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Newton's Laws: Friction: Inclined Plane

  1. Sep 27, 2007 #1
    My main question:
    Given an angle of incline, x, how can you find the coefficients of static and kinetic friction? Is it possible?

    My homework problem is:

    Someone wearing leather shoes is standing in the middle of a wooden plank. One end of the board is gradually raised until it makes an angle of 17 degrees with the floor, at which point the person begins to slide down the incline. Compute the coefficient of static friction.



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    I used the simple formula tan x = Mu s, and it seemed to work, giving me the correct answer of 0.31, although I have no idea why. Can anyone explain to me why tan x gives the coefficient, and whether it gives the static or kinetic coefficient? I know its hard to explain without a free body diagram, but I need help..
     
    Last edited: Sep 27, 2007
  2. jcsd
  3. Sep 27, 2007 #2

    learningphysics

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    you need to draw a freebody diagram... there are 3 forces in total acting on the person... the normal force... gravity and the frictional force...

    All these forces need to be divided into components along the plane and parallel to the plane...

    Normal force is perpendicular to the plane... frictional force is parallel to the plane... now you need to get the components of gravity perpendicular to the plane and parallel to the plane...
     
  4. Sep 27, 2007 #3
    Ohh. But how can I find any forces when all I am given is an angle?
    Sum of Fx = Fa- Ff and Fy= Fn-Fg right? I'm so confused..
     
  5. Sep 27, 2007 #4

    learningphysics

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    What is Fa?

    It's like the inclined plane... what is the component of gravity perpendicular to the plane? mg is the net gravitational force... what are the two components?
     
  6. Sep 27, 2007 #5
    Fg perpendicular is Fg(cos 17) right?
    So would it be Fy= Fg(cos 17) - mg
     
  7. Sep 27, 2007 #6
    Fy= mg(cos 17) - mg
    and Fa= Ff ?
     
  8. Sep 27, 2007 #7

    learningphysics

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    the mgcos(17) is right... but why do you have the "-mg" ? You need Fnormal. or rather use downwards into the perpendicular as negative and outwards as positive hence:

    Fy = Fnormal - mg(cos 17) = 0
     
    Last edited: Sep 27, 2007
  9. Sep 27, 2007 #8

    learningphysics

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    But what is Fa supposed to be? I don't understand what Fa is supposed to represent.
     
  10. Sep 27, 2007 #9
    Wait so what's the main equation I'm working with.. Ff= mu * Fn ?
     
  11. Sep 27, 2007 #10

    learningphysics

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    yes, you need to use that... but first get the x and y equations... just use Ff for now... we'll substitute mu*Fn afterwards...
     
  12. Sep 27, 2007 #11
    Oh I was saying Fa is applied force.. but I don't know why cause I guess there is no force applied, sorry
     
  13. Sep 27, 2007 #12

    learningphysics

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    You have Ff and mgsin(17) parallel to the plane.
     
  14. Sep 27, 2007 #13
    I'm sorry, what is mg*(cos 17) ? Isn't that Fgperpendicular, which is equal and opposite to the normal force?
     
  15. Sep 27, 2007 #14

    learningphysics

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    yes, in this particular case Fnormal = mgcos(17)

    So using [tex]\Sigma{F_y} = 0[/tex]

    then substituting in the left side:

    [tex]F_{normal} - mgcos(17) = 0[/tex]

    Hence Fnormal = mgcos17.

    Can you give a similar equation for [tex]\Sigma{F_x}[/tex]?
     
  16. Sep 27, 2007 #15
    so is it Fx= Ff -Fgparallel ?
     
  17. Sep 27, 2007 #16
    and Fx= 0, and Fgparallel= Fg (sin 17)
     
  18. Sep 27, 2007 #17
    so 0= Ff- mg(sin 17) therefore Ff= mg(sin 17) ?
     
  19. Sep 27, 2007 #18
    then since Ff= mu * Fn ,
    mg(sin 17) = mu * mg(cos 17)
    and mu = (sin 17)/(cos 17)
     
    Last edited: Sep 27, 2007
  20. Sep 27, 2007 #19
    Ohhhh and that equals 0.305!!!!!!
     
    Last edited: Sep 27, 2007
  21. Sep 27, 2007 #20
    Thanks soooo much I get it now. I guess I was trying to take a shortcut using that tangent thing and it messed me up. Thanks your amazing
     
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