Fishing on the Lake: Newsletter Challenge (06/02/09)

I guess the level of the lake remains unchanged. The 150 lb rock is already displacing water (raising the lake level) while inside the boat: tossing it overboard won't change the overall displacement.

The boat will be riding high - displacing less, after losing the 150 lb rock : unless swamped of course by some unexpected giant splashwave....

Aw, c'mon... who ever heard of boating with a 150 lb "rock". Is that your motherinlaw, really??

I would agree, but the actual level would go down an immeasurable amount because the splash caused by the rock will help a tiny portion of water to evaporate that would not had you not thrown the rock in.

Ah! Great point.

That is the exact accounting for the rock. Now how about the loss of weight in the boat? It woud weigh less, displace less water, thus raising the water level. Will the decrease due to the rock submerging offset the increase due to the boat loss of weight? But I am really more interested in what instrument you will use to measure the change.

ghdhdh

interferometry? How large s the lake?

The weight of the boat remains unchanged, so the amount of water displaced by the boat (due to the boat's weight) will remain unchanged. The net change in water level will only result from the change in the rock's disposition.

Good answer Getting right to the point.

The rock will displace 150 lbs of fluid weather in a boat or on the bottom of the body of fluid, or floating on the fluid if for some reason the density of the rock is lower then the density of the fluid. If the rock is in the boat, then the boat will ride lower in the lake and the level of the lake will be, say 10 feet. if the rock is in the lake, the boat will ride higher in the water and not displace the 150 lbs of fluid, but the rock will be in the lake displacing the 150 lbs of fluid so the displacement of the rock and the boat will be the same and the level of the lake will remain at 10 feet.

unless the fisherman throws the rock on the shore, then the level of the lake will drop.

Hello,

You have 150 lbs of Lead and 150 lbs of Aluminium. If immerced in water, do you think that both will displace equal amounts of water?

delicious

It seems to me that lessening the displacement of the boat will actually lower the level of the lake. The rock entering the water will increase the level of the lake. The question is one of the relationship between the two displacement factors.

What if the bottom of the lake is muck and the rock sinks into it thus displacing no water. Then the lake would be smaller

You never took into account the weight of the air in the boat. The rock in the boat displaces air so when the rock is removed there must be a corresponding amount of air added to the boat as a vacuum could not exist. You also have to take into account the altitude of the lake as the higher the lake the less the gravitational pull on the air mass

isn't the density of any material definid in the air?

or is the density defined (or measured) in the vacuum?

every materials density is measured (or alculated) as result of the measurement in the air.

The weight of the material is a result of the medium it is in, not the density. Density is mass per unit volume, not force per unit volume.

yes !!

What if the rock consists of a soluble substance, such as NaCl? The SG of solid NaCl is 2.165, so a 150 lb chunk of it would displace about 1.11 cubic feet (CF) of water. The SG of a saturated NaCl solution (26.4% w/w) is 1.204, which could be accomplished with 6.7 CF (418.2 lbs) of water mixed with 150 lbs of NaCl. The total weight of the solution would be 568.2 lbs and its volume would be 7.56 CF. This represents a 0.86 CF increase from the original 6.7 CF of water, which is less that the 1.11 CF displacement if the rock were to remain in solid form.

heh heh good hairsplitting there. Sadly nothing has been said about the meteorological conditions: temperature, humidity, that play into this.

You have the right perspective for Florida. But maybe not for Newfoundland.

If we're going to raise atmospheric issues, we also have to consider the butterfly effect. Is the act of heaving a boulder out of your boat sufficient to disrupt the weather, causing it to rain? thus, compensating for any evaporation and instead, further raising the level of the lake..

What we are ice fishing from the boat? Why not?

Are we fishing for Kingklip and who knowes about Kingklip

How big is the Rock? Is the rock as big as the boat? Granite is heavier than let's say scoria. If it still submerges completely. How much water does the boat displace with the rock. 150 lb worth of displacement. Need to know the size of the rock to complete the calculation. The rock could be twice the size of the boat and still submerge, it's the unknown that is scary.

