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Force at pipe bend

  1. Mar 22, 2016 #1
    1. The problem statement, all variables and given/known data
    why the force at outlet A shouldn't be 0.0707(12 cos120 - 12)

    2. Relevant equations


    3. The attempt at a solution
    as we can see from the figure , outlet A make an angle 120 with the horizontal line
     

    Attached Files:

    • 127.PNG
      127.PNG
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    • 128.PNG
      128.PNG
      File size:
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      Views:
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  2. jcsd
  3. Mar 23, 2016 #2

    haruspex

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    I think one of your attachments ('127') is for an earlier problem.
     
  4. Mar 23, 2016 #3
    ignore 128 , pls refer to the picture i upload now .
     

    Attached Files:

    • 126.PNG
      126.PNG
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      Views:
      28
  5. Mar 23, 2016 #4

    haruspex

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    You are confusing yourself. Both 126 and 127 were for an earlier problem. Neither has an angle 120 degrees in it. Only 128 matches the text of your post in this thread.
     
  6. Mar 23, 2016 #5
    sorr, pls refer to the picture i uploaded now , 129 and 126
     

    Attached Files:

    • 129.PNG
      129.PNG
      File size:
      11.9 KB
      Views:
      41
    • 126.PNG
      126.PNG
      File size:
      10.5 KB
      Views:
      32
  7. Mar 23, 2016 #6

    haruspex

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    No, the correct two images are 128 and 129.
    Again you are right. But if you look at the next line the correct numerical value is obtained.
     
  8. May 1, 2016 #7
    why Fx means the force of blade acts on water ? shouldnt it be force of water act on blade?
     
  9. May 1, 2016 #8

    haruspex

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    The author has chosen to define Fx and Fy as forces from the blade acting on the water.
     
  10. May 1, 2016 #9
    if i want to directly find the resultant force acting on the pipe bend on the water , how to find it ? what's the equation ?
     
  11. May 1, 2016 #10

    haruspex

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    I'm not sure what you are asking. If you mean, how to find force of water on blade, just define Fx and Fy in the opposite directions.
    If you mean how to find the resultant force without finding the x and y components, define your axes to be aligned with and perpendicular to the resultant, which is at some unknown angle theta to the incoming water. Then find the components of various lnown forces in those directions. It is just as much work, though.
     
  12. May 2, 2016 #11
    i mean find the resultant force acting on the pipe bend by the water , can we define Fx = resultant force acting on the pipe bend by the water instead of
    resultant force acting on the water by the pipe bend ? so , if we define Fx = resultant force acting on the pipe bend by the water, then Fx = 0.0707(12cos 120 -12) +0.1414(12 cos60 -12 ) = 2.12KN(to the left)

    is it correct to do so ?
    i just couldnt understand why the author define Fx as resultant force acting on water by the pipe bend ?
     
  13. May 2, 2016 #12

    haruspex

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    Sure, you can define it the other way round. It just flips its sign in every equation.
     
  14. May 2, 2016 #13
    Why we can't just not flip the equation and do as in the notes, but define fx as force acting on the bend by the water? I don't understand why the author define fx as force acting on the water by bend and do it this way
     
  15. May 2, 2016 #14

    haruspex

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    I can't think of a reason.
     
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