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Homework Help: Cable Tension problem

  1. Oct 28, 2007 #1
    An 8.9kg watermelon is placed at one end of a 6.1m, 238 N scaffolding supported by two cables. One supporting cable is at the opposite end of the scaffolding, and the other is 0.76 m from the watermelon. The acceleration of gravity is 9.8 m/s^2.

    How much tension is in the cable at the end of the scaffolding?
    How much tension is in the cable closest to the watermelon?

    I really have no clue how to go about this. Any relevant formulas I should know?
  2. jcsd
  3. Oct 28, 2007 #2


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    The system is in equilibrium. Sum of forces in y direction and sum of torques about any point must equal zero. Don't forget the torque from the scaffolding .
  4. Oct 28, 2007 #3
    For equilibrium of a body:
    1. ΣF = 0
    2. ΣT = 0
    where, F is the force and T is the torque.
    (bold ones are vector quantities.)
  5. Oct 28, 2007 #4
    ok well with the watermelon converted over to N i've got a total of 325.28N * 9.8m/s^2 = 3187.744 for my downward force. If that is correct then how do I apply the distance from the melon to figure the tension on the individual cables?
  6. Oct 28, 2007 #5
    Remember: Always try to be atleast dimensionally correct!
    How come, "325.28N * 9.8m/s^2 = 3187.744 for my downward force"?? You must be knowing, unit of force is N.. then how, N * m/s^2 = N??
    In fact, if tensions in the strings are T1 and T2, then
    T1 + T2 = 238N + 8.9kg*9.8m/s^2 = 325.22N.

    Now, take torque about any point and equate to zero to get one more equation involving T1 and/or T2. Then, solve for T1 and T2 from the two equations obtained. It would be quicker if you take torque about the point where the cable meets scaffolding (either of them).
  7. Oct 28, 2007 #6
    ah ok sorry about that. Thank yall very much for the assistance.
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