1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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,
  4. Feb 23, 2005 #3


    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