Challenge Questions

Unless of course the lake is frozen

Is the lake frozen under the boat, so water can lift the boat (assuming the sides do not taper in as they go up and the ice does not adhere to the boat)?

If there is a hole in the ice to fish through, and the rock falls through, isn't the problem about the same as summer?

Since ice floats on the water, the frozen exception only holds true if the boat is immobilized and the rock sits on the ice?

Would this be determined by the density of the rock and lake bottom? If this rock contained a good amount of iron ore the volume would be smaller and have a greater impact per square inch on the lake bed. Is the lake made mainly rock or muck. A rock will not release trapped air. A bed of muck will release the trapped air, gasses and organic material which will float

The level will fall as soon as the rock is released in the water. Bubbles rising to the surface ust means it can fall some more

Hi,

Volume displaced with stone inside the boat = Specific gravity of the stone / specific gravity of water. Let us say X cc. Since Specific gravity of the stone is greater than that of water, the the volume displaced droping the stone in the lake will be less tan X cc. Let us say that this volume is Y cc. Assuming that this is a large lake compared to the size of the boat, there will not be an appreciable difference in the top syrface area of the lake. And therefore the level will be decreased by (X - Y)/ Area of the lake.This could be only a few yactometers.

Job Thykkoottathil

It depends on ones frame of reference. With respect to the shore, the water level goes down. With respect to the water line of the boat the water level also goes down. With respect to a person in the boat next to you, the water level goes up. And with respect to the fish in the lake, the water level remains constant so you don't have to recalibrate your fish finder.

Assuming that the boat next to you is floating and you don't place the rock in that other boat,the level relative to the boat will remain constant. On the other hand, the surface will rise relative to the fish feeding on weeds attached to the bottom - and the fish feeding off insects that respond to the light or at the surface will also go upwards.

The only reason that you don't need to recalibrate your fish-finder is that the change is smaller than the resolution of the instrument.

The level of the lake rises. Rock has a higher specific gravity than water. When the rock is in the boat, the water displaced by the boat due to the rock is equal in weight to the 150 pound rock. When the rock is thrown in the water, it displaces only its own volume.

The level should drop when the rock is thrown overboard. The rock weighs the same whether in the boat or in the lake. Since the rock is in the boat initially it's obviously smaller than the boat, making the area of displacement while in the boat more than the area of displacement once out of the boat.

You are fishing on a beautiful lake. There is a 150 lb rock in your boat. It's in the way so you throw it overboard. What happens to the level of the lake?

Ans.:-

NOTHING WILL BE CHANGED, REMAIN SAME

PRADIP SARKAR

More amazing than the tiny drop in the lake water level, due to fewer pounds of water displaced, are the variety of answers and reasons so far! What if the rock landed in the outlet? Would that be laminar or turbulent flow? What if the rock was covering a hole in the bottom of the boat? What if the rock landed on the anchor line, pulling the bow down? What if the rock dissolves, releasing a fizz of gasses? Does this chain ever end? I guess it has to be eradicated at some point, or it will continue to grow like Kudzu (Great strip, but its artist/author is dead)!

If the rock is smooth laminar if the rock is rough turbulent

This is a trick question. But then again, aren't they all? If the rock floats the water level will be the same. If the rock sinks then the lake will rise. It would have to be an awfully small lake for the difference in height to be measurable though. It would depend on the composition of the rock.

The level of the lake remains the same. Before throwing the rock into the water, the boat displaced an amount of water equal to the mass of the boat and the rock. Placement of the rock in the water is a nett change of total displacement of 0. After the toss, the boat displaced less, but the rock displaced the same mass as it had in while the boat, thus the lake level remains the same.......ErikS

the water level has already been established by the diplacement of water from the bouyancy of the boat. when the rock is thrown over board the water level stays the same.

I guess that say's it all. GA jim.

Have to give you a GA for that well done illustration.

The figure/original author should get a GA but not me. I just borrowed it.

