Water Weight: Newsletter Challenge (November 2012)

Where in the problem statement does it say the weight even contacts the water? What if it is above the surface of the water, suspended by clip on the rim? If you assume that it is submerged, or displacing any water, than the other assumption can also be made that the flask is extremely tall, and there's plenty of room between the weight and the surface of the water.

Perfectly correct; and in that case the buoyancy can be neglected because it's 0; and the pressure doesn't change. However, that case is what mathematicians call a "trivial case": one takes note of it and moves on to the interesting cases. What makes this problem interesting is the balance of forces that make it possible for the system weight to remain unchanged even when the downward force of the liquid increases. When the object makes no contact with the liquid, all interest disappears. Is it not reasonable to assume that a problem called a "challenge" is intended to be interesting? And is it not reasonable to assume, therefore, that trivial cases are intended to be excluded?

Fair enough, but to read the statement, it says, "weight is moved inside the rim (making sure to not touch the bottom)".

Like I had posted earlier, the moment will change from off-axis to co-linear, so you will have a pressure change, but no mass change...

Simplicity can be interesting too.

The weight does not change but since the metal object partially fills the glass, it will raise the waterlevel and thereby the hydrostatic pressure on the bottom.

Kind regards,

couvrair

The overall mass remains the same, namely:-

The mass of the container (0.5kg)

The mass of the water (1 kg assuming fresh clean water)

The mass of the weight (0.25kg)

Total = 1.75kg.

Placing the mass inside the container won't change the overall mass, but it will displace its own volume in water, meaning the column of water will rise . . . and increase the pressure acting on the bottom of the container.

one quibble: to displace its own volume of water, you are assuming that the density of the metal object is higher than the water (no one has said the metal object is solid, it could be hollow and float). It will displace a volume of water equalling its weight if the density is less than (or equal to) the water.

Agreed. The weight will displace a minimum of 0.25kg of water.

Whether the mass floats or sinks, a minimum displacement will still mean a rise in the level of water in the container.

almost... If the density of the metal object is less than the water, the object will displace its weight in water. If the density is greater, then the object will displace a weight of water equal to the volume it occupies. The displacement will increase the water column height and will result in a higher pressure at the bottom but the density of the object establishes a maximum condition, not a minimum.

In either case, we both agree that the mass of the system does not change (unless as I indicated WAY up in the thread, someone is holding the weight up, in which case the whole problem changes as the system is no longer isolated).

If the glass is filled, some water will overflow when inserting the metal object.

If that water stays on the scale, the reading will not change (1.75kg).

If that water falls off, the reading will decrease by the amont of water displaced by the object, depending on the density of the object.

For aluminum, the reading will decrease to 1.75-(0.25/2.7) or 1.66kg

Francis

Good point about spilled water being on the scales (you seem to have sussed this duscussion very quick !).

I've not looked at the figures you quote, but good for you in inputting some sample data to argue over.

If the object remains within the water, what say you on the effect as it is raised and lowered ?

Welcome to the chat/madhouse.

The weight would stay the same. The metal object would weigh less within the water environment but it would displace its own weight in water, thus the net effect on the overall weight is zero.

Assumptions:

1) Metal object is moved to the inside of the glass, but still hanging from the inside edge (so it is the same height above the balance, can't touch the bottom, and is not supported externally).

2) Metal object is at least partially submerged in the water, but no water overflows the glass.

3) The balance is infinitely sensitive.

Result:

1) Water level rises due to displacement by the metal object. This raises the CG of the water, but not of anything else. Since a balance balances force, not mass (weight), the weight goes down due to the higher CG of the water and the reduction in force due to gravity acting at a greater distance.

2) Is the pressure on the bottom of the glass on the outside bottom or the inside bottom? Assuming outside bottom, then the pressure (force/area) goes down.

There is not enough information to answer the question.

If the glass is indeed "filled" - that is completely filled - water, to the volume of the object, will spill over the top when the object is immersed. The weight will decrease by the weight of this loss. Since the volume of the object is unknown, the water lost is unknown so the weight change is unknown. Since the water depth has not changed - the glass is still completely full - the pressure at the bottom of the container will not change.

If the glass is not "filled" and the water has room to rise when the object is immersed, then the weight will remain the same and the pressure at the bottom of the glass will increase because the depth has increased when the object was immersed.

Have you tried the question: can an aircraft take off from a conveyer belt driven at the same speed backwards as the airspeed of the airplane forward?

Norm Johnson

Best explanation so far.

Regarding the airplane, it will take off when it reaches the proper "airspeed".

If the conveyer is moving the aircraft backward, it will have to accelarate from

the negative airspeed to the positive take off airspeed (hoping it does not exceed

the maximum tire speed).

Francis.

Yes. This was wonderfully and throughly demonstrated by Myth Busters.

What weighs more, a pound of gold or a pound of feathers?

bandership

try to handle it

The aircraft is not trying to accelerate using friction with the ground.

The conveyer belt will generate extra drag as the aircraft attempts to accelerate to V2, but since the forward thrust is generated in the air rather than against the ground, the aircraft will indeed take off, airspeed will be normal, ground speed will be twice, power required and time to achieve lift off will be increased due to the additional drag

The weight remains the same as the gravitational force is the same, however the water level will rise and create a higher water pressure at the bottom of the glass, this will be in direct proportion to the change of height. The lower the SG of the metal weight; the greater the increase in pressure on the bottom of the glass.

Aquabob

the weight doesnt change, but the pressure at the bottom of the glass changes due to the increase in the height of the water

The term Filled comes into play

if the glass is full ,then water will be displaced and then the total mass will decrease by the volume of overflow. In this case the pressure at the bottom of the glass is still the head of water.

