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What is the tension in the right cable?

  1. Mar 12, 2015 #1
    1. The problem statement, all variables and given/known data
    2 cables are used to support a 24kg mass on a 1.6m long, 8kg uniform horizontal beam. The 24kg mass is 0.6m from the right cable. What is the tension in the right cable?

    2. Relevant equations
    torque = rFsinx

    3. The attempt at a solution

    Fy = ma = 24(9.8) = 235.2N
    Fx = ?
    resultant F - ?
     
  2. jcsd
  3. Mar 12, 2015 #2

    billy_joule

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    There is no Fx forces acting if I'm visualizing it correctly.A picture would help.
    Take moments about the left end, solve for the one unknown force - the rhs cable tension.
     
  4. Mar 12, 2015 #3
    Is the object moving? If so or if not, what does this tell you about the net force.
     
  5. Mar 12, 2015 #4

    BvU

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    And are the cables vertical? Where are they attached to the beam? At the ends ?
     
  6. Mar 12, 2015 #5

    haruspex

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    There is some confused thinking here:
    Fy = ma should be ##\Sigma F_y = ma_y##, i.e. the net force in the Y direction gives rise to an acceleration in the Y direction.
    It is not clear whether you are considering forces on the beam, forces on the mass, or forces on the two as a system. Whichever, 24(9.8) is only one force in the Y direction, not the net force.
    I assume the cables are vertical and attached to the ends of the beam. If so, follow billy_joule's advice. You can get the answer without using ##\Sigma F_y = ma_y##.
     
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