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AND THE ANSWER IS???

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tinob
Master Marine Consultant

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USA
1883 Posts

Initially Posted - 01/20/2003 : 15:48:22

Three guys go sailing on a very hot day(not the northeast)and when the wind finally dies altogether they decide to kick back and drain a beverage or two and try the fishing. Upon deciding on the right spot one fellow tosses the anchor and they bait up. QUESTION is, did the lake water level rise, fall or stay the same?
It was a very large anchor...just go with it.

Val on the hard DAGNABIT

Val Bisagni

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Douglas
Master Marine Consultant

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1595 Posts

Response Posted - 01/20/2003 : 17:24:23

The real question is does Val have a firm grip on reality or perhaps just to much time on his hands.

Doug&Ruth
Triska (Alberg 29)
Tacoma Wa.

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Mark Loyacano
Navigator

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USA
247 Posts

Response Posted - 01/20/2003 : 18:10:00

Val, does it matter what kind of bait they were using?
Mark L
(working on it ...)

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Brooke Willson
Admiral

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Response Posted - 01/20/2003 : 18:23:33

It depends on whether they used the head or the lake. . .

Seriously, no change at all. The displacement of the anchor on board equals the displacement of the anchor overboard.

I know that, Val, because I've noticed the Chesapeake doesn't rise when I lower my swing keel. <img src=icon_smile_big.gif border=0 align=middle>

Brooke

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dlucier
Master Marine Consultant

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Virgin Islands (United Kingdom)
7583 Posts

Response Posted - 01/20/2003 : 18:33:54

The lake level went down.

Picture a boat on a lake without an anchor.

Next place an anchor in the boat. When the anchor is in the boat, the full weight of the anchor is displaced in water, causing the lake to rise.

Now throw that anchor in the lake. When the anchor is thrown into the lake, it sinks and instead of displacing water equal to it's entire weight, it only displaces water equal to the anchor's volume, which is less. Since the total amount of water displacement is less, then the lake level will go down.

Don Lucier, 'North Star'
C25 SR/FK
Cradled on the hard, 200ft from Lake Erie

Edited by - on

jsummerfield
1st Mate

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USA
96 Posts

Response Posted - 01/20/2003 : 19:31:48

Is it salt water or fresh? Was it a closed lake below the dam level? Or was a river running into a lake with water just running over the spillway?

It was a very hot day, thus the lake water was evaporating. Had the anchor been in the sun, it would have caused further evaporation upon entering the water but that was not a given. Further, if the beverages were carbonated, the weight of the gas contained in the bottles would have escaped, and the gentlemen would have also dehydrated a bit as sweat. They would have lossed even more water if it was a malt beverage. Considering the heavy anchor, a similar affect as described above is affected by landing a fish. If they pulled a fish out of the the volume of the fish would have been reduced in the water but the weight added to displacement. Their density would about match the water.

The level fell.

What kind of beverage, cans or bottles, what is the humidity and why didn't I get to go with them?

John

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dlucier
Master Marine Consultant

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7583 Posts

Response Posted - 01/20/2003 : 19:55:34

John,

You forgot to determine if the beverage swilling guys have a head onboard or are going over the side! <img src=icon_smile_wink.gif border=0 align=middle>

Don Lucier, 'North Star'
C25 SR/FK
Cradled on the hard, 200ft from Lake Erie

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jsummerfield
1st Mate

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96 Posts

Response Posted - 01/20/2003 : 20:46:07

dlucier,
Brook Willson covered the liquid fluid exchange phenomenon in his first line.

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redwards
Deckhand

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13 Posts

Response Posted - 01/20/2003 : 22:00:53

I fiqure the water level won't go down until they pull the anchor and the water fills the hole left behind !!!!!!!!!!

Ray Edwards
C-25 #5079 Cruiz'n<img src=icon_smile_cool.gif border=0 align=middle>

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Gloss
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1916 Posts

Response Posted - 01/20/2003 : 22:51:28

did they bait the anchor?

Were they using minners, or werms?

Frank

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Ed Montague
Captain

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499 Posts

Response Posted - 01/20/2003 : 22:58:43

I vote for Don's answer, it may not be right but it sounds good.

Ed Montague on 'Yahoo'
1978 #765 SK, Stnd, Dinette ~_/)~

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jm
Captain

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Canada
290 Posts

Response Posted - 01/21/2003 : 09:37:31

Was the lake frozen, like ours is right now ! If so, then no change - the anchor is sitting on top of the ice.. (and don't stick your tongue on it ! ) A Catalina25 ice fishing hut.. hmmmm.

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Dave Bristle
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Djibouti
10005 Posts

Response Posted - 01/21/2003 : 12:24:54

Yup, Don's right. <img src=icon_smile_approve.gif border=0 align=middle>

Dave Bristle - 1985 C-25 #5032 SR-FK-Dinette-Honda "Passage" in SW CT

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jwilliams
Captain

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357 Posts

Response Posted - 01/21/2003 : 12:32:03

Hey all,

It's a zen thing, right?

