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Cable breaking under its own weight?

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data
    A scientist assembled large amount of cable. The longest piece of cable that is hung vertically and does not break under its own weight is of length k. He then places another piece of the cable on a friction-less table. A small part of the cable is hanging over one side and the cable begins to slide after the scientist released it. What is the maximum length of the piece of cable on the table if it is not to break while sliding?


    2. Relevant equations

    Force of gravity = m * a.

    Tension = Force of gravity

    3. The attempt at a solution

    I am at a total lost on this one. I assume that the part hanging over the table is infinitesimal at the beginning right?

    I don't need / want full answer, I need a hint on how to start on this problem.

    Thank you :)
     
    Last edited: Oct 24, 2011
  2. jcsd
  3. Oct 24, 2011 #2

    Delphi51

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    Welcome to PF!
    You know, this is very similar to the problem where you have one mass hanging from a string that goes over a pulley to another mass sitting on a table. You say the tension in the string is T and do F = ma for each mass, and so on. You can find the tension that way. In this case, you are pretty much given the maximum tension that the cable can handle. Hope this helps you get started!
     
  4. Oct 24, 2011 #3
    Can you tell me if I am heading in the right direction with this problem?

    Let the density of the cable be D, m be the mass of the cable of at length k.

    D = m / k

    m = D*k

    Force of gravity = m * g = D * k * g. Then this would be the breaking strength of the cable right?

    I am lost though on how this problem could be the same as a mass hanging from a string that goes over a pulley connected to another mass on the table?
     
  5. Oct 24, 2011 #4
    I got the length of the cable on the table is k, the same as the length of the cable as when it was hung vertically.

    That doesn't seem right does it?
     
  6. Oct 24, 2011 #5
    Ok so the length can't be k.

    But I have a question, would it be correct to say that when the cable breaks, the length of the cable that is still on the table has to be equal to the length of the cable hanging over the side?

    Thanks.
     
  7. Oct 24, 2011 #6
    As far as I can see, if the table is frictionless, it will never break, no matter how long it is, as it will be identical to freefall. Are you sure that's what the question said?
     
  8. Oct 24, 2011 #7

    Ibix

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    Jetwaterluffy - it's not identical to freefall because gravity is only accelerating the part that is hanging over the edge. That weight is pulling the part on the table.

    g.licata: It is very similar to the two-masses-and-a-string problem, except that as the cable slides the mass on the table gets lighter and the hanging mass gets heavier. Calculate the two masses as a function of length, and you look like you can handle it from there.
     
    Last edited: Oct 24, 2011
  9. Oct 24, 2011 #8
    Ibix, I'm not sure I understand what you just said.

    But if I were to assume that the length of the cable hanging over the table and the length of the cable still on the table to be the same, I get the maximum length to be 4k, can anyone confirm if this is correct?

    I have a feeling that the answer is wrong though :(
     
  10. Oct 24, 2011 #9

    gneill

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    Rather than ask for confirmation or denial for what could be "shot in the dark" guesses, why not post your method and working out of the problem so that it can be checked?
     
  11. Oct 24, 2011 #10
    I drew up a FBD and I got the equations:

    F_gravity - Tension = m_1 * a
    Tension = m_2 * a

    I solved for a, which is a = Tension / m_2

    Then I plugged that into the original equation.

    The breaking force for the cable is going to be Tension = density * k * g

    I plugged that into the same equation and subsituted m_1 and m_2 with density * length 1 and length 2 respectively

    The density then cancels out and I got:

    k = (length 1 * length 2) / (length 1 + length 2)

    That is where I am stuck, if I assume length 1 and length 2 are equal to each other, then I can continue and get the maximum length to be 4k.

    But like you said, I don't see a reason why the hanging part of the cable has to be equal to the part still on the table, so that assumption was a shot in the dark.
     
  12. Oct 24, 2011 #11

    gneill

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    Suppose that the total length of the cable is L and the part hanging over the edge is x. That leaves L-x on the table. You can then write your expression for the tension in the cable at the table's edge in terms of L and x. Can you determine the value of x where the tension is maximum?
     
  13. Oct 24, 2011 #12
    I tried using what you said but how would I solve for x? Do I take the derivative of Tension with respect to x?
     
  14. Oct 24, 2011 #13

    Ibix

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    Following gneill, you should be getting an expression for the tension in the cable in terms of x and L. Set this equal to the breaking strain of the cable, and solve for x. What is the lowest value of L for which there is a solution?
     
  15. Oct 24, 2011 #14

    gneill

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    Yes, take the derivative of the tension in the sliding cable with respect to x. Set to zero and solve for x. With that value of x the tension in the sliding cable will be maximum. So plug in that value of x into the equation for the tension yielding the maximum tension for a sliding cable of length L. Then it's time to equate that expression to the maximum allowable tension and solve for L in terms of k.
     
  16. Oct 24, 2011 #15

    Delphi51

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    Oh dear, I missed all the fun overnight. A fascinating problem with a nice simple final answer. This morning I tried it, made a dumb mistake and then read gneill's post instead of toughing it out, so I'll never know if I could have got it on my own.

    G.licata, you had a remarkable insight in post 8. Were you thinking the tension is due to the inertia of the horizontal part and the weight of the vertical part in equal measure? It might have been easier to work with just the masses, never mind the lengths.
     
    Last edited: Oct 24, 2011
  17. Oct 24, 2011 #16

    Ibix

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    I must admit that I didn't bother with the differentiating. I just set the tension as a function of x equal to the critical tension, solved for x, and noted that there are no real solutions for L<4k. But either way gets you there.
     
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