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Static force/Static friction

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose a block of mass = 10kg is on an incline plane that makes an angle = 45° from the horizontal. If a force of 10 N is being applied to the block, downward, and perpendicular to the surface, and the block is not moving, what is the static force of friction? What is the coefficient of static friction?



    2. Relevant equations

    [itex]\mu[/itex][itex]_{}s[/itex] = Fn

    3. The attempt at a solution


    [itex]\mu[/itex][itex]_{}s[/itex] = FN

    Am I on the right track?
     
    Last edited: Nov 2, 2011
  2. jcsd
  3. Nov 1, 2011 #2

    PhanthomJay

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    should be N -mgcos45 -10 = 0, N = mgcos45 +10. Don't confuse the normal force, N, with the force unit of Newtons (N)!!!
    This is the correct answer for the first part.
    you mean [itex]F_f [/itex] is less than or equal to [itex]\mu_sN[/itex], don't you?
     
  4. Nov 1, 2011 #3
    LOL Yes, I did mean that. So it will be [itex]\mu[/itex][itex]_{}s[/itex] = Ff / N?

    Which will be 69.3N/79.3N..correct?
     
  5. Nov 1, 2011 #4

    PhanthomJay

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    No-o. Since static friction force is always less than or equal to [itex]\mu N [/itex] , then [itex]\mu[/itex][itex]_{}s[/itex] is greater than or equal to Ff / N.
     
  6. Nov 1, 2011 #5
    Oh, okay! Thank you so much :biggrin:
     
  7. Nov 1, 2011 #6

    PhanthomJay

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    You're welcome. Your calculation for the Normal force is also correct. Note that only if the block was at rest but just on the verge of moving would the static friction coefficient be equal to Ff/N. Otherwise, it would be greater than Ff/N.
     
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