Pressure, bouyancy, mercury and steel. Oh my.

Factual Questions
1

My physics teacher posited this hypothetical situation (worth five points on my exam Friday).

Take a container partially filled with mercury and place a steel sphere in it. The sphere will float since steel is less dense than mercury. Now add double the water needed to cover the sphere. Which of the following will happen?

The sphere will move down (even a little bit) due to the pressure of the water.
The sphere will not move at all.
The sphere will move up (even a little bit) due to the bouyancy of the water.

2

WAG here: I think the steel sphere will either stay right where it is or maybe move up a bit…

I’m gonna place my bet on it staying right where it is. The downward pressure of the water will be applied equally to both the sphere and the mercury, balancing out (so it won’t go down). Can’t reason out why it won’t go up, but I have a gut feeling that it wouldn’t…

3

it wouldn’t go up, because the density of the steel sphere is greater than the density of the water.

it wouldn’t go down because it is still heavier than the mercury.

it is at it’s natural balance of buoyancy/density.

a comparison is the lego-eraser-hollow ball trick.

[partial hijack alert]

get a large glass. put the three objects in. then put in some thick substance- i think it was baby shampoo. the three objects will float. then add water. slowly, so the shampoo remains a distinct layer. the lego and the ball will rise, but the eraser stays where it is- it is at it’s natural balance. when you add some vegatable oil, the lego stays, and the ball rises. all of these things remain at the interface between their layers. they don’t move, except for turbulence, when you add another layer.

[/partial hijack alert]

there. an illustration of density and floating objects. please excuse any details that are incorrect- i got that bit of trivia a couple years ago and i can’t remember all the details. if anyone can correct some of my materials mistakes, please, do so! 6th grade science class was so long ago…

4

I’ll put my money on the sphere moving down into the mercury slightly from the increased pressure of the water upon it. Dry lab this by using the thought problem of your finger pushing upon the steel ball with the applied force of the water. The mercury will be displaced. This also presupposes that the mercury is compressible, which it is.

However, the small amount of water we are discussing would have an almost imperceptible effect. You might be able to measure this effect with ultrasonic level detection of the mercury/water interface to monitor the slight upward displacement of the mercury by the steel ball. The experiment would have to be conducted on a vibration isolation platform in order for the displacement to be detected over the noise floor of ambient vibrations.

5

I think that it’ll move up a tad, and I’m certain that it won’t move down. Think of it in terms of potential energy: The system will naturally tend towards the state of lowest energy. Now, we’ve got the full glass, with mercury, steel, and water, all in original positions. If the sphere moves down, then the potential energy of the steel and mercury combined is increasing, since it was at minimum before, and the bottom of the water, which was uneven before, is going to become more even, which means that some water is moving upwards, so the potential energy is increasing there, too. We don’t want the PE to increase, so it definitely won’t move down.

I just thought of another way to look at it: All that matters here is that water is less dense than steel, and steel is less dense than mercury, right? It doesn’t matter (qualitatively, at least) how much less dense the water is. Instead of water, let’s consider some hypothetical liquid with a density 99.99999% that of steel. What will that do? Well, we now have (almost) a layer of mercury, and a uniform layer of “heavy water”. The boundary between them should be (almost) level, which means that the ball has floated up.

6

I agree with Chronos.

I worked the situation out analytically, using a cube of steel rather than a sphere, for simplicity. (I’d rather not integrate the pressures over a non-flat surface tonight.)

I can’t use Greek characters, so I’ll use D to represent density. D[sub]S[/sub] is the density of steel; D[sub]M[/sub] that of mercury; and D[sub]W[/sub] that of water. We know D[sub]M[/sub] > D[sub]S[/sub] > D[sub]W[/sub].

First consider the steel cube (side length A) floating in mercury with air above. (Implicit in this is that air’s density is effectively zero. Since it’s several thousand times less, I think that’s probably OK.) Pretend that the cube is magically stable, and floats with top and bottom surface parallel to the mercury surface. The air above is at atmospheric pressure, p[sub]a[/sub].

The cube will sink by some distance L (0 < L < A) into the mercury. Balance the forces from:

Pressure on top: p[sub]a[/sub]A[sup]2[/sup]
Pressure on bottom: (p[sub]a[/sub] + D[sub]M[/sub]gL)A[sup]2[/sup]
Gravity on cube: D[sub]S[/sub]A[sup]3[/sup]g

Balance these (top pressure + gravity = bottom pressure) and solve for L, to get:

L = A * (D[sub]S[/sub]/D[sub]M[/sub])

In other words, the cube sinks into the mercury by a fraction of its own height proportional to the density ratio. That makes sense - if D[sub]S[/sub] were zero, it wouldn’t sink at all, and if D[sub]S[/sub] = D[sub]W[/sub], it sinks in all the way (neutral buoyancy).

Now imagine that we dump water in - enough to cover the steel cube completely (as it turns out, once it covers the cube, how much water doesn’t affect the result we’re after.) Say that, after things settle down, there’s a water depth of X above the surface of the mercury, with p[sub]a[/sub] above the surface of the water. How far does the steel cube sink into the mercury now? Call that L[sub]2[/sub].

