Need help deriving formula for rho-f in terms of rho-s, etc.

[the_chemist]

Hey there! This problem has been giving me some difficulty and I would really love some help. I will try to post what I have worked out and if anyway can tell me where my error lies or if they could help me out then...I would greatly appreciate it!

Question:
A right circular cone of base radius R, height H, and known density rho-s floats base down in a liquid of unknown density rho-f. A height h of the cone is above the liquid surface. Derive a formula for rho-f in terms of rho-s, R and h/H, simplifying it algebraically to the greatest possible extent. [Recall Archimedes' principle and note that the volume of a cone equals (base area)(height)/3.]

Alright! Now..what I've done (or tried to, lol): So first thing is first. Archimedes' principle! I understand it to mean that the buoyant force is equal to the weight of the displaced water. Mathematically, I understand this to mean Fb = F2 - F1. Where F2 = (rho-f)(gravity)(volume of the cone in the water) and
F1 = (rho-f)(gravity)(volume of the cone before it's placed in the water). This is assuming that F = (pressure)(area) where the volume is equal to the A times the height...so substituting in F = P*V.

The volume of the cone before being submerged, Vs, is [(pi)(R^2)h]/3.
The volume of the cone after being submerged, Vf, is [(pi)(R^2)H]/3.

Ok. So I pretty much lose myself after this point. Mostly because I don't know if I'm approaching this problem..practically or well. =/ Intuition tells me that this isn't the correct way to go about this problem but any help would really be appreciated! Many thanks in advance for ANY help.

 

[Deleted member 4993]

Re: Need help deriving formula for rho-f in terms of rho-s,

Hey there! This problem has been giving me some difficulty and I would really love some help. I will try to post what I have worked out and if anyway can tell me where my error lies or if they could help me out then...I would greatly appreciate it!

Question:
A right circular cone of base radius R, height H, and known density rho-s floats base down in a liquid of unknown density rho-f. A height h of the cone is above the liquid surface. Derive a formula for rho-f in terms of rho-s, R and h/H, simplifying it algebraically to the greatest possible extent. [Recall Archimedes' principle and note that the volume of a cone equals (base area)(height)/3.]

Alright! Now..what I've done (or tried to, lol): So first thing is first. Archimedes' principle! I understand it to mean that the buoyant force is equal to the weight of the displaced water. Mathematically, I understand this to mean Fb = F2 - F1. Where F2 = (rho-f)(gravity)(volume of the cone in the water) and
F1 = (rho-f)(gravity)(volume of the cone before it's placed in the water). This is assuming that F = (pressure)(area) where the volume is equal to the A times the height...so substituting in F = P*V.

The volume of the cone before being submerged, Vs, is [(pi)(R^2)h]/3.
The volume of the cone after being submerged, Vf, is [(pi)(R^2)H]/3.

Ok. So I pretty much lose myself after this point. Mostly because I don't know if I'm approaching this problem..practically or well. =/ Intuition tells me that this isn't the correct way to go about this problem but any help would really be appreciated! Many thanks in advance for ANY help.

                                A
                               /|\
                             /  |  \
    ~~~~~~~~~~~~~~~~~~~~~B /----|----\C~~~~~~~~~~
                          /     D     \
                        /       |       \ 
                   F  /_________|________\  E 
                               G

Let us assume that the fluid level is at BC

AD =h

AG = H

GF = R

BD = r

Density of fluid = Df

Density of solid = Ds

Mass of cone = 1/3 * (pi) * R^2 * H * Ds

Mass of displaced fluid = 1/3 * (pi) * {H*R^2 - h*r^2} * Df

Archemedes's Principle:

Mass of displaced fluid = Mass of cone

Now continue...

 

[the_chemist]

!!! Thank you so much! That was a very good explanation. I can see "r" and "R" and their proportion. One thing that is giving me a little trouble is why you would assume that there would be a varying radius. Is it for the same reason that one should assume there would be two volumes? One for before the cone gets in the water and then one when it's in it?

 

[Deleted member 4993]

!!! Thank you so much! That was a very good explanation. I can see "r" and "R" and their proportion. One thing that is giving me a little trouble is why you would assume that there would be a varying radius.

If you are refering to the fact that 'r' & 'R' are different - The picture (mental or otherwise) of a floating cone should make it obvious. The cone is floating partially - some part of it is submerged.

Is it for the same reason that one should assume there would be two volumes? One for before the cone gets in the water and then one when it's in it?