Hacker News Reader

Comment by awad | Hacker News Reader

ok i'll bite.

someone explain?

How about:

When the anchor is in the boat, it displaces its mass in water. When the anchor is in the water, the anchor displaces its volume in water.

That's basically right. I'd make it a bit more precise by noting that the anchor in the boat displaces a volume of water whose weight is equal to the anchor's weight (Archimedes' Principle); call this V_1. When in the water, the anchor simply displaces a volume of water equal to its own volume; call this V_2. Since the anchor is denser than water, V_1 > V_2, so the level of the lake goes down when the anchor gets thrown in the water.

Also, you need to replace "it's" with "its" in your solution. (In 2005, I edited the Definitive and Extended Edition of The Feynman Lectures on Physics for physics content and for spelling/grammar. Could you tell? :-)

Wouldn't it make a much smaller difference in volume if the anchor was unable to hit bottom? Then it would be applying a force downward on the boat (increasing volume submerged) in addition to the volume of the anchor.

Yeah, but then it wouldn't be called an anchor. Just as we reasonably assume the anchor is heavier than water, we can assume it will hit the bottom.

It is a reasonable assumption, but the term "anchor" can also reasonably be applied to an object expressly intended to anchor a boat, regardless of whether it is attached to a sufficiently long rope to work in all parts of the lake.

If the anchor doesn't hit bottom, the level of the lake shouldn't change at all; the boat plus the anchor should displace the same amount of water as they did when the anchor was inside the boat -- exactly enough water to equal the weight of the boat plus the anchor. The only difference is that since the anchor is now displacing a small amount water by itself, the boat doesn't need to displace as much, so it should rise a tiny bit.

where tiny bit = volume of anchor / surface area of lake. if the boat moves some because it isn't properly anchored, then presumably there would be more tension on the chain and thus a net rise in water level. of course, the problem initially never describes a chain & introducing the effects of one makes the problem far less tractable.

possibly - though the weight would be a lot less in the water than sat on the boat (aka in air) :)

i think what you are saying is true.

Would you mind sharing what were the corrections to the physics content? Have there been significant changes? Thanks!

There were no serious physics errors. Most corrections were typos and transcription errors (the original Lectures were a rush job, since there was no text for the course at the time). These are precisely the kind of errors that perplex beginners, though ("Hmm, where did that factor of 2 come from?"), so it was good to fix them.

Small world. I typeset the math for the Feynman Physics corrections at WestWords in Utah.

Thanks for the it's / its pointer, I should have caught that.

http://www.feynmanlectures.info/solutions/boat_anchor_lake_s...

To paraphrase, when the anchor is in the boat it's pushing the whole boat down, displacing exactly the volume of water that weights the same as the anchor.

When the anchor is on the floor of the lake it's only displacing the exact volume of the actual anchor. Since the anchor is more dense than water, it's less volume.

Instead of an anchor, imagine a lump of very dense material, where the density is 1000 kg/L. This is a 1000 times as dense as water, and a cube of this material with a length/width/height of 10 cm would weigh a tonne.

So the cube is in the boat, pushing the boat down and displacing 1 tonne of water (1 m^3).

You throw the cube off the boat and the boat rises, because it doesn't need to displace so much water. The weight now displaces hardly any water at all (actually 0.001m^3).

Reply on news.ycombinator.com