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How to find the force

  1. Sep 9, 2010 #1
    A pipe with a mouth 125cm2 of area, spraying water with a speed of 25cm s-1 in a Service Station. This water perpendicularly fall on to a wind screen of a car.
    Find the Force which act on wind screen by water.
    [Note:- consider that velocity of water become 0 after it hits the wind screen, that is mv2=0]

    The answer is given as 0.75N, I cannot to reach to this, please help me.
  2. jcsd
  3. Sep 9, 2010 #2


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    Hello Kalupahana,

    Welcome to Physics Forums!

    You'll have to show your attempted solution, and any relevant equations before we can help. For starters, in your relevant equations, how is momentum defined? What is the relationship between force and momentum?

    Also, could you double check the problem (and given answer) and make sure you wrote it in correctly? A few things about your problem statement seem suspicious. Water moving at 25 cm/sec isn't really "spraying." That's more of a trickle. A pipe that has an area of 125 cm2 means that it has a diameter of around a foot. That's a huge pipe. So you've described a huge pipe that's just trickling out a whole bunch of water (but not really "spraying" it per-se). And the answer, is that in units of Newtons (N) or kN? And are you sure the numerical value is 0.75 instead of 0.78?

    [Edit: Oh, and are there any changes in height involved (such as differences between the height of the pipe and the height of the windscreen) that are mentioned in the problem?]
  4. Sep 9, 2010 #3
    I also have a problem it practically can be happen. But question is given like that. No height between pipe & wind screen. The answer come in cubic meters (Dimensions L3T-1, Units m3s-1).
    The answer also given like 0.75N. It's in Newtons.

    How to get a relation with those things with force.
  5. Sep 9, 2010 #4


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    Well, it is possible that your textbook is giving you the wrong answer. It's rare, but I have seen it happen before.

    Anyway, go ahead and list your relevant equations, and show us your attempted solution. Then I(we) might be able to point you in the right direction and/or compare your answer with mine(ours).
    Try looking in your textbook for a relationship between force and momentum. This relationship is the key to solving this problem.
  6. Sep 10, 2010 #5
    kk i'll try
  7. Sep 10, 2010 #6
    The area of mouth of pipe = 0.012 m2.
    The speed of moving water = 0.25 ms-1
    So flow rate = 0.012 x 0.25 = 0.003 m3s-1

    I don't how to combine this with force. It;s not coming to my memory.:eek:

    The mass can be taken by eq of density, bt it's bot given in the question.
    [density of water = 1000 kg m-3]
    d = m/v
    so m = 3kg

    Then momentum can be taken by as
    mv = 3 x 0.25 = 0.75 kg ms-1

    NOw what to do?


    The water beam become 0 after stuck on wind screen.

    F = 0.75/1
    = 0.75 N

    Can I use time as 1sec in here.

    I was reached to answer like this.
  8. Sep 10, 2010 #7


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    Okay, but earlier you mentioned the area of the mouth of the pipe was 125 cm2 = 0.0125 m2. It makes a significant difference in this problem.
    Okay, but if we go with the original 125 cm2, the flow rate is 0.003125 m3/s.
    So close! But what you've calculated is

    (1000 [kg/m3])(0.003 [m3/sec]) = 3 [kg/s]

    The units are in kg/s, not kg.

    That is the change in mass per unit time. In other words,

    dm/dt = 3 kg/s

    (If we use the original 125 cm2 for the mouth area, the flow rate turns out to be dm/dt = 3.125 kg/sec)

    Multiplying those is not really the momentum. It's the change in momentum per unit time.

    (dm/dt)v = d(mv)/dt = (3 [kg/s])(0.25 [m/s]) = 0.75 [kg(m/s2)]

    (Or if we use the original 125 cm2 for the mouth area, the flow rate turns out to be dm/dt = (3.125 [kg/s])(0.25 [m/s]) = 0.781 [kg(m/s2)])
    The equation in red was the one I was looking for; there 'ya go! :smile:

    But there is no reason you need to divide by 1 sec. The division by unit time is already present in the above equations (you forgot to include it in your rate of mass flow). But you have already found that
    (dm/dt)v = d(mv)/dt = (3 [kg/s])(0.25 [m/s]) = 0.75 [kg(m/s2)]

    And you know that F = d(mv)/dt, so you've already found the answer!

    And take a look at the units. [kg(m/s2)] = [N]

    So you answer is 0.75 N.

    However, if we use the original 125 cm2 for the mouth area, the answer comes out to be 0.781 N. That is what threw me originally.

    If your book/coursework actually has 125 cm2 as the area (and everything else as is listed in the problem statement), but gives a final answer of 0.75 N, you book/coursework made a mistake in keeping track of significant figures.
  9. Sep 10, 2010 #8
    ok, thnx for all I got it

    Most of the times I forgot to put correct units.
    Thats my mistake, sorry for it.. kk
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