06 Nov 06
OK, another little physics puzzle. This is easy enough if you look at it in the right way. Sorry if it's been on here before.
You are in a boat floating in a swimming pool. There is a large rock in the boat with you. You throw the rock over the side, and it sinks to the bottom of the pool.
What happens to the level of water in the pool?
06 Nov 06
Originally posted by sugiezdNo, the water level should be lower if the rock is in the lake and higher if the rock is in the boat. This is because the rock's density is greater than that of water.
I would think that it remains the same as the rock was already displacing it's own volume of water whilst in the boat.
06 Nov 06
Originally posted by ketch90Lets expand the example a bit so you can understand it better:
I think it is the same
Lets say the boat has you and this rock. Lets say that you have several boats, one with stuff like rocks that all weigh say 100 Kg.
Now suppose some of them are made of ironwood ( I think the stuff sinks) and some of them are granite and some of it is 100 Kg of lead.
Now think about that. They all displace equal amounts of water. But if you throw them overboard, you can see the most dense ones will displace less than the less dense ones so it depends on how dense the individual stones are but they will not leave the height of the pool the same, each one will have its own level. Suppose you change the wooden one with depleted uranium, its what, twice as dense as lead?,
then it displaces less water than the lead weight so will lower the level of the pool compared to lead. So the more dense the object the lower the level of the pool when thrown over board. Its a volume ratio thing.
06 Nov 06
Originally posted by uzlessWell, yes. But if the rock were less dense than water, the opposite answer would be true. In the boat, the rock displaces a volume of water equal in weight to the rock (dense rock, lots of water displaced). In the lake, the rock displaces a volume of water equal to its volume (dense rock, small volume, less water displaced).
isn't it because the weight of the rock is pushing down on the bottom of the boat, thereby displacing more volume of water than if the rock was simply dropped in the lake?
07 Nov 06
Originally posted by PBE6Any of you heard of Archimedes?
Well, yes. But if the rock were less dense than water, the opposite answer would be true. In the boat, the rock displaces a volume of water equal in weight to the rock (dense rock, lots of water displaced). In the lake, the rock displaces a volume of water equal to its volume (dense rock, small volume, less water displaced).
The rock will displace its own volume of water whether in the boat or on the bottom of the lake.
That is how density is calculated, you know the weight of an object, you measure the volueme of water it displaces, divide one by the other and there you have it.
Remember the eureka can?
The water level remains the same.
07 Nov 06
Originally posted by sugiezdHere is an explanation of Archimedes' principle designed for children:
Any of you heard of Archimedes?
The rock will displace its own volume of water whether in the boat or on the bottom of the lake.
That is how density is calculated, you know the weight of an object, you measure the volueme of water it displaces, divide one by the other and there you have it.
Remember the eureka can?
The water level remains the same.
http://www.factmonster.com/ce6/sci/A0804583.html
Maybe you will be able to understand it.
07 Nov 06
1 edit
Originally posted by sugiezdYou'll notice I've already mentioned Archimedes' principle, so yes, I've heard of him 🙂
Any of you heard of Archimedes?
The rock will displace its own volume of water whether in the boat or on the bottom of the lake.
That is how density is calculated, you know the weight of an object, you measure the volueme of water it displaces, divide one by the other and there you have it.
Remember the eureka can?
The water level remains the same.
Archimedes principle goes like this. A body in water is subject to an upward force equal to the weight of the water displaced.
A corollary is that a floating body (i.e. one where the upward force exactly balances its weight) displaces its own mass of water. Not its own volume. So this is what's happening when the rock's in the boat.
You're quite right that once in the pool, the rock displaces its own volume of water. But the two cases are not the same, because the density of the rock is different to the density of water.
07 Nov 06
1 edit
Originally posted by sugiezdAs mentioned above, this is wrong. Any submerged body will displace a certain volume of water, and one of two things will happen:
Any of you heard of Archimedes?
The rock will displace its own volume of water whether in the boat or on the bottom of the lake.
That is how density is calculated, you know the weight of an object, you measure the volueme of water it displaces, divide one by the other and there you have it.
Remember the eureka can?
The water level remains the same.
(1) If the density of the object is less than that of water, it will float, and the weight of the displaced water will equal the weight of the object.
(2) If the density of the object is greater than that of water, it will sink, and the weight of the displaced water will be less than the weight of the object. In this case, the object displaces the maximum amount of water it can, which is equal to the volume of the object.
Since the denser object reaches a maximum in the amount of water it can displace by itself, it's easy to see that if you can get it to float (by, say, putting it in a boat) it will displace more water. Therefore the water level will be higher if the rock is in the boat.
07 Nov 06
Originally posted by sugiezdway to rock the boat suzie
Any of you heard of Archimedes?
The rock will displace its own volume of water whether in the boat or on the bottom of the lake.
That is how density is calculated, you know the weight of an object, you measure the volueme of water it displaces, divide one by the other and there you have it.
Remember the eureka can?
The water level remains the same.