Test your buoyancy understanding!

[elan]

The level of the canoe raising in the second case will also depend upon the density of the rock

 

[lowviz]

...//... - You have a canoe in a swimming pool with a rock inside the canoe. You throw the rock out of the canoe into the water. What happens to the level of the swimming pool?

- You have a canoe with a rock hanging off the front on a rope, completely submerged with water. If you cut the rope what happens to the level of the swimming pool? ...//...

OK.

Now I'm confused. The rock displaces the same amount of water whether it's in the canoe or on the bottom, no? But if the rock is on the bottom the canoe displaces less water, no?

I had a great instructor in the past who taught HOW to think about complicated problems:

1) Start with an extreme case that doesn't "break" the rules. Just like this problem, when all the values are small or close, it is way too easy to get confused.

How I figure this:

OK, we have a rock that is denser than water. The problem doesn't care about the exact density so let's pick one that displaces exactly one pound of water and, on land, the rock weighs exactly one thousand pounds. (very dense rock)

Put a big boat into the pool. Add the rock to the boat. The water level goes way up as the boat now got one thousand pounds heavier. Chalk the level of the pool as this is where we begin.

Throw the rock into the pool. The boat is exactly a thousand pounds lighter so the water level goes way down. But the thousand pound rock displaces about a pint of water, so the pool goes back up a tiny bit. Net effect, water level goes way down. So for any rock that is denser than water, the water level will go down in proportion to the rock's weight in air minus the weight of the water it displaces. In our case, 999 pounds of displaced water.

Next case:
The boat is being pulled down by the suspended rock that effectively weighs 999 pounds. One thousand pounds on land minus the one pound buoyant force. Here is where we start, chalk the side of the pool.

Cut the rope, and the boat is 999 pounds lighter. But, the rock is already under water and stays there, so no additional level change.

Result: Either way, the water in the pool goes down exactly the same amount.

 

[dmoore19]

The weight of the rock is less in the water than it is in air. The rock would weigh less in saltwater than it does in fresh water too. The rock, regardless of it's density, would displace the same amount of water if fully immursed. Because it is more dense than the water it displaces, it is negatively buoyant. The canoe would be providing buoyancy for the rock, it is suspended when attached to the rope. So in both cases the pool would drop.

 

[TMHeimer]

The weight of the rock is less in the water than it is in air. The rock would weigh less in saltwater than it does in fresh water too. The rock, regardless of it's density, would displace the same amount of water if fully immursed. Because it is more dense than the water it displaces, it is negatively buoyant. The canoe would be providing buoyancy for the rock, it is suspended when attached to the rope. So in both cases the pool would drop.

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?

 

[Guba]

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?

Consider "weight" to be interpreted "net weight". That's the commonly accepted term, hence "astronauts are weightless" is considered true. The real story, of course, is that there is a balance between the "weight" (mass of the astronaut times gravity) pulling down and centripetal acceleration pushing away from the earth (caused by his tangential velocity). When these two are combined, the astronaut is "weightless".

The same can be considered for a diver underwater (or a rock). The weight pulling down is offset or partially offset by the buoyant force. Don't believe me? I'll show you a photo of a diver at 45 feet below the surface standing on a set of bathroom scales. His weight? Zero pounds. Voila! He's "weightless".

(By the way...getting that shot drew a LOT of funny looks from passing divers!)

 

[TMHeimer]

Consider "weight" to be interpreted "net weight". That's the commonly accepted term, hence "astronauts are weightless" is considered true. The real story, of course, is that there is a balance between the "weight" (mass of the astronaut times gravity) pulling down and centripetal acceleration pushing away from the earth (caused by his tangential velocity). When these two are combined, the astronaut is "weightless".
The same can be considered for a diver underwater (or a rock). The weight pulling down is offset or partially offset by the buoyant force. Don't believe me? I'll show you a photo of a diver at 45 feet below the surface standing on a set of bathroom scales. His weight? Zero pounds. Voila! He's "weightless".

(By the way...getting that shot drew a LOT of funny looks from passing divers!)

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.

 

[NAM001]

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.

 

[Sly New Girl]

If its a backyard pool the level goes down. In order to achieve buoyancy you need to displace a volume of water equal to weight. The canoe is a bcd for the rock. To sink you only need displace a volume of water equal to the volume of the object and rock is denser than water.
You can't just sail through a rock ...

 

[PfcAJ]

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.

Mass is constant, weight is not.

 

[Tigerman]

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..