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Force Needed to be Exerted

  1. Aug 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Water is pumped through a hose-pipe at a rate of 90 kg per minute. It emerges from the hose-pipe horizontally with a speed of 20 m/s.What is the force is required from a person holding the hose-pipe to prevent it moving backwards?

    2. Relevant equations

    3. The attempt at a solution
    This is m/t
    So if we assume u=0m/s
    Then, F= 0.025 x 20 (using Ft=mv-mu)
    F= 0.5N
    The answer is 30N
    Can we assume that u is 0m/s?
  2. jcsd
  3. Aug 21, 2015 #2


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    {You might want to check this calculation again!}
  4. Aug 21, 2015 #3
    Even then,
    then the answer becomes,
    0.0004167 x 20
    = 0.0083 N
  5. Aug 21, 2015 #4


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    I repeat, check your mass flow calculation again. How many seconds are in 1 minute?
  6. Aug 21, 2015 #5


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    Can I recommend you work out the number of kg/s using a pencil and paper rather than a calculator.
  7. Aug 21, 2015 #6


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    90 kg/min you should be able to work out in your head. o_O
  8. Aug 21, 2015 #7
    What is Ft exactly?

    Try applying basic principles:
    F = ma is what we usually use when applying the 2nd Law, but a more appropriate form might be:
    F = d/dt (m*v) = v*(d/dt m) + m*(d/dt v)

    In the case of constant mass, this reduces to F = ma. In the case of constant velocity (this problem), this can be reduced to F = v*(d/dt m)

    You are given the rate of water being pumped, which is in units of kg/min (can be converted to kg/s). This looks a lot like.... dm/dt, right?

    So, you can calculate the force being pressed back against the person holding the hose. But you are asked how much force it takes to keep the hose from moving. If that's what you want, what is true about the total acceleration of the hose? Now apply that to the 2nd Law, and what do you get?
  9. Aug 21, 2015 #8
    Whoops! It's 1.5kg/s. It works now. Thank you!
  10. Aug 21, 2015 #9
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