I'm on a floating boat with an iron anchor at a swimming pool....?

Question

Firstly, i mark the water level of the swimming pool.
Then, I toss the iron anchor over...

So, does the water level rise or fall?
And does it matter if the anchor reaches the bottom? would there be a difference?

Answer 1

the water level will fall. when iron anchor is on board it is part of how much water is displaced by boat but the dropping anchor has no buoyancy in other words the anchor on board will displace more water than a dropped anchor.
anchor chains are also made of iron and the more chain you release the greater the water in the pool will fall.
In short anchor+boat will displace more water (pool higher)
than dropped anchor + boat (pool lower)
and more anchor dropped will make pool even lower.

Answer 2

Bouyancy is derived from displacement of water. When the anchor is in the boat, its entire weight is being displaced. The amount of water displaced is equal to the mass of the anchor.

When you let go of the anchor and it hits bottom, the boat is no longer supporting that weight. The density of an anchor is much much higher than water, so the water that the anchor displaces in the water compared to being in the boat is much less. Thus what happens is the water level in the swimming pool decreases.

However, if the chain/rope is short and the anchor doesn't hit the bottom, the boat is still supporting the weight of the anchor. So, the boat still has to displace water mass equal to that of the anchor. There will be no change in the water level if the anchor just hangs off the boat.

Answer 3

It rises. If you dip your finger into the water, the water level will rise, even if it is not by much. Anything that is added into the water will make the level rise. It's adding mass. Think about it this way, when that anchor is in the water, if the anchor was really just made of water, would there be more water in the pool than before? (yes) Would more water make the level rise? (yes)

Answer 4

The displacement by using the boat+anchor mixture does not replace, because of the fact the burden appearing is the comparable in the two circumstances. (This looks counter-intuitive yet right this is the reasoning: whilst thrown in the water, the anchor DOES displace an further quantity, however the corresponding upthrust REDUCES its useful weight to the boat, so those cancel out precisely. (in addition, i replaced into attempting to be a smartass, thinking the end results of the anchor chain or twine, yet that for the time of addition nulls itself out by using Archimedes' concept.) So the respond is: displacement and water point does not replace - as long as anchor is hooked as much as boat and does not hit backside or get help. (Now if the anchor replaced into no longer related to the boat, needless to say the quantity of water it displaced could in trouble-free terms be its quantity and not its weight, subsequently the boat could displace much less quantity by using (rho_anchor)*anchor_volume; and the anchor could displace further a million*anchor_volume, subsequently the entire displacement could shrink by using (rho_anchor-a million)*anchor_volume, so the water point could fall, assuming rho_anchor>a million i.e. you haven't any longer have been given a polystyrene anchor)

Answer 5

Hard question but I think that the water level would go down because when the anchor is in the boat the displacement would be due to its mass and when you throw over board the dispalcement is due to its volume and becaues it is more dense than water its displacement by volume would less than its displacement by mass

Answer 6

If the anchor does not reach the bottom there will be no change.

If the anchor does reach the bottom the watter level in the pool will fall.

Answer 7

since the weight of the anchor was already pressurising the water, I don't think it would make any difference.
But if you really want to find out, then just ask your closest olympic pool if you can drive your boat into their polished grounds:)

Answer 8

rise