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Impulse question given flow rate (litres per second)

  1. Apr 11, 2017 #1
    1. The problem statement, all variables and given/known data
    A fireman's hose acting as a water cannon, directs 100 litres/s (100 kg/s) of water horizontally at your chest at speed 5 ms-1. Assuming water essentially falls to the ground after hitting your chest (rather than say reflecting horizontally backwards), what horizontal force will you feel (in N)

    2. Relevant equations
    Impulse = change in momentum = Ft
    p = mv

    3. The attempt at a solution
    Hello. I do not know where to start from in this question. How do you get the mass from 100 litres/s and how would you to calculate force when not given a value for time for impulse?

    Help is greatly appreciated!
     
  2. jcsd
  3. Apr 11, 2017 #2

    kuruman

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    Better use ##F=\frac {dp}{dt}.##
     
  4. Apr 11, 2017 #3
    Im sorry i still dont know what to do
     
  5. Apr 11, 2017 #4

    kuruman

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    What is ##p## equal to? Take its derivative using the product rule.
     
  6. Apr 11, 2017 #5
    hello
    So would you go 100 kgms-1 * 5 ms-1 = 500 kgms-2
    does p = 500 kgms-2?

    Im sorry i dont know how to derive that number to get force. Also would deriving a plain number just be nothing???
    EG:
    derivative of x^2 is 2x.
    derivative of 2 x is 2
    derivative of 2 is nothing
     
  7. Apr 11, 2017 #6

    kuruman

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    Yes. Think of the derivative as how fast something is changing with respect to something else. A constant (plain number to you) does not change by definition, therefore its derivative is zero. Have you studied derivatives?

    I would go that way, but can you explain how you decided to multiply these two numbers?
    No. Here, p is momentum, p = m v.
     
  8. Apr 11, 2017 #7
    Hi thanks for the help btw.
    I multiplied them because of the momentum equation. Im still confused as how litres/s or kg/s converts to m (kg) which can be used for p = mv

    I studied some calculus but im very rusty on them. I know derivatives is just the function for rate of change of another function.
    So if impulse is change in momentum (and is so -500 kgms), if you derive it does it give you the force equation? would force just be 500N then?
     
  9. Apr 11, 2017 #8

    kuruman

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    Newton's second law is generally written as
    $$F=\frac{dp}{dt}=\frac{d(mv)}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$
    Look at the sum of the two products on the right. Can you identify what each of the four quantities represents? F is the force on your chest.
     
  10. Apr 11, 2017 #9
    Hello. So does...
    v * dm/dt = acceleration
    m * dv/dt = mass

    If so what does it have to do with the water hose then?
    Im even more confused as before, would you be able to give me an analogy/story to help understand whats happening.
     
    Last edited: Apr 11, 2017
  11. Apr 11, 2017 #10

    kuruman

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    Not what I meant.

    v = velocity of water just before it hits your chest (do you know what it is?)
    dm/dt = the amount of kg/s moving with velocity v that hits your chest every second (do you know what it is?)
    m = the total mass that has hit your chest (do you know what it is?)
    dv/dt = the rate of change of the speed of the water that hits your chest (do you know what it is?)

    Try to answer the questions with numbers. Two of them are given to you.

    Here is a story. When all these water molecules hit your chest, their speed drops from 5 m/s to zero. Your chest must exert a force on each one of them in order to stop them. By Newton's third law (action-reaction) all these molecules exert a force on your chest. Much like the force that air molecules exert on the palm of your hand if you stick it out of you car window as you barrel down the highway. That's what we are talking about.
     
  12. Apr 12, 2017 #11
    Ok i think im starting to get it now. I get the story aswell.
    Ok so are the values then...

    v = 5
    dm/dt = 100 kgs-1
    m = (depends on what time being hit for)
    dv/dt = -5

    * im not sure about the mass
    As F = v * dm/dt + m * dv/dt
    F = 5 * 100 + 0 * -5
    F = 500

    At t = 0, Force is 500. Is this right?
     
