Fluid/Buoyancy problem (three immiscible fluids and a ball that floats between them)

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  • #1
bw519
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Homework Statement:
There are 3 fluids, A, B and C, (with densities ρA, ρB and ρC) that when put together in a beaker, do not mix. If only A and B are put into a beaker, a ball floats at the interface between the layers with 1/3 of its volume in the lower layer and the other 2/3 completely submerged by the upper layer. When only B and C are in a beaker, the same ball will float at the top layer with 1/3 of its volume submerged in that top layer and the other 2/3 exposed in the air.
a) Rank the densities of the fluids
b) what is the density of the ball in terms of the densities of fluids A and B
Relevant Equations:
Fb=ρgV
a) I think in the beaker with A and B, A must be the top fluid and B must be the bottom. The ball sinks through the top layer but not through the bottom. In the second beaker (B and C), it does not sink through either layer. Therefore, both B and C are more dense than the ball. So B must be the bottom layer of the top beaker. However, in the second beaker, I am having more trouble. My first thought was that B was the top and C is the bottom. But would that make sense that the ball displaces the exact same amount of fluid B if in one beaker, the ball is also submerged in fluid A, while in the second it is exposed to the air? Also, using my answer to question B, I have an issue with either C or B being the top layer.
b) weight of the ball should equal the buoyant force (total buoyant force from both liquids)
Wball=FbA + FbB
weight of ball equals ρball*Vball*g
thefore, ρball*Vball*g=ρA*(2/3)Vball*g+ρB*(1/3)Vball*g canceling g and Vball gives

ρball=(2/3)ρA+1/3ρB

If that equation if correct, then I don't know how either B or C being the top layer of the second beaker makes sense. If B is the top layer for that beaker, wouldn't solving for the density of the ball just give 1/3ρB, since it is only submerged 1/3 of its volume in that layer? So the density of the ball in beaker 2 wouldn't match the density of the ball from beaker 1, even though it is the same ball. However, if I make fluid C the top layer, I get that the density of the ball is 1/3ρC. But if that is the case and I set the two equations for the density of the ball equal to each other and solve for the density of fluid C, it comes out to 2ρAB. But if that is the case, fluid C is more dense than fluid B and shouldn't be the top layer.
 
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Answers and Replies

  • #2
kuruman
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Remember that the buoyant force (BF) has (a) magnitude equal to the weight of the displaced fluid and (b) direction towards the interface. Draw free body diagrams for each situation each of which has three forces (1) BF due to the top fluid; (2) BF due to the bottom fluid; (3) mg due to gravity. Balance them.
 
  • #3
bw519
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So are you saying that the buoyant force liquid A in the first beaker is pointing downward? in which case the density of the ball for the first beaker would come out to ρball=1/3ρB- (2/3)ρA ?
 
  • #4
haruspex
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Sorry, forget that post... i deleted it too late.
I agree with your analysis, the question seems flawed.
 
  • #5
haruspex
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and (b) direction towards the interface
Not sure where you get that from. Bear in mind that the weight of the top fluid increases the pressure in the lower fluid, so in effect the ball is floating 1/3 submerged in the lower and 2/3 in the upper.
Another way to think of it is to imagine replacing the lower 1/3 of the ball with more of the fluid it is in, and the upper 2/3 with more of the fluid at that level. Clearly the system would still be in balance.
So the equation ρball=(2/3)ρA+1/3ρB is correct.
 
  • #6
bw519
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I am thinking it may just be a flawed problem like you said. And maybe I should just go with C being the top fluid in the second beaker since although either fluid being the top has a mathematical issue, the one with B being the top fluid might be easier to spot without actually doing the math (which the teacher may not have done)... There was a part C that I wouldn't know how to do unless C is the top fluid of the second beaker. However, using C as the top fluid, I think I figured a way to solve part C.
 
  • #7
kuruman
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I know where I got it. I posted impulsively without writing something down. I agree that this question is flawed.
 
  • #8
haruspex
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I am thinking it may just be a flawed problem like you said. And maybe I should just go with C being the top fluid in the second beaker since although either fluid being the top has a mathematical issue, the one with B being the top fluid might be easier to spot without actually doing the math (which the teacher may not have done)... There was a part C that I wouldn't know how to do unless C is the top fluid of the second beaker. However, using C as the top fluid, I think I figured a way to solve part C.
A possibility is that the 1/3 and 2/3 should have been the other way about in the second beaker.
 

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