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Simple Water Squirter

  1. Nov 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Area at the end with the plunger (A1) = 8cm^2
    Area at the nozzle (A2) = 0.5cm^2
    Velocity at the nozzle (V2) = 2m/s

    What is the force being applied to the plunger?

    2. Relevant equations

    A1*V1=A2*V2
    P + (1/2)pV^2 + pgy = P + (1/2)pV^2 + pgy
    P=F/A

    3. The attempt at a solution

    I used the first equation listed above to find the velocity of the plunger (V1) to be 0.125m/s. I'm having quite a difficult time finding any of the other variables (Pressure, Force, mass, density) for either side.
     
  2. jcsd
  3. Nov 26, 2007 #2

    Astronuc

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    Staff: Mentor

    How about changing the velocity (speed), which is acceleration, of the water?

    I presume one is ignoring the friction/viscous forces.
     
  4. Nov 26, 2007 #3
    I figured out that the density (p) is just 1 (because it's a water squirter) and the gauge pressure at the nozzle (P2) is 0. And the pgy terms are also 0 because the squirter is just horizontal. So using equation 2 from above (bernoulli's eqn) I got:

    P1 + 1/2(1)(V1^2) + 0 = 0 + 1/2(1)(V2^2) + 0
    P1=1/2(V2^2-V1^2)
    F1=P1*A1
    F1=1/2(2^2-.125^2)(8)=15.9375N

    And yes, I have no idea about fluid friction or viscosity at this point, hehe.
     
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