Physics Riddle: The Fate of the Balance

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The discussion revolves around a physics riddle concerning the balance of two containers, one with a steel ball and the other with a ping pong ball, submerged in water. Participants debate the effects of buoyancy and the forces acting on the balls, with many asserting that the buoyant force is equal for both due to their equal volumes. However, confusion arises regarding how the string attached to the ping pong ball affects the balance, leading to differing opinions on whether the scale will remain even or tilt. Ultimately, the consensus leans towards the idea that the upward force of the air inside the ping pong ball does not counterbalance the weight of the ball and string, suggesting the scale will tilt. The conversation highlights the complexities of buoyancy and the nuances of experimental setups in physics.
  • #51
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.
 
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  • #52
D H said:
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.
 
  • #53
Creator said:
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).

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.
 
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  • #54
rcgldr said:
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?
...
 
  • #55
Creator said:
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.
 
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  • #56
A.T. said:
But both scales are initially in balance, so the weight of both fluids is the same.

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) :confused:
 
  • #57
derek10 said:
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.
 
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  • #58
russ_watters said:
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.
 
  • #59
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.
 
  • #60
derek10 said:
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) :confused:
And what about the upper scale, on which the two identical balls hang?
 
  • #61
Dadface said:
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).
 
  • #62
OmCheeto said:
...

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...
 
  • #63
A.T. said:
And what about the upper scale, on which the two identical balls hang?

I want to play too! :)

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.
 
  • #64
rcgldr said:
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.
 
  • #65
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.
 
  • #66
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.
 

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  • #67
gmax137 said:
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.
 
  • #68
russ_watters said:
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.
 
  • #69
Doc Al said:
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.
 
  • #70
Vanadium 50 said:
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.
 
  • #71
You can make the glass arbitrarily thin.

I don't think this kind of quibbling is helpful. It's useful to be able to evolve a system from one that is difficult to understand to one that is simpler step by step.
 
  • #72
Vanadium 50 said:
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.
Wait, can we replace the post with a Han Solo action figure instead?
 
  • #73
Vanadium 50 said:
You can make the glass arbitrarily thin.

I don't think this kind of quibbling is helpful. It's useful to be able to evolve a system from one that is difficult to understand to one that is simpler step by step.

Wait ... I thought you were replacing the entire volume of the Ping-Pong ball with glass. If that's correct then I was not quibbling, but perhaps I'm missing what you are saying.
 
  • #74
No, the ping-pong ball is still hollow. If I wanted to replace it with something solid, that would be an extra step.
 
  • #75
Doc Al said:
...The buoyant force pushes up on the ping pong ball ... The upward pull of the ... string ... exactly balances the additional downward force due the increased depth of water.

Now I get it. Thanks Doc !

And on the RHS, the steel ball is not "pulling up" on the bottom, so the extra depth of water is what tips the balance down on the right.

The ball on the right could be lead or uranium, anything that doesn't float, its all the same. If the string on the right was a rigid rod holding another ping pong ball *down* it would have the same effect.

I guess previous posts have said all this; it just took awhile for the bulb to go off in my head.

Thanks again
 
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  • #76
gmax137 said:
The ball on the right could be lead or uranium, anything that doesn't float, its all the same.
or the ball on the right could also be a ping pong ball, and instead of a wire, a rod from above attached to the ball and pushing the ball down in order to hold the ball submerged in the water. The density of the ball doesn't matter as long as there's some external force that holds the ball in place and submerged in the water.

As posted earlier, if an external force holds the ball submerged, then the weight includes what would be water displaced by the ball and the balls weight doesn't matter. If an internal force holds the ball submerged, then the weight includes the weight of the water and the weight of the ball.
 
  • #77
Did anyone actually read Russ W's post (52)? There is really nothing more to add to it, yet there have been twenty more posts. Why? Doesn't proper Physics appeal to people any more? Would they rather do it all with arm waving?
 
  • #78
sophiecentaur said:
Did anyone actually read Russ W's post (52)? There is really nothing more to add to it, yet there have been twenty more posts. Why? Doesn't proper Physics appeal to people any more? Would they rather do it all with arm waving?
I think the problem was resolved (for most) way before post #52. Using Physics!

And yet...
 
  • #79
Doc Al said:
I think the problem was resolved (for most) way before post #52. Using Physics!

And yet...

But, but... we haven't analyzed the problem with a massless ping pong ball filled with helium. Nor have we determined if the floating ping pong ball scenario is affected by whether or not the material of the ping pong ball is hydrophilic or hydrophobic!

ps. I thought post #2 was adequate, but then PF happened. I've seen it before.
 
  • #80
sophiecentaur said:
Did anyone actually read Russ W's post (52)? There is really nothing more to add to it, yet there have been twenty more posts. Why? Doesn't proper Physics appeal to people any more? Would they rather do it all with arm waving?

I would take that even further back and ask if anyone actually read AlephZeros post (6).
 
  • #81
hehe, yeah, I think nathanael in post #2 was first to the answer, like cheeto said. But then different people find the answer using different methods, and I guess the rest of this thread has been a discussion of the various different methods. (and also people have mentioned extensions to the problem).
 
