I thought I would give you a physics puzzle to solve for a change. This puzzle was posed to me this week right before lunch and we had a fun discussion with my colleagues about it. The great thing about the puzzle, is that you can check for the solution in your kitchen if you have a machine to measure weights. Many cooks do, so you can become an experimental physicist for a day if you want to. The puzzle involves typical things that you can find in your home, like water, and a container for it. It also involves lead (in the original puzzle), which is substituted by a blue egg in the picture. If you have never seen blue eggs, go to a traditional chinese food store near your place and ask for duck eggs. So now you have all the ingredients.
So the idea is that you suspend the egg from a string (you can use superglue), and you lower it slowly into a container of water, but the string supports the weight of the egg. You can hang the string from your favorite support, like your finger. The water is seating atop an instrument that measures its weight. The question is: if you lower the egg in the water, will the weight marked in the register increase, stay the same, or decrease. There is more than one way to think about the puzzle, but in the end you should always arrive at the same answer. The really important question is if the weight on the measuring device changes, by how much does it change?
Incidentally, this reminds me of a time when I was in graduate school when lead was found in the water pipes of the university I was in, many many years ago. The water fountains in the building were closed with yellow warning tape, and right next to them a big water jug dispenser was placed so that people could still get water, but not from the tap. Someone with a good sense of humor (I imagine it was a physics graduate student) put signs on the water fountain from the pipes and the bottled water. The signs said “Leaded” and “Unleaded”.
Hey, I remember those water coolers! there were also dark rumors about the university administrators who bought those lead laden coolers, involving varying degrees of incompetence and corruption. No doubt they all found themselves in the Bush administration.
Here’s my answer (don’t read if you’re not done thinking yet!). The weight on the scale goes up as you drop the egg in. The amount it goes up by is 1N per liter of water the egg displaces.
Since the egg is just hanging there in the middle of the water, the bottom of the pot doesn’t know there’s an egg above it. It just knows there’s a certain amount of water pressure on it. That pressure depends only on height of the water level. So the scale must read the same as it would if the water level were at the same spot, with no egg.
Here’s my answer (stop reading now if you haven’t figured it out!): the weight on the scale will go up by an amount equal to the weight of the water displaced by the egg.
This is the same answer given by meichenl, but my reasoning proceeded differently. Without the water, the string exerts a force equal to the weight of the egg. When the egg is immersed in the water, it experiences buoyancy; the weight felt by the string is less by the weight of the water displaced by the egg. This means that the water is exerting an upward force on the egg that must be balanced by an equal downward force on the scale, which therefore registers an increase in weight.
I was also trying to see if I could come up with an argument based on the work needed to raise the level of the water, but off the top of my head I couldn’t quite get it to work.
It’s a fun puzzle! Speaking of which–I recently came across in my local public library a book called Physics for Entertainment by Yakov Perelman. It’s a translation of a popular book written during the early years of the Soviet Union.
While I’m not an admirer of the USSR, I’ve got to hand it to them for thinking that a book about physics was entertaining.
Maybe blogs like this will do the same for us.
I basically went along Todd’s thinking.
If the egg was supported via a Newton meter (or just a spring) and lowered into the water then the spring would contract. This is due to buoyancy, blah blah…
Hi,
Similiarly, put an empty glass jar + lid on the scale and weigh it.
Then, put a fly in the jar and cap the jar.
What does the scale read when the fly is alit on the jar ?
What does the scale read when the fly is flying around inside ?
Mass is conserved. (Mass)(gee)=weight. The closed system is conservative. If the egg supported in mid-air by your finger lowers its weight by displacement of fluid, the weight of the container, fluid, plus immersed egg increases in kind. Inertia of the closed system remains unaltered either way, then Equivalence Principle. (Possible EP parity violations – chiral egg proteins – are restricted to less than ~10^(-12) relative by thermodynamic considerations).
If the jar is sealed, add the weight of the fly whether perched or airborne. If the jar communicates with surrounding air, do not add the weight of the fly when flying (to first order). When a helicopter or plane buzzes low you can feel the pressure.
Oh, puzzles, I love puzzles!
Let’s go off a different path, and imagine the egg is filled with water instead of eggish stuff. You may not want to try this at home, specially during breakfast, but I always skip it anyhow
Imagine we put it down slowly again. Well this is not the exact same situation, but we get to ask the same question. Imagine somehow we have some invisible painting, (or Harry Potter’s hood) and we carefully painted the egg, and the string, with it before submerging it on the water, so what we see now is an eggy shaped ball of water *disapearing* into the pot. Let us assume we can ‘neglect’ the ‘small’ effect of the shell (and string) (remember physics is about approximations
). So, what happens now?
