Buoyant Force Problem: Fraction of Length in More-Dense Liquid | Solution

[amw2829]

Homework Statement

A less-dense liquid of density ρ1 floats on top of a more-dense liquid of density ρ2 . A uniform cylinder of length l and densityρ, with ρ1< ρ<ρ2, floats at the interface with its long axis vertical.

What fraction of the length is in the more-dense liquid?

Homework Equations

F=ρVg

The Attempt at a Solution

I thought I could use V_f=ρ/ρ₂ but the final answer should include ρ₁ and possibly l. Any help would be appreciated.

 

[ehild]

What is the magnitude of the buoyant force in general?

ehild

 

[amw2829]

What is the magnitude of the buoyant force in general?

ehild

ρ₁Vg + ρ₂Vg ?

 

[ehild]

What is V? ehild

 

[amw2829]

What is V?


ehild

∏r^2h...but I don't have r

 

[ehild]

Is V the volume of the whole rod?

ehild

 

[amw2829]

Is V the volume of the whole rod?

ehild

From what I gathered it is a cylinder.

 

[ehild]

The buoyant force is equal to the weight of the fluid displaced. The rod is immersed partly in one fluid and the other part is in the other fluid. It can not displace its whole volume from both fluids. Assume that x length is immersed in fluid 2, and the other piece is in fluid 1. What is the total buoyant force in terms of the volume of the rod?


ehild

 

[amw2829]

The buoyant force is equal to the weight of the fluid displaced. The rod is immersed partly in one fluid and the other part is in the other fluid. It can not displace its whole volume from both fluids. Assume that x length is immersed in fluid 2, and the other piece is in fluid 1. What is the total buoyant force in terms of the volume of the rod?


ehild

I'm still not sure how to calculate that with the given variables.

 

[ehild]

Write up Archimedes principle.

ehild

 

[amw2829]

Write up Archimedes principle.

ehild

F=weight of the fluid displaced.

 

[ehild]

F=weight of the fluid displaced.

The rod floats, it is in equilibrium. How is F related to the weight of the rod?
ehild

 

[bayan]

Hello, I am kind of stuck on this question too (well I have an answer which I am not too sure about)

http://img27.imageshack.us/img27/3685/lratio.jpg this is a simple diagram that I drew.

Let F₁ be the Buoyant force of less dense liquid and it is pointing downwards
while F₂ is the buoyant force of more dense liquid is pointing upwards.

Since it is floating in middle I have assumed it to be at equilibrium, and hence F₂= F₁+ F_G

I have used A as the area of cylinder since it is same for all regions. l₁ for length of cylinder in less dense liquid and l₂ for the length of cylinder in more dense liquid.

Since g appears on both sides of equation I have ignored it ( by divide the whole thing by g)

my equation ended up looking like the following;

ρ₂Al₂=ρ₁Al₁+ρAl

I have divided both sides by A leading to


ρ₂l₂=ρ₁l₁+ρl

I have divided both sides by ρ₂ which leads to

l₂=(ρ₁l₁+ρl)/ρ₂

the question asks what fraction is in more dense liquid which is l₂/l

(ρ₁l₁+ρl)/ρ₂l

simplifying this leads to
(ρ₁l₁/ρ₂l)+(ρ/ρ₂)

Does this seem correct?

Regards

 

[bayan]

any help would be appreciated

Regards

 

[ehild]

The hydrostatic pressure must be the same along the red line in the figure. The red line is at depth l=l₁+l₂ with respect to the top of block. The change of the hydrostatic pressure from zero to depth l is either ρgl or ρ₁gl₁+ρ₂gl₂. The two pressures must be equal.

You have to express l₂/l in terms of the densities. For that, use that l₁=l-l₂.

ehild

 

[bayan]

The hydrostatic pressure must be the same along the red line in the figure. The red line is at depth l=l₁+l₂ with respect to the top of block. The change of the hydrostatic pressure from zero to depth l is either ρgl or ρ₁gl₁+ρ₂gl₂. The two pressures must be equal.

You have to express l₂/l in terms of the densities. For that, use that l₁=l-l₂.

ehild

Hi and thank you for your reply.

I understand what you mean, but I can't seem to solve l₂/l easily, the answer I got was l₂/l=(ρ-ρ₁l+ρ₁l₂)/(ρ₂l₂)

I think I have made a mistake somewhere. I started by setting ρgl = ρ₁gl₁+ρ₂gl₂

divided both sides by g, took ρ₁l₁ to other side of equation giving me ρl-ρ₁l₁ = ρ₂l₂, then used l₁=l-l₂
and divided both sides by l to get l₂/l.

That lead me to l₂/l=(ρ-ρ₁l+ρ₁l₂)/(ρ₂l₂) but I still have l₂ on both sides of the equation.

Any obvious error sources that you can see?

Regards

 

[ehild]

I think I have made a mistake somewhere. I started by setting ρgl = ρ₁gl₁+ρ₂gl₂

divided both sides by g, took ρ₁l₁ to other side of equation giving me ρl+ρ₁l₁ = ρ₂l₂,

It is wrong from here. When you take a term to the other side of the equation, the sign will change. Substitute l-l2 for l1 first, collect the terms with l1 at one side of the equation, and the terms with l on the other side.

ehild

 

[bayan]

It is wrong from here. When you take a term to the other side of the equation, the sign will change. Substitute l-l2 for l1 first, collect the terms with l1 at one side of the equation, and the terms with l on the other side.

ehild

Hi, sorry I had - minus sign in my hand written workout, for some reason I must have put a plus sign, another one where I only put l₁=l-l₂ at the end lead me to l₂/l=(ρ-ρ₁(l-l₂))/(ρ₂)

Still end up with l & l₂ on both sides for some reason

I will upload hand written workout in a few hours

 

[ehild]

Hi, sorry I had - minus sign in my hand written workout, for some reason I must have put a plus sign,


another one where I only put l₁=l-l₂ at the end lead me to l₂/l=(ρ-ρ₁(l-l₂))/(ρ₂)

Still end up with l & l₂ on both sides for some reason

I will upload hand written workout in a few hours

You must collect the l2 terms on one side of the equation.

You have the equation

ρl=ρ₁l₁+ρ₂l₂.

Substitute l₁=l-l₂:

ρl=ρ₁(l-l₂)+ρ₂l₂

Expand:

ρl=ρ₁l-ρ₁l₂+ρ₂l₂

Collect the terms with l₂:

ρl=ρ₁l+(ρ₂-ρ₁)l₂

Bring ρ₁l to the left-hand side.

Can you proceed?


ehild

 

[bayan]

Hi and thank you for your reply.

I ended up with the following

ρ-ρ₁/ρ₂-ρ₁ = l₂/l

Does that seem ok?

 

[ehild]

Hi and thank you for your reply.

I ended up with the following

ρ-ρ₁/ρ₂-ρ₁ = l₂/l

Does that seem ok?

You need to add some parentheses.

ehild