Fluids, mass of object floating over oil above water

[Xenon]

Homework Statement

Wooden qube with 12cm faces floating inside water and oil in 2 layers.
The oil floats over the water.
Water density: 1gr/cm cubed
Oil density: 0.8 gr/cm cubed
What is the mass of the cube.

Homework Equations

The Attempt at a Solution

F(buoyancy) = g*(density)*volume , but how do i get water and oil in the equation.

Thank you for your help.

 

[paisiello2]

What if there was no water and only oil?

 

[jbriggs444]

There is a fairly easy way to solve this problem based on blindly applying the principle that the buoyant force on an object is equal to the total weight of the fluid that it displaces. That is not the approach that I am suggesting below...

The force called "buoyancy" is just another word for the difference between the pressure of the fluid on the bottom of an object (water in this case) and the pressure of the fluid on its top (air in this case). For irregularly shaped objects, it can be hard to compute buoyancy based on this. But in the case at hand we have a very nice cubical shape.

Can you calculate the pressure of the oil at a depth of 6 cm? Can you then calculate the pressure of the water 4 cm deeper than that? Multiply by the area of the bottom of the cube and what do you have?

 

[Xenon]

Pressure of oil at depth of 6cm:
P=phg
P=0.8*6*1000= 4800 Pa

Pressure of water 4cm deeper then that h=10 cm??
P=1*10*1000= 10000 Pa

48*(4800+10000)=

Iam getting lost with the units. what do you think so far?

 

[jbriggs444]

Pressure of oil at depth of 6cm:
P=phg
P=0.8*6*1000= 4800 Pa

Good. That's the pressure of the oil.

Pressure of water 4cm deeper then that h=10 cm??
P=1*10*1000= 10000 Pa

It looks like you got a little lost here. The water pressure is going to add to the oil pressure. But you don't have 10 cm of water. You only have 4.

 

[Xenon]

Pressure of water
P=1*4*1000= 4000 Pa

 

[jbriggs444]

That 4000 Pa is the extra pressure at the bottom of the 4 cm layer of water over and above the pressure at the oil/water boundary. You already calculated the pressure at the boundary as 4800 Pa. So what's the resulting fluid pressure at the 10 cm level?

 

[Xenon]

Pw+Po=4800 Pa+4000
P10cm level= 8800Pa
so, 8800Pa* area at the bottom (is it 12^2)?

And what about the pressure at the top, the 2 cm?

Thank you, iam really struggling with physics, and you are making it much easier.

 

[haruspex]

Pw+Po=4800 Pa+4000
P10cm level= 8800Pa
so, 8800Pa* area at the bottom (is it 12^2)?

And what about the pressure at the top, the 2 cm?

Thank you, iam really struggling with physics, and you are making it much easier.

You don't need to worry about air pressure. Whatever the air pressure is on the top surface, it also acts on the oil and ends up producing an equal upward pressure on the bottom surface, so it cancel out.
But pressure is not really the right way to solve this. You can do it for a rectangular block like this, but it could get very messy for a more complicated shape.
Trust Archimedes - find the weight of fluid displaced.
Maybe you are struggling to figure out exactly what volume of each fluid is displaced. The trick is to imagine undoing the displacement. If you wanted to replace the block by some quantities of oil and water in just such a way that the existing bodies of each stay where they are, which part of the block in the diagram would be replaced by water and which part by oil?

 

[Xenon]

How can i know how much place does each one take if i don't have a refernce point?

 

[haruspex]

How can i know how much place does each one take if i don't have a refernce point?

Suppose you have replaced the block by liquid as I said. The two bodies of fluid shown in the diagram must stay in place and be stable, right? So which bit will you have to fill in with water?