If the glass is not full then partial overflow may happen then mass will decrease and pressure will change

If no overflow occurs then only the pressure will change if all or part of the metal is immersed by the increase of water level

Depends where the water goes. If it all goes on the pan below it is still part of the system.

If part of it runs off, then some of the mass will no longer be part of the system.

If the water runoff gets into the balance and shorts out the balance or corrodes the balance bam you could be up for disciplinary action.

Neither changes.

Yes, the height increases, but the neck diameter has decreased. Pressure is the sum of the mass subject to gravity. Unlike a lake or barometer with increased depth, that has not changed in this circumstance.

This is still a simplification. A divet at the bottom of a shallow pond does not bear the whole weight of the pond. If we covered the pond, and added a high tube, only the tube diameter counts. The water in the middle still rests on a shelf. The bottom line is the gravational mass of water resting on water. Still though, the neck has decreased. Any increased pressure from height is still being redistributed as the container widens near the bottom. The quantity of water resting upon water has not increased.

Wrong!

Yeah, maybe, I am having my doubts, but it's getting too complex to calculate, and I don't know the relative impact of the two systems involved.

One can look at this system a level deeper, and the answer still does not change.

Is this a stack of bricks, a vessel of ideal gas, or something inbetween?

It is something inbetween. If it were an ideal gas, the pressure at the spout would be absorbed by the 'volume' of the gas. Water is roughly incompressible however.

If it were a stack of bricks with downward distributed vector forces (a series of horizontal slices), the answer is as the post above.

In this case however it is 'kind of' a pressure vessel, however as the fluid is non-compressible, the introduction of pressure from the spout is 'distributed' (as it always is), in this case by the 'surface area' of the vessel. Note however that in incorporating the weight to create a spout, we have increased the surface area of the vessel to which pressure, not weight, is being distributed.

The pressure at the bottom of the vessel is the sum of both the distributed weight it directly bears, plus the pressure distributed across the surface area. Also keep in mind that it is mostly only the pressure of the spout which is being distributed.

V = piR^2*h, SA = 2piR*h :: V is our constant. V*h is what creates distributable pressure. For convenience sake, lets say the weight is also a tube. We have two tubes: the glass with Rg, and the weight with Rw.

Volume with no weight = piRg^2*h1

Volume with weight = (piRg^2 - piRw^2) * h2

piRg^2*h1 = (piRg^2)*h2 - (piRw^2) * h2

(piRw^2) * h2 = (piRg^2)*(h2-h1) :: (Rw^2) * h2 = (Rg^2)*(h2-h1)

We are looking for V*h (specifically h2-h1), so substituting back in:

h2-h1 = ((Rw^2) * h2)/(Rg^2)

Vh = piRg^2*h1 x ((Rw^2) * h2)/(Rg^2) = pi*h1 x Rw^2) * h2 =

I'm spinnning gears for the moment, but you get the idea. Increased height is compensated for by decreased neck, in either a bricks or water situation, and also in an ideal gas situation with limited water at the spout.

Where did the tapered container idea come from?
Pressure is not what you have described.

By putting the weight inside the glass, the upper portion of the cylinder has become some sort of neck with increased water height and decreased diameter. It is not specified if the water broadens above it or not.

What I am bringing to the conversation, previously not considered (or considered and dismissed) is the narrower neck accompanying the higher head.

As I understand it, the whole concept of pressure at the bottom of cylinders is based on cumulative pressure by weight, not weightless distributed pressure as within gas systems. ..and yet, as when I mentioned vessel surface area, I can also see that omnidirectional pressure vectors would have some impact in such a system as well.

Neither though is it a pure omnidirectional vector system, otherwise shelves not adding to height might still add to system pressure, and certainly we would no longer have the proportional relationship between depth and pressure.

That's total baloney.

Offer me an alternative macro-level explanation of why the common fluid pressure-height laws behave as they do, and I'll consider it.

Take an elementary physics class, and maybe then we will consider you.

I see where I went wrong now. I was macroscopically modeling force, not pressure, as if it were a vase of marbles all weighing downward.

I was top of my college physics and calculus classes, and am a genius, but I took those back in HS which was quite a long time ago, and go unused as a programmer.

I was making the problem far too complicated, as I usually do.

I'm not an engineer so, help me with this one. So pressure at bottom of cylinder is function of height only. Okay. I have always taken that to mean, given a cylinder of fixed diameter. as additional water is added (thus more mass) to the cylinder, we only need to know the increase in depth (don't need to know mass) to calc the increase in pressure. For this challege depth increased but no additional water is added, and the surface area at bottom of cylinder remained the same.

You explained it yourself in your second sentence.

Read the first few posts on this thread- Usbport pointed out something I'd missed- what is the exact wording asking for? So lets look at it from the point of unit pressure as opposed to overall force, which seems to be what you're asking.

Mass is irrelevant in a sense. Pressure is UNIT pressure- pressure on a given area (eg. psi). Therefore, if you increase the mass ABOVE that given area, you increase the pressure; if you expand a system horizontally and add more mass so that the overall height of the system (and density) remain the same, there is no increase in unit pressure. This is true whether the system is liquid or solid (think of the soil bearing load of a 2 storey home Vs a skyscraper Vs a row house complex.

It's just that a liquid system has a few peculiarities- by adding height (or depth!) to the system without adding mass you increase still increase the head or column height, which is all that matters in deriving unit pressure for a given density. So you could even take a block of something very light (eg.Styrofoam), glue it to the bottom of your container and fill it back up with the initial volume of fluid. There would be an upwards vector created on the bottom of the container due to the buoyancy of the block, but at the same time the unit pressure of the fluid at the bottom beside the block would have increased due to the increased column height.

The hydrostatic pressure (inside) at the bottom of the glass increases but the pressure on the bottom of the glass that produces the weight force does not change because there has been no change in total mass.