If you take a big rock and drop it overboard from a canoe in a swimming pool, will the water in the pool rise?

Answer 1

Nobody seems to have figured this out yet. The rock in the canoe displaces a volume of water equal to its mass, as part of the total displacement of the canoe. When you drop the rock overboard, it displaces a volume of water equal to the volume of the rock. Since rock is denser than water (assuming we're not dealing with pumice here) the rock now displaces less water than it did in the canoe. So the water level in the pool will fall.

In the case of pumice or other floating rock, it displaces its mass in water in either case and the water level is unchanged.

Answer 2

When the rock is in the canoe, the canoe displaces an amount of water equal in weight to the weight of canoe and its cargo. For just the rock, that might be say 5 times its volume's worth of water. Once you throw the rock overboard, the volume of water displaced by the added weight of the rock inside the canoe will be replaced by only the volume of the rock, and therefore the water level will drop according to approximately 4 times the volume of the rock.

I guess...

Answer 3

If you put the canoe, with yourself and the rock into the pool, it displaces water. That means that, although the canoe floats, the weight of you and the rock in it would make it float lower into the water, displacing that volume of water.

If you throw the rock into the pool itself, it would displace no more water than it did while it was in the canoe with you. Therefore the water would not rise any more.

Answer 4

In fact the water level will fall. Instead of a rock, let's say it's a chuck of aluminum just so we can use numbers from a density table. My density table says Aluminum is 2.7*10^3 kg/m^3. A bar of aluminum .1m x .1m x 1m has a volume of .01 m^3. Its mass is .01 m^3 * 2.7*10^3 kg/m^3 = 27 kg. Because of its weight, when it's put in the canoe, 27 kg of water must be displaced by the canoe sinking a bit lower into the water. The water volume required for that mass of water is 27 kg/(10^3 kg/m^3) = .027 m^3.

When you toss the Al overboard, it only displaces .01 m^3 sitting on the bottom of the pool.

Answer 5

The rock was floating in the canoe, meaning that it displaced water volume with a weight equal to the weight of the rock.

However when you put the rock into the pool, it will sink. It will displace water equal to its *volume*, which is a lesser amount of water than the amount equal to its *weight*.

Therefore the water level in the pool will drop.

Answer 6

good quesion, does the rock inside the canoe displace as much water as the rock in the pool. I guess it depends on how heavy the rock is. in this case weight is more important than mass because the deeper the canoe sits in the water the more water it will displace, therefore if the rock is small and very heavy it will displace more water inside the canoe. whereas if the rock is big and light it will displace more being in the water.

Answer 7

Yes, since youre dropping something with mass, and it is more dense than water, It will sink to the bottom, and take place of the water that is already there, causing it to find more room, which would cause the water level to rise, depending on how big the rock is.

*Edit* @ Whidd, yes, but the canoe is more dense than water, so it doesn't affect the actaual water itself.

Answer 8

Yes, because it's taking up space that was previously taken up by water, so the water level rises a bit. But unless the pool is tiny, and the rock is huge, the change in water level won't be noticeable.

Answer 9

depend on how big the rock and pool are.The rock is displacing water however if there is a great size differential ,water tension will hold back a measurable rise
ha ..got me on the canoe part

Answer 10

Yes , because two things cannot occupy the same place and therefore the rock would replace where a large mass of water was and the water would have to occupy a different space (rise).