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Fluids: force, pressure, etc.

  1. Jul 19, 2009 #1
    1. The problem statement, all variables and given/known data
    The L-shaped tank shown in Fig. 14-33 is filled with water and is open at the top. If d = 3.24 m, what is the force due to the water (a) on face A and (b) on face B?

    IMAGE: http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c14/fig14_33.gif
    2. Relevant equations
    i know that F=PA (where P is pressure and A is surface area)
    P-Pi= ρgh (where ρ is desnity and Pi is initial atmospheric pressure)

    3. The attempt at a solution
    Other than those formulas, i don't know what else i need. But i can't see how i have enough information to use those formulas. I don't know how i would calculate initial atmospheric pressure or desnity without them being given.
     
  2. jcsd
  3. Jul 19, 2009 #2

    LowlyPion

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    The problem says it's filled with water, so you should have the density of that.

    As for atmospheric pressure ... look it up.
     
  4. Jul 28, 2009 #3
    okay so atmospheric pressure and water density i know now (didn't know that I was supposed to know those at the top of my head)....but i still don't know what to do with this problem...i know i somehow need to integrate something, but don't know what...really need guidance on this one
     
  5. Jul 28, 2009 #4

    djeitnstine

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    Pressure varies by depth. And is constant along any one level of water.

    So P= pgh is already evaluated, and A= w*h. where 'w' = width. so if it varies along the vertical direction how would your equation for force look like?
     
    Last edited: Jul 28, 2009
  6. Jul 28, 2009 #5
    okay, first tihng i'm confused with...should pressure not be p=p(atmospheric) +pgh???

    i used this to get p=(1.013e5)+(1000)(9.8)(6.48)=164804
    and then since p=F/A....(164804)(10.4976)=F
    so F=1730046.47 N

    is that not correct for part A? i would not be surprised if its wrong lol
     
  7. Jul 28, 2009 #6

    djeitnstine

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    If the tank is fully exposed to the atmosphere, then we can say atmospheric pressure acts on both sides of all the surfaces. So any effect of it is canceled out.
     
  8. Jul 28, 2009 #7
    hmmm...so you mean, since it is putting pressure on the walls from the outside as well??? so that it doesn't matter if the top is open?
     
  9. Jul 28, 2009 #8

    djeitnstine

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    No it wouldn't matter if the top is open, as long as the pressure above the water surface is atmospheric, and the pressure outside the tank is atmospheric also.

    Since pressure always acts normal to any surface, there will be no effect of the atmospheric pressure force since it is acting on both the inside and outside of the tank.

    Edit: from your diagram this would appear so, especially since no extra information about an additional pressure acting on any surface is given.
     
  10. Jul 28, 2009 #9
    ok i see! so then, p=(1000)(9.8)(6.48)=63504...yes?

    and then how would i go about part B? i'm pretty sure i need to integrate, so would i integrate F=pA=pghd^2 from d=0 to 3.24?? does tha make sense because i'm kinda guessing?....
     
  11. Jul 28, 2009 #10

    djeitnstine

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    Since pressure varies with depth, wouldn't force vary with depth also? Since at any one level the pressure and thus the force remains constant. Look at post #4 again.
     
  12. Jul 28, 2009 #11
    okay...so i would integrate pressure? so integrate pgh from h=2d to 3d?? and then whatever my answer is multiply it by the area?
     
  13. Jul 28, 2009 #12

    djeitnstine

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    No you already evaluated it. pgh is already an integral evaluated (Specifically it is [tex]\int_0^H (\rho g) dh[/tex] Where H= final height).

    So let P=pgh - which you already calculated. then all you're left with is [tex]\int_0^H (Pw) dh[/tex] , (Where w = width and dh is your differential height)
     
    Last edited: Jul 28, 2009
  14. Jul 28, 2009 #13
    i'm very sorry, i'm still confused...as you said, the pressure would be different as you vary the hieght....but how did i already evaluate pgh??? because for part a, the face was horizontal so it didn't vary, but for part B the face is vertical so i'm still not sure what do do for its pressure? and then in ur final formula, how did you get Pwdh??

    sorry if you feel like ur spoon feeding here, i'm just so confused!
     
  15. Jul 28, 2009 #14

    rl.bhat

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    Pressure at h = 2d is ρgh. Take a small strip length d and width dh on face B.
    Force on this strip df = ρgh*d*dh. d remains constant throughout B.
    So the total force = ρgd*Intg.h*dh from h = 2d to h = 3d.
     
  16. Jul 28, 2009 #15

    djeitnstine

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    In part A pressure is constant since the surface is at one height, so your Force acting on that is [tex]\rho g (2d) d^2[/tex] In part b as we said the force is not constant along this surface! since it encompasses a depth 2d - 3d. So you have to evaluate pressure [tex]\rho g h|^{2d}_{3d}[/tex].

    And since force again varies with depth because pressure is NOT the same when you change height! so let [tex]P = \rho g h|^{h=3d}_{h=2d}[/tex] (i e. pressure evaluated from height 3d to height 2d!), you now need to evaluate force in a similar fashion.

    Thus we get [tex]F_p = Pwh|^{h=3d}_{h=2d}[/tex]

    PS. I edited the integrals in post 12 to be clearer
     
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