A Balance of Power: Newsletter Challenge (February 2013)

If the steel ball in the left pan (not pane ) is suspended from the ceiling, then there is no weight gain in the left pan, but the water level will rise by the volume of the steel ball.

If the plastic ball is suspended in the same way, then the water in the right pan will rise by the same amount, and again there will be no additional force on the pan, so the two pans remain level.

If the plastic ball is anchored to the right hand tank in some way to keep it submerged, then you get the water level rise, plus the additional weight (force) of the plastic ball, so the right hand pan will be lower.

How much lower?

That depends on the contents of the hollow plastic ball. If, as one might assume, it is filled with air, then the difference will be difficult to spot (depending on the measurement system employed).

If the hollow centre is filled with, say, concrete, the difference will be marked.

If the hollow ball has, like most hollow plastic balls I've seen, a small hole in it, then it will fill with water, which means the water level will rise only by the volume of the plastic shell, rather than the total sphere, and the height difference between the pans will be the same as in the first case.

If the ball is filled with helium, I think the right hand pan will be higher.

Good explanation, and a GA.

I was having a little trouble seeing it. This clears it up nicely.

Hi ER

If both balls are suspended from over head so that the balls displace the same amount of water. Wouldn't the beam scale tip up if accurate enough. To the one that has the steel ball. The adhesive force of water to steel which it do believe is greater to most plastics. These adhesive forces helping to suspend some the weight of the water.

Lots of added assumptions, Concrete, Holes in balls and Helium. None of that is mentioned in the OP question. But then Chaos is always present.

Yes, if they overflow.

And the panes will get stained.

Okay. I want to be first to say this!

Obviously, there is not enough information!

1. Where is the balance in relation to the water?

2. Are the two containers of the same size and material?

3. Are the two containers of the same width?

4. Are the two immersed objects held with the same volume of material (one is suspended, yet the other is pushed into the water)?

5. Tides; where is the Moon in relation to the two containers?

6. Was this performed on a Tuesday?

7. Are the panes stained or clear?

8. Are the two containers filled with the same liquid and not one with say heavy water and the other with, say, seltzer?

9. Are the two containers and their liquids at the same temperature?

10. "Are they going to remain at the same height?" We have not defined the inertial frame of reference?

11. Have the Earth's, our solar system's, and the Milky Way's magnetic fields been taken into account?

12. What is the material of the steel ball and is it oxidized?

13. Exactly how is the plastic ball held under the water? Provide a proof of 500 words ±10 words with a certificate of conformance..

There are many other questions - too many to list here, but it is clear that we just can not go on any further without clarifying these questions.

.9 = see .2

3. Are the two containers of the same width?

Why is this important to know?

The maths show that what we need to know is the amount of the displaced water.

Even if the container is conical, with the pointy edge on the pan (i.e. width at contact virtually zero), still, the increased pressure will be felt by the inclined sides of the container and - being it a solid - the extra force will be carried all the way down and eventually appear at the tip.

"What will happen to the balance panes?"

1. What balance panes?

2. What is a "balance pane"?

Frying pan....

Pan creature mythological....


Pan......Am

Depends on how the plastic ball is held if it's weight will be contributory or not....

Hmm... but Q1 is still unanswered.

I assumed that the containers were on a bench, or maybe on the floor - then along comes this talk of "balance pans". They could be in the lab next door for all we know.

"Balance panes" has been corrected to "balance pans" - sorry for any confusion.

Well, that just steals all the fun out of this, doesn't it? :)

Although, one can always wonder if the steel balls were hanging from the rear bumper of the car in traffic ahead of you at one point.

Creates a new twist on the "oxidation" theory.

If both are KEPT immersed as you phrase it, regardless how this is accomplished, e.g. their own weight or a dimensionless and weightless (!) arm, only their immersed volume will have effect on fluid. So, two containers will still balance. But, If your "suspended" word is strict, AND hollow plastic ball's density is lower than water, it will NOT immerse, it will float, and things are different, the container with the hollow ball will have added weight equal to the percentage of ball's immersed volume, by the water density, so it will be lighter or equal to the first case, so it may unbalance upwards. S.M.

Nope.

Europium got it right. The container with the hollow ball will weigh more because there must be a net force applied downward on the ball to keep it submerged.