Jim Williams
Hey Jude C25fk 2958
Half Moon Bay, CA

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osmepneo
Past Commodore

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1420 Posts

Response Posted - 01/21/2003 : 17:28:30

Good question, but who really cares?

Don Peet
c25, 1665, osmepneo, sr/wk
The Great Sacandaga Lake, NY

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dlucier
Master Marine Consultant

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Virgin Islands (United Kingdom)
7583 Posts

Response Posted - 01/21/2003 : 20:10:28

"Good question, but who really cares?" - Don Peet

The people who really care about the lake level, are those who have a house on the lake or those whose lake is periodically drained! <img src=icon_smile_wink.gif border=0 align=middle>

Don Lucier, 'North Star'
C25 SR/FK
Cradled on the hard, 200ft from Lake Erie

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Brooke Willson
Admiral

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Response Posted - 01/21/2003 : 23:19:33

I'm not convinced, Don. I've been looking for an old Physics text. . . but this has to do with Achimedes' Law, which I think says a body immersed in water will displace a volume equal to its mass. It's all about density, not volume. Two objects of equal dimensions but unequal mass will displace different amounts of water.

I think. . .

What I can't figure out is the buoyancy part of the anchor being on the boat, since, as part of the boat, it's less dense.

It's been a long, long time since that one year in Engineering School!

Brooke

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dlucier
Master Marine Consultant

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Response Posted - 01/22/2003 : 01:32:06

"It's all about density, not volume." - Brooke

Brooke, you can't have Density without Volume, because

Density = Mass/Volume

Anyway, instead of me butchering an explanation of Archimedes' Principle and the Law of Floatation, I will just reprint an article from www.howstuffworks.com.

Question

Why can boats made of steel float on water when a bar of steel sinks?

Answer

The standard definition of floating was first recorded by Archimedes and goes something like this: An object in a fluid experiences an upward force equal to the weight of the fluid displaced by the object.

So if a boat weighs 1,000 pounds (or kilograms), it will sink into the water until it has displaced 1,000 pounds (or kilograms) of water. Provided that the boat displaces 1,000 pounds of water before the whole thing is submerged, the boat floats.
It is not very hard to shape a boat in such a way that the weight of the boat has been displaced before the boat is completely underwater. The reason it is so easy is that a good portion of the interior of any boat is air (unlike a cube of steel, which is solid steel throughout). The average density of a boat -- the combination of the steel and the air -- is very light compared to the average density of water. So very little of the boat actually has to submerge into the water before it has displaced the weight of the boat.

The next question to ask involves floating itself. How do the water molecules know when 1,000 pounds of them have gotten out of the way? It turns out that the actual act of floating has to do with pressure rather than weight. If you take a column of water 1 inch square and 1 foot tall, it weighs about 0.44 pounds depending on the temperature of the water (if you take a column of water 1 cm square by 1 meter tall, it weights about 100 grams). That means that a 1-foot-high column of water exerts 0.44 pounds per square inch (psi). Similarly, a 1-meter-high column of water exerts 9,800 pascals (Pa).

If you were to submerge a box with a pressure gauge attached (as shown in this picture) into water, then the pressure gauge would measure the pressure of the water at the submerged depth:

<img src="http://static.howstuffworks.com/gif/q254.jpg" border=0>

If you were to submerge the box 1 foot into the water, the gauge would read 0.44 psi (if you submerged it 1 meter, it would read 9,800 Pa). What this means is that the bottom of the box has an upward force being applied to it by that pressure. So if the box is 1 foot square and it is submerged 1 foot, the bottom of the box is being pushed up by a water pressure of (12 inches * 12 inches * 0.44 psi) 62 pounds (if the box is 1 meter square and submerged 1 meter deep, the upward force is 9,800 newtons). This just happens to exactly equal the weight of the cubic foot or cubic meter of water that is displaced!

It is this upward water pressure pushing on the bottom of the boat that is causing the boat to float. Each square inch (or square centimeter) of the boat that is underwater has water pressure pushing it upward, and this combined pressure floats the boat.

Here's the link.
http://www.howstuffworks.com/question254.htm

Don Lucier, 'North Star'
C25 SR/FK
Cradled on the hard, 200ft from Lake Erie

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osmepneo
Past Commodore

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Response Posted - 01/22/2003 : 05:49:32

Um, since I am on one of those lakes that periodically flood, and are drained for other than flood control reason, I guess I should be concerned. Right, Don? <img src=icon_smile.gif border=0 align=middle>

I am visuallizing a simple experiement to determin the answer to Val's question.

1. Fill a large container to the brim with water.
2. Place in the water a "boat" with an "anchor" in it. The boat and anchor will displace an amount of water.
3. Carefully remove the boat and anchor from the water and measure the level of the remaining water. Thois would allow one to calculate the amount of water displaced by the boat with the anchor in it. (If you care to do that <img src=icon_smile_blush.gif border=0 align=middle>
4. Return the boat to the water without the anchor, and very carefully lower the anchor into the water.
5. Observe whether any additional water spills over the edge of the container, or if there is additional space below the lip.

