Challenge Questions

I would say the block would have a tendency to rise a bit at the start then settle back to where it began. But I'm not so smart.

Nice job. Roger. I see why your GA score has been rising.

A little further on in the same Wikipedia citation they give the following.

The net force on the object is thus the sum of the buoyant force and the object's weight

As you can see 'g' now appears twice on the left side of the equation and so can not be simply canceled out. Unfortunately, there would be some minuscule effects on the buoyancy of the wood during acceleration and deceleration of the elevator.

Further wikipedia gives.

Commonly, the object in question is floating in equilibrium and the sum of the forces on the object is zero, therefore;

Since the acceleration/deceleration of the elevator causes disequilibrium conditions the formula you cite does not apply and should be amended such that

O=mg-(m(g+a)-pV(g+a)) where 'a' is the elevator induced acceleration. Integrating the result over time would give an even more precise description of the displacement of the silly little block of wood, but it would also cause enormous amounts of posterior pain and I am leaving on vacation instead.

Good day and God bless one and all,

Sincerely,

Mr. Gee

Hi Mr Gee,

You said:"The net force on the object is thus the sum of the buoyant force and the object's weight

As you can see 'g' now appears twice on the left [sic] side of the equation and so can not be simply canceled out."

You offer the following equation as the correct one:

O=mg-(m(g+a)-pV(g+a))

However that doesn't seem to make sense. Take for instance the situation where acceleration a=0, then your equation above yields:

O=mg-mg-pVg=-pVg (This can't be correct, right?)

Is that O in the beginning zero or a variable called O? I can't tell, either way it seems wrong, but maybe I'm missing something.

Roger

Well, I had a great vacation but, now I'm back.... Sigh....

Taking the case were a=0 is the static case mentioned above. They used F(net) I used mg. I was attempting to show that when the acceleration a is not equal to zero then the simple static case is changed so that it is different from the dynamic case.

in the above equations mg is the weight term and -pVg is the buoyancy term.

When the elevator moves the frame of reference for the static balance system described above is changed.

The statics equations do not show this change because the new acceleration is "outside" their frame of reference.

To understand the dynamic case it could be better to exaggerate the acceleration involved and ask what would "really" happen. If there was a 10 foot loop in the elevator cable bound by a tie wire then when the wire was cut the elevator would accelerate under g until it hit the end of the cable loop and rapidly stop.

Now the inertia of the to masses, the water (more dense) and the wood (less dense) come into play. During the initial drop the water conserves more of its static inertia than the wood so the water is more left behind than the wood. Then when the cable runs out the case is reversed. The water conserves more of its dynamic potential and will hit the bottom of the container hard, driving the wood upwards.

Since these are dynamic inertial effects over time successive iterations of the static model including the non-zero elevator acceleration a are needed to correctly model the problem. The equation I gave was a good first effort at expressing the reality of the modelled system. I have no idea if it is really correct. I just strongly suspect that it is a better model than the "good answers" above which do not account for the dynamics of an accelerated frame of reference in spite of the fact that the problem clearly states that one exists.

Sincerely,

Mr. Gee

Perhaps you should re-read the G.As? If they did not take account of variations in acceleration they didn't do anything. They didn't explicitly consider the effect of the rate of change of acceleration, but this is not relevant if the change happens relatively slowly (relative to the propagation of compressional forces in the water).

Of course the question specifically excluded the case of the lift being released into free-fall, because there is then no residual force to keep the water in the glass (or the block in the water)

Mr. Gee

I'm sorry, I still don't understand what you are saying. You don't have to use words, just please write your equations out more explicitly so that I can follow them. I assume you agree that the equation:

is correct. I think you are suggesting that where I assumed F_net=0, you assume some other value for F_net (mg?). If that's the case, please show all work so that I can try to understand what you are saying.

Roger

F(net)=mg-pVg is the agreed upon formula for the net forces, buoyant and weight in the unaccelerated frame of reference. Where F(net) is the net force acting on the buoyed block of wood. The mass of the wood is 'm', 'g' is gravitation, 'p' is the density of the water, V is the volume of water displaced by the wood according to Avogadro's principle.

Adding an acceleration 'a' from the elevator's motor changes the formula to;

F(net)=m(g+a) -pV(g+a)

expanding gives

F(net)=mg+ma-pVg-pVa

gathering terms

F(net)=mg-pVg+(ma-pVa)

in this form of the equation the net force is shown to be composed of the static gravitational component and the dynamic elevator acceleration component.

I found a citation that indicates 3ft/sec^2 is a reasonably smooth elevator acceleration with a max jerk of 5.6 ft/sec^2. I also found some real time data that gives peak acceleration deceleration values in the 0.5 to 0.9 m/s^2 range or lets say 1/10g.

The maximum inertial component then would be 1/10g*m the mass of the wood reduced by the buoyancy effect given by m*1/10g-pV*1/10g.

