|Mar17-12, 06:38 AM||#1|
We are preparing a physcs quiz and are planning to use the following problem
If we have a wooden block balanced at the junction of two immiscible liquids then we have the buoyancy of the two liquids balance the weight of the block.The lower liquid will exert your normal upthrust but the upper liquid also exerts an upthrust even though it is above the block.How does it happen?
We are thinking of a hypothetical situation in which we remove the upper liquid.If we now pour the other liquid then it tends to increase the hitting of the molecules of the lower liquid hence increasing the pressure.That is to say the upper liquid indirectly exerts that pressure using the lower liquid.
I am looking for an alternative approach or verification of the above explanation.
physics news on PhysOrg.com
>> Promising doped zirconia
>> New X-ray method shows how frog embryos could help thwart disease
>> Bringing life into focus
|Mar17-12, 07:51 AM||#2|
make your block have vertical sides so the geometry is simple, i.e. there's vertical force applied only to the bottom and top surfaces.
What's force up on bottom?
What's force down on top?
P X A both places , of course.
Pbottom = ρ1gh1 +ρ2gh2
Dont forget that the liquid above the block also exerts force down on the top of block, so raising level above block does not increase bouyant force. It remains equal to weight of fluid displaced.
|Mar18-12, 01:12 PM||#3|
first thing this is not a paradox at all.... u shud go thru d basics of fluid mechanics,,,, Archimedes principle is nothing but net upward force due to pressure difference.,,,
in ur problem the net force on the curved surface is zero (balanced),,, net force on the upper part is downward which is equal to p1xarea; where p1 is the pressure at upper surface,,,,
while net force on the lower surface is upward which is equal to p2xarea ; where p2 is the pressure at the lower surface,,,
the relation between p1 and p2 is :
p2 = p1+ρ1gh1+ρ2gh2
so eventually net upward force(thrust force) comes out is
Fthrust = Fupward - Fdownward
Fthrust = (ρ1gh1+ρ2gh2) x area
h1 = height of cylinder in liquid A (upper liquid)
h2 = height of cylinder in liquid B
take obvious meaning of other symbols
|Mar18-12, 03:28 PM||#4|
Sorry, Vik, I didn't read past that. I find the real newspeak to be double-plus ungood...
|Mar20-12, 01:28 PM||#5|
|Similar Threads for: Archimedes Paradox|
|Twins Paradox: The paradox within the paradox||Special & General Relativity||5|
|Archimedes problem||Introductory Physics Homework||2|
|Archimedes||Introductory Physics Homework||2|
|Resolving the Barber Paradox and the Russell's Paradox||General Discussion||18|
|What is the resolution of the The Bug-Rivet Paradox paradox in special relativity?||Special & General Relativity||4|