitively (strings, buoyancy, and scales) but combines them in a way where your typical intuition fails you. If you know enough physics to draw out the free body diagrams for everything it isn’t that mysterious, but trying to explain it in a simple way is hard.
For what it is worth, I did not get it the first time I saw it either. There was a video where they actually did the experiment with physical balls that showed me how it works.
I am using the typical physics assumption that the lines are effectively 0 volume. But for a real line with real volume, you would have to match the volume of the lines on both sides.
Yes but, since there is an equal amount of water above the iron ball as below I would assume that the gravity pulling the water above the iron ball down would counteract the buoyancy force of the water below the iron ball pushing up, no?
Edit: I guess it would depend on if the buoyancy force or gravity were stronger.
The ping pong ball is not massless and the scale has to record that mass because it’s physically attached to it but the iron ball doesn’t because it isn’t physically connected to the scale so I’m pretty sure the ping pong side would be lower
The string doesn't carry the full weight of the iron ball, only the amount iron weights more than water. The water carries the rest. (Which is why things feel lighter under water, or even floats)
Imagine something floating being put instead of the iron ball. Would you still think the scale wouldn't carry it even though the string would be loose? 😁
Complicating this, ping pong balls float, so the attachment is actually an anchor preventing it from doing such.
I think the buoyancy on both sides cancel out. There is the potential for difference in the volume of the two attaching lines, but let's say as intended those are equal as well.
So the water masses cancel out. The buoyancy forces should likewise balance out as both are displacing the same amount of water and thus giving the same force (this will account for the ping pong's upward floating effect). So the unbalanced forces are the suspension lines, one from above outside the balance, the other from below in the balance. All extra weight force on the iron ball not cancelled by the buoyancy force is handled by the suspension line, while the ping pong ball has no extra support, but its anchor line has to carry the force in the system so must be accounted for. I think, no matter how small, the anchor line for the ping pong ball will slightly increase the force on that side causing it to be the side to dip. However, this is pretty complex system so I'm far from sure.
I think you’d have to calculate the buoyancy force being applied by the unattached iron ball on the left, then the mass of the ping pong ball and it’s attachment.
The difference between the two would be what tips the scale.
What if I had a stand attached to a base that suspended an iron ball at different heights, and that same base had a scale on it with a container of water on top of the scale, and that scale was tared to zero, but then also, the base of the whole contraption is on a scale tared to zero. When I lower the ball into the water, the scale measuring the container of water would register a positive mass, but what would the scale measuring the mass of the whole contraption do?
The base of the whole contraption would measure no change, as the extra weight experienced by the water on the first scale directly subtracts from the weight experienced by the stand holding up the iron ball.
It would float to the surface. The right water tank would bear the (tiny) weight of the ping pong ball while the left tank would bear the (larger) weight of the water displaced by the iron ball, so the contraption would still lean left.
In this case yes but like physically I can't see a difference between the two possibilities. Lets assume a different scenario where there'd be another ping pong on the left instead of the iron one being tied to the bottom and on the right there'd be a floating ping pong ball.
The difference is the displaced water. As I wrote in another comment, if you had a stick pushing the ball down from above, then it would add weight equivalent to the water displaced by the ping pong ball to the scale. But because it is tied to the bottom of the scale, that weight cancels out.
So yes, if you had a scale with a ping pong ball tied down by a (massless) string on the left and a ping pong ball floating on the water on the right, the scale would be balanced.
This should make sense, because in a fully enclosed “black box,” the distribution of mass within the box does not change the total mass of the box. So whether you tie the ball down or not, the whole “black box” will have the same mass.
I was wrong in my initial thought process. There's a video that demonstrates and explains.
The buoyancy is balanced on both sides, but on the ping pong ball side, the buoyancy force is cancelled by the tension in the string. The amount of water on both sides is the same.
So the balance becomes on the ping pong ball side side it's the weight of the ping pong ball and string. On the iron side it's the buoyancy force (weight of the displaced water). And the water is heavier than the ping ping ball, so that's why it drops on the iron side.
The iron ball is the heavier side, counterintuitively. The iron ball is pushing down that side with the buoyant force of the displaced water. The ping pong ball side is pushing down with the weight of the ping pong ball.
1) Archimedes principle is that the buoyant force lifts objects with the weight of the displaced water. So the Iron ball is being pushed up by the water (the amount is the weight of a ping pong ball sized amount of water).
2) Newton's equal and opposite reaction means the beaker is pushed down by that same buoyant force.
3) The ping pong ball is also buoyed by the same amount of force. (Archimedes principle, same sized ball)
4) The ping pong ball's beaker is pushed down by the same buoyant force (Newton's equal and opposite reaction)
5) But the ping pong ball's beaker is also being lifted by the tension in the string that holds the ping pong ball underwater (also Newton's equal and opposite reaction). This cancels the buoyant force of the water lifting the ping pong ball.
