
A fisherman throws his anchor into the water. What happens to the water level of the lake? - tsally
http://www.feynmanlectures.info/exercises/boat_anchor_lake.html
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DanielBMarkham
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.

~~~
nuclear_eclipse
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!"

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bonsaitree
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.

~~~
lyesit
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.

~~~
bonsaitree
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.

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

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

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Xichekolas
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.

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RK
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...](http://www.amazon.com/Thinking-Physics-Understandable-Practical-
Reality/dp/0935218084/)

~~~
signa11
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)

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

~~~
bonsaitree
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.

~~~
mechanical_fish
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.

~~~
bonsaitree
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.

~~~
newsycaccount
At what local vapor-pressure?

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

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jbert
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.

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

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

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ardit33
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.

~~~
light3
I remember reading this question from sample Microsoft interview questions :P

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

(deliberately not spoiling)

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

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whatusername
who said there was a rope?

Cut out your assumptions and the problem becomes easier.

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awad
ok i'll bite.

someone explain?

~~~
antiismist
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
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? :-)

~~~
harpastum
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.

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nandemo
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.

~~~
jcl
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.

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

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

