Physics question involving a beaker, water, cork, glue and water

Quote: Doc

The final destination of the energy is in the form of internal energy of the water and the cork, which both will tend to be at very slightly higher temperatures than before.

I think that a very very very small part of the potential energy's destination is in the form of internal energy of the water and cork. The vast majority of the potential energy is "transformed" into kinetic energy of the water and cork.

When the cork is accelerating upward, a mass of water with a volume equal to the of the cork is accelerating downward. This picture suggests a simple model that could give some insight into the answer. Imagine two weights suspended in air from a pulley supported by a spring scale. One weight has a mass, m1, equal to a large cork, say about 2.4 grams, and the other weight, m2, has a mass equal to water having the same volume as the cork, say 10 grams. Initially the pulley is braked so it can't turn, and the spring scale would reflect the weight of the pulley plus (m1+m2)g.

If the pulley's brake is released m1 would accelerate upward and m2 would accelerate downward both with acceleration a. And the string connecting the weights would have a tension, T. Doing a force balance on m1 yields: T - m1 g =m1 a . And a force balance on m2 gives: T - m2 g = -m2 a . Subtracting the second equation from the first gives: (m2 - m1)g = (m1 + m2)a or a = g(m2 - m1)/(m1+m2) . So from the first equation T = m1(g + a) or T = 2 m1 m2 g /(m1 + m2). And then the spring scale would reflect a force equal to the weight of the pulley + 2 T, or pulley + 4 m1 m2 g / (m1 + m2).

The change in the reading of the spring scale from before to after unbraking the pulley would be 4 m1 m2 g / (m1 + m2) - (m1 + m2) g, which is negative. In the example for a 10 cc cork with a mass of 2.4 grams, the difference in readings would be -4.7 grams.

Of course this model neglects any of the fluid motion of the original puzzle, but it does shed some light on the result.

Quote: Wizard

I admit I'm not in your league when it comes to physics, but let me respectfully challenge you on this.

Suppose we build a device with a 100-foot rod holding a heavy ball, initially on top. All of this is on a scale. The stand weighs 100 pounds and the ball weighs 1000 pounds. Kind of like that ball that drops on New Year's Eve in Times Square, but this ball is capable of falling faster.

Then suppose we let the ball do a controlled fall at a constant rate of 1 foot per second. You seem to be saying that after the initial moment the ball starts moving that the total weight will revert back to 1,100 pounds? If yes, then what would happen if the speed of the fall were equal to its terminal velocity? Wouldn't the scale register 100 pounds until it hit the bottom?

I interpret your new problem to mean that the ball is initially at rest at the top, is released and begins to fall (accelerates), and then assumes a constant speed of 1 ft/sec until it contacts the base.

What keeps the ball from continuing to accelerate? The most likely explanation is some kind of interaction between the ball and the rod, such as friction. Alternatively, the New Year's Eve ball has a mechanism that restricts the falling of the ball.

Whether it is friction or from a mechanism, while that force on the ball balances the gravitational force on the ball and keeps the velocity uniform (no acceleration), the rod must be applying an upward force on the ball exactly equal to the force of gravity on the ball, i.e. 1,000 lbs. Otherwise the acceleration would continue.

When the rod is exerting this upward force on the ball, by Newton's 3rd law the ball is exerting an equal downward force on the rod. Assuming that the rod/base itself is not accelerating, the scale must provide an upward force on the rod equal to this 1,000 lbs. plus the 100 lbs weight of the rod/stand, i.e., the scale is pushing the rod with a force of 1,100 lbs. Again by Newton's third law, the rod is pushing on the scale with 1,100 lbs. while the ball is either stationary or moving at a constant speed.

During that initial acceleration, the ball places less than 1,000 lbs. force downward on the rod and the rod places less than 1,100 lbs. force downward on the scale. When the ball strikes the bottom and decelerates toward a stop, the force on the scale will be higher than 1,100 lbs. but will return to 1,100 lbs. when the ball comes to a complete stop.

