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Bulls, water-levels and other things

HS Thomas
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Joined: May 15, 2002
Posts: 3404
Maury the bull has just swallowed a time-bomb that is going to go off in 10 minutes.

How would you best describe the situation?

Angelo is in his boat in his pool about to lower a solid bronze sculpture into his pool. The sculpture has to be moved to it's throne which is at the bottom of the pool. Once placed on the throne, will the water in the pool have risen , fallen or stayed the same level.


Will the water level rise , fall, or stay the same when the sculpture is fully sub-merged but has not yet touched bottom ?

regards
[ September 11, 2003: Message edited by: HS Thomas ]
Joel McNary
Bartender

Joined: Aug 20, 2001
Posts: 1821

Ugh. That first joke/puzzle is just abominable!
Secondly, the water level in the lake will have fallen. While the scuplture is in the boat, it dislpaces water by weight. However, once it is on the throne, it will displace by volume. Since the statue sinks, the density is greater than water's, so displacement by volume is less than displacement by weight, so the water level falls.
While it is sinking, I guess that the level will remain constant (since it still displcing by weight), but I'm not sure...I can see it both ways. Time to get me a swimming pool, throne, statue, and boat and see what happens.


Piscis Babelis est parvus, flavus, et hiridicus, et est probabiliter insolitissima raritas in toto mundo.
Vinod John
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Joined: Jun 23, 2003
Posts: 162
Are these two diff questions ??? look like it is
For the second, the water level should not rise or fall because before the statue is lowered inside the pool, its weight has displaced equivalent volume of water (ie., weight of water equal to its waight is displaced).
Did my explaination make any sense ?
BTW, great picture
Vinod John
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Joined: Jun 23, 2003
Posts: 162
Originally posted by Joel McNary:

displacement by volume is less than displacement by weight

but the statue hasn't reach the bottom is it, so will this apply in this situation
[ September 11, 2003: Message edited by: Vinod John ]
HS Thomas
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Joined: May 15, 2002
Posts: 3404
I don't know. I haven't looked at the answer that closely yet. I'm still trying to get my head around the problem.
In case you missed Joel's comment.
Joel had the right answer to the joke.
A-bomb-in-a-bull!
regards
[ September 12, 2003: Message edited by: HS Thomas ]
HS Thomas
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Joined: May 15, 2002
Posts: 3404
The answer to the first half: The water level will fall.
Why ? When in the boat the statue pushes the boat down (thereby raising the water up around the boat) by an amount of water equal to the weight of the object. Once on the bottom and not attached to the boat, the sculpture will only cause the water to rise by an amount equal to the volume of water displaced. Bronze is extremeley dense so the weight of the sculpture displaces more water than the volume.
The answer to the bonus question : Believe it or not the water level will not change until the statue touches the bottom and the pressure on the rope is released.

Actually, I didn't show the rope.I should have added that the statue was gently lowered by some rope-pulley mechanism.
Hope this explanation is reasonable.

regards
[ September 13, 2003: Message edited by: HS Thomas ]
HS Thomas
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Joined: May 15, 2002
Posts: 3404
Just checking your answers again, Joel and Vinod, it seems you both are spot on. I think Joel answered the first and Vinod the second.

I still didn't get it. . I will one day.
regards
Bert Bates
author
Sheriff

Joined: Oct 14, 2002
Posts: 8829
    
    5
I haven't reviewed this in detail, but I have to take issue with the idea that density is related to displacement. A cubic foot of aluminum will displace the same amount of water as a cubic foot of lead.


Spot false dilemmas now, ask me how!
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Joel McNary
Bartender

Joined: Aug 20, 2001
Posts: 1821

Originally posted by Bert Bates:
I haven't reviewed this in detail, but I have to take issue with the idea that density is related to displacement. A cubic foot of aluminum will displace the same amount of water as a cubic foot of lead.

When they are underwater, yes, they will. However, when they are in the boat (not submerged), the lead will displace more than the aluminum. This is the basic principle that Archimedes discovered, causing him to streak Athens shouting "Eureka!" You see, he could say that because he had just come from a bath, and he didn't reek-a.
Don't laugh, it wasn't that funny.

----------
and the pressure on the rope is released.
Ah- he was using a rope to lower it! OK then, yes, the level will not change. I thought that he jsut threw it overboard....
Bert Bates
author
Sheriff

Joined: Oct 14, 2002
Posts: 8829
    
    5
Joel -
The boat idea is a mind bender...
If we assume REALLY STRONG hulls, a black hole loaded onto an aircraft carrier will displace a lot more water than that same black hole loaded onto a rowhoat...
So, does the answer to the original question depend on the dimensions of the boat? Umm, maybe not, I guess if the boat is still floating, then the weight displacement (rowboat or ship), should be the same...
While the statue is 'suspended' in the water the combination of the water displacement and the remaining weight displacement should leave the water level the same, and when the statue is placed on the throne, the boat's total weight will drop, to the water level will rise.
That's my answer and I'm sticking to it...
(I know, I know I'm late, but I just HAD to work it out for myself :roll: )
Jim Yingst
Wanderer
Sheriff

Joined: Jan 30, 2000
Posts: 18671
when the statue is placed on the throne, the boat's total weight will drop, to the water level will rise.
Are you sure you meant to say "rise"?


