Water Weight: Newsletter Challenge (November 2012)

Since the metal weight is initially hanging on the glass, the total force exerted on the bottom of the glass doesn't change, whether the metal weight is outside the glass or inside. Since the area of the bottom of the glass doesn't change, the pressure doesn't change.

Unless, as I pointed out in my first post, a chemical reaction takes place between the metal and the water. With sodium you could get an explosion that would temporarily increase the pressure, release gas, perhaps blow some water and metal out of the glass.

Since the metal weight is initially hanging on the glass, the total force exerted on the bottom of the glass doesn't change, whether the metal weight is outside the glass or inside. Since the area of the bottom of the glass doesn't change, the pressure doesn't change.

Pressure is a comparative measurement of force applied to a prescribed area- kpa, psi etc. The actual area the pressure is applied to is irrelevant to the pressure measurement (but very relevant to engineering!). You are referring to the overall force exerted on the bottom of the glass, not the unit pressure.

The glass is being 'weighed' on a balance. I'm saying that the 'weight' measured will be the same, regardless of whether the metal hangs outside the glass or inside the glass.

Since the 'weight' on the balance remains unchanged, the answer to the question 'Does the pressure on the bottom of the glass change?' is no. No change in weight, no change in surface area, no change in pressure, which is the weight divided by the surface area (pounds per square inch, newtons per square meter, whatever unit you want to use.) It didn't ask if the pressure of the water inside the glass changes.

True- the OP specs "on the bottom " not "at the bottom" which makes that valid, and then pressure can be construed as the exterior force. Ah, the joys of the English language!

The total force(weight) exerted on the scales does not change, provided, as you pointed out, that the metal is not one of those highly reactive elements, however if the metal object is placed in the glass such that it is partially or fully immersed in the water then the pressure on the bottom will change in that the water level in the glass will rise to the extent displaced by the object, and the pressure exerted on the bottom of the glass is caused by the height of the overlying water column. This has increased, therefore the pressure has increased. Some buoyancy is applied to the object which reduces its load on the glass lip, and this is exactly equal to the increased force applied to the base of the glass by the increased pressure.

a cup of coffee with a spoon of sugar?

It would make no difference to my answer, although the question did stipulate a metal object.

a cup of coffee with a spoon made of sugar?

You sound like Julie Andrews.

I agree with your answer the total mass on the scale will stay the same.

Regardless of the water being displaced if the metal is solid or hollow.

The pressure on the bottom of the glass would depend on the measurement being taken inside or outside of the glass container.

Inside the glass yes slightly increased pressure on the Outside no increase.

Can't add any more to #1 & #2 (both GA's but #2 much more complete) except to say that "filled" is a relative term; the OP does not specify that it is brimfull to overflowing or that the object is now actually submerged. Regardless 1 of 2 things will happen:

1. The glass was actually brimfull or close enough that some of the water overflows and is lost, and the mass changes by this amount (unless it is on the balance pan & not actually lost)
2. the glass was not actually brimfull and the water displaced by the object is not lost, therefore there is no change in recorded mass

In either case, the head pressure at the bottom of the glass changes according to the increase in column height (if any, see point 1);this is relevant to the displacement of the object and not to its density or location at any given point inside the glass even if the object is still not completely submerged in the second situation given.

No. Yes. (Presuming water is not displaced o/s glass)

"Does the pressure on the bottom of the glass change? Yes, if we are talking about the inside the bottom of the glass - by the increased height of water due to placing mass inside. Talking about the outside - no - tho as earlier poster said, distribution of pressure will.

I would imagine that the weight of the total unit, glass, liquid and weight, will decrease as the hanging weight has travelled towards the fulcrum of the balance.

The weight impacting on the bottom of the vessel should remain the same.

Thank you

That assumes that the object is moved either closer or further from the fulcrum point. If it is moved to the center from an edge that is equidistant from the edge there would be no change.

If memory serves, modern analytical balances do not discriminate as to the position on the balance plate.

S.G. = Mass of oject in air/mass of object in air - Mass of object in water

So yes it will be different

1 No, mass is conserved

2 Yes, pressure depends on depth of water

Making sure it doesn't touch the bottom, doesn't mean the object was actually submerged in the water, we cannot assume, we have to answer according only to supplied information.

Exactly! Let me clarify this a little bit further.

If the metal is not touching the bottom, then it is either:

1. Supported by something external to the system, or
2. Supported by the glass, or
3. It is floating (maybe a light metal can do this, although it may react with the water), or
4. It is sinking.

