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Mastering Physics: Kinetic and Static Friction

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data

    A box of textbooks of mass 24.5 rests on a loading ramp that makes an angle with the horizontal. The coefficient of kinetic friction is 0.260 and the coefficient of static friction is 0.360.

    As the angle is increased, find the minimum angle at which the box starts to slip.

    2. Relevant equations

    N+fs+W=0

    N=mg cos[tex]\vartheta[/tex]
    fs=mg sin[tex]\vartheta[/tex]

    3. The attempt at a solution

    In my notes, it gives an example of an object at rest on an inclined plane and sets up the two above equations with the angle and weight of the object both given. The problem is that when I add up the three forces N, fs, and W I do not get "0". This is leaving me quite confused.

    I have tried substituting the equations above into the top equation equal to zero, but I cannot figure out how to solve for theta since Weight is involved.

    Can anyone give me a push in the right direction. I feel like I am forgetting something.
     
  2. jcsd
  3. Feb 17, 2009 #2

    LowlyPion

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    Welcome to PF.

    You must remember that the frictional force is really only a maximum number that needs to be overcome in order for there to be motion along the incline.

    Hence what you need to consider is the point at which your

    m*g*sinθ = μ*m*g*cosθ
     
  4. Feb 19, 2009 #3
    Thank you for the quick response (sorry for my delay). So what does the symbol "μ" represent in the equation? Will I be able to set the equation equal to zero and factor out the common angle?

    Thanks,

    Rob
     
  5. Feb 19, 2009 #4
    [tex]\mu}[/tex] is the coefficient of friction. And the rest you got right.
     
  6. Feb 19, 2009 #5
    In this equation both mg are the same correct? So that would make μcosθ=sinθ

    I realize looking at this that I can't just factor out that angle. Is there a way to determine at what angle the book is no longer stationary short of plugging in random angles?
     
  7. Feb 19, 2009 #6

    LowlyPion

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    Have you run across the trig identity that Tanθ = Sinθ/Cosθ before?
     
  8. Feb 19, 2009 #7
    Yes, thanks for the reminder. That's one of those things that you end up smacking yourself in the head.
     
  9. Feb 19, 2009 #8

    LowlyPion

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    So ... for θ just take the arctan

    Tan-1(X) = θ
     
  10. Feb 19, 2009 #9
    Yep, got the answer and it was correct.

    Thank you for the help
     
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