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~christina~
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[SOLVED] block in oil
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
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
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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