That means there is a net upward force in the water (buoyancy) that acts in the opposite direction with equal magnitude, so the second container will weigh as much as it's twin + that of the downward force applied to the submerged ball.

Then again, like previous challenge questions, there is a lot of missing information. :)

The force you are talking about IS the one I imply in my first case (applied by a dimensionless and weightless arm) and IS exactly equal to the difference of the actual hollow ball weight (out of water of course), and an equal volume sphere water weight. So the net weight the container will see added, IS the total weiht of an equal volume sphere, filled with water plain and simple. (The actual plastic ball's weight, IS canceled out of calculations on the forced immersion case). Of course suspended steel ball will add exactly the same weight. Think about it again. It IS a bit tricky. (LOL) S.M.

There is no difference in the magnitude of force required to submerge the plastic ball vs the steel ball so the balance will remain the same. It is the same as placing the steel ball directly on the balance, then pushing down on the opposite pan using a plastic ball as the contact so they are level.

Unless the plastic ball is held submerged by a mechanism attached to the pan, then the applied force will not be external and the pan will be unbalanced by the added weight of the internal mechanism.

Correction; I should have wrote "...mechanism and ball"

Here is a simple experiment that will prove to you what happens in the second case where a buoyant object is forced down into a container of water.

Fill a bucket part way with water and hold it in the palm of your left hand.

Take a second, slightly smaller empty bucket, pan, or can and using your other hand push that second container down into the first.

As you push downward with your right hand, what do you feel with your left hand?

If you are not sure, attach the first bucket to a fish scale and repeat the experiment.

Now what do you say? :)

The downward force will increase by the change in depth of the water in the big bucket. this in no way invalidates my point. Use the same setup, now have someone carefully pour some ball bearings into the smaller bucket, they will reach a point where the smaller bucket is at the same depth but the amount of push from your hand is gone, gravity has taken over, put more ball bearings in and you will have to hold it up or it will sink in deeper.

In the original, the difference is that the metal ball would need to be held down if gravity wasn't doing the same job as the rod. otherwise, where does the pull in the rod for the metal ball come into the equation.

A side issue. If the buckets were tall, then compression in the rod for the plastic ball would vary with its depth in the water, flotation would increase with depth in the water.

It's late and I'm for my bed, I'll come back in the morning.

As long as you are pushing down, you will feel the force downward ( case: not reaching the bottom of the bucket). But when the immersion is complete and you stop pushing down, the force will disappear and you will not feel it any more. this experiment does not illustrate the problem completely.

In the challenge, We have, as fixed data:

1 plastic ball is immersed completely, but can be suspended from the ceiling or maybe anchored at the bottom of the container.

2 the steel ball is suspended from the ceiling Only. It can become immersed or not due to the movements of the pans.

3 the 2 containers are identical: implies the same types and the same liquids and the same initial quantity / height of liquid (presumed to be Water).

I think that ER has gone through it all, but it could be summarised clearly.

That is a problem, we do not know how the plastic ball is immersed in the liquid.

It takes force to do that. If it is tethered from the bottom, then the net change (from the scale's perspective) is essentially nil.

If the ball is pushed down into the fluid, that is another matter, because the force of buoyancy must be added to the "weight" imparted on the scale.

I agree (assuming that both balls are fixed to the ceiling by rigid rods - if the rods are identical, it doesn't matter how thick they are, within reason).

Since the weights of the balls can have no effect on the balance (they act on the ceiling only), it doesn't matter whether they're made of aerogel, platinum or unobtainium. They both displace the same volume of water, so the balance remains level.

GA

Yes.

When a body is suspended in a liquid, it loses a weight of the volume of body. It has nothing to do with the weight or density or any thing else

The balance pans will remain the same since there was never a procedure stated that the containers were placed in the pans. It's like asking, "How far can a dog run into the woods?"

RE: Dog Halfway, then he's running out.

Everything should remain equal, as long as the 2 balls are given the same mounting mechanism. Not using a string on the steel ball would be my suggestion, but a rod. The same with the plastic ball, mounted with the same type of rod, of course from above.

The interesting thing would be to see what happens when you start playing with the temperature and take it to just beyond the freezing point and/or boiling point.