  • #82
Wow, lots of posts in this thread since I was here last. I've been tied up with work and unable to follow up on my original post. I haven't read all the other posts in detail but I've skimmed through them. As it seems most people are finding different ways of understanding the puzzle I thought I would add mine.

After doing the following thought experiment, I came to the same conclusion as ZetaOfThree. I imagined the same setup as depicted in Micromass's illustration with the addition of a spring scale located under the stand that is holding the steel ball. The spring scale is adjusted so that it shows only the weight of the steel ball as it is held over the beaker of water. The beaker of water on the left of the balance scale contains the ping pong ball (either floating or suspended from the bottom by a string, makes no difference). In this configuration the balance scale is tipped to the left by the weight of the ping pong ball. Now the steel ball is slowly lowered into the water. The weight indicated by the spring scale begins to decrease as the steel ball is lowered into the water. The weight indicated by the spring scale will always read the weight of the steel ball minus the weight of the volume of water that it displaces. So as the steel ball is lowered into the water, weight is transferred from the spring scale to the right side of the balance scale. Once the volume of water displaced by the steel ball surpasses the weight of the ping pong ball the balance scale will tip to the right.

What seemed to be quite puzzling at first turns out to be quite simple.
 
  • #83
TurtleMeister said:
Wow, lots of posts in this thread since I was here last. I've been tied up with work and unable to follow up on my original post. I haven't read all the other posts in detail but I've skimmed through them. As it seems most people are finding different ways of understanding the puzzle I thought I would add mine.

After doing the following thought experiment, I came to the same conclusion as ZetaOfThree. I imagined the same setup as depicted in Micromass's illustration with the addition of a spring scale located under the stand that is holding the steel ball. The spring scale is adjusted so that it shows only the weight of the steel ball as it is held over the beaker of water. The beaker of water on the left of the balance scale contains the ping pong ball (either floating or suspended from the bottom by a string, makes no difference). In this configuration the balance scale is tipped to the left by the weight of the ping pong ball. Now the steel ball is slowly lowered into the water. The weight indicated by the spring scale begins to decrease as the steel ball is lowered into the water. The weight indicated by the spring scale will always read the weight of the steel ball minus the weight of the volume of water that it displaces. So as the steel ball is lowered into the water, weight is transferred from the spring scale to the right side of the balance scale. Once the volume of water displaced by the steel ball surpasses the weight of the ping pong ball the balance scale will tip to the right.

What seemed to be quite puzzling at first turns out to be quite simple.

That seems simple, what is hard to understand is what was mentioned earlier in the thread.
If you took a ping pong ball the same size as the steel ball (which is the same size of the left ping pong ball) and with a stiff rod instead of a thread which was of the same weight and dimension of the thread and pushed the ball with the rod into the water on the right side of the balance scale. The balance scale will tip to the right even though the amount of water displaced is the same in each of the beakers and the ping pong balls weigh the same and so do the thread and rod.
Presumably if the rod and ball was attached to the beaker after being pushed the balance would reach equalibrium as the person pushing is adding the uneven weight to one side.
Attaching the rod to the beaker makes both sides equal.
 
  • #84
Buckleymanor said:
That seems simple, what is hard to understand is what was mentioned earlier in the thread.
If you took a ping pong ball the same size as the steel ball (which is the same size of the left ping pong ball) and with a stiff rod instead of a thread which was of the same weight and dimension of the thread and pushed the ball with the rod into the water on the right side of the balance scale. The balance scale will tip to the right even though the amount of water displaced is the same in each of the beakers and the ping pong balls weigh the same and so do the thread and rod.
Presumably if the rod and ball was attached to the beaker after being pushed the balance would reach equalibrium as the person pushing is adding the uneven weight to one side.
Attaching the rod to the beaker makes both sides equal.

When the ping pong ball is held in a submerged position with a stiff rod, the applied force is from an external source not within the balance scale system (it is pushed down with a force equivalent to the weight of the displaced water). When the ping pong ball is held in a submerged position by a string attached to the bottom, the applied forces are all within the balance scale system (Newton's third law: they cancel out leaving only the weight of the ping pong ball itself).

Image the same thought experiment in my previous post, except this time the spring scale is reversed (to read weight in the opposite direction) and the steel ball is replaced with another ping pong ball. Remember, the spring scale, ping pong ball, and the stand are all external to the balance scale system. As the ping pong ball is pushed into the water it starts to displace water and the spring scale reading increases. The reading on the spring scale will indicate the weight of the volume of water that is being displaced. When the weight of the volume of water being displace is greater than the weight of the ping pong ball on the left of the balance scale, the balance scale will tip to the right.

In short, the difference is in the fact that the ball and stand on the right are not a part of the balance scale system. If I were to place the stand and ball on the right onto the right platten of the balance scale and balance it out, then moving the ball in and out of the water would have no effect.
 
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  • #85
micromass said:
What will happen to the balance:

abq50dE_460s_v4.jpg

Here the one above again:



Here a variant with pushing the ping pong ball down:

 
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  • #86
A.T. said:
Here a variant with pushing the ping pong ball down:



Answer:

 
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