Thanks,

Jim

I was certain someone would hit on displacement. Displacement of water mass is, after all, what allows a boat to float at all. Ask any Navy the size of a vessel, he will tell you in "Tons of Displacement". This is a great visual comfirmation of that principal.

I think you are right... I prob'ly saw that demonstration somewhere (prob'ly Mr. Welchlin's 7^th grade earth science class), and fergot!

G.A. Game over. There is nothing like a little experiment to validate analysis or a theory.

You just provided an example of what many good engineers practice. Always cross check designs with analysis and testing. When they agree, you have a good design. As the thread demonstrates, just having smart guys agreeing with each other proves nothing.

(By the way, I am not taking a cheap shot at smart guys as jim35848 proved the answer I was going to provide was also wrong.)

The water level will rise due to an increase in volume once the rock is submerged in the water. Simple example; place a penny on top of a glass of water were as the surface tension allows the penny to stay afloat. measure meniscus. now apply a slight amount of force so that the penny is fully submerged. measure meniscus. The water level rises ever so slightly.

Did you forget to remove your finger after pushing the penny down? Or were you measuring the height of the top of the meniscus relative to the height of the flat surface of the water (that would increase initially, because it takes time for a thin film to creep back down)?

asuming the rock had a perpose on the boat either its ballast or an anchor either way not wanting to lose it completly you secure a rope to the boat and the rock which isnt long enough to reach the bottem of the lake then the level would remain the same . although i agree with your basic science there are many veriables which cannot be allowed for as they havent been given as facts

correct jim35848. I was going to write the same thing but you illustrated it so perfectly that there is no longer any question. The mass of the rock is not the primary factor, it is the volumn of the boat (minus the rock) that controls the water level. go to the head of the class.

I think that with the rock still in the boat that the water will rise as you have well illustrated and the water will rise as not to be measured . But with the rock thrown out of the boat the lake will lower as not to be measued also with the rock in the lake it's porosity will displace some of the water causing the lake to lower it's level as not to be ......................

Where do you account for the water landing in the boat from the described big splash of a 150 lb. rock?

If the lake is frozen there will be no difference.

An excellent first example good score but please where is the maths ?

The level would go down. In the boat the rock displaces 150 pounds of water about 2.4 Cubic feet. When you throw it into the water it displaces the volume of the rock. Since almost all rocks are more dense than water the volume would be less than 2.4 Cubic feet. Since less water is displaced the level of the lake must go down.

Is a martian rock heavier than an earth rock?

no, less gravity on mars, if they are identical. but no boats on mars as far as i know

And if I take a martian rock back to earth - and find an earth rock from the same material and test it in the same gravitational field?

show me the martian rock and i will give you an answer

Oh no - it's time for my hollyday after all that work!

but there are huge banks! and no swimm suits!

Exactly, the water level will drop as the weight of the rock while in the boat is greater than the volume of the rock multiplied by the specific gravity of water.

Hans.

The lake level will not change assuming this is an average rock. Since matter cannot be created nor destroyed, the level will of course remain unchanged unless the rock has such a high temperature as to create a temporary minute expansion but will settle back to it's original state / level. Where he dropped it can determine a change; if at the mouth of a small stream, it could be enough to cause an increase of sediment backup as to cause the water to possibly be diverted to another body of water causing this lake level to fall from evaporation.

Another thing is the wave caused by this rock could splash over the bank lowering the level as well. All things are to be taken into consideration here including the water splashing up into the atmosphere causing some evaporation depending on ambient air temp. The lake could also be raised from ice crystals forming on the droplets if cold enough but only temporarily. Many other factors to consider.

The dense of the rock is higher than the dense of water (otherwise he can't sink to the ground (and have an effect to the lake-level)) - if the rock stay's in the boat the displacement of the rock's weight is the volume of the same weight of water, if the rock sinks to the ground than the dsiplacement is only the volume of the rock - and this volume is smaller then the volume of the same weight in water!

The level of the lake is (maybe unmeasurably) higher when the rock is on Board!