As to your final question, the scale would only register 100 lbs if the ball is not placing any downward force on the rod (no upward force from the rod on the ball either), meaning that the ball must be in free fall, not falling at any constant speed.

Did I explain that in a way that makes sense? Sometimes I jumble the words.

Quote: Fleaswatter

I think that a very very very small part of the potential energy's destination is in the form of internal energy of the water and cork. The vast majority of the potential energy is "transformed" into kinetic energy of the water and cork.

Absolutely correct, so long as the water and cork continue to be in motion with kinetic energy equal to the reduction in potential energy. I was trying to describe where the energy goes after the viscous forces dampen out all of the motion, leaving the water quite still with the cork floating on the surface. At that point, there has been a reduction in potential energy and there is no kinetic energy. I guess it is most proper to say that the potential energy is converted to kinetic energy which is further converted to internal energy of the masses, typically represented by a temperature increase.

Quote: Doc

I interpret your new problem to mean that the ball is initially at rest at the top, is released and begins to fall (accelerates), and then assumes a constant speed of 1 ft/sec until it contacts the base.

What keeps the ball from continuing to accelerate? The most likely explanation is some kind of interaction between the ball and the rod, such as friction. Alternatively, the New Year's Eve ball has a mechanism that restricts the falling of the ball.

Whether it is friction or from a mechanism, while that force on the ball balances the gravitational force on the ball and keeps the velocity uniform (no acceleration), the rod must be applying an upward force on the ball exactly equal to the force of gravity on the ball, i.e. 1,000 lbs. Otherwise the acceleration would continue.

When the rod is exerting this upward force on the ball, by Newton's 3rd law the ball is exerting an equal downward force on the rod. Assuming that the rod/base itself is not accelerating, the scale must provide an upward force on the rod equal to this 1,000 lbs. plus the 100 lbs weight of the rod/stand, i.e., the scale is pushing the rod with a force of 1,100 lbs. Again by Newton's third law, the rod is pushing on the scale with 1,100 lbs. while the ball is either stationary or moving at a constant speed.

During that initial acceleration, the ball places less than 1,000 lbs. force downward on the rod and the rod places less than 1,100 lbs. force downward on the scale. When the ball strikes the bottom and decelerates toward a stop, the force on the scale will be higher than 1,100 lbs. but will return to 1,100 lbs. when the ball comes to a complete stop.

As to your final question, the scale would only register 100 lbs if the ball is not placing any downward force on the rod (no upward force from the rod on the ball either), meaning that the ball must be in free fall, not falling at any constant speed.

Did I explain that in a way that makes sense? Sometimes I jumble the words.

1. I'm not a science guy, and I understood that.

2. Am I the only person here who thinks it'd be fun to watch a 1000 lb weight traveling at terminal velocity smash into a stationary scale?

Quote: Doc

Without going into speculative theories like gravitrons, I'll just say that gravitational attraction is a force, not an energy. Systems obtain potential energy (or have their potential energy increased) by doing work on them -- you move them a distance against the force of the gravitational field.

Something is turning the moon. As I understand things, changing the course of a 7.3459 x 10^22 kg rock hurtling through space at 3360 km/h takes energy (thank you, wolframalpha.com :). And the energy is constantly applied yet comes from nothing but the presence of another mass.

Baffling. When humanity understands and has control over gravitational forces, things will certainly be different. I doubt we survive that long.

Quote: Doc

Whether it is friction or from a mechanism, while that force on the ball balances the gravitational force on the ball and keeps the velocity uniform (no acceleration), the rod must be applying an upward force on the ball exactly equal to the force of gravity on the ball, i.e. 1,000 lbs. Otherwise the acceleration would continue.

When the rod is exerting this upward force on the ball, by Newton's 3rd law the ball is exerting an equal downward force on the rod. Assuming that the rod/base itself is not accelerating, the scale must provide an upward force on the rod equal to this 1,000 lbs. plus the 100 lbs weight of the rod/stand, i.e., the scale is pushing the rod with a force of 1,100 lbs. Again by Newton's third law, the rod is pushing on the scale with 1,100 lbs. while the ball is either stationary or moving at a constant speed.