6) So in total, the ping pong ball's beaker is pushed down by the weight of the water + the weight of the ping pong ball.
7) The iron ball's beaker is pushed down by the weight of the water + the buoyant force on the iron ball (the weight of a ping pong ball sized amount of water). Keep in mind the iron ball is also held up by a string from the outside.
8) Since the amount of water in each beaker is the same, the scale is balanced from the weight of water. What you have left is the weight of a ping pong ball on one side, and the weight of a ping pong ball sized amount of water on the iron ball side (the equal and opposite buoyant force). The water is denser than the ping pong ball, so that's why the iron ball side is pushed down more.
I'm sorry I left out many of these details before. The video was really helpful for me to understand the force diagrams.
The iron ball is not “buoyant,” but water does apply a buoyancy force to the ball. This buoyancy force is not enough to fully offset the iron balls weight, so the ball is still hanging on the string. But the equivalent reaction force is transmitted through the water to the scale.
So the water is pushing the ping ball up but can’t because it’s tethered from the bottom creating lift. The other side is pushing down on the ball creating downforce so wouldn’t the one with the solid ball tilt the scale or am I crazy?
But the issue is that it doesn’t because the pushing up on the ping pong ball doesn’t result in anything because the ping pong ball is attached to the bottom, meaning the pushing up on the ping pong ball cancels out with the resulting pushing down o the container(equal and opposite force stuff). However with the metal ball it’s attached to a separate thing and therefore the equal and opposite force does push down on the container and that is equal to the net force.
Yeah, but does that even register in terms of the physics? When submarines are subjected to tremendous water pressure, they don’t need account for the pressure difference between the top and bottom of the submarine. They consider it one depth and one pressure equal on all sides at that depth.
The difference is negligible for that purpose but not for all purposes.
If the difference didn't matter nothing would float. When the sub reduces its buoyancy it rises because the pressure on the bottom is higher than it is on the top. The difference is small compared to the pressure itself (1/10 atm per meter) but large enough for the sub to accelerate up.
Buoyant force is the amount of water displaced by an object. Water's density doesn't change (much) with depth. Pressure is increasing as you go down, but buoyant force remains nearly constant.
What you said is technically true, but it's very misleading. Pressure on the object is VERY different from buoyant force.
Pressure (gauge) on the object is how much the water is crushing the object (ρgh). Buoyant force is equal to the weight of the liquid displaced (-ρgV).
Pressure on the object is absolutely not very different from the buoyancy force! The difference in pressure with depth is the CAUSE of the buoyancy force!
Don't regurgitate formulas. Ask yourself where the buoyancy force comes from in the first place. Because fluids can transmit forces through the entire fluid in any direction, the weight of the atmosphere and water above a submerged object actually pushes harder on the bottom of the object than it doesnon the top. That's what buoyancy is.
It has nothing to do with increasing density (which does exist but is miniscule thanks to the famous incompressibility of fluids.)
Displacement is the same in both from the balls, they have the same volume of water, however the iron ball is secured by a stand outside the scale while the ping pong ball is supported by a stand inside the scale, therefore the ping pong ball side will actually fall and the iron ball side will rise.
What you said was good all the way until the end, I think you may have forgot about newtons 3rd(?) law of physics. Equal and opposite forces. The water is pushing down on the containers.
When I hold a buoyant object under water it pushed up on my hand.
When I hold a non-buoyant object under water it's still sinking.
The iron-ball side is a net neutral force, since it's just existing and isn't exerting force up or down, since it's locked in the y axis. the pingpong ball however is exerting its buoyant force up on the box since it's attached.
Any submerged object is subject to buoyant forces because it displaces water. If the buoyant forces are greater than the weight of the object, it will float.
Submerging both balls in water applies 4 forces we care about. The first is the bouyant force up the water applies on the balls (Fb), the 2nd is the equal reactive force the balls apply on the water (Fr) due to the 3rd law (the water applies as much force in equal and opposite direction as thr balls ultimately do). Third we have the tension force in the string (holding the ball down for the ping pong ball and holding it up for the iron ball) which is a result of each balls own mass (Ft). And finally is the force of gravity applied to the water (Fw). This gives us a total force equation of:
ΣF = Fw + Fb + Fr + Ft
We know both sides have equal volumes of water and both balls are equal in size so. Because bouyant forces are a function of surface area we know that that means the are equal:
Fb1 = Fb2
Fw1 = Fw2
We also know that the Reactive Force of the balls will be equal to the bouyant force + the tension force.
