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Wave travel along a tensioned rope

  1. Mar 21, 2007 #1
    Oliver (m=90.0 kg) uses a 4.60 m long rope to pull Jordan (m=53.6 kg) across the floor (μk=0.200) at a constant speed of 1.14 m/s. Jordan signals to Oliver to stop by "plucking" the rope, sending a wave pulse forward along the rope. The pulse reaches Oliver 149.0 ms later. What is the mass of the rope?

    I am doing m= ukmg*L\v*squared but it is not working, am I suppose to find a new velocity or something. Any help appreciated.
     
  2. jcsd
  3. Mar 21, 2007 #2
    Why do you say it is not working? Where did you get that equation and what do the variables mean?
     
  4. Mar 21, 2007 #3

    Mentz114

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    Work out the tension in the rope by calculating the force it takes to overcome the friction. Then use the relation between wave speed, mass, tension and length to get the mass.
     
  5. Mar 21, 2007 #4
    I did exactly what you are telling me and I am not getting the answer. I am doing M=ukmg*L\v*squared and ukmg is my Force tension that I calculated. These are my values that I am putting in
    uk= 0.2000
    m= 53.6 kg
    g= 9.8
    v= 1.14
    L=4.60

    It seems like a very easy question but I don't know where I am messing up.
     
  6. Mar 21, 2007 #5

    Mentz114

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    I'm confused by your symbols. What is M ? You must explain what your symbols mean, otherwise it's like a riddle, we have to guess. For instance I'm guessing the uk is not a country but a coefficient of friction.
     
  7. Mar 21, 2007 #6
    its mass...
     
  8. Mar 21, 2007 #7

    Mentz114

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    Well, that does not make sense because you've got m on both sides of your equation. I don't recognise the formula

    M=uk*m*g*L\v^2

    so I'm stuck also. You'll have to hope someone else does since you haven't explained it.
     
  9. Mar 21, 2007 #8
    ok i got that from the formula: vsquared = T\u and u= m\L (mass over length) and these are equations in my physics textbook used for waves.
     
  10. Mar 22, 2007 #9

    Doc Al

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    There's nothing wrong with that equation, which is derived from the formula for the speed of a wave (or pulse) on a stretched rope:
    [tex]v^2 = \frac{T}{\rho}[/tex]

    Where T is tension in the rope (which equals the friction force [itex]\mu m_{person} g[/itex]) and [itex]\rho = m_{rope}/L[/itex].

    I'll rewrite that equation like this:

    [tex]m_{rope} = \frac{\mu m_{person} g L}{v^2}[/tex]


    v should be the speed of the pulse, not the speed of the person. Calculate the speed of the pulse from the data given in the problem.
     
  11. Mar 22, 2007 #10
    there is no other way to find a new velocity it just takes me back to square one...I keep on getting 0.372 grams for my mass of rope...which other way can I find velocity
     
  12. Mar 22, 2007 #11
    Question: Would the speed of pulse to an observer watching Oliver and Jordan be the sum of the speed of the rope and the speed of the pulse if the rope where not moving?
     
  13. Mar 22, 2007 #12

    Doc Al

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    To find the speed of the pulse just use the distance it travels (the length of the rope) and the time it takes (which is given).

    Yes.
     
  14. Mar 22, 2007 #13
    k got it, just a calculational error on my part. Thanks again for the clarification...:)
     
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