You already covered case 1.

In case 2, the total weight will not change, as the increment of the water pressure will be counterbalanced by the decrement of the submerged object's weight due to buoyancy (I remind that: buoyant force = weight of displaced fluid).

In case 3, again there will be no change: On the one hand, the object is not anymore supported (i.e. there is no direct transfer of force to the balance plate), and on the other hand, the force induced by the water will increase for so much as the weight of the object. Tie!

In case 4, and until the object reaches the bottom, the total weight applied to the balance plate will DECREASE by an amount of : the object weight MINUS the displaced water weight.

If #3 it is still supported by the glass.

If it is #4 it is still supported by the glass.

In all three cases (2 through 4) the force is exactly the same upon the balance platter.

Case #2 is obvious.

Case #3 has the object suspended upon the fluid. The vectors of force will be downward upon the water for the object, upward for the water upon the object, but the water will now also impart an exactly equal to the object's downward force, but an opposite sense of force downward upon the balance pan. The mass of the water itself does not change, so its force is always downward.

Case #4 is just a variant of #3 where the object is in motion. If the motion is slow enough the effects of acceleration are negligible. Even if not, the total momentum is conserved, so the net difference over time is equal to the same force with a static model.

No, the object does not DIRECTLY apply any force to the plate. Direct contact is understood only among solid objects which microscopically-speaking use the electric forces of the molecule mesh to carry the force along the same direction it is applied.

Mind that the only thing that touches the plate is the bottom of the glass, and the only thing that touches the bottom of the glass is the water. The weight of the object itself appears only INDIRECTLY by causing the increment of the water pressure.

On the other hand, in case of the fluids, the direction of the force is not maintained but spread to all directions until it reaches a sort of equilibrium. Of course there will still be a pressure gradient due to gravity, but no other direction will be preferred. For this reason one cannot claim that the column of the water right beneath the object applies any different force than any other column nearby. It makes no difference, wherever the object floats.

Anyway, we agree that in cases 2-3 the total weight will be the same. I still disagree though about case 4. Imagine this scenario: You stand on the balance plate holding an object in your hand. Then you let it drop. Won't you think there will be a tilt in the balance? In this case there is no water, but air. Still a fluid, although lighter.

What is pulling up the floating object?

Think of it. If you suspend something into an empty glass, something has to hold it up.

Even if you erect a crane the objects mass will pull down on the crane. If the crane is on a scale and it picks up a car, that scale increases by the amount the car "weighs".

Partially fill a glass, put it on a scale, weigh it, then put in an ice cube, weigh it again, and report back.

What you are saying is that weighing a glass with ice cubes in it will weigh less than it does after the ice melts - ignoring evaporation. It won't!

Try that and prove me wrong.

First of all, let me repeat that we don't disagree on the fact that the weight will not change (leave case 4 aside for the time being as it is not a steady state). We could as well see the whole system as one object and say that as long the mass hasn't been changed, then the weight won't change. It's just that we tried to analyze the forces separately, for the sake of the discussion.

Of course I neither disagree that the glass with the ice cubes will change it's weight while the ice melts. I cannot see how I gave the impression that I support something like this.

Our disagreement lies only to the idea that it is the column of the water underneath that holds the object as if it were a solid column, while the rest of the water sits there happily minding its own business. Or that this "column" will push the bottom of the glass in any different way than any other water column. That's the idea I got from what you have written; correct me if I'm wrong.

Think it another way: If you are at the bottom of the lake, you don't give a damn whether there is a huge boat sailing right above you or it is a mile away. So if you ask me, what holds the boat, I would answer the water, and we would agree to that. If you ask me what force is applied to you, I would answer that it is the force which corresponds to of the depth you currently are, and I would forget anything about the boat (whose only contribution was to rise the water level).

Okay, the pressure inside the glass at its bottom is one thing and will change relative to the height of the fluid in the container.

The overall system mass does not change. I think we agree.

In the case of the ice in the glass, neither the inside pressure at the base of the glass nor the mass changes from the frozen state to the liquid state.

Making sure it doesn't touch the bottom, doesn't mean the object was actually submerged in the water, we cannot assume, we have to answer according only to supplied information.

The end result is the same. It is the total mass of the system whether it is hanging on the glass, floating in the water, or sitting on the bottom.

Yes exactly, only that during our previous discussion we left aside the case the object hasn't yet reached the bottom: In this case the balance plate knows nothing about the object's existence! In this case the only think that is sensed is the force due to the pressure of the water. It's only after the object reaches the bottom that the balance will realize it's existence. Mind that although the balance is a mass measurement device, it does not actually magically count the number of molecules! The mass is just indirectly perceived by measuring the gravitational forces that are applied on the plates; therefore, we need to see which forces are actually applied.

Let me reiterate what I said in my previous posts and sum up with this extra case:

IF the object floats (either on the surface, or inside the water in case it has the same density with it) then it displaces as much water as it's weight. This will result in the increment of the force applied to the bottom of the glass by the same amount due to the rise in water level (the math is trivial). But exactly this weight is the amount of weight subtracted as soon as the object got un-hung from outside the glass in order to be moved inside. So indeed there will be no change in the total weight, before and after the operation.

Now, if the object is more dense than the water, then it will sink and eventually reach the bottom (this case also covers the possibility to affix the object somewhere inside the glass). Again the water will increase it's force to the balance plate by the amount of the weight of the displaced water. On its turn, the object, by having direct solid contact with the glass will apply a force which is as much as it's weight MINUS the buoyant force, which ... da dah, equals the weight of the displaced water. It is easy to see that the total weight will again be the same, before and after moving the object to the interior of the glass.

Finally, let's cover the case where the object is in the course of sinking but hasn't yet reached the bottom: In this case, the balance plate has no damn knowledge of the existence of the object. As I said in a previous post, if you are submerged and a boat happens to pass over your head (or is sinking right above you!), you have no idea what's going on by the sense of touch alone. The water, being a fluid, takes care to smooth the pressure gradients almost instantaneously. You can only sense an increment of pressure at the moment the boat was put on water level, no matter when and where this happened. Back in our case, the balance will only sense an increment of weight induced by the increment of the water pressure equal to the weight of the displaced water. At the same time, it will sense a decrement of weight due to the fact that the object was un-hung. The total new weight sensed by the balance will thus be less by the amount:

weight of object - weight of displaced water, or if you like:

volume of object x difference of specific gravities of object and water

It is to be understood that even if there was a solid contact of some sort (e.g. the object is slipping down a pole affixed to the glass) again, as soon as there is acceleration, there would be a decrement of the force sensed by the balance. That's despite the fact that the overall mass doesn't change. In all accounts, such situations have to be handled differently.

"In this case the balance plate knows nothing about the object's existence! In this case the only think that is sensed is the force due to the pressure of the water. It's only after the object reaches the bottom that the balance will realize it's existence."

Let me see if I understand what you are saying. If you think that the phase where the object free falls through the liquid will be a time where the balance does not read the falling object's mass, I think you may be incorrect.

The liquid media is not friction free and you also need to consider buoyancy (if any). In other words, the liquid media still supports the falling object as it descends down into the container. If the glass contained a vacuum instead of a fluid the free fall phase would exclude the object's mass from being read by the balance, but in the case of the fluid, almost all of its mass is still part of the total system's mass.

The difference in the two scenarios (liquid vs. vacuum) depends on the difference in downward acceleration of the two objects.

As a thought experiment, imagine you are inside a submarine slowly descending/sinking through the depths. You still feel the force of gravity on your feet during the descent.

The liquid media is not friction free and you also need to consider buoyancy (if any).

Certainly there is friction, but in a fluid all pressure gradients tend to smooth out, therefore friction does not get transfered to the balance plate. Now, in the case you nervously splash the object you will create pressure waves that do reach the bottom and make the balance oscillate for a while. This is also the case if object descends so fast that creates turbulence. I assumed so far that we don't have such "violent" situations, but even in this case, the plate would make no difference whether there is a heavy metal object or rather a bubble of air of the same volume.

As for the buoyancy, I do account for it, but this is something that affects the object not the water or the balance plate!

In other words, the liquid media still supports the falling object as it descends down into the container. If the glass contained a vacuum instead of a fluid the free fall phase would exclude the object's mass from being read by the balance, but in the case of the fluid, almost all of its mass is still part of the total system's mass.

Mind that the total mass is irrelevant for the balance, because - as I already said in the previous post - the balance does not count molecules of any mass system. It just senses the forces that certain objects apply on its plates .

Moreover, imagine the fluid being neither vacuum or water, but something in between: air. Will this change anything? What if we have a lab (i.e. not absolute) vacuum?

As a thought experiment, imagine you are inside a submarine slowly descending/sinking through the depths. You still feel the force of gravity on your feet during the descent.

But we are not interested in the force that the sinking object feels for itself. You should rather consider the force that the ocean bottom (or a submerged diver) feels.

