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Newton's Second law: Tension on Cable

  1. Feb 27, 2017 #1
    Hey guys is my solution correct for the question below?

    1. The problem statement, all variables and given/known data

    Consider a 5.0 kg watermelon that is being accelerated at 2.0 m/s2 [up] by a cable. Find the tension in the cable.

    2. Relevant equations
    Fcable = m.a

    3. The attempt at a solution
    Fcable = m.a
    = 5.0 kg(2.0)
    = 10N
     
  2. jcsd
  3. Feb 27, 2017 #2

    kuruman

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    Your solution is incorrect. Newton's second law says Fnet = ma. Fnet is the sum of all the forces acting on the watermelon, not just the force from the cable.
     
  4. Feb 27, 2017 #3
    Ok I am confused now. I thought Fnet = m.a
     
  5. Feb 27, 2017 #4
    Or in this situation should it be Fcable +Fgravity
     
  6. Feb 27, 2017 #5

    kuruman

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    Yes, it should be as you say. Can you finish the problem now?
     
  7. Feb 27, 2017 #6
    Fcable = 10 N
    Fgravity = -49N

    Fcable + Fgravity = 10N - 49N
    = 39 N [down]

    Is this correst?
     
  8. Feb 27, 2017 #7

    kuruman

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    How do you figure Fcable = 10 N? It seems that you multiplied the acceleration by the mass and called that Fcable. We agreed that mass times acceleration is the net force. Write an expression for the net force as the sum of all the forces, set it equal to mass times acceleration and solve for Fcable.
     
  9. Feb 27, 2017 #8
    I think I get it now.

    So Fnet = Fcable + Fgravity
    10N = Fcable - 49N
    10N + 49N = Fcable
    Fcable = 59N
     
  10. Feb 27, 2017 #9

    kuruman

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    Yes, you got it. :smile:
     
  11. Feb 27, 2017 #10
    Thanks for your help.
     
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