This is a very good explanation. It can also be used to illustrate why you answer to the original question is wrong :)
Water is pushing the cork upwards, making it accelerate, and therefore an equal and opposite force is applied to the scale plate, pushing it down and increasing the weight.

Quote: QuadDeuces

Something is turning the moon. As I understand things, changing the course of a 7.3459 x 10^22 kg rock hurtling through space at 3360 km/h takes energy (thank you, wolframalpha.com :). And the energy is constantly applied yet comes from nothing but the presence of another mass.

It appears that you continue to use the term "energy" as equivalent to "force". Since you said "changing the course", I assume that you are referring to the Earth's effect on the Moon's orbit.

If the Moon were in a circular orbit, keeping it in that orbit would require a central force (mutual gravitational attraction) but would not require or consume energy. (Side note: I am ignoring frictional losses that the moon experiences, such as collisions with interplanetary particles and the drag losses from creating tides on the Earth -- I'm talking about the simplified problem of an orbiting body.)

The Moon, like essentially all orbiting bodies, really follows an elliptical path with an apogee and perigee. At the apogee (highest point) the Moon has its maximum gravitational potential energy and minimum kinetic energy (lowest speed). As it moves closer to the earth in that orbit, the speed increases: potential energy is converted to kinetic energy. As it passes perigee and starts moving to greater separation, some of the kinetic energy is converted back to potential energy.

A different-but-equivalent way of describing this is that, as the Moon comes closer to the Earth, the gravitational force acts through a distance (the change in radius) and adds energy to the moon, speeding it up. As the Moon passes perigee, some of the kinetic energy is consumed to do the work necessary to lift the moon to a higher altitude. Different descriptions, but the same phenomena. Gravity is a force, not an energy. When the force acts through a distance, it performs work and thereby adds or removes energy -- the force itself is not the energy.

Wait a minute! The problem specifically posed the question as "AT THE MOMENT the cork breaks free from the bottom of the beeker..." What's all this center-of-gravity-shifting-as-the-cork-moves-upward shit.

I say your answer is addressing a different problem and still believe the "MORE" camp is correct given the time frame of the proposed event.

I submit my highly uneducated layman's solution again:

I say the scale registers MORE weight, because the air in the cork provides a lifting force on the bottom of the beaker and once that force is removed, the beaker will be subject to a slight, temporary increase in gravitational force. However, since the cork is not being removed from the beaker, the inertial effect will only be temporary and the weight will quickly equalize to its original reading.

Quote: weaselman

This is a very good explanation. It can also be used to illustrate why you answer to the original question is wrong :)
Water is pushing the cork upwards, making it accelerate, and therefore an equal and opposite force is applied to the scale plate, pushing it down and increasing the weight.

Man, I've got too many conversations going on.

You seem to be losing track of the forces described by Newton's third law. If water is pushing the cork upward, the equal and opposite force is the cork pushing the water down. The scale is not directly involved in that application of the third law. If you want to talk about the scale, it is pushing up on the bottom of the beaker with a force equal and opposite to the force the beaker applies to the scale.

In contrast to the problem with the 1,000 lb. ball going down the rod at constant speed, the problem of the buoyant cork involves acceleration followed by deceleration (possibly but not necessarily with constant-velocity travel for some period in between.) We are interested in that period in which the cork is accelerating upward and an equal volume of water is accelerating downward. We could analyze the equal-and-opposite forces of Newton's third law, but they aren't the ones important to solving the problem. The important forces are the ones described in Newton's second law related to accelerating masses.

In this case, we can consider the entire system of beaker, water and cork. During the period of acceleration, not only is the center of mass of that system moving downward (as the Wizard stated) it is accelerating downward -- the cork is accelerating upward but there's an equal volume (greater mass) of water that is accelerating downward at the same rate, so the center of mass is accelerating downward.