If the density of the wood is 1/2 that of the water then when 'a' is negative – acting in the same vector orientation as gravitation - then the wood is maximally 1/20th less buoyed when the elevator starts to descend unaffected as the elevator has constant velocity and is 1/20 more buoyed when the elevator's descent is being halted as 'a' would have the opposite vector orientation as gravitation at that time.

This is exactly the behavior I would expect out of this system. As waves will form in the water due to the changing displacement of the block of wood the predictions of this model should be easy to check. Waves = changed displacement - no waves = no change. I don't have an elevator handy. I will await with some interest the report of someone who does have an handy elevator.

Please note that the above does not refer to air currents or anything of that sort. Merely a discussion of static and dynamic frames of reference as they effect buoyancy.

Sincerely,

Mr. Gee

PS. if anyone is having difficulty visualizing the intermittent nature of the external accelerations of an elevator here is a nice data set.

http://hypertextbook.com/facts/2005/elevator/acc-graph-1.gif

It remains impossible to be certain what you are saying. My guess is that you are effectively taking the specific gravity of the wood as unity in part of your calculations, and as 1/2 in others. Either that or you are ignoring the change in the force on the block that is needed to allow it to accelerate with the elevator.

Rather than trying to disentangle what you write, I'll re-derive the acceleration of the block using a format that is similar to yours (maybe Roger can do it better):

So we have (before acceleration):
Buoyancy = m_block.g

During acceleration (until block moves relative to water surface) we would have:
Buoyancy = m_block.(g+a)

So force on block due to buoyancy plus gravity combined is:
F_block = m_block.(g+a) - m_block.g = m_block.a

That results in the acceleration of the block being 'a' - exactly the same as the water - so the block moves exactly with the surface of the water.

Fyz

It should not require explanation that 'mg' as in F=mg is the formula for the weight of the block.

The buoyancy of the block is given by -pVg.

When the weight of the block is offset by buoyancy the net forces are given by;

F(net)=mg-pVg

When g is augmented by the acceleration or deceleration of the elevator the net forces acting on the block are;

F(net)=m*(g+a)-pV*(g+a)

I do not believe it is possible for me to be more clear.

Your derivation of Buoyancy=mg (weight) is simply not correct.

Roger, this means you are correct. F(net) and mg are not equivalent. In my attempt at a good equation I was working fast and loose trying to get away for my vacation. The second form is more rigorously derived and should correctly model the buoyancy of the block under changing external acceleration.

The greatest difficulty we seem to be facing here in our attempts to communicate about this problem are that the groups answer, that the block's buoyancy is unaffected is correct for the conditions when the elevator is standing still, raising or descending at a uniform rate, or at any altitude on this planet. Those are all steady state conditions.

The only time the blocks buoyancy is effected is when the elevator is starting or stopping - when its rate of motion is being effected by an external acceleration. Buoyancy is conservative under any and all steady state conditions. Inertia is always involved when the state of velocity is changing. My best estimate is that under the conditions given in the problem the level of the block of wood will vary by about 2% during the 3 seconds or so that a typical elevator uses to accelerate and then again during the 3 seconds it decelerates.

This change will be a transient change, as we are in agreement that the level is the same when the elevator is either at rest or moving at a uniform rate.

If any or all of you wish to remain convinced other wise I suggest we politely agree to disagree, although actually conducting the actual experiment remains another option...

Best wishes,

Mr. Gee

At least one thing is clear.

You write: "Your derivation of Buoyancy=mg (weight) is simply not correct"
however: if buoyancy when the elevator is not equal to mg (of the block), there will be a net force on the block. So the block will either rise or fall.

Let me rewrite F(net)=m*(g+a)-pV*(g+a) as F(net)=(m-pV)g +a(m-pV). For the elevator not moving (a = 0), F(net) = 0 so m-pV. For any value of a, F(net) is still zero. I still believe the block does not change position relative to the water's surface.

Thanks,

Jim

Excuse me but when F(net)=0 no acceleration occurs at all. You cannot have a≠0 and F(net)=0.

I believe that Jim is using F(net) as a source of relative acceleration. It's very difficult to convey on paper, even if everything is carefully set up, explained and consistently worked. The problem is even worse when the relativistic nature of the equations is not explicit - and in any case what is at issue here is that some do not accept the fundamental basis of this Archimedian-Newtonian relativity.

I think that net absolute forces and "proper" acceleration (i.e. relative to a non-accelerating frame) is more appropriate for explanation - and this is what I attempted in #101. It is clear that my attempt at clarity has not entirely succeeded - so I'd appreciate considered criticism of the way it is expressed. (I'll be happy to amend it a few times)

Thanks

Fyz

Can we at least agree that in an inertial reference frame, that F(net)=ma?

Well the reference frame of an observer outside the elevator is for this case an inertial reference frame. For the inertia of this observer does not change for this experiment. While an observer on the elevator will be in a non-inertial reference frame, for as the experiment proceeds the elevator rider does have his inertia change. Thus fictitious forces must be included for the rider to explain observed phenomena.

