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Homework Help: Friction - could you help me?

  1. Feb 14, 2010 #1
    A 40-N crate rests on a rough horizontal floor. A 12-N horizonatl force is then applied to it. If the coefficients of friction are s = 0.5 and k = 0.4, the magnitude of the frictional force on the crate is:

    what i did was :
    40*0.5=20 , 40*0.4=16 20+16=36 40/12=3.3 36/3=12

    how to solve this problem? could someone help?
    thank you.
     
  2. jcsd
  3. Feb 14, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi gracemir! Welcome to PF! :smile:

    (have a mu: µ :wink:)
    Nooo … you never use both µs and µk

    it's always either one or the other! :wink:

    Hint: first decide whether 12N is enough to move the crate, then decide what the friction force is. :smile:
     
  4. Feb 14, 2010 #3
    Re: Welcome to PF!

    how to decide whether 12N is enough to move or not?

     
  5. Feb 14, 2010 #4

    tiny-tim

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    oh come on … think …

    what is the definition of µs ?​
     
  6. Feb 14, 2010 #5
    the definition of mu s is F sub s / N?

     
  7. Feb 14, 2010 #6

    tiny-tim

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    No.

    µs is the maximum possible value of Fs/N.

    To find the actual value of Fs, we can use the fact that the acceleration (and velocity) is zero, so all the forces must add to zero. :smile:
     
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