In order for that system's center of mass to have started accelerating downward, there must have been some change in the net of the external forces imposed on it. Something is pushing/pulling that system down more than previously. Gravity hasn't changed. The only force that could change was the force applied by the scale. The scale is pulling the beaker system down more that it was before. Rather the scale is pushing the beaker system upward less than it was before. At that point, yes, we can look at the third law and find that the beaker is pushing back down on the scale less than it was before.

You are describing how things are, not why they are, which is my original question/observation.

Something is absorbing the kinetic energy and is keeping the moon from hurtling off into space. It is changing the big rock's course, overpowering inertia. That takes energy - or something else.

Quote: GameBoy

Wait a minute! The problem specifically posed the question as "AT THE MOMENT the cork breaks free ....

I think your comments may be related to the issues I discussed in my second post hidden by a spoiler button, several pages back. The Wizard specifically replied that he was not referring to those extremely-brief, extremely-small forces that he described as negligible. I complained that the problem seemed to be evolving, but I deferred to the Wizard's attempt to clarify the problem. I agree with him that the larger changes in the forces on the scale are likely to occur when there are actually accelerations of masses being considered, not just relaxation of internal stresses. You and I may have both been led down the wrong path by the comments about the moment that the cork breaks free, while the Wizard was intending to ask about when the cork was accelerating upward.

By the way, I know of no reason to think that there is any change in the gravitational forces during this process. All of the items have fixed masses and they remain in a uniform gravitational field.

Quote: QuadDeuces

Something is absorbing the kinetic energy and is keeping the moon from hurtling off into space. It is changing the big rock's course, overpowering inertia. That takes energy - or something else.

Nothing is absorbing the kinetic energy. In the case of a circular orbit, the kinetic energy remains constant. In the case of the elliptical orbit, kinetic energy is partially converted to potential energy (higher orbit) not absorbed by anything, and the potential energy is later partially converted back to kinetic energy. Where is energy being absorbed by anything? It is just changing from one energy form to another, and in this case they are both mechanical energy forms.

Quote: QuadDeuces

Something is absorbing the kinetic energy and is keeping the moon from hurtling off into space. It is changing the big rock's course, overpowering inertia. That takes energy - or something else.

There is a constant acceleration vector perpendicular to the velocity vector in a e=0 (perfectly circular) orbit. So the velocity is changing but the absolute value of it (speed) is not. We're talking about identical energy states in different physical positions, your assumption one is doing work on another just isn't true.

If something was absorbing the kinetic energy of the moon then it would be slowing down.

Edit: The moon is actually slowing down, but it has to do with tidal forces.

Quote: Doc

By the way, I know of no reason to think that there is any change in the gravitational forces during this process. All of the items have fixed masses and they remain in a uniform gravitational field.

How about a change in inertia? I think that's what I meant. The elimination of the lifting force on the bottom of the beaker by the cork creates a momentary downward inertial change resulting in more force being applied to the scale? Does that fall under the relaxation of internal stresses category?

Damn, I should've known going to my college physics classes stoned wouldn't work.

Quote: GameBoy

How about a change in inertia? I think that's what I meant. The elimination of the lifting force on the bottom of the beaker by the cork creates a momentary downward inertial change resulting in more force being applied to the scale? Does that fall under the relaxation of internal stresses category?

Damn, I should've known going to my college physics classes stoned wouldn't work.

I think you are just stumbling over some of the terms. No, I don't think the inertia of anything changed. An item's mass is a measure of its inertia, and nothing changed mass in this problem.

When I described something related to your answer a few pages back (behind a spoiler button), I noted that while the glue was still holding, the cork, the glue, and the bottom of the beaker were under tension. The cork and glue would be stretched, and the bottom of the beaker would be deformed in some manner. When the glue fails and the tension can no longer be supported, all of those deformations would quickly change. I speculated that the release of stress on the bottom of the beaker might cause it to move downward and cause the very first "twitch" of the scale in the direction indicating "more" force, perhaps much as you proposed. However, as the Wizard pointed out, and as I stated earlier, this would be a very small, very brief effect. You and I should both accept that the person who posed the question was asking about larger forces, even if they occurred nanoseconds later, rather than right "at the moment" of the glue breaking.