I believe Mr. Gee correctly follows the inertial reference frame of the stationary observer. As stated "p" is the density of water,the mass of the wood is "m", "g" is gravitation constant an according to Avogadro's principle "V" is the volume of water displaced by the wood. When the elevator floor sits at rest the buoyancy stated is:

F(net)=mg-pVg

When downward acceleration, "a" (which is 0<a<g) occurs from the elevator's motor then:

F(net)=m(g+a)-pV(g+a)

Remember, this F(net) cannot be zero for the block of wood clearly accelerates downward. Doing a partial expansion of this gives:

F(net)=mg+ma-pV(g+a)

ma=mg+ma-pV(g+a)

0=mg-pV(g+a)

Solving for the volume of water displaced by the wood gives

V=mg/(P(g+a))

Identifying which components change over time give this function:

V[t]=mg/(P(g+a[t])

so at t=0 a=0 then V[0]=mg/(P(g+0)=m/P

Exactly as expected, just a reconfiguration of Avogadro's expression.

However when moving down as defined 0<a<g so

V[0]/V[t]=(m/P)/(mg/P(g+a[t]))=(g+a[t])/g=1+a[t]/g

Since this must be greater than 1 for all downward accelerations (positive "a") then:

V[0]>V[t]

So the block rises out of the water. But remember again this happens only while the block is accelerating downward, not moving at the constant velocity potion of most of the downward travel.

You write:

"F(net)=m(g+a)-pV(g+a)"

But Fnet is meant to be the force on the block as viewed by the stationary observer. So I cannot see where the part of the expression m.(...a) can have come from.
I'm sure that the expression should read:
Fnet = m.g - ρ.V(g+a) = m.g - ρ.V.g - ρ.V.a
(N.B. note that this sign for the terms involving 'a' equation corresponds to measurement of position that increases with increasing height.)

Now, in the initial condition the block is at rest and not accelerating, so
m.g - ρ.V.g = 0 (this also gives Archimedes formula: m = ρ.V)
(I believe that this is the part of my previous post that Mr. Gee contested ...)
So we have
Fnet = - ρ.V.a = - m.g
That force will accelerate the block downwards at the rate 'a'
So the block is accelerating identically with the lift.

Do we have agreement yet?

As I cited, the initial equations come from Mr. Gee's initial post. I am certain his technique of adding the acceleration component "a" to the gravitational constant "g" to solve for the total force exhibited on the wood block is very straight forward and accurate. I do not have here my Physics reference book to cite chapter and verse but this method is correct. I only added some further analysis to show that the block must rise in reference to the water level for descent to occur.

Please review your equations, you wrote:

Now, in the initial condition the block is at rest and not accelerating, so
m.g - ρ.V.g = 0 (this also gives Archimedes formula: m = ρ.V)
(I believe that this is the part of my previous post that Mr. Gee contested ...)
So we have
Fnet = - ρ.V.a = - m.g
That force will accelerate the block downwards at the rate 'a'
So the block is accelerating identically with the lift.

You have a basic algebra error. At t=0 Fnet=0=mg-pVg. Only in complete free fall with no opposing force will |Fnet|=mg → |ma|=mg → |a|=g.

The block will accelerate at the same rate as the elevator. For the wood block to accelerate in any direction there must be a net force generated. The elevator does not pull the wood block down. The descent of the water must reduce the buoyancy force that opposes the gravitational force for a net force down to exist on the block. To reduce the buoyancy force, the only option is to reduce the volume of water being displaced.
Please excuse this blunt comment. I cannot make this any simpler.

I think we need to look at both the block and the water and probably even the glass holding the water. As the elevator accelerates downward, the force between the elevator and the glass will become less just enough for the glass and its contents to accelerate downward at the same rate as the elevator. The integral of the pressure between the glass and the water will become less just enough for the water and the block to accelerate downward at the same rate as the elevator. The integral of the pressure between the water and the block will become less just enough for the block to accelerate downward at the same rate as the elevator, the glass, and the water. If the elevator was to be accelerated upward instead of downward, the same phenomena would occur. The glass, water, and block would all move together. In all cases the total acceleration (gravity plus the linear acceleration of the elevator) of the block and the water are always the same and there is no relative motion between them.

You state: "To reduce the buoyancy force, the only option is to reduce the volume of water being displaced"

The buoyant force is the result of pressure pushing up on the bottom surface of the ball. There are TWO options (and of course an infinite number of combinations of the two): 1-reduce the surface area. 2-reduce the pressure.

When the elevator begins to move downward, the upward force on the bottom of the glass will be reduced, so the glass will accelerate downward. The downward motion of the glass will reduce the pressure on the bottom surface of the water. Neglecting the transient at initial acceleration, this reduced pressure is transferred to all surfaces of the water, including the bottom of the block. No reduction in surface area nor volume of water displaced is required.

