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Center of mass and Tension

  1. Feb 19, 2010 #1
    1. The problem statement, all variables and given/known data
    A man is standing 1 meter in from the left of a 4 meter long beam. The beam is held up by two wires perpendicular to the beam. The beam is 750 N and the man is 100kg what are the tensions of the two wires.

    2. Relevant equations

    F = ma

    Torque = F x d

    3. The attempt at a solution
    I know that because the beam is not accelerating the sum of the forces is 0

    so Fy = -750 - 980 + T1 + T2 = 0

    T1+T2 = 1730

    T2 = 1730 - T1

    Torque (I put my pivot point at the first tension wire making it 0)

    Torue = F x d = T2 * 4Sin(90) - 1730

    T2 *4 = 1730

    T2 = 423.5

    1730 - 423.5 = 1306.5 = T1

    Now the book says 1130 T1 and 610 T2

    I know for sure i'm missing something with center of mass im just not sure what.
     
  2. jcsd
  3. Feb 19, 2010 #2

    PhanthomJay

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    yes
    what is this? Are you taking the torque from the man and beam weights as (1730 N)*(1 m)? The man's weight acts at 1 m, but the beam's weight does not act there.
    You are not summing moments correctly. Where does the beam's weight of 980 N act on the beam ( how far from the left end?)?
     
  4. Feb 19, 2010 #3
    so i should use Torque = -980 N (man)*1 meter (away from my pivot point of the left end of the beam) * sin(90) making it -980

    and then add that with the -750N (beam) * 2meters (center of mass from the beam because it is uniform so it's l/2 or 4/2)

    and that would leave T2* 4 meters because it's on the other end of the beam?
     
  5. Feb 19, 2010 #4

    PhanthomJay

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    yes, good, add em up and solve for T2, then T1 will follow. As a check on your work, try summing moments about T2, the right end. The number you get for T1 should agree.
     
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