Cast It Into The Waters! -- A Classic
A man stuck in a small sailboat on a perfectly calm lake throws a stone overboard. It sinks to the bottom of the lake.
When the water again settles to a perfect calm, is the water level in the lake higher, lower, or in the same place compared to where it was before the stone was cast in?
Hint: You can use limits to solve this problem!
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7 solutions
This is a classic physics problem I have given in class many times. I understand the limits approach but all that is necessary is the definition of the bouyant force. The key concept is that to float, an object must displace a volume of water equal in mass to its own mass. E.g. a 1 kg stone, when placed on the boat, will cause the boat to displace an extra 1 kg of water. Since the water is less dense than the stone, the volume of this displaced water is greater than the stone's volume, i.e it occupies more space than the stone. When the stone is thrown overboard, it displaces only it's own volume of water. Since this volume is smaller than the volume it previously displaced while on the boat, the level of the lake goes down. Of course, if the rock is small, the level change is small, but it is never zero.
So you mean the stone pulling the entire BOAT down would cause a greater rise in water level, in the beginning?
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Exactly so. When the stone is in the boat, the extra amount of water displaced by the boat pushing down is equal in mass to the mass of the stone. When the stone is thrown into the water and sinks, the amount of water displaced is equal in volume to the volume of the stone.
How large and dense the rock is will determine the magnitude of the effect, but so long as the rock is denser than water, the shoreline will lower some amount when the rock is tossed out.
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"imagine it as a tiny pebble that weighs thousands of tons", the only logical thing about this is that you must have infinite fantasy to solve the problem this way...
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@Giuseppe Bresin – If you'd prefer to think about it abstractly, consider the buoyancy and conservation formulae used to model this scenario. Our goal is to discern the relative volumes of the amount of water displaced:
when the stone is in the boat The amount of water displaced ( = , ≤ , ≥ ? ) when the stone is in the water The amount of water displaced
However, in truth, we don't need to fully set up these equations, we need only find, from among all of the variables in the scenario, those unknowns which, when varied, affect the two sides of this equation differently (including those which affect only one side). On the LHS, I found that the weight of the stone was this property, and similarly, the volume of the stone was the important property on the RHS. Then I used the intuitive limit that I described above to understand the roles that these two properties must play in the equation, finding that as the density of the rock decreases towards 0, the shoreline rises, and as the density of the rock increase towards ∞ , the shoreline lowers.
In truth, at this point I looked at the teacher who first proposed this problem to me and asked, "Is the rock denser or less dense than water?" and, in return, he made a funny face at me. :) But, in my version of the problem, I specified that the rock sinks. Therefore, you know that that the density of the rock must be at least somewhat greater than the density of water.
@Zandra Vinegar Okay....I get it now.:)
same approach here is an upvote.
I understand this only if the stone plays a great role in the weight of the boat, but if the weight of the stone is minuscule it would depend on 1) the weight of the boat and 2) the importance of the water displacement. What if the stone took up a lot of water volume but weighed little? My answer was that it depends on the weight of the stone. This particular answer only touches on if the stone reaches a limit (or namely a ton). What if the stone was very small and had little effects on the boat or the displacement of water? Without knowing about the stone at all, could we assume that there's not enough information in the question?
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i agree and the size of the stone is miniscule
The direction that the shoreline moves doesn't depend on the rock's size or weight, but, rather only on its density. As long as the rock is more dense than water (this is a fact given in the problem by the part specifying that the rock sinks), the shoreline will lower some amount when the rock is tossed in.
i totally agree... this question is pretty much sensless... its like: heres 4 cups and if u choose right one u will get a candie...
Wow! Loved the approach!
The question elicits an answer that raises an issue of accuracy of measuring instruments. Every question has a context and the context here as I see it is the perception of someone who is in the boat that has vantage point of determining whether the water level is changed. And in this context, NO. The eyes are not trained to have a perception based on some calculations but on observable phenomenon. If the author would like to illustrate the concept of infinitissimal values or limiting conditions, he should have made a careful selection of the things that constitute the circumstance. It appears to me that the author is trying to make it appear that the problem has a practical scenario but I find it rather detached from practicality.
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"She," the author, doesn't have to be the observer; someone else could, or, Zandra could be on shore with a grad student in the boat tossing the stone overboard. Pick good nits.
This problem applies to any size rock and any density of rock that sinks in water. Maybe my answers have made you think that the effect would only be seen in extreme circumstances, but that's not the case. Limits aren't necessary to solve this problem at all, but that was my choice of approach.
When I wrote the problem, the idea was to illustrate that the size of the rock doesn't impact the direction of this effect seen. Instead, the density of the rock and how similar it is to water is the variable that is capable of affecting the direction of the shoreline's motion. The use of considering an extreme case to identify this relationship is, I think, a very powerful technique. But the answer holds even in moderate/practical cases of the scenario.
It depends on the size of the rock. If the rock is of infinite size you wouldn't be able to determine the water level where as if you had a "normal" sized rock you would be able to.