Furthermore, the first and last sentences of your last paragraph contradict each other. If the volume of water displaced were reduced, then the block must have risen with respect to the water, and with respect to the elevator, in which case the block would have a lower acceleration than the elevator.

Yes there was a single-character error in my typing (picture perhaps a bit redder than necessary for this sort of thing?). The line should have read:
Fnet = - ρ.V.a = - m.a

But you can see that the lines that followed were based on the correct version of the equation. However, as stated I was writing about the total forces on the block and not just the force due to buoyancy.

I agree that the elevator does not in this case pull the water or the object down. What happens is that the downwards acceleration of the elevator reduces the downwards force on the water, so the water accelerates downwards. The downwards acceleration of the water reduces the pressure gradient in the water, and this in turn reduces the pressure on the bottom of the block; that is the mechanism by which the supporting force on the block is reduced.

Where I disagree is when you write that "To reduce the buoyancy force, the only option is to reduce the volume of water being displaced". Do you really mean this? Or do you not consider that the equation [F(net)=ρ.V.(g+a)] relates to the buoyancy force applied between the water and the block? If you mean it, do you think that the forces would remain the same if the water and the block were placed on the moon - and if so what is the value of the Earth's gravitational constant doing in the expression?
Perhaps I'm being unfair here, but it appears that the problem lies in taking literally the way that Archimedes principle is usually expressed: "Any object, wholly or partly immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object". I think the limits of this statement become obvious if you place an immersed water-jet under the immersed object. Quite evidently the statement is only true in that simple form when the water is stationary. The equations you were using are an extension of this to a condition of uniform acceleration, and where there is no relative motion between the water and the supported object - and your equations show a reduction in buoyancy.

.

Erratum: I substituted Avegadro's for Archimedes in my post. Archimedes's principle relates to buoyancy.

I don't know if you realize it yet Jim but we are in good agreement now. Your algebra looks correct and your conclusion that when a=0 buoyancy is uneffected is also correct.

This is the relevant portion of my post,

"The only time the blocks buoyancy is effected is when the elevator is starting or stopping - when its rate of motion is being effected by an external acceleration. Buoyancy is conservative under any and all steady state conditions. Inertia is always involved when the state of velocity is changing. My best estimate is that under the conditions given in the problem the level of the block of wood will vary by about 2% during the 3 seconds or so that a typical elevator uses to accelerate and then again during the 3 seconds it decelerates."

This weekend I will get to town and I'll see if I can film a block of wood in a petri dish as an elevator moves. I'm predicting 2 waves. One on acceleration to descent and one on deceleration to stop.

Best wishes,

Mr. Gee

I'm finding it hard to cope - you say in one breath that you agree with Jim (who I think says "no change during acceleration"), and then talk about a 2% change. What do you think is the cause of the change? Which direction do you expect it to be?

The experiment will be fine if you can prevent the elevator wobbling - that will create waves, which will of course have nothing to do with vertical acceleration or a change in height.

"prevent the elevator wobbling"

Right! so the experiment needs at least two glasses or petri dshes, one with and one without the floating block, and an elevator that isn't in constant use, so the water can settle before starting the experiment.

Possibly hang the glass from a cradle? Not perfect, but it works to prevent coffee spills.

Right. I'm picturing something akin to a plant container hanging by three cords under a tripod. (Again, two identical setups, one with and one without the block)

I'm glad to see you guys are finally going to do a physical experiment with this. I've been gettig dizzy trying to follow the math.

Mr. Gee,

In your original post you gave the following equation:

O=mg-(m(g+a)-pV(g+a))

In your last post you've given:

F(net)=mg-pVg+(ma-pVa)

I don't see how these are equivalent unless you make F(net)=mg, which doesn't make any sense. So was there a mistake in your original post? Or are these two equations related somehow and if so could you show me please so I understand.

Roger

This really is over-complicating matters. But if you do the full algebra you will find that this predicts that the block will move with the fluid it is floating in.

Basically, Fnet is actually required. The required value for this change in force on the block (so that it accelerates with the elevator) is Mblock.d²V/dt². The initial mass of the displaced water is equal to Mblock. The change in force Fnet on the block is the change in force of the displaced fluid - or Mblock.d²V/dt².

In principle the height of the block will (as jim has stated) be independent of the acceleration. But there will be a slight dependence on the density and movement of the air within the elevator. Looking at these rather small effects:

For most practical elevators, one effect is likely to be a reduction in internal pressure while the elevator is moving, regardless of direction of movement - so the block will initially sink very slightly as the buoyancy due to the air reduces. However, as the elevator descends, the pressure of the surrounding air will rise, so the block will rise again (did someone say Easter?).

If the elevator behaves as a sealed rigid rectangular box, the behaviour will depend on the position of the glass within the elevator. The air density will be more uniform while the elevator is accelerating downwards, so the block will tend to rise if the glass is above the midpoint and sink if it is below.