  13. Apr 12, 2017 #12

    haruspex

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    Although widely quoted, that can mislead. See section 6 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-momentum-impacts/.
    I believe that would be infinite here.
    No.
    Yes (given the right units!) but not from the reasoning you gave.
    In applying kuruman's equation above, you need to be clear what part of the system is encompassed by the variables. We can consider the water in some leading section of the jet at some instant, and that part of it remaining in the jet thereafter. As the water advances, the rear of it advances at speed v, while the front always ends at your chest, so its mass is diminishing but its speed is constant. If p is the momentum of that piece, the rate at which that piece loses momentum is (v-0)dm/dt. The loss of momentum results from the force exerted on it by your chest.
    In terms of your equation, F = v * dm/dt + m * dv/dt, m is nonzero but dv/dt is zero.


    .
     
  14. Apr 12, 2017 #13
    If ##\dot{m}## is the mass flow rate, then, during the tiny time interval ##\Delta t##, the amount of mass striking the chest is ##\dot{m}\Delta t##. The horizontal velocity of this mass decreases from v to zero. So, $$\Delta (Mv)=-\dot{m}v\Delta t$$ This is equal to the impulse of the force ##F\Delta t##. So,$$F\Delta t=-\dot{m}v\Delta t$$or$$F=-\dot{m}v$$
     
  15. Apr 12, 2017 #14

    kuruman

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    This can also be viewed from the impulse standpoint. Given that mass of water ##\Delta m_{water}## hits you in time interval ##\Delta t##, we start with
    $$F_{on~you} =\frac { \Delta m_{water} ~\Delta v_{water}}{\Delta t}$$
    In other words, we "stretch" the mass over the time interval over which the collision takes place. Doing so does not change the impulse delivered to the target in time interval ##\Delta t##. From this it is easy to see that
    $$F_{on~you} = \Delta v_{water} \frac{\Delta m_{water}}{\Delta t}$$

    On edit: I hadn't seen Chestermiller's argument when I posted. It is the same except the time interval in that case is "smaller", whatever that means.
     
  16. Apr 12, 2017 #15
    Hi:
    From the data provided, there is no extrict solution for your question. Take into account, that, inside this context, Force is defined as the reason for bodies to change their motions. If there is no motion change, from the definition, Force cannot be computed.
    Solutions to this question are often given from the Energy point of view, it can also be derived an answer from the Work-Energy theorem, but the path of the water stream after the impact is required.
    Think on a ball hitting a wall. It can either penetrate or bounce after the impact. From the distance traveled by the ball at both cases, the force could be derived.

    In your question, neither a distance nor a time are given. In Nature, there is no instantaneous transition from a given speed to zero, that would give infinite force. At the same time, there is no zero motion for both bodies after the impact, that would return zero force.

    The key here, is that water would bounce backwards to a given distance and that the man would also move, returning a "system" mometum change.
    From that change in momentum, force can be derived.

    The Work - Energy theorem is the best approach from my point of view:

    Being Work= Force x distance ..... and ...
    Being Work= Change in energy...

    Elastic distortions have not been taken into account............
     
    Last edited: Apr 12, 2017
  17. Apr 12, 2017 #16
    I stand by what I said. This is just a straightforward fluid mechanics application of a macroscopic momentum balance to the material within a control volume, with a horizontal flow into the control volume and an exit flow at (basically) 90 degrees to the inlet flow out of the control volume. Analyses like this are in virtually all fluid mechanics textbooks. See Transport Phenomena by Bird, Stewart, and Lightfoot.
     
  18. Apr 12, 2017 #17
    No doubts about your knowledge. But I think the asker is not at a fluid dynamics level but at a classic mechanics one.
    Im not writting to show how much I know, but for helping as much as I can.
     
  19. Apr 12, 2017 #18
    The analyses that Kuruman, the OP, and I did form the basis for the more advanced fluid mechanics analysis. Plus, they use momentum balance, rather than your suggestion of energy balance. Momentum balance is the way to go, because the energy balance is incapable of determining the force. Now, please don't make me give you infraction points for giving misinformation. In my judgment as a PF Mentor, your post borders on misinformation.
     
  20. Apr 12, 2017 #19
    No menace required. I daily work on fluid mechanics and Im not for teaching anybody.

    Please ban me at this forum

    Bye bye all
     
  21. Apr 12, 2017 #20

    haruspex

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    So think of something else, like a lump of wet clay.
    That's easily resolved. Let the water bounce off to the sides, equally to left and right.
    Quite inappropriate since most or all of the energy is lost to heat.
    But you could learn something.
     
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