The water exerts pressure on the bottom of the bucket, this is the Head and is the density multiplied by the depth. The force is the head multiplied by the area under pressure. it doesn't matter what foreign article is floating or submerged as long as it is not in contact with the bottom, the only force on the bottom is the water. The only force the bucket can place on the scale pan is the water force plus it's own weight. You could float an ocean liner or run a submarine under the surface, in the bucket, the pressure on the bottom would still be the density by the depth. hence my contention that with equal volumes, there would be no difference. The pressure at the river bed does not change when an ocean liner passes over it,

In the problem, some are suggesting that the exterior force holding the plastic ball down is felt by the scale, but the metal ball is also acted on by an exterior force - gravity.

"The pressure at the river bed does not change when an ocean liner passes over it"

Then what holds the vessel up?

That's like saying putting a candle on a table does not change the force that the table exerts on the floor.

It is not the density, pressure, or head of the water that matters to scale, it is the mass of the water and whatever is floating on the water that impacts the scale.

In the case of suspending an object of equal or greater density into the container goes, the mass of that object is essentially not a factor (assuming it is not neutron star material) in the amount of force exerted on the pan of the scale.

However, if an object floats on or is pushed into the water, then there must be a change in force on the scale's pan because there is a net force exerted downward to submerge that object and a corresponding opposite and equal force pushing back up.

Indeed, and well said.

Immersing the ball in the water has exactly the same effect as adding that same volume of water to the container, with the result that the container now exerts an additional force on the pan over and above the force it already exerts.

Simply because the water is displaced is no reason to believe that the buoyant push it is giving the ball is not also felt elsewhere. That water is pushing back, and for it to push back, it, in turn, must push against something else: the container. That, in turn, pushes against the balance pan which, in turn, registers the force by moving downward.

Another way of looking at it is this: given that the water is only aware that it has been displaced, and given that it does not know (nor does it care) what the ball is made of, the ball might as well be made of water - the same amount of water that has been displaced. The water doesn't care how that ball is constructed, but only that the volume the ball now occupies has displaced an identical volume of water, nothing more.

GA

I think we are saying the same thing in different ways. I misread your #1 or I would not have bothered to post.

I get what you're saying now. By immersing the plastic ball in the water, even though you force the ball to a depth, being that the scale is not a stationary surface it will move down to a point that the plastic ball floats. This being said the side with the plastic ball will move down. The force used to submerge the ball must be constant, and with no resist until the pane bottoms out, or gets an equal pressure from the opposite pane it will continue to move.

Are we at cross purposes here? we may be mixing concepts and semantics.

Consider a canal, a boat is off to the left and moving slowly towards a pressure sensor, the water level is constant as the boat passes over the sensor and continues on it's journey. I maintain that the sensor shows no change during this pass. The water displaced by the boat is distributed over the whole body of water. If instead of the boat, we have a traveling crane supporting a wrecking ball submerged in the water, we would still see no change on the sensor. If the wrecking ball displaced the same amount of water as the boat, the surface levels would be the same and the one could replace the other and the sensor would read the same.

Your example is a little different than what I was thinking.

I am thinking of the container and the object as an isolated system with two different states; one is the steel ball suspended into the container, the second is a buoyant object pushed into the container's fluid instead of the suspended steel ball.

The ship in the canal is analogous to that buoyant object already in the container, but you are simply moving it around in horizontal circles.

That's what I suspected, when I realized I had misread #1 the first time round, plus I think I approached the solution from a different point having had to cope with the design of in-ground tanks and basements in floodplains. I thought I read that the push down to submerge the plastic ball would add to the downward force, I took this to be in addition to the raised water pressure.

I modified the problem to give me a better comparison, I thought of a plastic ball in both buckets, forced down by tubular struts from a fixed ceiling. To this point they are identical, I then considered what would be the effect of filling one ball with mercury via the tubular strut. It made no difference to the scale balance although the strut became a tension tie as gravity reached then overcame the buoyant force on the ball.