In practice the weight of the dust from the ceiling that lands on the block following the jerk as the elevator starts its descent might be more significant.

Are there any small relativistic effects?

Assuming that the observer is in the lift, wouldn't they be the same as added zero-curvature gravity? That would be identical relativistic effects on the block and on the water, so perhaps no consequent change in waterline?

Perhaps the next-most-important effect after the air would be reduction in internal pressure on the vessel (water and block would presumably rise together), followed by expansion of the block (that would bring a little more of the block out of the water due to air buoyancy...)

This is getting Scilly

Quote:

"Assuming that the observer is in the lift, wouldn't they be the same as added zero-curvature gravity?"

How do You know that the Observer isn't a "Plus-sized" Person? Why assume zero-curvature?

The observer didn't really need to be in the lift - (s)he's an artefact to simplify the thought experiment.

Zero curvature not an assumption: linear acceleration of the whole elevator (as specified) is equivalent to added zero-curvature gravity.

The acceleration was not labeled as linear in the question, only that it was accelerating at a slower rate than gravity would pull it down. A<G

If no one is in the elevator then an observation cannot be made... But My "Plus-Sized" observer was just a Crack aimed at the "Zero-Curvature" line.

I agree with most poster's that in relation to the surface of the Water the Wood would neither rise, or fall, but I have noticed not everyone that says this, agrees on why this is so.

I think it is because it would take another outside force "not mentioned" to jostle the Wood from it's resting point at the beginning of it's trek down the shaft.

Heisenbergs uncertiany principle has been shown to have an effect on such experiments. There is no way to seperate the experiment from the experimenter, who unconciously alters the results. I read somewhere that this was proven using bi-metal strips in a controlled environment, the only variable being the observers mind set.

I'm off to La La Land before I get thrashed for this one.

so the block will initially sink very slightly as the buoyancy due to the air reduces.

As block is floating in the water, buoyancy due to air doesn't come in picture. (Please Do not go in micro details of the portion of the block open in the air)

I think he's talking about the air pressure effecting the buoyancy of the water. The air is pushing down on the surface of the block and the water initially. When the air pressure lowers (as the elevator moves), the weight of the block and the buoyancy of the water changes unequally. If the surface area of the exposed water exceeds that of the top of the block, then the block sinks (because buoyancy is effected by the lower pressure more). If the surface area of the top of the block exceeds that of the exposed water, then the block will rise slightly (because the weight of the block is effected by the lower pressure more).

I should have read the posts more carefully - I'd made assumptions... What I meant was indeed what gsuhas called "microdetails". And, yes, gsuhas was right about micro - even for balsa-wood the maximum directly-caused change in submerged depth (assuming an 8' high elevator) would be in the region of parts in 10⁷.

Assuming that no air is pumped into or out of the porous wood, I agree that the buoyancy effect on the submerged part of the block would be the same as on the water.

However, if you are going to get pedantic on me...

The challenge states the block of wood is placed "in a glass of water" (note: not "in water"). You can either assume it sinks (-> no change other than due to elasticity of block and water), floats (i.e. some of the block is outside the water, and affected by air-density) or is quasi-neutral (potentially it moves about under the water).

If the wood is quasi-neutral, we then need to consider the effect of the pressure on the block and on the water. If the block is more compressible than water, its position will be unstable - eventually it would either float or sink. A change in air pressure could reverse this (though not in the time-scale that a practical elevator could continue accelerating). If the block is less compressible than water, it would (again eventually) move downward if the air pressure reduces.

Oh, did I mention the variation in water density due to adsorbed gas?

______________________________

To be sung to a well-known tune:
Ped-ant-ry you can do I can do worser
I can do pedantry worser than you
No you Kant
Yes I Cannes
...

I wish you had not used guest, so I could have given you a GA!

The challenge is VERY poorly worded.

Yes we can!

We may not be able to award good answers, but we sure can receive them.

On the other hand, the original level of the block of wood on the tenth floor will be even higher so in this context the wood lifted

If the elevator starts descending with an acceleration smaller than gravity (a < g)

Means the acceleration lower than the g force means no change (lower IMO means that the G force is stronger)

maybe if the elevator descended faster than the G force you would notice any change.

No change. It remains as it is.

A.Tahir DENGIZ

The block will stay in the same place. As gravity is reduced artificially inside the elevator, the effect will be the same for the wood and for the water in the glass. Therefore the wood and water would both now weigh less for the same volume of water replaced. If only the wood weighted less, then less volume would have been replaced and the boat would have risen. This however is not the case.

If there would be a change one would need vomit removalist's in every high rise around the planet. What a bloody stupid question!

If one would travel faster that 60miles an hour one would disintergrate into a huge mess. What a rediculaouse question. I cant get over it. It must be the drag of unfullfilled physlcal desires or pure bordom to call this a challange. Go and do some thing nice to your bodies, this stinks, Ky.

