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Tug of war... Finding oppostite person's force

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Don't know where to start cause this is part of Newtons laws apparently but seems to be more varied around his second law... Just can't figure out what equation to use.. Question : Rhys and Gareth are having a tug of war with 1.25kg of climbing rope. If Rhys pulls with a force of 24N and the rope accelerates away from him at 1.4m/s2, with what force is Gareth pulling the rope ?

    2. Relevant equations
    ΣNet = m*a
    Fnet= FA + FF

    3. The attempt at a solution
    I'm only assuming that you have to take the mass of Rhys and use that plus the weight of the rope to your advantage..
     
  2. jcsd
  3. Nov 19, 2015 #2

    SteamKing

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    Hmm ... It's too bad that you don't have the mass of Rhys (or Gareth) to plug into your equation. :frown:

    However, if you read the problem statement carefully, you will find that neither Rhys nor Gareth are doing any movement.

    When Rhys pulls with 24 N, what accelerates away from him? Is Rhys accelerating away from himself? Is that possible?
     
  4. Nov 20, 2015 #3
    I figured it out.. Don't know if its the right way.. But I did: 24N / 1.4m/s2 to find mass... Because m= Fnet/a... And seeing that this seems to be the only force mentioned.. And since it never mentioned them moving. That equaled to 17kg... 17kg+ 1.25kg( weight of rope) seeing that he is pulling the rope and Rhys. Then I got 18kg... 18kg * 1.4m/s2 = 26N... That was the answer in the book too.... Haha now I'm dealing with a much harder question
     

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  5. Nov 20, 2015 #4

    SteamKing

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    You were already given the mass of the rope (1.25 kg). You were asked to find out the force with which Gareth is pulling on the rope.

    What, it was too complicated to write a net force equation involving Rhys and Gareth?

    A stopped clock is right twice a day, too. That doesn't mean it's right all the time.
    Then you should start a new thread for this question rather than hijacking your own thread for the first problem.
     
  6. Nov 20, 2015 #5
    Ahhh.... Too many questions and a important test tmr... I don't have time to post now... I have to figure it out on my own
     
  7. Nov 20, 2015 #6

    CWatters

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    Your relevant equations are correct. You are given the mass and acceleration so you can work out fnet. You are given one of two forces so can work out the other.
     
  8. Nov 20, 2015 #7
    Have you drawn a free body diagram of the rope, or do you feel like you've advanced in physics beyond the need to draw free body diagrams?
     
  9. Nov 20, 2015 #8
    I just Started physics 3 months ago... II am only assuming it takes very much practice to move on from free body digrams
     
  10. Nov 20, 2015 #9
    Excellent. So, based on your free body diagram, please write out your Newton's 2nd law force balance for the rope.

    Chet
     
  11. Nov 20, 2015 #10
    I'm just confused with the weight of the rope... Cause isn't it possible that there is gravity beeing exerted onto it... And wouldn't that have an impact on the result
     

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  12. Nov 20, 2015 #11

    SteamKing

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    Haven't you ever been in a tug of war contest?

    The weight of the rope is irrelevant to the force of the pull being applied by the people on each end of the rope.
     
  13. Nov 20, 2015 #12
    No. Neglect the weight. That's in the vertical direction, and we're doing a force balance in the horizontal direction. So......
     
  14. Nov 20, 2015 #13
    By the way, your free body diagram is correct.

    Chet
     
  15. Nov 20, 2015 #14
    Ye so I will figure out when I can. I'm guessing it's just gonna be Netforce - Rhys = gareth in some form or not
     
  16. Nov 20, 2015 #15
    I have no idea what Rhys and gareth are.

    Chet
     
  17. Nov 20, 2015 #16
    Ye woops lol that wasn't really meaning that I meant more Fnet = F( of Gareth) - 24N
     
  18. Nov 20, 2015 #17
    Excellent. And, in terms of the mass of the rope m and the acceleration of the rope a, the net force is equal to ???
     
  19. Nov 20, 2015 #18
    Well Fnet = m*a
    a= Fnet/ m
    M= Fnet/a

    So Fnet = 0.400* 1.4 = 0.56
     
  20. Nov 20, 2015 #19

    SteamKing

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    Wrong problem, Mrchilko.
     
  21. Nov 20, 2015 #20
    The problem statement tells you that the mass of the rope is 1.25 kg and its acceleration is 1.4 m/s2. So what does this give you for the net force?

    Chet
     
    Last edited: Nov 20, 2015
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