Actually, since the rock is almost certainly more dense than water (since it probably sinks), it displaces the equivalent volume of water that matches its weight. While in the boat, it is this volume of water that is displaced. When the rock is thrown overboard (presumably into the water, which presumably is deep enough that the entire rock is submerged), the rock now displaces only its own volume which is less that what was displaced while in the boat. Hence, less volume is displaced and the level of the lake is infinitesimally lower.

if the lake isn't deep enough to put the whole rock under the surface of the water, the displacement of water is always less than the displacement of water while the rock is in the boat!

In this case the sea level falls a part!

And what if the boat sits on the ground with and without the rock on Board? Then the sea level can only rise!

But that is not a Boat - a Boat is swimming, not staying like a tree in the water!

And what you are fishing - flatscreens?

OK IF YOU'RE GOING TO GO THAT FAR WHY DON'T WE QUESTION THE AMOUNT OF WATER LOSS FROM THE RIPPLE AS A RESULT OF THROWING THE ROCK INTO THE LAKE CAUSING WATER TO LAP ONTO THE SORE AND EVAPORATING. OR THE AMOUNT OF WATER EVAPORATING AS WATER SPLASHED FROM THE LAKE WHEN THE ROCK IMPACTS THE WATER IS FRAGMENTED IN THE AIR. OR WE COULD QUESTION THE EFFECT OF OBSORBTION DEPENDING ON THE ROCKS POROSITY THE AMOUNT OF CHANGE INFANTESIMALY SMALLER YET. D.C.

The volume of the rock is what it displaces, not what it displaces when covered in shrink wrap. Absorbion is included, but if absorbtion is slow, this means that the soaking time of your porous rock begins to compete with the outflow in terms of the size of the transient effect of the rock bein repositioned (as opposed to the final effect of the rock being repositioned, where the outflow increase has time to reduce the lake level change into the measurement noise). The rock has an initial volume that is gradually decreasing. The rate of decrease probably varies with water pressure, temperature, variations and vibrations in water pressure, as the gradually narrowing pores decide how much water to admit and how much air to retain.

Finally!

Archimedes would have been proud. :)

But what happens if the rock is made from pumice?

And just for laughs - the apparently confusing question from, 'The world's most smarterest model,' or whatever it was, which rock would be heavier, the granite one or the pumice one?

heavier to handle?

which rock would be more voluminous by the same weight?

IF THE GRANITE WEIGHTS IN @ 150#, AND THE PUMICE WEIGHTS 150# ???????

exactly

Water level will drop because, while the rock is admittedly still in the lake, it is now only displacing 150# of 'rock' a smaller mass than 150# of water...

and while it is in the air?

If the rock is porous and admits water the level of the lake would rise less than than if it was solid . If the rock is soluble in water the level would become higher . If the rock is ice , it would at first float and the lake level would rise . If the ice rock melts the level of the lake would drop. CTC

The level would drop by an almost undetectable amount, the rock would no longer be weighting down the hull of the boat pushing against the water. there by displaysing less water area.

What I would like too know is did the person throwing the rock get wet or not because it's more fun to see how big a splash you get? The less wieght in the boat would offset the amount if displacement cause by the rock being dropped into the water.

The level of the lake will go down. In the boat, the mass of the rock (denser than water) will displace its mass in water. thrown into the lake, it will sink, displacing only its volume of water.

It will change by the same amount as when you pee over the transom.
Del

Beware of the dog - or the little girls!

If the rock floats!??! the water level would stay the same unless you chained it to the bottom of the lake. (Not accounting for the extra displacement of chain, lock and _{mother in-law} .) If the rock has neutral buoyancy the lake would be unaffected. If it sinks the water level would go down. If the lake was frozen it wouldn't matter until spring thaw. But what if the boat was sitting on the lake bottom and you threw it out to "float your boat"! Would you still have missed that large mouth bass hiding in the weeds?

GA Jim, Concrete is 150# per cu. ft. We may safely assume the rock has a similar density, if not more (granite 168#).

... non est disputandum.

Eureka!

GA Mr. 8, eh, 48, 35848, oh well, Jim.

The rock is presumably non-porous, to avoid disputes about the distinction between dry and wet densities. And I think the problem implies that it is dense enough to sink in water. But why does the lake have to be beautiful -- I wouldn't feel inclined to heave a rock into any remnant of natural beauty which has survived the attentions of biped monsters.

