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Moment about a point using graph paper

  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    the ans given is 95Nm , but i gt 80.9Nm , which part i did wrongly ?

    2. Relevant equations


    3. The attempt at a solution
    20(4) -39(5/surd 26(3)) +39(1/surd26)(1) -60(3/5)(1)+60(4/5)(3)
    =80.9
     

    Attached Files:

  2. jcsd
  3. Oct 7, 2015 #2

    BvU

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    What is a surd26 ?
    Could you tidy up your notation in general as well. 20(4) probably means 20 N * 4 m = 80 Nm ?

    [edit] o:) sorry, that 20(4) is a correct notation. Just had to get used to the sight.
     
    Last edited: Oct 7, 2015
  4. Oct 7, 2015 #3

    mjsd

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    firstly, I don't think

    20(4) -39(5/surd 26(3)) +39(1/surd26)(1) -60(3/5)(1)+60(4/5)(3)


    would yield 80.9

    The term 39(1/ Sqrt[26])(1) should not be there.... 60 (4/5)(3) the (3) should be a (2)

    however with these adjustments, the answer is still not 95 Nm clockwise by my calculations (more like 25-ish), I believe there might be typo somewhere??
     
  5. Oct 7, 2015 #4

    BvU

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    Sigh... after wasting some time on this, I conclude that surd26 is ##\sqrt { 26}## and you use it to decompose the 39 N into x and y components. For the ##\ \vec r\ ## in ##\ \vec \tau = \vec r \times \vec F\ ## you then take ##\ \vec r = (1,3)\ ##. Shouldn't that be ##\ \vec r = (0,3) \ ## ? It gives yet another answer, but I wouldn't trust a book answer in a book that lets 1 m be represented by a square ( :wink: )

    PS I get what mjsd gets. The bat types faster .... :smile:
     
  6. Oct 7, 2015 #5
    sorry , the book contain error , the 1m should be 12m on the 39N triangle . I have redo the question .
    20(4) -39(5/13)(3) +39(12/13)(1) -60(3/5)(1)+60(4/5)(3)
    =179N
     
  7. Oct 7, 2015 #6

    mjsd

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    As I said in my first post....
    39(12/13)(1) should NOT be there as the (1) is actually (0)...can you see that?
    and
    60(4/5)(3) is actually 60(4/5)(2)
    once you have got these
    you should get 95 Nm for your answers
     
  8. Oct 10, 2015 #7
    I know what u mean . In the above steps , I have resolved the 60N in this way( black and red) , whereas the green colour one is the resolution of force done by you .

    For the 39N force , I have resolved the force in this way ( black and red)
     

    Attached Files:

    • 39N.png
      39N.png
      File size:
      1.5 KB
      Views:
      48
    • 60N.png
      60N.png
      File size:
      1.7 KB
      Views:
      46
  9. Oct 12, 2015 #8
    yes , i know why it should be 39(12/13)(0), now my question is why cant I resolve the force in the way that i have posted earlier?
    I have resolved the 60N and 39N in this way( black and red) , whereas the green colour one is the resolution of force done by you .
     

    Attached Files:

    • 39N.png
      39N.png
      File size:
      1.5 KB
      Views:
      43
    • 60N.png
      60N.png
      File size:
      1.7 KB
      Views:
      46
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