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Pulley Problem!

  1. Jul 10, 2008 #1
    Sorry if my attempt is real pathetic, I'm not that great at physics...

    1. The problem statement, all variables and given/known data
    A girl sits in a basket supported by a string passing over a pulley. The girl pulls on the free end of the rope to lift herself.

    (a) If the mass of the girl and the basket together is 95 kg, with that force must she pull to raise herself at a constant speed?

    (b) With what force must she pull to achieve an upward acceleration of 1.3 m/s2?


    2. Relevant equations

    (a) F = ma = mg

    (b) F = ma

    3. The attempt at a solution

    (a) Fweight = mg = (95 kg)(9.8 m/s2) = 931 N
    Fapplied > Fweight (She must pull with a greater force than the 931 N)
    T = Fapplied - Fweight

    I just don't know where to go from here...

    (b) F = ma

    a = F/m

    1.3 m/s2 = F / 95 kg
    F = 123.5 N; but this seems too easy to be right...too good to be true for physics since its usually so much harder for me?
     
  2. jcsd
  3. Jul 10, 2008 #2

    rock.freak667

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    Force acting up is tension,T.
    Force down=weight=mg

    You want the resultant force to be upwards so that ma=T-mg.

    Now what does newton's first law say about a body moving at constant velocity?
     
  4. Jul 10, 2008 #3
    Newton's first law says that a body moving at constant velocity remains at constant velocity if the sum of the forces is equal to 0?

    OH, so the girl needs to pull with the same 931 N? But does that make sense? ...if she pulls with the same force, she's not going to go anywhere. I could be wrong though because like I said, physics is not my forte, haha!

    Also, can I have a bone thrown at me for part B as well?
     
  5. Jul 10, 2008 #4

    rock.freak667

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    Yes this is correct

    If she pulls with the same 931N, the resultant force on the system is zero. So she can still be moving upwards. 931N is like the minimum force she needs to pull with.

    Use the equation you made with the resultant force : ma=T-mg

    You want T. You know m(the mass),g and a(acceleration of the system)
     
  6. Jul 10, 2008 #5

    Thank you SO much for your help! I really appreciate it.

    I don't mean to nag, and I definitely don't question that you know 100x more than myself...but I still don't understand how she can be moving upwards with if she only exerts same force that is already exerted by her and the basket thats keeping her on the ground. Maybe its a concept I'm not realizing, but it just doesn't make sense to me. Wouldn't she be stationary since the forces are equal and moving in opposite directions?
     
  7. Jul 10, 2008 #6

    rock.freak667

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    Newton's First Law; A body at rest stays at rest,or if moving,continues to move in a straight line with constant velocity provided that the resultant force on the body is zero.

    Consider an object falling through the air. There are three forces on the object Weight,W, (downwards),Upthrust,U(upwards) and Friction,F (upwards to oppose the downward motion)

    [Friction given by F=kv]

    The downward resultant force on the object,ma=W-U-F.

    As the object continues to fall, Friction increases since velocity is increasing. Upthrust is a constant.

    At a certain velocity called the terminal velocity, the resulstant force on the body is zero.

    so that W=U+F. (Forces are balanced right?) . But even at this velocity, the object just doesn't stay there suspended in mid-air, it continues to fall! So even though the forces are balanced, it still moves.


    [I just explained terminal velocity if ever needed to know about it.]
     
  8. Jul 11, 2008 #7

    alphysicist

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    Hi spacethisride and rock.freak,

    I don't believe the answer to part a is 931N. Be sure to draw the force diagram with all forces included. What does it look like? What equation do you get from that force diagram?
     
  9. Jul 11, 2008 #8
    I made a force diagram, and I had a Normal force up, which is equal to the mg force down into the ground (931 N). Then I had an addition force of the tension T, also going up, resulting from the girl pulling the other end of the rope. However, I'm not sure about the equation...

    Fn = mg

    T=Fapplied

    Fnet=Fapplied - mg

    Fnet=T - mg

    I honestly don't know what to do from there, if thats even right.
     
  10. Jul 11, 2008 #9

    alphysicist

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    There is no normal force here. The problem states that she is moving upward at constant speed, so she is already in the air; once she loses contact with the ground there is no normal force.