Practical considerations** might require some ingenious contraption on board (possibly with swivelling boom and sliding counterweight) which can safely lift the rock and lower it gently into the water. A certain amount of non-figurative rocking-the-boat is nevertheless inevitable following the moment of release, if the rock is fairly dense.

But instead of all these complications, it would appear that the problem remains essentially unchanged if the rock were initially outside the boat. It could be clamped or hung, say beneath the CG (or should that be centre of buoyancy? or some other specimen of hydrostatic jargon?) so when it is released the "boat gets lighter" and displaces less water, leading to a fall in lake level, however imperceptible it may be.

The challenge seems almost trivial (at least so far), and satisfactory GAs have been posted. One could digress by considering a remotely related problem. A cylindrical block of wood weighing say 1 kg can be made to float in 100 ml of water (or even much less). How? Just lower it into a suitably dimensioned beaker containing the water, and lo and behold -- the wood will "displace" 1 litre of water which wasn't there to start with!

Hoping to see some wider discussion on floating bodies in this thread (while I get out of the way).

=TeeSquare=

** I've never attempted boating, or fishing for that matter (except perhaps for trouble, when I venture into alien domains such as CR4)

Quantum mechanics says that neither stone nor boat actually touches the water... I think I just confused myself...

Not a nano-scopic - just a makroskopic sight!

The level of the lake is dirupted! It's an impact the breaks the water tension, so in gaining forces to push up against the rock and to go down because of the rock it begins oscillating due to the grip of the water tension. There are pressure fluctuations, displacements, wave motion; all till finally the rock has begun descending due to density and the sucking water draft following the rock causes a central pressure that holds back on the outward transmitted wave, to slew these forces spirals usually break around the central tear. When finally the level compensates there is no difference in level due to displacement because the water tension just adds a little more pressure to the water that equals the possible displacement that would have happened in an inferior water vessel.

I want to know the area of the lake the depth of the lake it's underwater topography the hull W/L of the boat and why specifically a 150 lb rock so I can calculate how much that the water level rises and lowers it should keep us happy for hours if not days

Hello John,

I thing you are being 'greedy' now and, 'pushing your luck' for sure LOL :-)

Take care.........bb

Hi Baby Bear

All things are measureable but is it that important in this case there is not much evidence to go on i should imagine that the displacement of boat and rock is at the surface to begin with causing a rise of the surface of the lake and when the rock is submerged then providing the rock does sink into the mud at the bottom of the lake should cause a fall in the level of the The short of it is without going into any calculation as to the level of the lake is that i think rock in boat rise in lake rock out of boat lowering of the lake without any quantum. Cheers LOL John M

This is being made much more complicated than it needs to be. A constant volume of water is needed to displace a constant force, regardless if the force is floating or not. The only thing the density is important for is whether or not the rock floats.

Model it as a cone 1 kilometer in diameter and 150 meters deep, and go for it.

It all depends if it is Lake Logness. If so you have a very good chance that the water level will decrease when you throw the rock over board.

With the rock landing on the Logness monsters head,

he will be angry,

chasing after you.

You will try to get away,

thus with you and the monster out of the water…the water level will decrease again!

Lochness

Only another slight diversion. Should that be Loch Ness? Please do raise the ire of those wild people descended from the Celts, else they may throw the rock at us instead of over the side at Nessie

Probably, I was in a hurry. My apologies to any wild Celts whom I may have offended.

Nessie isn't a stone eater - isn't?

If it lands on the floating quai, the level remains unchanged. For more reasonable assumptions, its_me and jim##### have it right.

I don't have much time to answer this so I will be brief.

Whether the rock or even the boat floats is related to density but is not directly determined by density. Floating is the result of bouyancy. If the volume of water displaced by the boat weighs more than the boat (with all its contents) then the boat will float. If the volume of water displaced by the rock is lighter than the weight of the rock then the rock will sink.

Without doing any calculations, I would think the level of the water would remain unchanged because that water was already displaced while the rock was in the boat. In other words, the boat will displace less water without the rock but that rock would also displace its own water.

Not true cin-gold,

this is only true if the "rock" has a density that is less than or equal to the density of the water.