First we have an egg shaped ball of water and as we submerged it inside the pot all we ‘see’ is still a pot of water, but having more of it
. So, how much will the scale read now? I’ll let u figure that out… Remove the string and shell and think what happens again… 
I think Archimedes would have put it this way: Take the pot with water, submerge the egg inside as directed, freeze time, remove the egg. The mass of the original bulk of water and the one with the empty hole are obviously the same. So what is it that changes when we put back the egg or we fill it with water?
Imagine also we cut off the string, so what is the scale reading now? …. before the egg crashes at the bottom…and afterwards?
R
ehm… Archimedes…
Hello and congrats for your blog
Well i don’t know. What i know is that if the tension of the string is the same before and after the dip, then the weight will remain the same. It’s like holding it above the water.
BR
Hello. I’ve read your blog a couple of times, and this is interesting stuff.
If I had to say what would be the outcome (although I cannot see the picture) is that, it depends on how deep you’re submerging the egg and the lenght of the string.
At the begining, when the egg is not submerged, the only weight is due to the mass of the water.
If the egg where fully submerged, the weight would be that of the sum of both water and egg (because the buoyancy only acts “locally” or “internally”, changing the “shape of the system”), because if we apply Newton’s 3rd law, the egg will apply the same force downwards (the buoyancy it feels), so the total weight would be exactly the same.
On the other hand, if it touches water level, and until it reaches the point in which the spring becomes slack (time at which it will not be exerting any tension upwards), the weight will vary from the “empty” water and that of the sum. (We should take into account that the only variations due to this should come from the combined effect of buoyancy and gravity; gravity will stay the same, while buoyancy will depend on how deep we’re submerging the egg).
I hope that to be the correct solution :S
There is a story about Richard Feynmann and the guys at MIT arguing about what would happen if one put a rotating lawn sprinkler into a water tank. In normal operation, one would expect it to go round, just as it would in air (wouldn’t one?!), but what happens if the pump is reversed to suck in the other direction? Go “forward”, “backward”, or not rotate?
One of the MIT chaps complained that Feynmann on three successive days convinced him that each of the three answers was correct!
For physics-literate students, perhaps the most subtle use of these puzzles would be to generate a set of true and false arguments about what would happen, then ask the students to determine which is correct, which is incorrect, and where the fault is in the incorrect ones. Really subtle puzzles could have the right answer for the wrong reason!
Hi,
I agree with Todd – answer wise. I just drew a free body diagram (yes, you caught me I’m an engineering student not a physics one) and it seemed to provide me with said result.
On an unrelated note I read Y. Perelman’s “Physics can be fun”, “Math can be fun” and “Physics for entertainment” some time back (I guess I was in the 8th or so), and I loved it. I wish there were more books like this to allow me to get a better feel for the subject. I also remember George Gammmow’s book on physics which I read during my 8th I’d suggest it as a reference any day and I loved the way he brought the fundamentals into focus.
Nice puzzle. Your egg will displace a volume of water equal to the volume of egg that is submerged. The scales will register an increase in weight exactly equal to the weight of this water and the tension on the support string will decrease by the same amount.
It reminds me of a similar puzzle I heard a long time ago which can be solved in a similar way.
You are sitting in a boat in a small pond. You throw a lead brick out of the boat and into the water. Does the level of the water in the pond go up, down, or stay the same?
Lead brick out of boat into lake: The level of the lake will go down, because when the brick was in the boat (assuming it was floating) the boat displaced enough water to equal the weight of the brick. Tossing the brick out cause the boat to rise out of the water lowering the level of the lake, while the brick will displace (and slightly raise the level) water but not as much as when it was in the boat.
This brought to mind a time when I was stationed in Germany. I regularly traveled a road that went under a structure that I assumed was a trestle for railroad tracks, until one day I saw a large ship going over this. A surreal moment.
I asked my friend with me to consider that there was a scale to weigh the trestle just before the ship got to the trestle then again once it was in the middle – what difference would the two readings be?
He got it wrong.
Hi David
You won’t give the solution for this?
BR
dale,
is the answer that the weights are equal as the ship must displace from the trestle an amount of water equal to its own weight in order to stay afloat?
[...] I will go on a bit about flotation. As a matter of fact, some of you might remember a puzzle with an egg I wrote down a while ago. Of course, most of you have probably heard of Archimedes Principle as [...]