"I'm not back." - Bill Harding, Twister
Vinod John
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Joined: Jun 23, 2003
Posts: 162
There is three scenerio in the above problem
1) when the statue is on the boat, in this case the water displaced will be equivalent to the weight of the statue
2) when the statue is resting at the bottom on the pool or it being lowered by a supporting rope (and the statue is submerged) in this case the water displaced is equivalent to the volume of the statue.
3) when the statue is submered but is falling freely (no rope to lower it) ... this was the one that confused me ... though I hate to say this My answer might be wrong ... I might have to go with Joel answer because it this case too the statue's weight can only play a role in disturbing the water than displacing it. The water displace will solely depend on the volume occupied by the statue.
Bcos this is a solid bronze statue the water displaced by its weight will be greater than that displaced by its volume. so the level of water in the first scenerio will be higher than the next two
[ September 15, 2003: Message edited by: Vinod John ]
Jim Yingst
Wanderer
Sheriff

Joined: Jan 30, 2000
Posts: 18671
There is three scenerio in the above problem
1) when the statue is on the boat, in this case the water displaced will be equivalent to the weight of the statue

Plus the weight of the boat and its other contents, but that's constant throughout the problem, so we can ignore it. OK...
2) when the statue is resting at the bottom on the pool or it being lowered by a supporting rope (and the statue is submerged) in this case the water displaced is equivalent to the volume of the statue.
No, these are two different scenarios:
2.A) If the statue is being lowered (slowly) by a supporting rope, then the weight of the statue still pulls on the boat, and creates additional displacement. Ultimately, as long as the statue is being held up by the boat (at least in part) then the total amount of water displaced must have weight equal to the weight of the boat plaus the weight of the statue. So if the statue is being lowered on a rope, water displacement (and thus water level) is the same is in (1) above.
2.B) If the statue is resting on the bottom of the pool, then the ground now supporting some of the weight that was previously only offset by buoyancy. Which means that the total water being displaced by (statue + boat) is now less than it was in (1) and (2.A).
3) when the statue is submered but is falling freely (no rope to lower it) ... this was the one that confused me ... though I hate to say this My answer might be wrong ... I might have to go with Joel answer because it this case too the statue's weight can only play a role in disturbing the water than displacing it. The water displace will solely depend on the volume occupied by the statue.
Here the displacement and water level are exactly the same as in (2.B), though it's harder to see. But consider - the displacement by the boat is the same as in (2.B) because either way, it's just floating there with no connection to the statue. And the displacement by the statue is the same as in (2.B) because, either way, the statue is fully submerged, and its displacement is equal to the volume of the statue. So if the displacements by boat and statue are, individually, the same as in (2.B), the total displacement, and total water level, are the same in both cases.
Bcos this is a solid bronze statue the water displaced by its weight will be greater than that displaced by its volume.
But in this scenario the weight isn't being balanced by buoyancy - rather, the weight exceed buoyancy, and therefore the statue is accelerating downward. So "water displacement by weight" is no longer relevant here, because the excess weight isn't displacing anything, it's just driving the statue downward.
So, relative to (1), water levels are:
2.A) same
2.B) lower
3) lower, same as (2.B)
[Corrected "higher" to "lower" above - Jim]
[ September 15, 2003: Message edited by: Jim Yingst ]
Bert Bates
author
Sheriff

Joined: Oct 14, 2002
Posts: 8829
    
    5
Ah Jim -
Always catching me on those picky details... rise vs. lower...
I agree with your answer, assuming the lowering is happening at steady state, i.e. no acceleration.

How about this one, If a 100 lb. juggler is juggling 3, 5 lb balls (one is always in the air), while standing on a scale, how much weight will the scale register.
Oh BTW I don't know the official answer, so don't waste any time on this if you NEED to know the real answer from me!
Vinod John
Ranch Hand

Joined: Jun 23, 2003
Posts: 162
Originally posted by Jim Yingst:

Bcos this is a solid bronze statue the water displaced by its weight will be greater than that displaced by its volume.

But in this scenario the weight isn't being balanced by buoyancy - ...


Look like you are mistaken, the above posting is to give the reason, why scenerio 1 will have have higher water level (or displaces more volume of water) than other scenerios
Originally posted by Jim Yingst:


So, relative to (1), water levels are:
2.A) same
2.B) higher
3) higher, same as (2.B)

The water level in case of 2.B and 3 will be lower than 1 and 2 A because water displaced by the weight of the statue will be more than the water displaced by its volume.
and 2.B = 3 , 1 = 2 A
Jim Yingst
Wanderer
Sheriff

Joined: Jan 30, 2000
Posts: 18671
Look like you are mistaken, the above posting is to give the reason, why scenerio 1 will have have higher water level (or displaces more volume of water) than other scenerios
Ah, I thought ths comment was still in releation to scenario 3 since it was posted unter that. I see now it was a separate section. OK, nevermind then.
The water level in case of 2.B and 3 will be lower than 1 and 2 A
Correct - this was an error in the last couple lines of my post, now corrected. Thanks.
 
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