Back to the sub.

So, let's pretend you weigh 200 lbs.

As the sub slowly sinks, your feet impart a 200 lb force on the sub's floor, which imparts a force where?

Even if the sub is in a large tank, which is on a large scale, you would see virtually no difference in the "weight" of the whole system (tank, water, sub, and you) between the state where the sub is going down and when it reaches its resting position on the tank floor.

Note, this assumes that the rate of acceleration for that descent is a small fraction of G (9.8 m/sec•sec).

So, the sub can float, slowly sink, or rest on the bottom and the total system weight = the total system mass in all three cases.

If you still do not believe me, try this experiment. Stand on a scale with a 20 lb weight and hold that weight over your head. Note the reading on the scale.

Now, slowly lower the weight and watch the scale. Note the reading.

Lastly, when your arms are down, note the reading.

This is exactly the same thing as the water. In the experiment you represent the water slowly yielding to the weight, lowering it down. The vectors of force are identical.

I'm afraid you still fail to make the appropriate analogies. The thought experiment you propose with the object held at various positions has nothing to do with our case which employs a fluid.

Of course, we still agree on the generic notion that macroscopically the mass of the water holds the object. We still disagree on the way the weight force is transferred (i.e across a of water or not) and more importantly on the force analysis in the case the object is sinking.

I think I made clear the distinction between solid contact and being in a fluid. I repeat that solid objects transfer the force applied to them across their length preserving the direction. It's a molecule mess there that does the job. If you push a solid object against a second one, then it is as if you pushed the second object directly.

If on the other hand you push the air or water, then nobody will feel anything in some distance away. (Forget any turbulence, which anyway will only cause an oscillation back and forth the same initial resting point, and moreover all points in a imaginary sphere will be affected the same). The difference here is that the fluid hates pressure gradients and tries to smooth them in a speed depending on viscocity. (The only pressure gradient in our case - the same one that causes the buoyancy - is the one caused by gravity).

Here are some more appropriate analogies to use in our case (some have already been mentioned), so that you realize the difference between solids and fluids:

1. You are a diving in the ocean and suddenly a huge boat passes over your head. Again, if we forget the turbulence, (which anyway might reach you after the boat has been gone) you will feel no extra net increase of the vertical force applied on you. You would only have felt something (infinitecimal in the case of the ocean) the very time the boat was launched in the first place as the water level increased. But thereafter, the position of the boat wherever on earth wouldn't make any difference to you. You cannot just extend the boat's weight vector across any particular direction. The effect of the boat's weight is universal in all the water mass and what the bottom really senses is the increment of pressure due to the boat launch. Nothing else.

2. You submerge an object in a bucket with water. You will realize that even if you hold the object nearer to the surface or nearer to the bottom there would be no difference: In both cases there will be an upward and equal buoyant force. Can you make a similar analogy with a bucket full of sand? No! Not only there will be no buoyant force (unless the sand behaves as fluid e.g. caused by vibration, etc.) but the nearer to the bottom the object is the bigger the downward force it will experience, due to the column of the material above it.

3. You are on a balance holding an object in your hand. Then you drop it. Take these sub cases: a) You are in a vacuum, b) you are in the air, c) you are in the water. Is there any reason why these cases should be different? How could the balance know that the object is heading downwards and it will eventually hit it, or it will drop some place a centimeter off?

With all the above I think it is clear that analogies do not hold between solids and fluids in our case.

But if you like, take this as-close-as-possible solid analogy to our case: Imagine the object slipping down a pole that has direct contact with the bottom of the glass. There are solids all over the place. We can also forget the water whatsoever. If there is no acceleration then no problem: it is obvious that the balance plate has full knowledge of the object's weight. If nevertheless the object accelerates, then the plate will know something less of the object's weight.

I think this also makes clear that even if we consider the system macroscopically as a black box containing some masses (the perfectly valid approach you prefer to make), still, the case we have acceleration has to be handled differently.

I think we are arguing semantics now. I am not sure.

The bottom line is, the net weight of the system is the same as far as the scale goes, with the exception of the case where the object is sinking in the fluid. Then, there is a slight negative weight difference due to the acceleration of the object through the fluid.

Does that sum it up?

Exactly!

We can now rest our case

Good, I don't know about you, but I am up for a beer tonight. :)

"As the sub slowly sinks, your feet impart a 200 lb force on the sub's floor, which imparts a force where?"

The force is not transferred, because the sub is rigidly connected to nothing: the force accelerates the sub.

