Test your buoyancy understanding!

[xozzx]

In both instances the water level decreases slightly. The first because you're reducing the displacement of the canoe hull when you throw the rock overboard.

The 2nd is similar but the water level should decrease less because the weight of the rock in water will be slightly less as it's more buoyant (yes rocks are buoyant - just not very) so it is pulling the canoe hull down slightly less. It's less of a net change because the rock is already displacing it's volume in the water.

However, the way you wrote this - if the canoe itself is completely submerged in the water - the net change is zero...lol

Correct! if a little pedantic TMHeimer got there first tho.

 

[Tigerman]

I still think what actually happens is the tiles get broken by the rock and the maintenance department goes ballistic.
Ill check with them when I have a chance

 

[drinksfromtap]

The correct answer is that it decreases in the first case and stays the same in the second. The first case is explained below. In the second case the mass of the water displaced is exactly equal to the mass of the canoe, rock, paddler, and rope. The rock causes the same mass of water to be displaced whether it is sitting in the canoe or submerged in the water, assuming that the rock did not cause the canoe to sink in either case and the rock is not supported by the bottom. It doesn't matter if you displace a canoe-shaped volume or a canoe and rock- shaped volume. In both cases, volume = mass/density. Even if the rock were pumice and could float (assuming it wasn't forced under), this would be true. Even if the canoe sunk, the level wouldn't change

Edit: Regarding the rock pulling down the canoe hull less due to the buoyant force when it's submerged: yes, it is true that the canoe hull will displace less water due to the buoyant force on the rock reducing its weight. The amount of reduction in the canoe's water displacement will be exactly equal to the "new" volume displaced by the submerged rock.

EDIT: I was wrong about situation 1; changes to my original message are shown in italics. I agree with most others that the water level will decrease slightly, unless the density of the rock is equal to the density of the water. The short explanation is that when the weight of the rock is supported by the canoe, the canoe has to displace a volume of water around 2-3 times greater than the rock itself to stay afloat, assuming the rock is 2-3 times denser than water. Another way to think about this is that when the rock is resting on the bottom, the bottom of the pool is supporting a part of the rock's weight that was originally supported by the canoe. In the unusual situation that the rock density is equal to or less than (i.e. "floaty") wtar, the volume would be the same. Thank you, xozzx, for helping me see my error on this part The second part is correct as explained above.

 

[TMHeimer]

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?

 

[eponym]

Now I'm confused.

The key is that a rock, being denser than water, displaces a greater amount of water when in the boat than when in the water.

 

[Insta-Gator]

Volume to Weight conversions for common substances and materials (mobile)

1- cu/ft of pumice = 40.02 pounds

1- cu/ft of granite = 167.99 pounds

1- cu/ft of water (fresh) = 62.43 pounds

So, what kind of rock/stone are you using? Inside or outside the canoe makes a difference.

 

[xozzx]

The correct answer is that it stays the same. In both cases, the mass of the water displaced is exactly equal to the mass of the canoe, rock, paddler, and rope. The rock causes the same mass of water to be displaced whether it is sitting in the canoe or submerged in the water, assuming that the rock did not cause the canoe to sink in either case.

Edit: Regarding the rock pulling down the canoe hull less due to the buoyant force when it's submerged: yes, it is true that the canoe hull will displace less water due to the buoyant force on the rock reducing its weight. The amount of reduction in the canoe's water displacement will be exactly equal to the "new" volume displaced by the submerged rock.

You are bang on with the second part, but completely wrong with the first.

When the rock is in the boat, it displaces the weight of water equal to the weight of the rock. When it is in the water it is displacing the volume of the rock. Assuming the rock is denser than water (I have never seen a floating rock), then it will displace more water in the canoe than on the bottom of the pool.

 

[Tigerman]

I have seen floating rocks..

 

[The Chairman]

If the rock is denser than water, then the volume will decrease (albeit, it would have to be a massive rock to be able to measure this).

If the rock is less dense than water than the volume will remain the same.

The resultant force of the suspended rock (in the canoe, or floating) is buoyed up by the water. In other words the force would displace a volume of water equivalent to the weight of the rock.

The resultant vector force of a fully submerged rock is through the bottom of the pool. The rock would then displace only its actual volume. By definition of it being denser than water, that would be smaller than the volume that would be displayed by buoying up its weight.

 

[elan]

Hey, just a bit of fun, test your understanding of buoyancy / diving related physics.

2 (easyish) questions for now...

- 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?

Also explain your answer!

When the rock is released in both cases the level of the pool will drop. It soes not matter where the rock is - in the canoe or hanging on the rope as long as it is connecte to the canoe and does not touch the bottom somehow. However in the second case the canoe will be sitting at the higher level as the rock is already displacing the amount of water that otherwise would be displaced by the canoe.