>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?
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
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.
The anchor is already accounted for when inside the boat. It wouldn't displace any more water separately from the boat.
(boat with anchor) in water = (boat in water) + (anchor in water)
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.
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.