Thank of it this way:

If there are 2 identical boats on 2 lakes, each with different "rocks" of the same identical dimensions but made of different materials.

The first "rock" is made of aluminum and the second made of depleted uranium.

The boat with the heavier rock will draft deeper than the other boat lighter rock on board. Once the rocks are thrown overboard the boats will draft the same and the same amount of water will be displaced by the identically dimensioned rocks. The difference will be the water level will have dropped more for the boat that had the heavier rock.

Let me know what you think.

-A-

Dear Guest,

The original question stated the weight but not the volume of the rock, so there is no way of knowing what the buoyancy of the rock was. Therefor in my example I used two rocks of identical dimension but of different densities in order to illustrate that an object (e.g. rock) with a density greater than water will cause the water level to drop. Provided, of course, the rock is completely submerged.

I used two materials that I thought would be familiar to the engineers who read and comment on this site. Like it says at the top of the page: "The Engineer's Place for News and Discussion." Frankly I expected a more "African or European" type of question from the blokes on this forum. Something like, "5052 H32 or 6061 T6?" Knowing full well that the person asking the question would know it didn't make a calculable order of magnitude worth of difference which grade of aluminum when compared to the density of depleted uranium. Obviously that person would just be trying to sound clever and get a rise out of me.

But your question, I did not expect. And it serves to remind me that this forum is not exclusive to pointy headed, over educated, Star Trek fans. But that, anyone with the curiosity and interest in the subject of engineering can join in an even influence the conversation. Thank you, dear Guest.

As to your question, aluminum is heavier. Once uranium is completely depleted it is more like swish cheese and can even be a little squishy to the touch. Technically, aluminum has 832 angstroms per cubic mole to depleted uranium's 263.

I hope this helps.

-A-

Maybe you are the one missing the point of the question.

I think the real question here is why anyone would put a 150# rock in their boat. And how would you easily throw the rock overboard without rocking the boat?

One last question is what is the density of the water?

1. You're right.

2. You got me there too.

3. Out the back.

4. 0.998 g/cm³ (liquid at 20 °C, 1 atm)

-A-

Suppose the boat is 12 foot long and 4 feet wide and 2 foot deep. Because the boats prow is pointed then we will convert the boat to a shape of a brick. So the boats volume is 40 cubic feet. Weightof boat 150 lbs plus the 150 lb rock =300lb

Buoyancy Force

(Density)(Volume)

Weight of boat plus its load =300lb

Volume of boat 40 cubic feet

W=300lb Volume 40 Cu ft then W300 x V40 = 2496 lbs displacement and the boats submerged level is 1 Foot then the Cubic displacement = 20 Cu ft. The rock over the side displaces no volume as it is not pressing on the surface so the water level will not be altered

(What if the "rock" was actually my mother in law)

sometimes you need a rock in small boats to act like an anchor.

you know what you're talking about?

I suppose the swiss chees bit is, " . . . until melted", since they use this product's extreme characteristics to put points on and rods in artillery shells and bullets for armor peircing.

If the rock had the density of water, nothing changes but the waves made, and if more, lake less down, else lake less up.

Porosity does not change the rock's volume, it increases the surface area. If the pores are small enough, it may take time and pressure to get water to those surfaces, and if they are sealed inside, they are not part of the surface area (unless you are sealed inside there, I guess), they just deplete the density.

Regardless of the size of the lake, the outlets, surface tension, heat of the sun, wind, etc., when you suddenly put things in and take things out, the mean height changes at least for the moment. Lake water is made of tiny molecules always in motion, so it does not get stuck or move only an inch at a time. Figuring out how to measure the level precisely and in enough places is a problem, maybe something you could do with a laser on a stable foundation on the bedrock of a nearby mountain, by measuring several points repetitively and averaging. It needs to be a quick measurement, as the effect decays, but the fact that is is a step phenomenon means that in one fraction of a second, suddenly things are different. Evaporation and outflow and inflow will respond much more slowly, evaporation will be relatively constant and vary more on the wind, but again relatively slowly, and all these have no special dislike of step changes. Surface tension will help transmit the difference as a wave, but has no special dislike of step changes.

