5 ms·
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.
by antiismist 18y ago
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.
- mhartl 18y agoThat'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? :-)
- antiismist 18y agoThanks for the it's / its pointer, I should have caught that.
- harpastum 18y agoWouldn'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.
- sahaj 18y agoi think what you are saying is true.
- ErrantX 18y agopossibly - though the weight would be a lot less in the water than sat on the boat (aka in air) :)
- nandemo 18y agoYeah, 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.
- jcl 18y agoIt 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.
- jcl 18y agoIf 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.
- alphamule 18y agowhere 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.
- pfedor 18y agoWould you mind sharing what were the corrections to the physics content? Have there been significant changes? Thanks!
- mhartl 18y agoThere 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.
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- msg 18y agoSmall world. I typeset the math for the Feynman Physics corrections at WestWords in Utah.