Assumptions: 1. Both containers are on opposing pans of a 2-pan balance (one cantainer per pan) 2. When "filled" with water to the same height - the containers are not fully filled, i.e. they could continue to hold additional water at least equal to the volume of the balls without overflowing. 3. The water used to fill each container is from the same pool, i.e. the same composition, purity level, dissolved gas content and temperature. 4. Both containers are centered on their respective pans, i.e. no unequal lever action. 5. There are no discontinuities, such as electric or magnetic fields, or air currents due to wind or thermal effects that apply unequal upward or downward forces to either ball/container/pan.

Situation A. Stiff suspensions (one for each ball of the same diameter and length) are used to suspend both the steel and hollow plastic balls within their respective containers but above the surface of the water from a point external to the balance (the ceiling). Doing this will not disturb either of the balance pans. They will remain at the same height. Neither pan will "feel" any additional force, as the weight of both balls will be totally supported by the means of suspension from the ceiling.

Situation B. Starting from situation A, the "ceiling" is lowered sufficient to completely submerge both balls below the surface of the water. Now both pans will be equally supporting the weight/force of the water displaced by the equal volumes of each respective ball. Assuming this force does not exceed the capacity of the balance, the balance pans will continue to support the load and they will remain in balance (neither pan will have height advantage).

Situation C. Here is where "means of support" make a difference. Assume as stated in the problem that the steel ball is suspended from the ceiling and located below the surface of the water - say by a nylon fish line glued to the ball. Its volume will displace the weight of water equal to its volume. Now assume the hollow ball is attached to a nylon fish line with glue and that the other end of the fish line is glued to the inside bottom of its container, such that the length of nylon fish line is short enough to ensure the hollow ball stays submerged. By hollow I am assuming the cavity is "filled" with something lighter than water (like air or a vacuum), such that without the fish line the hollow ball would float on the surface of the water. I will also assume that the length of fish line between the bottom of the container and the bottom of its hollow ball is the same diameter and length as the fish line below the water surface that holds the steel ball in place, so that their displacement effect cancels each other. Now the pan with the hollow ball MIGHT drop slightly lower than the other pan due to the weight of the hollow ball that is fully supported by that pan of the balance, unless whatever "fills" the hollow ball makes that ball lighter than air, in which case the bouyancy force of the hollow ball would lift its pan of the balance slightly higher than the other pan.

Situation C could could also be thought of by reversing the experimental procedure a bit. Start by suspending a steel ball from the ceiling within an empty container atop its pan of a 2-pan balance. Then place an identical empty container with its hollow ball in it with fish line attaching the ball to the container. Now without the water present, the weight of the hollow ball in its container will be slightly greater than the other container with a steel ball supported by the ceiling and the pan with the hollow ball will be lower. Next add an equal mass of water to both containers, such that both balls are submerged, without overflowing the containers. Because the same weight (mass) of water has been added to each container, the effect of adding the water will be the same in each container, hence the pan with the hollow ball will continue to support the additional weight of the hollow ball and be positioned a bit lower, although with the water in each container the sensitivity of the two pans to further change/disruption will be reduced. As before however, if the hollow ball, unless tied down, were to rise in air, then both before and after the addition of water, its pan would be slightly higher than the one with the suspended steel ball (supported external to the balance).

The steel ball has zero bouyancy to overcome, the plastic ball does have bouyancy. Their lies the force not required on the steel ball side?

"Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object."

- Archimedes of Syracuse

I get that, and the plastic ball's weight does not displace enough liquid to submerge, but the metal balls weight overcomes the buoyancy. Once suspended beneath the waters surface the weight of the left container has increased in weight. That increase in weight is equal to the the weight of water have the same area as the ball? Is that right? If that is correct, then submerging the plastic ball would do the same, if it could be done without adding any force upon the water or container. They would be equal if this is possible. This is where I am at. How do you submerge the plastic ball without introducing any force or adding the weight of the plastic ball to the container? Is it possible?

Given that both submerged balls are held stationary, the water doesn't know how buoyant they are. They could both be hollow plastic balls or both steel. It makes no difference to the water. The water wants that space back and it is going to push against those balls to get it. Given that both balls occupy the same volume, the water pushes back at both equally. In the case of the steel ball, this push reduces the tension on the support rod but does not negate it completely, with the result that the steel ball appears lighter submerged than it does when dry. In the case of the hollow ball, the submerged ball now appears to have negative weight where, when dry, its weight is slightly positive. The submerged vs dry difference in weight for each ball is the same, however, and is equal to the weight of the water displaced by each.

