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Girl on a swing, find tension and force

  1. Apr 27, 2017 #1
    1. The problem statement, all variables and given/known data
    The girl weighs 30kg and her COM is at G. As she is passing the lowest position of the swing with rope length L=4.5m her speed is 3.6m/s. The swing and rope have no mass.

    Calculate
    a) the tension in the rope
    b) the force P that is aligned along the girl's arms

    see the picture below
    IMG_0032.jpg
    2. Relevant equations
    mechanics

    3. The attempt at a solution
    This seems too easy...
    IMG_0033.jpg
     
  2. jcsd
  3. Apr 27, 2017 #2
    If she was sitting on a stationary swing, the tension would be equal to the weight. However, this swing has a speed of 3.6 m/s so that equation is no longer true. The greater her speed, the greater the tension. You need something to relate those 2 parameters.
     
  4. Apr 28, 2017 #3
    Ah of course, the centripetal force! I haven't done any physics since I graduated college and I'm trying to remember all the stuff that I learned years ago :).

    If I call the centripetal force Fc= mv^2 / L

    Then T= Fc + mg

    and

    P = T / cos(30)

    correct? :)
     
  5. Apr 28, 2017 #4

    collinsmark

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    That looks like a good start. :smile:

    Be careful on part (b). Examine Figure 8 carefully. The force P -- the force aligned along the girl's arms -- is not the only force with an upwards component.
     
  6. Apr 28, 2017 #5
    Is the centripetal force also pointing up in case B then? This is where I always got confused with the centripetal force, its direction almost seems arbitrary.
     
    Last edited: Apr 28, 2017
  7. Apr 28, 2017 #6

    collinsmark

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    Yes, it's directed up; she is accelerating in the up direction.

    Regarding the tricky part I mentioned in my last post:

    Consider the forces above and below where her hands are on the rope.

    Above her hands, the tension on the rope is T. I think we all agree on that.

    But what about below her hands? You know that the force P, acting along the girl's arms, contributes to the tension. But is that all? Is there anything else?
     
  8. Apr 28, 2017 #7
    I think I was thinking about it wrong, I think I was approaching it as a statics problem where the sum of the forces is equal to zero when I should have thought about it as Fc=T-mg. It's starting to make more sense now.

    As for forces below her hands... Hmm, would it be mg? That's the only one I can think of.

    edit: but that would be down....

    I really don't know, I can't think of another force that's pointing up.
     
  9. Apr 28, 2017 #8

    collinsmark

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    :smile:
    What about the part of rope below her hands -- the part attached to the base of the swing? :wink:
     
  10. Apr 28, 2017 #9

    haruspex

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    Consider the joint where her hands meet the rope. The only acceleration at the lowest part of the swing is centripetal, which is straight up. So the net force acting on that joint must be straight up. The tension in the rope above the joint acts straight up, but that in her arms is not straight down. What force on that joint completes the picture?
     
  11. Apr 28, 2017 #10
    There would be another tension force in that rope, but wouldn't it be facing down?
     
  12. Apr 28, 2017 #11

    haruspex

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    It would be pulling down on that joint, yes. Why is that a problem? What will it do at the other end of that rope section?
     
  13. Apr 28, 2017 #12
    Equal but opposite. But collinsmark mentioned something about another force pointing up, I do not see it. And even with another force of tension, I now have too many unknowns to solve the problem. I don't know what the acceleration in x is. I have unknowns T2, P and Fx.
     
  14. Apr 28, 2017 #13

    haruspex

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    which is what that lower section tension will do at its lower end.
    You have more unknowns, but also more equations. There is a force balance at the joint with the hands. You can ignore the mass of the hands, so you do not need to consider centripetal force in that one.
     
  15. Apr 28, 2017 #14
    No I'm completely lost now. This doesn't make any sense to me anymore.

    At the point of contact with the hands, T2 would be pointing down, no?

    Seen from the perspective of just the swing, it would be pointing up, but would mg even exist since it's already taken up by the girl? And if the swing has no mass, why would there then even be a T2?
     
  16. Apr 28, 2017 #15

    collinsmark

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    Maybe it's easier to think of things in terms of tensions.

    The vertical component of the tension in her arms together with the vertical component of the tension in the rope below her hands must combine to a total, vertical force T.

    Recognizing that, you could draw a free body diagram showing the forces [on the girl].

    Does that help?
     
    Last edited: Apr 28, 2017
  17. Apr 28, 2017 #16
    qCPIsLl.jpg

    But I have too many unknowns now.
     
    Last edited by a moderator: Apr 28, 2017
  18. Apr 28, 2017 #17

    haruspex

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    I think you are confused by the dual role of the arms as both a tension member, whose tension is of interest, and as part of the girl. To clarify matters, replace the arms by another section of rope. So you have one vertical rope at the top joined to two angled ropes at a joint. Those two ropes go to different parts of what remains of the girl (which you can treat as a rigid body and the only mass present).
     
  19. Apr 28, 2017 #18
    Okay that's helpful, but I still don't see how the T2 helps me at all. The problem says that the swing and rope are massless. So in the FBD of just the lower part of the swing, what is the force pulling it down? What is counteracting T2? How do I find T2?
     
  20. Apr 28, 2017 #19

    collinsmark

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    I would draw the FBD considering the forces acting on the girl.

    GirlOnSwingFBD2.gif

    What must the vertical components of T2 and P add to? (Hint: it's not mg. She's accelerating up, remember. But it is something you've already calculated.)
     
  21. Apr 28, 2017 #20
    that makes a lot more sense. For some reason I was having a really hard time drawing a fbd of the girl.

    So then I get

    mv^2/L = T - T2cos30 -Pcos30
    mg=T2cos30 + Pcos30

    correct?
     
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