Glad you guys understand it. I obviously don't.

If a Helium balloon were tied to a brick on a scale, and the string is cut, the scale would show an increase in weight. The lifting power of the balloon is greater than its mass, thus the increase in weight.

I was surprised that the answer is debatable.

Somebody work in a lab somewhere?

Quote: Doc

You seem to be losing track of the forces described by Newton's third law. If water is pushing the cork upward, the equal and opposite force is the cork pushing the water down. The scale is not directly involved in that application of the third law.

Quote:

In contrast to the problem with the 1,000 lb. ball going down the rod at constant speed, the problem of the buoyant cork involves acceleration followed by deceleration (possibly but not necessarily with constant-velocity travel for some period in between.)

I was talking about the ball falling down with an acceleration that is less than g. Or, if you prefer, consider reversing the mechanism, and making it push the ball up with an acceleration.

Mythbusters !

Of course, we would need to add a step that would require something to get blown up.

Quote: weaselman

I was talking about the ball falling down with an acceleration that is less than g. Or, if you prefer, consider reversing the mechanism, and making it push the ball up with an acceleration.

Then maybe I can't figure out what your question is. I'll describe both problems again and see whether I cover it this time.

For the problem with the ball going down the rod:
Consider the ball as a system being acted upon by external forces that include gravitational attraction and some kind of interaction with the rod (friction or a mechanism.) If, as you suggest, the ball is falling at an acceleration greater than zero and less than g, then the rod is applying an upward force on the ball that is greater than 0 lbs. and less than the 1,000 lbs. with which gravity is pulling it down. The net force on the ball is the combination of the 1,000 lb. downward force of gravity and the non-specific 0<F<1000 upward force from the rod. Whatever the exact amount of that upward force, the ball is applying the exact same downward force on the rod. The scale will read the sum of (1) the 100 lbs. weight of the rod and (2) whatever is the amount of that downward force from the ball on the rod.

Now the Wizard described that problem as involving the ball dropping at a uniform 1 ft/sec after an initial period of acceleration. During the period of that uniform speed, the upward force of the rod on the ball must be 1,000 lbs. to balance with the downward force of gravity. In that case, the downward force from the ball on the rod is also 1,000 lbs., and the total force on the scale is (from the last sentence of my previous paragraph) 100 lbs. plus 1,000 lbs. = 1,100 lbs.

For the original problem of the cork in water in a beaker:
Consider the system as the beaker and all of its contents. It is possible to analyze the internal forces and movements within that system, and it is possible to analyze the external forces that act upon the entire system. If we are interested in the force as read by the scale, that is an external force on the beaker-water-cork-glue system. The net external force on the system consists of the combination of the force from gravitational attraction and the force from the scale. Depending on whether those two external forces balance or not, the center of mass of the system either (1) remains stationary or in a state of uniform motion (Newton's first law) or (2) accelerates, in which case we would consider Newton's second law.

During the period that the cork is floating upward and water is dropping, the center of mass of the beaker-cork-water-glue system is moving downward. For at least part of that time, the cork is accelerating upward, an equivalent volume (greater mass) of water is accelerating downward at the same rate, and the combination of those give the center of mass accelerating downward. In order for the center of mass to be accelerating downward, the upward force exerted on the system by the scale must be less than the downward force exerted on the system by gravity.

The scale reading, which is what the problem asked about, is equal in magnitude to both the upward force by the scale on the system and the downward force of the system on the scale, i.e., less than the original value that was exactly equal to the force of gravity on the system.

Side discussion to the beaker problem
Now before someone jumps on one comment I made, I should clarify. Because the beaker necessarily has a greater diameter than the cork, it is not really precise for me to say that when the cork is accelerating upward "an equivalent volume of water is accelerating downward at the same rate". It is more accurate to say that a greater volume of water is accelerating downward at a proportionately lower rate of acceleration, giving the same effect on the scale.

