A fisherman throws his anchor into the water. What happens to the water level of the lake? | Hacker News Reader

A fisherman throws his anchor into the water. What happens to the water level of the lake?

feynmanlectures.info

93 pointstsally45 comments

These questions bring out the detail-oriented in all of us.

- What if the fisherman throws the anchor into another lake besides the one he is in?

- What happens in a strong current when the anchor grabs the bottom? That pulls the boat down and increases water level.

I think we're so adept at nitpicking these things due to developing skills at debugging.

I love using that as an excuse when I get overly pedantic about how ideas or other daily-life things can go wrong. "After all these years of debugging code (or mentally debugging code paths), I'm just debugging life!"

Actually, if that anchor is attached to any sort of cabling who's density is less than that of water--such as nylon rope or cotton-based fiber attached to a rope, it all depends on the relative depth of the bottom of the lake's floor (i.e. the amount of rope payed out during the anchor's descent). For all but the thinnest cables, and shallowest lakes, the water will actually rise. For anyone who's been out on a boat in most lakes, the volume of rope attached to the anchor greatly exceeds the volume of the anchor.

If that anchor is attached to any cabling who's density is greater than that of water, the original reasoning still holds.

Btw, we're also assuming here that the lake itself has a constant volume over a sufficiently short time-scale relative to the sinking anchor. For some lakes which are essentially pond tributaries of large rivers, that is most definitely NOT the case.

The only way that the water level could rise is if the average density of the submerged rope and anchor were less then the density of the water. However if this were the case then the anchor would not reach the bottom due to having insufficient mass to submerge the rope, and the water level would remain the same.

For typical anchor and typical rope (i.e not metal chain) on a typical boat on a typical lake:

Scenario #1:

1. Negative buoyancy of anchor >> Positive buoyancy of rope. 2. Sufficient rope exists to reach the bottom with some additional rope remaining in the boat. 3. It's a calm enough pond without eddy currents such that, essentially, the only non-trivial forces acting on the rope once an anchorage has been established are isotonic (i.e water pressure).

Actually #3 completely obviates the need for an anchor in the first place, but I digress.

Imagine that one were to cut the rope at the anchor.

The rope begins to ascend to the surface due to its positive buoyancy and the amount of water being displaced will decrease over time until all of the rope is floating on surface and displacing its own weight.

Scenario #2:

The case is slightly different if the rope is under any kind of tension. Assuming a sufficiently strong rope and zero relative boat movement from its anchorage, the negative buoyancy forces due to the anchor are sufficient to overcome drag effects due to the current so the boat is very slightly being pulled under--hence an additional very slight increase in the water level of the pond.

Two additional edge-case visualizations that may be helpful. Think of the "perfect superdense anchor" scenario with a neutron-star and carbon nano-tube rope and a "worst-case" anchor consisting of aero-gel.

... but then it reaches equilibrium again because of the streams flowing in and out of the lake.

Ah reality, always in need of adjustment to our theories...

So if I'm on a boat and I happen to have a barrel of lake water in front of me and I dump it overboard, the water level stays the same?

Always good to know.

If you like these kinds of thought problems, I definitely recommend Thinking Physics by Lewis Carroll Epstein.

http://www.amazon.com/Thinking-Physics-Understandable-Practi...

or for more fun try irodov(Problems in general physics). most folks who have done iit would (should ?) already be intimately aware of it :o)

A similar question: What happens to the water level if a floating iceberg melts?

That is actually an easier version of the same question. The anchor version contains all of the same issues plus an extra curveball. If the iceberg had a giant boulder resting on top, you'd have the exact same question.

Ignoring surface tension effects, and assuming a much larger and relatively fixed volume and density of supporting water compared to that of the melting iceberg (i.e. not an edge-case of huge iceberg floating in a very small amount of water) and no loss/gain of iceberg mass over time to evaporation (which can be non-trivial in dry polar air) or precipitation, the water level will actually BOTH increase & decrease over time depending on the RATE of melting which depends mostly on the temperature of the surrounding water and the amount of wetted surface the ice has in contact with it. For most icebergs, the wetted surface also changes significantly over time as its center of buoyancy changes--hence iceberg rolling and sheer surface break-offs.

We're also assuming the perfect case of relatively pure fresh-water ice and sea-water salinity. When changes in salinity and "dirty-ice" are factored-in, it gets even more complex.

It's actually a very non-trivial solution which, among other things, is why predicting iceberg lifetimes and danger to shipping lanes in the open seas is still very much an art.

In the end, the iceberg WILL melt and a net additional volume of water will be added to the sea, but if one could measure things that precisely the sea-level change relative to the melting process will fluctuate up and down over the lifetime of the berg.

This is a prime example of why, back in my high-school trivia team days, I learned to dread questions about science. The problem isn't so much that the writers of trivia questions get the answers wrong (they do, but only very rarely). The problem is that they pose questions they think are simple, but they use real world examples ("an iceberg"), instead of the Platonic ideals they meant to use ("an ideal spherical iceberg composed of pure ice floating in a sea of pure water, neglecting evaporation, on a series of rainless days..." -- actually, come to think of it, it's pretty damn hard to construct the Platonic iceberg!)

When answering a science trivia question you need to work hard to cut your train of thought off at the level of the "obvious" answer before you get too far down the line to computing the "more correct" answer.

Reminds me of the math test given to one of the WWII characters in Cryptonimicon (Daniel Waterhouse's grandfather).

Agreed. Best to use something more familiar and in more controlled circumstances. An ice cube melting in a large glass of water at room temperature.

At what local vapor-pressure?

bonsaitree said water at room temperature. The vapor pressure of water at room temperature is around 25 millibar; nothing "local" about it.

Assuming the point of measurement is at the boat, the dominant effect is that it goes up and down in a decaying sinusoidal pattern due to the ripples caused by the anchor hitting the water (it was thrown, right?)

The centre point of the sinuisoidal decay is the new, lower, level as noted by other people. But I would imagine (someone going to calculate?) that effect would be at least 1 or 2 order of magnitude below the ripple effect, given reasonable assumptions.

I read the title and for some reason envisioned a fisherman on the bank.

Thousands upon thousands of internet KnowItAlls descend upon the hapless fisherman and spew hot air by the ton, evaporating the lake entirely.

I had this question in an interview. Good one, but since i have a physics minor, i was able to guess right, and walk through it by physics.

I remember reading this question from sample Microsoft interview questions :P

Grah, I guessed wrong without thinking! What happens is obvious with even a trivial amount of thought.

(deliberately not spoiling)

hindsight bias

not if the anchor doesn't hit the floor because the rope isn't long enough.

who said there was a rope?

Cut out your assumptions and the problem becomes easier.

ok i'll bite.

someone explain?

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.

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? :-)

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

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.

i think what you are saying is true.

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

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.

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.

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).

hmm what is the level of the water being measured against? the boat or the container of the water/lake?

ooops, considering that i just realised i should be focusing on the boat's volume rather than its mass, thats a pretty stupid question!

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