Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I found the problem described in the attached picture on the internet. In the comment sections there were two opposing solutions. So it made me wonder which of those would be the actual solution.

So basically the question would be the following. Assume we would have two identical beakers, filled with the same amount of the same liquid, lets say water. In the left beaker a ping pong ball would be attached to the bottom of the beaker with a string and above the right beaker a steel ball of the same size (volume) as the ping pong ball would be hung by a string, submerging the steel ball in the water as shown in the picture. If both beakers would be put on to a scale, what side would tip?

According to the internet either of the following answers was believed to be the solution.

  1. The left side would tip down, because the ping pong ball and the cord add mass to the left side, since they are actually connected to the system.
  2. The right side would tip down, because of buoyancy of the water on the steel ball pushing the steel ball up and the scale down.

Now what would the solution be according to physics?

enter image description here

share|improve this question
    
So the question becomes does submerging of a suspended steel ball affect the force balance on the scale. I think yes. –  ja72 Aug 12 at 16:57
1  
A few things appear to be missing. (1) The steel ball and ping pong ball presumably are same size. That's not mentioned. (2) Tip? Which way? I think you meant "tip down" rather than just "tip". –  David Hammen Aug 12 at 17:15
    
Consider the weight that the ball stand on the right has to support as the steel ball is lowered into the water: the weight gets lighter, because the water is holding up part of the weight. This far outweighs the ping pong ball's weight, showing why (1.)'s implication of "the steel ball isn't connected, so it's not important" is wrong. –  Tim S. Aug 12 at 18:03
1  
This is essentially the same as the well-known question: if you are in a canoe containing some rocks (but still afloat), after tossing the rocks into the pond does the water level rise, fall, or stay the same? You need to compare mass and volumes in each state. –  Carl Witthoft Aug 12 at 22:34
    
Jean, please carefully read the top two answers. They are correct. –  David Hammen Aug 13 at 12:07

7 Answers 7

Here is a free body diagram of the balls:

FBD of balls

… and one of the water volume:

FBD of water

The four balance equations are

$$ \begin{align} B_1 - T_1 - m_1 g & =0 \\ B_2 + T_2 - m_2 g & = 0 \\ F_1 + T_1 - B_1 - M g & = 0 \\ F_2 - B_2 - M g & = 0 \end{align} $$

where $\color{magenta}{B_1}$,$\color{magenta}{B_2}$ are the buoyancy forces, $\color{red}{T_1}$,$\color{red}{T_2}$ are the cord tensions and $M g$ is the weight of the water, $m_1 g$ the weight of the ping pong ball and $m_2 g$ the weight of the steel ball.

Solving the above gives

$$\begin{align} F_1 & = (M+m_1) g \\ F_2 & = M g + B_2 \\ T_1 & = B_1 - m_1 g \\ T_2 & = m_2 g - B_2 \end{align} $$

So it will tip to the right if the buoyancy of the steel ball $B_2$ is more than the weight of the ping pong ball $m_1 g$.

$$\boxed{F_2-F_1 = B_2 - m_1 g > 0}$$

This is the same answer as @rodrigo but with diagrams and equations.

share|improve this answer
    
What happens to the mass of the pingpong ball? –  Peter Raeves Aug 12 at 16:46
    
@ja72 - It looks like your last comment is out of line with your current answer. Glad you came to your senses and got the right answer! +1. Saying that it tips down to the left is equivalent to thinking that one can reach down, grab one's bootstraps, and lift oneself off the ground by those bootstraps. –  David Hammen Aug 12 at 20:33
    
"So it will tip to the right if the buoyancy of the steel ball B2 is more than the weight of the ping pong ball m2g" so if the ping pong ball was replaced by an equal sized ball that didn't float, the scales would go the other way? –  Trengot Aug 13 at 11:36
2  
@Trengot - Assuming the balls are the same size, the scales would balance if the ping pong ball was replaced with an identically-sized ball with the same density as that of water. If the ping pong ball was instead replaced with an identically-sized steel ball, the scale would tilt down to the left rather than the right. –  David Hammen Aug 13 at 12:23
1  
@David Hammen - The weight of the ball on the right does not matter, as long as the ball is denser than water the same net force on the bottom of the right beaker will apply, any extra weight is born by the cable. Only the height of the water matters on the right, and since that additional height is identical to adding a ball sized volume of water, the scale will always tip right as a ball of water is heavier than a Ping pong ball. –  Zach Johnson Aug 13 at 14:24

The weight on the left bowl would be the weight of the water plus vase plus ping-pong ball (plus thread, ignored).

