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What am I doing wrong?

  1. Sep 14, 2004 #1
    I found this question to be answered twice on this site, both by hallsofivy. However, I still don't quite get it. Please help.

    A window washer pulls herself upward using the bucket pulley appartus. a. how hard must she pull downward to raise herself slowly at constant speed? b. if she increase this force by 15% what will her acceleration be? The mass of the person plus the bucket is 65 kg.

    It was posted that I need to use the formula F=ma, i understand this part and I understand that there is no acceleration because there is constant speed. But how do I use what I know? The bucket and HEr is equal to 65. I multiplied this by 9.8 to find the weight which equals 637 kg. But this isn't the correct force. The answer should be 320, or as I found about half of the answer I obtained. What am I doing wrong?
     
  2. jcsd
  3. Sep 14, 2004 #2
    are u doing part a?
    are u sure they are asking for the force, or for the work done...?
     
  4. Sep 14, 2004 #3
    Because the problem was in a chapter dealing with forces, I'm assuming that's what they're asking for. Also by asking how hard must she pull downwards, doeson't that ask for the force she has must exert onto the rope to pull herself down? :bugeye:
     
  5. Sep 14, 2004 #4

    Doc Al

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    Draw yourself a picture of what's going on. Let the "object" be the "bucket + woman". Draw all the forces acting on that object. Hint: How many ropes are pulling on the object? What must be the tension in the ropes? How hard must she pull on her rope?
     
  6. Sep 14, 2004 #5
    Oh I see.. So because we are using a pulley system, there is tension on each side of the pulley . So would I solve 2T-mg = ma
    2T-637 = 0
    =318.5 or about 320.
    So is it okay to assume that 1 pulley will always divide the force needed in half?
     
  7. Sep 14, 2004 #6
    exactly...
     
  8. Sep 14, 2004 #7
    Got part a. Thanks -

    For part b. I take .15 x 320 = 48
    320+48=368.
    F= ma
    368/65 = a
    a= 5.66? What am I doing wrong here?
    It should be 1.5 ...
     
    Last edited: Sep 14, 2004
  9. Sep 14, 2004 #8

    Doc Al

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    That's the force she exerts. So what's the new force on "bucket + woman"? Don't forget to consider all the forces: the ropes pulling up, gravity pulling down. Then apply F = ma using the net force.
     
  10. Sep 14, 2004 #9

    NateTG

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    Also, remember that the tension is equal on both sides of the pulley, so if the woman pulling harder increases the tension on one side...
     
  11. Sep 14, 2004 #10
    will I use Newton's formula to figure out this change in tension?
     
  12. Sep 15, 2004 #11

    Doc Al

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    Hint1: You already figured out the change in tension in the rope. It was given.

    Hint2: Before the increase in tension, the forces on the "bucket + woman" added to zero since she was not accelerating. So what additional force has been added? So the net force is now what?

    Once you realize what the net force is, then use Newton's F = ma to find the acceleration.
     
  13. Sep 15, 2004 #12
    T= 320 x .15 = 48 + 320 = 368 N
    F=ma
    368= 320a
    a= 1.15 m/s^2 . This is correct! However, I have another quick question. For part a and for the first part of part b, I solved for the tension when they asked for her force. How is her force equal to the tension of the rope? Shouldn't it be different... ?
     
  14. Sep 15, 2004 #13

    Doc Al

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    not so fast...

    No it isn't! First, 368 N is just the tension in the rope, not the net force on the "object". Second, why did you put 320 as the mass? You know the mass of the "bucket + woman" is 65 kg.

    Answer these questions, in order: (1) what is the tension in the rope? (2) what force do the ropes exert on the bucket + woman? (3) what is the weight of the bucket + woman? (4) what is the net force on the bucket + woman? (5) what is the acceleration?
    The woman pulls on the rope and the rope pulls back on the woman. (Recall Newton's 3rd law.) What allows a rope to pull is the tension in the rope. If the woman pulls the rope with 10 N of force, then the tension in the rope pulls back on her with 10 N of force.
     
  15. Sep 15, 2004 #14
    OHh I see. Though I did it incorrectly I got the same answer. Everything makes sense now. Thanks for the help!
     
  16. Sep 15, 2004 #15

    Doc Al

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    Not sure what you mean, since a= 1.15 m/s^2 is not the correct answer.
     
  17. Sep 16, 2004 #16
    Yes, sorry about that... I actually got 1.5 m/s^2.
     
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