Before looking at other posts - it makes no difference. On Earth's surface the wood floats at an immersion where the downward force on the wood (= mass of wood x g) = the upthrust due to water displaced (= mass of water displaced x g). Both are proportional to g so the value of g makes no difference. In the descending elevator the effective gravity is g - a but still cancels. This ignores any effects like compression of the wood or water.

Codey

If a < g, then the block should remain at the same level, because the water and the wood are both acted on equally by g ??

Yes - the "net" G (g-a) is the same for the wood and the displaced water. Unless someone can show differently, it does not seem that the result would be different if a=g or a>g. Thoughts?

Actually "Guest" has the most correct answer.

"

If the elevator decends from the tenth floor to the first floor, then the original water surface (tenth floor) will be far above the small block of wood now found in a glass of water on the first floor.

"

I agree: "Guest" provided an excellent perspective and caught the questioner"s weakness of not properly explaining that they meant the water level relative to where it was originally with respect to the block of wood. GA well deserved!

At a=g the object would be in free fall and experience no gravity whatsoever. In this scenario I'm not sure what would happen. Certainly there is no buoyancy or mg anymore. There really is no reason for the glass to stay on the floor either since it would be weightless.

At a>g the elevator would be moving downward faster than things could fall, meaning everything not tied down would fall towards the ceiling of the elevator.

http://en.wikipedia.org/wiki/Weightlessness

Thanks,

See what you mean: the bodies would be weightless, but unless acted upon by another force they would tend to stay where they are? For a>g we should then expect water, glass (possibly broken) and the block of wood on the ceiling of the elevator!

The question asks if the block of wood will move below the ORIGINAL level of the glass of water. The ORIGINAL level of the glass of water was floor level + X. Therefore when the glass and wood decend, the wood will decend below the ORIGINAL water level.

Quote:

"The ORIGINAL level of the glass of water was floor level + X. Therefore when the glass and wood decend, the wood will decend below the ORIGINAL water level."

Whoa, You're saying if the Wood was at an elevation of say "100 feet" compared with ground level, & the Water @ "99.9 feet", when the Elevator moves down the Wood descends below the "99.9" figure.

Question Solved!!! While everyone else is thinking about Air Pressure, and Bouyancy on an unknown type of Wood, Brilliant Abstrat02!

What's done is done with. This was covered in post #7 - which already got a GA.

Assuming the changes are not too fast, surface tension should take over - I should have included this in the earlier list of possible modifying factors earlier. If the water wets the block, reduced gravity would cause the block to sink - and a non-wetted block would rise. At zero gravity the water would presumably spread along everything it wets and the surface would (eventually) form similar shapes to constrained soap-bubbles.

Thanks, I appreciate the response and it makes sense now (even though we may be heading off topic from the original question).

The pull of gravity is equal to both water and wood, so my guess is that the wood will stay at the same level whether the elevator is moving or not. If you perform the actual experiment (which I haven't), any movement of water and wood would be due to the horizontal movements caused by imperfections in the elevator construction.

No Change

There seems to be some "confusion" about what "the original water surface" means. Some think it is relative to the glass. Some think it is relative to the floor location where it started. Maybe it is relative to a fixed coordinate system centered at the initial point of the big bang.

My choice it relative to the glass. I'm going to be in the elevator. If I'm on the original floor, I can not even see the glass. I don't' even know where the center of the big bang was.

I personally like a little ambiguity in the challenge questions. It can lead to lots of interesting discussions. For those who like to complain about poorly worded challenge questions, I suggest you submit a few to see just how many ways people can misunderstand your intent.

Thanks,

Jim

I would think it would sink,

Can not say why exactly, but I have in my mind the ultimate test of bouyancy in my mind.

Videos from astronauts in the space station, where they make a large water droplet say 4" in diameter,, It's sutrface tension makes it into a nice sphere then they stick a straw into it and blow a bubble of air into it. Here is a GIANT ifference in density and mass yet the airbubble because of no weight so the bouyany is neutral it remains inside the water,,

Quote:

"I have in my mind the ultimate test of bouyancy in my mind."

It all depends on if Your Mind is buoyant?

First, to whomever posed this challenge question, bravo for a simple elegant poser that is devoid of ambiguities.

Now onto the question itself.