Suppose the rock has a specific gravity of 3 and water SG =1, then as the rock sits in the boat, it will displace a volume 3 times its weight while in the boat. If the rock is thrown into the lake, it will displace a volume of water only equal to the volume of the rock. Or 1/3 less than it does as it sits in the boat. Another way to look at the problem is to think of the rock as a cubic foot of volume. In the boat it will displace 3 cubic feet of water. In the water, the rock will displace 1 cubic foot of water. Thus the lake actually will rise by 2 cubic feet of displacement dispersed over the lake's area.

Dude, think of a bowl floating in a tub of water. if you press the bowl down into the water (without swamping it), the water level rises. Lets say the bowl, when pushed almost all the way down without swamping, represents the volume of water displaced by the weight of the rock if it were inside the bowl. Lets also say that 1/3 the volume of the bowl represents the "actual" volume of the rock. If you push the bowl down only 1/3 into the water, the level in the tub rises less than if you push the bowl almost all the way down. Therefore, the level is higher when the rock is in the boat, and lower when it is in the water.

This is very easy to check experimentally. Go to the kitchen and find two mixing bowls - a large one and a medium one. Fill the large one with water. Place the medium one in the large one to act as a boat. You'll need a "rock" so look in the kitchen for something that is heavy, say a full jar of mustard. Put the "rock" in the "boat" and let the water in the "lake" run over the edge till it's stable. Now, put the "rock" on the bottom of the "lake", allowing the "boat" to float empty. Check the level of the water.

Put the two bowls in the sink for your spouse to wash up. Pour out any water that leaked into the mustard and put it back in the fridge, reminding yourself to let one of the kids try it first to see if it's still OK. If asked about the water on the floor, lie and say the dog peed.

QED

great misery - i've no dog!

Hi,

What happens to the level of the lake?

Answer: Nothing.................. In the boat the rock displaced a certain volume of water, as the boat floats lower. When over-board, it displaces the same amount of water.

bb

Think more deeply...

Hello CrAzY-MiKe,

If the "think more deeply" was aimed at me. I realize now I was wrong, after seeing the two pics of the tank of water.

I cannot think any deeper................ After a breakdown I am still not up to full running speed yet. Sometimes even the apparently simple things are to much for me at the moment. ..........................

You may have noticed I am not answering so many 'technical' questions as I used to? The reason is my breakdown. Take care...................

bb

Actually, bb, I was going more for the pun!

Hello CrAzY-MiKe,

No problems OK?

bb

Hi,

I have seen I have been voted 'off topic',..............WHY? I ANSWERED THE QUESTION!

bb

did anyone question the SIZE of the rock? if the rock was a "Big Jesus Rock" the amount of water displaced would not be proportional to say a hunk of rock with Man.....answering this question is like running in the special olympics, even if you win, you are still retarded... We will discuss this at our morning meeting and report back.....

Because the rock in the boat displaced an equal weight of water and the rock overboard will displace an equal volume of water (and the rock is more dense than water) the level of the lake will rise.

by even that reasons the level of the lake should fall

Nothing happens to the level of the lake because any change to the level was already made when the boat was launched. Meaning that weight is just weight no matter if it is in the boat or thrown overboard after the boat was launched.

Once again ... It Is all about displacement... The rock while in the boat displaces only that exact weight in water... That IS why boats float.

The "Rock" being denser than water has less volume than 150# of water thus takes up less room in the lake, than its weight displaces (in Water) while in the boat...

A serious but simple answer:

The boat floats because of the water it displaces, according to its mass and the boat with the rock within it, to me would seem to displace water according to the mass/weight of the rock within the boat and not according to its volume. If the rock is removed then the water would rise a small amount to compensate for the change / removal of the rock, but when the rock is dropped in the water the water it then displaces would be a lesser amount because of the large mass and smaller volume. I believe without any maths etc that the lake would rise when the rock was placed directly fully beneath the water

If the rock magically disappeared, then:

1. the boat rises appreciably, being less loaded and
2. the lake falls minutely, being less displaced by boat.