My one Good Answer, and it's wrong.

The right answer (when the object sinks) is in #52, succintly summarized in #58, and laboriously explained in #105.

I think that someone should do the experiment of letting the metal mass drop slowly down into the water and check to see if the measured weight, during it's descent, is equal or different than the starting weight with the metal object outside the glass (or the water).

I tested it and the only thing you notice, is a temporary increase of the weight when the object touches the bottom. This comes back to the normal unchanged weight. This pulse is due to the kinetic energy of the mass dissipating when it hits the bottom.

The accuracy: well maybe someone with more sophisticated equipment can try it to confirm or not.

Myriads of experiments already done support the very basic physics sufficient to analyze this configuration.

Falling through water, the mass of the object exhibits an acceleration less than that of gravity. f=ma tells us the reduced acceleration results from a reduced net downward force on the object. This results in turn from the upward buoyant force on the object which is equal to the weight of the water the object displaces. The displaced water raises the head (of water) inside the container by some delta. The effect of the increased head is to add a downward force equal to the weight of water of the delta-tall cylinder (of water) that appears above the former surface of the water when the object is immersed. That cylinder of water is exactly the water displaced by the object. The downward force it adds is the weight of the displaced water. The upward force of buoyancy on the object is also the weight of the displaced water. These forces net to zero because they are antiparallel.

Thus the net downward force measured by the scale does not change, even though the pressure on the inside bottom of the container increases.

That is what the challenge invites us to assert and to explain.

If you think you understand, you might consider a cork attached to and near the bottom of a water filled-container on a scale. What happens to the scale-weight and the water pressure on the bottom when the attachment is broken? At first? While the cork is rising? When the system reaches equilibrium?

I was wrong to pooh-pooh this experiment.

The cork (of my previous post) revealed the missing detail

If a denser-than-water object is suspended immersed in water from a hanger attached to the water's container and then released, the weight shown by a scale supporting the container will immediately decrease by the immersed weight of the object, that is by its weight less the weight of water it displaces (because the sinking object no longer has a rigid connection to the container). The scale will read this reduced weight until the object reaches the bottom, when the reading will return to its starting value.

The scale-weight increase when the object reaches the bottom has nothing to do with kinetic energy: scales read force.

The object sinks because a downward force acts on it. That force is the net of the object's weight and buoyancy: its immersed weight. As the object sinks, the force of its immersed weight is not transferred to the container--since there is no rigid connection between them--and so is not seen by the scale. When the object reaches the bottom of the container and is no longer free to be accelerated by its immersed weight, that weight, that force, is transmitted by the rigid container to the scale.

In sum: (1) immersed object suspended from the top of the container: scale reads system weight; (2) object sinking: scale reading reduced by the force driving the object to the bottom; (3) object on the bottom: scale reads system weight.

Worth noting: if the object is tethered to a support connected neither to the container nor the scale, so that it's supported before it reaches the bottom, the scale reading does not change when the object reaches the end of the tether.

Well back in the olden times and in the day of the platform scales and washtubs and the days we noodeled a lot of big fish. We wouild put the tub on the scales, fill it with water 3/4 full and mark the weight and then put in a 25 pound flathead. The total weight would not increase 25 pounds. While this confused a 12 year old in the 1940's and now 'mass' makes the difference. What is confuseing is how we get the steel bolt to float.

It's early and I didn't get much sleep last night so someone please tell me what I'm doing wrong here.

My answer is Yes the weight will change.

It took me a while to wrap my head around this but the only thing that's different in the two scenarios is that the metal object in the first scenario had the upward buoyant force of air acting on it. In the second scenario, the metal has the upward buoyant force of water acting on it.

The water will exert a higher buoyant force so when I do force balance, the normal force on the glass bottom goes down.

Am I missing a force in my FBD?

It doesn't matter what the relative buoyancy is of the metal. Even if it is less the game plays out the same.

Look at my reply in #30 and the post I am responding to.

In the end, however much mass is on the balance platter is exactly what will be read by the balance. It really does not matter if the object is attached to the glass, floating on the water, in the water, or sitting on the bottom.

The only time you would have a discrepancy is if the block is in motion (accelerating up or down) in the glass of fluid.

The pressure of water (plus metal) on the inside bottom of the glass increases.
The pressure of the scale platform against the outside bottom remains the same.

Yes and Yes

At least you're not a nay sayer !

Niether the weight or pressure will change.

