Test your buoyancy understanding!

[elan]

The REAL question here - other than how to explain the broken tiles to the maintenance dudes - is how big of a rock would you need for this to be visible in the swimming pool. Or how small the pool would need to be..

I think i all depends upon how deep your pockets are and the amount of space that your wallet takes in the canoe relative to the amount of space taken by the rock.
In the best case scenario the amount of canoe displacement change after paying to the dudes will be the least!

 

[Guba]

Yeah, I understand all that--it's relative/a balance, etc. The mass is the same though, no? When you study all those examples of "How much air must you put in your lift bag(s) to raise a 100 lb. outboard motor that sits in 100 fsw"--- it's still a 100 lb. motor, in water or on land.

Yes, the mass is constant. As for raising the motor, a complete answer concerning how much air would be needed to lift it would include figuring out the buoyancy of the motor itself. That 100 pound motor (in air) wouldn't actually weigh 100 pounds in water---it wouldn't have much buoyancy, but it would have some. THEN you would have to calculate how much air would be required to lift it in water (something less than 100 pounds of lift).

 

[Bob DBF]

The charateristics of the rick is moot when the boat. If the rock is the size of one pint and weighs 20 lb's it will make the boat displace 20 lbs of water. When thrown over the side the boat becomes 20 lbs lighter and displaces 20 lbs less water and in the water the pint size rock displaces 1 lb of water so in the boat the rock will have the result of adding 20 pints of water to the pool and in the water it will be 1. So in the water the rock displaces 1/20th it does in the boat. In the boat weight coults in the water displacement counts.

Now if you could explain the 500# canary in the foward torpedo room.



Bob
------------------
There is no problem that can't be solved with a liberal application of sex, tequila, money, duct tape, or high explosives, not necessarily in that order.

 

[Tigerman]

I think that might be a penguin rocket?

 

[TMHeimer]

Yes, the mass is constant. As for raising the motor, a complete answer concerning how much air would be needed to lift it would include figuring out the buoyancy of the motor itself. That 100 pound motor (in air) wouldn't actually weigh 100 pounds in water---it wouldn't have much buoyancy, but it would have some. THEN you would have to calculate how much air would be required to lift it in water (something less than 100 pounds of lift).

Understood. My thinking is that the difference in the rock's weight in water, as you said, is not much from what it weighs when in the canoe, as it has very little buoyancy. And no difference in weight whether tied to the boat (in water) or on the bottom. So that leaves displacement. I figure the rock on the bottom displaces very little water. The rock in the canoe causes the canoe to displace more water than the rock on the bottom does because the rock's weight in the canoe becomes spread out over the whole bottom of the canoe--when it's removed, the canoe displaces a fair bit less water now, thus the pool water level drops. That make any sense?

 

[Rogersea]

This Thread reminds me of an old Styxs song Titled "Too Much Time On My Hands" .......

Cheers,
Roger

 

[Guba]

Understood. My thinking is that the difference in the rock's weight in water, as you said, is not much from what it weighs when in the canoe, as it has very little buoyancy. And no difference in weight whether tied to the boat (in water) or on the bottom. So that leaves displacement. I figure the rock on the bottom displaces very little water. The rock in the canoe causes the canoe to displace more water than the rock on the bottom does because the rock's weight in the canoe becomes spread out over the whole bottom of the canoe--when it's removed, the canoe displaces a fair bit less water now, thus the pool water level drops. That make any sense?

Yup, that pretty much sums it up. In essence, your explanation is roughly the same as those of several posters. I think we have a concensus! (Hey, that's a rarity here on SB, isn't it?)

 

[elan]

Yup, that pretty much sums it up. In essence, your explanation is roughly the same as those of several posters. I think we have a concensus! (Hey, that's a rarity here on SB, isn't it?)

Had a spare air been used in place of the rock and a Jacket BC in place of the canoe in the OP and all this was posted in the DIR forum we would already had more than a 1k posts and no consensus

 

[dmoore19]

Hmm.. You're saying the weight of something that's in water changes from what it was in air? It is easier to lift and maneuver in water, as we all have found out, but the weight actually changes?

If you weigh the rock while it is in water it will weigh less, yes that is what I am saying.

 

[TMHeimer]

If you weigh the rock while it is in water it will weigh less, yes that is what I am saying.

OK, I agree. We're basically discussing the word "weight". I see what you say and agree with you that the weight of an object is a combination of its mass, it's buoyancy, how it's affected by gravity, density of what's around it (water, air, etc.), etc. So weight can change, though mass doesn't. I guess most lay people like myself think of weight as the same as mass. Ei: If I'm walking on land, or scuba diving, or taking a space walk, I still "weigh" 200 pounds, even if a scale may or may not say so. Or would it be more correct to say I still have 200 lbs. of mass? I assume people living in Denver would "weigh" a bit more because the atmosphere is a little less dense.