# Physics Riddle

by micromass
Tags: physics, riddle
PF Gold
P: 6,500
 Quote by Vanadium 50 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.
 Emeritus Sci Advisor PF Gold P: 16,462 No, the ping-pong ball is still hollow. If I wanted to replace it with something solid, that would be an extra step.
P: 861
 Quote by Doc Al ...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 doesnt 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
HW Helper
P: 7,176
 Quote by gmax137 The ball on the right could be lead or uranium, anything that doesnt 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.
 Sci Advisor Thanks PF Gold P: 12,256 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?
Mentor
P: 41,568
 Quote by sophiecentaur 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...
PF Gold
P: 1,435
 Quote by Doc Al 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.
PF Gold
P: 2,031
 Quote by sophiecentaur 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).
 HW Helper P: 3,464 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).
 P: 759 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.
P: 504
 Quote by TurtleMeister 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.
P: 759
 Quote by Buckleymanor 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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