No, the overall weight measured is the overall mass as affected by graviational acceleration. The same mass is present. The average pressure on the bottom of the glass does not change since the mass supported is constant, and the area of the glass bottom is the same as before. The liquid pressure at the bottom of the glass will change somewhat, depending on the volume of the 0.25 Kg object, as this will increase the overall column height bearing force against the area of the bottom of the glass.

Assumptions:

1. The glass is not filled to the rim.
2. The balance has an overhead balance beam with hanging pans.
3. The glass is placed with its CG directly under the pan attachment to the beam before the weight is added.
4. The weight is shaped like a boat so it can float.
   

The weight will not change.

The pressure of the glass on the pan will have a gradient when the weight is outside the glass (the pan will sway so that the new CG is under the attachment point). This gradient will disappear after the weight is floating in the water in the glass (the glass will swing back to its original position).

If the weight were simply hung on the inside of the glass instead of the outside, the weight would be the same but the pressure gradient would reduce.

I am reminded of a ship canal aqueduct that spans over a road. You can drive along the road and under a ship crossing the bridge. The question then was,"Does the aqueduct feel a heavier load when the ship is on it?."

I consider this to be on topic.

Since the Total Mass will not change we will have the same reading on the scale.

The Down Force will be the same in both cases: F1 = F2

The Pressure issue: { To simplify the writing, I will omit the acceleration and S.G. etc}

Let us take S=area of the base of the Glass; P1 pressure at the bottom of the filled glass in case 1; P2 = same for the second case.

Case 1: F1 = 0.5 (Glass) + 1 (Water) + 0.25 (Metal hanging outside)

OR F1 = 0.5 (Glass) + S x P1 + 0.25 (Metal)

Case 2: F2 = 0.5(Glass) + S x P2 +(0.25 - V (Displaced Volume of the Metal).

Since F1 = F2, we can see that we can write:

S*P1 + 0.25 = S*P2 +0.25 - V

and S*P1 - S*P2 = -V thus, P2 <> P1 and that is why the pressure will vary while the Mass remains the same.

Note that the value of V can be = 0, meaning that the Metal is not immersed, and then P1 = P2 while the Mass remains the same.

I think that this explains, simply, all the cases.

You are in a track!

Abe

No...Yes.

I finally read all the posts and see that more detail is required in my response.

We all(mostly) agree that the weight indicated by the balance will not change provided the point of attachment of the string supporting the weight is not changed. So, when the weight is on the outside of the glass, it's weight(and that of the string, if you must) is applied to the rim of the glass, and so is part of the [glass, string, weight, water] system.

So, why would the force on the bottom of the glass(inside) change when the weight is moved inside?

1. If the metal weight does not touch the water, the situation is no different than when it was outside the container. A good analytical balance can accomodate positional changes in the material being weighed provided it remains on the balance pan and does not touch anything external to the system and balance pan.

2. If the metal weight does touch the water, the amount of water displaced will raise the water level, which will increase the bottom(inside) pressure. This would seem to make the system register an increase in weight, but the tension on the suspending string and consequently the downward force on the rim of the glass will be decreased by the same amount.

Voila! Need I say more?

The scale will weigh .25kg less because the weight does not hang on the side any more ;I presume it is now being held by hand ...say on a string??

The pressure at the bottom of he glas will be more because the water height is more because the weight displaced some water.

If the weight no longer hangs from the side of the glass, then what you say is true.

BTW, an assertion based on a presumption is a tenuous thing. I would expect it from a politician, but not from an engineer.

I'm okay with any engineer making and assumption and a corresponding analysis if they clearly state that there is an assumption that needs to be made.

I see it as a flag that some requirement is missing and needs to be added or clarified. Better that simply replying, "Insufficient data."

If one could assume it was set up like this then the answer to both of the questions is:

1. yes, the object will weigh less in the water, but its mass remains the same

2, yes the force on the bottom of the container will increase due to a higher head pressure caused by the displaced water

After reading the responses I wonder if we were correct in some of our assumptions. If I take literally, "A tall 0.5 kg glass filled with 1 L water" and "and the weight taken again" I might think that the glass was full, was removed from the balance, and the weight placed inside, spilling some water and then placed back on the scale. Ignored because there was then, no way to answer the challenge.

I might also wonder if "0.5 kg glass" is one that weighs 0.5kg or one sized to hold 0.5kg of preserves or jelly or peanut butter. Such a one would not have held a litre of water. Again, the challenge would be meaningless.