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NO IT WONT DEPEND ON SIZE OF ROCK WHATEVER THE SIZE OF ROCK BE......IF IT IS DENSER THAN WATER THE ANSWER WOULD BE THE SAME ie THE WATER LEVEL WILL LOWER.
Well just as we can have a dense stone we can have a very light stone and then it doesn't pull the boat down much and when it sinks(it is just a little denser than water) it will take more space in the water. Am I wrong? If so, why?
I disagree. You stated : "consider the limiting case where the stone is much, much denser than water." But this was not stated in the question.
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It is clearly mentioned in the problem statement that the stone sinks to the bottom....
That means the stone is denser than water .................
As long as this holds, the answer wont change..ie
the water level will be lowered as long as the stone is denser than water even by slightest of the difference.
The stone was asked to consider much much denser than water only for imagination purpose..
Solving this with Archimedes' principle:
Boat without rock: F B = V B ∗ ρ w ∗ g
Boat with rock: F B + F R = ( V B + Δ V B ) ∗ ρ w ∗ g
Subtracting these equations: − F R = − Δ V B ∗ ρ w ∗ g
Newton's second law: F R = m R ∗ g
By substution: m R ∗ g = Δ V B ∗ ρ w ∗ g
By definition: m R = V R ∗ ρ R
By substitution: V R ∗ ρ R = Δ V B ∗ ρ w
Rewriting: Δ V B V R = ρ R ρ w
As stated the rock sinks thus its density must be greater than the water density: ρ R ρ w < 1
Therefore: V R < Δ V B
This means the volume displaced by the boat only due to the extra weight of the rock is greater than the volume of the rock and thus the water level will drop when the rock is thrown overboard.
Excelent explanation! Thanks
Wow nice approach
Does your solution account for the mitigating force of water tension? Because the displacement equal to the stone's weight is spread across the entire hull of the boat, the stone will displace less water relative to its weight. once thrown in the water, water tension is eliminated and displacement equal to its weight will occur.
Still, it is more a physics problem than a math problem (especially, a math problem on limits).
@Connor Bryan a good explanation.
as the stone is denser than water, it sinks and the base of the river provides some normal force to the rock rather than only buoyancy force ... while in boat it was completely supported by buoyancy force itself , hence the water is more displaced when rock is in boat.
The rock is in the water, the water is displaced due to its mass. As long as its mass does not change the displacement of the water does not change. The normal force on the rock is only keeping it form sinking even more.
Water pressure at depth is greater but the mass of rock stays the same The deeper the water the more the water around the rock is compressed
Water Evaporates Over Time So The Level Is Lower.
But that, besides being irrelevant and much untrue (vapor pressure / humidity), completely ignores the problem of the rock.
This isnt a trivia question.
This is negligible; see the above answer
A solution is using archimedian principal.
as we throw stone in lake weight of boat decreases.So as we know by archimedian principal that displacement of water due to external object is relative to weight of that object so as weight decreases so is relative displacement
When the rock is inside the floating boat, Archimedes' Principle dictates the volume of water displaced is equal to the mass of the stone. When the stone is thrown into the water the volume of water displaced is just the volume of the rock. As the specific gravity of the stone is obviously > 1 (it sinks) the volume of water in the lake displaced is less when the rock is thrown out, hence the lake level drops.
Before throwing the stone into the lake, volume of water displaced by the stone will be (Weight of water displaced(say Ww)/Specific weight of water(say Sw)=weight of the stone(say Ws) /specific weight of water(Sw)=Ws/Sw) since Archimedes' Principle states that Upthrust=Weight of water displaced=Ww which balances the weight of the stone(Ws) giving that Ww=Ws.
When the stone sinks and water settles to a perfect calm, Volume of water displaced by the stone=Weight of the stone (Ws)/Specific weight of the stone(say Ss)=Ws/Ss
As Ss>Sw (Stone is denser than water), (Ws/Ss)<(Ws/Sw) i.e. the volume of water displaced in the 2nd case is less than that in the 1st case and hence the water level becomes lower as compared to 1st case.
When the stone is on the boat, it exert pression in the water with his weight force rising the water level, in the bottom of lake the stone's weight force is canceld by normal force.
This ignores the volume of water possibly placed up by the volume of the rock.
A Solution Using Limits
The rock sinks, therefore it is denser than water, but few people have good intuition for using that knowledge to guess at what will happen when the stone is cast out of the boat.
So, instead of dealing with small forces and slight changes that we don't have strong intuition for, consider the limiting case where the stone is much, much denser than water: imagine it as a tiny pebble that weighs thousands of tons! When it's in the boat, the boat would be pulled deeply into the water, pushing the shoreline up as the water is displaced by the boat. (An amount of water equal to the stone's weight would be displaced.)
When the stone is thrown overboard, the boat rises in the water because the weight of the stone is no longer pulling it down. Meanwhile, the stone sinks to the bottom of the lake and displaces only a small volume of water.
In summary, the stone displaces far more water when it's in the boat, pulling it down, compared to the amount of water it displaces when resting on the bottom of the lake. Therefore, when the stone is thrown out of the boat, the shoreline lowers.
Relaxing the limit: this logic holds so long as the stone is denser than water. We know it is denser than water because it sinks.