So did that cover everything? Is everyone suitably bored now?

Quote: Doc

The net force on the ball is the combination of the 1,000 lb. downward force of gravity and the non-specific 0<F<1000 upward force from the rod. Whatever the exact amount of that upward force, the ball is applying the exact same downward force on the rod. The scale will read the sum of (1) the 100 lbs. weight of the rod and (2) whatever is the amount of that downward force from the ball on the rod.

Exactly. Now think about the cork in the beaker in the same terms.
The water is pushing the cork up with a force, larger than the cork's weight. The cork is pushing water down with the same force. The scale will read the sum of the weights of the beaker and water plus the force with which the cork is being pushed, which is greater than it's weight. Therefore the reading of the scale must be higher while th ecork is accelerating upward then when it is stationary.

I have a scale at the office that measures down to the nearest gram. I can get a jat, water and a cork easily. I'm less certain about water-soluble glue. any known brands or kinds? How about the common white glue, either liquid or on stick? is the scale sensitive enough?

Since there was no indicated end time for this puzzle, I kinda expected a new thread to announce the answer. I gotta send Thanks to Doc for mentioning in the Chip of the Day thread that Mike gave his answer, sigh, yesterday.


And the answer is...

Quote: Wizard

So what I believe to be the correct answer....

Wait a sec.

You believe the answer is....?

I CALL FOUL!

If you're gonna ask a math puzzle or trivia question, you better have an answer - or admit up front that you don't know.

Now I realize that your area of expertise is probability and statistics, and that this was a physics question, I still call FOUL!

Quote: DJTeddyBear

I CALL FOUL!

This is the Wizard's ballgame here, and he is entitled to a certain degree of respectful demeanor from us thereby, but in this case I have to say I concur with you. Pretty bad.

Quote: weaselman

The water is pushing the cork up with a force, larger than the cork's weight. The cork is pushing water down with the same force. The scale will read the sum of the weights of the beaker and water plus the force with which the cork is being pushed, which is greater than it's weight. Therefore the reading of the scale must be higher while th ecork is accelerating upward then when it is stationary.

Nope. You are looking at forces internal to the system, many of which cancel out when the system is considered as a whole. You also left out some of them. Do you acknowledge that water is accelerating downward? Have you even considered the fact that as the water accelerates downward, the force it exerts on the beaker has changed? I have a feeling you haven't even thought about those internal forces.

Doesn't matter -- the only important aspect of the internal forces is that as the internal stresses are released, the center of mass moves. That is only important if the center of mass accelerates, which it does for at least some of the time. Knowing the direction that the center of mass is accelerating is all that is necessary to determine the direction of the net external force. Since the CM has begun accelerating downward, the net external force is downward. The force of gravity has not changed, so upward force from the scale must have decreased.

As the cork reaches the surface, both the cork and the water will decelerate, resulting in the center of mass decelerating its downward movement (accelerating upward), and the force on the scale will be higher than it was originally. When everything comes to a stop, the scale force will be back at its original value.

Quote: DJTeddyBear

And the answer is...

Wait a sec.

You believe the answer is....?

I CALL FOUL!

If you're gonna ask a math puzzle or trivia question, you better have an answer - or admit up front that you don't know.

Now I realize that your area of expertise is probability and statistics, and that this was a physics question, I still call FOUL!

I understand that after all of the various viewpoints were being expressed, the Wizard may not have wanted to declare an absolute answer like a dictator. That's fine and perhaps prudent.

What does bother me is this:

The Wizard posted his answer yesterday at 1:21 PM PDT.

I submitted a supportive response at 2:34, then at 3:05 posted an "oops!" with a clarification to my position, still supportive of the Wizard's answer.

The Wizard then made a post in which he address me and said, "I admit I'm not in your league when it comes to physics, but let me respectfully challenge you on this. " That post from the Wizard appears to have been deleted from the thread!

I quoted that now-deleted post in my reply at 4:56, and the Wizard has been absent from the thread ever since, while I have been scrambling to defend his answer and my posts and explain Newton's laws to people who don't seem to follow them.