It’s a bit more complicated, but we can still just balance the pressure and gravitational forces:

Pressure on top: {p[sub]a[/sub] + D[sub]W[/sub]g(X-A+L[sub]2[/sub])} * A[sup]2[/sup]
Pressure on bottom: {p[sub]a[/sub] + D[sub]W[/sub]Xg + D[sub]M[/sub]L[sub]2[/sub]g} * A[sup]2[/sup]
Gravity on cube: D[sub]S[/sub]A[sup]3[/sup]g

Again, balance as before and solve for L[sub]2[/sub]:


          D[sub]S[/sub]-D[sub]W[/sub]
L[sub]2[/sub] = A * -------
          D[sub]M[/sub]-D[sub]W[/sub]

Happily, this degenerates to the result we got without the water if we set D[sub]W[/sub] to zero, as it should. Actually, I could have just skipped the first derivation and gotten it straight from here.

Anyway, what does this show? That the steel cube will rise slightly. Looking up the actual densities :: rummaging ::, I find: D[sub]W[/sub] = 1 g/cm[sup]3[/sup]; D[sub]S[/sub] = 7.86 g/cm[sup]3[/sup]; and D[sub]M[/sub] = 13.546 g/cm[sup]3[/sup].

This means that L (no water) will be 0.58A, and L[sub]2[/sub] is 0.54A.

Add water, and a steel cube will rise by about 4% of its side length. A steel sphere will behave similarly, although the math to show it would be a little more involved.

7

I go with Chronos as well though my opinion seems irrelevent following brads explanation. The apparent weight of the sphere will be lower when it is immersed.

Its easy to see this if you consider a sphere with an equal density to water. Its a no brainer to see that this sphere will no longer sink into the mercury at all.

8

I used to teach physics…

disclaimer: I didn’t go through all the above math.

Something to consider. A floating object will displace the volume of liquid in which it is floating equal to that objects weight. Thereby, a 4 lb. block of wood will displace 4 lbs of water when floating. If the weight of the object is more than its volume of the liquid (in this case water) it will sink. At this point, bouyancy is more likely to be dicussed. (not that it’s not important with floating objects)

An object underwater (not floating) will displace it’s volume of water. The weight of the displaced water will be less than the weight of the object. In this example, lets say that a 10 lb block of iron sinks in water and displaces 6 lbs of water. If you were to weigh the iron block underwater, it would have an apparent weight of 10 - 6 or 4 lbs.

Now to the steel ball. Let’s say it weighs 10 lbs. It is floating in mercury. It displaces its weight in mercury, which is 10 lbs. It ain’t moving, so all forces are equal. When water is poured over it, it now displaces water. It "sinks’ in the water. In this case, it displaces less than it’s volume in water, since some of the sphere is in mercury. The apparent weight of the sphere decreases due to the water displaced. The question is: will the decrease in apparent weight alter the weight of mercury displaced? the answer is “yes”. The apparent weight is truly the gravitational force acting downward. Since there is less force pushing/pulling down on the steel ball into the mercury, less mercury will be displaced and the ball will rise a bit.

9

::fluttering eyelashes::

Spritle is my hero!

::sigh::

10

Bing bing bing! I think we have a winner. Spritle, you da man!

Intuitively I was going to argue no change, but buoyancy comes into effect. (I screwed someone up on that before.) The problem I saw involved a block of ice with a steel weight in it sitting in a tub of water. If the ice floats, it displaces the weight of the total. If the ice sinks to the bottom, it displaces its volume. Same principle applies here.

11

Just to toss another nail into the coffin here:
The weights used to standardize analytical balances come with information about the density of the materials from which they are made. Knowing the temperature and the barometric pressure (ie air density) allows you to calculate the relative bouyancy of the weights you have vs the weights used in the balance.
-Bouyancy counts !

12

I agree with the prevailing opinion, although my reasoning was slightly different.

If you set up the experiment with sufficient volume units of Liquid[sub]a[/sub] with a density of 100, object (O) with a density of 50 and liquid [sub]b[/sub] with a variable density of 99 to 1 and run the numbers, it comes out even enough to be obvious that the weight of the object is entirely supported by the densest liquid[sub]b[/sub], so O rises all the way out of liquid A. For each reduction of density in liquid[sub]b[/sub] (well, each one below 50 anyway) a larger portion of O will drop into liquid[sub]a[/sub] However, even at density 1, liquid[sub]b[/sub]will have some effect, so long as O is less dense than liquid[sub]a[/sub].

Not really different, perhaps, but it helped me to get variable numbers.

13

I like simple visual explanations, here is another.

Imagine a container of mercury with a top floating on it. In the center we have a hole with a steel cylinder, rather than a sphere, in it floating in the mercury. O rings are between the top and sides of the container and the hole and cylinder. Obviously when we press on the top the cylinder will rise unless we exert the same psi on the cylinder as is on the top.

When we pour water on top the depth over the top will be greater than the depth over the cylinder resulting in unequal psi, so the top will go down slightly and the cylinder will rise.