Is water level the same after an anchor is thrown from a boat?

AI Thread Summary
When an anchor is thrown from a boat into a pond, the water level initially stays the same due to the buoyant force acting on the boat. The boat displaces a volume of water equal to the weight of the anchor while it is onboard. Once the anchor is thrown overboard, it sinks and displaces an equal volume of water based on its own mass, which matches the weight it displaced while in the boat. However, the boat rises slightly due to the reduced weight, leading to a decrease in the volume of water displaced by the boat. Ultimately, the water level remains unchanged because the displaced volume by the anchor equals the volume displaced by the boat when the anchor was onboard.
jamesbiomed
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Homework Statement



(5) A person sitting in a boat floating on a small pond throws a heavy anchor overboard. Does the level of the pond rise, fall, or stay the same? Explain.


Homework Equations



FB=mass of fluid displaced times g
Fnety=Fb-mg



The Attempt at a Solution




Theoretically, the pond level should stay the same. For the boat/man system after the anchor toss, the force of gravity will be less, because the mass of the boat/man system is less without the anchor. The buoyant force since it is floating is equal to the weight of the fluid displaced. As the anchor is tossed from the boat, there will be a shift in the net force up in the positive y-direction (assuming up is positive) causing the boat to rise relative to the water, and a portion of the water once displaced by the boat to return. This will cause an overall drop in water level.
However, when the anchor is thrown in, it will sink, until it reaches the bottom, and its exact mass will displace the same mass of water, which will cause a rise in the pond level exactly the mass of the anchor.

...I wrote that based on my verbal understanding of buoyant force. But what I'm struggling with is exactly how buoyant force will affect the boat after the anchor is tossed. Like I said, it will rise so that the amount of fluid it displaces is less. But how much less? The buoyant force will decrease, because the mass of the water displaced will decrease. I know F net y=Fb-fg. So fg is going to decrease, which will cause F net y to increase. But Fb will also decrease. So how can I better explain this, and determine precisely if the amount displaced by the anchor will equal the change due to the boat being lighter?

Sorry for the length...
 
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Consider the boat by itself (no anchor yet). If you put an anchor in the boat, how much additional water is displaced? (Consider Archimedes' principle.) Compare that to how much water is displaced by the anchor when it's just tossed into the water.
 
The additional water displaced--would be equal in mass to the added anchor. Does this mean we could set Fnet y =0?
 
jamesbiomed said:
The additional water displaced--would be equal in mass to the added anchor.
Yes, when the anchor is put on the boat, the additional water displaced equals the mass of the anchor.

What about when the anchor is just put in the water directly?
Does this mean we could set Fnet y =0?
Not sure what you mean. For something in equilibrium, such as the floating boat, Fnet = 0.
 
The mass of the water displaced by the anchor will equal the mass of the anchor for sure.

Okay, I wasn't thinking right about the boat being equilibrium.

I think knowing this, I should be able to conclude it.

Thanks a lot for the help!
 
jamesbiomed said:
The mass of the water displaced by the anchor will equal the mass of the anchor for sure.
Really? Why do you think that?
 
James this would mean the mass of an aluminum anchor would have the same mass as that of a lead anchor. A "water anchor" would have the same mass as the water it displaced.

You see where the good Doctor is leading you...
 
The water displaced is equal to the VOLUME of the anchor.

Archimedes Principle, Fbuoyant = ρliquidVg
 
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