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Can anyone check my working on the tension upcoming test question

  1. Nov 22, 2013 #1
    1. The problem statement, all variables and given/known data

    [URL=http://s1345.photobucket.com/user/Duk_Bato/media/Untitled_zps84f60bfd.png.html][PLAIN]http://i1345.photobucket.com/albums/p679/Duk_Bato/Untitled_zps84f60bfd.png[/URL][/PLAIN]

    2. Relevant equations

    ƩMA = Ʃ r x F

    3. The attempt at a solution

    1.6 * T+ 0.8 * sin 40 *T = 1.2 * 250
    T (1.6 + 0.8 * sin 40 ) = 300
    T = 300 / ( 1.6 + 0.8 * sin 40 )
    T = 141.9 N

    RX= 108.701

    FY=141.9+141.9+250=533.8

    RF= Root(533.8^2 + 108.701^2) = 544.76N
     
  2. jcsd
  3. Nov 22, 2013 #2

    tiny-tim

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    hi cracktheegg! :smile:

    your T looks ok

    is your Rx positive or negative?

    check the signs in your Ry :wink:
     
  4. Nov 22, 2013 #3

    PhanthomJay

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    Looks good to here, call it Ax
    You have some errors here. You are looking for Ay by summing forces in the y direction = 0. You have signage errors and you forgot to calculate the vertical component of one of your tension forces. Assume Ay acts up, and try summing forces in the y direction = 0 again.
    Make correction above and this will make the magnitude of the reaction force at A, correct.
     
  5. Nov 22, 2013 #4
    I find the pulley Resultant force= 257.21

    0=141.9cos40 - Rdy (up)
    0=-141.9sin40 - 141.9 +Rdx (right)
    root(Rdy^2+rdx^2)= 257.21N

    tan^-1(Rdy/rdx) =65

    -250+257.21sin65+Ray=0
    Ray=16(up)

    RX= 108.701

    Rf= root(108.701^2+16^2)= correct?
     
  6. Nov 22, 2013 #5

    tiny-tim

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    hi cracktheegg! :smile:
    i think you have your Rdx and Rdy the wrong way round (though that won't affect your final result)

    but i don't understand why you're doing it this way …

    the question doesn't ask for the pulley resultant, and finding it takes much longer and gives more opportunity for making mistakes :redface:

    why didn't you simply do Ry for the beam?
     
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