Another Physics Brain Teaser

Easy on there PI, I'm more dense than Mercury, how the stuff would I know.

Now Mercury is 1.7 more dense than steel. If it was 1:1 there would be neutral bouyancy 1:1.7 indicates to me that .7 of the steel balls volume would be sticking out of the mercury. Now that's when I thought I would have to work out the volume of a sphere, then calculate .7 of that volume and then work out what thickness of sphere there would be from a slice that was of the required .7 of volume.

Then I thought I need a beer and then come back to this, but upon a bit of pondering I decided that 2 beers are better than one.

Then I thought, " I wonder what reduction in mass there would be in a carton of beer if I drank 6 of them," lot more fun than working out this sphere ****.

If you had a steel ball sitting on a concrete floor and added a metre of water, I would think that the metre of water will have a bouyant force on the ball pushing up ie making the ball lighter. If the ball is floating in mercury and a metre of water is added making it lighter I would expect the bouyancy, making the ball lighter would decrease the draft of the ball.

Time for another beer.

Poida

 

Yes, the water will also add some buoyancy.

 

My original thinking was that the water would add buoyancy and the ball would float higher out of the mercury. But it is counter-intuitive that the ball would swap buoyancy provided by the mercury for buyancy provided by the water. I think it would still 'choose' to float in the mercury as this is the denser medium. I can't think how to put it in mathematical terms, but my gut instinct is that it will float at the same level within the mercury. Not very scientific, I know, but I need one of Poida's beers to get my juices flowing.

 

I share your intuitive reservations, but buoyancy is buoyancy, although my experience of things floating in Mercury is limited to lighthouses.

I agree with your draft, (although I get .562m when I do the calculations by hand !). I would suggest that the .215m^3 of steel sphere above the Hg level would still provide 215kg buoyancy when inundated with the water.

As the immersion is 105 kg/cm, I would think it might reduce its immersion in the Hg by a couple of cms.

Just to tie it in with the discussions in the Stability and Seaworthiness, the figures for it look dreadful despite being made of steel.

But it does have a prismatic of .685 so would be suitable for high speed service. Perhaps it's an IMOCA sphere.

 

(stupid and silly)

Are loadlines dependant of barometric pressure ?

 

Lighthouses floating in mercury???

I understand your logic completely, it just seems "wrong" that it will float away from the denser medium. If it floats up, it will gain buoyancy from the water, but lose more from the mercury, and so sink back down again.

 

Floating the huge and heavy fresnel lenses in a circular trough of mercury was the traditional way to get them to rotate friction free.

I'm not sure there would be any bobbing up and down. Surely equilibrium would be reached when the mass of the sphere is supported in the two mediums combined?

 

I kind of like this one, the more water added the deeper it sinks, but as mercury might be compressed by the weight of the water it will become more dense and float the sphere higher, also I think that the pressure of the mercury acting on it self will have a effect of more density every mm. down.
Well its Friday tomorrow, and I will test this out by taking a stainless steel tube say 4” in diam. And 6 feet tall , weld 3 attachments and put one a hose, 1st 1’ down, 2nd at 3’ and one at the bottom, and with 2 mates will test who gets the most effect from the beer due to “pressure difference”

Any mods around, I need to get my password for my old ady back, the email I used to register is no more.

 

Crag,

Cunning.

Yeah, one side of me agrees completely. I just can't picture the ball floating up out of the mercury.
At any rate, calculating the exact draft is a tedious and quite tricky problem due to the fact it ain't wall sided (so the 105kg/cm is only an average), and the ball is leaving one medium and entering another.

Johtaja,

Nice experiment.
How come though, that the e-mail alert message has slightly different text to what appears on the website? It does this a lot, and I don't understand why?

 

Depends on your experience. For some it is a challenge because they haven't spent the time considering the implications of dispalacement. Not everybody here is an Architect or engineer.
Don't be sorry, just leave the problem to those who are challenged by it to consider it without trying to insult them because you have more schooling on the topic.
Thanks

 

That's the definition of a brain teaser
(Raggi, now with wine and cheese)

Think about it this way:
First the sphere is floating partly in mercury, partly in air (density 1.2kg/m3?).
If you added more mercury (replaced some of the air), would the sphere rise? Of course. So I say, by intuition, that if we replace the air around the sphere with something heavier, like water, the sphere will float higher.

 

so if A = the volume in air or water and V is the total volume, then

floating in Hg

(V-A)*13.6 = V*8 therefore A = .412V

floating in Hg and water

(V-A)*13.6 +A(1) =V*8 therefore A =.444V

so the sphere rises out of the Hg when the water is added.

The rest is just solid geometry makework that I can look up if some one wants me too.

 

So, in reality, the water is pressing down on the mercury (just like air) meaning it takes more to displace, plus the sphere weighs less, and thus it is lifted?

Is that understood somewhat close to what it's supposed to?

 

Basicaly, because the water is pressing on the mercury, we need only consider the wet weight of the sphere. In truth, if you wanted the TRUE answer you would have had to use the density of air (SG 0.00013 @ 5C) in the first equation also.

 

Ah, I see. But for this learning example, I think I will make do, and think of air as having no density. No need to over-stimulate the grey matter