CONCLUSION:

IF additional water spills over the edge of the container, then the answer to Val's question would be "The lake level would rise."

IF there is additional space below the lip of the container, then the answer to Val's question would be, "The lake level would drop."

Or IF, the level of the water in the container is full to the brim, the answer to Val's question is "It would have no effect on the level of the lake."

I'd be interested in whether someone does this simple little experiement.

Don Peet
c25, 1665, osmepneo, sr/wk
The Great Sacandaga Lake, NY

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osmepneo
Past Commodore

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Response Posted - 01/22/2003 : 05:53:11

Mmmmm, just another thought, "one fellow tosses the anchor". It doesn't say he tossed the anchor into the water, perhap he tossed the anchor onto the shore. In that case the lake level would rise. <img src=icon_smile_approve.gif border=0 align=middle>

Don Peet
c25, 1665, osmepneo, sr/wk
The Great Sacandaga Lake, NY

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osmepneo
Past Commodore

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Response Posted - 01/22/2003 : 16:37:10

D*&*& I can't get this thing out of my mind.

The basic laws are: (I'm digging back probably even further than Brooke to get this):

1. A floating object displaces water equal to the weight of the object - with some kind of allowances for the volume of a hull.

2. A sunk object displaces water equal to the volume of the object.

Now Val's boat anchor displaces a certain amount of H2O. When they decide to fish, and throw the anchor overboard, the boat no longer displaces as much water and rides higher in the water. At the same time the lake no longer needs to displace as much weight so the lake will go down abit.

But the anchor is thrown into the water, so the above analysis works only for the time that the anchor is off the boat and in the air. Once the anchor sinks, the lake rises again. The question is whether it rises by as much as it lowered when the anchor was taken off the boat?

I'm pretty comfortable to this point, the question seems to be does the displaced water of the anchor weigh more than the anchor? Obviously the anchor weighs more, or it wouldn't sink when tossed into the side. So, I'm drawn to the conclusion that lake level will be "slightly" lower when the anchor is in the water, than when the anchor is in the boat.

I think this what don said. It just takes me awhile to figure this stuff out.

Don Peet
c25, 1665, osmepneo, sr/wk
The Great Sacandaga Lake, NY

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Champipple
Master Marine Consultant

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Response Posted - 01/22/2003 : 17:31:14

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>
Was the lake frozen, like ours is right now ! If so, then no change - the anchor is sitting on top of the ice.. (and don't stick your tongue on it ! ) A Catalina25 ice fishing hut.. hmmmm.

<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

This answer got me thinking, when lakes freeze over there is less evaporation....If in deed it is a very hot day, then the lake level would lower because of evaporation. Now if the guys drank a lot of beer and hung it over the side to eliminate, then they may be able to keep the level even with the evaporation. However, lets say they caught a fish, if they remove that from the lake, then the lake level would go down. Dropping the anchor, per an above post wouldn't change displacement, since that weight was in the boat to begin with.....but....there would be a portion of the anchor rode that is no longer in the boat and no longer in the water, so maybe the lake level would go up....Then again, if its a hot day, like Val's post says, the heat would melt the ice in the mountains and fill the lake back up, considering there may or may not be mountains of course.

So in conclusion, how much beer really is on the boat, cuz its going to take that, a few bar napkins for illustrations, and a night at the Richland Tavern with Harv and the boys to work this one out.

dw

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Derek Crawford
Master Marine Consultant

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Response Posted - 01/22/2003 : 17:46:31

AAAAARRRRGGHH! My brain has gone into overload mode....<img src=icon_smile_blackeye.gif border=0 align=middle><img src=icon_smile_sad.gif border=0 align=middle>
Derek

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Bryan Beamer
Master Marine Consultant

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Response Posted - 01/22/2003 : 18:13:42

Duane,

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote> However, lets say they caught a fish, if they remove that from the lake, then the lake level would go down. Dropping the anchor, per an above post wouldn't change displacement, since that weight was in the boat to begin with<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

If this is true then catching a fish and putting it on the boat would not change the lake level. The weight of the fish was in the water now it's on the boat that wouldn't change displacement.

This is kind of like if you were driving down the road at 55mph and you had a bird in the car how fast would the bird have to fly to keep from hitting the rear window of the car.

Bryan Beamer
Daylight Again
C250wk #495

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Brooke Willson
Admiral

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Response Posted - 01/22/2003 : 18:20:48

We surrender, Val. Give us the answer before the net and our brains melt down.

Brooke

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dlucier
Master Marine Consultant

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Virgin Islands (United Kingdom)
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Response Posted - 01/22/2003 : 18:35:00

Think about this one,

"Can an object, without altering it's physical properties or any other molecular trickery, both float on water and sink like a rock?"

And I'm not referring to Derek's boat that he sank!

Don Lucier, 'North Star'
C25 SR/FK
Cradled on the hard, 200ft from Lake Erie

Edited by - dlucier on 01/22/2003 18:36:56

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