Okay, I have learned much here today. just to satisfy my curiosity: If the steel ball had 2 times the area of the plastic ball would they balance at that point?

It is not the area of the object but the volume of water displaced. Were one ball instead a cube - one having the same volume as the original ball but, being a cube, having a somewhat larger surface area - the result will be the same. The object could have thousands of times the area, say, a fibrous or porous object for example, but one having the same volume, the result will still be the same. Not the area, the volume, only.

So buoyancy is a constant based on volume? If an object submerged has a volume of 10 cc and a dry weight of 50 grams, in water its weight is 40 grams. Then a different object with a the same volume of 10cc and a dry weight of 30 grams, in water it weighs 20 grams? That is if 1cc of water weighs 1 gram. If this is what you're saying then each container (if it was this was applied in the scenario) would equally increase by 10 grams and remain balanced? If this is correct, I have got it, if it is incorrect I give up.

You've got it, but do you understand why it works out this way?

Yes, because if the plastic ball is secured by a rod it might as well be solid like the steel ball. I was thinking it through as if instead of the balls, if there were to smaller open top containers rigidly mounted from above, and they were half submerged in the water. Filling the either side with water would impart nothing onto the water in the lower container. The additional weight is carried by the mounting mechanism. Also I figured out that if I suspend a weight from a rubber air filled ball that it will float on top of the water. If it is submerged to the point that pressure of the water crushes the ball to have a volume less than the volume of the equal weight of water, it will continue to sink. Is that correct?

no the right one will weigh more as the weight of the steel ball is suspended from above while the plastic must be anchored from below in the container thus adding the weight of the ball and anchor material

Why must the plastic ball be anchored from below? Both balls could be suspended by rods from above.

If both suspended from ceiling with rigid rods, balance stays level. Otherwise container with plastic ball is heavier, whether ball is immersed or floating.

"If both suspended from ceiling with rigid rods, balance stays level."

I think it would not. The case of the plastic ball would be different from the steel ball because of buoyancy.

Think of a large bucket 3/4 filled with water sitting on a scale. Now add a second, smaller, bucket that is empty so that it sits or floats it on top of the water.

You expect the scale to now include the mass of the second bucket, and it will.

Next, using your hand, force the smaller bucket downward into the water of the second bucket. You will feel an opposite force as you try to forcibly submerge the second bucket.

The scale will now read even higher. The exact amount read by the scale will be the mass of the large bucket + the mass of the water + the force applied by your hand downward.

In the puzzle, your arm is replaced by a ridged rod and the smaller bucket with a hollow plastic ball, but the scenario is the same.

The steel ball being the same volume as the plastic ball, has the same buoyancy. The weight of the steel ball just exceeds it's buoyancy. If it was 10 grams, the 2 gram plastic ball would float, the 50 gram steel ball would sink. In the case of the steel ball in would weigh 40 grams. Now if you add 9 grams to the plastic ball without changing it's volume (like putting 9 grams of lead inside) it would sink, but it would still have the same buoyancy. It would weigh 1 gram in the water, and 11 grams out of the water. The the steel ball is the same, weighing 40 grams in the water and 50 grams out of the water. The water supports this weight weather it is held by a string or a rod. Europium explains this well, here in his posts, it took some beating, but I finally got it.

A.H. Here is where we parted. In your bucket example, the increase on the scale, when you push the small bucket down, is the weight of the water displaced by the small bucket. As the small bucket is pushed further in, the water level around it rises. The scale sees the bucket + the new depth of water by the cross section, the upward force on your hand is the weight of the displaced water.

Imagine that instead of a bucket, you have a sealed container except that it has a tube to it. If you pour mercury into it, it becomes the metal ball, you don't have to push because gravity is doing it for you.

I see both of your points, I just need to construct an experiment that would prove/disprove it, I guess.

In both cases (suspended steel ball and forced buoyant ball) the scale increases in value.

It is not clear from the puzzle how the plastic ball is held submerged; just that it is, and so in my solution (at any rate) I covered both cases. For the case of the plastic ball's being fastened to its own container, any means will work provided identical volumes of water are displaced on both sides of the balance.

