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You are currently reading a thread in /sci/ - Science & Math

Thread replies: 16
Thread images: 1
>a man is sitting in a boat, which is floating in a swimming pool filled with water. There is an anchor in the boat. He marks the level of the water in the pool.
>The anchor is then removed from the boat and dropped into the pool.

Does the water level drop or increase?
>>
stays the same faggot
>>
>>7786477
Wouldn't that not be true if the pool was shallow enough that the anchor couldn't reach its equilibrium depth?
>>
>>7786943
doesnt matter
as long as its deep enough to put the entire anchor and chain in the pool without any part poking out
>>
>>7786980
>>7786477
it does matter. putting the anchor in the boat will displace the amount of water needed to counter the weight. putting the anchor in the water will displace the volume of the anchor which is lower
>>
>>7786474
Implying the anchor sets at the bottom and the chain isn't taut then the water level would drop due to the decrease in water displacement would cause. Although if the anchor never hit the bottom it would stay the same
>>
>>7786474
you can literally solve for the problem NOW, but you're too lazy intellectually to formalize the problem on your own and expect us to do it for you.

There is really no difficulty so off you go.
>>
It will decrease. Anyone who doesn't think so needs to get the fuck off of /sci/.
>>
>>7786474
It depends on the density of the anchor. An extremely dense anchor would make the boat displace more volume than the anchor would in the water on it's own.
>>
>>7787056
No it doesn't. By definition, the density of the anchor is greater than the density of water, otherwise it wouldn't sink or be an anchor. And that's all you need to know.
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>>7787056
as long as the anchor has mass if will effect it, it doesn't matter what that mass is.
>>
>>7787081
this
>>
>>7787122
What if it has negative mass?
>>
>>7786474

The anchor is already accounted for when inside the boat. It wouldn't displace any more water separately from the boat.

In shorthand:
(boat with anchor) in water = (boat in water) + (anchor in water)
>>
>>7786474

Okay, two things will happen.

One: The water will rise according to the volume displaced by the anchor.

Two: The water will fall, because the anchor will not be pushing down on the boat as much, so the boat will not be displacing so much water.

IIRC...

The weight of displaced water is always equal to the weight of the boat pushing down on it.

Since the anchor, in the boat, is displacing its equivalent weight in water, but out of the boat, displaces its equivalent volume, the water level of the pool will drop, because the anchor is denser than the water.
>>
>>7786474
nigger frogposter
Thread replies: 16
Thread images: 1
Thread DB ID: 433913



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