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Pressure & Force

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data
    How much force does the fluid exert on the end of the cylinder at A?


    2. Relevant equations

    3. The attempt at a solution

    p= po + density*g*d
    Can I say the pressure is (10kg)(9.80)/pi*(.02m)^2 + (900 kg/m^3)*(9.80)(.70m) and set that equal to F/pi*(.1m)^2 to solve for force? Does the tube enter the container at halfway up?
  2. jcsd
  3. Apr 28, 2008 #2
    I cannot see anything wrong with your answer assuming as you have that the tube enters the container at halfway up. Without that assumption its not possible to give a final answer.
    Normal atmospheric pressure can be ignored because its effect will be zero on the final force exerted.
  4. Apr 28, 2008 #3
    I did Pressure= (10kg)*(9.80)/pi*(.02m)^2 + (900kg/m^3)(9.80)(.70m) = 84159.92 N/m^2

    84159.92 = F/pi * (.1m)^2 F= 2600 N rounded off with two significant figures because the question tells me to, but this is wrong. What's not quite right?
  5. Apr 28, 2008 #4


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    Hi bcjochim07,

    I think you need to account for the atmospheric pressure that acts above the piston.
  6. Apr 28, 2008 #5
    ok, I didn't understand why apelling told me I didn't

    so Pressure = (10kg)(9.80)/pi*(.02m)^2 + 101300N/m^2 + (900kg/m^3)(9.80)(.70) = 185460

    185460 = F/pi* (.1m)^2 F= 5800 N ?
  7. Apr 28, 2008 #6
    also the second part of the question asks for force at b, so I would do the same sort of thing only with a depth of 1.3m?
  8. Apr 28, 2008 #7


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    As long as it's safe to assume that these pipes and things are circular in cross section (and that your original assumption about the halfway mark is correct), I think that's right.
  9. Apr 28, 2008 #8
    Ok so here's what I did for the force at b:

    (10kg)(9.80)/pi*(.02m)^2 + 101300 + (1.3m)(9.80)(900 kg/m^3) = P

    P= 190752 = F/ pi * (.1m)^2 F= 5992.65 N. This is mastering physics homework, and it says to round it to two significant figures, anybody know how I would do that on a computer?
  10. Apr 29, 2008 #9
    Oh ok.... scientific notation of course. Does my answer for part b seem logical?
  11. Apr 29, 2008 #10
    The reason why I suggested ignoring atmospheric pressure is because I was thinking in terms of the resultant force on the cylinder end, whereas you required the force from the liquid only. Sorry to have mislead you.
  12. Apr 29, 2008 #11
    Does my answer for part b seem ok?
  13. Apr 29, 2008 #12
    Any thoughts anybody?
  14. Apr 29, 2008 #13


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    If you're having doubts about it, you can use the pressures at A and B to check that they are consistent, using the same equation:

    P_B = P_A +\rho g h

    where h is the depth of B relative to A.

    Or you can compare the forces at A and B. The force at B is the force at A (that is the force the top of the container is pusing down on the fluid) plus the weight of the fluid above point B:

    F_B &= F_A + m g \nonumber\\
    &= F_A + \rho g (\pi r^2) h\nonumber

    because [itex]\pi r^2 h[/itex] is the volume of the liquid above point B. (Be sure to use the full number for [itex]F_A[/itex], though, not the amount after you rounded off to two digits.)
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