If the plastic ball is floating on top (as suggested) that container should be lighter than the container with the submerged steel ball, as the plastic ball is lighter than the water displaced by the steel ball. Well that is what I think have taken from this discussion.

The submerged steel ball contributes nothing to the weight acting on that side (container + water). Adding a plastic ball to the other one will increase the weight (container + water + plastic ball).

How would you get the plastic ball to submerge? You'd force it down, correct? that force would be applied to the container and the water. The force it would take would be equal to the weight of water displaced. The weight of the steel ball applies this force without your assistance. If the steel ball weighed 50 grams in the air and 40 grams in the water. The the scale will see that 10 gram difference. When you put the steel ball in the water it was like adding more water, it got deeper.

If you applied force to submerge the plastic ball, then tethered it to the bottom of the container, you would do work (= force X distance moved), but once it's in and tethered, that's it. Total weight is container plus water plus ball. The weight of the steel ball acts on the ceiling; it cannot add to what the balance pan "sees".

That is what I was thinking before, but then where did the 10 grams go? If the steel ball is tethered from the ceiling on a string, the tension on the string is 50 grams while the ball is in the air. When it submerges into the water it becomes 40 grams. Why? because 10 grams is supported by the water. So the water weight increases by ten grams and the depth increases the same as if you had added 10 grams of water.

Trying to sort this out in the pub - now on my 5th pint. Best leave it til tomorrow. Sorry.

Imagine a U-shaped glass tube with tall sides and a built-in stand. Fill the 2-pint tube to the one-pint mark with your favorite grog, taking care not to drink any. :) Place the U-shaped tube on the scale which I just happened to have here in my wallet and have placed on top the bar next to your replacement pint which I've placed on my tab in compensation for my making you suffer through this epic tome.

Next, open the pressure-relief valve on the hollow plastic frictionless piston which magically (being of German manufacture) fits perfectly into either side of the U-shaped tube. Pick a side, any side. Using the attached rod, slide the piston into the tube until it is just flush with the surface of the grog and close the valve.

Whilst holding the rod in a manner which exactly counters the weight of the piston/rod assmbly, tare the scale.

Now apply downward pressure to the rod, forcing the piston downward until the top of the piston is even with the one-pint mark. Hold the piston in that position and read the scale. Is it still zero?

Next, using the supplied magic wand (also of German manufacture), wave it before the piston, causing it magically expand and lock in place, then release your grip on the rod.

What does the scale read now?

Next, open the valve, unlock the piston and remove it and, finally, drink the grog. Read the scale if you like, but I wouldn't bother. If I were you I'd shut off this damn phone right now and check out that HOT redhead wot's been lookin' at you all this time :-)))

I just performed a simple experiment for you Ichapah. I took a 14 ounce water glass and filled it 7/8 full of water and placed it onto my kitchen single pan electronic balance. I then tared the balance so that the digital display read zero. Then I pushed one of my fingers into the water, being careful not to touch the side of the glass anywhere. As I extended my finger deeper into the water I observed an increasing positive value on the balance display, showing that even though my finger was supported by my hand/arm etc. it was still effectively communicating a force to the balance equal to the weight of the volume of water displaced by my finger. For the record approximately an inch of my finger below water corresponded to a 10 gram addition to the balance reading.

I did some experiments too. I used a couple of measuring cups, a ping pong ball, and a solid plastic ball nearly the same size. I marked the glass, captured water coming out of the pour spout, and used a letter scale. I played with this a little yesterday and today. I went far beyond what is discussed here. I thank you for confirming that though. I just learned all this from Europium on this blog.

The problem statement explicitly states that the hollow plastic ball is 'immersed'; that is, submerged completely. It is not 'floating on top'.

Yes I got that. I was just pointing out that adding the the weight of a ball that floats would be less added weight than a steel ball submerged. (I replied to the wrong post)

Ah. No prob.

This is not wrong. See post #68. I actually did this.

Yes , of Coarse!

On second thought - the answer would only be the same if the steel ball was heavy enough that it was below the surface - not floating - then the displacement would be the same!

There is some missing information, so I'm going to assume a couple of things.

