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Overcoming Friction Problem

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A 75kg box slides down a 25.0 degree ramp with an acceleration of 3.60m/s[tex]^{2}[/tex]
    a.) Find [tex]\mu[/tex]k between the box and the ramp.
    b.) What acceleration would a 175kg box have on this ramp?

    2. Relevant equations
    Fapplied,y=(?N)(sin[STRIKE]0[/STRIKE])
    Fapplied,x=(?N)(cos[STRIKE]0[/STRIKE])
    Fg=mg
    [tex]\sum[/tex]Fy=Fn+Fapplied,y-Fg-0
    Fk=[tex]\mu[/tex]kFn
    [tex]\sum[/tex]Fx=Fapplied,x-Fk=ma[tex]_{2}[/tex]

    3. The attempt at a solution
    Given: ax=3.60m/s
    a.) Fg=(75kg)(9.81m/s[tex]^{2}[/tex])=736N
    b.) m=175kg

    I do not understand how to get Fk or Fn without first finding [tex]\mu[/tex], so I am stuck for now...
     
    Last edited: Oct 12, 2009
  2. jcsd
  3. Oct 11, 2009 #2

    Delphi51

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    You can't find the normal force or solve for the coefficient without separating the mg into components parallel and perpendicular to the ramp.
    ramp.jpg
     
  4. Oct 11, 2009 #3

    rl.bhat

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    The acceleration of the sliding box is given by
    ma = mgsinθ - μmgcosθ
    Solve for μ.
     
  5. Oct 11, 2009 #4
    So in that case, I should change it into
    Fapplied,y=736cos25 down
    Fapplied,x=736sin25 right
    and continue from there? Or are those symbols for something else?

    Is this how it should look in graph form?
    untitled-2.jpg

    @rl.bhat ah, so it is that simple... is the acceleration constant or 9.81m/s^2
     
    Last edited: Oct 11, 2009
  6. Oct 12, 2009 #5

    rl.bhat

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    a is given in the problem and it is constant. g is 9.8 m/s^2
     
  7. Oct 12, 2009 #6
    So 0 = mgsinθ - μmgcosθ
    Ok, thank you both for the help, now to lock this thread...
     
  8. Oct 12, 2009 #7

    rl.bhat

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    It is not correct.
    3.6 m/s^2 = mgsinθ - μmgcosθ
     
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