If the rock now reappears at the bottom of the lake, then:

1. the lake (and boat) rises even more minutely, since the rock displaces less than 150 pounds of water, but still a net drop.
2. temporarily, the lake outflow decreases, both downstream and pressure sensitive outflows into ground water, and any pressure sensitive inflows like underground springs (not surface streams) increase, until a new equilibrium is achieved. Because of continual disturbances to the lake surface and level that are more significant than this level change, this is not an infinitely decaying asymptotic curve, although an sum of impulses analysis could model its contribution as such. The decrease in lake area may also decrease evaporation outflow, but that is probably less even than the the outflow change.

The level of the lake remains the same, only the boat raises, as the water displaced by the rock is removed from the boat and transferred to the lake.

It depends on the size and density of the rock ,also absorbancy qualities and characteristics of this rock.(sandstone,granite etc)

NOTHING. actually there will be a slight disturbance to the water level as the rock touches the water but it will soon be back to its original state. probably, there will be some changes due to the volume of the rock being added into the water but this can be neutralized by the reduction of displacement of water by the boat. remember, the heavier the load, the more water being displaced by the boat... if there will be any change at all, it will be very insignificant...

Archimedes, won't he have the right answer?

It all depends. If I throw it overboard and the stone lands on the shore, the water level will go down.

If the stone's density is greater than that of the water, then it will replaces 150 lb of water while in the boat. This volume of water will be greater than the volume of the stone. So if I throw the stone overboard the stone will sink and replaces only its own volume which is less than the volume of water it replaces while in the boat. Therefore the level of the water will go down.

Following the same argument if the stone is the same density as the water, the level will stay the same.

And if the density of the stone is less than that of the water, the stone while in the boat will rise the level of the water by a volume of water of 150lb. When thrown overboard the stone will not sink and again replaces 150lb of water. Therefore the level will stay the same.

I know I'm party pooping, but can anyone tell me how many times this has been a Challenge Question? Twice to my knowledge.

Good to see we Engineers are heeding the call to recycle and reuse!!

Perhaps its time to burn it!

I'll have a go;

Simple algorithm with many missing anomalies;

If the ratio of events equals: Boat lakes equal at least Mega volume then water tension holding forces times volume dimensions equals the pressure ,

plus aeration and impurity at pressure in water equals state of ratio throughout areas a to z;

all events have ratio in events a through y where area z has zero point zero repeater x rock volume and displacement event, plus current and convection displacements, stored heat energy, chemical reactions.

In a system whose friction comprised of layer rubbing, surface friction opposition and areas where liquid motion is pressure carried over obstacles in all areas a through y except z which has a boat surface displacement of zero point zero repeater x.

Probable conclusion any mega volume lake will not move it's level in this event.

Some are still finding it an overwhelming challenge. Weight is not the same as volume. Talking/writing is not the same as listening/reading, never mind understanding.

The lake level drops minutely. When the rock is in the boat it is displacing 150 lbs of water. In the water it only displaces the volume of the rock. Assuming a density of 300 lb/cubic foot, the rock will displace 32 lbs of water. 120 lbs of water displaced in a lake will be impossible to detect.

Maybe it could be detected, but it might be a race between the time needed to average out noise of surface ripples and the tendency of the outflow/inflow to reset the level of the lake. Brownian motion would lubricate the move to full equilibrium, but the surface is continually being disturbed. Wind might be a major variable, since it can reshape the surface with a mini storm surge downwind. Then there are all the other potential objects entering and leaving the lake, like boaters and animals, possibly at a time very close to the rock entering the water. Well, that's why the waves on the ocean have such irregular patterns, and why there are rogue waves out there.

Additions to your list for continuing conditions that lower the water level: evaporation & leakage into ground water. And for water continually raising the water: feed streams.

I guess that it follows that when a fish is caught and placed in the boat that the water level will lower. (This, preserving the water level, must be the reason limits are put on how many fish one can catch.) And when the fisherman pulls his boat from the water, the water level will again fall.