• The equilibrium prior to descent:
• The gravitational attraction between the earth and the objects on the elevator sum up to produce a downward vector equal to (m_glass + m_water + m_wood)*g. An equal and opposite normal force from the surface of the elevator cancels this force thus producing no movement in the z direction of our inertial reference frame.
• Like wise this happens at the glass water boundary minus the gravitational force created by the glass. At the water, wood boundary the net force must still be zero for the wood block is not moving. (Yes, it maybe bobbing some but I won't cover that until later.) The downward force is now just m_wood*g. The opposing force now generated is called buoyancy, this force is produced by displacing some of the water with the wood. The mass of volume of water displaced by the wood block equals the mass of the entire wood block during equilibrium. Thus it floats.
• Conditions during downward acceleration of less than g (The non-equilibrium condition of the challenge):
• During the downward acceleration the total downward forces at the floor glass boundary are the same as before. But since the objects are now accelerating downward the net force cannot be zero. If they were then no change in movement would occur. So since the downward forces are the same, the upward normal force must have reduced.
• Similarly for the wood block water interface the downward force stays the same but to permit downward acceleration the upward forces must be less so the block must rise out of water.
• Conditions during downward constant velocity, zero acceleration:
• Identical conditions for the glass floor boundary conditions as prior to motion.
• Identical conditions for the block of wood water boundary conditions prior to motion. The mass of the water displaced equals the mass of the wood block.
• Transitioning from each of the above conditions:
• The glass to floor interface will exhibit little to no transition phase condition.
• The wood block to water interface will now add viscous friction forces to the dynamic system. This will cause the block to bob up and down for a brief, damped period of time after the transition. This effect will be very tiny with a→0 but still present. As a→g this bob and damped oscillation will become more prominent.

So on the exact transition to moving downward acceleration, the inertia of the wood block will cause the block to apparently rise in the water. But actually the water will be receding away from the block. This effect will be dependent on the actual rate of downward deceleration. Once acceleration ceases though both the block and water move at the same dynamics and water levels.

"During the downward acceleration the total downward forces at the floor glass boundary are the same as before. But since the objects are now accelerating downward the net force cannot be zero. If they were then no change in movement would occur. So since the downward forces are the same, the upward normal force must have reduced."

If the force on the bottom of the glass is unchanged, why doesn't it just stay put, instead of falling with the elevator?

Are you having trouble differentiating between downward forces and upward forces? You include my comment that no change in movement would occur if all forces stayed the same, but then complain that movement occurs. Since the ceiling of the elevator is not pushing the glass, water or block down, no new downward only forces exist. Therefore, since acceleration does happen, less upward only forces must occur.

Clearly you are no Physicist.

I was merely trying to interpret what you wrote. I have so far failed to find a way to read it that does not contradict what you write in this later post. If you would care to write it so that mere mortals (Physicist?s or not) can follow your logic, maybe we can have a discussion that does not involve personal invective.

So far as the intended aspects of the problem are concerned, Roger Pink wrote a perfectly satisfactory solution (with a good explanation) at post #3. I commend it to your attention. You may also be interested in some of the related discussion in the challenge of 31 March, entitled "the Balance".

I do apologize for my flippant remark. That was uncalled for. Please check my additional comment for hopefully a clearer explanation of my analysis technique.

Although I think I've traced down one red herring, I still can't follow the total. I just think you've got so much detail you've missed the crucial point, which (typically) Roger and SlideRuler address simply and directly in their different ways. Another way of looking at this is that measuring the mass of water displaced by a floating object is equivalent to using any other form of balance. They all use forces to compare masses. Therefore, so long as the accelerations and gravities are equal for all objects, equivalent masses remain equivalent regardless of the level of acceleration/gravity.

Although I think I've traced down one red herring, I still can't follow the total.

Who added a Fish to the Water???

the upward force on the bottom of the glass is reduced when the elevator moves. Due to the acceleration being a<g this allows gravity to keep the glass on the floor of the elevator.

Serves me right for trying to use Socratic methods!

Are you saying that less cratic methods would be better?

Please do not disinter my gramma

You Wrote: "During the downward acceleration the total downward forces at the floor glass boundary are the same as before."

This isn't correct. The total "downward" forces at the floor glass boundary is:

F_{Total,Downward} = m_glass(a_total) = m_glass(g - a_elevator)

The total downward force at the floor glass boundary certainly doesn't stay the same when the elevator accelerates downward. The total downward force is reduced as the elevator accelerates downward as can be seen in the equation above.

So, isn't that the same as saying that the overall gravitational pull is reduced, as if the glass, water and wood were all at rest on the Moon? Can we ignore the motion of the system and just ask whether the buoyancy of the wood is affected by the reduced gravity? Ignoring the obvious problem of the whole thing freezing solid in about 0.2 seconds!

People,people please read my write up carefully. I separate the downward forces from the upward forces to describe the components that sum up to make the net force on the system. The downward gravitational force on a one kilogram mass is one Newton. If that one kilogram mass rests on a table, the normal force from the table is one Newton up. Making a net force of zero Newtons. Thus the mass stays at rest on the table. If the kilogram mass falls off the table the downward force on the mass still is one Newton. However, now no opposing force exists, so now the net force is solely the one Newton downward. The initial acceleration is one g.

Now, I admit that my analysis of the buoyancy forces might be off the mark. But for the wood block to accelerate downward an imbalance of forces must exist on the wood block. So I believe that the block should from the perspective of the water level effectively rise above the water. (More accurately begin to fall a little later than the water does.)

