What equations are needed to solve a block in oil puzzle?

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    Block Oil Puzzle
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Homework Help Overview

The problem involves a cubical block of wood floating at the interface between oil and water, with specific measurements and densities provided. The questions focus on calculating gauge pressures at the block's surfaces and determining the block's density.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations relevant to gauge pressure and buoyancy, questioning the volume of the block and the effects of the interface between oil and water. There is uncertainty about whether to consider buoyancy in the force balance and how to compute pressures at the top and bottom of the block.

Discussion Status

Some participants have offered guidance on how to approach the calculation of gauge pressures and the volume of the block. There is ongoing exploration of the relationship between pressure, depth, and the densities of the fluids involved. Multiple interpretations of the setup are being considered.

Contextual Notes

Participants are navigating the implications of the block's position relative to the fluid interface and the definitions of gauge pressure. There is a mention of ignoring atmospheric pressure for simplification, and some confusion remains regarding the volume calculation and the specific contributions of oil and water to the pressures experienced by the block.

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[SOLVED] block in oil

Homework Statement


A cubical block of wood 10.0cm on a side floats at the interface between oil and water with it's lower surface 3.00cm below the interface. The density of the oil is 0.790kg/m^3.
a) what is the gauge pressure at the upper face of the block?
b) what is the gauge pressure on the lower surface of the block?
c) find the density of the block.

http://img156.imageshack.us/img156/6067/81479390rw0.th.jpg


Homework Equations





The Attempt at a Solution



do I use these equations ?
I'm not sure about the volume of the block. If according to the picture (I redrew it), it's 2cm below the interface what would that change?

I thought I would use these equations but not sure after the interface issue arises in this problem

a) [tex]B=(P_{bottom}-P_{top})A = \rho g V= mg[/tex]

I don't have the buoyant force though...and I thought that the Pbottom - Ptop would be the gauge pressure but is this correct?

b) I think that since it is below the interface of the oil then I would take the density of water instead of the oil
[tex]B=(P_{bottom}-P_{top})A = \rho g V= mg[/tex]

c) not sure once again but thinking of using this equation

[tex]B-F_g= (\rho _{fluid}- \rho_{object})g V_{object}[/tex]

However I don't have the mass of the block...hm..

Help please.
could someone tell me if I'm going in the right direction?

Thank you
 
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You do NOT have to consider buoyancy in the force balance because the buoyancy is caused by the pressure on the faces.:)
 
so how would I solve to find the gauge pressure?
 
can someone Please help me out with this problem??
 
The gauge pressure is just the pressure relative to atmospheric pressure.

You need to compute Ptop and Pbottom first, using just how much oil or water is above the depth of the top or the bottom of the block.

For c) you can use the equation you quoted at a). What is the volume of a block with height 10 cm and top area A?
 
kamerling said:
The gauge pressure is just the pressure relative to atmospheric pressure.

so it would be P-P(atmosphere)?
You need to compute Ptop and Pbottom first, using just how much oil or water is above the depth of the top or the bottom of the block.
so would there be 3cm of oil?

For c) you can use the equation you quoted at a). What is the volume of a block with height 10 cm and top area A?

volume of a block?

well it would be 10cm^3
 
You can ignore atmospheric pressure because it gets added to all the pressures, and just treat the experiment as if it's done in vacuum.

There would be 3 cm of oil at the top.

10 cm^3 isn't correct. I asked for the volume with A still in it, because A also appears in
[tex] B=(P_{bottom}-P_{top})A = \rho g V= mg [/tex]
 
Thanks for your help kamerling
 

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