1. This is a question about buoyant forces on balls, and not a trick question like "Can you stick out your tongue and touch your nose?" or "Kits, cats, sacks and wives, how many were going to Saint Ives?"

2.The identical containers are actually sitting on the balance pans of a working scale and they are at the same height before the balls are added.

3. The challenge says that the steel ball is hung so that it remains inside the left container, but it does not say that it is hung so that it is immersed in the water. It could be suspended above the water. It also does not say that the steel ball is at the same height as the plastic ball. It does specify that the plastic ball is indeed immersed in the water. In this answer I will assume that the steel ball is suspended at the same height as the plastic ball and that they are both immersed in water.

4. The means by which the steel ball and the plastic ball are held may differ in volume, and that is actually likely. The steel ball could be held by a string, but the plastic ball would need something rigid like a rod to push it under the water. For this answer I will assume that they are the same volume.

My answer is that the pans will not move. The movement of the pans depends only on the volume displaced by the balls in the two containers. It does not matter what they are made of or how much they weigh. If they displace identical volumes of water, the water inside the containers will rise an identical amount.

The buoyancy of the two balls, the upward force exerted by the water in them, is the same. The fact that neither ball will actually float is important, but they will not float for different reasons. The steel ball will not float because it is denser than the water but it "loses" weight equal to the amount of water that it displaces. The plastic ball will not float because it is being held down by a strong force exerted through the rod, but it likewise "loses" weight equal to the amount of water it displaces, and its weight actually becomes negative. In other words, it would float if not held under the water. The effect of the two balls is the same.

The upward buoyant force is the same for the two balls. That means that the containers are pushing them both upward with the same force. That also means that there is an equal force pushing both containers downward, and the pans will not move.

Another way of stating this is that the water level rises an equal amount in both containers, and that additional amount is what the scale is weighing. It does not depend on the weight of the balls themselves.

Dear Canary,

The method of support of the hollow ball does matter in whether the balance "sees" the weight of the hollow ball or not. If a rigid rod external to the container holds the hollow ball under the water, the weight of the hollow ball does not matter. That ball's weight (like that of the steel ball) is not seen by the balance. Only the displacement force of the water displaced by the hollow sphere which equals the volume (and displacement force) of the steel sphere comes into play. If however, the hollow ball is held under the water by a string or rod attached to the ball and the inside bottom of the container, then the weight of the support and the hollow ball are "seen" by that pan in addition to the volume displacement.

The weight of both balls much be accounted for. In the case of the steel ball, the problem states it is supported external to the balance. Hence ONLY its volume affects its pan position, whereas we are not told where/how the hollow ball is supported, meaning its weight must EITHER be supported external to the balance or BY the balance, in which case, its pan would be supporting more weight and its height would drop.

Dear ARH,

Yes, you are absolutely correct. When I read "keep it immersed inside the right container" I understood it as using an external means of support pushing it into the container from the top. Thanks for catching that assumption.

However, I think I disagree with your conclusion. I think the plastic side would rise instead of drop. First, the buoyancy of the steel ball is pushing its side down by the weight of water equal to the volume of the steel ball. Call that 1 unit of force.

Next, the plastic ball is not being supported externally, so its weight is added to the right container. Since it is hollow, it weighs much less than the equivalent volume of water. Call that 0.1 unit of force. However, its buoyancy is not added to the right container. In fact, it could be immersed or floating, and it would not make any difference to the weight of the container. Its weight of 0.1 unit exerts less downward force than the 1.0 unit buoyancy of the steel ball does. Therefore it will rise and the steel ball container will fall.

Canary

That would be correct as the scale is made to measure the force of gravity only. The force of the buoyancy would be negated by tension on the tether of the floating ball. When suspended from above it was compression on the rod suspending the floating ball. A clearer way of seeing this would be 2 balls that float but have different weights. They displace different amounts of water and add only a downward force equal to the effect of gravity. So a ball floating half submerged only adds its weight in air to the container. A ball that floats 3/4 submerged adds only it's weight in air to the container. If you submerge these balls from below with a tether there is no gravity or external force being added in a downward direction. Just tension on the tether.