Where are you Wizard? I assume that either he has been very busy or he is checking with that physics resource that he trusts more than he trusts me. I have no problem with that, but it would be nice to know whether he is still around following this discussion.

Doesn't he have his radio show on Thursdays? Let's cut the guy some slack.

This is fascinating. To me the question sounded easy, and after a little thought I figured that the volume of water equal to the size of the cork being drawn downward by gravity as it replaces the cork would temporarily 'add weight' as measured by the scale. I really don't 'know' that I am correct. I have read all the other answers and can be convinced that almost all of them make some sense. There appear to be no physicists on this forum. What would interest me is what would happen if we were able to ask 100 university professor physicists this exact question? Would we get 100 answers, all the same, and some comment about how simple this question was? Or a third picking heavier, a third picking lighter, and a third thesame, with a similar potpourri of answers that our forum members proposed? Or most giving the correct answer and the few who get it wrong hanging their heads in shame?
I've asked son of SOOPOO to weigh in here, and if he doesn't know the answer to find a physics professor at his University to weigh in. I like using the phrase 'weigh in' on this thread.

Quote: SOOPOO

There appear to be no physicists on this forum.

The Wizard's dad has a PhD in physics. In the past, when the Wizard has been skeptical of some answers provided in the forum, he has gone to Papa for an answer in which he can feel confident. That seems like a reasonable course of action this time, since there are such strong, dissenting opinions being expressed here.

I'd certainly be willing to have the Wizard's dad review and comment on my interpretation of either the cork-in-a-beaker problem or the falling-ball-on-a-rod problem (in the now-vanished post .) I don't know that he would be interested enough to read all of this stuff.

I haven't read all the responses here in detail so my apologies in advance if someone has already stated or responded to this. I voted for the scale measuring a temporary increase in weight. Here's why:

1) Before the glue dissolves it is acting like a spring to keep the cork down against the buoyant force. The glue must have "stretched" itself over some distance "x" to achieve this tensile force. At the moment it dissolves and releases the cork, the spring must "unstretch" itself and by Newton's Third Law must also push down on the beaker even if for only a brief instant.

2) In addition if the water's center of mass is accelerating downward (as I think everyone agrees on) then this would be the equivalent of dropping this volume of water down on the scale. Therefore it seems the scale should read an additional force equal to the buoyant force (less the water’s drag friction) for the duration of the cork's trip to the surface.

3) At the instant the cork emerges from the water sufficiently so that it's self-weight is equal to the buoyant force then the scale would only be reading the weight of the water, the weight of the cork (less the water's drag force), and the beaker's weight.

4) The cork would still have some velocity and would therefore continue to emerge from the water until gravity caused it to have a negative velocity. The scale, in addition to the weight of the water and beaker, would therefore only register a portion of the weight of the cork equal to the volume of water displaced at any given instant (plus or minus the water's drag force).

5) Finally the cork after some bobbing up and down would come to equilibrium and the scale also after wavering back and forth would eventually settle on the final reading: weight of water + weight of beaker + weight of cork. Same as we started with before the glue dissolved.

In all the above I am assuming that the scale itself is relatively an infinitely stiff and sufficiently sensitive enough spring to register the forces essentially instantaneously. I don't know if there is such a readily available scale to be able to perform this experiment easily enough, however.

Shall we put a wager on it?

The question of how a downward acceleration of the center of mass of an object on a scale (due to changes in the internal mass distribution of the object and not an external force on the object) affects the force on the scale can be resolved by a simple experiment that almost anyone can do.

1. Stand on a scale (with a bathroom scale it will help to stand sideways so that your knees don't block your view of the scale's display).
2. Keeping your eyes on the scale's display, squat down quickly and stop in a squatted position.
3. Observe that your measured weight decreases as you start to squat and increases as you reach a squatted position.

When you start to squat, your center of mass accelerates downward, decreasing the force on the scale. As you reach a squatted position, your center of mass decelerates, which is to say it accelerates upward, increasing the force on the scale.

