Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Fluid motion

  1. Feb 22, 2005 #1
    Hey.. I was looking over a physics team test, and I can't figure out how to solve this problem. I've asked my teachers and they can't help me either. :grumpy:


    Water runs out of a horizontal drainpipe at a rate of 120 kg per minute. It falls 3.2 m to the ground. Assuming the water doesn't splash up, what average force does it exert on the ground?


    I'm not sure where to start with this- I believe that if the diameter of the pipe was given, you could find the velocity of the water, but without it I'm not sure where to begin. Just a little help would be greatly appriciated. :smile:
     
  2. jcsd
  3. Feb 23, 2005 #2
    The change in the momentum of a particle of fluid in the vertical direction equals the negative of (its mass)* sqrt(2* acceleration due to gravity* 3.2m). This momentum is transfered to the ground ( assuming, as you said, the water doesn't splash around). Since the particles weighing 120kg fall on the ground per minute, the average force is
    - (120)*sqrt( 2*acc. due to gravity* 3.2) Newton per min.
    A fluid particle may have a velocity in the horizontal direction, but it remains unchanged throughout the fall & the fluid continues to flow along the ground. I assumed, based on the few drainpipes I've seen in my life :smile: ,that initially the vertical component of the velocity was negligible.
    I am, with great respect,
    Einstone.
     
  4. Feb 23, 2005 #3

    xanthym

    User Avatar
    Science Advisor

    To clarify Einstone's solution, we have for the momentum of fluid mass "m":
    p = m*v
    Δp = m*Δv
    {Average Force During Time ΔT} = Δp/ΔT = m*Δv/ΔT

    The kinetic energy change of fluid mass "m" is related to the work performed by gravity over distance "d" on "m" by:
    Δ{(1/2)*m*(v^2)} = (1/2)*m*{(vf)^2 - (vi)^2} = F*d
    Or in this case, since vi=(0):
    (1/2)*m*(vf)^2 = m*g*d
    vf= sqrt{2*g*d}
    Δv = sqrt{2*g*d} - 0 = sqrt{2*g*d}

    {Average Force On Ground During Time ΔT} = -Δp/ΔT =
    = -m*Δv/ΔT = -m*sqrt{2*g*d}/ΔT =
    = -(120 kg)*sqrt{2*(9.8 m/sec^2)*(3.2 m)}/(60 sec)
    = (-15.84 N)


    ~~
     
    Last edited: Feb 23, 2005
  5. Feb 23, 2005 #4
    Thank you so much, 16 N was the answer given. :)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook