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Friction Force

  1. Oct 4, 2007 #1

    klm

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    A dancer is standing on one leg on a drawbridge that is about to open. The coefficients of static and kinetic friction between the drawbridge and the dancer's foot are mu_s and mu_k, respectively. n represents the normal force exerted on the dancer by the bridge, and Fg represents the gravitational force exerted on the dancer, as shown in the drawing For all the questions, we can assume that the bridge is a perfectly flat surface and lacks the curvature characteristic of most bridges.



    Before the drawbridge starts to open, it is perfectly level with the ground. The dancer is standing still on one leg. What is the x component of the friction force, Ff?



    Express your answer in terms of some or all of the variables n, mu_s, and/or mu_k.

    i thought the answer should be -mu_s X N , but that is wrong.
     
    Last edited: Oct 4, 2007
  2. jcsd
  3. Oct 4, 2007 #2

    PhanthomJay

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    Have you tried applying Newton's laws in the horizontal direction?
     
  4. Oct 4, 2007 #3

    klm

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    yeah i have friction force pointing horizontally on the -x axis. so then i did Fnetx=-Fs=max=0 and Fnet y= n-w=0, so n=mg. but since the answer is suppose to be in terms of Us, Uk , or n. i thought -UsN
     
  5. Oct 4, 2007 #4

    PhanthomJay

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    Yes, I also don't understand how you can express 0 as a function of any of those variables.
     
  6. Oct 4, 2007 #5

    klm

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    i dont mean that the variables are zero, i just mean that there is no net force. so -Fs=0 and n-w=0
     
  7. Oct 4, 2007 #6

    PhanthomJay

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    So are you sayinfg F_f = 0?
     
  8. Oct 4, 2007 #7

    klm

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    actually yes i think that would be right b/c there is no Ff
     
  9. Oct 4, 2007 #8

    PhanthomJay

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    Well maybe, but why not no magnitude at all?
     
  10. Oct 4, 2007 #9

    klm

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    yeah i think you are right, it shouldnt have a magnitude
     
  11. Oct 4, 2007 #10

    PhanthomJay

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    Well, maybe yes, and maybe no, I may have spoken too quickly. In general, what is the static friction force formula?
     
  12. Oct 4, 2007 #11

    klm

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    Fs=UsN
    i think i understand, thank you
     
  13. Oct 4, 2007 #12

    PhanthomJay

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    No, that is close, but that is not the formula.
     
  14. Oct 4, 2007 #13

    klm

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    that is the formula that we learned
     
  15. Oct 4, 2007 #14

    klm

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    that is the formula that we learned fsmax=usn
     
  16. Oct 4, 2007 #15

    PhanthomJay

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    Yes fsmax=usN. But fs in general is ???
     
  17. Oct 4, 2007 #16

    klm

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    i dont know
     
  18. Oct 4, 2007 #17

    PhanthomJay

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    Well, you think in this problem that fs might be zero. And you already know that in this problem fs=usN is the wrong answer. So can you venture to guess what the formula for Fs is in general? (HINT: think in between).
     
  19. Oct 4, 2007 #18

    klm

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    do you mean like fs=ma ? so fs=0 ? i dont really understand what you are getting at sory
     
    Last edited: Oct 4, 2007
  20. Oct 4, 2007 #19

    PhanthomJay

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    Oh OK, I guess you can look it up and you will find that the static friction force is less than or equal to usN. In other words, it can vary from 0 to usN. So back to the problem at hand, is the friction force 0, usN, or somewhere in between (or none of the above).
     
  21. Oct 4, 2007 #20

    klm

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    oh um okay. yes i know that it can vary, but thank you for clarifying what you mean. i understand how to do the problem now, so thank you very much for all your help.
     
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