The weight on the right bowl would be the weight of the water plus vase plus the buoyancy of the steel ball (plus the buoyancy of the submerged thread, ignored). That buoyancy is the weight of an equivalent volume of water.

Since the ping-pong ball is lighter than water, the scale will tip to the right.

Why is that the weight on the right bowl? Look at it this way: the ball is in equilibrium, so the sum of all forces on it will be 0. These forces are weight, tension on the thread and buoyancy. So the tension on the thread is $tension = ball - buoyancy$ (obvious?). And the weight on the right plate is the sum of all weights minus the tension on the thread. That is $water + vase + ball - tension$, which is the same as $water + vase + buoyancy$.

share|improve this answer
    
I agree, except the buoyancy of the ping pong ball also acts on the left side, but it can be lumped in the cord tension. –  ja72 Aug 12 at 17:25
3  
@ja72 - No, it doesn't. The water exerts an upward buoyant force B on the ball, so by Newton's third, the ball exerts a downward force $B$ on the water. The water too is in equilibrium, so the upward force exerted on the water by the bottom of the beaker is $B+W_w$. The thread is pulls the bottom of the beaker upward with force $B-W_b$. The total force on the bottom of the beaker is thus $(B+W_w)-(B-W_b)=W_w+W_b$. The buoyancy term cancels out. Another way to look at this term: It's an internal force. It doesn't count because of Newton's third. –  David Hammen Aug 12 at 17:33
2  
Another way to get the same result is to look at the ping pong ball+water+thread as one collective object. The external forces on this system are air pressure on the top (which we're ignoring), gravitational force $g m_w + g m_b = W_w+W_b$, and the upward force from the bottom of the beaker. The ball+water+thread system stationary, so the net force is zero, and thus the upward force exerted by the bottom of the beaker is $W_w+W_b$. –  David Hammen Aug 12 at 17:36
1  
@DavidHammen -Indeed. You could even crush the ping-pong ball and put the pieces under the vase and it will weight the same. Ignoring any compression the water pressure is doing to the submerged ball, of course. –  rodrigo Aug 12 at 17:54
    
@jean - And the weight of the steel ball and the force tramferred to the floor through the right stand... –  rodrigo Aug 13 at 11:28

A Thought Experiment

We can arrive at an intuitive explanation without any special knowledge of physics. The strategy is to re-create the setup as closely as possible while keeping the two sides in balance.

Imagine that you start with two identical beakers, filled with the same amount of water, no balls. Placed on the scale, they balance.

On the left, place a ping-ping ball with a thread dangling down. Let's pretend that the thread and the walls of the ball are of negligible weight. With that approximation, the scales remain balanced. (After all, all we have done is given a name to an arbitrary sphere of air above the water.)

Next, pretend that there is a water sprite at the bottom of the left beaker, operating a winch, tightening the string. Again, this has no effect on the scale, as the configuration change to the left beaker is self-contained. The ball sinks, and the water level rises.

On the right, lower a permeable ball into the water, suspended from a string. (Pretend that the walls of the ball are of negligible weight.) The ball fills up with water that was already in the beaker. Again, the scale remains balanced, since all we have done is given a name to an arbitrary sphere of water.

Suppose that there is a King Midas inside the right ball, turning water into gold, or steel, or whatever denser material. It makes no difference, since any additional weight will be borne by the string that suspends the right ball.

So far, the scales remain balanced. But what is the difference between the scenario so far and the one in your question? The water level on the right did not rise when we lowered the porous ball into the right beaker, the way it would have had we lowered a solid steel ball.

So, pour an amount of water in the right beaker equivalent to the volume of the steel ball, and you have recreated the setup! Of course, the scale would then tip to the right.

share|improve this answer
    
Great answer - another approach for the Pingpong-Ball would be to deform the left glassmake a Dent to the inside, widen it to a sphere, so you have a glas-sphere in the inside of the Glas, now make the connection from the outer glass to the inner glas-bubble very thin and you have the scenario. And the form of the Glas cannot change the weight! –  Falco Aug 15 at 10:54
1  
This is all very good but you seem to be assuming that the water sprite has negligible mass and I'm not sure that's physically justifiable! :-) –  David Richerby Aug 17 at 10:19

I'm amazed that this is so confounding to some. This is too long to be a comment, so I'm making it an answer. The TL;DR version: The answers that say the scale will tilt down to the right are correct. The beaker full of water with the steel ball suspended from above is heavier than is the beaker that contains the ping pong ball anchored from below.