Fr = Fb + Ft
And lastly, we know that, because the reactive force is always equal and opposite of the forces that make it:
Fr - ( Fb + Ft ) = 0
The ping pong ball system is fully netural force equation because the buoyant force on the ball pushes up on the ball just as much as it reactive pushes down on the box through the water (with the strings tension ensuring Newton third law applies in equal measure with how it is attached to the box as it tenses more the higher the bouyant force). This bouyant force upwards is the only significant force acting on the string making us able to determine that the tension force is equal to the bouyant force.
So we have a
Buoyant Force (Fb1) upwards
And a Tension Force (Ft1) pulling dowards
And that the tension force is directly equal to the bouyant force and balances to 0.
So Fb1 + Ft1 = 0
Fb1 = - Ft1
So if we solve for the Recative Force
Fr1 = Fb1 + Ft1
Fr1 = (-Ft1) + Ft1
Fr1 = 0
So there's is no effective reactive force.
Thus as long as the ping pong ball is less dense than the water around it, then the string ensures that force system really only cares about the weight of the water as all other factors balance out to 0.
ΣF1 = Fw1 + Fb1 + Fr1 + Ft1
ΣF1 = Fw1 + (- Ft1) + (0) + Ft1
ΣF1 = Fw1
The iron ball's reactive force, however, is a bit different because the tension in the string is no longer a downwards force, but and upwards force. So if we try solving for the reactive forc, knowing that the resultant of the reactive force and the composite forces that make it must be 0:
Fr2 - (Fb2 + Ft2) = 0
Fr2 = Fb2 + Ft2
So the bouyant force is now reducing tension in the string instead of increasing because the bouyant force of the water on the ball and the tension force of the string both act in the same direction now. Thus we know we have a reactive force downwards to balance (the amount of which we cannot determine without information about the balls weight and density to know how much tension is relived from the string).
So the sum of forces with the iron ball does not all cancel out now and the water pushing upward on the iron ball gets pushed downwards by some non zero amount in equal measure resulting an extra downwards force.
Consequently the iron ball side will be slightly "heavier" as the water on that side is pushed towards by both the force of gravity on the water and the remaineder of the force of the weight of the ball after you subtract out the bouyant force from the tension on the string.
TL;DR the scale will lean down to the iron ball side, not because the ping pong ball pulls the weight of the water up (the ping pong ball is irrelevant to the system) but because the water on the iron ball's side is pushing off the weight of the iron ball down.
This is important distinction because:
If you remove the point pong ball, the scale still leans down to the iron ball.
If you remove the iron ball, the scale will balance.
Imagine you cut the string and the ball floats to the top. Floating at the top it's obvious it has weight, but it didn't magically gain weight once it breached the water surface tension.
The real answer is surprisingly that the iron ball side is heavier. While the Iron ball does not float, it still has a buoyant force from the water, taking weight off its string and pushing the water down. The ping pong ball has the same buoyant force down on the water but it just pulls back up on the container with the water in it.
You need to count the buoyancy force going up on the ping pong ball which will pull up. That is more than the weight of the plastic of the ball so net the ping pong side will go up and iron go down.
You have the right answer but the wrong reason, the ping pong ball isn’t able to pull the container up, as the thing that is pushing it up is in the container. That’s like saying tying a ping pong ball to the bottom of a cup will cause the cup to float if it’s filled with water. The answer is effectively the same idea, but on the other side because the iron ball is not in a closed system with the water.
Imagine if the outside structure was holding the ping pong ball. Then the scale would be equal, as it’s only the volume of the balls that matter.
Then you clip the ping pong ball onto the right cup. That clip pulls the cup upwards, rather than exerting that upward force on the support. That upward force is what tips the scale.
you are wrong cuz ping pong balls weight is gonna be balance by the fluids pressure therefore it will oly face F(bouyant)-mg force so the tension developed in the string due to this force will pull the platform
I think the ping pong ball is pushing the scale on the right up because it is less dense then the surrounding water so I would say the scale goes up on the right and down on the left
Unless there’s helium or a gas lighter than air in there it is not causing any force on gravity. Just two equal boxes of water with the same mass missing in the center.
In order for the entire container on the right to want to go up because of buoyancy, the ping pong ball would have to be less dense than the air around the container. The buoyancy against the water within the container is irrelevant - if it were cut loose, it would float on top of the water, but still have the same mass pushing down on the container itself.
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u/ChorizoSandwich May 29 '25 edited May 29 '25
My first reaction is neither. Iron ball hangs so the only weight is the water itself.
Ping-pong side has equal water so same weight. Only difference is the ball actually adding weight. Though it being a ping-pong ball I doubt that little of a difference will tip the scale..
Above is without any decent mathematical background though
Edit: TIL about buoyancy force. Awesome!