I now think the problem lies at least partly in your use of English. If by "downwards forces at a boundary" you mean "the sum total of all forces that are applied to the object above that boundary" it might begin to make some sort of sense.

But it overcomplicates the situation and misses the basic features: once the lift and its contents are in constant acceleration everything is the same for the water as for the block; the apparent weight of a fixed volume of water changes by the same proportionate amount as the apparent weight of the block - so in that respect the block needs to displace the same volume of water while accelerating as it did originally.

I highlight here exactly my point from within your comment. "once the lift and its contents are in constant acceleration..." The contents are not in constant acceleration throughout this process. I believe the transition to constant acceleration for the block of wood occurs slightly later than the water. This is why when a roller coaster rider going over the crest of a hill lifts up a little from their seat.

Going up from your seat is caused by by delay in reducing the support from your legs, or the elasticity of the seating or of your bottom or all combined. When the force from the glass on the water reduces, there will be a compressional wave (actually decompression, but what's in a name) that propagates through the water. It will reach the base of the block before it reaches the surface of the water. The sides of the glass will also flex. So, if the change is sudden there will be a small amount of bobbing. Without all sorts of extra detail there's no way of knowing whether the initial relative movement at the water-line will be up or down.

But I read you to say that "during downward acceleration of less than g the block would rise". I think we now agree that steady acceleration will neither cause the block to rise nor sink (relative to the surface of the water). I'm also saying that we have no way to know whether transients will initially cause the block to rise or fall. That probably depends on the distance between the block and the sides of the vessel (and also on the rigidity of the vessel and what type of wood is used and ...).
____________________________________________
According to Igor Transientz, the well-known mathematical physicist,

First, I've been away so my reply is delayed here.

Yes, I do agree that once water and wood come to a constant acceleration equilibrium (now there's an oxymoron for you) the water meniscus line will reside in the same place on the wood as before the descent. But the instance in question appears to be once the elevator reaches this constant acceleration, not the glass, nor the water nor the block of wood. The latter objects lag behind in responding to the elevator's accelerating descent, the simplified principle of inertia. So prior to equilibrium occurring, lower objects will be descending at higher rates of speed than above objects. Therefore, the water line descends below the previous level on the wood.

A thought experiment: replace the block of wood with an idealised zero-thickness weightless flexible bag part-filled with water and air (the bag is loose so the air is not compressed). Assume for now that the glass is perfectly rigid. Does the water inside the bag behave any differently to the water outside it?

Now we may be ready to think what happens with a wooden block. If we ignore the flexibility of the glass, we can see that the only changes are due to differences in stress propagation times through the water and through the block (and through the air if we are counting that). So we are already moving away from intrinsic effects, and considering differences between materials.
. The speed of sound in water is about 1500m/s, and the speed of sound in wood will be between 900m/s and 5300m/s (depending on the type of wood and on the orientation). If the block lies 15-cm deep in the glass (that's already a very big glass) the timescale for the differences in propagations through the block and water will be between -7 and +7 microseconds. That is the sort of timescale in which you would need to change the acceleration for this effect to be significant.
. The other effect is that internal pressure on the glass reduces, so the walls will contract. This will cause (among other effects) a surface wave in the water in the glass, that would initially cause the water surface to rise initially relative to the block. Unless the glass is exceptionally thick, this will be the slowest effect (as well as the largest if the glass were started suddenly). However, that could well be countermanded by the greater reduction in dimension further down where the pressure change is greater.

So, yes there will likely in practice be some very slight bobbing (albeit without the reasonable parameters of this challenge), but we've no idea which way it will be - even initially. Moreover, the time-scale of the bobbing will be quite short, so it could have reversed several times by the time the elevator has come to constant acceleration.

P.S. If local dynamic equilibrium is an oxymoron, then so is any form of equilibrium on this earth, and possibly even throughout the universe.

I quoted you exactly in my response by copying and pasting from your original post. Where is the confusion? Please read post #38 again.

Have you considered the glass water interface?

Archimedes strikes again, and the answer is "stays the same".

I want to point out however, that water is isotropic (having the same properties in all directions), whereas the wood is orthotropic (having distinct independent properties along the three axes). For example, White ash wood, 12% moist, has the following values for Elastic modulus E , and Poisson's ratio v:

EL=12000MPa; ER=1500MPa; and ET=960MPa;

vLr=.371; vLt=.440; vRt=.689; vTr=.360; vRl=.059; vTl=.051

(Upper case = stress axis; lower case = displacement axis; Radial, Longitudinal, Tangential) which is quite a variation.

The first conclusion we can draw is "the behaviour is different depending on which is the g axis".

Secondly, since gravitational changes affect the dimensions of the wood, it cannot remain at the same immersion level.

To determine the degree of these effects requires more time and energy than I have, especially when I am sure the answer will be "stays the same"