Assumptions

  • The two flasks are identical. To make this so, down to the splitting of hairs, let's attach a connector to the bottom of both flasks. The connector will be used to anchor the ping pong ball on the left to the bottom. We need that same connector, unused, on the right so as to make the flasks identical.
  • The two flasks contain identical quantities of water.
  • The ping pong ball and steel ball are the same size and are fully suspended in the water.
  • The ping pong ball is less dense than water while the steel ball is of course more dense than water.
  • The strings are of negligible mass.
  • The scales are very sensitive and can detect differences in mass to the sub-centrigram level.

Experiment #1: Anchored ping pong ball on the left, no steel ball on the right

This one's easy: The left side is heavier. A simple explanation is to look at the water+ping pong ball on the left as a system. This system is static, so the net force is zero. The mass of the system is the sum of the masses of the water and the ball: $m_{w+b}=m_w + m_b$. Gravity exerts a downward force of $g m_{w+b} = g(m_w+m_b)$. Ignoring atmospheric pressure, the only other force is that of the bottom of the flasks on the water. This must exactly oppose the weight of the water+ball system to have a net force of zero. Thus the force transmitted to the left side of the scale is $W_l = g(m_f+m_w+m_b)$ where $m_f$ is the mass of the flask. On the right, there's only the mass of the water and the flask, so the force transmitted to the scale on the right is $g(m_f+m_w)$, or $g m_b$ less than that on the left. The scale tips down to the left.

Note that I ignored the forces on the string, buoyancy, and pressure. Invoking these results in the same answer as above, but with a lot more effort. The ball has three forces acting on it, gravitation ($W_b = g m_b$, downward), buoyancy ($B=g \rho_w V_b$, upward), and tension ($T$, downward). The ball is static, so $T+W_b = B$, or $T=B-W_b$. The water has three forces acting on it, gravitation ($W_w = g m_w$, downward), the third law counter to the buoyant force the water exerts on the ball ($B = g \rho_w V_b$, but now directed downward rather than upward), and the force by the bottom of the flask on the water ($F_p$ upward). The net force on the water is zero, so $F_p = W_w + B$. The forces on the bottom of the flask are the tension in the string, directed upward, and the pressure force from the water, directed downward: $F_f = F_p - T = (W_w + B) - (B-W_b) = W_w + W_b = g(m_w + m_b)$.

Some will say "but how does the reaction force to buoyancy transmit to the bottom of the flask?" Note that I didn't invoke Newton's third law in the context of the counter to the buoyant force eventually acting on the bottom of the flask. I used static analysis. The explanation of how this force is eventually transmitted to the bottom of the flask is via pressure. The force by the flask on the water is equal but opposite to the force by the water on the flask, and this is pressure times area. The presence of the ball raises the height of the top of the water by an amount needed to accommodate the volume of the ball, and this increases the pressure at the bottom of the flask. If the flask is cylindrical, this is a fairly easy computation: $\Delta h = V_b/A$, and thus $\Delta P = \rho g \Delta h A = \rho g V_b$. That's the magnitude of the buoyancy force.

Experiment #2: No ping pong ball on the left, suspended steel ball on the right

Now the scale will tilt down to the right. There's an easy way, a hard way, and a harder way to solve this. The harder way involves pressure, and the result will be the same as the other two approaches. I'll bypass pressure. The easy way is a static analysis. The water exerts a buoyant force on the ball, which exerts an equal but opposite force on the water. The water is static, so the bottom of the flask exerts a force on the water equal to the sum of its weight and the magnitude of the buoyant force: $W_w + B = g m_w + B$. Adding the weight of the flask gives the total weight on the right: $W_r = g(m_f + m+w) + B$. On the left, all we have is the weight of the flask and the water. The scale tilts down to the right.

Experiment #3: Anchored ping pong ball on the left, suspended steel ball on the right

Now we know the weight registered by the flask+water+ping pong ball system on the left and the weight registered by the flask+water+suspended steel ball system ob the right. It's a simple matter of comparing the two: $W_r - W_l = g(m_f + m+w) + B - g(m_f+m_w+m_b) = B - g m_b$. Since the ping pong ball floats, $B>g m_b$, so the scale tilts down to the right.

