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Homework Help: Couple of liquid problems, need help desperately

  1. Jan 20, 2005 #1
    First Problem
    A flat plate moves normally toward a discharging jet of water at the rate of 3 m/s. The jet discharges water at the rate of .1 m^3/s and at a speed of 18 m/s. (A) Find the force on the plate due to the jet. (B) Compare it with if the plate were stationary.

    In part A the V toward each other is 21, 3 to the right, 18 to the left. Just unsure on how to calculate the force which is obviously caused by the plate being hit with the water.

    Second Problem
    When at rest, a liquid stands at the same level. A height difference occurs when the system is given an acceleration toward the right. Show that H= aL/g.

    Im really poor at proving these, just some helpful steps to start me off would be greatly appreciated
  2. jcsd
  3. Jan 20, 2005 #2


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    First one: force is equal to momentum change. Momentum is a vector quantity, which simply means that the amplitude of the momentum is dependant on its direction. So if the velocity of the water jet goes from 18 m/s to 0 m/s by shooting out sideways, that is different from the velocity going 18 m/s in one direction, then being turned around and going 18 m/s in the opposite direction. You don't have direction change in your problem, so if you assume the jet goes off sideways (ie: 90 degrees from the initial direction) then the dV = 18 m/s. Except the flat plate is coming towards it at 3 m/s so the total dV is 21 m/s (did I understand you right?)... Force is the change in momentum = m(dot) * dV

    Calculate the mass flow rate, m(dot) by taking the density of water and multiplying it by the volumetric flow rate you gave (0.1 m3/s). You should have a mass flow rate in kg/s.

    Now multiply kg/s times the 21 m/s and you'll get the force in Newtons.

    Second problem. It says some body of liquid is accelerating in the horizontal direction. So you might imagine the liquid piling up towards the rear end of this container it might be in. At the back of the container, the liquid is so much higher than at the front. That higher level exerts a pressure of rho*g*H (density * acceleration due to gravity * height of liquid). That higher pressure balances (is equal to) the force due to acceleration which is rho*a*L
    (look familiar? density * horizontal acceleration * length of liquid column in the horizontal direction).

    End result:
    rho*g*H = rho*a*L

    Solve for H.
  4. Jan 21, 2005 #3
    Thanks a lot, understand both of them completely now, i really appreciate it
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