Beam me up
You have two *square-section beams of wood. One has a density greater than half that of water, the other less than half that of water. You throw them in a pond. What do you notice about the orientation of the floating beams?
*In the original posting, I unintentionally wrote "rectangular".
*In the original posting, I unintentionally wrote "rectangular".
Labels: funphysics, SharedPuzzle
posted by Chris at
3:30 PM
24 Comments:
They line up.
Just kidding.
The one greater than half that of water, sinks to more than half way.
The one less than half that of water, sinks to less than half way.
That's true. But it's not what the question asked for.
i would want to say they float, since i could claim neither is greater than water's density... so they are horizontal. unless they are very skinny slabs of wood, then they will be vertical in the pond, referring to the 'up' part of 'beam me up'.
The beams have a rectangular cross section.
Take them to be 4" x 4" x 6' if that helps you visualise better.
since it doesn't say how much greater than half the density of water, it could be argued that one of them is located at the bottom of the pond.
Wait, you said their "orientation"?
I think they're straight.
JT - they both definitely float.
Ragknot - I wondered who'd spot the double-entendre. If they were bent, I'd want my money back ;)
Anonymous Aug 19, 7 pm is getting the right idea.
To Chris,
The "orientation" of straight did not refer to any physical attribute.
Ragknot - DUH! Isn't it gob-smackingly obvious that I understood that. Are we on the same wavelength?
I am still hunting for what you mean with the word "Orientation".
Ok, so you didn't mean their sexual orientation. I assumed a beam would be straight.
Evidently you didn't mean the depth that they would sink to.
I first thought they would be parallel, if there were any flow direction, but with no mention of flow, who knows?
Beam me up? Maybe they are tied together, and if they are about equal size, the dense one would be below water. and the light one is UP.
Kid Sister rule. However, I did mean sexual orientation "straight","bent" in the last post or 2, but not in the problem.
I agree that the word "orientation" doesn't seem to have the meaning that it is often used to mean. (Ain't that the truth). I mean which surfaces are at the top, are they horizontal or vertical. The depth seems a reasonable parameter to mention. The problem is supposed to be a sensible physics problem. Assume the beams are well separated, non-parallel (but that doesn't matter) not stuck in the mud, or singing an ode to an albatross :) I won't normally post silly trick questions.
It's a business doing pleasure with you :)
Ragknot. You gave a pretty good answer to the Archimedes Principle question (even the is was spelt incorrectly). It was partly because of that, that I decided to post this harder problem.
Well, they will be horizontal, not standing up on end if the density is similiar from one end to the other. one end would have to be much more dense for them to stand.
Are you saying that the dense one would be vertical and the less dense would lay horizonatl?
I can see that, but I think that might be what you mean.
That's better. But you're still complicating it. Of course you should assume that the beams have a perfectly constant uniform density throughout. Make reasonable assumptions. If I were to cover every single possible assumption explicitly, it would take weeks to write the problem. e.g. Assume that you can neglect effects due to the rotation of the Earth around it's own axis, possible modification of the DNA structure of the beams due to interactions with cosmic rays causing the density to violently oscillate and send out showers of sap blinding you so that you can't determine their orientation, that one part of the pond may be warmer than the other, that a halibut doesn't decide to set up a hotel business on one of them, that we are nowhere near the Bermuda triangle, they don't become waterlogged, etc.:)
The indentation on the surface of the water is larger around the more dense piece as it is effectimg the surface tension of the water more than the less dense piece??
Although surface tension is a fair thing to consider - neglect it.
A hint - think about the torques acting on the beams.
The lighter beam will float with a flat surface at the top.
The heavier beam will float with a corner at the top.
The detailed analysis turns out to be very fiddly. The following link will give you an inkling about how to understand the result:
http://www.engineeringtoolbox.com/centre-gravity-buoyancy-d_1286.html
You should now try to work out why metastabilty works in the two cases.
In the case of the beams, recognise that the length is unimportant (as long as its a fair bit greater than the beam width) and you can look at it as a 2D problem viewable from the square ends of the beam. THe centre of mass is in the geometric centre of the beam and the cenre of buoyancy is in the geometric centre of the immersed part of the beam. The immersed area of the beam is fixed for a given density. Consider what happens when you disturb the beam from it's presumed stable position. You'll begin to see why I haven't published a full analysis.
okay, i can neglect a few things, but whats about the pond ground limiting the wave length of virtual particles against the much higher atmosphere limit causing more virtual particles to be created because of heisenbergs uncertainty in the lower part of the wood beam and creating a small pressure through the casimir effect, what about that?? Lol =)
well i think more logical than explicitly founded, that the lighter beam would lay flat because his center of mass is above the water surface, while the heavier wood would stay vertically since his CoM is below the water surface, and both CoM try to get as low as possible.
Oh no! That was one of the most important assumptions to make. How could I have left that out???? I'll never be able to hold my face up in public again. My reputatation's in tatters. The shame of it all. The utter folly. I'm ruined. Sob :)
Yes, CoM wants be low, but CoB wants to move high. As the beam changes angle, the CoB moves and the CoM changes height/depth with respct to the water surface. It really is quite tedious.
As I implied, a detailed analysis would be very difficult to undertake on these pages (no pics and loads very tedious trig and geom to be replaced with text only). It would take several pages and all that work for a problem that nobody gives a rats about anyway.
I'll confess that I've only just about convinced myself that I've got it right.
Also, I've (finally) learnt my lesson; no more probs from me before I've written up the solution first.
My claim is only true for square (or nearly square) beams. Flat beams will float big side up regardless of density.
I think a similar result applies to equilateral triangle cross-section beams. But I'm not going there.
Ok, now I see you meant to say square beams. Each of my visulations were of load bearing beams.... which is what a beam is to me. Beams have a much larger depth vs. width in my thinking. A 4x4 is not a beam in my head. A 4" x 12" is a beam, but that probably just my habit. I can see a 4x4 would be different.
I am going to try a pencil soon.
I had said "rectangular" beams in the initial problem in error. I've only just noticed; I meant and thought "square" - I obviously hadn't checked the question thoroughly - again. But I had suggested using 4" x 4" x 6' in my second comment. My apologies.
It's unfortunate that I called them "beams" (although I can't think of a better word), obviously I chose it for the strong double-entendre in the title.
Anyway, I had a good laugh writing some of my responses. I hope you enjoyed them.
More shame - the problem reminds of the "stable 1" and "stable 2" orientations of space capsules when the land in the sea. (Oh yes, I'm that old). It may me that the denser beam has two stable positions.
When I set the problem, I was sure that I had a text book that covered it - I seem to be wrong about that.
Hi Ragknot, because of the recent attacks that have been made on me, I'm going through the postings to check for just cause.
I realise that some of the stuff, particularly the long list of assumptions could be interpreted as me being rotten to you.
I lapsed into a sort of Monty Python mode of humour when editing the post, and decided that the humourous intent was clear enough.
t:.b:H had acknowledged that humour, but I notice you haven't. I want to assure that I had no malicious intent, I just got carried away with the MP idea once I had started it.
I unreservedly apologise for any possible upset that I may have caused you.
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