# Question about bouyancy... Looking for a hint

Gweedz
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Topic 190756

This one has been bugging me for a long time. I can come up with two very convincing answers, but with opposing results.

A solid iron sphere is floating in a bath of mercury. You pour water over the sphere and cover it with water. Does the sphere rise, sink, or stay at the same height?

Can someone please give me a small hint of what I should be looking for in answering this? I don't mind researching and digging for info, but I've exhausted all my resouces with no luck. Please don't give me the entire answer - I want to work at it. Thanks.

wumpus
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### Question about bouyancy... Looking for a hint

Equilibrium is what you are after. Volume of mercury displaced * density of mercury = iron sphere mass. When you add water, what changes?

Gweedz
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### Let's see... the mass of the

Let's see... the mass of the displaced water is deducted from the mass of the sphere, so a sphere with less mass will sit higher in the fluid?

It's what I would write on a test - but I still don't understand why.

Does the weight of the water have any effect? It is pushing down on the sphere and the mercury, so if they have the same area then the downward force would be the same, but if the sphere had much more "top" area than the mercury surface area, wouldn't the weight of the water push the sphere a bit lower? I know the questions doesn't mention surface area, but doesn't that come into play?

Is the phrase "the water COVERS the sphere" important?

David Hammer
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### RE: Equilibrium is what you

Message 25091 in response to message 25089

Quote:
Equilibrium is what you are after. Volume of mercury displaced * density of mercury = iron sphere mass. When you add water, what changes?

I think this assumes the density of air is zero, which is a good approximation.
Volume of mercury displaced * density of mercury + Volume of air displaced * density of air = iron sphere mass.

This should be a good hint.

Gweedz
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### RE: RE: Volume of mercury

Message 25092 in response to message 25091

Quote:
Quote:
Volume of mercury displaced * density of mercury + Volume of air displaced * density of air = iron sphere mass.

Since density of water is more than density of air, you need to diminish the volume of mercury displaced to end up with the same result (mass of sphere). In other words, the sphere will rise.

I was leaning towards this answer but my reasoning was different. I figured with air on the sphere, it doesn't move. If you pour mercury it rises to the top. So any liquid with a density between air and mercury will cause it to rise. That was my reasoning, but I had no proof. Now I do.

And my thinking of pressure and weight was nowhere close.

Thanks!

If you like these types of questions I have a few more that I can post for entertainment. Unless you think they're too easy and a waste of space.

Mike Hewson
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### RE: Does the sphere rise,

Quote:
Does the sphere rise, sink, or stay at the same height?

I'll assume the question means: If you replace the air that surrounds the upper part of the iron sphere ( and the mercury ) with water then will the ball change position with respect to the top surface of the mercury?
Would this alter the either the density of the iron or the mercury? Thus does it alter the difference in the densities of iron and mercury? Would it change the force exerted upward by the mercury on the iron sphere? Would it change the downward force of gravity on the iron sphere ( or the mercury for that matter )?
Hint: Pump all the air away to vacuum.....
Hint: Pour in some more mercury....

....Eureka! ;-)

( It doesn't change .... )

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

David Hammer
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### Here is a thought experiment

Here is a thought experiment that may make the solution clearer.

If you weighed an iron ball in a vacuum would it weigh the same as it weighs at the bottom of a swimming pool? By weight I mean the force the ball would apply to a scale.

In a vacuum the iron ball would weigh
[mass of iron * acceleration due to gravity]
Gravity pulls the ball down and there is no force pushing up.

On the other hand at the bottom of a pool it would weigh
[mass of iron * acceleration due to gravity - mass of displaced water * acceleration due to gravity]
Gravity pulls the ball down but there is now a buoyancy force pushing up on the ball.

Would the iron ball need to sink as far into mercury to support its weight at the bottom of a swimming pool as it would in a vacuum?

edit to improve clarity ???

Jquake
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### Rise. Instead of a sphere,

Rise.

Instead of a sphere, assume the iron is box shaped: much wider than tall, so that it floats with the top surface horizontal. Add water up to the level of the top surface (so the top stays dry). Does the iron fall, rise, or not change?

Because the sides of the box are vertical planes, the water does not push up or down. The water does not push down on the (dry) top surface. So far, no effect.

But the water pressure at the surface of the mercury pushes down on it (in the area outside the box). This pressure is transmitted everywhere, including the bottom surface of the iron, lifting it a bit.

Alternately, mercury is about 13 times as dense as water. If the top of the box was 13 inches above the mercury -- instead of adding 13 inches of water, add one inch of mercury. Obviously, the iron would rise.

The shape of the iron is not relevant. Whatever the shape, adding water will cause the iron to rise. Once the water reaches the hightst point of the iron object, adding water will no longer change its hight.

