 #1
bw519
 5
 2
 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 2ρA+ρB. But if that is the case, fluid C is more dense than fluid B and shouldn't be 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 2ρA+ρB. 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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