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Hydrostatics: long, bent tube

  1. Aug 22, 2014 #1
    A uniform long tube is bent into a circle of radius R and it lies in vertical plane. Two liquids of same volume but densities ρ and δ fill half the tube. Show that the angle θ = arctan{(ρ-δ)/(ρ+δ)}
     

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  3. Aug 23, 2014 #2

    NascentOxygen

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    How would you approach a problem like this?
     
  4. Aug 23, 2014 #3
    I tried taking length of equal arcs of two diff liquid col of same volume as "l" and arc length corresponding to angle θ as x and equating masses of liq columns of both sides of vertical line I got
    x/(l-x) =(ρ-δ)/(ρ+δ)

    if it is correct what to do next ?
     
  5. Aug 23, 2014 #4

    Orodruin

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    What is the condition for hydrostatic equilibrium? Is it that the masses on both sides are equal or is it something else?
     
  6. Aug 23, 2014 #5
    In hydrostatic equilibrium the pressures must be equal .we are to measure pressures which depend on height. How can height be measured in terms of R and θ?
     
  7. Aug 23, 2014 #6

    Orodruin

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    Correct, pressure = height x density.

    It is now time to do your trigonometry homework :wink:
     
  8. Aug 23, 2014 #7
    please help to solve it.
     
  9. Aug 23, 2014 #8

    Orodruin

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    I can try to guide you, but I will not solve he problem for you as per forum rules (which are reasonable in order to encourage you to think).

    Well, to start you know the geometry of the problem. Each fluid takes up a full right angle in the pipe. I would start by drawing circles and triangles and try to relate the height of the right column in the figure to the angle θ. Then continue by finding the height of the heavy liquid on the left side and last the height of the light liquid. It will also help to draw a horizontal and a vertical line through the center of the circle.
     
  10. Aug 23, 2014 #9
    please draw the fig and show the heights
     
  11. Aug 23, 2014 #10

    Orodruin

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    You already have a figure that you attached to your first post. I suggest you start from there and draw the vertical distance from the bottom to the surface of one of the liquids. I would like to help you, but you have to show some effort and make an attempt, after that I can tell you whether it is right and point you in the right direction if it is not.
     
  12. Aug 23, 2014 #11
    Here is the construction. where I get vertical length of lighter liq col at left = Rcos θ +Rsin θ
    horizontal length of denser liq col at left = Rsin θ
    horizontal length of denser liq col at right =Rcos θ
    then shall i take total length of horizontal liq col ?
    this can give the result . but height should be vertical. Is the proceeding correct?
     

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    Last edited: Aug 23, 2014
  13. Aug 23, 2014 #12

    Orodruin

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    Very good. Yes, you must find the verical height of the heavy liquid.

    So now it is a matter of finding the heights above the bottom of the heavy liquid. Let us start with the right side, the vertical line through the center of the circle has length R. It must be equal to the height of the fluid on that side plus the height of one of the triangles. Which triangle and what is the height of that triangle expressed in R and θ?
     
  14. Aug 23, 2014 #13
    If the vertical ht of denser liq at left be taken as Rsinθ and at right as Rcosθ then result will come. but how is it possible

    If the difference of vertical ht between rt and left of denser liq is taken as the resultant ht( ie Rcos θ -Rsin θ ) balancing the the vertical ht of lighter liq of ht Rcos θ +Rsin θ ,ht measured from the junction of the two liquids then the it will satisfy the result.

    I think that I have done right thing.

    please comment
     
    Last edited: Aug 23, 2014
  15. Aug 23, 2014 #14

    Orodruin

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    Because this gives the same difference between the heights as computing the actual heights. This is also a valid way of making the computation. The difference in height of the heavy liquid times ρ has to equal the height of the light liquid times δ. The reason it gives the same difference is related to a 90 degree rotation.

    Let us try the following: You know the distance from thte bottom to the center of the circle. You have obviously managed to compute the horizontal distance from the light-heavy interface to the horizontal line through the center of the circle. What is then the height of the heavy liquid to the left? A similar argument can be made on the right side.
     
  16. Aug 23, 2014 #15
    thanks for encouraging ,
     
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