This can make sense intuitively. A falling object (accelerating downward) is exerting a downward force of less than the force of gravity on it--that's why it's not hovering or accelerating upward.

Or, consider standing on a scale holding a barbell. As you lift the barbell, wouldn't you expect the extra force you're applying to the barbell to be passed down to your feet, making you apply more force to the scale? Conversely, when you lower the barbell, the temporary decrease in force will apply to the scale as well.

Quote: Paisello

Shall we put a wager on it?

Do you have a proposal as to how the wager would be resolved?

Quote: paisiello

I haven't read all the responses here in detail so my apologies in advance if someone has already stated or responded to this. I voted for the scale measuring a temporary increase in weight. Here's why:
...
2) In addition if the water's center of mass is accelerating downward (as I think everyone agrees on) then this would be the equivalent of dropping this volume of water down on the scale. Therefore it seems the scale should read an additional force equal to the buoyant force (less the water’s drag friction) for the duration of the cork's trip to the surface.
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Shall we put a wager on it?

paisiello:

With regard to your point (1), I have described some possible effect that may exist like that. I don't think I have thought it through fully. However, the duration of such a force would be extremely short and the impulse negligible. In his comments earlier in the thread, the Wizard specifically ruled out these negligible effects. He eventually made it clear that he is interested in the changes in the scale reading associated with the cork moving upward and water moving downward to displace the cork.

With regard to your point (2), we agree that the water's center of mass is moving downward. In fact, I have said that the center of mass of the entire system consisting of the beaker and its contents is moving downward, even though the cork is moving upward. As I have stated several times, it is not important that the CM is moving down -- that could be associated with a force on the scale that is either higher, lower, or the same as its original value. The important factor is that at least early in the period of motion (cork up, water down) the center of mass is not just moving downward, it is accelerating downward. It is the acceleration that is important. This is not at all "the equivalent of dropping this volume of water down on the scale", if you mean letting the water strike the scale and decelerate its downward motion. It is precisely the opposite. I think I have explained the issue of external forces and the associated affect on the acceleration of the CM several times previously in this thread, so I won't repeat myself again here.

Your points (3), (4), and (5) describe events later in the process, and I agree that there will be times with forces on the scale that are greater than the original force, and a final, equilibrium state where the force on the scale returns to its original value.

If your suggestion that we put a wager on it applies specifically to the force on the scale during the time period represented by your point (2), when the cork is accelerating upward and the water is accelerating downward, then Yes, I would be willing to wager that the force on the scale is less than the original value, as opposed to your suggestion that it is higher than the original value.

As ZPP asked, how to you propose we resolve the question to determine the correct analysis and result? Perhaps that should be established before setting specific wagers. Do you like the idea of presenting your analysis and mine to the Wizard's PhD-in-Physics father for review? Would that provide a suitably independent and qualified authority to rule on such a matter? I have no idea whether he would be willing to fill that role.

No, no, you guys have connvinced me. Until tomorrow when I change my mind again.

Quote: Doc

The Wizard's dad has a PhD in physics.

Technically, I do to (although, it is in astrophysics, not very close to Newtonian mechanics, and I have not been working in the field for years, but did do some teaching of elementary physics to kids lately).

Having said that, I have to admit, that in this thread I was wrong.

Wizard and Doc are correct claiming that the weight will decrease. I made a mistake when analyzing this problem, by replacing a cork floating up in water with a fly flying up in the air. I thought it was simpler to analyze it that way, but it turns out, that such "simplification" changes the problem qualitatively, and makes the answer opposite to the correct one.

Quote: weaselman

I have to admit, that in this thread I was wrong.

That is a statement that far too many people in this forum, and in society in general, are not willing to say out loud (or put in writing). Many won't even admit it to themselves. I am pleased and a bit impressed to see it here. Well done, weaselman.

As for myself, I have so much experience being wrong, that acknowledging it is routine. I made several incorrect statements in this thread that I later had to revise. For my posts, "Ooops!" is a common exclamation.