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Forces necessary to hold a board steady (double check my work?)

  1. Nov 25, 2012 #1
    This is the problem:
    http://oi47.tinypic.com/6qij4o.jpg

    I set fx, fy, and B equal to zero.
    'x' forces: Fx - Bsin(30) = 0
    'y' forces: Fy - mg + B(cos(30)) = 0
    lower end: -0.5*mg(cos(30)) + 0.4*B = 0

    Using the third equation I solved for B, and got -477.4.
    I then plugged B into the first two equations and calculated Fy=854.4 and Fx=-238.7.

    I don't feel confident in these numbers, so I wanted to double check with someone who has a better understanding. Did I go wrong somewhere, or does this look okay?
    Thanks in advance, I really appreciate your help.
     
  2. jcsd
  3. Nov 25, 2012 #2

    SammyS

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    Why set all those to zero?
    B should be positive !
     
  4. Nov 26, 2012 #3
    I set them equal to zero because I thought the net force should be equal to 0 in order to hold the board still... is that wrong?
     
  5. Nov 26, 2012 #4
    This is what I have now...

    (-0.5)(m)(g)(cos(30))+(B)(0.4)=0
    B=-(-.5)(45)(9.8)(cos(30))/0.4
    B= 477.4

    Fx-B(sin(30))=0
    Fx=(477.4)(sin(30))
    Fx= 238.7

    Fy-mg+B(Cos(30))=0
    Fy=-477.4(Cos(30))+(45)(9.8)
    Fy=27.6


    Am I anywhere close to being right?
     
    Last edited: Nov 26, 2012
  6. Nov 26, 2012 #5

    SammyS

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    The net force is zero, but that doesn't mean that each individual component making up the net force is zero.
     
  7. Nov 26, 2012 #6

    SammyS

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    The method is right.

    The numbers look reasonable.

    Of course, all of those answers are in units of Newtons.
     
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