Then again,there is "Does the pressure on the bottom of the glass change?" I took it to actually be the bottom of the glass, not the bottom of the water as some of us did. I did it because there was no way to know, when the weight was moved inside, whether it was hung on the glass above the water or immersed in the water to some extent. I could only answer for the bottom of the glass.

Then, as I pointed out if it is an overhead beam type of balance, the glass must be perfectly centered under the suspension point or the pan will swing sideways until the CG of the pan and glass is directly under the suspension point, and the glass will be on a sloping pan with a pressure gradient across it. This did not remove the possibility of an answer because we were not asked for the pressure, just if it would change.

I read the challenge as saying the object is suspended from a hanger hooked over the rim of the glass, and the object is submerged when moved inside. As many have said, neither the weight of the system nor the pressure on the outside of the bottom changes; but the pressure on the inside increases, a change exactly nulled by the change in the force exerted on the rigid glass by the hanger. Modulo the interchangeablity of mass and pressure--interchangeable because the relevant area and the acceleration of gravity are constant--the magnitude of both changes equals the difference in the mass of the object and the mass of water it displaces.

The total weight will remain the same as nothing has been added or removed. The pressure on the bottom of the glass will be increased as will the pressure on the sides of the glass by the amount of weight/mass of the object orginally above the water line.

The question asks if the weight changes. Weight is a measure of the force of gravity on a mass (w=mg). The mass will not change therefore the weight will not change, however the balance will register a different weight.

This is because for a balance to read correctly, the center of gravity must be in the center of the platen. Fundamental law of levers: F1*D1 = F2*D2, where D is the distance from the fulcrum to the center of the load. When the object is ON THE OUTSIDE of the glass, the center of gravity is moved slightly away from the fulcrum of the balance. The resulting increase in distance from the fulcrum to the glass makes the object read heavier than it actually is.

When the object is moved to the INSIDE of the glass, the center of gravity is moved toward the balance fulcrum thereby reducing the distance to the fulcrum. The weight will now read as lighter.

Example:

Total mass of object = .5kg(glass) + 1 kg(mass of 1L of water) + .25kg(object)=1.75kg

Weight = M*G = 1.75 kg * 9.8m/s/s = 17.15 kN

let D1 = distance from fulcrum to weights = 10 cm

Let F1 = Measured weight

Let D2 = distance from fulcrum to center of gravity of glass, water & object assembly. = 10 cm

Let F2 = Actual weight of assembly. = 17.15 kN

F1*D1 = F2*D2

therefore

F1 = (F2*D2)/D1 = (17.15kN*10cm)/10cm = 17.15kN measured weight

Now if we move the object to the inside of the glass reducing D2 to 9cm:

F1 = (F2*D2)/D1 = (17.15kN*9cm)/10cm = 15.43kN measured weight.

Now for the pressure question:

The problem says that it is a TALL glass. When the object is moved inside the glass, water will be displaced causing the level in the glass to rise. Pressure in a column is proportional to the height of water in the column: P(pressure) = d(weight per unit volume)*H(depth). Therefore when the water level rises, the pressure at the bottom of the glass also rises.

Example:

Let us assume that the water level in the glass before putting the object in is 1 foot.

Let p = pressure at the bottom of glass

Let d = specific weight of water = 62.4 lb/cubic foot

Let H = depth of water in glass = 1 foot

P = d*H = (62.4 lb/cu.ft.)*1 foot = 62.4 lb/sq.ft. (psi)

Now we add the object. The water level is raised to 1.5 feet:

P= d*H = (62.4 lb/cu.ft.)*1.5 feet = 93.6 psi

Conclusion:

Yes the weight changes. Yes the pressure changes.

You just joined this forum.

We tend to read all the previous entries by other colleagues and, if we need to add a new idea, then it makes sense to contribute it.

All what you wrote was already debated in all directions including what you just wrote.

please read the other entries. It makes life easier for we (others) trying to follow the discussion and maybe learn something we might have omitted.

Thanks for your contribution, and you are welcome.

Sorry. All I did was click the link I got in my email and started typing. I thought everyone gets a chance to answer. Forget this whack forum. I'm outta here.

there are 2 cases do distinguish:

1.) the glass is complete filled with water onto the upper corner of the glass:

- then the water flows from the glass to the pan of the balance if you move the metal object inside the glass (if the water stays on the pan, the blance shows the same weigth like before; if the water stays not on the pan the weigth decreases)

- the pressure at the bottom of the glass is not changing

2.)the glass is not complete filled with water:

two cases again:

2.1) the volume of the glass is greater than the sum of the volume of the water(1liter) and the volume of the metal object

- the weight doesn't change

-the pressure at the bottom increases

2.2) the volume of the glass is less than the sum of the volume of water (1 liter) and the volume of the metall object

- water flows from the glass onto the pan (see 1.))