That must be HUGE Macromolekules - or HUGE Hydrogen and Oxygen Atoms for a Brown Motion of a 150 Pound-Rock!

The level of water in the lake stays the same because a floating body displaces its own weight of water. As the 150 lb rock leaves the boat, the boat rises the equivalent of same volumetric amount of water that the rock displaces when it is submerged in the lake.

The rising Boat creates a vcacuum between the Boat and the Water Surface. Filling up this vacuum lets fall the level of the lake!

Keep your day jobs, all you non-small-drop theorists who defy the graphical evidence above! Mental kudzu!

Assuming that temperature of the water remaining the same and no evapouration has taken place (OR no water or other substaces were added or taken from the lake) and specific gravity of the lake water equals one, the new water level will be reduced by 150 / (Specific gravity of the rock X surface ares of the lake)

Archimedes principle says that the rock in the boat will displace a volume of water that is equal to the weight of the rock. Throw the rock overboard and it will displace a volume of water equal to its volume. Since I assume the rock is denser than water (8.33 lb/gal), it displaces more water when it is in the boat. Therefore, the lake level will go down since less water is displaced now that the rock is overboard.

The water level will lower, since the density of the rock is greater than water and the rock is no longer displacing 150 lbs. of water. I see this happen when I toss in the anchor.

Now if this had been an ugly lake and the "rock" was actually a chunk of sodium the water level would rise and dissipate into the atmosphere.

But if this had been a slime green swimming pool there would be no effect on the water level whatsoever as proven by the following two photographs...

/p]

By the way, you just scared off the fish!

The level of the lake would change based upon the difference (more or less) between the mass of the rock in the water, and the effect of the 150 lbs. of pressure exerted on the surface of the water, transmitted through the hull of the boat.

There would be a fraction of a second while the rock was in the air before hitting the water, when the force of the hull on the surface would decline (unable to calculate without the size and shape of the bottom of the boat, or the dimensions of the lake).

. . . not to mention knowing the vertical acceleration of the throw and the effective vertical moving weight of the thrower's body, the total mass of the boat and remaining contents, . . . . Of course, the body parts eventually come back down.

Wait, most or all of the bottom of the boat is under the surface and so does not exert force on the surface of the lake, only the sides.

If you define every point of contact of the lake as its surface, then the total force is the weight of all the water and all objects floating on it. I suppose air pressure on the lake's upper surface might also be in some equations.

Well, your rock seems to be in the set of rocks that sink. Once the rock hits bottom, the mass works on the earth under the lake. Are you suggesting the bottom of the lake deforms? I think it is the rock's volume that affects the level of the lake.

Well, opinions obviously vary as to the effect of rocks on boats and water.

I quite agree. The volume of the rock is what has to be considered here. Even if you were to hold a beach ball under the water that was the same volume as the rock, it would displace the same amount of water.

Also, when I think of the "surface " of the water, I think of any point where the hull of the boat touches it. If you were to freeze the lake and then remove the boat, the volume of the remaining indentation would equal the volume of water displaced.

The original question is: will the water level in the lake rise or fall. Post #49 is the best response.

But you brought up the idea of freezing the lake. Let's see, the freezing would cause the water to expand raisng the level of the top of the ice. Since the ice is also water, the top of the ice would have to be consider to be the water level. Now with the lake frozen, what will happen when the rock is all tossed overboard. If the ice is thick and the boat is stuck in the ice, there would be no net change in the water level of the lake. But if the rock were to break through the ice and sink, the water level would go down because as others have stated, and post#49 illustrates, the density of the rock is greater than that of water and therefore consumes less volume fully submerged than when on the boat displacing 120# of water (lower density & greater volume).

The sameweight of water has 9/10 the volume of ice!

The Ice-Level is (dependend from the banks) 1/9 higher then the liquid level!

If i throw the rock ovreboard to the ice surface and the rock breaks the ice and thinks down to the bottom - the boat s frozen in the icy surface - than the rock counts double (the displacement when he is in the frozen and immobile boat) and the volume when he is in the water, the level rises!

so larger the bank ( and so smaller the angle of inclination of the bank to the horizontal) so easier is the measurment of the level rising or falling! it just needs an inchmeter!

