# Homework Help: A vessel filled with two liquid how to determine boant force?

1. Dec 30, 2011

### vkash

A vessel filled with two liquid how to determine boyant force?

A vessel contains oil of density(800kgm-3) over mercury (density 13600kgm-3) A homogeneous Sphere is floats with half of its volume immersed in mercury and other half in oil. What is density of the material of the sphere?

Last edited: Dec 30, 2011
2. Dec 30, 2011

### ehild

What does Archimedes' Principle say?

ehild

3. Dec 31, 2011

### vkash

It says that buoyant force is equal to weight of liquid displaced.

4. Dec 31, 2011

### ehild

There are two kinds of liquid displaced. What is the volume and the weight of each when the volume of the sphere is V?

ehild

5. Dec 31, 2011

### vkash

But lighter liquid seems to push the sphere downward rather than pushing it upward. and lower liquid seems to push with greater power.
even after this. can i apply that principle?

6. Dec 31, 2011

### ehild

You get upward force at the bottom of the immersed volume, and an upward force at the top.
Pascal's law states that the force a liquid exerts on a surface is normal to the surface and F=PA.
As the pressure is greater at the bottom of the immersed object then at the top, the liquid pushes the bottom surface upward by a greater force than it pushes the top surface downward. The resultant force is always upward.

ehild

7. Dec 31, 2011

### vkash

thnank's sir(are u male) for helping me,

8. Dec 31, 2011

### ehild

Can you proceed from here?

ehild

9. Dec 31, 2011

### vkash

Actually i have already(before posting) done this question by this method. But I did not understand that is it correct or not. that's why i post it here.

10. Dec 31, 2011

### ehild

Do you understand it now?

Take a block instead of the sphere. The area of the top and bottom sides is A. The pressure at depth d in the oil is
Ptoilgd +P0, so the downward force on the top surface of the block is Ft= A(gρoilgd +P0).

At the bottom of the block, the pressure is that of a column of oil of height d+h/2 and a column of mercury of height h/2. Pboilg(d+h/2)+ ρHgg(h/2)+P0, so the force Fb =A(ρoilg(d+h/2)+ ρHgg(h/2)+P0)
The resultant is the buoyant force: BF=Ag( ρoil(d+h/2)+ ρHg(h/2)-ρoild)=Ag(ρoil(h/2)+ ρHg(h/2))=g(V/2)(ρoil+ ρHg).

ehild

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