It's the myriad of different disturbances in the ocean that make the surface such a crazy quile of waves and angles, and occasionally result in "rogue waves" of great height and danger, when all the little ups go up together some lucky where, some lucky time (and nearby, go cown all together, too). Happy sailing!

The buoyancy force on the boat is equal to the weight of water displaced.For a floating boat this buoyancy force equals the weight of the boat and its load, the force up up (buoyancy) equals the force directed downwards the force directed downwards (weight).

Consider a small ishing boat 12 feet long 4 feet wide and 2 foot deep, we would like to represent it similar to a brick in shape since the boat is rather pointed let us assume the average length is 10 feet long If we squared it up. To be safe let us assume that that the waves on the water will not swamp us and sink the boat . Such a boat displaces a volume of water 10 feet long 4 feet wide and 1 foot deep.

The volume of water displaced by our loaded boat is 40 Cu feet

Since the weight of the boat and its load

= Buoyancy Force

(Density) (Volume)

So 0ur small boat Say its weight is weight is 150 lbs plust the 150 lb rock

= 300 lbs so = 300 pounds x 40 cubic feet =1200 pounds of displaced water not accounting for the weight of the occupant of the boat I have not worked out how many cubic feet the boat displaces yet

Go metric!

Use a metric boat 4 meters by 2 meters submerged .2 M, displacing 1.6 cubic meters of water at 1000 kilograms per cubic meter, so it is displacing 1600 kilograms, and a 50 kilogram rock (~110 lb.) that is .2 M cube shaped (8"x8"x8"). When the rock is removed, the boat rises .2 x 50 / 1600 and the lake falls very little. Assume the lake is i kilometer square with vertical banks. When the rock is removed from the boat, the lake falls ( .2 x 50 / 1600 ) * 8 / 1,000,000 ) = 0.00000005 M. If the rock sinks, the lake rises .2 x .2 x .2 / 1,000,000 = 0.000000008 M. Correct me if I botched this!

The Boat rises 6.25mm relatively to the level of the lake, the lake lavel falls 50nm to the ground by removing the rock - so the Boat rises 6.25mm-50nm to the ground.

sinking rock lets the lake level 80pm risen; the Boat rises 6.25mm -50nm+80pm to the ground and the lake level rises now 80pm, subtract the 50nm falling by removing the rock, the lake level falls 49.92nm over all!

any question?

not 80 pm - just 8nm and so 42nm

if what u say is true then the rock which is floating in the first instance ie(in the boat ) would it not have to be floating (only partialy submerged ) in the second instance for what your saying to be correct ?

It undulates a lot to start with, then settles at a lower level. When the rock was in the boat it was displacing its own weight in water. Unless it was a pumice rock, it sank and displaced less than its own weight.

Nothing. The amount of water displaced is the same whether the rock resides in the boat or rests on the bottom of the lake.

THE WATER LEVEL WILL BE LESS.

THINK OF IT THIS WAY.

YOU HAVE A HUGE BOAT. 1200sqft BOTTOM SURFACE AREA AND THERE IS A 500,000 POUND 1'" DIAMETER ROCK IN THE WATER. IF YOU PUT THAT ROCK IN THE BOAT. IT WOULD PUSH DOWN ON THE BOAT MUCH MORE, IN TURN DISPLACING MORE WATER, THEN THE 1" DIAMETER ROCK WOULD EVER DISPLACE.

I THINK THERE ISNT ENOUGH INFORMATION HERE. THEY DO NOT TELL YOU THE DIAMETER OF THE ROCK.

The dimensions:

The Boat should be greater than the rock!

And the Lake should be greater than the Boat!

And the thickness of all these things?

THE WATER LEVEL WILL BE LESS.

THINK OF IT THIS WAY.

YOU HAVE A HUGE BOAT. 1200sqft BOTTOM SURFACE AREA AND THERE IS A 500,000 POUND 1'" DIAMETER ROCK IN THE WATER. IF YOU PUT THAT ROCK IN THE BOAT. IT WOULD PUSH DOWN ON THE BOAT MUCH MORE, IN TURN DISPLACING MORE WATER, THEN THE 1" DIAMETER ROCK WOULD EVER DISPLACE.

