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Hydrostatic pressure - barrel vs. small cylinder of water

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  1. Sep 7, 2015 #1

    Rectifier

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    This problem was translated from Swedish, sorry for any grammatical errors present.

    The problem

    FX0WyuW.jpg
    The grey is liquid is water. The small container is a pipe made of glas and the big container is a barrel. Water is poured into the glas pipe. When water reaches 12m over the barrel top(lid) the barrel breaks.

    Calculate
    a) the mass of the water in the glass pipe (over the lid)
    b) the total force on the lid when the barrel breaks

    Relevant equations
    Liquid - water
    h = 12m
    R = 0.20m
    r = 0.003m
    (^^values from below the figure in my book)
    ##p=dgh##

    The attempt

    a)
    Density is described by ## d = \frac{m}{V} ##
    Volume of a cylinder is ##V = \pi r^2 h##
    density for water is ##d = 1000 kg/m^3##

    Mass is thus:

    ## m = d \cdot V = d \cdot \pi r^2 h= 1000 kg/m^3 \cdot \pi (0.003)^2 12 = 0.339292... kg ##​

    b)
    There is a hole in the lid so I calculate the area of the doughnut.
    The area of the lid is

    ##A_L = \pi r_n^2 = \pi (R-r)^2 = \pi (0.20 - 0.003)^2 ##
    The pressure is

    ##p = \frac{F}{A_L} \\ F=pA_L = \pi (0.20 - 0.003)^2p ##​

    I have only one more interesting formula in the book on this chapter - namely:

    ## p=dgh ##
    where
    d = density
    g = gravity at the surface of the material
    h = height

    I know that

    d for water is = 1000 kg/m^3
    g = 9.82
    h = 12m

    then
    ##F=pA = A_Ldgh = 1000 \cdot 9.82 \cdot 12 \cdot \pi (0.20 - 0.003)^2 = 14367.3 N ##​

    It feels like I should add atmospheric pressure somwhere but I am not 100% sure.
    Please help :,(


    EDIT:
    Apparently ## F_{total} = F_{water} + F_{air}##

    and the foce from the air is
    ## F=p_{atm}A_L = 100kPa A_L = 100 \cdot 10^3 \pi (0.20 - 0.003)^2 = 12192.2...##

    Thus ##F_{total} = F_{water} + F_{air} = 14367.3 N + 12192.2N =26559.5 N##
    Does it look right?
     
    Last edited: Sep 7, 2015
  2. jcsd
  3. Sep 7, 2015 #2

    SteamKing

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    You have added the force due to air pressure acting on the lid


    Is this the correct formula for the area of the donut? Remember, (R2 - r2) ≠ (R - r)2
    You should make and examine a free body diagram of the lid of the barrel to see if adding in the force due to air pressure is valid.
     
  4. Sep 7, 2015 #3

    Rectifier

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    Thank you for your help! I will be back with another try soon.

    Congrats on almost 10k posts! :D
     
  5. Sep 7, 2015 #4

    Rectifier

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    Attempt 2

    b)

    Donut
    ##A_L = A_1 - A_2 = \pi R^2 - \pi r^2 = \pi( R^2 - r^2)= \pi (0.20^2 - 0.003^2) ##
    The pressure is

    ##p = \frac{F}{A_L} \\ F=pA_L = \pi (0.20^2 - 0.003^2)p ##

    ## p=dgh ##​

    ##F=pA = A_Ldgh = 1000 \cdot 9.82 \cdot 12 \cdot \pi (0.20^2 - 0.003^2) = 14804.9N ##​
    and the foce from the air is

    ## F=p_{atm}A_L = 100kPa A_L = 100 \cdot 10^3 \pi (0.20^2 - 0.003^2) =12563.5 N##​

    Thus
    ##F_{total} = F_{water} - F_{air} = 14804.9N -12563.5 N = 2241 N ## up towards the sky
    Does it look right now :D?
     
    Last edited: Sep 7, 2015
  6. Sep 7, 2015 #5

    SteamKing

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    You overlooked my suggestion about drawing a free body diagram of the barrel lid to see if it is valid to include the force due to air pressure in your calculations...

    Your current calculations assume that the force due to air pressure is acting against the force due to hydrostatic pressure on the lid. Is this assumption really true?

    Remember, air pressure is acting on the all exterior surfaces of the barrel, and air pressure is also acting on the column of water in the pipe.
     
  7. Sep 7, 2015 #6

    Rectifier

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    Sorry, will try again.
     
  8. Sep 7, 2015 #7

    Rectifier

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    I am not that good at drawing forces, sorry. Here is my attempt:
    vrHUF6R.jpg
    orange - water's force on the lid
    black - air's force on the lid
    grey - forces of no interest

    I guess that the main problem here is to realize wether that air pressure on the column of water translates into more waterpressure on the lid.

    I am sorry but I have no idea how to turn that into a calculation. Thank you for your help so far!
     