Gweedz
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### RE: Once the water reaches

Message 25096 in response to message 25095

Quote:
Once the water reaches the hightst point of the iron object, adding water will no longer change its hight.

Why not? Wouldn't the pressure on the top surface of the object cancel out the pressure on the bottom surface, then you're left with the little bit extra pressure on the bottom surface created by the weight of the water on the mercury (area outside the iron object)?

This is what initially confused me - I wasn't sure whether to use density or weight & pressure in figuring out an answer. From what I read above, I understand the process better using density.

MarkF
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### Guido: It would be the

Guido:
It would be the difference between the pressures at the bottom from the top that might change the position of the block/sphere. But once the block/sphere is covered the pressure difference is constant, at least as long as the second fluid is incompressible.

Mike Hewson
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### RE: From what I read above,

Quote:
From what I read above, I understand the process better using density.

I've been thinking hard about a good microscopic explanation of Archimede's Law, and why the principle doesn't change with the materials you use. I think you need four things:
(1) A gravitational field. Can't have bouyancy without it. Gravity pulls your solid body down, and creates a density gradient in your fluid ( air or water ) with the highest density at the bottom.
(2) A fluid medium. Fluids can't support shear forces, hence 'hydrostatic' pressure is exerted equally in all directions ( isotropic ) at a given point within the liquid. This does not mean the pressure is the same at all points within the fluid. For most reasonable equations of state, a fluid will increase it's pressure with increasing density ( compositon and temperature constant ).
(3) A solid body. This gives a constant size and shape, that is constant surface areas to present to the surrounding fluid.
(4) Nothing else. No thermal gradients, explosions, adsorptions, absorbtions, etc...
Now any small element of the body's surface facing toward the fluid will have a force exerted on it, which resolves to a normal or perpendicular direction to that surafce ( fluids don't support shear ). This force will point inwards, being balanced/resisted by whatever holds the body in shape, so there's no collapse/expansion. Bouyancy is a upward force which may or may not exceed the weight of the body. This bouyant force is the total of all the perpendicular forces over all of the surface that is in contact with the fluid. This is a vector summation. Any surface elements that have a normal in the horizontal plane ( thus the tangent plane to the surface is aligned with a vertical plane ) will only experience lateral forces, not up or down, and thus are not relevant to bouyancy. Surface elements with a normal directed somewhere below the horizontal plane will experience an upward force from the fluid, which resolves to the component of the force from the fluid in the vertical direction ( the horizontal component not contributing ). Those with a normal directed somewhere above the hozizontal plane will experience a downward force ( components resolved similiarly ). Thus for a sphere, which has a helpful symmetry, the force summation ( nett direction down ) from the upper hemisphere is pushed upon by less dense/pressurised fluid than the force summation ( nett direction up ) from the more dense/pressurised lower hemisphere. Phew!!!
Now if we define ( in our imagination ) some volume, within the greater fluid body, which has the exact outer surface ( say a spherical shell ) as the solid body we are actually using, then we conclude that our solid is displacing a mass = volume*density of that fluid. The surrounding fluid is supporting our real body to the same degree as the volume of fluid it has displaced. If our body has the same density as the fluid, it will remain in situ as did the fluid within our imaginary shell in the absence of the solid. If our body has greater density then we gain no more bouyancy, but we have more mass within that volume, so gravity wins and we sink. If the body is less dense then bouyancy is still unchanged, but we have less mass within the volume, so gravity loses and we rise.
Replacing air by water in the original question thus increases the density of the fluid that our solid is replacing ( above the mercury level ), we will thus have an increase in the bouyant force from that. The sphere was initially sitting to some level upon or partly within the mercury fluid. It was displacing a fraction of the sphere's worth of mercury, and it had reached some equilibrium between gravity ( down ) and bouyancy ( up ) from the mercury prior to adding the water. So you know what happens when we add the water ....... :-)

(edit) I've thus implied some shape dependence here, which could introduce some torque as well.

(edit) I've implied that if there is not a surface element which is in some way downward facing which is in contact with the general fluid body then Archimedes is negated. ( 'Suction' cups used by glaziers to handle glass for instance, or underwater caissons used in construction ).

(edit) 'Eureka' in ancient Greek means 'I need a towel!' ;0)

(edit) The 'swim bladder' of some fish is an internal balloon to which gas is added to, or subtracted from, by chemical means. It changes the volume of water that the fish displaces. Adjust this to suit the water conditions at a given depth for neutral bouyancy on a given day, and you reduce the energy to be expended by swimming to stay at a certain depth.

I have made this letter longer than usual because I lack the time to make it shorter ...

... and my other CPU is a Ryzen 5950X :-) Blaise Pascal

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