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Homework Help: Tension on a rope from jumping tightwalker

  1. Mar 3, 2009 #1
    A 65.0 kg tightrope walker stands at the center of a rope. The rope supports are 10 m apart and the rope sags 10.0^\circ at each end. The tightrope walker crouches down, then leaps straight up with an acceleration of 7.70 {\rm m}/{\rm s}^{2} to catch a passing trapeze. What is the tension in the rope as he jumps?

    F=ma
    Forces, basically.

    So, based on the information, I think it's correct to assume because of the tightwalker jumping upwards then the Force from the tightwalker is slackened to 9.8(gravity) - 7.70(acceleration upwards).
    From there: F=Fysin([tex]\theta[/tex])+Fysin([tex]\theta[/tex]) - ma (from the tightwalker) = 0

    I plug it in and get 180.71 and it's not the correct answer. I'm doing something wrong here.
     
  2. jcsd
  3. Mar 3, 2009 #2
    "because of the tightwalker jumping upwards then the Force from the tightwalker is slackened"

    They want the tension as he jumps, before he leaves the rope.
    Hint: The tightrope walker pushes down to jump up, putting more tension on the rope.
     
  4. Mar 3, 2009 #3
    Maybe I am wrong, but I think the 7.7 m/s^2 that is given is the absolute acceleration of the person. So, the acceleration and mass are both given, which makes it a simple calculation. Using those numbers do you get the correct answer?
     
  5. Mar 3, 2009 #4
    ty Imperitor i got it

    it doesn't slacken as he jumps - it adds more force to the equation
    therefore, it = ma(jumper) + m(7.7)(jumper adding force)

    and if i plug those in it yields the correct answer :D
     
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