0
$\begingroup$

A thin-walled container of mass $m$ floats vertically at the separation surface of the two liquids of density $ρ_1$ and $ρ_2$ . The whole mass of the container is concentrated in the part of height $h$.

The question is to determine the immersion depth $h'$of the container in the lower liquid if the bottom of the container has a thickness $h$ and an area $S$ and if the container itself is filled with the liquid of density $ρ_1$.

enter image description here

I applied the fundamental principle of the statics on the container

$Balance$ $Sheet$:

The weight of the container ($-mg$)

The buoyancy force applied by the fluid with density $ρ_2$ ($ρ_2gSh$)

The weight of the fluid with density $ρ_1$ (-$ρ_2gSh'$)

I summed up the forces and set them to zero and find my $h'$

but then realised that the fluid with density $ρ_1$ applied also a buoyoncy force on the top of the container , and the fluid with density $ρ_2$ applied a buoyoncy force on the other fluid.

Where's the problem ?

$\endgroup$
  • $\begingroup$ $m$ is the mass pf the container which is concentrated in the black part with thickness $h$ $\endgroup$ – Brophys Nov 22 '16 at 20:09
0
$\begingroup$

You can completely neglect the part of the container that sticks out into the liquid with density $\rho_1$ because its weight and its buoyancy cancel each other out exactly.

The balance for the rest of the container becomes:

$$\text{weight}=\text{buoyancy}$$

Assume the container has constant cross-section $S$, then with $mg$ the weight of the container plus the weight of the material between the bottom and $h$ ($^*$ proof below the fold): $$h'S\rho_1g+mg=(h'+h)S\rho_2g$$


$^*$

Buoyancy

$$W=mg+(h'+h'')\rho_1Sg$$ $$B=h''\rho_1Sg+(h'+h)\rho_2Sg$$ $$W=B$$ $$mg+(h'+h'')\rho_1Sg=h''\rho_1Sg+(h'+h)\rho_2Sg$$ Now decompose $W$ and $B$ into part above and below the liquid separation line: $$W_1=h'\rho_1Sg+mg\tag{1}$$ $$B_1=(h+h')\rho_2Sg$$ $$W_2=h''\rho_1Sg$$ $$B_2=h''\rho_1Sg$$ $$\implies W_2=B_2$$ With: $$W=W_1+W_2$$ $$B=B_1+B_2$$ $$W=B$$ Or: $$W_1+W_2=B_1+B_2$$ $$\implies W_1=B_1$$ Which with substitution gives us the expression above the fold.

$\endgroup$
  • $\begingroup$ $(h'+h)Aρ_2g$ is $buoyancy$ and the weight of fluid 1 is $h′Aρ_1g$ what about $hAρ_2g$ ? and wheres the weight of the container ($mg$) $\endgroup$ – Brophys Nov 22 '16 at 19:25
  • $\begingroup$ Your problem states: "The whole mass of the container is concentrated in the part of height $h$." I therefore assumed the container was massless. If not, add $mg$ to the LHS of the equation but then you need to know $A$ to get a solution. $\endgroup$ – Gert Nov 22 '16 at 19:47
  • $\begingroup$ yes but where did $hAρ_2g$ came from ? $\endgroup$ – Brophys Nov 22 '16 at 19:50
  • $\begingroup$ I've edited the answer accordingly. $\endgroup$ – Gert Nov 22 '16 at 19:51
  • $\begingroup$ I still have one problem , I don't understand where did $hSρ_2g$ ? what is it $buoyancy=(h′+h)Sρ_2g$ and the total weight is the weight of the container plus the weight of the fluid 1 so $weight= $ $\endgroup$ – Brophys Nov 22 '16 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.