What is the pressure at the bottom of the flask?

[NasuSama]

Homework Statement

Liquid 1 (density $ρ_{1}$ = 8290 kg/m³) is poured into a large open cylinder to a depth of $d_{1}$ = 85.9 cm. Liquid 2 (density $ρ_{2}$ = 5695 kg/m³) is then poured on top of liquid 1 to a depth of $d_{2}$ = 70.3 cm. Then liquid 3 (density $ρ_{3}$ = 2933 kg/m³) is poured on top of liquid 2 to a depth of $d_{3}$ = 66.9 cm. Assume the liquids are immiscible (do not mix at all), so the total depth of the fluid is $d_{TOT}$ = 223.1 cm. Find P, the pressure at the bottom of the flask.

Homework Equations

I would assume that it's this formula, the Pascal's principle...

$\Delta P = \rho g(\Delta h)$

The Attempt at a Solution

I combined all the values altogether to obtain the pressure, but it's incorrect. I've heard that someone said that the pressure at the bottom will be lower after adding the liquids on top. I am not sure how to show this. I can only think that Pascal's principle is the only formula to start.

 

[NasuSama]

I am totally incorrect for this equation...

$P = \rho_{3}d_{3}g - (\rho_{2}d_{2}g - \rho_{1}d_{1}g)$

 

[Chestermiller]

I am totally incorrect for this equation...

$P = \rho_{3}d_{3}g - (\rho_{2}d_{2}g - \rho_{1}d_{1}g)$

Hi again. The contributions of the three fluids add together to give the pressure at the bottom. All the minus signs in your equation should be plus signs.

 

[NasuSama]

Sorry for triple post. I do not know how this problem is approached. I have heard that someone said that the pressure at the bottom must be lower after the liquids are added to the top, but I still don't get how to show this by formula.

 

[NasuSama]

Hi again. The contributions of the three fluids add together to give the pressure at the bottom. All the minus signs in your equation should be plus signs.

Then, I have...

$Total = \rho_{1}d_{1}g + \rho_{2}d_{2}g + \rho_{3}d_{3}g$

However, that is not the pressure at the bottom. It's the pressure of the whole liquids, but not at the bottom.

 

[Chestermiller]

Then, I have...

$Total = \rho_{1}d_{1}g + \rho_{2}d_{2}g + \rho_{3}d_{3}g$

However, that is not the pressure at the bottom. It's the pressure of the whole liquids, but not at the bottom.

Yes, it is the pressure at the bottom, over and above atmospheric pressure. What make you think it is not? How do you define the term "the pressure of the whole liquids?"

 

[NasuSama]

Yes, it is the pressure at the bottom, over and above atmospheric pressure. What make you think it is not? How do you define the term "the pressure of the whole liquids?"

I don't know.

Here is the answer I have. It's 1.28 * 10^(5) Pa, which is the previous answer I have. Did you get that answer with your calculator?

 

[Chestermiller]

I don't know.

Here is the answer I have. It's 1.28 * 10^(5) Pa, which is the previous answer I have. Did you get that answer with your calculator?

Yes. This is the same answer I get for the gage pressure (the pressure over and above atmospheric pressure ) on the bottom. The atmospheric pressure is 1 * 10 ^5 Pa. So the total absolute pressure on the bottom is 2.28 * 10^5 Pa.