Experiment #4: Same as experiment #2, but now add water on the left

We can simply add water to the flask on the left in experiment #2 to make the scale balance. When we do that, we'll see that the scales balance when the water levels in the two flasks are at exactly the same height above the bottom of the flask. (This is the pressure argument.) If we measure the amount of water added, it will be equal in volume to the volume of the ball. (This is the buoyancy argument).

Experiment #5: Anchored ping pong ball on the left, left flask from experiment #4 on the right

Since the two flasks in experiment #4 have the same weight, the scale will still tilt down to the right, just as in experiment #3. If we look at the two flasks, we'll see that the water level in them is the same.

Experiment #6: Anchored ping pong ball on the left, anchored crushed ping pong ball on the right

Here we replace the steel ball in experiment #3 with a crushed ping pong ball anchored from the bottom. Since the buoyant force cancels out in the ping pong ball+water system (see experiment #1), one might think that the intact versus crushed ping pong ball test will balance. That is not the case. The intact ping pong ball weighs a tiny bit more. It has about 4 centigrams of air inside it. This is part of the measurement on the left but not on the right. The system with the intact ping pong ball is slightly heavier. Since our scale is accurate to the sub-centigram level, the scale will tilt down to the left in this experiment.

The above is incorrect. The water level will be a bit lower on the crushed ping pong ball side. Unless ping pong balls are inflated to considerably more than atmospheric pressure, the slightly increased pressure on the crushed ping pong ball side will more or less compensate for the reduction in mass.

Experiment #7: Replace the steel ball in experiment #2 with an intact ping pong ball

Finally, let's replace the string attached to the tower that suspends the steel ball in the water with a rigid rod attached to the tower that forces a ping pong ball to be immersed underwater. The result will be identical to experiment #2. The buoyant force is equal to volume, not mass. It doesn't matter what kind of ball we use so long as the volume remains the same. The effect on the tower will of course be markedly different, but the tower is not a part of the systems we are weighing.

share|improve this answer

Well I got this badly wrong, and grovellingly apologise to those I traduced.

It seemed easy: the water in both is the same weight, so I thought that removing it would make no difference to the balance. This was wrong: removing the water from the right hand beaker does have an effect, the presence of the suspended ball does add extra weight to it, so the right hand pan descends.

I did some experiments to check this out, using a plastic drinking cup on a sensitive digital weighing scale, I was limited by the maximum weight the scale would show, to 200 gm total, which constrained how I did the tests. I photographed the results (apologies for the backgrounds, ignore the green label) :

photos of the results here. The first photo (top left) shows the cup with water and a piece of plastic tethered to the bottom of it. The second (top right) has the plastic removed and hung on the rim of the cup, above the water and shows that there is no difference in the weight. This was what I expected. The third image (bottom left) shows the water alone (the hook came off and I discarded it) , note the weight, and the final photo has a 100 gm steel test weight suspended in the water, and, to my initial surprise, the weight shown on the scale has gone up. So the correct conclusion is that the right hand pan will descend.

As a final experiment, not photographed, I noted the water level and scale reading before I lowered the steel weight in. After I lowered the weight under the surface I removed water back to the same level. That is I removed the water that was displaced by the weight, and the scale reading went back to the original. To me this shows that the additional weight on the pan when the heavy mass is submerged, is equal to the weight of the water displaced.

This leads to a simple explanation of why the right hand pan dips. Remove the steel ball and imagine it leaving behind a hole in the water exactly the same size as the ball, so that the overall level of the water surface is what it was with the ball still immersed. Picture this hole filling up with extra water: then the forces on the spherical blob of water that has replaced the ball are exactly the same as those that acted on the suspended ball. To me this shows that the presence of the ball adds a weight equal to that of the volume of water displaced.

It also shows that the only two things that matter about the suspended object are its volume, and that it is more dense than water. Its weight and shape are immaterial (so long as it does not trap any additional air as it is lowered below the surface.)