    So you have the tension force upward from the girl pulling on the rope, and the weight force downward. However, there is one other force. What is it?
     
  11. Jul 11, 2008 #10
    Hmm...is it like an additional downward force from her arm? That's the only thing I can think of...Air resistance? but I think I'm supposed to ignore air resistance. I haven't the slightest clue...another hint, perhaps? Haha I'm sorry :
     
  12. Jul 11, 2008 #11

    alphysicist

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    One end of the rope is pulling on her (because she is pulling down on it, it pulls up on her). Where is the other end of the rope at?
     
  13. Jul 11, 2008 #12
    Oh! That makes sense! The other end of the rope is going to be going down as she pulls herself up. I don't understand how I would incorporate that into the equation though..So if the tension pulls up on her, that means whatever she pulls with also adds to the net force?

    Fnet=T - mg + Fshe pulls with

    But then I still can't figure out T or Fshe pulls with...aren't they essentially the same?
     
  14. Jul 11, 2008 #13

    alphysicist

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    That's right, the tension and her pulling force are the same; since the pulley is massless, you assume that the tension all along the rope is the same. So one end of the rope pulls up on her, and the other end pulls up on the basket. So what do you get for the answer?
     
  15. Jul 11, 2008 #14
    She's sitting inside the basket (she and basket combined exert a downward force of 931 N), so isn't the only thing pulling her up the force applied by her arm, which equals T?

    That's what I don't understand about the problem is how you can solve for T when it equals the force applied by her arm. That's just 0. Thus, you can't subtract anything from 931 N.
     
  16. Jul 11, 2008 #15

    alphysicist

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    Imagine she's moving upwards, and someone comes along and cuts the rope that's connected to her basket; she and the basket would fall, right?. She's still holding onto her end of the rope, but the end tied to the basket is also pulling up on her, so there are two tension forces pulling upwards. The fact that those two tension forces are coming from the same rope does not make any difference.

    So from the force diagram, with those three forces (two tensions and the weight) you can find the tension. And then, as you said the force applied from her arm is equal to the tension.
     
  17. Jul 11, 2008 #16
    Wait, I think there might be a miscommunication on which end she is pulling on...I have it like this: (the corners of the rope are obviously curved around the pulley, but being on a computer has its limitations by drawing with normal symbols. periods are only for spacing purposes)

    ........... __ __ __
    ...........|.(pulley).|
    ...........|.............|
    ...........|.............| <- I thought the girl was pulling up on this free end
    GIRL IN BASKET
     
  18. Jul 11, 2008 #17

    alphysicist

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    Doesn't she have to pull down on the rope? Which means the rope is pulling up on her?

    But yes, if I'm seeing your diagram correctly I agree with it. The rope is attached to the object (girl+basket) at two places, so acting on the object are two upward tension forces, and also the weight force downwards. As you said earlier, the acceleration is zero for this part so these three forces must sum to zero. What equation do you get from the force diagram, including just these three forces?
     
  19. Jul 11, 2008 #18
    Eep, I meant to have my diagram only show that one end of the rope is attached to the girl inside the basket and the other end is hanging free...does that change anything?
     
  20. Jul 11, 2008 #19
    Ah, it's almost 4 AM where I am and I have an exam tomorrow morning...I was just wondering if there's any other hints you could give me just because I don't think I'm grasping what I need to figure out. I realize thats the whole point of a physics problem but I've been working on this for hours which is way longer than what it should be. I'd really appreciate it. Thanks
     
  21. Jul 11, 2008 #20

    alphysicist

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    Here is what you originally had that was giving the incorrect answer:

    The tension in the rope is equal to her pulling force. On the object (girl+basket), there is one tension force T upwards, and a weight force of 931 N downwards, so we say T-931=0 (since her acceleration is zero). So T=931, and therefore her pulling force is 931 N.

    Now the above work is not right, but it's very close. The only problem in the reasoning is that there is not one tension, there are two tension forces upwards, and so both tensions have to appear in the equation. With just a few changes to the above work, you can get the right answer. What do you get?
     
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