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Cable car system : Tension in the pull cable

  1. May 14, 2014 #1
    1. The problem statement, all variables and given/known data

    Figure shows a section of a cable car system. The maximum permissible mass of each car with occupants is 2800 kg. The cars, riding on a support cable are pulled by a second cable attached to the support tower on each car. Assume that the cables are taut and inclined at θ= 35°. What is the difference in tension between adjacent sections of pull cable if the cars are at the maximum permissible mass and are being accelerated up the incline at .81 m/s2.

    2. Relevant equations

    FNET=ma

    3. The attempt at a solution

    From figure,

    T2 = T1 +Mg sin θ
    T1 = Mg sin θ
    Thus, T2 = 2 Mg sin θ

    And Tn = n Mg sin θ

    Thus, T5 - T2 = (5-2) Mg sin θ = 3*9.8*2800*sin(35°)= 47216.8 N

    Is this answer correct?
     

    Attached Files:

  2. jcsd
  3. May 14, 2014 #2

    BvU

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    I read the exercise somewhat differently: adjacent sections of pull cable means you have to take e.g. T2 - T1, not T5 -T2 as you seem to be doing. Then: the pull cable is the one that provides the acceleration, so I would expect the acceleration should appear somewhere in the answer, isn't it ?
     
  4. May 14, 2014 #3
    Oh of course! I thought the same about the T2-T1 thing and then saw this
    https://www.physicsforums.com/showthread.php?t=291753
    and redid it.
    But yeah! The acceleration is in the answer, It skipped my mind soo bad when I redid it with the section things and all. :cry:
    So the free body diagram goes to this,
    and so the equation for Tn = n(ma+mg sinθ)
    And T5-T3= 2(2800)(.80+9.8 sin(35°))
    For Tn-Tn-1 = 2800(.80+9.8 sin(35°))

    So, i guess we could answer both! But also refer to the question,
    Figure shows a section of a cable car system. The maximum permissible mass of each car with occupants is 2800 kg. The cars, riding on a support cable are pulled by a second cable attached to the support tower on each car. Assume that the cables are taut and inclined at θ= 35°. What is the difference in tension between adjacent sections of pull cable if the cars are at the maximum permissible mass and are being accelerated up the incline at .81 m/s2.

    But then, there's be a chance to get at the implied meaning if someone has a solution manual of some sort to the Fundamentals of Physics- 8th edition extended by Halliday, Resnick and Walker..o have had it in class or something.. I guess there's no better way to know it than that!
     

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  5. May 14, 2014 #4

    BvU

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    I see. This has nothing to do with physics any more. It is clear you completely understand the situation. The formulation of the exercise is confusing, not intentionally, I suppose, but they could have written "tension between adjacent sections of three cars each" to keep it clear...

    They mention an acceleration of .81 m/s2. Is there a reason you write 0.80 ?
     
  6. May 15, 2014 #5
    No! It has to be .81! Clearly saturated with the problem i am now! :tongue2:
     
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