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Finding the maximum tension in a cable

  1. Jul 17, 2012 #1
    I was trying to solve this question:

    In a scrap yard, a car of mass 1200 kg is suspended from a crane using an electromagnet of mass 1400 kg.When the electricity is turned off the car is dropped. Assuming the spring constant for the crane and the supporting cable is 3.5 MN/m, find the maximum tension in the cable after the car is dropped.

    Please, correct me if I am wrong, the tension in the wire after the car is dropped is acting upwards right?
    So the free body diagram would have the tension pointing upwards and the weight of the magnet pointing downwards?
    My lecturer has done this Fmax = kx + m(of the magnet)g
    I dont understand how he got this.
    An explanation would be very helpful
    Thanks!!
     
  2. jcsd
  3. Jul 17, 2012 #2

    TSny

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    Yes, the tension force acting on the magnet is upwards immediately before and immediately after the car is dropped.

    Yes

    I don't understand it either. If Fmax is intended to represent the total force on magnet, then it appears that there's a sign error because, as you noted, the tension force (kx) and the weight act in opposite directions.

    Let's forget that formula for now and start with a basic question: What is the tension in the cable just before the car is dropped? You have the data to determine this and I think the answer to this question will place us well on the way to getting the final answer to the problem.
     
    Last edited: Jul 17, 2012
  4. Jul 17, 2012 #3
    Thanks TSny...
    So basically what you are saying is that Fmax is equal to the difference between the tension right before the car is dropped and the weight of the magnet??
    Since that tension is greater than the weight of the magnet, the resultant(Fmax) would point upwards like we agreed
     
  5. Jul 17, 2012 #4
    In case you misunderstood. what my lecturer got for Fmax is actually the maximum tension that he found. thats all he had done.
     
  6. Jul 17, 2012 #5

    TSny

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    Still can't see where he's going with it.

    Anyway, can you answer the question I gave you? It really will get you very close to the final answer (unless I'm misunderstanding the statement of the problem.)
     
  7. Jul 17, 2012 #6
    Well I calculated the tension just before the car is released to be 25506 N.
    i did g*(mass of magnet + mass of car)
     
  8. Jul 17, 2012 #7

    TSny

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    Good. Now, what do you think the tension in the cable will be immediately after the car is released?
     
  9. Jul 17, 2012 #8
    Immediately after its released the tension would be the same as before for a split second. this is probably why the magnet would be suddenly pulled upwards. Then the cable will become slack...is that correct?
     
  10. Jul 17, 2012 #9

    TSny

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    Yes, the tension right after release will still be 25.5 kN and then the mass will start moving upward with the tension decreasing. Can you describe the subsequent motion? (The cable will not actually go slack.)

    Can you now see the answer to the original question?
     
  11. Jul 18, 2012 #10
    The motion is Simple Harmonic Motion. So the final answer is 25.5 kN?
     
  12. Jul 18, 2012 #11

    TSny

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    Yes, I believe that's the answer unless I've misunderstood the problem. The value of the spring constant is not needed.

    I still don't understand the equation that your lecturer wrote. Not sure what value of x he intended to substitute into the equation. To get an answer of 25.5 kN, x would need to be the distance that the car alone would stretch the cable.
     
  13. Jul 18, 2012 #12
    Oh my lecturer has found the value for x in an earlier question.

    X = F/k = (1200*9.81)/3.5*10^6 = 3.36 mm
    and he said this value is the static deflection of the loaded system from equilibrium. Which means he is considering the cable and magnet to be one system. So yeah he is considering X to be the elongation of the cable solely due to the weigh of the car.
     
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