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Stretch of Rope

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A Tugboat T, having a mass of 19 Mg is tied to a barge B having a mass of 75 Mg. If the rope is elastic such that it has a stiffness k=600kN/m, determine the maximum stretch oin the rope during the initial towing. Original both the tugboat v = 15 km/hr, and the barge v = 10 km/h. Neglect resistance of water

    2. Relevant equations

    3. The attempt at a solution

    F=ma yield 0 for T and B. Not sure what to do
  2. jcsd
  3. Apr 15, 2009 #2


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    From the "neglect resistance of water" comment, I'm guessing this is a conservation-of-energy problem.
  4. Apr 27, 2009 #3
    ok Conservatin of Energy

    T(t)+V(t) = T(b)+V(b)

    T(t) = .5mv^2 where m = 19Mg & v = 15 km/h
    = 2137.5
    V(t) = .5ks^2 where k = 600 kN/m & s = unknown

    T(b) = .5mv^2 where m = 75 Mg & v = 10 km/h
    = 3750
    V(b) = .5ks^2 where k = 600 kN/m & s = unknown

    I'm lost after this. I dont belive my V's are correct. Plus which variable is the answer the problem question - How much will the rope stretch?
  5. Apr 27, 2009 #4


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    You've taken a small step in the right direction.

    There's no reason for setting the energies of the tugboat and barge equal to each other, as it seems you're trying to do. Instead, set the total energy of everything (tugboat, barge, + rope) before the rope is stretched, equal to the total energy of everything once the rope has reached it's maximum stretched length.

    Take a look at the ".5 k s2" you wrote. What does that term, as well as k and s, represent?
  6. Apr 27, 2009 #5
    ok your saying

    initial T(t)+T(b)+V(t)+V(b) = final T(t) + T(b) + V(t) + V(b)

    V = .5ks^2 is the potential energy where k = the rope stretch and s = the length of the rope

    Since there is no mention of the ropes length, would the initial s values =0, and the final s value is the answer to the problem.
  7. Apr 27, 2009 #6


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    That's better, but the rope's potential energy should be part of the equation too.

    Yes, it is the potential energy of the rope. But no, k and s are not what you are saying. Try looking up the potential energy of a spring in your textbook, k and s have the same meaning for the rope as they do for a spring.

  8. Apr 28, 2009 #7


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    Another hint: s is the stretch of the rope (change in it's length), so the problem is asking for the maximum s.
  9. Apr 28, 2009 #8
    I was under the assumption that the V's where the ropes potential energy.

    What is the equation for V(t) & V(b) then?

    So to find the stretch of the rope i would need to find the initial s and final s. the difference of the two is the answer?
  10. Apr 28, 2009 #9


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    What are "t" and "b" referring to in V(t) and V(b), if the V's are supposed to be for the rope?

    The rope's potential energy is ½ks2, as you said before, where s is the stretch of the rope.

    Initally the rope is unstretched, so s=____? initially.
  11. Apr 29, 2009 #10


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    Perhaps another way to look at this problem is as follows:

    What specific forms does energy take in this scenario? Answer to that is: kinetic energy for the tugboat, kinetic energy for the barge, and potential energy due to the rope stretching. So there will be three terms contributing to the total energy.
  12. Apr 29, 2009 #11
    Ok, i think this is how i should set it up

    t = tugboat, b = barge, T = kinetic energy, .5mv^2, and V = potential of rope, .5ks^2

    Initial T(t) + T(b) + V = final T(t) + T(b) + V
    But wouldnt everything just cancel out except for the V's.

    I believe that the unstretched ropes has a s=0, which would make the final s = 0 too. what am i doing wrong
  13. Apr 29, 2009 #12

    Doc Al

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    Why would everything cancel? Are the initial and final KEs the same?
    The final state is to have the rope maximally stretched, so how could the final s = 0?

    Hint: You'll need more than energy conservation to solve this. What else is conserved?
  14. Apr 29, 2009 #13
    Momentum is conserved as well. That what i thought from the begging but redbelly said it was a conservation of energy problem. I guess i took it to literaly.

    Conservation of Momentum

    initial m(t)v(t) + m(b)+v(b) = final m(t)v(t) + m(b)v(b)

    19*15 + 75*10 = 19v + 75v

    final v = 11.0106... km/h

    So now i can used Conservation of Energy with my new final


    T(t) = .5mv^2 = 2137.5
    T(b) = 3750
    V = .5ks^2 = 0

    Finals (using the new v)

    T(t) = 1151.71647...
    T(b) = 4546.24921...
    V = 300s^2

    sum of initial = sum of final
    solve for s

    5887.5 = 5697.96568 + 300s^2 therefor s = .7948...m (my fingers are corssed)
  15. Apr 29, 2009 #14

    Doc Al

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    I suggest that you redo your calculations using standard units: m in kg, v in m/s, k in N/m.
  16. May 1, 2009 #15
    ok good call, I got a much different answer

    m = 19 Mg and 75 Mg.. 19000 kg and 75000kg
    v = 15 and 10 km/h... 4.1666666 and 2.777777 m/s
    k = 600 kN/m... 600000 N/m

    19000(4.166666) + 75000(2.77777) = v(19000+75000)
    v = 3.0585106 m/s

    .5*19000*4.1666^2 + .5*75000*2.7777^2 + .5*600000*0 = .5*19000*3.059^2 + .5*75000*3.059^2 + .5*600000*s^2

    s = .2207 m is this right
  17. May 2, 2009 #16

    Doc Al

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    Looks OK to me.
  18. May 2, 2009 #17
    great, thanks for all ur help
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