I THINK THERE ISNT ENOUGH INFORMATION HERE. THEY DO NOT TELL YOU THE DIAMETER OF THE ROCK.

i forgot to login. haha

There will be no change.

The level of the lake is already determined by other factors: a spill way for example. Neither you, the boat, the rock, a case of beer or your Ford Explorer will change that.

L.J.

And there is no drain of that lake?

If the lake is deeper than the settled rock the lake level would remain unchanged, however if the lake is shallower than the settled rock the lake level would be lower. The fact that the lake is beautiful has no logical effect on the outcome.

"however if the lake is shallower than the settled rock"

By this do you imply that the rock sank into the soil on the bottom of the lake?

I love this question as it was one I floundered on at my university application interview. Naturally I now agree that the level falls as displacement when in the boat is of water equal to weight of rock but only volume of rock once on the lake bed assuming SG (rock) > 1.0

I am intrigued by the block of wood question. Assuming SG =0.667 then the block of wood will be 1.5 ltr or 12.4cm (5in) diameter and height. If the container is 12.45cm diameter would the block of wood float in effectively neglible water. Do you only need sufficient water to ensure the block doesnt touch the container. I am not sure myself - in practice unless the beaker is perfectly straight and the block perfectly balanced I can see problems of tipping.

Looking at it this another way suppose a once we have thrown the rock out our boat is floating on the lake (not a huge conceptual leap). Suppose you then brought a boat shaped container up under it. Whilst the container was say 10 x the size of the boat I have no problem that the boat continues to float but as you shrink the containerwould there come a point if there was insufficient water to displace that the boat would ettle on the bottom of the container?

I think so but then I am not really sure of the floating mechanism

A 5"x5" cylinder of uniform SG 0.667 should float upright, so (other than the effects of surface tension which increase as the edges get closer) you should only need enough water to come 2/3 of the way up the sides of the block, plus a tiny bit extra to fill underneath it

Ah but what would happen if it was a salt lake? Wouldn't the natural bouancy be different thus the boat would ride higher in the water?

How bout that huh?

Heck I don't know but it's intriguing to see all the theoretical values present in the previous comments.

Unless the density of the rock was intermediate between pure and salt water, the direction of the effect would be unchanged.

Or did you mean that the rock was rock salt? In which case the level would drop initially when the salt sinks, and then drop some more as the salt dissolves (at least at low concentrations, salt dissolved in water is slightly more compact than the two separately).

And if i lift the rock to throw him overboard - there is an additional force that makes the Boat sink äquivalent to an additional weight - if that lake has an drain to external there is not the same volume of water after i have thrown the rock.

How about this?

Lake Level drops subtly

Because the wave caused by the rock being thrown overboard will ultimately wash some distance up and beyond the original shoreline (how far dependant upon size of the lake) and thus leaving some amount of water on the shoreline, therfore no longer in the lake, thus lake level is some minute amount lower.

Doesn't Density matter for anything anymore? Think 150lbs of pumice in a lake large enough to 'float' it.

Take a 1" cube with a specific gravity greater than water and place it into a 3" 'boat' where it floats on top of the water in a bowl. The weight of the boat and the cube pushes against the water as gavity pulls them towards earth. TAKING SURFACE TENSION into account, the water rises against the walls of the bowl...

Take out the cube and dump it into the water. Splashes aside, evaporation aside (these are minimal), the cube sinks to the bottom of the bowl and causes VOLUMETRIC displacement of water, so the water rises against the confines of the walls of the bowl.

Throw a 150lb pumice rock into a lake out of a boat instead, AND ASSUMING THERE IS ENOUGH WATER TO FLOAT THE ROCK, there is still going to be more volumetric displacement than with the rock in the boat.

No matter what you do, the water level will rise!

The cube in the floating boat also displaced water. In fact it displaced its weight of water. When resting on the bottom, it displaced its volume. As the SG is greater than that of water, the first volume will exceed the second (some of the weight of the block is now supported by the botom of the bowl/lake). Try it for yourself, perhaps using a cream carton and a marble in a small bowl, and then report what you see...

Assuming the rock density is less than or equal to water nothing happens to the lake level.

If the rock is more dense then the lake level falls and either way I scream Eureka!