  9. Sep 7, 2015 #8

    SteamKing

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    Take the calculation in steps.

    How would you calculate the force due to air pressure acting on the outside of the barrel (just the top), assuming that there was no water inside the barrel?

    Why not? You did it when the fluid in the column was water.

    You already know what the pressure of the atmosphere is. The pressure of the atmosphere and the pressure of the water in the pipe can be added together, since pressure is a scalar quantity.

    [EDIT] Remember, we are interested only in the forces acting on the lid of the barrel. The forces acting on the sides of the barrel and of the pipe cancel out and are of no interest anyway, since they are not acting on the barrel lid.
     
  10. Sep 7, 2015 #9
    Your equation for the fluid pressure should include the atmospheric pressure on the fluid at the top of the glass pipe :
    $$p=p_{atm}+dgh$$

    Chet
     
  11. Sep 7, 2015 #10

    Rectifier

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    ##p = \frac{F}{A} \\ F=pA \\##
    A is the area of the lid and p is the atmospheric pressure
     
  12. Sep 7, 2015 #11

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    Attempt 3

    b)

    Donut
    ##A_L = A_1 - A_2 = \pi R^2 - \pi r^2 = \pi( R^2 - r^2)= \pi (0.20^2 - 0.003^2) ##
    The pressure is

    ##p_L = \frac{F_L}{A_L} \\ F_L=p_LA_L = \pi (0.20^2 - 0.003^2)p ##

    -------> ## p=p_{atm} + dgh ##​

    ##F=A_Lp = A_L(dgh+p_{atm}) = \pi (0.20^2 - 0.003^2) (1000 \cdot 9.82 \cdot 12 + 100 \cdot 10^3 ) = 27368.4 ##

    Is that it :eek: ?​
     
    Last edited: Sep 7, 2015
  13. Sep 7, 2015 #12
    I don't know. The rhs of your equation is not equal to the left side.

    What SteamKing and I have been trying to point out is that the atmospheric pressure cancels out from the results for the net force on the lid.

    Chet
     
  14. Sep 7, 2015 #13

    SteamKing

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    I think we have established that you know the basics. The question about whether to include atmospheric pressure in your force calculations does not really require any calculation, but a more subtle thing called analysis.

    If we look at the drawing you made of the pressure acting on the barrel:

    proxy.php?image=http%3A%2F%2Fi.imgur.com%2FvrHUF6R.jpg
    We see that the orange arrows represent the hydrostatic pressure acting on the lid, but since the pipe is open at the top, the atmospheric pressure must be added to hydrostatic pressure produced by the column of water in the pipe. So, your orange arrows should also have a bit extra added to them due to atmospheric pressure.

    However, the black arrows show atmospheric pressure acting on the outside of the lid of the barrel, separate from what internal pressure is produced by the water + air pressure acting from the pipe. What to do?
     
  15. Sep 7, 2015 #14

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    What equation are you referreing to?

    Hmm, interesting. I will try to look a bit more into it.
     
  16. Sep 7, 2015 #15

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    My thoughts now are to calculate the black force on the lid and the hydrostatic pressure on the other side of the lid.
    ## F_T = F_O - F_B \\ F_B=A_Lp_{atm} \\ F_O=A_L(dgh+p_{atm}) \\ F_T = F_O - F_B = A_L(dgh+p_{atm}) - A_Lp_{atm} = A_Ldgh ##

    Am I on the right track?
     
  17. Sep 7, 2015 #16
    The downward force on the top of the lid is ##p_{atm}A_L##. The upward force on the bottom of the lid is ##(p_{atm}+dgh)A_L##. What is the net of these forces?

    Chet
     
  18. Sep 7, 2015 #17

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    I think I got it :D take a look on my comment above. Thank you for your tip :)
     
  19. Sep 7, 2015 #18
    Yes. I posted it right before you posted your result. Your answer is correct.

    Chet
     
  20. Sep 7, 2015 #19

    Rectifier

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    Yey! Thank you to both of you!

    I will post the solution to the last step in a moment for anyone having problems with ths problem in the future. Thanks once again!
     
  21. Sep 7, 2015 #20

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    Solution b)
    ( hopefully :D )

    b)

    Area of the lid
    ##A_L = A_1 - A_2 = \pi R^2 - \pi r^2 = \pi( R^2 - r^2)= \pi (0.20^2 - 0.003^2) ##

    The pressure is

    ##p = \frac{F}{A} ##

    ## p_h=p_{atm} + dgh ##

    ## F_T = F_O - F_B \\ F_B=A_Lp_{atm} \\ F_O=A_Lp_h = A_L(dgh+p_{atm}) \\ F_T = F_O - F_B = A_L(dgh+p_{atm}) - A_Lp_{atm} = A_Ldgh \\ F_T = A_Ldgh = 1000 \cdot 9.82 \cdot 12 = 117840 N ##


     
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