5 ms·
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
by mhartl 18y ago
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? :-)
- 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.