From what I see I think everyone is making it more complex than it really is. The volume of the water displaced is the same if the 2 spheres are identical, but the weight of the steel one is suspended from an external location so it is not added to the total weight of the container. The weight of the hollow ball is added to the other container. Therefore the container with the hollow ball will weight more. This is assuming that the ball is totally immersed and not floating and that was a given condition.

The container with the hollow ball will have more weight and the balance pan will be lower to reflect that.

"It can scarcely be denied that the supreme goal of all theory is to make the irreducible basic elements as simple and as few as possible without having to surrender the adequate representation of a single datum of experience."

~ Albert Einstein, from On the Method of Theoretical Physics, the Herbert Spencer Lecture, Oxford, June 10, 1933

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"In every field of inquiry, it is true that all things should be made as simple as possible - but no simpler. (And for every problem that is muddled by over-complexity, a dozen are muddled by over-simplifying.)"

~ Syndicated columnist Sydney J. Harris (paraphrasing Einstein), 1964

In the case of the hollow ball being attached to the ceiling, is the method of attachment not in compression? If so, removing the attachment (rod, if you like) must surely mean that there's less force acting on that side of the balance? (I thought it was all sewn up too, once :-) )

The pan with the empty ball would weight more. since the steel ball is suspended in the other container, the weight of it means nothing.

Dear Mr. Big,

You are right if the hollow ball is supported by its container. But if like the steel ball, it is supported from the ceiling or some other point external to the balance and containers via a rigid support, then its weight like that of the steel ball is of no consequence and then neither pan would weigh more.

True, that was the assumption. i was picturing the steel ball being supported by the ceiling. and as for the hollow ball i was picturing it being attached in the same manner. but attached to the floor of the container. and since the liquid remained at the same level. i pictured the same size support was used for both test subjects. so basically, i was imagining a mirror image of the set up with the liquid surface being the starting point

Gravity is a downward force opposing buoyancy. To have an object sink gravity (weight) must overcome buoyancy. For an object to float buoyancy must overcome gravity. Both are always there. You added gravity to the floating ball side (weight of floating ball). You have to add buoyancy to the sinking ball side (buoyancy of the sinking ball).

The balance pan stays inside a gravitational environment!

Then the left pan with the steel ball sinks down!

a balance pan measures the mass in a gravitational environment or the force of the gravittional environment to the mass; not the volume of this mass!

because the mass of the steel ball adds to the mass of the watervolume the force to the gtravitational center is higher than the force of the water and the hollow plastic ball!

How much mass does the suspended steel ball add? If all the mass were added, there would be no tension on the string from which the ball is suspended, and so the string would go slack. But the string remains taut, and so this added mass must be less than the mass of the steel ball, yes? If, on the other hand, the steel ball added no mass at all, then your assertion is incorrect. What remains, then? A mass somewhere in-between All or Nothing, yes? If so, what determines this mass which is added? Is any ol' mass good enough, or does this mass have a specific value, based on some physical property of the steel ball? The mass of water displaced by the ball, perhaps? What displaces water? What IS displacement, anyhow? *Volume!* How would we determine that volume? By the mass added divided by the density of water perhaps? So it would seem that mass and volume are inextricably mixed here! True, a balance does not measure volume, not directly, anyway, but the volume is determined easily enough by measuring the increase in mass when we immerse the ball. It is such a good method to determine volume that Archimedes used it to discover the King's jeweler had robbed the King! Seems the jeweler had substituted a lesser metal in making the King's crown, in place of the portion of gold he secretly kept for himself - and was beheaded for his theft! Volume plays a key role in solving this problem, as well. Well enough in fact that I could estimate the ring size of another poster's index finger based solely on the results of his experiments here. :-))

if the steel ball sinks onto the ground, the difference weight is the difference of the density of the steel ball (7.85 kg/dm³) and the plastic ball (<1) multiplied by the volume of the ball; if the steel ball sinks not onto the griund, then the density of the steel ball is the same as the density of the water (1 kg/dm³); so the forces of these two cases are different by a value of 6.85 kg/dm³ multiplied by the volume of the ball!

By suspending the two balls in question from the ceiling, there will be no weight gain to either of the liquid filled containers so it does not matter what gas is inside the hollow sphere. The liquid will be displaced an equal amount due to identical spherical volumes within identical containers... Therefore the balance pans will not change in height.