- the pressure at the bottom increases

3.) the metal object doesn't touch the water level if it is changed inside the glass than thewre will be no change of weight and no change of pressure at the bottom

Correct! All the mass that was there before the change is there after the change, so the weight remains the same.

But off hand, don't agree with the increase pressure on bottom of glass. Why?

Keeping area at bottom of glass constant, if additional water was added for additional height, of course pressure increases due to increase mass of water. If the glass was made taller and thinner, thus increasing water column height and decreasing area at bottom of glass, then again, pressure increases because mass per square inch increases.

But in this case the water is displaced by the submerged object, so total mass of water remains unchanged and area at bottom of glass remains unchanged, so mass per square inch remains unchanged.

Remember:

in a liquid, the height of the liquid and its S.G. determines the pressure at the bottom!

The total mass (weight) of the liquid has nothing to do with the pressure, here on earth (at least).

refer to Torrichelly's experimet: a Drum of water + a Vertical pipe of a small diameter attached hermetically to it, and having a height of # 10m, filled with water, will probably burst the drum ... revise the whole concept.

Sorry, but this is a lot of talk but none addresses the subject matter.

Your comparison with a car transporter is out of its place here.

the weight (metal piece) was out of the water and then we assume it gets completely into the water The water surface was at a height H1, in the glass. After immersion of the metal piece, the water surface goes up to a new height = H2, with H2 > h1.

Can you now say that the pressure exerted, at the new H2, on the inside bottom of the glass is bigger than before immersion? If yes, the we have no argument. If No, then you have a problem in this field of physics.

google' Water Pressure Law' or visit the link http://en.wikipedia.org/wiki/Pascal's_law

just to refresh your memory.

Bad metaphor. My reference was specifically to unit pressure (psi, whatever.), not to overall force exerted as pointed out in the first sentence. Your comparison equates to lifting the entire glass itself, not increasing the height of the water column.

Here's how it works:

1. Pressure is the sum of the force exerted on a given area
2. This force is quantified by the WEIGHT (not mass, as the pressure would be different on Mars or the Moon) that is acting directly on that area
3. As we are dealing with a liquid under ambient conditions, we can reference to "gauge" pressure (pressure above ambient) instead of absolute pressure
4. Also, as we are dealing with a liquid the pressure is not only directed downwards but creates an equivalent pressure laterally
5. Pressure varies in a linear fashion. It goes from zero gauge at the surface level of the fluid up to whatever the pressure is at the bottom of the vessel containing the fluid (incidentally, this is the pressure that submarines- of the diving sort, not edible- are designed to with stand).
6. Any displacement of fluid that raises the height of the fluid column INCREASES the amount (& therefore both weight & mass) of the fluid directly over & therefore acting on a given unit area under that fluid. Therefore it increases the pressure at that point. Think of as running your 4 cars on 2 carriers. Now put all 4 on 1. Which has the highest pressure exerted on the bridge.
7. Keep in mind that an object has to exert force (and therefore have mass) to displace a fluid. The weight measurement of that force is what displaces the fluid; the pressure measurement at the given base area (or any point below the surface of the fluid) increases as a result of the additional force up (but not exceeding) to the amount of the displacement, not the weight of the object.

The formula for static pressure is:

P = h x rho x g

Where P = Pressure
Where h = height (or depth from the top surface)
Where rho = density of the liquid
Where g = force of gravity (9.8 m/s/s)

Obviously, the deeper you descend into the ocean the more pressure you encounter. It matters not whether that "container" of liquid is thousands of miles wide or 1" wide. The formula holds true.

while lifting the rock ...

The problem statement doesn't say that the weight is at the bottom, merely that the weight is moved inside the rim of the glass, so it would be a safe assumption that the effects of buoyancy are negligible.

F=PA. The A hasn't changed, and all that's happened is whatever moment there was from the weight hanging at some distance off-axis, is now on-axis, (or closer). So the scale won't read any differently, but the pressure distribution around the bottom of the flask will be different.

Bouyancy cannot be neglected: it's the reason the weight of the system doesn't change even though the pressure increases. The increased pressure increases the downward force on the scale; but the net force on the scale doesn't change, because another downward force is decreased by exactly the same amount. That force is the weight of the object, and what decreases it is the buoyancy.