I realise now that something very similar has been said in comments and answers already given, and although I arrived at this on my own, I appreciate and acknowledge their prior insight.

share|improve this answer
2  
Consider a similar case where you support the balls on the balance using springs, where the weight of the steel ball is partly supported by the spring and partly by the tension. The springs are both the same weight, but would removing them make no difference to the balance? –  Emilio Pisanty Aug 14 at 19:41
    
@200_success. Could you explain that further please, in particular how it applies to this problem?? –  Harry Weston Aug 15 at 10:40
    
@Emilio. Not the same case, on the left side the spring would be held up by the balance pan, and on the right side it is disconnected from the balance. Unlike the water which is supported by the balance on both sides. –  Harry Weston Aug 15 at 10:42
1  
Please do read the other answers before replying. If the answer was as easy as you're suggesting, somebody else would surely have posted it. So your answer is either wrong or a duplicate; it turns out to be the former which, again, you'd have discovered if you'd read the other answers. On the other hand, feel free to post a separate question about why your answer is wrong: somebody should be able to explain it to your satisfaction. –  David Richerby Aug 17 at 10:34
1  
@David Richerby. I hope that my apology and extended editing of my original answer covers this. I have learned the lesson. –  Harry Weston Aug 18 at 11:25

For some height $h$ of the liquid, the pressure of the water on the bottom of the beaker is $P = h \rho g$ where $\rho$ is the density of the water*. Since $h$ is the same for both beakers, $P$ is the same.

Net force on the bottom of the left hand beaker is sum of the water pressure $PA$, minus tension of string pulling on the bottom:

$$F_{left} = P A - T$$

On the right hand side, the only force on the bottom of the beaker is

$$F_{right} = P A$$

The difference is the tension in the string: as long as the tension is positive (that is, the pingpong ball would float if we cut the string), the right hand side will tip down.


* Note - further to comment below (which may disappear), I do mean "density of the water", and not "effective density of water which is lighter because mass / volume is smaller as there is a bit of volume without water in it". That is just how hydrostatics works: the pressure will be that caused by the column of water, and it will be the same at every point along the bottom. To state otherwise is to perpetuate the confusion.

share|improve this answer
    
but the density of the liquid isn't the same in both sides. Density is mass/volume, the volume is the same but there is more mass in the right than in the left. So the effective density of liquid in the left is lower. The effective density of the right beaker is greater than regular water, but that is when the tension of the string on the right acts to cancel out part of the force. That is what the longer answers basically try to explain –  Jim Aug 26 at 19:47
    
@jim I disagree. The pressure is the same at every point at the bottom of the beaker (otherwise liquid would flow). At a point on the bottom that has no pingpong (or steel) ball above it, the weight is clearly the weight of the entire column of water. Ergo there is no "effective density" argument and pressure at the bottom of both beakers is the same. It would be different if the ball was not tethered because it would float up and the dynamics of that (liquid) motion would complicate things. –  Floris Aug 26 at 19:51
    
The pressure is the same across the bottom. No argument. But density=mass/volume. The mass of matter within the volume is definitely less than the mass of the same volume of pure water. Thus, the average or effective density must be less. Both methods provide the correct answer, they are simply two ways of visualizing the same problem –  Jim Aug 26 at 19:57
    
@jim - agree there is less water. But there is a differential pressure across the ping pong ball (for any vertical column) that exactly provides the additional force (countered by the tension in the string). This is why I reduced the argument to "look at the pressure on the bottom" at which point the answer to this interesting riddle becomes very obvious. –  Floris Aug 26 at 20:03

Hm... I believe the scale will remain in a level position, without any tipping to either side. How I see this is this:

First, I've distinguished two individual bodies of mass: the first being a steel ball attached to a string, further attached to a stand. Since this combined weight culminates onto the stand itself, and not on the scale, the only interaction to the scale boils down to the actual volume of the steel ball. This volume is the only factor that will influence how the scale would tip.

And since the volume of the steel ball is equal to the ping-pong ball, the scale should remain level, because it's about the equal volume of the two balls, displacing an equal amount of liquid, and not about what the volume of each ball is contrived of.

So, I believe the scale will remain level because the stand is canceling out the weight of the steel ball, but not its volume.

share|improve this answer
2  
The stand on the right does not carry the entire weight of the steel ball. Part of the weight is carried by buoyancy, which is transmitted to the scale. –  Mark Aug 13 at 1:12
2  
According to this, doesn't it matter if I replace the ping-pong ball with an unenriched uranium ball of the same volume? You are right that the mass of the steel ball is not so important, but the mass of the ping-pong certainly is. –  rodrigo Aug 13 at 6:52

protected by Qmechanic Aug 13 at 0:28

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.