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Homework Help: Equillibrium on a beam

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data

    A 20.0 m long uniform beam weighing 670 N is supported on walls A and B, as shown in Fig. 9-77.
    http://www.webassign.net/giancoli5/9_77.gif

    (a) Find the maximum weight a person can be to walk to the extreme end D without tipping the beam.
    (b) Find the forces that the walls A and B exert on the beam when the same person is standing at point D.
    (c) Find the forces that the walls A and B exert on the beam when the same person is standing at a point 2.0 m to the right of B.
    (d) Find the forces that the walls A and B exert on the beam when the same person is standing 2.0 m to the right of A.

    I figured out A + B but not C and D.


    2. Relevant equations

    sigma Fy = 0
    sigma FTorque = 0


    3. The attempt at a solution


    FA = beam A
    FB= beam B
    mg= mass of beam * 9.8
    FC= the person


    a) 670N
    b) 0 N from wall A and 1340N from wall B
    C) using FA as the pivot: 0FAsin90 + 12FBsin90 -8.5mgsin90 - 15FCsin90 = 0
    12FB - 15745 = 0
    FB = 1312.08 WHICH IS NOT CORRECT.
    D)
     
  2. jcsd
  3. Jan 17, 2010 #2

    PhanthomJay

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    You have your dimensions wrong. If summing torques about A, the distance fron the cg of the beam to A, for example, is 7 feet. Also, you don't need m and g, since the force units of weight are already given.
     
  4. Jan 17, 2010 #3

    ideasrule

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    What's FC? Where did you factor in the weight of the beam or the weight of the person?
     
  5. Jan 17, 2010 #4
    FC = the force of the person on the beam as mentioned in the question.

    The weight of the beam is mg.

    Thanks.
     
  6. Jan 17, 2010 #5
    Thanks jay. Your correct the mg is 7 feet to the right of the FA, or beam A. When i wrote mg i was referring to the Force units.

    How do i go from here?

    Thanks!
     
  7. Jan 17, 2010 #6

    PhanthomJay

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    You do have parts 'a' and 'b' correct. For part 'c', try summing torques about point A, and solve for the unknown reaction force at B. Then solve for the unknown reaction force at A by summing forces in the y direction = 0. As a check, sum torques about point B = 0 to determine the reaction force at A. You can solve part 'd' in a similar fashion.
     
  8. Jan 23, 2010 #7
    Thanks Phanthom Jay!!

    Could you take a look at this: https://www.physicsforums.com/showthread.php?p=2544626

    Thanks again.
     
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