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Statics Problem - Moment on a Bridge Help Please!

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data

    The uniform work platform, which has a mass per unit length of 28kg/m, is simply supported by cross rods A and B the 90-kg construction worker starts from point B and walks to the right. At what location S will the combined moment of the weights of the man and platform about point B be zero?

    Platform illustration:

    __A____________B___man (going to the right)___

    Mass per unit length of bridge: 28kg/m
    Mass of man: 90kg

    Distance from far left end of bridge to A: 1m
    Distance between A and B: 4m
    Distance from B to far right end of bridge: 3m
    Distance from B to man: S

    2. Relevant equations

    M=d*F or M=r*F (cross product)

    3. The attempt at a solution

    What I did was that I set point B to be (0,0) so any distances to the left of B is (-) and to the right will be (+). Also, I set all the weights to be (-) since it's going down the (-) y-axis.

    Next I assigned the F from B to Man to be (90+28S) and the F from B to the remaining distance from Man to far right end of the bridge to be 28(3-S).

    For the forces on the left side:

    left end to A: 28
    between A and B: 112

    Next, I used M=d*F and set the moment of B from the left and right side to equal each other which leaves me with:

    588=-90S-28S^2+(3-S)(-8.4+28S)
    840=78S-56S^2
    0=78S-56S^2-840
    0=9.8(78S-56S^2-840) <-- to calculate weight
    0=548.8S^2-764.4S+8232

    When I try to solve for S, I use the quadratic formula but I can't solve for S because the number in the sqrt is (-) thus I can't solve for S. Am doing something wrong with the signs or is my approach completely wrong? Thanks for your help!
     
  2. jcsd
  3. Sep 5, 2009 #2
    Do you know the solution? Maybe you should look at the platform as a seesaw without reaction in A. In that situation you would only have reaction in B plus the weight of the platform and the man. It would be easy to calculate the moment around B.
     
  4. Sep 5, 2009 #3

    I don't have the solution sorry. What do you mean by reaction in B? If I disregard A does that mean that I'll just have a combined distance of 5 from the left side to point B and just calculate the total weight from that side? Thanks again.
     
  5. Sep 6, 2009 #4
    I've attached a picture to explain what I mean. On my picture, force A i faced downward. That is a situation when this guy walks far enough to the right (over distance S as I have assumed). If this guy stands close enough to the point B, force A is facing upward. So I guessed that there has to be a situation when force A becomes zero before shifting it's orientation. My assumption is that that will happen when distance is S. I hope that somebody will answer with better idea.
     

    Attached Files:

  6. Sep 7, 2009 #5
    Oh ok, at this point I thought that I can just ignore the forces on A b/c the question was asking about B. I'm still a little confused...sorry. It's a little hard to explain this via text only...lol
     
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