Ice is a wonder, as it floats at the top of your drink so the warmest liquid is cooled. It's almost as if God wanted cocktails to be perfect. However, when fishing on a lake, ice tea enjoys the same wonderful access to cooling, and refutes any proof that He wants us swilling alcohol in the hot sun while operating a dangerous vehicle.

The behaviors of solids are much more difficult to predict than fluids, which is why they are less the focus of interview questions and brain teasers. Some are crystals, and have properties predicted by the geometry of their lattice, like the tetrahaedron bonds of a diamond, and some are amorphous solids, like glass (never a crystal except to sell glassware). Their behavior is also very different when impure, like case hardened steel due to carbon, and impurities creating conduction in semiconductors.

The most amazing thing to me about this Challenge Question is that now I can lift a 150 lb. rock and throw it overboard! I didn't have to be born on the Planet Krypton, get bit by a radioactive spider, or spend hours in the gym. I can do this just because CR4 says so!

Seriously, let me frame the issue a different way. Let's say you weigh 200 lbs., and you are in the habit of bathing in a tub. You always fill the tub to the same level before you get in. Then, you go on a weight loss program, and because of all the time you spend exercising, you take showers instead of sitting in a tub. When you get down to 150 lbs., you decide to treat yourself to a regular bath, including a good long soak. You fill the tub to the same level you always do before you get in. Will the water level with you in the tub be the same as when you weighed 200 pounds? Will it be higher or lower?

do you really await an answer to that?

It will remain the same as before. As the weight of the rock while it was inside the boat already caused an equal amount of water to be displaced - the same amount of water will be displaced when the rock is thrown into the lake.

Bob the mariner

The weight of the rock while it was inside the boat already caused an equal weight of water to be displaced - the rock's volume of water will be displaced when the rock is thrown into the lake. Eureka, run naked through the streets!

well - done!

the level of the lake would go down.

when the rock is in the boat the water has to support its weight displacing ~2.4 cubic feet of water. when you throw it in the rock will only displace its volume.

so unless this is a very low density rock 1 gram per cubic centimeter or less (in which case it will float and still displace ~2.4 cubic feet of water) it will sink so you take its volume and subtract it from ~2.4 cubic feet what you are left with is the amount of volume change. to determine the amount of level change you would need to know the amount of surface the lake has.

You are fishing on a beautiful lake. There is a 150 lb rock in your boat. It's in the way so you throw it overboard. What happens to the level of the lake?

After any wave action has settled, the lake level would drop. The weight of the rock was previously displaced by the boyancy of the boat equivalent to about 15 gallons of water. The displaced volume of a 150 pound rock on the lake bottom would be equivalent to only 3 to 4 gallons of water.

Nothing,As the water displacement is the same inside as outside of the boat.

The water will obviously go down......

1) With the rock in the boat there needs to be an equal amount of water displaced (by weight) to equal the total weight of the boat with the rock in it. (or it won't float)

2) The boat/rock combo is "pushing down" moving an equal amount of water by weight out of the way. This makes the water level rise. THIS raised water level is the "starting point".

3) Throwing the rock overboard lightens the boat which will then displace less water so the level of the lake will drop. The 150lb rock also displaces less water (because it is no longer floating), so again the level of the lake will drop.

Must remember that in the beginning the maximum amount of water was displaced causing the highest water level.

The total density of rock plus boat has not changed, and therefore the displacement will not change, and thus, the level will not change.

Chris

Archimedes says the lake rises

archimredes did much more than his principle!

Yep he ran naked through the streets shouting eureka and then complained he was dam cold

And what do we today?

Sitting on a chair and waiting for the l(a)unch time?

And freeze if someone cuts the elctricity for the heating?

The rock will displace its volume of water on the lake. However, since volume of lake is considerably much larger than the volume of the rock, the change will not be appraised.

Aljoursan