Simple fluids but is the application correct

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The discussion revolves around a physics problem involving a U-tube filled with two liquids of different densities, accelerated to the right. Participants explore the relationship between the heights of the liquids before and after acceleration, emphasizing the need to apply Newton's second law and pressure equations. A key point raised is the angle of the fluid surface during acceleration, with hints suggesting that the surface remains horizontal initially. The conversation also touches on the dynamics of fluid behavior on inclined surfaces and the effects of acceleration on liquid levels. The focus remains on deriving the height difference before acceleration based on the given conditions.
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Homework Statement



problem states that ... there is a u tube and filled with liquid of densities ρ and 2ρ. the base length of the tube is 'l'. now it is accelerated towards right with accl.=a. if the HEIGHT DIFFERENCE AFTER ACCELERATING THE SYSTEM IS '0'. then what was the height difference before acceleration??

Homework Equations



P=ρgh; F=ma ; F=(ρ.V)a ;

The Attempt at a Solution


i tried to find pressure at bottom most level at interface of the two liquids, then the net force, then applied Newtons 2nd law, but could'nt get the solution...
 
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hi phyeinstein_c! :smile:

hint: what will be the angle of the surface of the fluid (in either half of the 'U') ? :wink:
 
lets see...if the u tube is moving towards right the major driving force is on the left side of the tube. since the heights are equal the 2p liquid is in the left tube .

first find relations between height h1 and h2 in the two sides of tube when at rest...then find height h when accelerated.. then keeping volume constant use h1 + h2 = 2h
 
tiny-tim said:
hi phyeinstein_c! :smile:

hint: what will be the angle of the surface of the fluid (in either half of the 'U') ? :wink:
they're not at angle. initially they are at a height difference. that's it is wat is given.
 
i know, i mean, suppose there was only one pipe (or a jar or vase), what would be the angle of the water surface? :wink:
 
it wud be horizontal obviously
 
nope! :wink:
 
why...?
 
suppose the water was ice, and the surface was at an angle, and a drop of water was on the surface …

what angle would the surface have to be for the drop of water to stay where it is? :wink:
 
  • #10
<=45 i guess
 
  • #11
phyeinstein_c said:
<=45 i guess

how do you get that? :confused:
 
  • #12
:P just guessing... mgsinθ is equal to the friction acting.. of surface tension watever... unless u describe the surface of ice to be smooth...
 
  • #13
phyeinstein_c said:
… unless u describe the surface of ice to be smooth...

yes, of course it's smooth! :rolleyes:
 
  • #14
ok... then it can't stay where it is..
 
  • #15
why not?

(if the block of ice is accelerating)
 
  • #16
oh yeah that's there... true.. then we can find the component of ma and normal reaction in horizontal direction and solve..
 

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